INSTRUCTOR'S SOLUTIONS MANUAL
FOR
SERWAY AND JEWETT'S
PHYSICS
FOR SCIENTISTS AND ENGINEERS
SIXTH EDITION
Ralph V. McGrew
Broome Community College
James A. Currie
Weston High School
Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States
1
Physics and Measurement
CHAPTER OUTLINE
1.1
1.2
1.3
1.4
1.5
1.6
1.7
ANSWERS TO QUESTIONS
Standards of Length, Mass,
and Time
Matter and Model-Building
Density and Atomic Mass
Dimensional Analysis
Conversion of Units
Estimates and Order-ofMagnitude Calculations
Significant Figures
Q1.1
Atomic clocks are based on electromagnetic waves which atoms
emit. Also, pulsars are highly regular astronomical clocks.
Q1.2
Density varies with temperature and pressure. It would be
necessary to measure both mass and volume very accurately in
order to use the density of water as a standard.
Q1.3
People have different size hands. Defining the unit precisely
would be cumbersome.
Q1.4
(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms
Q1.5
(b) and (d). You cannot add or subtract quantities of different
dimension.
Q1.6
A dimensionally correct equation need not be true. Example:
1 chimpanzee = 2 chimpanzee is dimensionally correct. If an
equation is not dimensionally correct, it cannot be correct.
Q1.7
If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about
10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on
vacation.
Q1.8
On February 7, 2001, I am 55 years and 39 days old.
55 yr
F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10
GH 1 yr JK
H 1d K
9
s ~ 10 9 s .
Many college students are just approaching 1 Gs.
Q1.9
Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.
Q1.10
The mass of the forty-six chapter textbook is on the order of 10 0 kg .
Q1.11
With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
1
2
Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1
Standards of Length, Mass, and Time
No problems in this section
Section 1.2
P1.1
Matter and Model-Building
From the figure, we may see that the spacing between diagonal planes is half the distance between
diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the
Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance
L = 0.200 nm , the diagonal planes are separated by
Section 1.3
*P1.2
1 2
L + L2 = 0.141 nm .
2
Density and Atomic Mass
Modeling the Earth as a sphere, we find its volume as
4 3 4
π r = π 6.37 × 10 6 m
3
3
e
j
3
= 1.08 × 10 21 m 3 . Its
m 5.98 × 10 24 kg
=
= 5.52 × 10 3 kg m3 . This value is intermediate between the
V 1.08 × 10 21 m 3
tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to
3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density material
must be down below the surface.
density is then ρ =
P1.3
a
fb
g
e j
With V = base area height V = π r 2 h and ρ =
ρ=
a
fa
ρ = 2.15 × 10 kg m
3
3
9
3
.
m
for both. Then ρ iron = 9.35 kg V and
V
19.3 × 10 3 kg / m3
= 23.0 kg .
= 9.35 kg
7.86 × 10 3 kg / m3
Let V represent the volume of the model, the same in ρ =
ρ gold =
P1.5
F 10 mm I
f GH 1 m JK
1 kg
m
=
2
π r h π 19.5 mm 2 39.0 mm
4
*P1.4
m
, we have
V
m gold
V
V = Vo − Vi =
ρ=
. Next,
ρ gold
ρ iron
4
π r23 − r13
3
e
=
m gold
9.35 kg
and m gold
F
GH
j
FG IJ e
H K
e
4π ρ r23 − r13
m
4
, so m = ρV = ρ π r23 − r13 =
V
3
3
j
j
.
I
JK
Chapter 1
P1.6
3
4
4 3
π r and the mass is m = ρV = ρ π r 3 . We divide this equation
3
3
for the larger sphere by the same equation for the smaller:
For either sphere the volume is V =
m A ρ 4π rA3 3 rA3
=
=
= 5.
m s ρ 4π rs3 3 rs3
a f
Then rA = rs 3 5 = 4.50 cm 1.71 = 7.69 cm .
P1.7
*P1.8
Use 1 u = 1.66 × 10 −24 g .
F 1.66 × 10
GH 1 u
F 1.66 × 10
= 55.9 uG
H 1u
F 1.66 × 10
= 207 uG
H 1u
-24
I = 6.64 × 10
JK
gI
JK = 9.29 × 10
gI
JK = 3.44 × 10
g
−24
g .
−23
g .
−22
g .
(a)
For He, m 0 = 4.00 u
(b)
For Fe, m 0
(c)
For Pb, m 0
(a)
The mass of any sample is the number of atoms in the sample times the mass m 0 of one
atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is
-24
−24
m 0 = 27.0 u = 27.0 u × 1.66 × 10 −27 kg 1 u = 4.48 × 10 −26 kg .
Then the mass of 6.02 × 10 23 atoms is
m = Nm 0 = 6.02 × 10 23 × 4.48 × 10 −26 kg = 0.027 0 kg = 27.0 g .
Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be
written m = Nm 0 .
0.027 0 kg = 6.02 × 10 23 m 0 , so m 0 =
0.027 kg
6.02 × 10 23
= 4.48 × 10 −26 kg ,
in agreement with the first assertion.
(b)
The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u ,
where M is the numerical value of the atomic mass. It divides out exactly for all substances,
giving 1.000 000 0 × 10 −3 kg = N 1.660 540 2 × 10 −27 kg . With eight-digit data, we can be quite
sure of the result to seven digits. For one mole the number of atoms is
N=
F 1 I 10
GH 1.660 540 2 JK
−3 + 27
= 6.022 137 × 10 23 .
(c)
The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one
molecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .
(d)
For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.
b
g
b
g
4
Physics and Measurement
P1.9
b gFGH 101 kgg IJK = 4.5 × 10
kg I
JK = 3.27 × 10 kg .
Mass of gold abraded: ∆m = 3.80 g − 3.35 g = 0.45 g = 0.45 g
F 1.66 × 10
GH 1 u
−27
Each atom has mass m 0 = 197 u = 197 u
3
−4
kg .
−25
Now, ∆m = ∆N m 0 , and the number of atoms missing is
∆N =
∆m
=
m0
4.5 × 10 −4 kg
3.27 × 10 −25 kg
= 1.38 × 10 21 atoms .
The rate of loss is
∆N
∆t
∆N
∆t
P1.10
P1.11
=
FG
H
1 yr
1.38 × 10 21 atoms
365.25 d
50 yr
IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ
K H 24 h K H 60 min K H 60 s K
= 8.72 × 10 11 atoms s .
e
je
j
(a)
m = ρ L3 = 7.86 g cm 3 5.00 × 10 −6 cm
(b)
N=
(a)
The cross-sectional area is
3
= 9.83 × 10 −16 g = 9.83 × 10 −19 kg
9.83 × 10 −19 kg
m
=
= 1.06 × 10 7 atoms
m 0 55.9 u 1.66 × 10 −27 kg 1 u
e
j
a
f a
fa
fa
A = 2 0.150 m 0.010 m + 0.340 m 0.010 m
= 6.40 × 10
−3
2
m .
f.
The volume of the beam is
ja
e
f
V = AL = 6.40 × 10 −3 m 2 1.50 m = 9.60 × 10 −3 m3 .
Thus, its mass is
e
FIG. P1.11
je9.60 × 10 m j = 72.6 kg .
F 1.66 × 10 kg I = 9.28 × 10
The mass of one typical atom is m = a55.9 ufG
H 1 u JK
3
m = ρV = 7.56 × 10 kg / m
(b)
−3
3
3
−27
0
m = Nm 0 and the number of atoms is N =
−26
kg . Now
72.6 kg
m
=
= 7.82 × 10 26 atoms .
−26
m 0 9.28 × 10
kg
Chapter 1
P1.12
(a)
F 1.66 × 10
GH 1 u
−27
The mass of one molecule is m 0 = 18.0 u
kg
I = 2.99 × 10
JK
−26
5
kg . The number of
molecules in the pail is
N pail =
(b)
1.20 kg
m
=
= 4.02 × 10 25 molecules .
m 0 2.99 × 10 −26 kg
Suppose that enough time has elapsed for thorough mixing of the hydrosphere.
N both = N pail
F m I = (4.02 × 10
GH M JK
pail
25
F 1.20 kg I ,
GH 1.32 × 10 kg JK
molecules)
total
21
or
N both = 3.65 × 10 4 molecules .
Section 1.4
P1.13
Dimensional Analysis
The term x has dimensions of L, a has dimensions of LT −2 , and t has dimensions of T. Therefore, the
equation x = ka m t n has dimensions of
e
L = LT −2
j aTf
m
n
or L1 T 0 = Lm T n − 2 m .
The powers of L and T must be the same on each side of the equation. Therefore,
L1 = Lm and m = 1 .
Likewise, equating terms in T, we see that n − 2m must equal 0. Thus, n = 2 . The value of k, a
dimensionless constant, cannot be obtained by dimensional analysis .
*P1.14
(a)
Circumference has dimensions of L.
(b)
Volume has dimensions of L3 .
(c)
Area has dimensions of L2 .
e j
Expression (i) has dimension L L2
1/2
= L2 , so this must be area (c).
Expression (ii) has dimension L, so it is (a).
Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .
e j
6
Physics and Measurement
P1.15
*P1.16
(a)
This is incorrect since the units of ax are m 2 s 2 , while the units of v are m s .
(b)
This is correct since the units of y are m, and cos kx is dimensionless if k is in m −1 .
(a)
a f
a∝
∑F
or a = k
∑F
represents the proportionality of acceleration to resultant force and
m
m
the inverse proportionality of acceleration to mass. If k has no dimensions, we have
a = k
(b)
P1.17
In units,
M ⋅L
T
2
=
kg ⋅ m
s2
F
F
L
M ⋅L
, F =
, 2 =1
.
m T
M
T2
, so 1 newton = 1 kg ⋅ m s 2 .
Inserting the proper units for everything except G,
LM kg m OP = G kg
Ns Q m
2
2
Multiply both sides by m
Section 1.5
*P1.18
.
m3
2
and divide by kg ; the units of G are
kg ⋅ s 2
.
Conversion of Units
a
fa
f
Each of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have area
e
4 96.0 ft 2
P1.19
2
2
jFGH 3.128mft IJK
2
= 35.7 m 2 .
Apply the following conversion factors:
1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 m
FG 1
H 32
IJ b2.54 cm inge10 m cmje10
K
86 400 s day
−2
in day
9
j=
nm m
9.19 nm s .
This means the proteins are assembled at a rate of many layers of atoms each second!
*P1.20
8.50 in 3 = 8.50 in 3
FG 0.025 4 m IJ
H 1 in K
3
= 1.39 × 10 −4 m 3
Chapter 1
P1.21
Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should
expect the area to be about A ≈ 30 m 50 m = 1 500 m 2 .
a
fa
f
Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the
conversion: 1 m = 3.281 ft .
a
Analyze: A = LW = 100 ft
1m I
F 1 m IJ = 1 390 m
150 ft fG
f FGH 3.281
a
J
K
H 3.281 ft K
ft
2
= 1.39 × 10 3 m 2 .
Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper
units of m 2 . Unit conversion is a common technique that is applied to many problems.
P1.22
(a)
a
fa
fa
f
V = 40.0 m 20.0 m 12.0 m = 9.60 × 10 3 m 3
3
3
b
g
V = 9.60 × 10 m 3.28 ft 1 m
(b)
3
= 3.39 × 10 5 ft 3
The mass of the air is
e
je
j
m = ρ air V = 1.20 kg m3 9.60 × 10 3 m3 = 1.15 × 10 4 kg .
The student must look up weight in the index to find
e
je
j
Fg = mg = 1.15 × 10 4 kg 9.80 m s 2 = 1.13 × 10 5 N .
Converting to pounds,
jb
e
g
Fg = 1.13 × 10 5 N 1 lb 4.45 N = 2.54 × 10 4 lb .
P1.23
(a)
Seven minutes is 420 seconds, so the rate is
r=
(b)
30.0 gal
= 7.14 × 10 −2 gal s .
420 s
Converting gallons first to liters, then to m3 ,
e
r = 7.14 × 10 −2 gal s
jFGH 3.1786galL IJK FGH 10 1 Lm IJK
−3
3
r = 2.70 × 10 −4 m3 s .
(c)
At that rate, to fill a 1-m3 tank would take
t=
F 1m
GH 2.70 × 10
3
−4
m
3
IF 1 h I =
G J
s JK H 3 600 K
1.03 h .
7
8
Physics and Measurement
*P1.24
(a)
(b)
(c)
(d)
P1.25
FG 1.609 km IJ = 560 km = 5.60 × 10 m = 5.60 × 10 cm .
H 1 mi K
F 0.304 8 m IJ = 491 m = 0.491 km = 4.91 × 10 cm .
Height of Ribbon Falls = 1 612 ftG
H 1 ft K
F 0.304 8 m IJ = 6.19 km = 6.19 × 10 m = 6.19 × 10 cm .
Height of Denali = 20 320 ftG
H 1 ft K
F 0.304 8 m IJ = 2.50 km = 2.50 × 10 m = 2.50 × 10 cm .
Depth of King’s Canyon = 8 200 ftG
H 1 ft K
5
Length of Mammoth Cave = 348 mi
7
4
3
5
3
5
From Table 1.5, the density of lead is 1.13 × 10 4 kg m 3 , so we should expect our calculated value to
be close to this number. This density value tells us that lead is about 11 times denser than water,
which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ =
ρ=
23.94 g
2.10 cm 3
F 1 kg I FG 100 cm IJ
GH 1 000 g JK H 1 m K
3
m
. We must convert to SI units in the calculation.
V
= 1.14 × 10 4 kg m3
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our
result is indeed close to the expected value. Since the last reported significant digit is not certain, the
difference in the two values is probably due to measurement uncertainty and should not be a
concern. One important common-sense check on density values is that objects which sink in water
must have a density greater than 1 g cm 3 , and objects that float must be less dense than water.
P1.26
It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile
in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to
mI
a1 acrefFGH 6401 miacres IJK FGH 1 609
J
mi K
2
*P1.27
P1.28
The weight flow rate is 1 200
FG
H
ton 2 000 lb
h
ton
2
= 4.05 × 10 3 m 2 .
IJ FG 1 h IJ FG 1 min IJ =
K H 60 min K H 60 s K
667 lb s .
1 mi = 1 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609.
(a)
1 mi h = 1.609 km h
(b)
55 mi h = 88.5 km h
(c)
65 mi h = 104.6 km h . Thus, ∆v = 16.1 km h .
P1.29
Chapter 1
(a)
F 6 × 10 $ I F 1 h I FG 1 day IJ F 1 yr I =
GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK
(b)
The circumference of the Earth at the equator is 2π 6.378 × 10 3 m = 4.01 × 10 7 m . The length
12
190 years
e
j
of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 × 10 11 m. Thus, the
6 trillion dollars would encircle the Earth
9.30 × 10 11 m
= 2.32 × 10 4 times .
4.01 × 0 7 m
1.99 × 10 30 kg
mSun
=
= 1.19 × 10 57 atoms
m atom 1.67 × 10 −27 kg
P1.30
N atoms =
P1.31
V = At so t =
V 3.78 × 10 −3 m 3
=
= 1.51 × 10 −4 m or 151 µm
2
A
25.0 m
b
a
fe
13.0 acres 43 560 ft 2 acre
1
V = Bh =
3
3
7
3
= 9.08 × 10 ft ,
P1.32
g
j a481 ftf
h
or
e
V = 9.08 × 10 7 ft 3
jFGH 2.83 ×110ft
−2
m3
3
I
JK
B
FIG. P1.32
= 2.57 × 10 6 m3
P1.33
*P1.34
9
b
ge
jb
g
Fg = 2.50 tons block 2.00 × 10 6 blocks 2 000 lb ton = 1.00 × 10 10 lbs
The area covered by water is
a fe
j a fa fe
j
2
= 0.70 4π 6.37 × 10 6 m
A w = 0.70 AEarth = 0.70 4π REarth
2
= 3.6 × 10 14 m 2 .
The average depth of the water is
a
fb
g
d = 2.3 miles 1 609 m l mile = 3.7 × 10 3 m .
The volume of the water is
e
je
j
V = A w d = 3.6 × 10 14 m 2 3.7 × 10 3 m = 1.3 × 10 18 m 3
and the mass is
e
je
j
m = ρ V = 1 000 kg m3 1.3 × 10 18 m3 = 1.3 × 10 21 kg .
10
P1.35
Physics and Measurement
(a)
d nucleus, scale = d nucleus, real
e
Fd
I = 2.40 × 10 m FG 300 ft IJ = 6.79 × 10
jH 1.06 × 10 m K
GH d
JK e
ft jb304.8 mm 1 ft g = 2.07 mm
atom, scale
−15
−10
atom, real
d nucleus, scale = 6.79 × 10 −3
(b)
Vatom
=
Vnucleus
3
4π ratom
3
3
4π rnucleus
3
=
FG r
Hr
IJ = FG d
K Hd
3
atom
nucleus
IJ = F 1.06 × 10
K GH 2.40 × 10
3
atom
nucleus
−10
−15
m
m
I
JK
3
= 8.62 × 10 13 times as large
*P1.36
scale distance
between
P1.37
=
FG real IJ FG scale IJ = e4.0 × 10
H distanceK H factorK
13
jFGH 71..04××1010 mm IJK =
−3
km
200 km
9
The scale factor used in the “dinner plate” model is
0.25 m
S=
5
1.0 × 10 lightyears
= 2.5 × 10 −6 m lightyears .
The distance to Andromeda in the scale model will be
e
je
j
Dscale = Dactual S = 2.0 × 10 6 lightyears 2.5 × 10 −6 m lightyears = 5.0 m .
P1.38
P1.39
FG
H
F e6.37 × 10 mjb100 cm mg I
=G
GH 1.74 × 10 cm JJK = 13.4
IJ = FG e6.37 × 10 mjb100 cm mg IJ = 49.1
K GH 1.74 × 10 cm JK
(a)
2
AEarth 4π rEarth
r
=
= Earth
2
A Moon 4π rMoon
rMoon
(b)
VEarth
=
VMoon
3
4π rEarth
3
3
4π rMoon
3
Fr
=G
Hr
IJ
K
8
Moon
3
6
3
Earth
2
6
2
8
To balance, m Fe = m Al or ρ FeVFe = ρ Al VAl
ρ Fe
FG 4IJ π r
H 3K
Fe
3
FG 4 IJ π r
H 3K
FG ρ IJ = a2.00 cmfFG 7.86 IJ
H 2.70 K
Hρ K
= ρ Al
Al
1/3
rAl = rFe
Fe
Al
3
1/3
= 2.86 cm .
−3
ft , or
Chapter 1
P1.40
11
The mass of each sphere is
m Al = ρ Al VAl =
4π ρ Al rAl 3
3
m Fe = ρ FeVFe =
4π ρ Fe rFe 3
.
3
and
Setting these masses equal,
4π ρ Al rAl 3 4π ρ Fe rFe 3
ρ
=
and rAl = rFe 3 Fe .
3
3
ρ Al
Section 1.6
P1.41
Estimates and Order-of-Magnitude Calculations
Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball
as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m3 , while the volume of
one ball is
FG
H
4π 0.038 m
3
2
IJ
K
3
= 2.87 × 10 −5 m3 .
48
~ 10 6 ping-pong balls in the room.
2.87 × 10 −5
As an aside, the actual number is smaller than this because there will be a lot of space in the
room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best
1
packing fraction” is π 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the
6
above estimate reduces to 1.67 × 10 6 × 0.740 ~ 10 6 .
Therefore, one can fit about
P1.42
A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,
b
gb
gb
g
the tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 × 10 7 rev ~ 10 7 rev .
P1.43
In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least
1
in 2 = 43 × 10 −5 ft 2 . Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a
16
quarter-acre plot of land is about
n=
a
fe
j
0.25 acre 43 560 ft 2 acre
total area
=
= 2.5 × 10 7 blades ~ 10 7 blades .
area per blade
43 × 10 −5 ft 2 blade
12
P1.44
Physics and Measurement
A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then
approximately 4 × 10 −3 in 3 . Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to a
depth of 1 inch is
ft
a1 acrefa1 inchf = a1 acre ⋅ infFGH 431560
acre
2
I F 144 in I ≈ 6.3 × 10
JK GH 1 ft JK
2
2
6
in 3 .
The number of raindrops required is
n=
*P1.45
volume of water required 6.3 × 10 6 in 3
=
= 1.6 × 10 9 ~ 10 9 .
volume of a single drop
4 × 10 −3 in 3
Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then
a fa
fa
fa
f
V = 0.5 1.3 m 0.5 m 0.3 m = 0.10 m3 .
The mass of this volume of water is
e
je
j
m water = ρ water V = 1 000 kg m3 0.10 m3 = 100 kg ~ 10 2 kg .
Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The
mass of copper required is
e
je
j
mcopper = ρ copper V = 8 920 kg m3 0.10 m3 = 892 kg ~ 10 3 kg .
P1.46
The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum
cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250
million people, and 365 days in a year, so
e250 × 10
6
jb
g
cans day 365 days year ≅ 10 11 cans
are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we
estimate this represents
e10
P1.47
11
jb
gb
gb
g
cans 0.1 oz can 1 lb 16 oz 1 ton 2 000 lb ≈ 3.1 × 10 5 tons year . ~ 10 5 tons
Assume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about
1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year).
Therefore,
# tuners ~
F 1 tuner I F 1 piano I (10
GH 1 000 pianos JK GH 100 people JK
7
people) = 100 .
Chapter 1
Section 1.7
*P1.48
13
Significant Figures
METHOD ONE
We treat the best value with its uncertainty as a binomial 21.3 ± 0.2 cm 9.8 ± 0.1 cm ,
a
f a
A = 21.3a9.8f ± 21.3a0.1f ± 0.2a9.8 f ± a0.2 fa0.1f cm
f
2
.
The first term gives the best value of the area. The cross terms add together to give the uncertainty
and the fourth term is negligible.
A = 209 cm 2 ± 4 cm 2 .
METHOD TWO
We add the fractional uncertainties in the data.
a
f FGH 210..23 + 90..18 IJK = 209 cm
fa
A = 21.3 cm 9.8 cm ±
P1.49
a
f
π r 2 = π 10.5 m ± 0.2 m
(a)
2
± 2% = 209 cm 2 ± 4 cm 2
2
= π (10.5 m) 2 ± 2(10.5 m)(0.2 m) + ( 0.2 m) 2
= 346 m 2 ± 13 m 2
a
f
2π r = 2π 10.5 m ± 0.2 m = 66.0 m ± 1.3 m
(b)
3
P1.50
(a)
P1.51
a
f a
m = a1.85 ± 0.02f kg
4
(b)
3
(c)
(d)
f
r = 6.50 ± 0. 20 cm = 6.50 ± 0.20 × 10 −2 m
ρ=
m
c hπ r
4
3
3
also,
δ ρ δ m 3δ r
=
+
.
ρ
m
r
In other words, the percentages of uncertainty are cumulative. Therefore,
a f
δ ρ 0.02 3 0.20
=
+
= 0.103 ,
6.50
ρ 1.85
ρ=
and
a
1.85
c hπ e6.5 × 10
4
3
f
−2
j
m
3
= 1.61 × 10 3 kg m 3
a
f
ρ ± δ ρ = 1.61 ± 0.17 × 10 3 kg m3 = 1.6 ± 0.2 × 10 3 kg m3 .
2
14
P1.52
*P1.53
Physics and Measurement
(a)
756.??
37.2?
0.83
+ 2.5?
796.5/ 3/ = 797
(b)
0.003 2 2 s.f. × 356.3 4 s.f. = 1.140 16 = 2 s.f.
(c)
5.620 4 s.f. × π > 4 s.f. = 17.656 = 4 s.f.
a
f
a
a
f a
P1.55
a
f
a
f
f
1.1
17.66
We work to nine significant digits:
1 yr = 1 yr
P1.54
f
F 365.242 199 d I FG 24 h IJ FG 60 min IJ FG 60 s IJ =
GH 1 yr JK H 1 d K H 1 h K H 1 min K
31 556 926.0 s .
The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m , but this answer must be
rounded to 115.9 m because the distance 19.5 m carries information to only one place past the
decimal. 115.9 m
b
V = 2V1 + 2V2 = 2 V1 + V2
g
a
fa fa f
V = a10.0 mfa1.0 mfa0.090 mf = 0.900 m
V = 2e1.70 m + 0.900 m j = 5.2 m
U
δA
0.12 m
=
= 0.0063 |
19.0 m
A
|| δ V
δw
0.01 m
=
= 0.010 V
= 0.006 + 0.010 + 0.011 = 0.027 =
w
1.0 m
V
|
δt
0.1 cm
|
=
= 0.011 |
t
9.0 cm
W
V1 = 17.0 m + 1.0 m + 1.0 m 1.0 m 0.09 m = 1.70 m 3
3
2
3
3
3
FIG. P1.55
1
1
1
3%
1
1
1
Additional Problems
P1.56
It is desired to find the distance x such that
1 000 m
x
=
100 m
x
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that
a
fb
g
x 2 = 100 m 1 000 m = 1.00 × 10 5 m 2
and therefore
x = 1.00 × 10 5 m 2 = 316 m .
Chapter 1
*P1.57
Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass
a
m 0 = 197 u
fFGH 1.66 ×110u
−27
kg
I = 3.27 × 10
JK
−25
kg .
So, the number of atoms in the cube is
N=
19 300 kg
3.27 × 10 −25 kg
= 5.90 × 10 28 .
The imagined cubical volume of each atom is
d3 =
1 m3
= 1.69 × 10 −29 m 3 .
5.90 × 10 28
So
d = 2.57 × 10 −10 m .
P1.58
A total
P1.59
F I
a fe j FG VV IJ e A j = GG V JJ e4π r j
H K
H K
F 3V IJ = 3FG 30.0 × 10 m IJ = 4.50 m
=G
H r K H 2.00 × 10 m K
Atotal = N A drop =
total
−6
total
2
total
4π r 3
3
drop
drop
3
2
−5
One month is
b
gb
gb
g
1 mo = 30 day 24 h day 3 600 s h = 2.592 × 10 6 s .
Applying units to the equation,
e
j e
j
V = 1.50 Mft 3 mo t + 0.008 00 Mft 3 mo 2 t 2 .
Since 1 Mft 3 = 10 6 ft 3 ,
e
j e
j
V = 1.50 × 10 6 ft 3 mo t + 0.008 00 × 10 6 ft 3 mo 2 t 2 .
Converting months to seconds,
V=
e
1.50 × 10 6 ft 3 mo
2.592 × 10 6 s mo
j e
t+
j
0.008 00 × 10 6 ft 3 mo 2
e2.592 × 10
Thus, V [ft 3 ] = 0.579 ft 3 s t + 1.19 × 10 −9 ft 3 s 2 t 2 .
6
s mo
j
2
t2.
15
16
P1.60
P1.61
Physics and Measurement
af
af
α ′(deg)
α (rad)
tan α
sin α
difference
15.0
20.0
25.0
24.0
24.4
24.5
24.6
24.7
0.262
0.349
0.436
0.419
0.426
0.428
0.429
0.431
0.268
0.364
0.466
0.445
0.454
0.456
0.458
0.460
0.259
0.342
0.423
0.407
0.413
0.415
0.416
0.418
3.47%
6.43%
10.2%
9.34%
9.81%
9.87%
9.98%
10.1%
24.6°
2π r = 15.0 m
r = 2.39 m
h
= tan 55.0°
r
h = 2.39 m tan( 55.0° ) = 3.41 m
a
f
h
55°
r
FIG. P1.61
*P1.62
Let d represent the diameter of the coin and h its thickness. The mass of the gold is
m = ρV = ρAt = ρ
F 2π d
GH 4
2
I
JK
+ π dh t
where t is the thickness of the plating.
LM a2.41f
MN 4
m = 19.3 2π
2
a fa
+ π 2.41 0.178
= 0.003 64 grams
fOPPe0.18 × 10 j
Q
−4
cost = 0.003 64 grams × $10 gram = $0.036 4 = 3.64 cents
This is negligible compared to $4.98.
P1.63
The actual number of seconds in a year is
b86 400 s daygb365.25 day yrg = 31 557 600 s yr .
The percent error in the approximation is
eπ × 10
7
j b
s yr − 31 557 600 s yr
31 557 600 s yr
g × 100% =
0.449% .
Chapter 1
P1.64
V = L3 , A = L2 , h = L
(a)
V = A h
L3 = L2 L = L3 . Thus, the equation is dimensionally correct.
(b)
e
Vrectangular object
P1.65
(a)
j
= Awh = aAw fh = Ah , where
Vcylinder = π R 2 h = π R 2 h = Ah , where A = π R 2
The speed of rise may be found from
v=
(b)
aVol rate of flowf = 16.5 cm
(Area:
(a)
a
π D2
4 )
3
π 6 .30 cm
4
f
s
2
= 0.529 cm s .
Likewise, at a 1.35 cm diameter,
v=
P1.66
A = Aw
16.5 cm 3 s
a
π 1.35 cm
4
f
2
= 11.5 cm s .
1 cubic meter of water has a mass
e
je
je
j
m = ρV = 1.00 × 10 −3 kg cm3 1.00 m 3 10 2 cm m
(b)
3
= 1 000 kg
As a rough calculation, we treat each item as if it were 100% water.
cell:
m = ρV = ρ
FG 4 π R IJ = ρFG 1 π D IJ = e1 000 kg m jFG 1 π IJ e1.0 × 10
H3 K H6 K
H6 K
3
3
3
−6
j
m
3
= 5.2 × 10 −16 kg
kidney: m = ρV = ρ
FG 4 π R IJ = e1.00 × 10
H3 K
3
−3
kg cm3
jFGH 34 π IJK ( 4.0 cm)
3
= 0.27 kg
fly:
m=ρ
FG π D hIJ = e1 × 10
H4 K
2
−3
kg cm 3
= 1.3 × 10 −5 kg
P1.67
V20 mpg =
(10 8 cars)(10 4 mi yr )
= 5.0 × 10 10 gal yr
20 mi gal
V25 mpg =
(10 8 cars)(10 4 mi yr )
= 4.0 × 10 10 gal yr
25 mi gal
Fuel saved = V25 mpg − V20 mpg = 1.0 × 10 10 gal yr
jFGH π4 IJK a2.0 mmf a4.0 mmfe10
2
−1
j
cm mm
3
17
18
P1.68
P1.69
Physics and Measurement
F
GH
IF
JK GH
IF
JK GH
IF
JK GH
I FG
JK H
IJ FG
KH
I
JK
furlongs
220 yd
0.914 4 m 1 fortnight 1 day
1 hr
= 8.32 × 10 −4 m s
fortnight 1 furlong
1 yd
14 days
24 hrs 3 600 s
This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth.
v = 5.00
The volume of the galaxy is
e
j e10
π r 2 t = π 10 21 m
2
19
j
m ~ 10 61 m3 .
If the distance between stars is 4 × 10 16 m , then there is one star in a volume on the order of
e4 × 10
The number of stars is about
P1.70
10
50
3
m star
The density of each material is ρ =
Al:
Cu:
Brass:
Sn:
Fe:
P1.71
10 61 m 3
(a)
(b)
ρ=
a
b
4 51.5 g
g
f a3.75 cmf
4b56.3 g g
ρ=
=
π a1.23 cmf a5.06 cmf
4b94.4 g g
ρ=
=
π a1.54 cmf a5.69 cmf
4b69.1 g g
ρ=
=
π a1.75 cmf a3.74 cmf
4b 216.1 g g
ρ=
=
π a1.89 cmf a9.77 cmf
π 2.52 cm
j
~ 10 11 stars .
2
9.36
2
8.91
2
7.68
2
7.88
g
cm
3
g
cm3
FG g IJ is
H cm K
F g IJ is
The tabulated value G 8.92
H cm K
The tabulated value 2.70
This would take
e
2% smaller.
3
5% smaller.
cm3
g
cm3
g
cm
3
FG
H
The tabulated value 7.86
3.16 × 10 7 s yr
3
4 3 4
π r = π 5.00 × 10 −7 m = 5.24 × 10 −19 m3
3
3
1 m3
=
= 1.91 × 10 18 micrometeorites
5.24 × 10 −19 m3
Vmm =
3
g
b3 600 s hrgb24 hr daygb365.25 days yrg =
Vcube
Vmm
3
m ~ 10 50 m 3 .
4m
m
m
.
=
=
V π r 2h π D2h
= 2.75
2
16
j
1.91 × 10 18 micrometeorites
3.16 × 10 7 micrometeorites yr
= 6.05 × 10 10 yr .
g
cm3
IJ is
K
0.3% smaller.
Chapter 1
ANSWERS TO EVEN PROBLEMS
5.52 × 10 3 kg m3 , between the densities
of aluminum and iron, and greater than
the densities of surface rocks.
P1.34
1.3 × 10 21 kg
P1.36
200 km
P1.4
23.0 kg
P1.38
(a) 13.4; (b) 49.1
P1.6
7.69 cm
P1.8
(a) and (b) see the solution,
N A = 6.022 137 × 10 23 ; (c) 18.0 g;
(d) 44.0 g
P1.2
P1.40
P1.10
(a) 9.83 × 10 −16 g ; (b) 1.06 × 10 7 atoms
P1.12
(a) 4.02 × 10 25 molecules;
(b) 3.65 × 10 4 molecules
P1.14
(a) ii; (b) iii; (c) i
P1.16
(a)
P1.18
35.7 m 2
P1.20
1.39 × 10 −4 m3
P1.22
P1.24
M ⋅L
; (b) 1 newton = 1 kg ⋅ m s 2
T2
5
3
4
(a) 3.39 × 10 ft ; (b) 2.54 × 10 lb
5
rAl = rFe
FG ρ IJ
Hρ K
13
Fe
Al
P1.42
~ 10 7 rev
P1.44
~ 10 9 raindrops
P1.46
~ 10 11 cans; ~ 10 5 tons
P1.48
a209 ± 4f cm
P1.50
(a) 3; (b) 4; (c) 3; (d) 2
P1.52
(a) 797; (b) 1.1; (c) 17.66
P1.54
115.9 m
P1.56
316 m
P1.58
4.50 m 2
P1.60
see the solution; 24.6°
P1.62
3.64 cents ; no
P1.64
see the solution
P1.66
(a) 1 000 kg; (b) 5.2 × 10 −16 kg ; 0. 27 kg ;
2
7
(a) 560 km = 5.60 × 10 m = 5.60 × 10 cm ;
(b) 491 m = 0.491 km = 4.91 × 10 4 cm ;
(c) 6.19 km = 6.19 × 10 3 m = 6.19 × 10 5 cm ;
(d) 2.50 km = 2.50 × 10 3 m = 2.50 × 10 5 cm
P1.26
4.05 × 10 3 m 2
P1.28
(a) 1 mi h = 1.609 km h ; (b) 88.5 km h ;
1.3 × 10 −5 kg
(c) 16.1 km h
P1.68
8.32 × 10 −4 m s ; a snail
P1.30
1.19 × 10 57 atoms
P1.70
see the solution
P1.32
2.57 × 10 6 m3
19
2
Motion in One Dimension
CHAPTER OUTLINE
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Position, Velocity, and
Speed
Instantaneous Velocity and
Speed
Acceleration
Motion Diagrams
One-Dimensional Motion
with Constant Acceleration
Freely Falling Objects
Kinematic Equations
Derived from Calculus
ANSWERS TO QUESTIONS
Q2.1
If I count 5.0 s between lightning and thunder, the sound has
traveled 331 m s 5.0 s = 1.7 km . The transit time for the light
is smaller by
b
ga f
3.00 × 10 8 m s
= 9.06 × 10 5 times,
331 m s
so it is negligible in comparison.
Q2.2
Yes. Yes, if the particle winds up in the +x region at the end.
Q2.3
Zero.
Q2.4
Yes. Yes.
Q2.5
No. Consider a sprinter running a straight-line race. His average velocity would simply be the
length of the race divided by the time it took for him to complete the race. If he stops along the way
to tie his shoe, then his instantaneous velocity at that point would be zero.
Q2.6
We assume the object moves along a straight line. If its average
x
velocity is zero, then the displacement must be zero over the time
interval, according to Equation 2.2. The object might be stationary
throughout the interval. If it is moving to the right at first, it must
later move to the left to return to its starting point. Its velocity must
be zero as it turns around. The graph of the motion shown to the
right represents such motion, as the initial and final positions are
the same. In an x vs. t graph, the instantaneous velocity at any time
t is the slope of the curve at that point. At t 0 in the graph, the slope
of the curve is zero, and thus the instantaneous velocity at that time
is also zero.
t0
t
FIG. Q2.6
Q2.7
Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the
velocity of the particle is unchanging, or is a constant.
21
22
Motion in One Dimension
a
f
Q2.8
Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A common
misconception is that immediately after the doughnut is released, both the velocity and acceleration
are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut
floating at rest in mid-air.
Q2.9
No: Car A might have greater acceleration than B, but they might both have zero acceleration, or
otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the
recent past.
Q2.10
Yes. Consider throwing a ball straight up. As the ball goes up, its
v
velocity is upward v > 0 , and its acceleration is directed down
v0
a < 0 . A graph of v vs. t for this situation would look like the figure
to the right. The acceleration is the slope of a v vs. t graph, and is
always negative in this case, even when the velocity is positive.
a
a f
f
t
FIG. Q2.10
Q2.11
(a)
Accelerating East
(b)
Braking East
(c)
Cruising East
(d)
Braking West
(e)
Accelerating West
(f)
Cruising West
(g)
Stopped but starting to move East
(h)
Stopped but starting to move West
Q2.12
No. Constant acceleration only. Yes. Zero is a constant.
Q2.13
The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall,
and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken
as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is
taken as the bottom of the cliff, then the maximum height would be 30 m.
The velocity is independent of the origin. Since the change in position is used to calculate the
instantaneous velocity in Equation 2.5, the choice of origin is arbitrary.
Q2.14
Once the objects leave the hand, both are in free fall, and both experience the same downward
acceleration equal to the free-fall acceleration, –g.
Q2.15
They are the same. After the first ball reaches its apex and falls back downward past the student, it
will have a downward velocity equal to vi . This velocity is the same as the velocity of the second
ball, so after they fall through equal heights their impact speeds will also be the same.
Q2.16
With h =
1 2
gt ,
2
a
f
1
2
g 0.707t . The time is later than 0.5t.
2
(a)
0.5 h =
(b)
The distance fallen is 0.25 h =
a f
1
2
g 0.5t . The elevation is 0.75h, greater than 0.5h.
2
Chapter 2
Q2.17
Above. Your ball has zero initial speed and smaller average speed during the time of flight to the
passing point.
SOLUTIONS TO PROBLEMS
Section 2.1
Position, Velocity, and Speed
P2.1
v = 2.30 m s
*P2.2
(a)
(b)
v=
∆x 57.5 m − 9.20 m
=
= 16.1 m s
∆t
3.00 s
(c)
v=
∆x 57.5 m − 0 m
=
= 11.5 m s
∆t
5.00 s
(a)
v=
∆x 20 ft
1m
=
∆t 1 yr 3.281 ft
FG
H
IJ FG 1 yr IJ = 2 × 10 m s or in particularly windy times
K H 3.156 × 10 s K
1 yr
∆x 100 ft F 1 m I F
IJ = 1 × 10 m s .
=
v=
G
J
G
H
K
H
∆t
1 yr 3.281 ft 3.156 × 10 s K
−7
7
7
(b)
The time required must have been
∆t =
P2.3
P2.4
−6
FG
H
3 000 mi 1 609 m
∆x
=
v 10 mm yr 1 mi
(a)
v=
∆x 10 m
=
= 5 ms
∆t
2s
(b)
v=
5m
= 1.2 m s
4s
(c)
v=
x 2 − x1 5 m − 10 m
=
= −2.5 m s
4 s−2 s
t 2 − t1
(d)
v=
x 2 − x 1 −5 m − 5 m
=
= −3.3 m s
7 s−4 s
t 2 − t1
(e)
v=
x 2 − x1 0 − 0
=
= 0 ms
8−0
t 2 − t1
x = 10t 2 : For
af
xamf
ts
= 2.0
2.1
3.0
=
44.1
90
40
(a)
v=
∆x 50 m
=
= 50.0 m s
∆t 1.0 s
(b)
v=
∆x 4.1 m
=
= 41.0 m s
∆t
0.1 s
IJ FG 10 mm IJ =
KH 1 m K
3
5 × 10 8 yr .
23
24
P2.5
Motion in One Dimension
(a)
Let d represent the distance between A and B. Let t1 be the time for which the walker has
d
the higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip in
t1
d
d
d
−3.00 m s = − . Then the times are t1 =
and t 2 =
. The average speed
t2
5.00 m s
3.00 m s
is:
b
v=
v=
(b)
Section 2.2
P2.6
(a)
e
2 15.0 m 2 s 2
8.00 m s
j=
b
b
d+d
=
+ 3 .00d m s
d
5.00 m s
g
2d
b8.00 m sgd
g e15.0 m s j
g b
2
3.75 m s
Instantaneous Velocity and Speed
e
Thus, at t = 3.00 s: x = e3.00 m s ja3.00 sf =
j
At any time, t, the position is given by x = 3.00 m s 2 t 2 .
2
2
i
ja
e
27.0 m .
f
2
At t f = 3.00 s + ∆t : x f = 3.00 m s 2 3.00 s + ∆t , or
b
ja f
g e
x f = 27.0 m + 18.0 m s ∆t + 3.00 m s 2 ∆t
(c)
∆t → 0
(a)
F x − x I = lim e18.0 m s + e3.00 m s j∆tj =
GH ∆t JK
f
i
2
∆t → 0
x f − xi
t f − ti
=
a2.0 − 8.0f m = − 6.0 m =
a4 − 1.5f s 2.5 s
18.0 m s .
−2.4 m s
The slope of the tangent line is found from points C and
D. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,
b
g
b
g
v ≅ −3.8 m s .
(c)
.
at ti = 1.5 s , x i = 8.0 m (Point A)
at t f = 4.0 s , x f = 2.0 m (Point B)
v=
(b)
2
The instantaneous velocity at t = 3.00 s is:
v = lim
P2.7
2
She starts and finishes at the same point A. With total displacement = 0, average velocity
= 0 .
i
(b)
Total distance
=
Total time
g
The velocity is zero when x is a minimum. This is at t ≅ 4 s .
FIG. P2.7
Chapter 2
P2.8
(a)
(b)
P2.9
*P2.10
58 m
≅
2.5 s
54 m
At t = 4.0 s, the slope is v ≅
≅
3s
49 m
At t = 3.0 s, the slope is v ≅
≅
3.4 s
36 m
At t = 2.0 s , the slope is v ≅
≅
4.0 s
At t = 5.0 s, the slope is v ≅
23 m s .
18 m s .
14 m s .
9.0 m s .
∆v 23 m s
≅
≅ 4.6 m s 2
∆t
5.0 s
(c)
a=
(d)
Initial velocity of the car was zero .
(a)
v=
(b)
v=
(c)
v=
(d)
v=
(5 − 0 ) m
(1 − 0) s
= 5 ms
(5 − 10) m
(4 − 2) s
= −2.5 m s
(5 m − 5 m)
(5 s − 4 s)
0 − (−5 m)
(8 s − 7 s )
= 0
= +5 m s
FIG. P2.9
Once it resumes the race, the hare will run for a time of
t=
x f − xi
vx
=
1 000 m − 800 m
= 25 s .
8 ms
In this time, the tortoise can crawl a distance
a
f
x f − xi = 0.2 m s ( 25 s)= 5.00 m .
25
26
Motion in One Dimension
Section 2.3
P2.11
Acceleration
Choose the positive direction to be the outward direction, perpendicular to the wall.
v f = vi + at : a =
P2.12
(a)
a
f
∆v 22.0 m s − −25.0 m s
=
= 1.34×10 4 m s 2 .
∆t
3.50 ×10−3 s
Acceleration is constant over the first ten seconds, so at the end,
c
h
v f = vi + at = 0 + 2.00 m s 2 (10.0 s)= 20.0 m s .
Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the
velocity changes to
c
h
v f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s .
(b)
In the first ten seconds,
1 2
1
2
at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m .
2
2
c
x f = x i + vi t +
h
Over the next five seconds the position changes to
x f = xi + vi t +
a
f
1 2
at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m .
2
And at t = 20.0 s ,
x f = x i + vi t +
*P2.13
(a)
a
f
1 2
1
2
at = 200 m + 20.0 m s (5.00 s)+ −3.00 m s 2 (5.00 s) = 262 m .
2
2
c
h
distance traveled
. During the first
∆t
quarter mile segment, Secretariat’s average speed was
The average speed during a time interval ∆t is v =
v1 =
0.250 mi 1 320 ft
=
= 52.4 ft s
25.2 s
25.2 s
b35.6 mi hg .
During the second quarter mile segment,
v2 =
1 320 ft
= 55.0 ft s
24.0 s
b37.4 mi hg .
For the third quarter mile of the race,
v3 =
1 320 ft
= 55.5 ft s
23.8 s
b37.7 mi hg ,
and during the final quarter mile,
v4 =
continued on next page
1 320 ft
= 57.4 ft s
23.0 s
b39.0 mi hg .
Chapter 2
(b)
27
Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the race
was
v f − vi
a=
P2.14
(a)
=
total elapsed time
57.4 ft s − 0
= 0.598 ft s 2 .
( 25. 2 + 24.0 + 23.8 + 23.0) s
Acceleration is the slope of the graph of v vs t.
a (m/s2)
2.0
For 0 < t < 5.00 s, a = 0 .
1.6
For 15.0 s < t < 20.0 s , a = 0 .
For 5.0 s < t < 15.0 s , a =
a=
v f − vi
t f − ti
1.0
.
8.00 − (−8.00)
15.0 − 5.00
0.0
= 1.60 m s 2
t (s)
0
5
a=
v f − vi
t f − ti
(i)
For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = −8.00 m s ,
t f = 15.0 s
v f = 8.00 m s
a=
(ii)
x = 2.00 + 3.00t − t 2 , v =
At t = 3.00 s :
v f − vi
t f − ti
=
a
f
8.00 − −8.00
= 1.60 m s 2 .
15.0 − 5.00
ti = 0 , vi = −8.00 m s , t f = 20.0 s , v f = 8.00 m s
a=
P2.15
15
FIG. P2.14
We can plot a(t ) as shown.
(b)
10
v f − vi
t f − ti
=
8.00 − (−8.00)
20.0 − 0
dx
dv
= 3.00 − 2.00t , a =
= −2.00
dt
dt
(a)
x = ( 2.00 + 9.00 − 9.00) m = 2.00 m
(b)
v = (3.00 − 6.00) m s = −3.00 m s
(c)
a = −2.00 m s 2
= 0.800 m s 2
20
28
P2.16
Motion in One Dimension
(a)
2
At t = 2.00 s , x = 3.00( 2.00) − 2.00( 2.00)+ 3.00 m = 11.0 m.
a f
At t = 3.00 s , x = 3.00 9.00
2
a f
− 2.00 3.00 + 3.00 m = 24.0 m
so
v=
(b)
∆x 24.0 m − 11.0 m
= 13.0 m s .
=
3.00 s − 2.00 s
∆t
At all times the instantaneous velocity is
v=
d
3.00t 2 − 2.00t + 3.00 = (6.00t − 2.00) m s
dt
c
h
At t = 2.00 s , v = 6.00( 2.00)− 2.00 m s = 10.0 m s .
At t = 3.00 s , v = 6.00(3.00)− 2.00 m s = 16.0 m s .
P2.17
a=
(d)
At all times a =
(a)
a=
(b)
Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 .
(c)
a = 0 , at t = 6 s , and also for t > 10 s .
(d)
Maximum negative acceleration is at t = 8 s, and is approximately −1.5 m s 2 .
Section 2.4
P2.18
∆v 16.0 m s − 10.0 m s
=
= 6.00 m s 2
∆t
3.00 s − 2.00 s
(c)
d
(6.00 − 2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ).
dt
∆v 8.00 m s
=
= 1.3 m s 2
∆t
6.00 s
Motion Diagrams
(a)
(b)
(c)
(d)
(e)
continued on next page
Chapter 2
(f)
Section 2.5
P2.19
29
One way of phrasing the answer: The spacing of the successive positions would change
with less regularity.
Another way: The object would move with some combination of the kinds of motion shown
in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude
and direction.
One-Dimensional Motion with Constant Acceleration
c
From v 2f = vi2 + 2 ax , we have 10.97 ×10 3 m s
h
2
= 0 + 2 a( 220 m) , so that a = 2.74×10 5 m s 2
which is a = 2.79 ×10 4 times g .
P2.20
P2.21
(a)
x f − xi =
(b)
a=
v f − vi
t
a
f
1
1
vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s .
2
2
c
=
h
2.80 m s − 6.61 m s
= −0.448 m s 2
8.50 s
Given vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = −5.00 cm ,
1
1 2
2
at : −5.00 − 3.00 = 12.0( 2.00)+ a( 2.00)
2
2
32.0
a =−
= −16.0 cm s 2 .
−8.00 = 24.0 + 2 a
2
x f − x i = vi t +
*P2.22
(a)
Let i be the state of moving at 60 mi h and f be at rest
d
0 = b60 mi hg
2
+ 2 a x x f − xi
v xf2 = v xi
ax
(b)
Similarly,
fFGH 5 1280mift IJK
−3 600 mi F 5 280 ft I F 1 h I
=
G
JG
J = −21.8 mi h ⋅ s
242 h H 1 mi K H 3 600 s K
F 1 609 m IJ FG 1 h IJ = −9.75 m s
= −21.8 mi h ⋅ s G
H 1 mi K H 3 600 s K
2
a
+ 2 a x 121 ft − 0
2
b
0 = 80 mi h
ax = −
(c)
i
b
g
2
6 400 5 280
b
a
+ 2 a x 211 ft − 0
422 3 600
g
g
2
.
f
mi h ⋅ s = −22.2 mi h ⋅ s = −9.94 m s 2 .
Let i be moving at 80 mi h and f be moving at 60 mi h .
d i
b60 mi hg = b80 mi hg + 2a a211 ft − 121 ftf
2 800b5 280 g
a =−
mi h ⋅ s = −22.8 mi h ⋅ s = −10.2 m s
2a90fb3 600g
2
+ 2 a x x f − xi
v xf2 = v xi
2
2
x
x
2
.
30
*P2.23
Motion in One Dimension
(a)
Choose the initial point where the pilot reduces the throttle and the final point where the
boat passes the buoy:
x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = −3.5 m s 2 , t = ?
x f = xi + v xi t +
1
axt 2 :
2
a
f
100 m = 0 + 30 m s t +
1
−3.5 m s 2 t 2
2
c
h
c1.75 m s ht − a30 m sft + 100 m = 0 .
2
2
We use the quadratic formula:
t=
t=
c
−b ± b 2 − 4ac
2a
h
30 m s ± 900 m 2 s 2 − 4 1.75 m s 2 (100 m)
c
2 1.75 m s
2
h
=
30 m s ± 14.1 m s
3.5 m s 2
= 12.6 s or 4.53 s .
The smaller value is the physical answer. If the boat kept moving with the same acceleration,
it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.
P2.24
e
j
(b)
v xf = v xi + a x t = 30 m s − 3.5 m s 2 4.53 s = 14.1 m s
(a)
Total displacement = area under the v , t curve from t = 0
to 50 s.
a f
b
b
ga f b
ga f
ga
f
1
50 m s 15 s + 50 m s 40 − 15 s
2
1
+ 50 m s 10 s
2
∆x = 1 875 m
∆x =
(b)
From t = 10 s to t = 40 s , displacement is
(c)
b
ga f b
ga f
1
50 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m .
2
∆v (50 − 0) m s
0 ≤ t ≤ 15 s : a1 =
=
= 3.3 m s 2
∆t
15 s − 0
15 s < t < 40 s : a 2 = 0
∆x =
40 s ≤ t ≤ 50 s : a 3 =
continued on next page
∆v (0 − 50) m s
=
= −5.0 m s 2
∆t
50 s − 40 s
FIG. P2.24
Chapter 2
(d)
1
1
a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 2
2
2
c
h
c
h
(i)
x1 = 0 +
(ii)
1
x 2 = (15 s) 50 m s − 0 + 50 m s (t − 15 s) or x 2 = 50 m s t − 375 m
2
(iii)
For 40 s ≤ t ≤ 50 s ,
a
x3 =
f
a
f
FG area under v vs t IJ + 1 a (t − 40 s) + a50 m sf(t − 40 s)
H from t = 0 to 40 sK 2
2
3
or
x 3 = 375 m + 1 250 m +
ja
1
−5.0 m s 2 t − 40 s
2
e
f + b50 m sgat − 40 sf
2
which reduces to
b
g e
j
x 3 = 250 m s t − 2.5 m s 2 t 2 − 4 375 m .
P2.25
total displacement 1 875 m
=
= 37.5 m s
total elapsed time
50 s
(e)
v=
(a)
Compare the position equation x = 2.00 + 3.00t − 4.00t 2 to the general form
x f = xi + vi t +
1 2
at
2
to recognize that x i = 2.00 m, vi = 3.00 m s, and a = −8.00 m s 2 . The velocity equation,
v f = vi + at , is then
c
h
v f = 3.00 m s − 8.00 m s 2 t .
The particle changes direction when v f = 0 , which occurs at t =
time is:
a
x = 2.00 m + 3.00 m s
(b)
fFGH 38 sIJK − c4.00 m s hFGH 38 sIJK
2
3
s . The position at this
8
2
= 2.56 m .
2v
1 2
at , observe that when x f = xi , the time is given by t = − i . Thus,
a
2
when the particle returns to its initial position, the time is
From x f = xi + vi t +
t=
c
a
−2 3.00 m s
−8.00 m s
and the velocity is v f = 3.00 m s − 8.00 m s 2
2
f=3 s
hFGH 34 sIJK =
4
−3.00 m s .
31
32
*P2.26
Motion in One Dimension
The time for the Ford to slow down we find from
1
v xi + v xf t
2
2 250 m
2 ∆x
t=
=
= 6.99 s .
v xi + v xf 71.5 m s + 0
x f = xi +
d
a
i
f
Its time to speed up is similarly
t=
2(350 m)
0 + 71.5 m s
= 9.79 s .
The whole time it is moving at less than maximum speed is 6.99 s + 5.00 s + 9.79 s = 21.8 s . The
Mercedes travels
a
fb ga
1
1
v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s
2
2
= 1 558 m
d
x f = xi +
i
f
while the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m − 600 m = 958 m .
P2.27
(a)
c
h
vi = 100 m s , a = −5.00 m s 2 , v f = vi + at so 0 = 100 − 5t , v 2f = vi2 + 2 a x f − xi so
2
c
h
0 = (100 ) − 2(5.00) x f − 0 . Thus x f = 1 000 m and t = 20.0 s .
P2.28
(b)
At this acceleration the plane would overshoot the runway: No .
(a)
Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = −2.00 m s 2 . Use these
values in the general equation
x f = xi + vi t +
to find
a
1 2
at
2
f 12 c−2.00 m s ht
2
x f = 0 + 30.0t m s +
when t is in seconds
c
2
h
x f = 30.0t − t 2 m .
e
j
To find an equation for the velocity, use v f = vi + at = 30.0 m s + −2.00 m s 2 t ,
v f = (30.0 − 2.00t ) m s .
(b)
The distance of travel x f becomes a maximum, x max , when v f = 0 (turning point in the
motion). Use the expressions found in part (a) for v f to find the value of t when x f has its
maximum value:
From v f = (3.00 − 2.00t ) m s , v f = 0 when t = 15.0 s. Then
c
h
2
x max = 30.0t − t 2 m = (30.0)(15.0)−(15.0) = 225 m .
Chapter 2
P2.29
33
In the simultaneous equations:
R|
S|x
T
v xf = v xi + a x t
f
− xi =
R|v = v − c5.60 m s h(4.20 s)U|
U|
ht V|W we have S|T 62.4 m = 12 cv + v h(4.20 s) V|W .
xf
1
v xi + v xf
2
c
So substituting for v xi gives 62.4 m =
2
xi
xi
xf
1
v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s)
2
c
h
14.9 m s = v xf +
1
5.60 m s 2 ( 4.20 s).
2
c
h
Thus
v xf = 3.10 m s .
P2.30
Take any two of the standard four equations, such as
substitute into the other: v xi = v xf − a x t
x f − xi =
R|
S|x
T
v xf = v xi + a x t
f
− xi =
1
v xi + v xf
2
c
U|
ht V|W. Solve one for v
1
v xf − a x t + v xf t .
2
c
h
Thus
1
x f − xi = v xf t − a x t 2 .
2
Back in problem 29, 62.4 m = v xf ( 4.20 s)−
v xf =
P2.31
v f − vi
(a)
a=
(b)
x f = vi t +
t
=
632
e j=
5 280
3 600
1.40
a
fFGH
1
2
−5.60 m s 2 ( 4. 20 s)
2
c
h
62.4 m − 49.4 m
= 3.10 m s .
4.20 s
−662 ft s 2 = −202 m s 2
I a f a fa f
JK
5 280
1 2
1
1.40 − 662 1.40
at = 632
2
3 600
2
2
= 649 ft = 198 m
xi ,
and
34
P2.32
Motion in One Dimension
(a)
The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yields
t=
v f − vi
a
=
20.0 m s − 0 m s
2.00 m s 2
= 10.0 s .
The total time is thus
10.0 s + 20.0 s + 5.00 s = 35.0 s .
(b)
The average velocity is the total distance traveled divided by the total time taken. The
distance traveled during the first 10.0 s is
x 1 = vt =
FG 0 + 20.0 IJ(10.0)= 100 m .
H 2 K
With a being 0 for this interval, the distance traveled during the next 20.0 s is
x 2 = vi t +
1 2
at = ( 20.0)( 20.0)+ 0 = 400 m.
2
The distance traveled in the last 5.00 s is
x 3 = vt =
FG 20.0 + 0 IJ(5.00)= 50.0 m.
H 2 K
The total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity is
x 550
= 15.7 m s .
given by v = =
t 35.0
P2.33
We have vi = 2.00 ×10 4 m s, v f = 6.00 ×10 6 m s , x f − xi = 1.50 ×10−2 m .
c
c
h
h
2 1.50 ×10−2 m
2 x f − xi
1
=
= 4.98 ×10−9 s
vi + v f t : t =
4
6
2
vi + v f
2.00 ×10 m s + 6.00 ×10 m s
c
h
(a)
x f − xi =
(b)
v 2f = vi2 + 2 a x x f − xi :
d
ax =
i
v 2f − vi2
2( x f − xi )
e6.00 × 10
=
6
ms
j − e2.00 × 10
2
2(1.50 × 10 −2 m)
4
ms
j
2
= 1.20 × 10 15 m s 2
Chapter 2
*P2.34
(a)
c
h
c
2
v xf2 = v xi
+ 2 a x x f − x i : 0.01 3 ×10 8 m s
ax
(b)
c3×10
=
6
h
2
ms
80 m
= 0 + 2 a x ( 40 m)
h
2
= 1.12 ×10 11 m s 2
We must find separately the time t1 for speeding up and the time t 2 for coasting:
x f − xi =
x f − xi =
1
1
v xf + v xi t1 : 40 m = 3 × 10 6 m s + 0 t1
2
2
t1 = 2.67 × 10 −5 s
d
i
e
j
1
1
v xf + v xi t 2 : 60 m = 3 × 10 6 m s + 3 × 10 6 m s t 2
2
2
t 2 = 2.00 × 10 −5 s
d
i
e
j
total time = 4.67 ×10−5 s .
*P2.35
(a)
Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t gives
v c = 0 + am tm
am =
(b)
vc
.
tm
The displacement between 0 and t m is
x f − xi = v xi t +
1 vc 2
1
1
axt 2 = 0 +
t m = v c tm .
2 tm
2
2
The displacement between t m and t 0 is
x f − xi = v xi t +
a
f
1
a x t 2 = v c t0 − tm + 0 .
2
The total displacement is
∆x =
FG
H
1
1
v c t m + v c t 0 − v c t m = v c t 0 − tm
2
2
IJ
K
.
(c)
For constant v c and t 0 , ∆x is minimized by maximizing t m to t m = t 0 . Then
v t
1
∆x min = v c t 0 − t 0 = c 0 .
2
2
(e)
This is realized by having the servo motor on all the time.
(d)
We maximize ∆x by letting t m approach zero. In the limit ∆x = v c t 0 − 0 = v c t 0 .
(e)
This cannot be attained because the acceleration must be finite.
FG
H
IJ
K
a
f
35
36
*P2.36
Motion in One Dimension
Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its
motion from entry to exit,
1
axt 2
2
1
A = 0 + vi ∆t d + a∆t d2 = v d ∆t d
2
1
v d = vi + a∆t d
2
x f = xi + v xi t +
(a)
The speed halfway through the photogate in space is given by
2
v hs
= vi2 + 2 a
FG A IJ = v
H 2K
2
i
+ av d ∆t d .
v hs = vi2 + av d ∆t d and this is not equal to v d unless a = 0 .
(b)
The speed halfway through the photogate in time is given by v ht = vi + a
FG ∆t IJ and this is
H 2K
d
equal to v d as determined above.
P2.37
(a)
Take initial and final points at top and bottom of the incline. If the ball starts from rest,
vi = 0 , a = 0.500 m s 2 , x f − xi = 9.00 m .
Then
d
i
ja
e
f
v 2f = vi2 + 2 a x f − xi = 0 2 + 2 0.500 m s 2 9.00 m
v f = 3.00 m s .
(b)
x f − x i = vi t +
1 2
at
2
1
0.500 m s 2 t 2
2
t = 6.00 s
e
9.00 = 0 +
(c)
Take initial and final points at the bottom of the planes and the top of the second plane,
respectively:
vi = 3.00 m s, v f = 0 , x f − xi = 15.00 m.
c
h
v 2f = vi2 + 2 a x f − xi gives
a=
(d)
j
v 2f − vi2
c
2 x f − xi
h
=
a
0 − 3.00 m s
2(15.0 m)
f
2
= −0.300 m s 2 .
Take the initial point at the bottom of the planes and the final point 8.00 m along the second:
vi = 3.00 m s, x f − xi = 8.00 m , a = −0.300 m s 2
d
i b
v 2f = vi2 + 2 a x f − xi = 3.00 m s
v f = 2.05 m s .
g + 2e−0.300 m s ja8.00 mf = 4.20 m
2
2
2
s2
Chapter 2
P2.38
37
Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue’s car.
For her we have x is = 0 , vis = 30.0 m s , a s = −2.00 m s 2 so her position is given by
x s (t )= x is + vis t +
a
f
1
1
a s t 2 = 30.0 m s t + −2.00 m s 2 t 2 .
2
2
c
h
For the van, x iv = 155 m, viv = 5.00 m s , a v = 0 and
x v (t )= xiv + viv t +
a
f
1
a v t 2 = 155 + 5.00 m s t + 0 .
2
To test for a collision, we look for an instant t c when both are at the same place:
30.0t c − t c2 = 155 + 5.00t c
0 = t c2 − 25.0t c + 155 .
From the quadratic formula
2
tc =
25.0 ± ( 25.0) − 4(155)
2
= 13.6 s or 11.4 s .
The smaller value is the collision time. (The larger value tells when the van would pull ahead again
if the vehicles could move through each other). The wreck happens at position
a
f
155 m + 5.00 m s (11.4 s)= 212 m .
*P2.39
As in the algebraic solution to Example 2.8, we let t
represent the time the trooper has been moving. We graph
x car = 45 + 45t
x (km)
1.5
car
1
and
2
x trooper = 1.5t .
They intersect at
police
officer
0.5
10
t = 31 s .
20
30
FIG. P2.39
40
t (s)
38
Motion in One Dimension
Section 2.6
P2.40
Freely Falling Objects
a
f
Choose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Then
yi = 0 , vi = 0 , and a = −g = −9.80 m s 2 . The position and the velocity at time t become:
y f − yi = vi t +
1
1
1 2
at : y f = − gt 2 = − 9.80 m s 2 t 2
2
2
2
e
j
and
c
h
v f = vi + at : v f = −gt = − 9.80 m s 2 t .
1
2
9.80 m s 2 (1.00 s) = −4.90 m
2
1
2
at t = 2.00 s : y f = − 9.80 m s 2 ( 2.00 s) = −19.6 m
2
1
2
at t = 3.00 s : y f = − 9.80 m s 2 (3.00 s) = −44.1 m
2
h
h
h
at t = 1.00 s : y f = −
(b)
at t = 1.00 s : v f = − 9.80 m s 2 (1.00 s)= −9.80 m s
at t = 2.00 s : v f
at t = 3.00 s : v f
P2.41
c
c
c
(a)
c
h
= −c9.80 m s h( 2.00 s)= −19.6 m s
= −c9.80 m s h(3.00 s)= −29.4 m s
2
2
Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is
that due to gravity, a = −g = −9.80 m s 2 . During the flight, Goff went 1 mile (1 609 m) up and then
1 mile back down. Determine his speed just after launch by considering his upward flight:
d
i
v 2f = vi2 + 2 a y f − yi :
jb
e
0 = vi2 − 2 9.80 m s 2 1 609 m
vi = 178 m s .
g
His time in the air may be found by considering his motion from just after launch to just before
impact:
y f − yi = vi t +
a
f
1
1 2
at : 0 = 178 m s t − −9.80 m s 2 t 2 .
2
2
c
h
The root t = 0 describes launch; the other root, t = 36.2 s , describes his flight time. His rate of pay
may then be found from
pay rate =
b
gb
g
$1.00
= 0.027 6 $ s 3 600 s h = $99.3 h .
36.2 s
We have assumed that the workman’s flight time, “a mile”, and “a dollar”, were measured to threedigit precision. We have interpreted “up in the sky” as referring to the free fall time, not to the
launch and landing times. Both the takeoff and landing times must be several seconds away from
the job, in order for Goff to survive to resume work.
Chapter 2
P2.42
39
1
We have y f = − gt 2 + vi t + yi
2
h a
c
f
0 = − 4.90 m s 2 t 2 − 8.00 m s t + 30.0 m .
Solving for t,
t=
8.00 ± 64.0 + 588
.
−9.80
Using only the positive value for t, we find that t = 1.79 s .
P2.43
1 2
2
at : 4.00 = (1.50)vi −(4.90)(1.50) and vi = 10.0 m s upward .
2
(a)
y f − yi = vi t +
(b)
v f = vi + at = 10.0 −(9.80)(1.50) = −4.68 m s
v f = 4.68 m s downward
P2.44
The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity).
Thus, in 0.20 s it will fall a distance of
∆y = vi t −
1 2
2
gt = 0 − 4.90 m s 2 (0. 20 s) = −0.20 m .
2
c
h
a
f
This distance is about twice the distance between the center of the bill and its top edge ≅ 8 cm .
Thus, David will be unsuccessful .
*P2.45
(a)
From ∆y = vi t +
1 2
at with vi = 0 , we have
2
t=
a f=
2 ∆y
2(−23 m)
a
−9.80 m s 2
c
= 2.17 s .
h
(b)
The final velocity is v f = 0 + −9.80 m s 2 ( 2.17 s)= −21.2 m s .
(c)
The time take for the sound of the impact to reach the spectator is
t sound =
∆y
v sound
=
23 m
= 6.76 ×10−2 s ,
340 m s
so the total elapsed time is t total = 2.17 s + 6.76 × 10 −2 s ≈ 2.23 s .
40
P2.46
Motion in One Dimension
At any time t, the position of the ball released from rest is given by y1 = h −
1 2
gt . At time t, the
2
1 2
gt . The time at which the
2
h
1
h
first ball has a position of y1 = is found from the first equation as = h − gt 2 , which yields
2
2
2
h
h
. To require that the second ball have a position of y 2 = at this time, use the second
t=
g
2
position of the ball thrown vertically upward is described by y 2 = vi t −
equation to obtain
F I
GH JK
h
h 1
h
= vi
− g
. This gives the required initial upward velocity of the second
2
g 2 g
ball as vi = gh .
P2.47
(a)
v f = vi − gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore,
c
h
vi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s .
(b)
y f − yi =
1
v f + vi t
2
c
h
y f − yi =
*P2.48
(a)
b
f
ga
1
29.4 m s 3.00 s = 44.1 m
2
Consider the upward flight of the arrow.
d i
0 = b100 m sg + 2e −9.8 m s j∆y
2
2
v yf
= v yi
+ 2 a y y f − yi
2
∆y =
(b)
2
10 000 m 2 s 2
19.6 m s 2
= 510 m
Consider the whole flight of the arrow.
y f = yi + v yi t +
b
1
ayt 2
2
g
0 = 0 + 100 m s t +
1
−9.8 m s 2 t 2
2
e
j
The root t = 0 refers to the starting point. The time of flight is given by
t=
P2.49
100 m s
4.9 m s 2
= 20.4 s .
Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m =
1
9.80 m s 2 t 2 , t = 0.782 s.
2
c
h
(a)
With the horse galloping at 10.0 m s, the horizontal distance is vt = 7.82 m .
(b)
t = 0.782 s
Chapter 2
P2.50
Take downward as the positive y direction.
(a)
While the woman was in free fall,
∆y = 144 ft , vi = 0 , and a = g = 32.0 ft s 2 .
Thus, ∆y = vi t +
before impact is:
1 2
at → 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just
2
c
h
c
h
v f = vi + gt = 0 + 32.0 ft s 2 (3.00 s)= 96.0 ft s .
(b)
While crushing the box, vi = 96.0 ft s , v f = 0 , and ∆y = 18.0 in. = 1.50 ft . Therefore,
a=
(c)
v 2f − vi2
a f
2 ∆y
=
a
0 − 96.0 ft s
f
2
2(1.50 ft )
Time to crush box: ∆t =
= −3.07 ×10 3 ft s 2 , or a = 3.07 ×10 3 ft s 2 upward .
2(1.50 ft)
∆y
∆y
= v +v =
or ∆t = 3.13 ×10−2 s .
f
i
v
0 + 96.0 ft s
2
P2.51
a f
y = 3.00t 3 : At t = 2.00 s , y = 3.00 2.00
3
= 24.0 m and
vy =
A
dy
= 9.00t 2 = 36.0 m s .
dt
If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is
y b = y bi + vi t −
1
1 2
gt = 24.0 + 36.0t − (9.80)t 2 .
2
2
Setting y b = 0 ,
0 = 24.0 + 36.0t − 4.90t 2 .
Solving for t, (only positive values of t count), t = 7.96 s .
*P2.52
Consider the last 30 m of fall. We find its speed 30 m above the ground:
y f = yi + v yi t +
1
ayt 2
2
a f 12 e−9.8 m s ja1.5 sf
2
0 = 30 m + v yi 1.5 s +
v yi =
2
−30 m + 11.0 m
= −12.6 m s .
1.5 s
Now consider the portion of its fall above the 30 m point. We assume it starts from rest
d
i
b−12.6 m sg = 0 + 2e−9.8 m s j∆y
2
2
v yf
= v yi
+ 2 a y y f − yi
2
∆y =
2
160 m 2 s 2
−19.6 m s 2
Its original height was then 30 m + −8.16 m = 38.2 m .
= −8.16 m .
41
42
Motion in One Dimension
Section 2.7
P2.53
(a)
Kinematic Equations Derived from Calculus
J=
da
= constant
dt
da = Jdt
z
a = J dt = Jt + c 1
but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + ai
dv
dt
dv = adt
a=
z zb
v = adt =
g
Jt + ai dt =
but v = vi when t = 0, so c 2 = vi and v =
1 2
Jt + ai t + c 2
2
1 2
Jt + ai t + vi
2
dx
dt
dx = vdt
v=
z z FGH
x = vdt =
IJ
K
1 2
Jt + ai t + vi dt
2
1 3 1 2
Jt + ai t + vi t + c 3
6
2
x = xi
x=
when t = 0, so c 3 = xi . Therefore, x =
(b)
a
a 2 = Jt + ai
c
f
2
= J 2 t 2 + ai2 + 2 Jai t
a 2 = ai2 + J 2 t 2 + 2 Jai t
FG
H
1 3 1 2
Jt + ai t + vi t + xi .
6
2
h
1 2
a 2 = ai2 + 2 J
Jt + ai t
2
IJ
K
Recall the expression for v: v =
a
f
1 2
1
Jt + ai t + vi . So v − vi = Jt 2 + ai t . Therefore,
2
2
a
a 2 = ai2 + 2 J v − vi
f
.
Chapter 2
P2.54
(a)
See the graphs at the right.
Choose x = 0 at t = 0.
At t = 3 s, x =
a
f
1
8 m s (3 s)= 12 m .
2
a
f
At t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m .
At t = 7 s, x = 28 m +
P2.55
a
f
1
8 m s ( 2 s)= 36 m .
2
8 ms
= 2.67 m s 2 .
3s
For 3 < t < 5 s, a = 0 .
(b)
For 0 < t < 3 s, a =
(c)
For 5 s < t < 9 s , a = −
(d)
At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m .
(e)
At t = 9 s, x = 36 m +
(a)
16 m s
= −4 m s 2 .
4s
a
a=
f
a
f
1
−8 m s ( 2 s)= 28 m .
2
FIG. P2.54
dv d
= −5.00 ×10 7 t 2 + 3.00 ×10 5 t
dt dt
c
h
a = − 10.0 ×10 7 m s 3 t + 3.00 ×10 5 m s 2
Take x i = 0 at t = 0. Then v =
dx
dt
z ze
t
t
0
0
x − 0 = vdt =
j
−5.00 × 10 7 t 2 + 3.00 × 10 5 t dt
t3
t2
+ 3.00 × 10 5
3
2
3 3
7
x = − 1.67 × 10 m s t + 1.50 × 10 5 m s 2 t 2 .
x = −5.00 × 10 7
e
(b)
j e
c
h
The bullet escapes when a = 0 , at − 10.0 ×10 7 m s 3 t + 3.00 ×10 5 m s 2 = 0
t=
(c)
j
c
hc
New v = −5.00 ×10 7 3.00 ×10−3
3.00 ×10 5 s
= 3.00 ×10−3 s .
10.0 ×10 7
h + c3.00×10 hc3.00×10 h
2
5
−3
v = −450 m s + 900 m s = 450 m s .
(d)
c
hc
x = − 1.67 ×10 7 3.00 ×10−3
h + c1.50×10 hc3.00×10 h
3
x = −0.450 m + 1.35 m = 0.900 m
5
−3 2
43
44
P2.56
Motion in One Dimension
a=
dv
= −3.00 v 2 , vi = 1.50 m s
dt
Solving for v,
dv
= −3.00 v 2
dt
z
v
z
t
v −2 dv = −3.00 dt
v = vi
−
When v =
t =0
1 1
1 1
+ = −3.00t or 3.00t = − .
v vi
v vi
vi
1
, t=
= 0.222 s .
2
3.00 vi
Additional Problems
*P2.57
a f
The distance the car travels at constant velocity, v 0 , during the reaction time is ∆x 1 = v 0 ∆t r . The
time for the car to come to rest, from initial velocity v 0 , after the brakes are applied is
t2 =
v f − vi
a
=
0 − v0
v
=− 0
a
a
and the distance traveled during this braking period is
a∆xf
2
= vt 2 =
Fv
GH
+ vi
f
2
I t = FG 0 + v IJ FG − v IJ = − v .
JK H 2 K H a K 2 a
0
2
0
2
0
Thus, the total distance traveled before coming to a stop is
a f + a ∆x f
sstop = ∆x
*P2.58
(a)
1
2
= v 0 ∆t r −
v 02
.
2a
v 02
(See the solution to Problem 2.57) from the
2a
intersection of length s i when the light turns yellow, the distance the car must travel before
the light turns red is
v2
∆x = sstop + si = v 0 ∆t r − 0 + si .
2a
If a car is a distance sstop = v 0 ∆t r −
Assume the driver does not accelerate in an attempt to “beat the light” (an extremely
dangerous practice!). The time the light should remain yellow is then the time required for
the car to travel distance ∆x at constant velocity v 0 . This is
v2
∆t light
(b)
v
s
∆x v 0 ∆t r − 20a + si
=
=
= ∆t r − 0 + i .
v0
v0
2 a v0
With si = 16 m, v = 60 km h , a = −2.0 m s 2 , and ∆t r = 1.1 s ,
∆t light = 1.1 s −
F 0.278 m s I + 16 m F 1 km h I =
G
J
G
J
2e −2.0 m s j H 1 km h K 60 km h H 0.278 m s K
60 km h
2
6. 23 s .
Chapter 2
*P2.59
(a)
(b)
As we see from the graph, from about −50 s to 50 s
Acela is cruising at a constant positive velocity in
the +x direction. From 50 s to 200 s, Acela
accelerates in the +x direction reaching a top speed
of about 170 mi/h. Around 200 s, the engineer
applies the brakes, and the train, still traveling in
the +x direction, slows down and then stops at
350 s. Just after 350 s, Acela reverses direction (v
becomes negative) and steadily gains speed in the
−x direction.
200
∆v
100
∆t
100 200 300 400
0
–50 0
t (s)
–100
FIG. P2.59(a)
The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent
to the v versus t curve in this interval. From the tangent line shown, we find
a = slope =
(c)
45
∆v (155 − 45) mi h
=
= 2. 2 mi h s = 0.98 m s 2 .
(100 − 50) s
∆t
a
Let us use the fact that the area under the v versus
t curve equals the displacement. The train’s
displacement between 0 and 200 s is equal to the
area of the gray shaded region, which we have
approximated with a series of triangles and
rectangles.
f
200
∆x 0 → 200 s = area 1 + area 2 + area 3 + area 4 + area 5
5
100
0
4
1 2
0
b
ga f b
ga f
+ b160 mi hga100 sf
1
+ a50 sfb100 mi hg
2
1
+ a100 sfb170 mi h − 160 mi hg
2
= 24 000bmi hgasf
3
100 200 300 400
≈ 50 mi h 50 s + 50 mi h 50 s
FIG. P2.59(c)
Now, at the end of our calculation, we can find the displacement in miles by converting
hours to seconds. As 1 h = 3 600 s ,
∆x 0 → 200 s ≈
F 24 000 mi I asf =
GH 3 600 s JK
6.7 mi .
t (s)
46
*P2.60
Motion in One Dimension
Average speed of every point on the train as the first car passes Liz:
∆x 8.60 m
=
= 5.73 m s.
1.50 s
∆t
The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway
8.60 m
through the next 1.10 s, the speed of the train is
= 7.82 m s . The time required for the speed
1.10 s
to change from 5.73 m/s to 7.82 m/s is
1
1
(1.50 s)+ (1.10 s)= 1.30 s
2
2
so the acceleration is: a x =
P2.61
∆v x 7.82 m s − 5.73 m s
=
= 1.60 m s 2 .
∆t
1.30 s
The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then
mm d
v xi = 1.04 mm d and a x = 0.132
. The increase in the length of the hair (i.e., displacement)
w
during a time of t = 5.00 w = 35.0 d is
FG
H
∆x = v xi t +
b
IJ
K
1
axt 2
2
f 12 b0.132 mm d ⋅ wga35.0 dfa5.00 wf
ga
∆x = 1.04 mm d 35.0 d +
or ∆x = 48.0 mm .
P2.62
Let point 0 be at ground level and point 1 be at the end of the engine burn. Let
point 2 be the highest point the rocket reaches and point 3 be just before
impact. The data in the table are found for each phase of the rocket’s motion.
a f
v 2f − 80.0
(0 to 1)
2
a fb
= 2 4.00 1 000
g
120 = 80.0 +( 4.00)t
c
2
(1 to 2)
0 −(120) = 2(−9.80) x f − xi
h
so
v f = 120 m s
giving
t = 10.0 s
giving
x f − xi = 735 m
0 − 120 = −9.80t
giving
This is the time of maximum height of the rocket.
a
fb
v 2f − 0 = 2 −9.80 −1 735
(2 to 3)
g
v f = −184 = (−9.80)t
(a)
t total = 10 + 12.2 + 18.8 = 41.0 s
(b)
cx
f
− xi
h
total
= 1.73 km
continued on next page
t = 12.2 s
giving
t = 18.8 s
FIG. P2.62
Chapter 2
(c)
v final = −184 m s
0
#1
#2
#3
P2.63
P2.64
t
0.0
10.0
22.2
41.0
Launch
End Thrust
Rise Upwards
Fall to Earth
a
x
0
1 000
1 735
0
f
v
80
120
0
–184
a
+4.00
+4.00
–9.80
–9.80
Distance traveled by motorist = 15.0 m s t
1
Distance traveled by policeman = 2.00 m s 2 t 2
2
c
h
(a)
intercept occurs when 15.0t = t 2 , or t = 15.0 s
(b)
v(officer)= 2.00 m s 2 t = 30.0 m s
(c)
x(officer )=
c
h
1
2.00 m s 2 t 2 = 225 m
2
c
h
Area A1 is a rectangle. Thus, A1 = hw = v xi t .
1
1
Area A 2 is triangular. Therefore A 2 = bh = t v x − v xi .
2
2
The total area under the curve is
b
A = A1 + A 2 = v xi t +
bv
x
g
vx
vx
A2
vxi
g
A1
− v xi t
2
0
and since v x − v xi = a x t
t
FIG. P2.64
A = v xi t +
1
axt 2 .
2
The displacement given by the equation is: x = v xi t +
same result as above for the total area.
1
a x t 2 , the
2
t
47
48
P2.65
Motion in One Dimension
(a)
Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is
achieved in time t1 we can use the following three equations:
x=
a
a
f
f
1
v + vi t1 , 100 − x = v 10.2 − t1 and v = vi + at1 .
2
The first two give
FG 1 t IJ v = FG10.2 − 1 t IJ at
H 2 K H 2 K
200
a=
b20.4 − t gt .
100 = 10.2 −
1
1
For Maggie: a =
1
1
200
a18.4fa2.00f =
200
For Judy: a =
a17.4fa3.00f =
(b)
1
5.43 m s 2
3.83 m s 2
v = a1 t
a fa f
Judy: v = a3.83fa3.00f =
Maggie: v = 5.43 2.00 = 10.9 m s
(c)
11.5 m s
At the six-second mark
x=
f
a fa f + a10.9fa4.00f = 54.3 m
a fa f + a11.5fa3.00f = 51.7 m
1
5.43 2.00
2
1
Judy: x = 3.83 3.00
2
Maggie: x =
a
1 2
at1 + v 6.00 − t1
2
2
2
Maggie is ahead by 2.62 m .
P2.66
a1 = 0.100 m s 2
1
1
x = 1 000 m = a1 t12 + v1 t 2 + a 2 t 22
2
2
at
at
1
1
1 000 = a1 t12 + a1 t1 − 1 1 + a 2 1 1
2
a2
2
a2
FG
H
IJ
K
FG IJ
H K
a 2 = −0.500 m s 2
t = t1 + t 2 and v1 = a1 t1 = −a 2 t 2
2
1 000 =
t1 =
t2 =
a1 t1
12.9
=
≈ 26 s
−a 2 0.500
FG
H
IJ
K
a
1
a1 1 − 1 t12
2
a2
20 000
= 129 s
1.20
Total time = t = 155 s
Chapter 2
P2.67
Let the ball fall 1.50 m. It strikes at speed given by
c
h
2
+ 2 a x f − xi :
v xf2 = v xi
c
h
v xf2 = 0 + 2 −9.80 m s 2 (−1.50 m)
v xf = −5.42 m s
and its stopping is described by
d
2
v xf2 = v xi
+ 2 a x x f − xi
b
0 = −5.42 m s
ax =
g
2
e
−2.00 × 10
j
+ 2 a x −10 −2 m
−29.4 m 2 s 2
−2
i
m
= +1.47 × 10 3 m s 2 .
Its maximum acceleration will be larger than the average acceleration we estimate by imagining
constant acceleration, but will still be of order of magnitude ~ 10 3 m s 2 .
*P2.68
(a)
x f = xi + v xi t +
1
a x t 2 . We assume the package starts from rest.
2
−145 m = 0 + 0 +
t=
(b)
x f = xi + v xi t +
1
−9.80 m s 2 t 2
2
c
2(−145 m)
−9.80 m s 2
h
= 5. 44 s
1
1
2
a x t 2 = 0 + 0 + −9.80 m s 2 (5.18 s) = −131 m
2
2
c
h
distance fallen = x f = 131 m
e
j
(c)
speed = v xf = v xi + a x t = 0 + −9.8 m s 2 5.18 s = 50.8 m s
(d)
The remaining distance is
145 m − 131.5 m = 13.5 m .
During deceleration,
v xi = −50.8 m s, v xf = 0, x f − xi = −13.5 m
c
h
2
v xf2 = v xi
+ 2 a x x f − xi :
a
0 = −50.8 m s
ax =
f
2
+ 2 a x (−13.5 m)
−2 580 m 2 s 2
= +95.3 m s 2 = 95.3 m s 2 upward .
2 −13.5 m
a
f
49
50
P2.69
Motion in One Dimension
(a)
1
1 2
at = 50.0 = 2.00t + (9.80)t 2 ,
2
2
4.90t 2 + 2.00t − 50.0 = 0
y f = v i1 t +
t=
−2.00 + 2.00 2 − 4( 4.90)(−50.0)
2( 4.90)
Only the positive root is physically meaningful:
t = 3.00 s after the first stone is thrown.
(b)
1 2
at and t = 3.00 − 1.00 = 2.00 s
2
1
2
substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) :
2
y f = vi 2 t +
vi2 = 15.3 m s downward
(c)
v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downward
v 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downward
P2.70
(a)
1
d = (9.80)t12
2
t1 + t 2 = 2.40
4.90t 22
d = 336 t 2
a
336t 2 = 4.90 2.40 − t 2
− 359.5t 2 + 28.22 = 0
t2 =
359.5 ± 358.75
= 0.076 5 s
t2 =
9.80
(b)
P2.71
(a)
(d)
359.5 ± 359.5 2 − 4( 4.90)( 28.22)
9.80
d = 336 t 2 = 26.4 m
In walking a distance ∆x , in a time ∆t , the length
of rope A is only increased by ∆x sin θ .
∆x
sin θ .
∴ The pack lifts at a rate
∆t
∆x
x
sin θ = v boy = v boy
∆t
A
x
2
x + h2
FG IJ
HK
d 1
dv v boy dx
=
+ v boy x
dt A
dt
A dt
v boy v boy x dA
x
dA
= v = v boy
a = v boy
− 2
, but
A
dt
A
dt
A
2
2
2 2
2
2
v boy
v boy h
h v boy
x
∴ a=
=
1− 2 =
2
3 2
A
A A
A
x 2 + h2
a=
F
GH
(c)
2
1
2
Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% .
2
v=
(b)
so
f
2
v boy
h
,0
v boy , 0
I
JK
c
h
FIG. P2.71
Chapter 2
P2.72
h = 6.00 m, v boy = 2.00 m s v =
However, x = v boy t : ∴ v =
(a)
(b)
v boy x
∆x
x
sin θ = v boy =
.
12
A
∆t
x 2 + h2
c
2
v boy
t
c
2
v boy
t2
+h
(a)
=
c 4t
4t
2
+ 36
h
12
.
a f vb m s g
ts
0
0
0.5
0.32
1
0.63
1.5
0.89
2
1.11
2.5
1.28
3
1.41
3.5
1.52
4
1.60
4.5
1.66
5
1.71
FIG. P2.72(a)
From problem 2.71 above, a =
a f aem s j
P2.73
h
2 12
h
2
h 2 v boy
cx
2
+h
h
2 3 2
=
2
h 2 v boy
c
2
v boy
t2
+h
h
2 3 2
=
c4t
144
2
+ 36
h
32
.
2
ts
0
0.67
0.5
0.64
1
0.57
1.5
0.48
2
0.38
2.5
0.30
3.
0.24
3.5
0.18
4.
0.14
4.5
0.11
5
0.09
FIG. P2.72(b)
We require x s = x k when t s = t k + 1.00
jb
1
3.50 m s 2 t k + 1.00
2
t k + 1.00 = 1.183t k
xs =
e
t k = 5.46 s .
ja
1
4.90 m s 2 5.46 s
2
e
f
2
(b)
xk =
(c)
v k = 4.90 m s 2 5.46 s = 26.7 m s
vs
e
ja f
= e3.50 m s ja6.46 sf =
2
= 73.0 m
22.6 m s
g
2
=
jb g
1
4.90 m s 2 t k
2
e
2
= xk
51
52
P2.74
Motion in One Dimension
Time
t (s)
0.00
Height
h (m)
5.00
0.25
5.75
0.50
6.40
0.75
6.94
1.00
7.38
1.25
∆h
(m)
∆t
(s)
v
(m/s)
midpt time
t (s)
0.75
0.25
3.00
0.13
0.65
0.25
2.60
0.38
0.54
0.25
2.16
0.63
0.44
0.25
1.76
0.88
0.34
0.25
1.36
1.13
0.24
0.25
0.96
1.38
0.14
0.25
0.56
1.63
0.03
0.25
0.12
1.88
–0.06
0.25
–0.24
2.13
–0.17
0.25
–0.68
2.38
–0.28
0.25
–1.12
2.63
–0.37
0.25
–1.48
2.88
–0.48
0.25
–1.92
3.13
–0.57
0.25
–2.28
3.38
–0.68
0.25
–2.72
3.63
–0.79
0.25
–3.16
3.88
–0.88
0.25
–3.52
4.13
–0.99
0.25
–3.96
4.38
–1.09
0.25
–4.36
4.63
–1.19
0.25
–4.76
4.88
7.72
1.50
7.96
1.75
8.10
2.00
8.13
2.25
8.07
2.50
7.90
2.75
7.62
3.00
7.25
3.25
6.77
3.50
6.20
3.75
5.52
4.00
4.73
4.25
3.85
4.50
2.86
4.75
1.77
5.00
0.58
TABLE P2.74
acceleration = slope of line is constant.
a =−1.63 m s 2 = 1.63 m s 2 downward
FIG. P2.74
53
Chapter 2
P2.75
The distance x and y are always related by x 2 + y 2 = L2 .
Differentiating this equation with respect to time, we have
y
B
x
dy
dx
2x + 2y
=0
dt
dt
L
y
dy
dx
= −v .
is v B , the unknown velocity of B; and
dt
dt
From the equation resulting from differentiation, we have
Now
O
FG IJ
H K
FG
H
A
x
dy
x dx
x
= − (−v).
=−
dt
y dt
y
But
v
α
FIG. P2.75
IJ
K
y
v
v 3
1
=
= 0.577 v .
= tanα so v B =
v . When α = 60.0° , v B =
tan 60.0°
3
x
tan α
ANSWERS TO EVEN PROBLEMS
(a) 2 × 10 −7 m s ; 1 × 10 −6 m s ;
P2.24
(b) 5 × 10 yr
(a) 1.88 km; (b) 1.46 km;
(c) see the solution;
(d) (i) x 1 = 1.67 m s 2 t 2 ;
P2.4
(a) 50.0 m s ; (b) 41.0 m s
(ii) x 2 = 50 m s t − 375 m ;
P2.6
(a) 27.0 m ;
2
(b) 27.0 m + 18.0 m s ∆t + 3.00 m s 2 ∆t ;
(iii) x 3
P2.2
8
b
g e
e
ja f
(c) 18.0 m s
P2.8
(a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0
P2.10
5.00 m
P2.12
(a) 20.0 m s ; 5.00 m s ; (b) 262 m
P2.14
P2.16
j
b
g
= b 250 m sgt − e 2.5 m s jt
2
2
− 4 375 m ;
(e) 37.5 m s
P2.26
958 m
P2.28
(a) x f = 30.0t − t 2 m; v f = 30.0 − 2t m s ;
e
j
a
f
(b) 225 m
1
a x t 2 ; 3.10 m s
2
P2.30
x f − xi = v xf t −
(a) see the solution;
(b) 1.60 m s 2 ; 0.800 m s 2
P2.32
(a) 35.0 s; (b) 15.7 m s
(a) 13.0 m s; (b) 10.0 m s; 16.0 m s;
P2.34
(a) 1.12 × 10 11 m s 2 ; (b) 4.67 × 10 −5 s
P2.36
(a) False unless the acceleration is zero;
see the solution; (b) True
(c) 6.00 m s 2 ; (d) 6.00 m s 2
P2.18
see the solution
P2.20
(a) 6.61 m s; (b) −0. 448 m s 2
P2.38
Yes; 212 m; 11.4 s
P2.22
(a) −21.8 mi h ⋅ s = −9.75 m s 2 ;
P2.40
(a) −4.90 m ; −19.6 m; −44.1 m;
(b) −9.80 m s; −19.6 m s; −29.4 m s
P2.42
1.79 s
(b) −22.2 mi h ⋅ s = −9.94 m s 2 ;
(c) −22.8 mi h ⋅ s = −10.2 m s 2
54
Motion in One Dimension
P2.44
No; see the solution
P2.60
1.60 m s 2
P2.46
The second ball is thrown at speed
vi = gh
P2.62
(a) 41.0 s; (b) 1.73 km; (c) −184 m s
P2.48
(a) 510 m; (b) 20.4 s
P2.64
v xi t +
P2.50
(a) 96.0 ft s ;
P2.66
155 s; 129 s
P2.68
(a) 5.44 s; (b) 131 m; (c) 50.8 m s ;
3
2
(b) a = 3.07 × 10 ft s upward ;
(c) ∆t = 3.13 × 10 −2 s
P2.52
1
a x t 2 ; displacements agree
2
(d) 95.3 m s 2 upward
38.2 m
P2.70
(a) 26.4 m; (b) 6.82%
2
P2.54
(a) and (b) see the solution; (c) −4 m s ;
(d) 34 m; (e) 28 m
P2.72
see the solution
P2.56
0.222 s
P2.74
see the solution; a x = −1.63 m s 2
P2.58
(a) see the solution; (b) 6.23 s
3
Vectors
CHAPTER OUTLINE
3.1
3.2
3.3
3.4
Coordinate Systems
Vector and Scalar
Quantities
Some Properties of Vectors
Components of a Vector
and Unit Vectors
ANSWERS TO QUESTIONS
Q3.1
No. The sum of two vectors can only be zero if they are in
opposite directions and have the same magnitude. If you walk
10 meters north and then 6 meters south, you won’t end up
where you started.
Q3.2
No, the magnitude of the displacement is always less than or
equal to the distance traveled. If two displacements in the same
direction are added, then the magnitude of their sum will be
equal to the distance traveled. Two vectors in any other
orientation will give a displacement less than the distance
traveled. If you first walk 3 meters east, and then 4 meters
south, you will have walked a total distance of 7 meters, but
you will only be 5 meters from your starting point.
Q3.3
The largest possible magnitude of R = A + B is 7 units, found when A and B point in the same
direction. The smallest magnitude of R = A + B is 3 units, found when A and B have opposite
directions.
Q3.4
Only force and velocity are vectors. None of the other quantities requires a direction to be described.
Q3.5
If the direction-angle of A is between 180 degrees and 270 degrees, its components are both
negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite
signs.
Q3.6
The book’s displacement is zero, as it ends up at the point from which it started. The distance
traveled is 6.0 meters.
Q3.7
85 miles. The magnitude of the displacement is the distance from the starting point, the 260-mile
mark, to the ending point, the 175-mile mark.
Q3.8
Vectors A and B are perpendicular to each other.
Q3.9
No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction,
not magnitude.
55
56
Vectors
Q3.10
Any vector that points along a line at 45° to the x and y axes has components equal in magnitude.
Q3.11
A x = B x and A y = B y .
Q3.12
Addition of a vector to a scalar is not defined. Think of apples and oranges.
Q3.13
One difficulty arises in determining the individual components. The relationships between a vector
and its components such as A x = A cos θ , are based on right-triangle trigonometry. Another problem
would be in determining the magnitude or the direction of a vector from its components. Again,
A = A x2 + A y2 only holds true if the two component vectors, A x and A y , are perpendicular.
Q3.14
If the direction of a vector is specified by giving the angle of the vector measured clockwise from the
positive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by the
magnitude of the vector.
SOLUTIONS TO PROBLEMS
Section 3.1
P3.1
P3.2
Coordinate Systems
a f
a fa f
y = r sin θ = a5.50 mf sin 240° = a5.50 mfa −0.866f =
x = r cos θ = 5.50 m cos 240° = 5.50 m −0.5 = −2.75 m
(a)
−4.76 m
x = r cos θ and y = r sin θ , therefore
x1 = 2.50 m cos 30.0° , y1 = 2.50 m sin 30.0° , and
a
f
a
f
bx , y g = a2.17 , 1.25f m
x = a3.80 mf cos 120° , y = a3.80 mf sin 120° , and
bx , y g = a−1.90, 3.29f m .
1
1
2
2
(b)
P3.3
2
2
d = ( ∆ x) 2 + ( ∆ y) 2 = 16.6 + 4.16 = 4.55 m
The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m.
(a)
We can use the Pythagorean theorem to find the distance from the origin to the fly.
distance = x 2 + y 2 =
(b)
θ = tan −1
FG 1 IJ = 26.6° ; r =
H 2K
a2.00 mf + a1.00 mf
2.24 m, 26.6°
2
2
= 5.00 m 2 = 2.24 m
Chapter 3
P3.4
(a)
d=
bx
2
− x1
g + by
2
2
− y1
g
2
=
c2.00 − −3.00 h + a−4.00 − 3.00f
2
2
d = 25.0 + 49.0 = 8.60 m
(b)
a2.00f + a−4.00f = 20.0 =
F 4.00 IJ = −63.4°
= tan G −
H 2.00 K
2
r1 =
θ1
r2 =
2
4.47 m
−1
a−3.00f + a3.00f
2
2
= 18.0 = 4.24 m
θ 2 = 135° measured from the +x axis.
P3.5
We have 2.00 = r cos 30.0°
r=
2.00
= 2.31
cos 30.0°
and y = r sin 30.0° = 2.31 sin 30.0° = 1.15 .
P3.6
We have r = x 2 + y 2 and θ = tan −1
(a)
FG y IJ .
H xK
The radius for this new point is
a− x f
2
+ y2 = x2 + y2 = r
and its angle is
tan −1
(b)
FG y IJ =
H −x K
180°− θ .
b g
( −2 x) 2 + ( −2 y) 2 = 2r . This point is in the third quadrant if x , y is in the first quadrant
b g
or in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180°+ θ .
(c)
b g
( 3 x) 2 + ( −3 y) 2 = 3r . This point is in the fourth quadrant if x , y is in the first quadrant
b g
or in the third quadrant if x , y is in the second quadrant. It is at an angle of − θ .
57
58
Vectors
Section 3.2
Vector and Scalar Quantities
Section 3.3
Some Properties of Vectors
P3.7
x
100 m
x = 100 m tan 35.0° = 70.0 m
tan 35.0° =
a
f
FIG. P3.7
P3.8
R = 14 km
θ = 65° N of E
R
13 km
θ
6 km
1 km
FIG. P3.8
P3.9
− R = 310 km at 57° S of W
(Scale: 1 unit = 20 km )
FIG. P3.9
P3.10
(a)
Using graphical methods, place the tail of
vector B at the head of vector A. The new
vector A + B has a magnitude of
6.1 at 112° from the x-axis.
y
A
A+B
(b)
The vector difference A − B is found by
placing the negative of vector B at the
head of vector A. The resultant vector
A − B has magnitude 14.8 units at an
angle of 22° from the + x-axis.
—B
B
A—B
x
O
FIG. P3.10
Chapter 3
P3.11
(a)
d = − 10.0 i = 10.0 m since the displacement is in a
C
straight line from point A to point B.
(b)
The actual distance skated is not equal to the straight-line
displacement. The distance follows the curved path of the
semi-circle (ACB).
s=
(c)
P3.12
59
5.00 m
d
B
A
FIG. P3.11
b g
1
2π r = 5π = 15.7 m
2
If the circle is complete, d begins and ends at point A. Hence, d = 0 .
Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1 . The resultant force
vector F1 + F2 is of magnitude 9.5 N and at an angle of 57° above the x -axis .
y
F1 + F2
F2
F1
x
0 1 2 3 N
FIG. P3.12
P3.13
(a)
The large majority of people are standing or sitting at this hour. Their instantaneous foot-tohead vectors have upward vertical components on the order of 1 m and randomly oriented
horizontal components. The citywide sum will be ~ 10 5 m upward .
(b)
Most people are lying in bed early Saturday morning. We suppose their beds are oriented
north, south, east, west quite at random. Then the horizontal component of their total vector
height is very nearly zero. If their compressed pillows give their height vectors vertical
components averaging 3 cm, and if one-tenth of one percent of the population are on-duty
nurses or police officers, we estimate the total vector height as ~ 10 5 0.03 m + 10 2 1 m
a
3
~ 10 m upward .
f
a f
60
P3.14
Vectors
N
Your sketch should be drawn to scale, and
should look somewhat like that pictured to
the right. The angle from the westward
direction, θ, can be measured to be
4° N of W , and the distance R from the
1m
W
15.0 meters
θ
R
sketch can be converted according to the
scale to be 7.9 m .
3.50
meters
30.0°
8.20
meters
E
S
FIG. P3.14
P3.15
To find these vector expressions graphically, we
draw each set of vectors. Measurements of the
results are taken using a ruler and protractor.
(Scale: 1 unit = 0.5 m )
(a)
A + B = 5.2 m at 60°
(b)
A – B = 3.0 m at 330°
(c)
B – A = 3.0 m at 150°
(d)
A – 2B = 5.2 m at 300°.
FIG. P3.15
*P3.16
The three diagrams shown below represent the graphical solutions for the three vector sums:
R 1 = A + B + C , R 2 = B + C + A , and R 3 = C + B + A . You should observe that R 1 = R 2 = R 3 ,
illustrating that the sum of a set of vectors is not affected by the order in which the vectors are
added.
100 m
C
B
A
A
B
R1
A
R2
C
B
FIG. P3.16
R3
C
Chapter 3
P3.17
The scale drawing for the graphical solution
should be similar to the figure to the right. The
magnitude and direction of the final displacement
from the starting point are obtained by measuring
d and θ on the drawing and applying the scale
factor used in making the drawing. The results
should be
(Scale: 1 unit = 20 ft )
d = 420 ft and θ = −3°
FIG. P3.17
Section 3.4
P3.18
Components of a Vector and Unit Vectors
Coordinates of the super-hero are:
a f a f
y = a100 mf sina −30.0°f =
x = 100 m cos −30.0° = 86.6 m
−50.0 m
FIG. P3.18
P3.19
A x = −25.0
A y = 40.0
A = A x2 + A y2 =
a−25.0f + a40.0f
2
2
= 47.2 units
We observe that
Ay
tan φ =
.
Ax
FIG. P3.19
So
φ = tan −1
F A I = tan 40.0 = tan a1.60f = 58.0° .
GH A JK 25.0
y
−1
x
The diagram shows that the angle from the +x axis can be found by subtracting from 180°:
θ = 180° − 58° = 122° .
P3.20
a
f
The person would have to walk 3.10 sin 25.0° = 1.31 km north , and
a
f
3.10 cos 25.0° = 2.81 km east .
61
62
P3.21
P3.22
Vectors
x = r cos θ and y = r sin θ , therefore:
j
b g e
j
b g e
j
x = 12.8 cos 150° , y = 12.8 sin 150° , and x , y = −11.1i + 6.40 j m
(b)
x = 3.30 cos 60.0° , y = 3.30 sin 60.0° , and x , y = 1.65 i + 2.86 j cm
(c)
x = 22.0 cos 215° , y = 22.0 sin 215° , and x , y = −18.0 i − 12.6 j in
a
a
f a f
f a f
a−25.0 mfi + a43.3 mfj
x = d cos θ = 50.0 m cos 120 = −25.0 m
y = d sin θ = 50.0 m sin 120 = 43.3 m
d=
*P3.23
b g e
(a)
(a)
Her net x (east-west) displacement is −3.00 + 0 + 6.00 = +3.00 blocks, while her net y (northsouth) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant
displacement is
R=
b x g + by g
net
2
net
2
=
a3.00f + a4.00f
2
2
= 5.00 blocks
and the angle the resultant makes with the x-axis (eastward direction) is
θ = tan −1
FG 4.00 IJ = tan a1.33f = 53.1° .
H 3.00 K
−1
The resultant displacement is then 5.00 blocks at 53.1° N of E .
(b)
*P3.24
The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .
Let i = east and j = north. The unicyclist’s displacement is, in meters
N
280 j + 220 i + 360 j − 300 i − 120 j + 60 i − 40 j − 90 i + 70 j .
R
R = −110 i + 550 j
=
a110 mf + a550 mf
2
2
at tan −1
= 561 m at 11.3° west of north .
110 m
west of north
550 m
The crow’s velocity is
v=
∆ x 561 m at 11.3° W of N
=
∆t
40 s
= 14.0 m s at 11.3° west of north .
E
FIG. P3.24
Chapter 3
P3.25
63
+x East, +y North
∑ x = 250 +125 cos 30° = 358 m
∑ y = 75 +125 sin 30°−150 = −12.5 m
c∑ xh + c∑ yh = a358f + a−12.5f
c∑ yh = − 12.5 = −0.0349
tan θ =
c∑ xh 358
2
d=
2
2
2
= 358 m
θ = −2.00°
d = 358 m at 2.00° S of E
P3.26
The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of
the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to
Chicago. In equation form:
d DC east = d DA east + d AC east = 730 cos 5.00°−560 sin 21.0° = 527 miles.
d DC north = d DA north + d AC north = 730 sin 5.00°+560 cos 21.0° = 586 miles.
By the Pythagorean theorem, d = ( d DC east ) 2 + ( d DC north ) 2 = 788 mi .
Then tan θ =
d DC north
= 1.11 and θ = 48.0° .
d DC east
Thus, Chicago is 788 miles at 48.0° northeast of Dallas .
P3.27
(a)
See figure to the right.
(b)
C = A + B = 2.00 i + 6.00 j + 3.00 i − 2.00 j = 5.00 i + 4.00 j
C = 25.0 + 16.0 at tan −1
FG 4 IJ =
H 5K
6.40 at 38.7°
D = A − B = 2.00 i + 6.00 j − 3.00 i + 2.00 j = −1.00 i + 8.00 j
a−1.00f + a8.00f at tan FGH −81.00.00 IJK
D = 8.06 at b180° − 82.9°g = 8.06 at 97.2°
2
D=
P3.28
2
−1
bx + x + x g + by + y + y g
= a3.00 − 5.00 + 6.00f + a 2.00 + 3.00 + 1.00 f
F 6.00 IJ = 56.3°
θ = tan G
H 4.00 K
d=
1
2
3
2
2
1
2
−1
2
3
2
= 52.0 = 7.21 m
FIG. P3.27
64
P3.29
Vectors
We have B = R − A :
A x = 150 cos120° = −75.0 cm
A y = 150 sin 120° = 130 cm
R x = 140 cos 35.0° = 115 cm
R y = 140 sin 35.0° = 80.3 cm
FIG. P3.29
Therefore,
a f
e
j
B = 115 − −75 i + 80.3 − 130 j = 190 i − 49.7 j cm
B = 190 2 + 49.7 2 = 196 cm
FG
H
θ = tan −1 −
P3.30
IJ
K
49.7
= −14.7° .
190
A = −8.70 i + 15.0 j and B = 13. 2 i − 6.60 j
A − B + 3C = 0 :
3C = B − A = 21.9 i − 21.6 j
C = 7.30 i − 7.20 j
or
C x = 7.30 cm ; C y = −7.20 cm
P3.31
(a)
aA + Bf = e3 i − 2 jj + e− i − 4jj =
2 i − 6 j
(b)
aA − Bf = e3i − 2jj − e− i − 4jj =
4i + 2 j
(c)
A + B = 2 2 + 6 2 = 6.32
(d)
A − B = 4 2 + 2 2 = 4.47
(e)
θ A+B = tan−1 −
θ A−B = tan−1
P3.32
(a)
FG 6 IJ = −71.6°=
H 2K
FG 2 IJ = 26.6°
H 4K
288°
D = A + B + C = 2 i + 4j
D = 2 2 + 4 2 = 4. 47 m at θ = 63.4°
(b)
E = − A − B + C = −6 i + 6 j
E = 6 2 + 6 2 = 8.49 m at θ = 135°
Chapter 3
P3.33
e
j
d1 = −3.50 j m
e
j
d 2 = 8.20 cos 45.0° i + 8.20 sin 45.0° j = 5.80 i + 5.80 j m
e
j
d 3 = −15.0 i m
a
f a
f e−9.20i + 2.30jj m
R = d1 + d 2 + d 3 = −15.0 + 5.80 i + 5.80 − 3.50 j =
(or 9.20 m west and 2.30 m north)
The magnitude of the resultant displacement is
R = R x2 + R y2 =
FG 2.30 IJ =
H −9.20 K
The direction is θ = arctan
P3.34
2
2
= 9.48 m .
166° .
A = 10.0
Refer to the sketch
R = A + B + C = −10.0 i − 15.0 j + 50.0 i
= 40.0 i − 15.0 j
a f + a−15.0f
R = 40.0
a−9.20f + a2.30f
2
2 12
R
B = 15.0
C = 50.0
= 42.7 yards
FIG. P3.34
P3.35
F = F1 + F2
(a)
a
a
a f
F = 60.0 i + 104j − 20.7 i + 77.3 j = e39.3 i + 181 jj N
f
f
a
f
F = 120 cos 60.0° i + 120 sin 60.0° j − 80.0 cos 75.0° i + 80.0 sin 75.0° j
F = 39.3 2 + 181 2 = 185 N
θ = tan −1
(b)
P3.36
F3 = − F =
East
x
0m
1.41
–0.500
+0.914
R=
FG 181 IJ =
H 39.3 K
77.8°
e−39.3 i − 181jj N
West
y
4.00 m
1.41
–0.866
4.55
2
2
x + y = 4.64 m at 78.6° N of E
65
66
P3.37
Vectors
A = 3.00 m, θ A = 30.0°
B = 3.00 m , θ B = 90.0°
A x = A cos θ A = 3.00 cos 30.0° = 2.60 m
A y = A sin θ A = 3.00 sin 30.0°= 1.50 m
e
j
A = A x i + A y j = 2.60 i + 1.50 j m
Bx = 0 , By = 3.00 m
e
j
A + B = 2.60 i + 1.50 j + 3.00 j =
P3.38
B = 3.00 j m
so
e2.60i + 4.50jj m
Let the positive x-direction be eastward, the positive y-direction be vertically upward, and the
positive z-direction be southward. The total displacement is then
e
j
e
j
e
j
d = 4.80 i + 4.80 j cm + 3.70 j − 3.70k cm = 4.80 i + 8.50 j − 3.70k cm .
P3.39
2
2
2
(a)
The magnitude is d = ( 4.80) +(8.50) + (−3.70) cm = 10.4 cm .
(b)
Its angle with the y-axis follows from cos θ =
8.50
, giving θ = 35.5° .
10. 4
B = Bx i + By j + Bz k = 4.00 i + 6.00 j + 3.00k
B = 4.00 2 + 6.00 2 + 3.00 2 = 7.81
α = cos −1
β = cos −1
γ = cos −1
P3.40
FG 4.00 IJ =
H 7.81 K
FG 6.00 IJ =
H 7.81 K
FG 3.00 IJ =
H 7.81 K
59.2°
39.8°
67.4°
The y coordinate of the airplane is constant and equal to 7.60 ×10 3 m whereas the x coordinate is
given by x = vi t where vi is the constant speed in the horizontal direction.
is
At t = 30.0 s we have x = 8.04×10 3 , so vi = 268 m s. The position vector as a function of time
b
g e
j
P = 268 m s t i + 7.60 × 10 3 m j .
At t = 45.0 s , P = 1. 21 × 10 4 i + 7.60 × 10 3 j m. The magnitude is
P=
c1.21×10 h + c7.60×10 h
4 2
3 2
m = 1.43 ×10 4 m
and the direction is
θ = arctan
F 7.60×10 I =
GH 1.21×10 JK
3
4
32.2° above the horizontal .
Chapter 3
P3.41
P3.42
(a)
A = 8.00 i + 12.0 j − 4.00k
(b)
B=
(c)
C = −3A = −24.0 i − 36.0 j + 12.0k
A
= 2.00 i + 3.00 j − 1.00k
4
R = 75.0 cos 240° i + 75.0 sin 240° j + 125 cos 135° i + 125 sin 135° j + 100 cos 160° i + 100 sin 160° j
R = −37.5 i − 65.0 j − 88.4i + 88.4j − 94.0 i + 34.2 j
R = −220 i + 57.6 j
2
R = (−220 ) + 57.6 2 at arctan
FG 57.6 IJ above the –x-axis
H 220 K
R = 227 paces at 165°
P3.43
(a)
e5.00 i − 1.00j − 3.00k j m
C=A+B=
2
2
2
C = (5.00) +(1.00) +(3.00) m = 5.92 m
(b)
D = 2A − B =
e4.00i − 11.0j + 15.0k j m
2
2
2
D = ( 4.00) +(11.0) +(15.0) m = 19.0 m
P3.44
The position vector from radar station to ship is
e
j
e
j
S = 17.3 sin 136° i + 17.3 cos 136° j km = 12.0 i − 12.4 j km.
From station to plane, the position vector is
e
j
P = 19.6 sin 153° i + 19.6 cos 153° j + 2.20k km,
or
e
j
P = 8.90 i − 17.5 j + 2.20k km.
(a)
To fly to the ship, the plane must undergo displacement
D = S− P =
(b)
e3.12 i + 5.02 j − 2.20k j km .
The distance the plane must travel is
2
2
2
D = D = (3.12) +(5.02) +( 2.20) km = 6.31 km .
67
68
P3.45
Vectors
The hurricane’s first displacement is
is
FG 41.0 km IJ(3.00 h) at 60.0° N of W, and its second displacement
H h K
FG 25.0 km IJ(1.50 h) due North. With i
H h K
representing east and j representing north, its total
displacement is:
FG 41.0 km cos 60.0°IJ a3.00 hfe− ij + FG 41.0 km sin 60.0°IJ a3.00 hfj + FG 25.0 km IJ a1.50 hfj = 61.5 kme− ij
H h
K
H h
K
H hK
+144 km j
2
2
with magnitude (61.5 km) +(144 km) = 157 km .
P3.46
(a)
a
E=
(b)
y
e15.1i + 7.72 jj cm
a
a
f
e−7.72i + 15.1jj cm
f
a
a
f
e+7.72 i + 15.1jj cm
f
27.0° 27.0°
F
FIG. P3.46
A x = −3.00 , A y = 2.00
(a)
A = A x i + A y j = −3.00 i + 2.00 j
(b)
A = A x2 + A y2 = (−3.00) +( 2.00) = 3.61
2
tan θ =
Ay
Ax
=
2
2.00
= −0.667 , tan−1 (−0.667)= −33.7°
(−3.00)
a
f
θ is in the 2 nd quadrant, so θ = 180°+ −33.7° = 146° .
(c)
G
E
G = + 17.0 cm sin 27.0° i + 17.0 cm cos 27.0° j
G=
P3.47
f
F = − 17.0 cm sin 27.0° i + 17.0 cm cos 27.0° j
F=
(c)
a
f
E = 17.0 cm cos 27.0° i + 17.0 cm sin 27.0° j
R x = 0 , R y = −4.00 , R = A + B thus B = R − A and
Bx = R x − A x = 0 − (−3.00)= 3.00 , By = R y − A y = −4.00 − 2.00 = −6.00 .
Therefore, B = 3.00 i − 6.00 j .
27.0°
x
Chapter 3
P3.48
Let +x = East, +y = North,
x
300
–175
0
125
P3.49
y
0
303
150
453
y
= 74.6° N of E
x
(a)
θ = tan−1
(b)
R = x 2 + y 2 = 470 km
(a)
R x = 40.0 cos 45.0°+30.0 cos 45.0° = 49.5
y
R y = 40.0 sin 45.0°−30.0 sin 45.0°+20.0 = 27.1
A
R = 49.5 i + 27.1j
(b)
a49.5f + a27.1f = 56.4
F 27.1 IJ = 28.7°
θ = tan G
H 49.5 K
2
R=
2
O
−1
B
45°
x
45°
C
FIG. P3.49
P3.50
Taking components along i and j , we get two equations:
6.00 a − 8.00b + 26.0 = 0
and
−8.00 a + 3.00b + 19.0 = 0 .
Solving simultaneously,
a = 5.00 , b = 7.00 .
Therefore,
5.00A + 7.00B + C = 0 .
69
70
Vectors
Additional Problems
P3.51
Let θ represent the angle between the directions of A and B. Since
A and B have the same magnitudes, A, B, and R = A + B form an
isosceles triangle in which the angles are 180°−θ ,
magnitude of R is then R = 2 A cos
R
θ /2
θ
θ
, and . The
2
2
FG θ IJ . [Hint: apply the law of
H 2K
cosines to the isosceles triangle and use the fact that B = A .]
Again, A, –B, and D = A − B form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity
B
θ
A
θ
A
D
–B
FIG. P3.51
a1 − cosθ f = 2 sin FGH θ2 IJK
2
FG θ IJ .
H 2K
gives the magnitude of D as D = 2 A sin
The problem requires that R = 100D .
Thus, 2 A cos
FG θ IJ = 200 A sinFG θ IJ . This gives tanFG θ IJ = 0.010 and
H 2K
H 2K
H 2K
θ = 1.15° .
P3.52
Let θ represent the angle between the directions of A and B. Since
A and B have the same magnitudes, A, B, and R = A + B form an
isosceles triangle in which the angles are 180°−θ ,
magnitude of R is then R = 2 A cos
θ
θ
, and . The
2
2
FG θ IJ . [Hint: apply the law of
H 2K
cosines to the isosceles triangle and use the fact that B = A . ]
Again, A, –B, and D = A − B form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity
a1 − cosθ f = 2 sin FGH θ2 IJK
2
FG θ IJ .
H 2K
Fθ I
Fθ I
The problem requires that R = nD or cosG J = n sinG J giving
H 2K
H 2K
F 1I
θ = 2 tan G J .
H nK
gives the magnitude of D as D = 2 A sin
−1
FIG. P3.52
Chapter 3
P3.53
(a)
R x = 2.00 , R y = 1.00 , R z = 3.00
(b)
R = R x2 + R y2 + R z2 = 4.00 + 1.00 + 9.00 = 14.0 = 3.74
(c)
cos θ x =
cos θ y =
cos θ z =
*P3.54
F I
GH JK
F R I = 74.5° from + y
GH R JK
F R I = 36.7° from + z
GH R JK
Rx
R
⇒ θ x = cos−1 x = 57.7° from + x
R
R
Ry
R
y
⇒ θ y = cos−1
Rz
⇒ θ z = cos−1
R
z
Take the x-axis along the tail section of the snake. The displacement from tail to head is
a
a
f
f
240 m i + 420 − 240 m cos 180°−105° i − 180 m sin 75° j = 287 m i − 174 mj .
2
2
Its magnitude is (287) +(174) m = 335 m . From v =
distance
, the time for each child’s run is
∆t
a fa
a fb
fb
g
Inge: ∆t =
distance 335 m h 1 km 3 600 s
=
= 101 s
v
12 km 1 000 m 1 h
Olaf: ∆t =
420 m ⋅ s
= 126 s .
3.33 m
ga f
Inge wins by 126 − 101 = 25.4 s .
*P3.55
The position vector from the ground under the controller of the first airplane is
a
fa f a
= e17.4i + 8.11j + 0.8k j km .
fa
f a
f
a
fa f a
= e16.5 i + 6.02 j + 1.1k j km .
fa
f a
f
r1 = 19.2 km cos 25° i + 19.2 km sin 25° j + 0.8 km k
The second is at
r2 = 17.6 km cos 20° i + 17.6 km sin 20° j + 1.1 km k
Now the displacement from the first plane to the second is
e
j
r2 − r1 = −0.863 i − 2.09 j + 0.3k km
with magnitude
2
2
2
(0.863 ) +( 2.09) +(0.3) = 2.29 km .
71
72
*P3.56
Vectors
3
Let A represent the distance from island 2 to island 3. The
displacement is A = A at 159° . Represent the displacement from 3
to 1 as B = B at 298° . We have 4.76 km at 37° +A + B = 0 .
A
28° B
For x-components
a4.76 kmf cos 37°+ A cos 159°+B cos 298° = 0
37°
1
69°
C
2
N
E
3.80 km − 0.934 A + 0.469B = 0
FIG. P3.56
B = −8.10 km + 1.99 A
For y-components,
a4.76 kmf sin 37°+ A sin 159°+B sin 298° = 0
2.86 km + 0.358 A − 0.883B = 0
(a)
We solve by eliminating B by substitution:
a
f
2.86 km + 0.358 A − 0.883 −8.10 km + 1.99 A = 0
2.86 km + 0.358 A + 7.15 km − 1.76 A = 0
10.0 km = 1.40 A
A = 7.17 km
*P3.57
(b)
B = −8.10 km + 1.99(7.17 km)= 6.15 km
(a)
We first express the corner’s position vectors as sets of components
a
a
f
f
a
a
f
f
A = 10 m cos 50° i + 10 m sin 50° j = 6.43 m i +7.66 mj
B = 12 m cos 30° i + 12 m sin 30° j = 10.4 m i +6.00 mj .
The horizontal width of the rectangle is
10.4 m − 6.43 m = 3.96 m .
Its vertical height is
7.66 m − 6.00 m = 1.66 m .
Its perimeter is
2(3.96 + 1.66) m = 11.2 m .
(b)
The position vector of the distant corner is Bx i + A y j = 10.4 mi +7.66 mj = 10.4 2 + 7.66 2 m at
7.66 m
= 12.9 m at 36.4° .
tan−1
10.4 m
Chapter 3
P3.58
Choose the +x-axis in the direction of the first force. The total force, y
in newtons, is then
31 N
12.0 i + 31.0 j − 8.40 i − 24.0 j =
e3.60 ij + e7.00jj N
8.4 N
The magnitude of the total force is
2
x
R
.
12 N
35.0°
horizontal
24 N
2
(3.60) +(7.00) N = 7.87 N
and the angle it makes with our +x-axis is given by tan θ =
(7.00)
(3.60)
θ = 62.8° . Thus, its angle counterclockwise from the horizontal is
35.0°+62.8° = 97.8° .
P3.59
FIG. P3.58
,
d 1 = 100 i
d 2 = −300 j
a
a
f
f
a
a
f
f
d 3 = −150 cos 30.0° i − 150 sin 30.0° j = −130 i − 75.0 j
d = −200 cos 60.0° i + 200 sin 60.0° j = −100 i + 173 j
4
R = d1 + d 2 + d 3 + d 4 =
e−130 i − 202 jj m
a−130f + a−202f =
F 202 IJ = 57.2°
φ = tan G
H 130 K
2
R=
2
240 m
FIG. P3.59
−1
θ = 180 + φ = 237°
P3.60
e
j
dr d 4 i + 3 j − 2 t j
=
= 0 + 0 − 2 j = − 2.00 m s j
dt
dt
b
g
The position vector at t = 0 is 4i + 3 j . At t = 1 s , the position is 4i + 1j , and so on. The object is
moving straight downward at 2 m/s, so
dr
represents its velocity vector .
dt
P3.61
a
f a
f
v = v x i + v y j = 300 + 100 cos 30.0° i + 100 sin 30.0° j
e
j
v = 387 i + 50.0 j mi h
v = 390 mi h at 7.37° N of E
73
74
P3.62
Vectors
(a)
e
j
You start at point A: r1 = rA = 30.0 i − 20.0 j m.
The displacement to B is
rB − rA = 60.0 i + 80.0 j − 30.0 i + 20.0 j = 30.0 i + 100 j .
e
j
You cover half of this, 15.0 i + 50.0 j to move to r2 = 30.0 i − 20.0 j + 15.0 i + 50.0 j = 45.0 i + 30.0 j .
Now the displacement from your current position to C is
rC − r2 = −10.0 i − 10.0 j − 45.0 i − 30.0 j = −55.0 i − 40.0 j .
You cover one-third, moving to
1
r3 = r2 + ∆r23 = 45.0 i + 30.0 j + −55.0 i − 40.0 j = 26.7 i + 16.7 j .
3
e
j
The displacement from where you are to D is
rD − r3 = 40.0 i − 30.0 j − 26.7 i − 16.7 j = 13.3 i − 46.7 j .
You traverse one-quarter of it, moving to
r4 = r3 +
b
g
1
1
rD − r3 = 26.7 i + 16.7 j + 13.3 i − 46.7 j = 30.0 i + 5.00 j .
4
4
e
j
The displacement from your new location to E is
rE − r4 = −70.0 i + 60.0 j − 30.0 i − 5.00 j = −100 i + 55.0 j
of which you cover one-fifth the distance, −20.0 i + 11.0 j, moving to
r4 + ∆r45 = 30.0 i + 5.00 j − 20.0 i + 11.0 j = 10.0 i + 16.0 j .
The treasure is at (10.0 m, 16.0 m) .
(b)
Following the directions brings you to the average position of the trees. The steps we took
numerically in part (a) bring you to
rA +
ar
then to
ar
then to
A
+ rB
f+r
2
+
rB + rC
A
C
−
ar
A +rB
2
3
f+r
−
f
=
ar
a
f FGH
r + rB
1
rB − rA = A
2
2
IJ
K
rA + rB + rC
3
A + rB + rC
f
r + rB + rC + rD
= A
3
4
4
rA + rB +rC + rD f
a
rA + rB + rC + rD
r −
r + rB + rC + rD + rE
4
+ E
= A
and last to
.
4
5
5
a
D
3
f
This center of mass of the tree distribution is the same location whatever order we take the
trees in.
Chapter 3
*P3.63
(a)
75
Let T represent the force exerted by each child. The xcomponent of the resultant force is
af a f a f
T cos 0 + T cos 120°+T cos 240° = T 1 + T −0.5 + T −0.5 = 0 .
The y-component is
T sin 0 + T sin 120 + T sin 240 = 0 + 0.866T − 0.866T = 0 .
FIG. P3.63
Thus,
∑ F = 0.
P3.64
(b)
If the total force is not zero, it must point in some direction. When each child moves one
360°
space clockwise, the total must turn clockwise by that angle,
. Since each child exerts
N
the same force, the new situation is identical to the old and the net force on the tire must still
point in the original direction. The contradiction indicates that we were wrong in supposing
that the total force is not zero. The total force must be zero.
(a)
From the picture, R 1 = a i + bj and R 1 = a 2 + b 2 .
(b)
R 2 = ai + bj + ck ; its magnitude is
2
R1 + c 2 = a 2 + b 2 + c 2 .
FIG. P3.64
76
P3.65
Vectors
Since
A + B = 6.00 j ,
we have
bA
x
g e
j
+ Bx i + A y + B y j = 0 i + 6.00 j
FIG. P3.65
giving
A x + B x = 0 or A x = −Bx
[1]
A y + B y = 6.00 .
[2]
and
Since both vectors have a magnitude of 5.00, we also have
A x2 + A y2 = Bx2 + By2 = 5.00 2 .
From A x = −Bx , it is seen that
A x2 = Bx2 .
Therefore, A x2 + A y2 = Bx2 + By2 gives
A y2 = By2 .
Then, A y = By and Eq. [2] gives
A y = By = 3.00 .
Defining θ as the angle between either A or B and the y axis, it is seen that
cos θ =
Ay
A
=
By
B
=
3.00
= 0.600 and θ = 53.1° .
5.00
The angle between A and B is then φ = 2θ = 106° .
Chapter 3
*P3.66
Let θ represent the angle the x-axis makes with the horizontal. Since
angles are equal if their sides are perpendicular right side to right
side and left side to left side, θ is also the angle between the weight
and our y axis. The x-components of the forces must add to zero:
x
y
0.127 N
Ty
θ
θ
−0.150 N sin θ + 0.127 N = 0 .
0.150 N
θ = 57.9°
(b)
(a)
FIG. P3.66
The y-components for the forces must add to zero:
a
f
+Ty − 0.150 N cos 57.9° = 0 , Ty = 0.079 8 N .
(c)
P3.67
The angle between the y axis and the horizontal is 90.0°−57.9°= 32.1° .
The displacement of point P is invariant under rotation of
2
the coordinates. Therefore, r = r ′ and r 2 = r ′ or,
bg
y
b g + by ′g . Also, from the figure, β = θ −α
F y′ I
F yI
∴tan G J = tan G J − α
H xK
H x′ K
y ′ e j − tan α
=
x ′ 1 + e j tan α
x 2 + y2 = x′
2
P
y′
2
−1
r
β
−1
α
y
x
x′
α
θ
β
t
O
y
x
FIG. P3.67
Which we simplify by multiplying top and bottom by x cosα . Then,
x ′ = x cosα + y sinα , y ′ = −x sinα + y cosα .
ANSWERS TO EVEN PROBLEMS
P3.2
a
fa
f
(a) 2.17 m, 1.25 m ; −1.90 m, 3.29 m ;
(b) 4.55 m
P3.16
see the solution
P3.18
86.6 m and –50.0 m
P3.4
(a) 8.60 m;
(b) 4.47 m at −63.4° ; 4.24 m at 135°
P3.20
1.31 km north; 2.81 km east
P3.6
(a) r at 180°− θ ; (b) 2r at 180°+ θ ; (c) 3r at –θ
P3.22
−25.0 m i + 43.3 m j
P3.8
14 km at 65° north of east
P3.24
14.0 m s at 11.3° west of north
P3.10
(a) 6.1 at 112°; (b) 14.8 at 22°
P3.26
788 mi at 48.0° north of east
P3.12
9.5 N at 57°
P3.28
7.21 m at 56.3°
P3.14
7.9 m at 4° north of west
P3.30
C = 7.30 cm i − 7.20 cm j
x
77
78
Vectors
P3.50
a = 5.00 , b = 7.00
P3.52
2 tan −1
P3.54
25.4 s
P3.56
(a) 7.17 km; (b) 6.15 km
P3.58
7.87 N at 97.8° counterclockwise from a
horizontal line to the right
P3.32
(a) 4.47 m at 63.4°; (b) 8.49 m at 135°
P3.34
42.7 yards
P3.36
4.64 m at 78.6°
P3.38
(a) 10.4 cm; (b) 35.5°
P3.40
1.43 × 10 4 m at 32.2° above the horizontal
P3.42
−220 i + 57.6 j = 227 paces at 165°
P3.44
(a) 3.12 i + 5.02 j − 2.20k km; (b) 6.31 km
P3.60
e
j
(b) e −7.72 i + 15.1jj cm;
(c) e +7.72 i + 15.1jj cm
P3.62
b−2.00 m sgj ; its velocity vector
(a) a10.0 m, 16.0 mf ; (b) see the solution
P3.64
(a) R 1 = a i + bj ; R 1 = a 2 + b 2 ;
(a) 74.6° north of east; (b) 470 km
P3.66
P3.46
P3.48
e
j
(a) 15.1i + 7.72 j cm;
FG 1 IJ
H nK
(b) R 2 = ai + bj + ck ; R 2 = a 2 + b 2 + c 2
(a) 0.079 8N; (b) 57.9°; (c) 32.1°
4
Motion in Two Dimensions
CHAPTER OUTLINE
4.1
4.2
4.3
4.4
4.5
4.6
ANSWERS TO QUESTIONS
The Position, Velocity, and
Acceleration Vectors
Two-Dimensional Motion
with Constant Acceleration
Projectile Motion
Uniform Circular Motion
Tangential and Radial
Acceleration
Relative Velocity and
Relative Acceleration
Q4.1
Yes. An object moving in uniform circular motion moves at a
constant speed, but changes its direction of motion. An object
cannot accelerate if its velocity is constant.
Q4.2
No, you cannot determine the instantaneous velocity. Yes, you
can determine the average velocity. The points could be widely
separated. In this case, you can only determine the average
velocity, which is
v=
Q4.3
(a)
a
v
a
(b)
v
a
v
a
v
∆x
.
∆t
a
v
a v
a
v
a
v
a v
10 i m s
−9.80 j m s 2
Q4.4
(a)
Q4.5
The easiest way to approach this problem is to determine acceleration first, velocity second and
finally position.
Vertical: In free flight, a y = − g . At the top of a projectile’s trajectory, v y = 0. Using this, the
(b)
d
i
maximum height can be found using v 2fy = viy2 + 2 a y y f − yi .
Horizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximum
height, one needs the flight time, t. Using the vertical information found previously, the flight time
can be found using v fy = viy + a y t . The horizontal position is x f = vix t .
If air resistance is taken into account, then the acceleration in both the x and y-directions
would have an additional term due to the drag.
Q4.6
A parabola.
79
80
Motion in Two Dimensions
Q4.7
The balls will be closest together as the second ball is thrown. Yes, the first ball will always be
moving faster, since its flight time is larger, and thus the vertical component of the velocity is larger.
The time interval will be one second. No, since the vertical component of the motion determines the
flight time.
Q4.8
The ball will have the greater speed. Both the rock and the ball will have the same vertical
component of the velocity, but the ball will have the additional horizontal component.
Q4.9
(a)
Q4.10
Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at the
top of the trajectory.
Q4.11
No. The projectile with the larger vertical component of the initial velocity will be in the air longer.
Q4.12
The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of
gravity. Its horizontal component of acceleration is zero.
Q4.13
(a)
Q4.14
60°. The projection angle appears in the expression for horizontal range in the function sin 2 θ . This
function is the same for 30° and 60°.
Q4.15
The optimal angle would be less than 45°. The longer the projectile is in the air, the more that air
resistance will change the components of the velocity. Since the vertical component of the motion
determines the flight time, an angle less than 45° would increase range.
Q4.16
The projectile on the moon would have both the larger range and the greater altitude. Apollo
astronauts performed the experiment with golf balls.
Q4.17
Gravity only changes the vertical component of motion. Since both the coin and the ball are falling
from the same height with the same vertical component of the initial velocity, they must hit the floor
at the same time.
Q4.18
(a)
yes
no
(b)
(b)
no
no
yes
(c)
(c)
no
yes
(b)
(d)
(d)
yes
(e)
no
no
yes
In the second case, the particle is continuously changing the direction of its velocity vector.
Q4.19
The racing car rounds the turn at a constant speed of 90 miles per hour.
Q4.20
The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc.
Q4.21
(a)
The velocity is not constant because the object is constantly changing the direction of its
motion.
(b)
The acceleration is not constant because the acceleration always points towards the center of
the circle. The magnitude of the acceleration is constant, but not the direction.
(a)
straight ahead
Q4.22
(b)
in a circle or straight ahead
Chapter 4
Q4.23
v
v
v
a
a
v
a
a
v
Q4.24
a
a v
a
81
v
r
a
r
r
a aa
v
v
a
v
r
r
v
a
Q4.25
The unit vectors r and θ are in different directions at different points in the xy plane. At a location
along the x-axis, for example, r = i and θ = j, but at a point on the y-axis, r = j and θ = − i . The unit
vector i is equal everywhere, and j is also uniform.
Q4.26
The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward of
the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat
is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.
Q4.27
(a)
The ball would move straight up and down as observed by the passenger. The ball would
move in a parabolic trajectory as seen by the ground observer.
(b)
Both the passenger and the ground observer would see the ball move in a parabolic
trajectory, although the two observed paths would not be the same.
(a)
g downward
Q4.28
(b)
g downward
The horizontal component of the motion does not affect the vertical acceleration.
SOLUTIONS TO PROBLEMS
Section 4.1
The Position, Velocity, and Acceleration Vectors
a f
a f
xm
0
P4.1
ym
−3 600
−3 000
0
−1 270
−4 270 m
(a)
1 270
−2 330 m
Net displacement = x 2 + y 2
= 4.87 km at 28.6° S of W
FIG. P4.1
b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =
(b)
Average speed =
(c)
Average velocity =
180 s + 120 s + 60.0 s
4.87 × 10 3 m
= 13.5 m s along R
360 s
23.3 m s
82
P4.2
*P4.3
Motion in Two Dimensions
e
j
(a)
r = 18.0t i + 4.00t − 4.90t 2 j
(b)
v=
(c)
a=
b18.0 m sgi + 4.00 m s − e9.80 m s jt j
2
(d)
e−9.80 m s j j
ra3.00 sf = a54.0 mf i − a32.1 mf j
(e)
v 3.00 s = 18.0 m s i − 25.4 m s j
(f)
a 3.00 s =
2
a
f b
a
f e−9.80 m s j j
g b
g
2
The sun projects onto the ground the x-component of her velocity:
a
f
5.00 m s cos −60.0° = 2.50 m s .
P4.4
(a)
From x = −5.00 sin ω t , the x-component of velocity is
vx =
FG IJ b
H K
g
dx
d
=
−5.00 sin ω t = −5.00ω cos ω t
dt
dt
and a x =
dv x
= +5.00ω 2 sin ω t
dt
similarly, v y =
and a y =
FG d IJ b4.00 − 5.00 cos ω tg = 0 + 5.00ω sin ω t
H dt K
FG d IJ b5.00ω sin ω tg = 5.00ω
H dt K
2
cos ω t .
e5.00ω i + 0jj m s
e0 i + 5.00ω jj m s .
At t = 0 , v = −5.00ω cos 0 i + 5.00ω sin 0 j =
and a = 5.00ω 2 sin 0 i + 5.00ω 2 cos 0 j =
(b)
2
a4.00 mfj + a5.00 mfe− sin ω t i − cos ω t jj
a5.00 mfω − cos ω t i + sin ω t j
a5.00 mfω sin ω t i + cos ω t j
r = x i + yj =
v=
a=
(c)
2
2
a
f
The object moves in a circle of radius 5.00 m centered at 0 , 4.00 m .
Chapter 4
Section 4.2
P4.5
Two-Dimensional Motion with Constant Acceleration
v f = vi + a t
(a)
v f − vi
a=
=
t
r f = ri + v i t +
(b)
e9.00i + 7.00jj − e3.00i − 2.00jj =
3.00
e2.00i + 3.00jj m s
2
1 2
1
a t = 3.00 i − 2.00 j t + 2.00 i + 3.00 j t 2
2
2
e
e
j
j
e
j
e
j
x = 3.00t + t 2 m and y = 1.50t 2 − 2.00t m
P4.6
FG IJ e
j
H K
dv F d I
a=
= G J e −12.0t jj = −12.0 j m s
dt H dt K
v=
(a)
e
2
j
r = 3.00 i − 6.00 j m; v = −12.0 j m s
(b)
P4.7
dr
d
3.00 i − 6.00t 2 j = −12.0t j m s
=
dt
dt
e
a f e
j
j
v i = 4.00 i + 1.00 j m s and v 20.0 = 20.0 i − 5.00 j m s
∆ v x 20.0 − 4.00
=
m s 2 = 0.800 m s 2
∆t
20.0
∆ v y −5.00 − 1.00
ay =
=
m s 2 = −0.300 m s 2
∆t
20.0
(a)
ax =
(b)
θ = tan −1
(c)
At t = 25.0 s
FG −0.300 IJ = −20.6° =
H 0.800 K
339° from + x axis
a f a fa f
a f a fa f
1
1
2
a x t 2 = 10.0 + 4.00 25.0 + 0.800 25.0 = 360 m
2
2
1
1
2
y f = yi + v yi t + a y t 2 = −4.00 + 1.00 25.0 + −0.300 25.0 = −72.7 m
2
2
v xf = v xi + a x t = 4 + 0.8 25 = 24 m s
x f = xi + v xi t +
a f
v = v + a t = 1 − 0.3a 25f = −6.5 m s
Fv I
F −6.50 IJ = −15.2°
θ = tan G J = tan G
H 24.0 K
Hv K
yf
yi
y
−1
y
x
−1
83
84
P4.8
Motion in Two Dimensions
a = 3.00 j m s 2 ; v i = 5.00 i m s ; ri = 0 i + 0 j
(a)
r f = ri + v i t +
v f = vi + a t =
(b)
LM5.00ti + 1 3.00t jOP m
2
N
Q
e5.00i + 3.00tjj m s
1 2
at =
2
2
a f
a fa f e
1
2
t = 2.00 s , r f = 5.00 2.00 i + 3.00 2.00 j = 10.0 i + 6.00 j m
2
so x f = 10.0 m , y f = 6.00 m
j
a f e
j
+ v = a5.00f + a6.00f
v f = 5.00 i + 3.00 2.00 j = 5.00 i + 6.00 j m s
2
v f = v f = v xf
*P4.9
(a)
2
2
yf
2
For the x-component of the motion we have x f = xi + v xi t +
e
j
0.01 m = 0 + 1.80 × 10 7 m s t +
e4 × 10
t=
=
14
j e
= 7.81 m s
1
axt 2 .
2
1
8 × 10 14 m s 2 t 2
2
e
j
j
m s 2 t 2 + 1.80 × 10 7 m s t − 10 −2 m = 0
e1.8 × 10 m sj − 4e4 × 10
2e 4 × 10 m s j
−1.80 × 10 7 m s ±
2
7
14
14
je
j
m s 2 −10 −2 m
2
−1.8 × 10 7 ± 1.84 × 10 7 m s
8 × 10 14 m s 2
We choose the + sign to represent the physical situation
t=
Here
y f = yi + v yi t +
4.39 × 10 5 m s
8 × 10
14
ms
2
= 5.49 × 10 −10 s .
1
1
a y t 2 = 0 + 0 + 1.6 × 10 15 m s 2 5.49 × 10 −10 s
2
2
e
e
je
j
2
= 2.41 × 10 −4 m .
j
So, r f = 10.0 i + 0.241 j mm .
(b)
e
je
v f = v i + at = 1.80 × 10 7 m s i + 8 × 10 14 m s 2 i + 1.6 × 10 15 m s 2 j 5.49 × 10 −10 s
e
j e
j e
m sj i + e8.78 × 10 m sj j
j
= 1.80 × 10 7 m s i + 4.39 × 10 5 m s i + 8.78 × 10 5 m s j
=
(c)
(d)
e1.84 × 10
7
5
e1.84 × 10 m sj + e8.78 × 10 m sj =
F 8.78 × 10 I = 2.73°
Fv I
θ = tan G J = tan G
v
H K
H 1.84 × 10 JK
2
7
vf =
−1
y
x
−1
5
5
7
2
1.85 × 10 7 m s
j
Chapter 4
Section 4.3
P4.10
Projectile Motion
x = v xi t = vi cos θ i t
b
ga
fa
x = 300 m s cos 55.0° 42.0 s
f
3
x = 7. 23 × 10 m
y = v yi t −
b
1 2
1
gt = vi sin θ i t − gt 2
2
2
ga
fa
f 12 e9.80 m s ja42.0 sf
2
y = 300 m s sin 55.0° 42.0 s −
P4.11
(a)
2
= 1.68 × 10 3 m
The mug leaves the counter horizontally with a velocity v xi
(say). If time t elapses before it hits the ground, then since there
is no horizontal acceleration, x f = v xi t , i.e.,
t=
xf
v xi
=
a1.40 mf
v xi
FIG. P4.11
In the same time it falls a distance of 0.860 m with acceleration
downward of 9.80 m s 2 . Then
y f = yi + v yi t +
1
1
a y t 2 : 0 = 0.860 m + −9.80 m s 2
2
2
e
jFGH 1.40v m IJK
xi
Thus,
e4.90 m s je1.96 m j =
2
v xi =
(b)
2
0.860 m
3.34 m s .
The vertical velocity component with which it hits the floor is
e
v yf = v yi + a y t = 0 + −9.80 m s 2
1.40 m I
= −4.11 m s .
jFGH 3.34
m s JK
Hence, the angle θ at which the mug strikes the floor is given by
θ = tan −1
F v I = tan FG −4.11 IJ =
GH v JK
H 3.34 K
yf
xf
−1
−50.9° .
2
.
85
86
P4.12
Motion in Two Dimensions
The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking
the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are
x f = v xi t +
1
1
1
a x t 2 = v xi t + 0 and y f = v yi t + a y t 2 = 0 − g t 2 .
2
2
2
When the mug reaches the floor, y f = − h so
−h = −
1 2
gt
2
which gives the time of impact as
t=
(a)
2h
.
g
Since x f = d when the mug reaches the floor, x f = v xi t becomes d = v xi
2h
giving the
g
initial velocity as
v xi = d
(b)
g
.
2h
Just before impact, the x-component of velocity is still
v xf = v xi
while the y-component is
v yf = v yi + a y t = 0 − g
2h
.
g
Then the direction of motion just before impact is below the horizontal at an angle of
θ = tan −1
Fv
GG v
H
yf
xf
I
JJ = tan
K
−1
Fg
GG
Hd
2h
g
g
2h
I
JJ =
K
tan −1
FG 2 h IJ
HdK
.
Chapter 4
P4.13
(a)
The time of flight of the first snowball is the nonzero root of y f = yi + v yi t1 +
b
gb
g
0 = 0 + 25.0 m s sin 70.0° t1 −
t1 =
2( 25.0 m s) sin 70.0°
9.80 m s 2
1
a y t12
2
1
9.80 m s 2 t12
2
e
j
= 4.79 s .
The distance to your target is
b
a
g
f
x f − xi = v xi t1 = 25.0 m s cos 70.0° 4.79 s = 41.0 m .
Now the second snowball we describe by
y f = yi + v yi t 2 +
b
g
1
ayt2 2
2
e
j
0 = 25.0 m s sin θ 2 t 2 − 4.90 m s 2 t 22
a
f
t 2 = 5.10 s sin θ 2
x f − xi = v xi t 2
b
g
a
f
a
f
41.0 m = 25.0 m s cos θ 2 5.10 s sin θ 2 = 128 m sin θ 2 cos θ 2
0.321 = sin θ 2 cos θ 2
Using sin 2θ = 2 sin θ cos θ we can solve 0.321 =
1
sin 2θ 2
2
2θ 2 = sin −1 0.643 and θ 2 = 20.0° .
(b)
a
f
a
t1 − t 2 = 4.79 s − 1.75 s = 3.05 s .
P4.14
f
The second snowball is in the air for time t 2 = 5.10 s sin θ 2 = 5.10 s sin 20° = 1.75 s , so you
throw it after the first by
From Equation 4.14 with R = 15.0 m , vi = 3.00 m s , θ max = 45.0°
∴g =
vi2 9.00
=
= 0.600 m s 2
R 15.0
87
88
P4.15
Motion in Two Dimensions
h=
so
b
g
vi2 sin 2θ i
vi2 sin 2 θ i
; R=
; 3h = R ,
g
2g
b
2
3 vi2 sin 2 θ i vi sin 2θ i
=
g
2g
g
2 sin 2 θ i tan θ i
=
=
3 sin 2θ i
2
4
= 53.1° .
thus θ i = tan −1
3
or
FG IJ
HK
*P4.16
(a)
To identify the maximum height we let i be the launch point and f be the highest point:
d
i
+ 2b− g gb y
2
2
v yf
= v yi
+ 2 a y y f − yi
0=
y max =
vi2
vi2
2
sin θ i
max
−0
g
2
sin θ i
.
2g
To identify the range we let i be the launch and f be the impact point; where t is not zero:
1
ay t 2
2
1
0 = 0 + vi sin θ i t + − g t 2
2
2 vi sin θ i
t=
g
y f = yi + v yi t +
b g
1
axt 2
2
2 v sin θ i
+ 0.
d = 0 + vi cos θ i i
g
x f = xi + v xi t +
For this rock, d = y max
vi2 sin 2 θ i 2 vi2 sin θ i cos θ i
=
g
2g
sin θ i
= tan θ i = 4
cos θ i
θ i = 76.0°
(b)
Since g divides out, the answer is the same on every planet.
(c)
The maximum range is attained for θ i = 45° :
d max vi cos 45° 2 vi sin 45° g
=
= 2.125 .
d
gvi cos 76° 2 vi sin 76°
So d max =
17d
.
8
Chapter 4
P4.17
a f
(a)
x f = v xi t = 8.00 cos 20.0° 3.00 = 22.6 m
(b)
Taking y positive downwards,
y f = v yi t +
1 2
gt
2
a f 12 a9.80fa3.00f
y f = 8.00 sin 20.0° 3.00 +
(c)
89
a
f
10.0 = 8.00 sin 20.0° t +
2
= 52.3 m .
a f
1
9.80 t 2
2
4.90t 2 + 2.74t − 10.0 = 0
t=
*P4.18
a2.74f
−2.74 ±
2
+ 196
= 1.18 s
9.80
We interpret the problem to mean that the displacement from fish to bug is
a
a
f
a
f
f a
f
2.00 m at 30° = 2.00 m cos30° i + 2.00 m sin30° j = 1.73 m i + 1.00 m j.
If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above the
bug. The initial velocity of the water then is directed through the point with displacement
a1.73 mfi + a1.03 mfj = 2.015 m at 30.7°.
For the time of flight of a water drop we have
1
axt 2
2
x f = xi + v xi t +
b
g
1.73 m = 0 + vi cos 30.7° t + 0 so
t=
1.73 m
.
vi cos 30.7°
The vertical motion is described by
y f = yi + v yi t +
1
ayt 2 .
2
The “drop on its path” is
1
−3.00 cm = −9.80 m s 2
2
e
.73 m I
jFGH v 1cos
J
30.7° K
2
.
i
Thus,
vi =
1.73 m 9.80 m s 2
= 2.015 m 12.8 s −1 = 25.8 m s .
cos30.7° 2 × 0.03 m
e
j
90
P4.19
Motion in Two Dimensions
(a)
We use the trajectory equation:
y f = x f tan θ i −
gx 2f
2 vi2 cos 2 θ i
.
With
x f = 36.0 m, vi = 20.0 m s, and θ = 53.0°
we find
e9.80 m s ja36.0 mf = 3.94 m.
= a36.0 mf tan 53.0°−
2b 20.0 m sg cos a53.0°f
2
2
yf
2
2
The ball clears the bar by
a3.94 − 3.05f m =
(b)
0.889 m .
The time the ball takes to reach the maximum height is
t1 =
b
f
ga
20.0 m s sin53.0°
vi sin θ i
=
= 1.63 s .
g
9.80 m s 2
xf
The time to travel 36.0 m horizontally is t 2 =
t2 =
vix
36.0 m
= 2.99 s .
( 20.0 m s) cos 53.0°
a
f
Since t 2 > t1 the ball clears the goal on its way down .
P4.20
b
g
The horizontal component of displacement is x f = v xi t = vi cos θ i t . Therefore, the time required to
d
reach the building a distance d away is t =
. At this time, the altitude of the water is
vi cos θ i
y f = v yi t +
FG
H
IJ FG
K H
g
d
d
1
a y t 2 = vi sin θ i
−
vi cos θ i
2
2 vi cos θ i
IJ
K
Therefore the water strikes the building at a height h above ground level of
h = y f = d tan θ i −
gd 2
2 vi2 cos 2 θ i
.
2
.
Chapter 4
*P4.21
(a)
For the horizontal motion, we have
1
axt 2
2
24 m = 0 + vi cos 53° 2.2 s + 0
x f = xi + v xi t +
a
fa
f
vi = 18.1 m s .
(b)
As it passes over the wall, the ball is above the street by y f = yi + v yi t +
b
fa
ga
f 12 e−9.8 m s ja2.2 sf
2
y f = 0 + 18.1 m s sin 53° 2.2 s +
2
1
ayt 2
2
= 8.13 m .
So it clears the parapet by 8.13 m − 7 m = 1.13 m .
(c)
Note that the highest point of the ball’s trajectory is not directly above the wall. For the
whole flight, we have from the trajectory equation
g FGH 2v
b
y f = tan θ i x f −
g
2
i
cos
2
Ix
θ JK
i
2
f
or
F
f GG 2 18.19m.8 sm scos
g
Hb
a
2
6 m = tan 53° x f −
2
2
I
Jx .
53° JK
Solving,
e0.041 2 m jx
−1
2
f
− 1.33 x f + 6 m = 0
and
xf =
a
fa f .
1.33 ± 1.33 2 − 4 0.0412 6
e
2 0.0412 m
−1
j
This yields two results:
x f = 26.8 m or 5.44 m
The ball passes twice through the level of the roof.
It hits the roof at distance from the wall
26.8 m − 24 m = 2.79 m .
2
f
91
92
*P4.22
Motion in Two Dimensions
When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by
xf =
a3.25 kmf − a2.15 kmf
2
y f = x f tan θ −
2
= 2.437 km
gx 2f
2 vi2 cos 2 θ i
e9.8 m s jb2 437 mg
−2 150 m = b 2 437 mg tan θ −
2b 280 m sg cos θ
∴−2 150 m = b 2 437 mg tan θ − a371.19 mfe1 + tan θ j
2
2
i
2
2
i
2
i
i
2
∴ tan θ − 6.565 tan θ i − 4.792 = 0
∴ tan θ i =
F
H
1
6.565 ±
2
a6.565f − 4a1fa−4.792f IK = 3.283 ± 3.945 .
2
Select the negative solution, since θ i is below the horizontal.
∴ tan θ i = −0.662 , θ i = −33.5°
P4.23
The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from
y f = yi + v yi t +
1
ayt 2
2
1
−9.80 m s 2 t 2
2
t = 2.86 s .
−40.0 m = 0 + 0 +
e
j
The extra time 3.00 s − 2.86 s = 0.143 s is the time required for the sound she hears to travel straight
back to the player.
It covers distance
b343 m sg0.143 s = 49.0 m =
a
where x represents the horizontal distance the rock travels.
x = 28.3 m = v xi t + 0t 2
∴ v xi =
f
x 2 + 40.0 m
28.3 m
= 9.91 m s
2.86 s
2
93
Chapter 4
P4.24
From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His
2
2
= v yi
+ 2 a y y f − yi .
vertical velocity and vertical displacement are related by the equation v yf
Applying this to the upward part of his flight gives 0 =
2
v yi
d i
+ 2e −9.80 m s ja1.85 − 1.02f m . From this,
2
v yi = 4.03 m s . [Note that this is the answer to part (c) of this problem.]
ja
e
f
2
For the downward part of the flight, the equation gives v yf
= 0 + 2 −9.80 m s 2 0.900 − 1.85 m .
Thus the vertical velocity just before he lands is
v yf = −4.32 m s.
(a)
His hang time may then be found from v yf = v yi + a y t :
e
j
−4.32 m s = 4.03 m s + −9.80 m s 2 t
or t = 0.852 s .
(b)
Looking at the total horizontal displacement during the leap, x = v xi t becomes
a
2.80 m = v xi 0.852 s
f
which yields v xi = 3.29 m s .
(c)
v yi = 4.03 m s . See above for proof.
(d)
The takeoff angle is: θ = tan −1
F v I = tan F 4.03 m s I =
GH v JK
GH 3.29 m s JK
yi
−1
50.8° .
xi
(e)
Similarly for the deer, the upward part of the flight gives
2
2
v yf
= v yi
+ 2 a y y f − yi :
d
i
ja
e
f
2
+ 2 −9.80 m s 2 2.50 − 1.20 m
0 = v yi
so v yi = 5.04 m s .
d
i
ja
e
f
2
2
2
For the downward part, v yf
= v yi
+ 2 a y y f − yi yields v yf
= 0 + 2 −9.80 m s 2 0.700 − 2.50 m
and v yf = −5.94 m s.
e
j
The hang time is then found as v yf = v yi + a y t : −5.94 m s = 5.04 m s + −9.80 m s 2 t and
t = 1.12 s .
94
*P4.25
Motion in Two Dimensions
The arrow’s flight time to the collision point is
t=
x f − xi
v xi
=
b
150 m
= 5.19 s .
45 m s cos 50°
g
The arrow’s altitude at the collision is
y f = yi + v yi t +
b
1
ayt 2
2
f
ga
= 0 + 45 m s sin 50° 5.19 s +
(a)
ja
1
−9.8 m s 2 5.19 s
2
e
f
2
= 47.0 m .
The required launch speed for the apple is given by
d
2
2
v yf
= v yi
+ 2 a y y f − yi
e
i
ja
2
0 = v yi
+ 2 −9.8 m s 2 47 m − 0
f
v yi = 30.3 m s .
(b)
The time of flight of the apple is given by
v yf = v yi + a y t
0 = 30.3 m s − 9.8 m s 2 t
t = 3.10 s .
So the apple should be launched after the arrow by 5.19 s − 3.10 s = 2.09 s .
*P4.26
For the smallest impact angle
θ = tan −1
Fv I,
GH v JK
yf
xf
we want to minimize v yf and maximize v xf = v xi . The final y-component
2
2
of velocity is related to v yi by v yf
= v yi
+ 2 gh , so we want to minimize v yi
and maximize v xi . Both are accomplished by making the initial velocity
horizontal. Then v xi = v , v yi = 0 , and v yf = 2 gh . At last, the impact
angle is
θ = tan −1
Fv I =
GH v JK
yf
xf
tan −1
F
GH
2 gh
v
I
JK
.
FIG. P4.26
Chapter 4
Section 4.4
P4.27
P4.28
Uniform Circular Motion
b
g
v2
, T = 24 h 3 600 s h = 86 400 s
R
2π R 2π ( 6.37 × 10 6 m)
v=
=
= 463 m s
T
86 400 s
b
a=
g
2
6
6.37 × 10 m
= 0.033 7 m s 2 directed toward the center of Earth .
r = 0.500 m;
vt =
a=
P4.30
2
20.0 m s
v2
ac =
=
= 377 m s 2
r
1.06 m
The mass is unnecessary information.
b463 m sg
a=
P4.29
95
ac =
a
f
2π r 2π 0.500 m
=
= 10. 47 m s = 10.5 m s
60 .0 s
T
200 rev
a
10. 47
v2
=
R
0.5
v2
r
e
f
2
= 219 m s 2 inward
ja
f
v = a c r = 3 9.8 m s 2 9.45 m = 16.7 m s
a
f
Each revolution carries the astronaut over a distance of 2π r = 2π 9.45 m = 59.4 m. Then the rotation
rate is
16.7 m s
P4.31
(a)
v = rω
a
a
fb
fb
FG 1 rev IJ =
H 59.4 m K
gb
gb
0.281 rev s .
g
g
At 8.00 rev s , v = 0.600 m 8.00 rev s 2π rad rev = 30.2 m s = 9.60π m s .
At 6.00 rev s , v = 0.900 m 6.00 rev s 2π rad rev = 33.9 m s = 10.8π m s .
6.00 rev s gives the larger linear speed.
b
(b)
9.60π m s
v2
=
Acceleration =
r
0.600 m
(c)
At 6.00 rev s , acceleration =
g
2
= 1.52 × 10 3 m s 2 .
b10.8π m sg
0.900 m
2
= 1.28 × 10 3 m s 2 .
96
P4.32
Motion in Two Dimensions
The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration.
ac = g
so
v2
= g.
r
Solving for the velocity, v = rg =
a6 ,400 + 600fe10 mje8.21 m s j =
3
v=
2
7.58 × 10 3 m s
2πr
T
and
e
T = 5.80 × 10 3 s
Section 4.5
P4.33
j
3
2π r 2π 7 ,000 × 10 m
=
= 5.80 × 10 3 s
v
7.58 × 10 3 m s
T=
FG 1 min IJ = 96.7 min .
H 60 s K
Tangential and Radial Acceleration
We assume the train is still slowing down at the instant in question.
ac =
v2
= 1.29 m s 2
r
at =
−40.0 km h 10 3 m km
∆v
=
15.0 s
∆t
b
ge
1h
3 600 s
j = −0.741 m s
e1.29 m s j + e−0.741 m s j
F a I = tan FG 0.741 IJ
GH a JK
H 1.29 K
2 2
a = a c2 + a t2 =
at an angle of tan −1
je
t
2 2
−1
c
a = 1.48 m s 2 inward and 29.9 o backward
P4.34
(a)
at = 0.600 m s 2
(b)
4.00 m s
v2
ar =
=
r
20.0 m
(c)
a = a t2 + a r2 = 1.00 m s 2
b
θ = tan −1
g
2
= 0.800 m s 2
ar
= 53.1° inward from path
at
2
FIG. P4.33
Chapter 4
P4.35
r = 2.50 m , a = 15.0 m s 2
(a)
(b)
ja
e
f
a c = a cos 30.0 o = 15.0 m s 2 cos 30° = 13.0 m s 2
v2
r
so v 2 = ra c = 2.50 m 13.0 m s 2 = 32.5 m 2 s 2
ac =
e
j
FIG. P4.35
v = 32.5 m s = 5.70 m s
(c)
a 2 = a t2 + a r2
so a t = a 2 − a r2 =
P4.36
97
e15.0 m s j − e13.0 m s j =
2 2
2
7.50 m s 2
(a)
See figure to the right.
(b)
The components of the 20.2 and the 22.5 m s 2 along the rope together
constitute the centripetal acceleration:
e
j a
f e
j
a c = 22.5 m s 2 cos 90.0°−36.9° + 20.2 m s 2 cos 36.9° = 29.7 m s 2
(c)
*P4.37
v2
so v = a c r = 29.7 m s 2 1.50 m = 6.67 m s tangent to circle
r
v = 6.67 m s at 36.9° above the horizontal
a
ac =
f
FIG. P4.36
at
Let i be the starting point and f be one revolution later. The curvilinear motion
with constant tangential acceleration is described by
∆ x = v xi t +
1
axt 2
2
1 2
at t
2
4π r
a
θ
ar
2π r = 0 +
at =
and v xf = v xi + a x t , v f = 0 + a t t =
Then tan θ =
4π r t 2
at
1
= 2
=
a r t 16π 2 r 4π
t
FIG. P4.37
2
v 2f 16π 2 r 2
4π r
=
. The magnitude of the radial acceleration is a r =
.
r
t
t 2r
θ = 4.55° .
98
Motion in Two Dimensions
Section 4.6
P4.38
(a)
Relative Velocity and Relative Acceleration
e
a
j
v H = 0 + a H t = 3.00 i − 2.00 j m s 2 5.00 s
e
j
f
v H = 15.0 i − 10.0 j m s
e
a
j
v J = 0 + a j t = 1.00 i + 3.00 j m s 2 5.00 s
e
f
j
v J = 5.00 i +15.0 j m s
e
j
v HJ = v H − v J = 15.0 i − 10.0 j − 5.00 i − 15.0 j m s
e
j
v HJ = 10.0 i − 25.0 j m s
v HJ = (10.0) 2 + ( 25.0) 2 m s = 26.9 m s
(b)
a
1
1
a H t 2 = 3.00 i − 2.00 j m s 2 5.00 s
2
2
rH = 37.5 i − 25.0 j m
e
rH = 0 + 0 +
e
f
2
j
a
f e
1
2
1.00 i + 3.00 j m s 2 5.00s = 12.5 i + 37.5 j m
2
= r − r = 37.5 i − 25.0 j − 12.5 i − 37.5 j m
e
rJ =
rHJ
j
H
j
j
e
J
j
e
j
rHJ = 25.0 i − 62.5 j m
rHJ =
(c)
2
2
67.3 m
e
j
a HJ = a H − a J = 3.00 i − 2.00 j − 1.00 i − 3.00 j m s 2
a HJ =
*P4.39
a25.0f + a62.5f m =
e2.00i − 5.00 jj m s
2
v ce = the velocity of the car relative to the earth.
v wc = the velocity of the water relative to the car.
v we = the velocity of the water relative to the earth.
These velocities are related as shown in the diagram at the right.
(a)
(b)
Since v we is vertical, v wc sin 60.0° = v ce = 50.0 km h or
v wc = 57.7 km h at 60.0° west of vertical .
Since v ce has zero vertical component,
b
vce
vwe
60°
vwc
v we = v ce + v wc
FIG. P4.39
g
v we = v wc cos 60.0° = 57.7 km h cos 60.0° = 28.9 km h downward .
Chapter 4
P4.40
The bumpers are initially 100 m = 0.100 km apart. After time t the bumper of the leading car travels
40.0 t, while the bumper of the chasing car travels 60.0t.
Since the cars are side by side at time t, we have
0.100 + 40.0t = 60.0t ,
yielding
t = 5.00 × 10 −3 h = 18.0 s .
P4.41
Total time in still water t =
d 2 000
=
= 1.67 × 10 3 s .
v 1. 20
Total time = time upstream plus time downstream:
1 000
= 1.43 × 10 3 s
(1.20 − 0.500 )
1 000
=
= 588 s .
1.20 + 0.500
t up =
t down
Therefore, ttotal = 1.43 × 10 3 + 588 = 2.02 × 10 3 s .
P4.42
v = 150 2 + 30.0 2 = 153 km h
θ = tan −1
P4.43
99
FG 30.0 IJ =
H 150 K
11.3° north of west
For Alan, his speed downstream is c + v, while his speed upstream is c − v .
Therefore, the total time for Alan is
t1 =
2L
L
L
c
+
=
2
c+v c−v
1 − v2
.
c
For Beth, her cross-stream speed (both ways) is
c2 − v2 .
Thus, the total time for Beth is t 2 =
Since 1 −
2L
c2 − v2
=
2L
c
1−
v2
c2
.
v2
< 1 , t1 > t 2 , or Beth, who swims cross-stream, returns first.
c2
100
P4.44
Motion in Two Dimensions
(a)
To an observer at rest in the train car, the bolt accelerates downward and toward the rear of
the train.
a=
tan θ =
b2.50 m sg + b9.80 m sg
2
2.50 m s 2
9.80 m s 2
2
= 10.1 m s 2
= 0.255
θ = 14.3° to the south from the vertical
(b)
P4.45
a = 9.80 m s 2 vertically downward
Identify the student as the S’ observer and the professor as
the S observer. For the initial motion in S’, we have
v ′y
v ′x
= tan 60.0° = 3 .
Let u represent the speed of S’ relative to S. Then because
there is no x-motion in S, we can write v x = v ′x + u = 0 so
that v x′ = −u = −10.0 m s . Hence the ball is thrown
backwards in S’. Then,
v y = v ′y = 3 v ′x = 10.0 3 m s .
Using v y2 = 2 gh we find
e10.0 3 m sj =
h=
2e9.80 m s j
FIG. P4.45
2
2
15.3 m .
The motion of the ball as seen by the student in S’ is shown in diagram (b). The view of the professor
in S is shown in diagram (c).
*P4.46
Choose the x-axis along the 20-km distance. The ycomponents of the displacements of the ship and
the speedboat must agree:
b26 km hgt sina40°−15°f = b50 km hgt sin α
11.0
= 12.7° .
α = sin −1
50
x
N
25°
40°
15°
α
E
y
The speedboat should head
15°+12.7° = 27.7° east of north .
FIG. P4.46
Chapter 4
101
Additional Problems
*P4.47
b
g
(a)
The speed at the top is v x = vi cos θ i = 143 m s cos 45° = 101 m s .
(b)
In free fall the plane reaches altitude given by
d
i
0 = b143 m s sin 45°g + 2e −9.8 m s jd y − 31 000 ft i
F 3.28 ft IJ = 3.27 × 10 ft .
y = 31 000 ft + 522 mG
H 1m K
2
2
v yf
= v yi
+ 2 a y y f − yi
2
2
3
f
(c)
f
For the whole free fall motion v yf = v yi + a y t
e
j
−101 m s = +101 m s − 9.8 m s 2 t
t = 20.6 s
(d)
ac =
v2
r
e
j
v = a c r = 0.8 9.8 m s 2 4,130 m = 180 m s
P4.48
At any time t, the two drops have identical y-coordinates. The distance between the two drops is
then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,
af b g b
g
d = 2 x t = 2 v xi t = 2 vi cos θ i t = 2 vi t cos θ i .
P4.49
After the string breaks the ball is a projectile, and reaches the ground at time t: y f = v yi t +
−1.20 m = 0 +
1
−9.80 m s 2 t 2
2
e
j
so t = 0.495 s.
Its constant horizontal speed is v x =
x 2.00 m
=
= 4.04 m s
t 0.495 s
b
g
4.04 m s
v2
so before the string breaks a c = x =
r
0.300 m
2
= 54.4 m s 2 .
1
ayt 2
2
102
P4.50
Motion in Two Dimensions
(a)
b
gd i
y f = tan θ i x f −
g
2 vi2
x 2f
cos 2 θ i
Setting x f = d cos φ , and y f = d sin φ , we have
b
gb
g
d sin φ = tan θ i d cos φ −
Solving for d yields, d =
or d =
(b)
P4.51
Setting
b
g
2 vi2
bd cos φ g .
2
2
cos θ i
FIG. P4.50
2 vi2 cos θ i sin θ i cos φ − sin φ cos θ i
2 vi2 cos θ i sin θ i − φ
g cos 2 φ
g
2
g cos φ
.
b
v 2 1 − sin φ
dd
φ
= 0 leads to θ i = 45°+
and d max = i
dθ i
2
g cos 2 φ
g
.
Refer to the sketch:
(b)
b
g
∆ x = v xi t ; substitution yields 130 = vi cos 35.0° t .
1
∆ y = v yi t + at 2 ; substitution yields
2
b
a
g
f
1
−9.80 t 2 .
2
20.0 = vi sin 35.0° t +
Solving the above gives t = 3.81 s .
(a)
vi = 41.7 m s
(c)
v yf = vi sin θ i − gt , v x = vi cos θ i
FIG. P4.51
a fa f
At t = 3.81 s , v yf = 41.7 sin 35.0°− 9.80 3.81 = −13.4 m s
a
f
v x = 41.7 cos 35.0° = 34.1 m s
2
v f = v x2 + v yf
= 36.7 m s .
Chapter 4
P4.52
(a)
The moon’s gravitational acceleration is the probe’s centripetal acceleration:
(For the moon’s radius, see end papers of text.)
a=
v2
r
v2
1
9.80 m s 2 =
6
1.74 × 10 6 m
e
j
v = 2.84 × 10 6 m 2 s 2 = 1.69 km s
(b)
P4.53
(a)
2π r
T
2π r 2π (1.74 × 10 6 m)
T=
=
= 6.47 × 10 3 s = 1.80 h
v
1.69 × 10 3 m s
v=
b
5.00 m s
v2
ac =
=
r
1.00 m
g
2
= 25.0 m s 2
at = g = 9.80 m s 2
(b)
See figure to the right.
(c)
a = a c2 + a t2 =
φ = tan −1
e25.0 m s j + e9.80 m s j
FG a IJ = tan
Ha K
t
2 2
−1
c
P4.54
9.80 m s 2
25.0 m s 2
2 2
= 26.8 m s 2
= 21.4°
FIG. P4.53
x f = vix t = vi t cos 40.0°
10.0 m
.
vi cos 40.0°
At this time, y f should be 3.05 m − 2.00 m = 1.05 m .
Thus, when x f = 10.0 m , t =
Thus, 1.05 m =
bv sin 40.0°g10.0 m + 1 e−9.80 m s jLM 10.0 m OP .
v cos 40.0°
2
N v cos 40.0° Q
2
i
2
i
From this, vi = 10.7 m s .
i
103
104
P4.55
Motion in Two Dimensions
The special conditions allowing use of the horizontal range equation applies.
For the ball thrown at 45°,
D = R 45 =
vi2 sin 90
.
g
For the bouncing ball,
v 2 sin 2θ
D = R1 + R 2 = i
+
g
vi 2
2
e j
sin 2θ
g
where θ is the angle it makes with the ground when thrown and when bouncing.
(a)
We require:
vi2 vi2 sin 2θ vi2 sin 2θ
=
+
g
g
4g
4
5
θ = 26.6°
sin 2θ =
(b)
FIG. P4.55
The time for any symmetric parabolic flight is given by
1 2
gt
2
1
0 = vi sin θ i t − gt 2 .
2
y f = v yi t −
If t = 0 is the time the ball is thrown, then t =
2 vi sin θ i
is the time at landing.
g
So for the ball thrown at 45.0°
t 45 =
2 vi sin 45.0°
.
g
For the bouncing ball,
2 v sin 26.6° 2
t = t1 + t 2 = i
+
g
e j sin 26.6° = 3v sin 26.6° .
vi
2
i
g
The ratio of this time to that for no bounce is
3 vi sin 26.6 °
g
2 vi sin 45 .0 °
g
=
1.34
= 0.949 .
1.41
g
Chapter 4
P4.56
105
Using the range equation (Equation 4.14)
R=
vi2 sin( 2θ i )
g
vi2
. Given R, this yields vi = gR .
g
If the boy uses the same speed to throw the ball vertically upward, then
the maximum range occurs when θ i = 45° , and has a value R =
v y = gR − gt and y = gR t −
gt 2
2
at any time, t.
R
, and so the maximum height reached is
g
At the maximum height, v y = 0, giving t =
y max
P4.57
R g
= gR
−
g 2
F RI
GH g JK
2
= R−
R
R
=
.
2
2
Choose upward as the positive y-direction and leftward as the
positive x-direction. The vertical height of the stone when released
from A or B is
a
vi
f
yi = 1.50 + 1.20 sin 30.0° m = 2.10 m
(a)
B
A
The equations of motion after release at A are
a
30° 1.20 m
f
v y = vi sin 60.0°− gt = 1.30 − 9.80t m s
30°
vi
v x = vi cos 60.0° = 0.750 m s
e
∆ x = a0.750t f m
−1.30 ± a1.30f
When y = 0 , t =
j
y = 2.10 +1.30t − 4.90t 2 m
FIG. P4.57
A
2
+ 41.2
−9.80
(b)
a
fa
f
= 0.800 s. Then, ∆ x A = 0.750 0.800 m = 0.600 m .
The equations of motion after release at point B are
a
f
a
f
v y = vi − sin 60.0° − gt = −1.30 − 9.80t m s
v x = vi cos 60.0 = 0.750 m s
e
j
yi = 2.10 − 1.30t − 4.90t 2 m .
When y = 0 , t =
b
+1.30 ±
1.50 m s
v2
=
r
1.20 m
a−1.30f
−9.80
g
2
+ 41.2
a
fa
f
= 0.536 s. Then, ∆ x B = 0.750 0.536 m = 0.402 m .
2
= 1.87 m s 2 toward the center
(c)
ar =
(d)
After release, a = − g j = 9.80 m s 2 downward
106
P4.58
Motion in Two Dimensions
The football travels a horizontal distance
R=
b g = a20.0f sina60.0°f = 35.3 m.
vi2 sin 2θ i
g
2
9.80
Time of flight of ball is
FIG. P4.58
2 v sin θ i 2( 20.0) sin 30.0°
t= i
=
= 2.04 s .
g
9.80
The receiver is ∆ x away from where the ball lands and ∆ x = 35.3 − 20.0 = 15.3 m. To cover this
distance in 2.04 s, he travels with a velocity
v=
P4.59
(a)
15.3
= 7.50 m s in the direction the ball was thrown .
2.04
1 2
g t ; ∆ x = vi t
2
Combine the equations eliminating t:
∆y= −
FG IJ
H K
1 ∆x
∆y= − g
vi
2
b g = FGH −2g∆ y IJK v
From this, ∆ x
2
2
.
2
i
FIG. P4.59
thus ∆ x = vi
−2 ∆ y
−2( −300 )
= 275
= 6.80 × 10 3 = 6.80 km .
g
9.80
(b)
The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be
3 000 m directly above the bomb when it hits the ground.
(c)
When φ is measured from the vertical, tan φ =
therefore, φ = tan −1
F ∆ x I = tan F 6 800 I =
GH ∆ y JK
GH 3 000 JK
−1
∆x
∆y
66. 2° .
Chapter 4
*P4.60
(a)
We use the approximation mentioned in the problem. The time to travel 200 m horizontally is
∆x
200 m
t=
=
= 0.200 s . The bullet falls by
v x 1,000 m s
(c)
ja f
1
1
a y t 2 = 0 + −9.8 m s 2 0.2 s
2
2
e
∆ y = v yi t +
(b)
107
The telescope axis must point below the barrel axis
0.196 m
= 0.056 1° .
by θ = tan −1
200 m
t=
50.0 m
= 0.050 0 s . The bullet falls by only
1 000 m s
∆y=
ja
1
−9.8 m s 2 0.05 s
2
e
a
f
2
2
= −0.196 m .
barrel axis
bullet path
scope axis
50
150 200 250
FIG. P4.60(b)
= −0.0122 m .
a
f
f
1
1
200 m , the scope axis points to a location 19.6 cm = 4.90 cm above the
4
4
barrel axis, so the sharpshooter must aim low by 4.90 cm − 1.22 cm = 3.68 cm .
At range 50 m =
(d)
t=
150 m
= 0.150 s
1 000 m s
ja
a
f
1
2
−9.8 m s 2 0.15 s = 0.110 m
2
150
19.6 cm − 11.0 cm = 3.68 cm .
Aim low by
200
∆y=
(e)
t=
e
f
250 m
= 0.250 s
1 000 m s
∆y=
ja
1
−9.8 m s 2 0.25 s
2
e
Aim high by 30.6 cm −
f
2
= 0.306 m
a
f
250
19.6 cm = 6.12 cm .
200
(f), (g) Many marksmen have a hard time believing it, but
they should aim low in both cases. As in case (a) above,
the time of flight is very nearly 0.200 s and the bullet
falls below the barrel axis by 19.6 cm on its way. The
0.0561° angle would cut off a 19.6-cm distance on a
vertical wall at a horizontal distance of 200 m, but on a
vertical wall up at 30° it cuts off distance h as shown,
where cos 30° = 19.6 cm h , h = 22.6 cm. The marksman
barrel axis
scope
30°
19.6 cm
must aim low by 22.6 cm − 19.6 cm = 3.03 cm . The
answer can be obtained by considering limiting cases.
Suppose the target is nearly straight above or below
you. Then gravity will not cause deviation of the path
of the bullet, and one must aim low as in part (c) to
cancel out the sighting-in of the telescope.
30°
h
19.6 cm
scope axis
bullet hits here
FIG. P4.60(f–g)
108
P4.61
Motion in Two Dimensions
(a)
From Part (c), the raptor dives for 6.34 − 2.00 = 4.34 s
undergoing displacement 197 m downward and
10.0 4.34 = 43.4 m forward.
a fa f
∆d
v=
∆t
P4.62
(b)
α = tan −1
(c)
197 =
FG −197 IJ =
H 43.4 K
a197f + a43.4f
2
2
= 46.5 m s
4.34
−77.6°
1 2
gt , t = 6.34 s
2
FIG. P4.61
Measure heights above the level ground. The elevation y b of the ball follows
yb = R + 0 −
with x = vi t so y b = R −
(a)
gx 2
2 vi2
1 2
gt
2
.
The elevation yr of points on the rock is described by
y r2 + x 2 = R 2 .
We will have y b = y r at x = 0 , but for all other x we require the ball to be above the rock
surface as in y b > y r . Then y b2 + x 2 > R 2
F R − gx I
GH 2 v JK
2
2
+ x2 > R2
2
i
R2 −
gx 2 R
vi2
+
g 2x4
4vi4
g 2x4
4vi4
+ x2 > R2
+ x2 >
gx 2 R
vi2
.
If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s
parabolic trajectory has large enough radius of curvature at the start, the ball will clear the
gR
whole rock: 1 > 2
vi
vi > gR .
(b)
With vi = gR and y b = 0 , we have 0 = R −
gx 2
2 gR
or x = R 2 .
The distance from the rock’s base is
x−R=
e
j
2 −1 R .
Chapter 4
P4.63
(a)
While on the incline
v 2f − vi2 = 2 a∆x
v f − vi = at
a fa f
v 2f − 0 = 2 4.00 50.0
20.0 − 0 = 4.00t
v f = 20.0 m s
t = 5.00 s
(b)
FIG. P4.63
Initial free-flight conditions give us
v xi = 20.0 cos 37.0° = 16.0 m s
and
v yi = −20.0 sin 37.0° = −12.0 m s
v xf = v xi since a x = 0
a
fa
f a
v yf = − 2 a y ∆ y + v yi 2 = − 2 −9.80 −30.0 + −12.0
a16.0f + a−27.1f
2
v f = v xf 2 + v yf 2 =
(c)
t1 = 5 s ; t 2 =
v yf − v yi
ay
=
2
f
2
= −27.1 m s
= 31.5 m s at 59.4° below the horizontal
−27.1 + 12.0
= 1.53 s
−9.80
t = t1 + t 2 = 6.53 s
(d)
P4.64
a f
∆x = v xi t1 = 16.0 1.53 = 24.5 m
y 2 = 16 x
Equation of bank:
Equations of motion:
x = vi t
1
y = − g t2
2
a1f
a 2f
a3 f
F I
GH JK
1 x2
g
. Equate y
2 vi2
from the bank equation to y from the equations of motion:
Substitute for t from (2) into (3) y = −
L 1 Fx
16 x = M− g G
MN 2 H v
From this, x = 0 or x
3
FIG. P4.64
I OP ⇒ g x − 16x = xF g x − 16I = 0 .
JK PQ 4v
GH 4v JK
F 10 I = 18.8 m . Also,
64v
=
and x = 4G
g
H 9.80 JK
1 Fx I
1 a9.80fa18.8f
y = − gG J = −
2 Hv K
2
a10.0f = −17.3 m .
2
2
2
2 4
2
i
4
i
2 3
4
i
4
i
1/3
4
2
2
2
2
i
2
109
110
P4.65
Motion in Two Dimensions
(a)
a f
1 2
1
at ; 70.0 = 15.0 t 2
2
2
70.0 = vi t
Roadrunner: ∆ x = vi t ;
Coyote:
∆x =
Solving the above, we get
vi = 22.9 m s and t = 3.06 s.
(b)
At the edge of the cliff,
a fa f
v xi = at = 15.0 3.06 = 45.8 m s .
Substituting into ∆ y =
1
a y t 2 , we find
2
a
f
1
−9.80 t 2
2
t = 4.52 s
−100 =
∆ x = v xi t +
a fa
f a fa
f
1
1
2
a x t 2 = 45.8 4.52 s + 15.0 4.52 s .
2
2
Solving,
∆ x = 360 m .
(c)
For the Coyote’s motion through the air
a f
t = 0 − 9.80a 4.52f =
v xf = v xi + a x t = 45.8 + 15 4.52 = 114 m s
v yf = v yi + a y
P4.66
−44.3 m s .
Think of shaking down the mercury in an old fever thermometer. Swing your hand through a
circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through onequarter of a circle of radius 60 cm in 0.1 s. Its speed is
1
4
a2π fa0.6 mf ≅ 9 m s
0.1 s
and its centripetal acceleration is
v 2 ( 9 m s) 2
≅
~ 10 2 m s 2 .
r
0.6 m
The tangential acceleration of stopping and reversing the motion will make the total acceleration
somewhat larger, but will not affect its order of magnitude.
Chapter 4
P4.67
(a)
∆ x = v xi t , ∆ y = v yi t +
1 2
gt
2
a
111
f
d cos 50.0° = 10.0 cos 15.0° t
and
a
f
− d sin 50.0° = 10.0 sin 15.0° t +
a
f
1
−9.80 t 2 .
2
Solving, d = 43.2 m and t = 2.88 s .
(b)
Since a x = 0 ,
FIG. P4.67
v xf = v xi = 10.0 cos 15.0° = 9.66 m s
a f
v yf = v yi + a y t = 10.0 sin 15.0°−9.80 2.88 = −25.6 m s .
Air resistance would decrease the values of the range and maximum height. As an airfoil, he
can get some lift and increase his distance.
*P4.68
For one electron, we have
y = viy t , D = vix t +
1
1
a x t 2 ≅ a x t 2 , v yf = v yi , and v xf = v xi + a x t ≅ a x t .
2
2
The angle its direction makes with the x-axis is given by
θ = tan −1
v yf
v xf
= tan −1
v yi
axt
= tan −1
v yi t
axt
2
= tan −1
y
.
2D
FIG. P4.68
Thus the horizontal distance from the aperture to the virtual source is 2D. The source is at
coordinate x = − D .
*P4.69
(a)
The ice chest floats downstream 2 km in time t, so that 2 km = v w t . The upstream motion of
the boat is described by d = ( v − v w )15 min. The downstream motion is described by
2 km
d + 2 km = ( v + v w )( t − 15 min) . We eliminate t =
and d by substitution:
vw
bv − v g15 min + 2 km = bv + v gFGH 2vkm − 15 minIJK
v
va15 minf − v a15 minf + 2 km =
2 km + 2 km − va15 minf − v a15 minf
v
v
va30 minf =
2 km
v
w
w
w
w
w
w
w
2 km
vw =
= 4.00 km h .
30 min
(b)
In the reference frame of the water, the chest is motionless. The boat travels upstream for 15 min
at speed v, and then downstream at the same speed, to return to the same point. Thus it travels
for 30 min. During this time, the falls approach the chest at speed v w , traveling 2 km. Thus
vw =
∆x
2 km
=
= 4.00 km h .
∆ t 30 min
112
*P4.70
Motion in Two Dimensions
Let the river flow in the x direction.
(a)
To minimize time, swim perpendicular to the banks in the y direction. You are in the
water for time t in ∆ y = v y t , t =
(b)
80 m
= 53.3 s .
1.5 m s
b
g
The water carries you downstream by ∆ x = v x t = 2.50 m s 53.3 s = 133 m .
K
vw
(c)
K
vw
K
vw
K
vs
K
vs
K
K
vs + vw
K
K
vs + vw
K
vs
To minimize downstream drift, you should swim so that
K
K
your resultant velocity v s + v w is perpendicular to your
K
swimming velocity v s relative to the water. This condition
is shown in the middle picture. It maximizes the angle
between the resultant velocity and the shore. The angle
1.5 m s
K
between v s and the shore is given by cos θ =
,
2.5 m s
K
K
vs + vw
K
vs
K
K
vs + vw
K
v w = 2.5 m/s i
θ = 53.1° .
(d)
Now v y = v s sin θ = 1.5 m s sin 53.1° = 1.20 m s
t=
∆y
80 m
=
= 66.7 s
v y 1.2 m s
b
g
∆ x = v x t = 2.5 m s − 1.5 m s cos 53.1° 66.7 s = 107 m .
θ
Chapter 4
*P4.71
113
Find the highest firing angle θ H for which the projectile will clear the mountain peak; this will
yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield
the maximum range under these conditions if both θ H and θ L are > 45° ; x = 2500 m, y = 1800 m ,
vi = 250 m s .
a f
1 2
1
gt = vi sin θ t − gt 2
2
2
x f = v xi t = vi cos θ t
y f = v yi t −
a
f
Thus
t=
xf
vi cos θ
.
Substitute into the expression for y f
x
IJ
1 F x
− gG
a f v cos
θ 2 H v cos θ K
y f = vi sin θ
f
f
i
i
2
= x f tan θ −
gx 2f
2 vi2 cos 2 θ
gx 2f
1
2
= tan θ + 1 so y f = x f tan θ − 2 tan 2 θ + 1 and
but
cos 2 θ
2 vi
e
0=
gx 2f
2 vi2
j
tan 2 θ − x f tan θ +
gx 2f
2 vi2
+ yf .
Substitute values, use the quadratic formula and find
tan θ = 3.905 or 1.197 , which gives θ H = 75.6° and θ L = 50.1° .
b
g
Range at θ H =
vi2 sin 2θ H
= 3.07 × 10 3 m from enemy ship
g
3.07 × 10 3 − 2 500 − 300 = 270 m from shore.
b
g
Range at θ L =
vi2 sin 2θ L
= 6.28 × 10 3 m from enemy ship
g
6.28 × 10 3 − 2 500 − 300 = 3.48 × 10 3 from shore.
Therefore, safe distance is < 270 m or > 3.48 × 10 3 m from the shore.
FIG. P4.71
114
*P4.72
Motion in Two Dimensions
We follow the steps outlined in Example 4.7, eliminating t =
d cos φ
to find
vi cos θ
vi sin θ d cos φ gd 2 cos 2 φ
− 2
= − d sin φ .
vi cos θ
2 vi cos 2 θ
Clearing of fractions,
2 vi2 cos θ sin θ cos φ − gd cos 2 φ = −2 vi2 cos 2 θ sin φ .
To maximize d as a function of θ, we differentiate through with respect to θ and set
a
f
2 vi2 cos θ cos θ cos φ + 2 vi2 sin θ − sin θ cos φ − g
dd
= 0:
dθ
a
f
dd
cos 2 φ = −2 vi2 2 cos θ − sin θ sin φ .
dθ
We use the trigonometric identities from Appendix B4 cos 2θ = cos 2 θ − sin 2 θ and
sin φ
1
sin 2θ = 2 sin θ cos θ to find cos φ cos 2θ = sin 2θ sin φ . Next,
= tan φ and cot 2θ =
give
cos φ
tan 2θ
a
f
cot 2φ = tan φ = tan 90°−2θ so φ = 90°−2θ and θ = 45°−
φ
.
2
ANSWERS TO EVEN PROBLEMS
P4.2
e
j
(b) v = 18.0 i + a 4.00 − 9.80t f j ;
(c) a = e −9.80 m s j j ;
(d) a54.0 mf i − a32.1 mf j ;
(e) b18.0 m sg i − b 25.4 m sg j ;
(f) e −9.80 m s j j
(a) r = 18.0t i + 4.00t − 4.90t 2 j ;
P4.8
e
P4.4
e
j
e
e
P4.10
e7.23 × 10
3
j
m, 1.68 × 10 3 m
g
horizontally;
2h
2h
below the horizontal
(b) tan −1
d
(a) d
P4.14
0.600 m s 2
P4.16
(a) 76.0°; (b) the same; (c)
(c) a circle of radius 5.00 m centered at
0 , 4.00 m
P4.18
25.8 m s
(a) v = −12.0t j m s ; a = −12.0 j m s 2 ;
(b) r = 3.00 i − 6.00 j m; v = −12.0 j m s
P4.20
d tan θ i −
e
j
+5.00 m − sin ω t i − cos ω t j ;
e
j
esin ω t i + cos ω t jj ;
v = 5.00 m ω − cos ω t i + sin ω t j ;
a = 5.00 m ω 2
a
P4.6
j
FG IJ
H K
j
a = 0 i + 5.00ω 2 j m s 2 ;
(b) r = 4.00 m j
j
(b) r = 10.0 i + 6.00 j m; 7.81 m s
P4.12
(a) v = −5.00ω i + 0 j m s ;
j
v = 5.00 i + 3.00tj m s ;
2
2
e
(a) r = 5.00t i + 1.50t 2 j m ;
f
e
j
e2 v
gd 2
2
i
cos 2 θ i
j
17d
8
Chapter 4
P4.22
33.5° below the horizontal
P4.48
2vi t cos θ i
P4.24
(a) 0.852 s; (b) 3.29 m s ; (c) 4.03 m s;
(d) 50.8°; (e) 1.12 s
P4.50
(a) see the solution;
−1
F
GH
2gh
I
JK
P4.26
tan
P4.28
0.033 7 m s 2 toward the center of the
Earth
v
P4.30
0.281 rev s
P4.32
7.58 × 10 3 m s; 5.80 × 10 3 s
P4.34
(a) 0.600 m s 2 forward;
(b) 0.800 m s 2 inward;
φ
(b) θ i = 45°+ ; d max =
2
(a) see the solution; (b) 29.7 m s 2 ;
(c) 6.67 m s at 36.9° above the horizontal
P4.38
(a) 26.9 m s ; (b) 67.3 m;
(c) 2.00 i − 5.00 j m s 2
e
j
P4.40
18.0 s
P4.42
153 km h at 11.3° north of west
P4.44
(a) 1.69 km s ; (b) 6.47 × 10 3 s
P4.54
10.7 m s
P4.56
R
2
P4.58
7.50 m s in the direction the ball was
thrown
P4.60
(a) 19.6 cm; (b) 0.0561°;
(c) aim low 3.68 cm; (d) aim low 3.68 cm;
(e) aim high 6.12 cm; (f) aim low;
(g) aim low
P4.62
(a)
P4.64
a18.8 m; − 17.3 mf
P4.66
see the solution; ~ 10 2 m s 2
P4.68
x = −D
P4.70
(a) at 90° to the bank; (b) 133 m;
(c) upstream at 53.1° to the bank; (d) 107 m
P4.72
see the solution
2
(a) 10.1 m s at 14.3° south from the
vertical; (b) 9.80 m s 2 vertically
downward
P4.46
27.7° east of north
g
2
g cos φ
P4.52
(c) 1.00 m s 2 forward and 53.1° inward
P4.36
b
vi2 1 − sin φ
115
gR ; (b)
e
j
2 −1 R
5
The Laws of Motion
CHAPTER OUTLINE
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
The Concept of Force
Newton’s First Law and
Inertial Frames
Mass
Newton’s Second Law
The Gravitational Force
and Weight
Newton’s Third Law
Some Applications of
Newton’s Laws
Forces of Friction
ANSWERS TO QUESTIONS
Q5.1
(a)
The force due to gravity of the earth pulling down on
the ball—the reaction force is the force due to gravity
of the ball pulling up on the earth. The force of the
hand pushing up on the ball—reaction force is ball
pushing down on the hand.
(b)
The only force acting on the ball in free-fall is the
gravity due to the earth -the reaction force is the
gravity due to the ball pulling on the earth.
Q5.2
The resultant force is zero, as the acceleration is zero.
Q5.3
Mistake one: The car might be momentarily at rest, in the
process of (suddenly) reversing forward into backward motion.
In this case, the forces on it add to a (large) backward resultant.
Mistake two: There are no cars in interstellar space. If the car is remaining at rest, there are
some large forces on it, including its weight and some force or forces of support.
Mistake three: The statement reverses cause and effect, like a politician who thinks that his
getting elected was the reason for people to vote for him.
Q5.4
When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seat
on her body. Clark is standing, however, and the only force on him is the friction between his shoes
and the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, but
the rest of his body experiences almost no accelerating force (only that due to his being attached to
his accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newton’s
first law, relative to the ground. Relative to Claudette, however, he is moving toward her and falls
into her lap. (Both performers won Academy Awards.)
Q5.5
First ask, “Was the bus moving forward or backing up?” If it was moving forward, the passenger is
lying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If the
bus was backing up at any reasonable speed, a sudden stop could not make a suitcase fly far. Fine
her for malicious litigiousness.
Q5.6
It would be smart for the explorer to gently push the rock back into the storage compartment.
Newton’s 3rd law states that the rock will apply the same size force on her that she applies on it. The
harder she pushes on the rock, the larger her resulting acceleration.
117
118
The Laws of Motion
Q5.7
The molecules of the floor resist the ball on impact and push the ball back, upward. The actual force
acting is due to the forces between molecules that allow the floor to keep its integrity and to prevent
the ball from passing through. Notice that for a ball passing through a window, the molecular forces
weren’t strong enough.
Q5.8
While a football is in flight, the force of gravity and air resistance act on it. When a football is in the
process of being kicked, the foot pushes forward on the ball and the ball pushes backward on the
foot. At this time and while the ball is in flight, the Earth pulls down on the ball (gravity) and the ball
pulls up on the Earth. The moving ball pushes forward on the air and the air backward on the ball.
Q5.9
It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass,
gravity pulls it downward. A vertical component of the tension must balance the weight for the
cable to be in equilibrium. If the cable were completely horizontal, then there would be no vertical
component of the tension to balance the weight.
Some physics teachers demonstrate this by asking a beefy student to pull on the ends of a
cord supporting a can of soup at its center. Some get two burly young men to pull on opposite ends
of a strong rope, while the smallest person in class gleefully mashes the center of the rope down to
the table. Point out the beauty of sagging suspension-bridge cables. With a laser and an optical lever,
demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bends
the bench. Give them “I make the floor sag” buttons, available to instructors using this manual.
Estimate the cost of an infinitely strong cable, and the truth will always win.
Q5.10
As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and the
scale reads the larger upward force that the floor exerts on them together. Around the top of the
weight’s motion, the scale reads less than average. If the iron is moving upward, the lifter can
declare that she has thrown it, just by letting go of it for a moment, so our answer applies also to this
case.
Q5.11
As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the mass
causes an increase in the acceleration.
Q5.12
As the rocket takes off, it burns fuel, pushing the gases from the combustion out the back of the
rocket. Since the gases have mass, the total remaining mass of the rocket, fuel, and oxidizer
decreases. With a constant thrust, a decrease in the mass results in an increasing acceleration.
Q5.13
The friction of the road pushing on the tires of a car causes an automobile to move. The push of the
air on the propeller moves the airplane. The push of the water on the oars causes the rowboat to
move.
Q5.14
As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of
the Earth on his foot. In the second case, the action is the force exerted on the girl’s back by the
snowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is the
force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the
force exerted on the window by the air molecules; the reaction is the force on the air molecules
exerted by the window. We could in each case interchange the terms ‘action’ and ‘reaction.’
Q5.15
The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then the
resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were
winning with the rope steadily moving in their direction or if the contest was even, then the tension
would still be 9 200 N. In all of these case, the acceleration is zero, and so must be the resultant force
on the rope. To win the tug-of-war, a team must exert a larger force on the ground than their
opponents do.
Chapter 5
119
Q5.16
The tension in the rope when pulling the car is twice that in the tug-of-war. One could consider the
car as behaving like another team of twenty more people.
Q5.17
This statement contradicts Newton’s 3rd law. The force that the locomotive exerted on the wall is
the same as that exerted by the wall on the locomotive. The wall temporarily exerted on the
locomotive a force greater than the force that the wall could exert without breaking.
Q5.18
The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since the
athlete and the sack of sand have the same weight, the acceleration of the system must be zero.
Q5.19
The resultant force doesn’t always add to zero. If it did, nothing could ever accelerate. If we choose a
single object as our system, action and reaction forces can never add to zero, as they act on different
objects.
Q5.20
An object cannot exert a force on itself. If it could, then objects would be able to accelerate
themselves, without interacting with the environment. You cannot lift yourself by tugging on your
bootstraps.
Q5.21
To get the box to slide, you must push harder than the maximum static frictional force. Once the box
is moving, you need to push with a force equal to the kinetic frictional force to maintain the box’s
motion.
Q5.22
The stopping distance will be the same if the mass of the truck is doubled. The stopping distance will
decrease by a factor of four if the initial speed is cut in half.
Q5.23
If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the
tires and the road is less than the maximum static friction force. Anti-lock brakes work by “pumping”
the brakes (much more rapidly that you can) to minimize skidding of the tires on the road.
Q5.24
With friction, it takes longer to come down than to go up. On the way up, the frictional force and the
component of the weight down the plane are in the same direction, giving a large acceleration. On
the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the
incline is frictionless, it takes the same amount of time to go up as it does to come down.
Q5.25
(a)
The force of static friction between the crate and the bed of the truck causes the crate to
accelerate. Note that the friction force on the crate is in the direction of its motion relative to
the ground (but opposite to the direction of possible sliding motion of the crate relative to
the truck bed).
(b)
It is most likely that the crate would slide forward relative to the bed of the truck.
Q5.26
In Question 25, part (a) is an example of such a situation. Any situation in which friction is the force
that accelerates an object from rest is an example. As you pull away from a stop light, friction is the
force that accelerates forward a box of tissues on the level floor of the car. At the same time, friction
of the ground on the tires of the car accelerates the car forward.
120
The Laws of Motion
SOLUTIONS TO PROBLEMS
The following problems cover Sections 5.1–5.6.
Section 5.1
The Concept of Force
Section 5.2
Newton’s First Law and Inertial Frames
Section 5.3
Mass
Section 5.4
Newton’s Second Law
Section 5.5
The Gravitational Force and Weight
Section 5.6
Newton’s Third Law
P5.1
For the same force F, acting on different masses
F = m 1 a1
and
F = m2 a2
(a)
m1
a
1
= 2 =
3
m2
a1
(b)
F = m1 + m 2 a = 4m1 a = m1 3.00 m s 2
a
f
c
a = 0.750 m s
*P5.2
v f = 880 m s, m = 25.8 kg , x f = 6 m
v 2f = 2 ax f = 2 x f
F=
P5.3
2
mv 2f
2x f
FG F IJ
H mK
= 1.66 ×10 6 N forward
m = 3.00 kg
e
j
a = 2.00 i + 5.00 j m s 2
e6.00i + 15.0jj N
a6.00f + a15.0f N =
∑ F = ma =
∑F =
2
2
16.2 N
h
Chapter 5
P5.4
Fg = weight of ball = mg
v release = v and time to accelerate = t :
a=
(a)
∆v v v = = i
∆t t t
Distance x = vt :
x=
(b)
FG v IJ t =
H 2K
Fg v Fp − Fg j =
i
gt
Fg v i + Fg j
gt
Fp =
P5.5
vt
2
e
j
m = 4.00 kg , v i = 3.00 i m s , v 8 = 8.00 i + 10.0 j m s , t = 8.00 s
∆v 5.00 i + 10.0 j
a=
=
m s2
t
8.00
F = ma = 2.50 i + 5.00 j N
e
j
2
2
F = ( 2.50) +(5.00) = 5.59 N
P5.6
(a)
Let the x-axis be in the original direction of the molecule’s motion.
e
v f = vi + at: −670 m s = 670 m s + a 3.00 × 10 −13 s
j
a = −4. 47 × 10 15 m s 2
(b)
For the molecule,
∑ F = ma . Its weight is negligible.
e
j
Fwall on molecule = 4.68 × 10 −26 kg −4.47 × 10 15 m s 2 = −2.09 × 10 −10 N
G
Fmolecule on wall = +2.09 × 10 −10 N
121
122
P5.7
The Laws of Motion
(a)
∑ F = ma and v 2f = vi2 + 2 ax f or a =
v 2f − vi2
2x f
.
Therefore,
∑F = m
ev
2
f
− vi2
j
2x f
∑ F = 9.11 × 10 −31
(b)
LMe7.00 × 10
kg N
5
j − e3.00 × 10
2b0.050 0 mg
m s2
2
5
m s2
j OPQ
2
= 3.64 × 10 −18 N .
The weight of the electron is
c
hc
h
Fg = mg = 9.11×10−31 kg 9.80 m s 2 = 8.93 ×10−30 N
The accelerating force is 4.08 ×10 11 times the weight of the electron.
P5.8
P5.9
a
f
(a)
Fg = mg = 120 lb = 4.448 N lb (120 lb)= 534 N
(b)
m=
Fg
g
=
534 N
= 54.5 kg
9.80 m s 2
Fg = mg = 900 N , m =
900 N
= 91.8 kg
9.80 m s 2
cF h
g
P5.10
on Jupiter
c
h
= 91.8 kg 25.9 m s 2 = 2.38 kN
Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just
canceled out by a glass of tomato juice. By subtraction, Fg = mg p and Fg = mg C give
c h
c
p
c h
C
h
∆Fg = m g p − g C .
For a person whose mass is 88.7 kg, the change in weight is
b
g
∆Fg = 88.7 kg 9.809 5 − 9.780 8 = 2.55 N .
A precise balance scale, as in a doctor’s office, reads the same in different locations because it
compares you with the standard masses on its beams. A typical bathroom scale is not precise enough
to reveal this difference.
Chapter 5
P5.11
(a)
∑ F = F1 + F2 = e 20.0 i + 15.0 jj N
∑ F = ma:
20.0 i + 15.0 j = 5.00a
a = 4.00 i + 3.00 j m s 2
e
j
or
a = 5.00 m s 2 at θ = 36.9°
(b)
F2 x = 15.0 cos 60.0° = 7.50 N
FIG. P5.11
F2 y = 15.0 sin 60.0° = 13.0 N
e
j
F2 = 7.50 i + 13.0 j N
∑ F = F1 + F2 = e27.5 i + 13.0 jj N = ma = 5.00a
a=
P5.12
e5.50i + 2.60jj m s
2
= 6.08 m s 2 at 25.3°
We find acceleration:
r f − ri = v i t +
1 2
at
2
a
f
j
1
2
4.20 m i − 3.30 mj = 0+ a 1.20 s = 0.720 s 2 a
2
a = 5.83 i − 4.58 j m s 2 .
e
Now ∑ F = ma becomes
Fg + F2 = ma
e
b
j
ge
j
F2 = 2.80 kg 5.83 i − 4.58 j m s 2 + 2.80 kg 9.80 m s 2 j
F2 =
P5.13
(a)
e16.3 i + 14.6 jj N
.
You and the earth exert equal forces on each other: m y g = M e a e . If your mass is 70.0 kg,
a70.0 kg fc9.80 m s h =
=
2
ae
(b)
5.98 ×10 24 kg
~ 10−22 m s 2 .
You and the planet move for equal times intervals according to x =
50.0 cm high,
2xy
ay
xe =
=
1 2
at . If the seat is
2
2xe
ae
a
f
my
70.0 kg 0.500 m
ae
~ 10 −23 m .
xy =
xy =
ay
me
5.98 × 10 24 kg
123
124
P5.14
The Laws of Motion
∑ F = ma reads
e−2.00 i + 2.00j + 5.00i − 3.00j − 45.0ij N = me3.75 m s ja
2
where a represents the direction of a
e−42.0 i − 1.00jj N = me3.75 m s ja
2
FG 1.00 IJ below the –x-axis
H 42.0 K
∑ F = 42.0 N at 181° = mc3.75 m s ha .
∑F =
2
2
(42.0) +(1.00) N at tan−1
2
For the vectors to be equal, their magnitudes and their directions must be equal.
(a)
∴ a is at 181° counterclockwise from the x-axis
(b)
m=
(d)
v f = v i + at = 0 + 3.75 m s 2 at 181° 10.0 s so v f = 37.5 m s at 181°
42.0 N
= 11.2 kg
3.75 m s 2
e
j
v f = 37.5 m s cos 181° i + 37.5 m s sin 181° j so v f =
P5.15
(c)
v f = 37.5 2 + 0.893 2 m s = 37.5 m s
(a)
15.0 lb up
(b)
5.00 lb up
(c)
0
Section 5.7
P5.16
e−37.5 i − 0.893 jj m s
Some Applications of Newton’s Laws
dy
dx
= 10t , v y =
= 9t 2
dt
dt
dv y
dv x
= 10 , a y =
= 18t
ax =
dt
dt
vx =
At t = 2.00 s , a x = 10.0 m s 2 , a y = 36.0 m s 2
∑ Fx = ma x : 3.00 kg e10.0
∑ Fy = ma y : 3.00 kg e36.0
j
m s j = 108 N
m s 2 = 30.0 N
2
∑F=
Fx2 + Fy2 = 112 N
Chapter 5
P5.17
m = 1.00 kg
50.0 m
mg = 9.80 N
α
0.200 m
tan α =
25.0 m
α = 0.458°
0.200 m
T
T
Balance forces,
mg
2T sin α = mg
T=
P5.18
9.80 N
= 613 N
2 sin α
FIG. P5.17
T3 = Fg
(1)
T1 sin θ 1 + T2 sin θ 2 = Fg
(2)
T1 cos θ 1 = T2 cos θ 2
(3)
θ2
θ1
Eliminate T2 and solve for T1
T1 =
bsinθ
Fg cos θ 2
1
cos θ 2 + cos θ 1 sin θ 2
Fg
g
=
Fg cos θ 2
b
sin θ 1 + θ 2
T3 = Fg = 325 N
FG cos 25.0° IJ = 296 N
H sin 85.0° K
F cos θ IJ = 296 NFG cos 60.0° IJ =
=T G
H cos 25.0° K
H cos θ K
P5.19
1
1
2
See the solution for T1 in Problem 5.18.
T1
θ1
T1 = Fg
T2
g
θ2
T3
163 N
FIG. P5.18
T2
125
126
P5.20
The Laws of Motion
(a)
An explanation proceeding from fundamental physical principles will
be best for the parents and for you. Consider forces on the bit of string
touching the weight hanger as shown in the free-body diagram:
Horizontal Forces:
Vertical Forces:
∑ Fx = ma x : −Tx + T cos θ = 0
∑ Fy = ma y : −Fg + T sin θ = 0
FIG. P5.20
You need only the equation for the vertical forces to find that the tension in the string is
Fg
. The force the child feels gets smaller, changing from T to T cos θ , while
given by T =
sin θ
the counterweight hangs on the string. On the other hand, the kite does not notice what you
are doing and the tension in the main part of the string stays constant. You do not need a
level, since you learned in physics lab to sight to a horizontal line in a building. Share with
the parents your estimate of the experimental uncertainty, which you make by thinking
critically about the measurement, by repeating trials, practicing in advance and looking for
variations and improvements in technique, including using other observers. You will then
be glad to have the parents themselves repeat your measurements.
P5.21
Fg
e
(b)
T=
(a)
Isolate either mass
sin θ
=
0.132 kg 9.80 m s 2
sin 46.3°
j=
1.79 N
T + mg = ma = 0
T = mg .
The scale reads the tension T,
so
FIG. P5.21(a)
e
j
T = mg = 5.00 kg 9.80 m s 2 = 49.0 N .
(b)
Isolate the pulley
T2 + 2T1 = 0
T2 = 2 T1 = 2mg = 98.0 N .
(c)
∑ F = n + T + mg = 0
FIG. P5.21(b)
Take the component along the incline
n x + Tx + mg x = 0
or
0 + T − mg sin 30.0° = 0
T = mg sin 30.0° =
= 24.5 N .
a f
mg 5.00 9.80
=
2
2
FIG. P5.21(c)
Chapter 5
P5.22
127
The two forces acting on the block are the normal force, n, and the
weight, mg. If the block is considered to be a point mass and the xaxis is chosen to be parallel to the plane, then the free body
diagram will be as shown in the figure to the right. The angle θ is
the angle of inclination of the plane. Applying Newton’s second
law for the accelerating system (and taking the direction up the
plane as the positive x direction) we have
FIG. P5.22
∑ Fy = n − mg cos θ = 0: n = mg cos θ
∑ Fx = −mg sin θ = ma : a = −g sin θ
(a)
When θ = 15.0°
a = −2.54 m s 2
(b)
Starting from rest
d
i
v 2f = vi2 + 2 a x f − xi = 2 ax f
e
ja
f
v f = 2 ax f = 2 −2.54 m s 2 −2.00 m = 3.18 m s
P5.23
Choose a coordinate system with i East and j North.
∑ F = ma = 1.00 kg e10.0
j
m s 2 at 30.0°
a5.00 Nfj + F = a10.0 Nf∠30.0° = a5.00 Nfj + a8.66 Nfi
1
∴ F1 = 8.66 N (East )
*P5.24
FIG. P5.23
First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, ∑ Fx = ma
a f
T = 5 kg a
n
(1)
Next consider the block that moves vertically. The forces on it are
the tension T and its weight, 88.2 N.
We have
5 kg
+x
T
49 N
T
+y
9 kg
Fg = 88.2 N
FIG. P5.24
∑ Fy = ma
a f
88.2 N − T = 9 kg a
(2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give 88.2 N = 14 kg a. Then
b
g
a = 6.30 m s 2 and T = 31.5 N .
128
P5.25
The Laws of Motion
After it leaves your hand, the block’s speed changes only
because of one component of its weight:
∑ Fx = ma x
Taking v f
− mg sin 20.0° = ma
d
i
= 0 , v = 5.00 m s, and a = −g sina 20.0°f gives
v 2f = vi2 + 2 a x f − xi .
i
a
2
fc
0 = (5.00) − 2(9.80) sin 20.0° x f − 0
h
or
xf =
P5.26
FIG. P5.25
25.0
= 3.73 m .
2(9.80) sin 20.0°
a
f
m1 = 2.00 kg , m 2 = 6.00 kg , θ = 55.0°
(a)
∑ Fx = m 2 g sin θ − T = m 2 a
and
T − m1 g = m1 a
a=
*P5.27
a
m 2 g sin θ − m1 g
= 3.57 m s 2
m1 + m 2
FIG. P5.26
f
(b)
T = m1 a + g = 26.7 N
(c)
Since vi = 0 , v f = at = 3.57 m s 2 ( 2.00 s)= 7.14 m s .
c
h
We assume the vertical bar is in compression, pushing up
on the pin with force A, and the tilted bar is in tension,
exerting force B on the pin at −50° .
∑ Fx = 0:
−2 500 N cos 30°+ B cos 50° = 0
∑ Fy = 0:
−2 500 N sin 30°+ A − 3.37 × 10 3 N sin 50° = 0
B = 3.37 × 10 3 N
A = 3.83 × 10 3 N
Positive answers confirm that
B is in tension and A is in compression.
30°
2 500 N
50°
A
B
2 500 N cos30°
B cos50°
A
2 500 N sin30°
FIG. P5.27
B sin50°
Chapter 5
P5.28
First, consider the 3.00 kg rising mass. The forces on it are
the tension, T, and its weight, 29.4 N. With the upward
direction as positive, the second law becomes
∑ Fy = ma y : T − 29.4 N = a3.00 kg fa
(1)
The forces on the falling 5.00 kg mass are its weight and T,
and its acceleration is the same as that of the rising mass.
Calling the positive direction down for this mass, we have
∑ Fy = ma y : 49 N − T = a5.00 kg fa
FIG. P5.28
(2)
Equations (1) and (2) can be solved simultaneously by adding them:
a
f a
f
T − 29.4 N + 49.0 N − T = 3.00 kg a + 5.00 kg a
(b)
This gives the acceleration as
a=
(a)
19.6 N
= 2.45 m s 2 .
8.00 kg
Then
a
fc
h
T − 29.4 N = 3.00 kg 2.45 m s 2 = 7.35 N .
The tension is
T = 36.8 N .
(c)
Consider either mass. We have
y = vi t +
*P5.29
1
1 2
2
at = 0 + 2.45 m s 2 (1.00 s) = 1.23 m .
2
2
c
h
As the man rises steadily the pulley turns steadily and the tension in
the rope is the same on both sides of the pulley. Choose man-pulleyand-platform as the system:
T
∑ Fy = ma y
+T − 950 N = 0
T = 950 N .
The worker must pull on the rope with force 950 N .
950 N
FIG. P5.29
129
130
*P5.30
The Laws of Motion
Both blocks move with acceleration a =
a=
(a)
FG m
Hm
IJ
K
− m1
g:
2 + m1
2
F 7 kg − 2 kg I 9.8 m s
GH 7 kg + 2 kg JK
2
= 5.44 m s 2 .
Take the upward direction as positive for m1 .
d
i
2
+ 2 a x x f − xi :
v xf2 = v xi
b
0 = −2.4 m s
xf = −
g + 2e5.44 m s jdx − 0i
2
2
5.76 m 2 s 2
e
2 5.44 m s 2
j
f
= −0.529 m
x f = 0.529 m below its initial level
(b)
ja
e
v xf = v xi + a x t: v xf = −2.40 m s + 5.44 m s 2 1.80 s
f
v xf = 7. 40 m s upward
P5.31
Forces acting on 2.00 kg block:
T − m1 g = m 1 a
(1)
Forces acting on 8.00 kg block:
Fx − T = m 2 a
(a)
(2)
Eliminate T and solve for a:
a=
Fx − m1 g
m1 + m 2
a > 0 for Fx > m1 g = 19.6 N .
(b)
Eliminate a and solve for T:
T=
a
m1
Fx + m 2 g
m1 + m 2
f
FIG. P5.31
T = 0 for Fx ≤−m 2 g = −78.4 N .
(c)
Fx , N
ax , m s 2
–100
–78.4
–50.0
0
50.0
100
–12.5
–9.80
–6.96
–1.96
3.04
8.04
Chapter 5
*P5.32
(a)
For force components along the incline, with the upward direction taken as positive,
∑ Fx = ma x :
− mg sin θ = ma x
e
j
a x = − g sin θ = − 9.8 m s 2 sin 35° = −5.62 m s 2 .
For the upward motion,
d
2
+ 2 a x x f − xi
v xf2 = v xi
b
0= 5 m s
xf =
(b)
i
g + 2e−5.62 m s jdx − 0i
2
2
25 m 2 s 2
e
j
2 5.62 m s 2
f
= 2.22 m .
The time to slide down is given by
x f = xi + v xi t +
1
axt 2
2
0 = 2.22 m + 0 +
t=
a
1
−5.62 m s 2 t 2
2
e
j
f = 0.890 s .
2 2.22 m
5.62 m s
2
For the second particle,
x f = xi + v xi t +
1
axt 2
2
a
f e
ja
0 = 10 m + v xi 0.890 s + −5.62 m s 2 0.890 s
v xi =
−10 m + 2.22 m
= −8.74 m s
0.890 s
speed = 8.74 m s .
f
2
131
132
P5.33
The Laws of Motion
First, we will compute the needed accelerations:
a1f
a 2f
a3 f
a4f
ay = 0
v yf − v yi 1. 20 m s − 0
=
ay =
t
0.800 s
= 1.50 m s 2
Before it starts to move:
During the first 0.800 s:
While moving at constant velocity: a y = 0
v yf − v yi 0 − 1.20 m s
=
During the last 1.50 s:
ay =
1.50 s
t
= −0.800 m s 2
Newton’s second law is:
FIG. P5.33
∑ Fy = ma y
b
ge
j b
g
S = 706 N + b72.0 kg ga
+S − 72.0 kg 9.80 m s 2 = 72.0 kg a y
P5.34
(a)
When a y = 0 , S = 706 N .
(b)
When a y = 1.50 m s 2 , S = 814 N .
(c)
When a y = 0 , S = 706 N .
(d)
When a y =−0.800 m s 2 , S = 648 N .
(a)
Pulley P1 has acceleration a 2 .
Since m1 moves twice the distance P1 moves in the same
time, m1 has twice the acceleration of P1 , i.e., a1 = 2 a 2 .
(b)
From the figure, and using
∑ F = ma:
m 2 g − T2 = m 2 a 2
T1 = m1 a1 = 2m1 a 2
T2 − 2T1 = 0
y
.
a1f
a 2f
a3 f
FIG. P5.34
Equation (1) becomes m 2 g − 2T1 = m 2 a 2 . This equation combined with Equation (2) yields
FG
H
IJ
K
T1
m
2m1 + 2 = m 2 g
m1
2
T1 =
(c)
m1 m 2
m1 m 2
g and T2 =
g .
2m1 + 12 m 2
m1 + 14 m 2
From the values of T1 and T2 we find that
a1 =
m2 g
T1
=
2m1 + 12 m 2
m1
and a 2 =
m2 g
1
a1 =
.
2
4m 1 + m 2
Chapter 5
Section 5.8
*P5.35
Forces of Friction
+y
+y
n ground = Fg /2 = 85.0 lb
n tip
22.0° 22.0°
F2
F1
f
+x
+x
F = 45.8 lb
Fg = 170 lb
Free-Body Diagram of Person
22.0°
Free-Body Diagram of Crutch Tip
FIG. P5.35
From the free-body diagram of the person,
∑ Fx = F1 sina22.0°f − F2 sina22.0°f = 0 ,
which gives
F1 = F2 = F .
Then, ∑ Fy = 2 F cos 22.0°+85.0 lbs − 170 lbs = 0 yields F = 45.8 lb.
(a)
Now consider the free-body diagram of a crutch tip.
∑ Fx = f −( 45.8 lb) sin 22.0°= 0 ,
or
f = 17. 2 lb .
∑ Fy = n tip −( 45.8 lb) cos 22.0°= 0 ,
which gives
n tip = 42.5 lb .
For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so
f
17.2 lb
f = f s max = µ s n tip and µ s =
=
= 0.404 .
n tip 42.5 lb
a f
(b)
As found above, the compression force in each crutch is
F1 = F2 = F = 45.8 lb .
133
134
P5.36
The Laws of Motion
For equilibrium: f = F and n = Fg . Also, f = µ n i.e.,
µ=
µs =
f
F
=
n Fg
75.0 N
= 0.306
25.0 9.80 N
a f
FIG. P5.36
and
µk =
P5.37
∑ Fy = ma y :
60.0 N
= 0.245 .
25.0(9.80) N
+n − mg = 0
fs ≤ µ sn = µ s mg
This maximum magnitude of static friction acts so long as the tires roll without skidding.
∑ Fx = ma x :
− f s = ma
The maximum acceleration is
a = −µ s g .
The initial and final conditions are: x i = 0 , vi = 50.0 mi h = 22.4 m s , v f = 0
d
i
v 2f = vi2 + 2 a x f − xi : − vi2 = −2 µ s gx f
(a)
xf =
vi2
2 µg
xf =
a22.4 m sf =
2(0.100 )c9.80 m s h
xf =
vi2
2 µg
2
(b)
2
a22.4 m sf =
2(0.600)c9.80 m s h
256 m
2
xf =
2
42.7 m
Chapter 5
P5.38
If all the weight is on the rear wheels,
(a)
F = ma: µ s mg = ma
But
∆x =
so µ s =
2 ∆x
:
gt 2
µs =
(b)
*P5.39
(a)
at 2 µ s gt 2
=
2
2
a
fb
2 0.250 mi 1 609 m mi
e9.80 m s ja4.96 sf
2
2
g=
3.34 .
Time would increase, as the wheels would skid and only kinetic friction would act; or
perhaps the car would flip over.
The person pushes backward on the floor. The floor pushes forward
on the person with a force of friction. This is the only horizontal
force on the person. If the person’s shoe is on the point of slipping
the static friction force has its maximum value.
∑ Fx = ma x :
∑ Fy = ma y :
f = µ sn = ma x
n − mg = 0
ma x = µ s mg
x f = xi + v xi t +
1
ax t 2
2
e
e
j
j
a x = µ s g = 0.5 9.8 m s 2 = 4.9 m s 2
1
3 m = 0 + 0 + 4.9 m s 2 t 2
2
FIG. P5.39
t = 1.11 s
(b)
P5.40
xf =
2x f
2(3 m)
1
µ s gt 2 , t =
=
= 0.875 s
2
µs g
(0.8) 9.8 m s 2
c
h
m suitcase = 20.0 kg , F = 35.0 N
∑ Fx = ma x :
∑ Fy = ma y :
(a)
−20.0 N + F cos θ = 0
+n + F sin θ − Fg = 0
F cos θ = 20.0 N
cos θ =
20.0 N
= 0.571
35.0 N
θ = 55.2°
(b)
n = Fg − F sin θ = 196 − 35.0(0.821) N
n = 167 N
FIG. P5.40
135
136
P5.41
The Laws of Motion
m = 3.00 kg , θ = 30.0° , x = 2.00 m, t = 1.50 s
(a)
x=
1 2
at :
2
a
f
1
2
a 1.50 s
2
4.00
a=
= 1.78 m s 2
2
1.50
2.00 m =
FIG. P5.41
a f
∑ F = n + f + mg = m a :
Along x: 0 − f + mg sin 30.0° = ma
b
f = m g sin 30.0°− a
g
Along y: n + 0 − mg cos 30.0° = 0
n = mg cos 30.0°
a
f
f m g sin 30.0°−a
a
, µ k = tan 30.0°−
=
= 0.368
n
mg cos 30.0°
g cos 30.0°
(b)
µk =
(c)
f = m g sin 30.0°−a , f = 3.00 9.80 sin 30.0°−1.78 = 9.37 N
(d)
v 2f = vi2 + 2 a x f − xi
a
a
f
c
f
h
where
x f − xi = 2.00 m
a fa f
v 2f = 0 + 2 1.78 2.00 = 7.11 m 2 s 2
v f = 7.11 m 2 s 2 = 2.67 m s
Chapter 5
*P5.42
First we find the coefficient of friction:
∑ Fy = 0:
n
+n − mg = 0
f = µ sn = µ s mg
∑ Fx = ma x :
− µ s mg = −
v 2f
mvi2
2 ∆x
=
vi2
+ 2 a x ∆x = 0
n
f
b
g
ja
e
f
mg
mg sin10°
2
88 ft s
v2
µs = i =
= 0.981
2 g∆x 2 32.1 ft s 2 123 ft
f
mg cos10°
FIG. P5.42
Now on the slope
∑ Fy = 0:
∑ Fx = ma x :
+n − mg cos 10° = 0
f s = µ sn = µ s mg cos 10°
− µ s mg cos 10°+ mg sin 10° = −
∆x =
=
P5.43
mvi2
2 ∆x
vi2
2 g µ s cos 10°− sin 10°
b
g
b88 ft sg
=
2e32.1 ft s ja0.981 cos 10°− sin 10°f
2
2
152 ft .
T − f k = 5.00 a (for 5.00 kg mass)
9.00 g − T = 9.00 a (for 9.00 kg mass)
Adding these two equations gives:
a f
137
a fa f
9.00 9.80 − 0.200 5.00 9.80 = 14.0 a
a = 5.60 m s 2
∴ T = 5.00 5.60 + 0.200 5.00 9.80
a f
a fa f
= 37.8 N
FIG. P5.43
138
P5.44
The Laws of Motion
Let a represent the positive magnitude of the acceleration −aj of
m1 , of the acceleration −a i of m 2 , and of the acceleration +aj of m 3 .
Call T12 the tension in the left rope and T23 the tension in the cord
on the right.
For m1 ,
∑ Fy = ma y
+T12 − m1 g = −m1 a
For m 2 ,
∑ Fx = ma x
−T12 + µ k n + T23 = −m 2 a
and
∑ Fy = ma y
n − m2 g = 0
for m 3 ,
∑ Fy = ma y
T23 − m 3 g = +m 3 a
n
T12
T23
f = µ kn
m2 g
T12
T23
m1 g
m3 g
we have three simultaneous equations
b
g
− 0.350a9.80 N f − T = b1.00 kg ga
+T − 19.6 N = b 2.00 kg ga .
−T12 + 39.2 N = 4.00 kg a
+T12
23
23
(a)
FIG. P5.44
Add them up:
a
f
+39.2 N − 3. 43 N − 19.6 N = 7.00 kg a
a = 2.31 m s 2 , down for m1 , left for m 2 , and up for m 3 .
(b)
a
fc
Now −T12 + 39. 2 N = 4.00 kg 2.31 m s 2
h
T12 = 30.0 N
a
fc
and T23 − 19.6 N = 2.00 kg 2.31 m s 2
h
T23 = 24.2 N .
P5.45
(a)
See Figure to the right
(b)
68.0 − T − µm 2 g = m 2 a (Block #2)
T − µm1 g = m1 a (Block #1)
T
m1
n1
m1
Adding,
g b
g
68.0
a=
bm + m g − µg =
1
1.29 m s 2
2
T = m1 a + µm1 g = 27. 2 N
T
m2
f2 = µ k n 2
m1 g = 118 N
68.0 − µ m1 + m 2 g = m1 + m 2 a
F
n2
T
f1 = µ k n 1
b
m2
m2 g = 176 N
FIG. P5.45
F
Chapter 5
P5.46
(Case 1, impending upward motion)
Setting
∑ Fx = 0:
P cos 50.0°−n = 0
fs , max = µ sn:
fs , max = µ s P cos 50.0°
a
f
= 0.250 0.643 P = 0.161 P
Setting
∑ Fy = 0:
a f
P sin 50.0°−0.161P − 3.00 9.80 = 0
Pmax = 48.6 N
(Case 2, impending downward motion)
As in Case 1,
FIG. P5.46
fs, max = 0.161P
Setting
∑ Fy = 0:
a f
P sin 50.0°+0.161P − 3.00 9.80 = 0
Pmin = 31.7 N
*P5.47
y
When the sled is sliding uphill
∑ Fy = ma y :
+n − mg cos θ = 0
∑ Fx = ma x :
+ mg sin θ + µ k mg cos θ = ma up
n
f = µ k n = µ k mg cos θ
mg sin θ
v f = 0 = vi + a up t up
mg cos θ
vi = − a up t up
1
vi + v f t up
2
1
1
2
∆x = a up t up + 0 t up = a up t up
2
2
∆x =
d
e
x
i
FIG. P5.47
j
When the sled is sliding down, the direction of the friction force is reversed:
mg sin θ − µ k mg cos θ = ma down
∆x=
1
2
a down t down
.
2
Now
t down = 2t up
1
1
2
= a down 2t up
a up tup
2
2
a up = 4a down
e j
b
2
g sin θ + µ k g cos θ = 4 g sin θ − µ k g cos θ
5 µ k cos θ = 3 sin θ
µk =
FG 3 IJ tanθ
H 5K
g
f
139
140
*P5.48
The Laws of Motion
Since the board is in equilibrium, ∑ Fx = 0 and we see that the normal
forces must be the same on both sides of the board. Also, if the
minimum normal forces (compression forces) are being applied, the
board is on the verge of slipping and the friction force on each side is
a f
f = fs
max
n
f
f
n
= µ sn .
The board is also in equilibrium in the vertical direction, so
∑ Fy = 2 f − Fg = 0 , or
f=
Fg
2
.
Fg = 95.5 N
FIG. P5.48
The minimum compression force needed is then
n=
*P5.49
(a)
f
µs
a
=
Fg
2µ s
=
95.5 N
= 72.0 N .
2(0.663)
f
n + F sin 15°− 75 N cos 25° = 0
n
F
15°
f s, max
∴ n = 67.97 − 0.259 F
fs , max = µ s n = 24.67 − 0.094F
25°
For equilibrium: F cos 15°+24.67 − 0.094F − 75 sin 25°= 0 .
This gives F = 8.05 N .
75 N
FIG. P5.49(a)
(b)
n
F cos 15°−( 24.67 − 0.094F )− 75 sin 25° = 0 .
F
This gives F = 53.2 N .
15°
f s, max
25°
75 N
FIG. P5.49(b)
(c)
f k = µ k n = 10.6 − 0.040 F . Since the velocity is constant, the net
force is zero:
F cos 15°−(10.6 − 0.040 F )− 75 sin 25° = 0 .
This gives F = 42.0 N .
n
F
15°
fk
25°
75 N
FIG. P5.49(c)
Chapter 5
*P5.50
We must consider separately the disk when it is in contact with the roof
and when it has gone over the top into free fall. In the first case, we take
x and y as parallel and perpendicular to the surface of the roof:
∑ Fy = ma y :
+n − mg cos θ = 0
n = mg cos θ
then friction is f k = µ k n = µ k mg cos θ
∑ Fx = ma x :
FIG. P5.50
− f k − mg sin θ = ma x
a
f
a x = − µ k g cos θ − g sin θ = −0.4 cos 37°− sin 37° 9.8 m s 2 = −9.03 m s 2
The Frisbee goes ballistic with speed given by
i b
d
2
v xf2 = v xi
+ 2 a x x f − xi = 15 m s
g + 2e−9.03 m s ja10 m − 0f = 44.4 m
2
2
2
v xf = 6.67 m s
For the free fall, we take x and y horizontal and vertical:
d
2
2
= v yi
+ 2 a y y f − yi
v yf
b
i
g + 2e−9.8 m s jdy
b4.01 m sg = 6.84 m
= 6.02 m +
0 = 6.67 m s sin 37°
2
2
f
− 10 m sin 37°
i
2
yf
19.6 m s 2
Additional Problems
P5.51
(a)
see figure to the right
(b)
First consider Pat and the chair as the system.
Note that two ropes support the system, and
T = 250 N in each rope. Applying ∑ F = ma
2T − 480 = ma , where m =
480
= 49.0 kg .
9.80
FIG. P5.51
Solving for a gives
a=
(c)
500 − 480
= 0.408 m s 2 .
49.0
∑ F = ma on Pat:
320
∑ F = n + T − 320 = ma , where m = 9.80 = 32.7 kg
n = ma + 320 − T = 32.7(0.408)+ 320 − 250 = 83.3 N .
s2
141
142
P5.52
The Laws of Motion
∑ F = ma gives the object’s acceleration
∑ F = e8.00 i − 4.00tjj N
a=
2.00 kg
m
dv
a = 4.00 m s 2 i − 2.00 m s 3 tj =
.
dt
e
j e
j
Its velocity is
z
z
v
t
dv = v − v i = v − 0 = adt
vi
0
ze
t
j e
v = e 4.00t m s ji − e1.00t
v=
0
2
(a)
j
m s jj .
4.00 m s 2 i − 2.00 m s 3 tj dt
2
3
2
We require v = 15.0 m s , v = 225 m 2 s 2
16.0t 2 m 2 s 4 + 1.00t 4 m 2 s 6 = 225 m 2 s 2
1.00t 4 + 16.0 s 2 t 2 − 225 s 4 = 0
t2 =
−16.0 ±
a16.0f − 4a−225f = 9.00 s
2
2.00
t = 3.00 s .
Take ri = 0 at t = 0. The position is
z ze
t
t
0
0
r = vdt =
e
r = 4.00 m s 2
at t = 3 s we evaluate.
e18.0i − 9.00jj m
(c)
r=
(b)
So r = (18.0) +(9.00) m = 20.1 m
2
2
j e
j
4.00t m s 2 i − 1.00t 2 m s 3 j dt
2
3
j t2 i − e1.00 m s j t3 j
3
2
Chapter 5
*P5.53
(a)
y
Situation A
∑ Fx = ma x :
∑ Fy = ma y :
FA + µ sn − mg sin θ = 0
+n − mg cos θ = 0
a
FA = mg sin θ − µ s cos θ
fs
FA
mg cos θ
FIG. P5.53(a)
f.
y
Situation B
∑ Fx = ma x :
∑ Fy = ma y :
FB cos θ + µ s n − mg sin θ = 0
FB
fs
a
a
mg sin θ − µ s cos θ
mg cos θ
mg sin θ
FIG. P5.53(b)
FB cos θ + µ s mg cos θ + µ s FB sin θ − mg sin θ = 0
FB =
f
cos θ + µ s sin θ
f
FA = 2 kg 9.8 m s 2 sin 25°−0.16 cos 25° = 5.44 N
FB =
a
f
19.6 N 0.278
= 5.59 N
cos 25°+0.16 sin 25°
Student A need exert less force.
(d)
FB =
FA
F
= A
cos 25°+0.38 sin 25° 1.07
Student B need exert less force.
b g
P − Q = b3 kg ga
Q = b 4 kg ga
P5.54
18 N − P = 2 kg a
a f
Adding gives 18 N = 9 kg a so
FIG. P5.54
a = 2.00 m s 2 .
(b)
e
j
Q = 4 kg 2 m s 2 = 8.00 N net force on the 4 kg
e
j
P − 8 N = 3 kg 2 m s 2 = 6.00 N net force on the 3 kg and P = 14 N
e
j
18 N − 14 N = 2 kg 2 m s 2 = 4.00 N net force on the 2 kg
continued on next page
x
n
− FB sin θ + n − mg cos θ = 0
Substitute n = mg cos θ + FB sin θ to find
(c)
x
n
mg sin θ
Eliminate n = mg cos θ to solve for
(b)
143
144
P5.55
The Laws of Motion
(c)
From above, Q = 8.00 N and P = 14.0 N .
(d)
The 3-kg block models the heavy block of wood. The contact force on your back is
represented by Q, which is much less than the force F. The difference between F and Q is
the net force causing acceleration of the 5-kg pair of objects. The acceleration is real and
nonzero, but lasts for so short a time that it never is associated with a large velocity. The
frame of the building and your legs exert forces, small relative to the hammer blow, to bring
the partition, block, and you to rest again over a time large relative to the hammer blow.
This problem lends itself to interesting lecture demonstrations. One person can hold a lead
brick in one hand while another hits the brick with a hammer.
(a)
First, we note that F = T1 . Next, we focus on the
mass M and write T5 = Mg . Next, we focus on the
bottom pulley and write T5 = T2 + T3 . Finally, we
focus on the top pulley and write T4 = T1 + T2 + T3 .
Since the pulleys are not starting to rotate and are
frictionless, T1 = T3 , and T2 = T3 . From this
Mg
information, we have T5 = 2T2 , soT2 =
.
2
Then T1 = T2 = T3 =
Mg
3 Mg
, and T4 =
, and
2
2
T5 = Mg .
(b)
Since F = T1 , we have F =
Mg
.
2
FIG. P5.55
P5.56
We find the diver’s impact speed by analyzing his free-fall motion:
c
h
v 2f = vi2 + 2 ax = 0 + 2 −9.80 m s 2 (−10.0 m) so v f = −14.0 m s.
Now for the 2.00 s of stopping, we have v f = vi + at :
a
0 = −14.0 m s + a 2.00 s
f
2
a = +7.00 m s .
Call the force exerted by the water on the diver R. Using
e
j
∑ Fy = ma ,
e
+ R − 70.0 kg 9.80 m s 2 = 70.0 kg 7.00 m s 2
R = 1.18 kN .
j
Chapter 5
P5.57
(a)
145
The crate is in equilibrium, just before it starts to
move. Let the normal force acting on it be n and
the friction force, fs .
Resolving vertically:
n = Fg + P sin θ
FIG. P5.57
Horizontally:
P cos θ = fs
But,
fs ≤ µ sn
i.e.,
c
P cos θ ≤ µ s Fg + P sin θ
h
or
a
f
P cos θ − µ s sin θ ≤ µ s Fg .
Divide by cos θ :
a
f
P 1− µ s tan θ ≤ µ s Fg sec θ .
Then
Pminimum =
(b)
P=
µ s Fg sec θ
1 − µ s tan θ
.
0.400(100 N ) sec θ
1 − 0.400 tan θ
b g
Pa N f
θ deg
0.00
15.0
30.0
45.0
60.0
40.0
46.4
60.1
94.3
260
If the angle were 68.2° or more, the expression for P would go to infinity and motion would
become impossible.
146
P5.58
The Laws of Motion
(a)
Following the in-chapter Example about a block on a frictionless incline, we have
c
h
a = g sin θ = 9.80 m s 2 sin 30.0°
a = 4.90 m s 2
(b)
The block slides distance x on the incline, with sin 30.0° =
c
h
c
0.500 m
x
h
x = 1.00 m: v 2f = vi2 + 2 a x f − xi = 0 + 2 4.90 m s 2 (1.00 m)
v f = 3.13 m s after time t s =
(c)
Now in free fall y f − yi = v yi t +
2x f
vf
=
2(1.00 m)
3.13 m s
= 0.639 s .
1
ayt 2 :
2
b
g
−2.00 = −3.13 m s sin 30.0° t −
e4.90 m s jt + b1.56 m sgt − 2.00 m = 0
2
1
9.80 m s 2 t 2
2
e
j
2
t=
−1.56 m s ±
b1.56 m sg − 4e4.90 m s ja−2.00 mf
2
9.80 m s 2
Only one root is physical
t = 0.499 s
b
g
a
f
x f = v x t = 3.13 m s cos 30.0° 0.499 s = 1.35 m
(d)
total time = t s + t = 0.639 s + 0.499 s = 1.14 s
(e)
The mass of the block makes no difference.
2
Chapter 5
P5.59
With motion impending,
n + T sin θ − mg = 0
b
f = µ s mg − T sin θ
g
and
T cos θ − µ s mg + µ sT sin θ = 0
FIG. P5.59
so
T=
µ s mg
.
cos θ + µ s sin θ
To minimize T, we maximize cos θ + µ s sin θ
b
g
d
cos θ + µ s sin θ = 0 = − sin θ + µ s cos θ .
dθ
*P5.60
(a)
θ = tan−1 µ s = tan−1 0.350 = 19.3°
(b)
T=
a
fc
0.350 1.30 kg 9.80 m s 2
cos 19.3°+0.350 sin 19.3°
(a)
See Figure (a) to the right.
(b)
See Figure (b) to the right.
(c)
For the pin,
∑ Fy = ma y :
h=
4.21 N
a
fc
h
mg = 36.4 kg 9.8 m s 2 = 357 N
C cos θ − 357 N = 0
357 N
.
C=
cos θ
For the foot,
FIG. P5.60(a)
∑ Fy = ma y :
FIG. P5.60(b)
+n B − C cos θ = 0
n B = 357 N .
(d)
For the foot with motion impending,
∑ Fx = ma x :
+ f s − C sin θ s = 0
µ sn B = C sin θ s
357 N cos θ s sin θ s
C sin θ s
µs =
=
= tan θ s .
nB
357 N
b
(e)
The maximum coefficient is
µ s = tan θ s = tan 50.2° = 1.20 .
g
147
148
P5.61
The Laws of Motion
∑ F = ma
For m1 :
For m 2 :
T = m1 a
T − m2 g = 0
Eliminating T,
a=
m2 g
m1
For all 3 blocks:
FIG. P5.61
a
f
F = M + m1 + m 2 a =
af
P5.62
e j
1
2
2
1
a f
t2 s2
ts
aM + m + m fFGH mm g IJK
xm
0
0
0
1.02
1.04 0
0.100
1.53
2.34 1
0.200
2.01
4.04 0
0.350
2.64
6.97 0
0.500
3.30
10.89
0.750
3.75
14.06
1.00
FIG. P5.62
From x =
1 2
1
at the slope of a graph of x versus t 2 is a , and
2
2
e
j
a = 2 × slope = 2 0.071 4 m s 2 = 0.143 m s 2 .
From a ′ = g sin θ ,
a ′ = 9.80 m s 2
FG 1.77 4 IJ = 0.137 m s
H 127.1 K
2
, different by 4%.
The difference is accounted for by the uncertainty in the data, which we may estimate from the third
point as
b
ga f = 18%.
0.350 − 0.071 4 4.04
0.350
Chapter 5
P5.63
a
f
(1)
m1 a − A = T ⇒ a =
T
+A
m1
(2)
MA = R x = T ⇒ A =
T
M
(3)
m2 a = m2 g − T ⇒ T = m2 g − a
(a)
b
g
FIG. P5.63
Substitute the value for a from (1) into (3) and solve for T:
LM F T + AI OP .
N GH m JK Q
T = m2 g −
1
Substitute for A from (2):
LM F T + T I OP =
N GH m M JK Q
T = m2 g −
(b)
1
From (2), A =
1
1
2
a− A =
a
m 2 g m1 + M
a
f
m1 M + m 2 M + m1
f
.
T
, Substitute the value of T:
M
A=
(d)
LM
m M
N m M + m am
Solve (3) for a and substitute value of T:
a=
(c)
m2 g
Mm 2 g
m1 M + m 2 m1 + M
a
f
m1 m 2 g
m1 M + m 2 m1 + M
a
f
.
1
OP
+ Mf Q
.
149
150
P5.64
The Laws of Motion
(a), (b) Motion impending
n = 49.0 N
n = 49.0 N
f s1
P
5.00 kg
15.0 kg
f s1
f s2
Fg = 49.0 N
147 N
196 N
a
fs1 = µn = 14.7 N
f
fs2 = 0.500 196 N = 98.0 N
FIG. P5.64
P = f s1 + f s 2 = 14.7 N + 98.0 N = 113 N
(c)
Once motion starts, kinetic friction acts.
a
f
a
f b
g
112.7 N − 0.100 49.0 N − 0.400 196 N = 15.0 kg a 2
a 2 = 1.96 m s 2
a
f b
g
0.100 49.0 N = 5.00 kg a1
a1 = 0.980 m s 2
*P5.65
(a)
Let x represent the position of the glider along the air track. Then z 2 = x 2 + h02 ,
12
−1 2
dx 1 2
dz
dz
is the rate at which string passes
= z − h02
2z
x = z 2 − h02
, vx =
. Now
dt 2
dt
dt
over the pulley, so it is equal to v y of the counterweight.
e
j
e
j a f
c
v x = z z 2 − h02
(b)
(c)
ax =
h
−1 2
v y = uv y
dv y
dv x
du
d
at release from rest, v y = 0 and a x = ua y .
= uv y = u
+ vy
dt
dt
dt
dt
80.0 cm
, z = 1.60 m , u = z 2 − h02
z
For the counterweight
sin 30.0° =
e
∑ Fy = ma y :
j
−1 2
e
z = 1.6 2 − 0.8 2
j a1.6f = 1.15 .
−1 2
T − 0.5 kg 9.8 m s 2 = −0.5 kga y
a y = −2T + 9.8
For the glider
∑ Fx = ma x :
a
f
T cos 30° = 1.00 kg a x = 1.15 a y = 1.15 −2T + 9.8 = −2.31T + 11.3 N
3.18T = 11.3 N
T = 3.56 N
Chapter 5
*P5.66
The upward acceleration of the rod is described by
y f = yi + v yi t +
1
ayt 2
2
1
a y 8 × 10 −3 s
2
a y = 31.2 m s 2
e
1 × 10 −3 m = 0 + 0 +
j
2
The distance y moved by the rod and the distance x
moved by the wedge in the same time are related
y
y
. Then their speeds and
by tan 15° = ⇒ x =
x
tan 15°
accelerations are related by
FIG. P5.66
dy
dx
1
=
dt tan 15° dt
and
d2x
dt
2
=
FG
H
IJ
K
d2y
1
1
=
31.2 m s 2 = 117 m s 2 .
2
tan 15° dt
tan 15°
The free body diagram for the rod is shown. Here H and H ′ are forces exerted by the guide.
∑ Fy = ma y :
n cos 15°− mg = ma y
e
j
e
n cos 15°−0.250 kg 9.8 m s 2 = 0.250 kg 31.2 m s 2
10.3 N
= 10.6 N
n=
cos 15°
j
For the wedge,
∑ Fx = Ma x :
e
−n sin 15°+ F = 0.5 kg 117 m s 2
a
f
j
F = 10.6 N sin 15°+58.3 N = 61.1 N
*P5.67
(a)
Consider forces on the midpoint of the rope. It is nearly in
equilibrium just before the car begins to move. Take the y-axis
in the direction of the force you exert:
∑ Fy = ma y :
−T sin θ + f − T sin θ = 0
T=
(b)
T=
100 N
= 410 N
2 sin 7°
f
.
2 sin θ
FIG. P5.67
151
152
P5.68
The Laws of Motion
Since it has a larger mass, we expect the 8.00-kg block to move
down the plane. The acceleration for both blocks should have the
same magnitude since they are joined together by a non-stretching
string. Define up the left hand plane as positive for the 3.50-kg
object and down the right hand plane as positive for the 8.00-kg
object.
∑ F1 = m1 a1 :
∑ F2 = m 2 a 2 :
− m1 g sin 35.0°+T = m1 a
FIG. P5.68
m 2 g sin 35.0°−T = m 2 a
and
a fa f
a8.00fa9.80f sin 35.0°−T = 8.00a .
− 3.50 9.80 sin 35.0°+T = 3.50 a
Adding, we obtain
a
f
+45.0 N − 19.7 N = 11.5 kg a .
(b)
Thus the acceleration is
a = 2.20 m s 2 .
By substitution,
a
fc
h
−19.7 N + T = 3.50 kg 2.20 m s 2 = 7.70 N .
(a)
The tension is
T = 27.4 N .
P5.69
a = 5.00 m s 2
Choose the x-axis pointing down the slope.
a
v f = vi + at: 30.0 m s = 0 + a 6.00 s
2
f
a = 5.00 m s .
Consider forces on the toy.
∑ Fx = ma x :
e
mg sin θ = m 5.00 m s 2
j
θ = 30.7°
∑ Fy = ma y : − mg cos θ + T = 0
T = mg cos θ = 0.100 9.80 cos 30.7°
T = 0.843 N
a
fa f
FIG. P5.69
Chapter 5
*P5.70
153
Throughout its up and down motion after release the block has
∑ Fy = ma y :
+n − mg cos θ = 0
n = mg cos θ .
Let R = R x i + R y j represent the force of table on incline. We have
∑ Fx = ma x :
+ R x − n sin θ = 0
R x = mg cos θ sin θ
∑ Fy = ma y : − Mg − n cos θ + Ry = 0
R y = Mg + mg cos 2 θ .
e
j
R = mg cos θ sin θ to the right + M + m cos 2 θ g upward
*P5.71
FIG. P5.70
Take +x in the direction of motion of the tablecloth. For the mug:
∑ Fx = ma x
0.1 N = 0.2 kg a x
a x = 0.5 m s 2 .
Relative to the tablecloth, the acceleration of the mug is 0.5 m s 2 − 3 m s 2 = −2.5 m s 2 . The mug
reaches the edge of the tablecloth after time given by
∆ x = v xi t +
1
axt 2
2
1
−2.5 m s 2 t 2
2
t = 0.490 s .
e
−0.3 m = 0 +
j
The motion of the mug relative to tabletop is over distance
ja
1
1
a x t 2 = 0.5 m s 2 0.490 s
2
2
e
The tablecloth slides 36 cm over the table in this process.
f
2
= 0.060 0 m .
154
P5.72
The Laws of Motion
∑ Fy = ma y : n − mg cos θ = 0
or
a f
n = a82.3 N f cos θ
n = 8.40 9.80 cos θ
∑ Fx = ma x : mg sin θ = ma
or
a = g sin θ
e
j
a = 9.80 m s 2 sin θ
θ , deg n , N
0.00
82.3
a, m s 2
0.00
5.00
82.0
0.854
10.0
81.1
1.70
15.0
79.5
2.54
20.0
77.4
3.35
25.0
74.6
4.14
30.0
71.3
4.90
35.0
67.4
5.62
40.0
63.1
6.30
45.0
58.2
6.93
50.0
52.9
7.51
55.0
47.2
8.03
60.0
41.2
8.49
65.0
34.8
8.88
70.0
28.2
9.21
75.0
21.3
9.47
80.0
14.3
9.65
85.0
7.17
9.76
90.0
0.00
9.80
FIG. P5.72
At 0°, the normal force is the full weight and the acceleration is zero. At 90°, the mass is in free fall
next to the vertical incline.
Chapter 5
P5.73
(a)
Apply Newton’s second law to two points
where butterflies are attached on either half
of mobile (other half the same, by symmetry)
(1)
(2)
(3)
(4)
T2 cos θ 2 − T1 cos θ 1 = 0
T1 sin θ 1 − T2 sin θ 2 − mg = 0
T2 cos θ 2 − T3 = 0
T2 sin θ 2 − mg = 0
Substituting (4) into (2) for T2 sin θ 2 ,
T1 sin θ 1 − mg − mg = 0 .
FIG. P5.69
Then
T1 =
2mg
.
sin θ 1
Substitute (3) into (1) for T2 cos θ 2 :
T3 − T1 cos θ 1 = 0 , T3 = T1 cos θ 1
Substitute value of T1 :
T3 = 2mg
2mg
cos θ 1
=
= T3 .
sin θ 1
tan θ 1
From Equation (4),
T2 =
(b)
mg
.
sin θ 2
Divide (4) by (3):
mg
T2 sin θ 2
.
=
T2 cos θ 2
T3
Substitute value of T3 :
tan θ 2 =
FG
H
mg tan θ 1
tan θ 1
, θ 2 = tan−1
2
2mg
IJ
K
.
Then we can finish answering part (a):
T2 =
(c)
mg
sin
b
tan−1 12
tan θ 1
g
.
D is the horizontal distance between the points at which the two ends of the string are
attached to the ceiling.
D = 2A cos θ 1 + 2A cos θ 2 + A and L = 5A
D=
RS
T
LM
N
FG
H
1
L
2 cos θ 1 + 2 cos tan−1 tan θ 1
2
5
IJ OP + 1UV
KQ W
155
156
The Laws of Motion
ANSWERS TO EVEN PROBLEMS
P5.2
1.66 × 10 6 N forward
P5.4
P5.42
152 ft
Fg v vt
(a) ; (b)
i + Fg j
gt
2
P5.44
(a) 2.31 m s 2 down for m1 , left for m 2 and
up for m 3 ; (b) 30.0 N and 24.2 N
(a) 4.47 × 10 15 m s 2 away from the wall;
(b) 2.09 × 10 −10 N toward the wall
P5.46
Any value between 31.7 N and 48.6 N
P5.48
72.0 N
P5.8
(a) 534 N down; (b) 54.5 kg
P5.50
6.84 m
P5.10
2.55 N for an 88.7 kg person
P5.52
(a) 3.00 s; (b) 20.1 m; (c) 18.0 i − 9.00 j m
P5.12
e16.3i + 14.6jj N
P5.54
(a) 2.00 m s 2 to the right;
(b) 8.00 N right on 4 kg;
6.00 N right on 3 kg; 4 N right on 2 kg;
(c) 8.00 N between 4 kg and 3 kg;
14.0 N between 2 kg and 3 kg;
(d) see the solution
P5.6
F I
GH JK
e
j
P5.14
(a) 181°; (b) 11.2 kg; (c) 37.5 m s ;
(d) −37.5 i − 0.893 j m s
P5.16
112 N
P5.18
T1 = 296 N ; T2 = 163 N ; T3 = 325 N
P5.56
1.18 kN
P5.20
(a) see the solution; (b) 1.79 N
P5.58
P5.22
(a) 2.54 m s 2 down the incline;
(b) 3.18 m s
(a) 4.90 m s 2 ; (b) 3.13 m s at 30.0° below
the horizontal; (c) 1.35 m; (d) 1.14 s; (e) No
P5.60
(a) and (b) see the solution; (c) 357 N;
(d) see the solution; (e) 1.20
P5.24
see the solution; 6.30 m s 2 ; 31.5 N
P5.62
see the solution; 0.143 m s 2 agrees with
P5.26
(a) 3.57 m s 2 ; (b) 26.7 N; (c) 7.14 m s
P5.28
(a) 36.8 N; (b) 2.45 m s 2 ; (c) 1.23 m
P5.30
(a) 0.529 m; (b) 7.40 m s upward
P5.32
(a) 2.22 m; (b) 8.74 m s
P5.34
(a) a1 = 2 a 2 ;
m1 m 2 g
m m g
; T2 = 1 2m ;
(b) T1 =
m2
2m1 + 2
m1 + 42
m2 g
m2 g
; a2 =
(c) a1 =
m2
4
m
2m1 + 2
1 + m2
e
j
P5.36
µ s = 0.306 ; µ k = 0.245
P5.38
(a) 3.34; (b) Time would increase
P5.40
(a) 55.2°; (b) 167 N
0.137 m s 2
P5.64
(a) see the solution;
(b) on block one:
49.0 N j − 49.0 N j + 14.7 N i ;
on block two: −49.0 N j − 14.7 N i − 147 N j
+196 N j − 98.0 N i + 113 N i ;
(c) for block one: 0.980 i m s 2 ;
for block two: 1.96 m s 2 i
P5.66
61.1 N
P5.68
(a) 2.20 m s 2 ; (b) 27.4 N
P5.70
mg cos θ sin θ to the right
e
j
+ M + m cos 2 θ g upward
P5.72
see the solution
6
Circular Motion and Other
Applications of Newton’s Laws
CHAPTER OUTLINE
6.1
6.2
6.3
6.4
6.5
Q6.4
Newton’s Second Law
Applied to Uniform Circular
Motion
Nonuniform Circular Motion
Motion in Accelerated
Frames
Motion in the Presence of
Resistive Forces
Numerical Modeling in
Particle Dynamics
ANSWERS TO QUESTIONS
Q6.1
Mud flies off a rapidly spinning tire because the resultant force
is not sufficient to keep it moving in a circular path. In this case,
the force that plays a major role is the adhesion between the
mud and the tire.
Q6.2
The spring will stretch. In order for the object to move in a
circle, the force exerted on the object by the spring must have a
mv 2
. Newton’s third law says that the force exerted on
size of
r
the object by the spring has the same size as the force exerted
by the object on the spring. It is the force exerted on the spring
that causes the spring to stretch.
Q6.3
Driving in a circle at a constant speed requires a centripetal
acceleration but no tangential acceleration.
(a)
The object will move in a circle at a constant speed.
(b)
The object will move in a straight line at a changing speed.
Q6.5
The speed changes. The tangential force component causes tangential acceleration.
Q6.6
Consider the force required to keep a rock in the Earth’s crust moving in a circle. The size of the
force is proportional to the radius of the circle. If that rock is at the Equator, the radius of the circle
through which it moves is about 6400 km. If the rock is at the north pole, the radius of the circle
through which it moves is zero!
Q6.7
Consider standing on a bathroom scale. The resultant force on you is your actual weight minus the
normal force. The scale reading shows the size of the normal force, and is your ‘apparent weight.’ If
you are at the North or South Pole, it can be precisely equal to your actual weight. If you are at the
equator, your apparent weight must be less, so that the resultant force on you can be a downward
force large enough to cause your centripetal acceleration as the Earth rotates.
Q6.8
A torque is exerted by the thrust force of the water times the distance between the nozzles.
157
158
Circular Motion and Other Applications of Newton’s Laws
Q6.9
I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one
would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in
a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the
space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface.
Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and
is in free fall.
Q6.10
This is the same principle as the centrifuge. All the material inside the cylinder tends to move along
a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving
around in a circular path.
Q6.11
The ball would not behave as it would when dropped on the Earth. As the astronaut holds the ball,
she and the ball are moving with the same angular velocity. The ball, however, being closer to the
center of rotation, is moving with a slower tangential velocity. Once the ball is released, it acts
according to Newton’s first law, and simply drifts with constant velocity in the original direction of
its velocity when released—it is no longer “attached” to the rotating space station. Since the ball
follows a straight line and the astronaut follows a circular path, it will appear to the astronaut that
the ball will “fall to the floor”. But other dramatic effects will occur. Imagine that the ball is held so
high that it is just slightly away from the center of rotation. Then, as the ball is released, it will move
very slowly along a straight line. Thus, the astronaut may make several full rotations around the
circular path before the ball strikes the floor. This will result in three obvious variations with the
Earth drop. First, the time to fall will be much larger than that on the Earth, even though the feet of
the astronaut are pressed into the floor with a force that suggests the same force of gravity as on
Earth. Second, the ball may actually appear to bob up and down if several rotations are made while
it “falls”. As the ball moves in a straight line while the astronaut rotates, sometimes she is on the side
of the circle on which the ball is moving toward her and other times she is on the other side, where
the ball is moving away from her. The third effect is that the ball will not drop straight down to her
feet. In the extreme case we have been imagining, it may actually strike the surface while she is on
the opposite side, so it looks like it ended up “falling up”. In the least extreme case, in which only a
portion of a rotation is made before the ball strikes the surface, the ball will appear to move
backward relative to the astronaut as it falls.
Q6.12
The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and
around in a circle.
Q6.13
There is no such force. If the passenger slides outward across the slippery car seat, it is because the
passenger is moving forward in a straight line while the car is turning under him. If the passenger
pushes hard against the outside door, the door is exerting an inward force on him. No object is
exerting an outward force on him, but he should still buckle his seatbelt.
Q6.14
Blood pressure cannot supply the force necessary both to balance the gravitational force and to
provide the centripetal acceleration, to keep blood flowing up to the pilot’s brain.
Q6.15
The person in the elevator is in an accelerating reference frame. The apparent acceleration due to
gravity, “g,” is changed inside the elevator. “g”= g ± a
Q6.16
When you are not accelerating, the normal force and your weight are equal in size. Your body
interprets the force of the floor pushing up on you as your weight. When you accelerate in an
elevator, this normal force changes so that you accelerate with the elevator. In free fall, you are
never weightless since the Earth’s gravity and your mass do not change. It is the normal force—your
apparent weight—that is zero.
Chapter 6
Q6.17
159
From the proportionality of the drag force to the speed squared and from Newton’s second law, we
derive the equation that describes the motion of the skydiver:
m
dv y
dt
= mg −
Dρ A 2
vy
2
where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist’s
body. At terminal speed,
ay =
dv y
dt
F 2mg I
= 0 and V G
H Dρ A JK
T
12
.
When the parachute opens, the coefficient of drag D and the effective area A both increase, thus
reducing the speed of the skydiver.
Modern parachutes also add a third term, lift, to change the equation to
m
dv y
dt
= mg −
Dρ A 2 Lρ A 2
vy −
vx
2
2
where v y is the vertical velocity, and v x is the horizontal velocity. The effect of lift is clearly seen in
the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute.
Q6.18
The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that
terminal speed is proportional to the square root of radius. When moving with terminal speed, an
object is in equilibrium and has zero acceleration.
Q6.19
Lower air density reduces air resistance, so a tank-truck-load of fuel takes you farther.
Q6.20
Suppose the rock is moving rapidly when it enters the water. The speed of the rock decreases until it
reaches terminal velocity. The acceleration, which is upward, decreases to zero as the rock
approaches terminal velocity.
Q6.21
The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot be
predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our
sense of free will, of being able to make choices for ourselves that can appear to be random, may be
an illusion. It may have nothing to do with the subatomic randomness described by quantum
mechanics.
160
Circular Motion and Other Applications of Newton’s Laws
SOLUTIONS TO PROBLEMS
Section 6.1
P6.1
Newton’s Second Law Applied to Uniform Circular Motion
m = 3.00 kg , r = 0.800 m. The string will break if the tension exceeds
the weight corresponding to 25.0 kg, so
a f
Tmax = Mg = 25.0 9.80 = 245 N .
When the 3.00 kg mass rotates in a horizontal circle, the tension
causes the centripetal acceleration,
T=
Then
v2 =
and
0 ≤ v ≤ 65.3
or
P6.2
a f
3.00 v 2
mv 2
=
.
r
0.800
so
a
f a
a f
f
0.800 T
0.800 Tmax 0.800 245
rT
=
≤
=
= 65.3 m 2 s 2
m
3.00
3.00
3.00
FIG. P6.1
0 ≤ v ≤ 8.08 m s .
v2
, both m and r are unknown but remain constant. Therefore, ∑ F is proportional to v 2
r
2
18.0
and increases by a factor of
as v increases from 14.0 m/s to 18.0 m/s. The total force at the
14.0
higher speed is then
In
∑F = m
FG IJ
H K
F 18.0 I a130 Nf = 215 N .
∑ Ffast = GH 14.0 JK
Symbolically, write
Dividing gives
F mI
∑ Fslow = GH r JK b14.0
∑ Ffast
∑ Fslow
=
FG 18.0 IJ
H 14.0 K
ms
g
2
2
and
F mI
∑ Ffast = GH r JK b18.0
g
2
ms .
2
, or
F 18.0 I
2
F 18.0 I a130 Nf =
∑ Ffast = GH 14.0 JK ∑ Fslow = GH 14.0 JK
2
215 N .
This force must be horizontally inward to produce the driver’s centripetal acceleration.
Chapter 6
P6.3
P6.4
e
je
2
(a)
(b)
2.20 × 10 6 m s
v2
=
= 9.13 × 10 22 m s 2 inward
a=
r
0.530 × 10 −10 m
e
j
Neglecting relativistic effects. F = ma c =
(b)
mv 2
r
e2.998 × 10 m sj
kg j
a0.480 mf
7
e
(a)
= 8.32 × 10 −8 N inward
2
F = 2 × 1.661 × 10
P6.5
j
9.11 × 0 −31 kg 2.20 × 10 6 m s
mv 2
=
F=
r
0.530 × 10 −10 m
−27
2
= 6.22 × 10 −12 N
static friction
e j
ma i = f i + nj + mg − j
∑ Fy = 0 = n − mg
v2
= f = µn = µmg .
r
2
50.0 cm s
v2
=
= 0.085 0 .
Then µ =
rg
30.0 cm 980 cm s 2
thus n = mg and
∑ Fr = m
b
a
P6.6
(a)
P6.7
fe
∑ Fy = ma y , mg moon down =
v = g moon r =
(b)
g
mv 2
down
r
e1.52 m s je1.7 × 10
2
e
j
6
j
m + 100 × 10 3 m = 1.65 × 10 3 m s
j
2π 1.8 × 10 6 m
2πr
v=
,T=
= 6.84 × 10 3 s = 1.90 h
T
1.65 × 10 3 m s
n = mg since a y = 0
The force causing the centripetal acceleration is the frictional force f.
From Newton’s second law f = ma c =
mv 2
.
r
But the friction condition is f ≤ µ sn
i.e.,
FIG. P6.7
mv 2
≤ µ s mg
r
a
fe
v ≤ µ s rg = 0.600 35.0 m 9.80 m s 2
j
v ≤ 14.3 m s
161
162
Circular Motion and Other Applications of Newton’s Laws
2
v
=
r
b86.5 km hge
1h
3 600 s
je
1 000 m
1 km
2
j F 1g I=
GH 9.80 m s JK
P6.8
a=
P6.9
T cos 5.00° = mg = 80.0 kg 9.80 m s 2
2
61.0 m
b
ge
0.966 g
j
a68.6 Nfi + a784 Nfj
(a)
T = 787 N : T =
(b)
T sin 5.00° = ma c : a c = 0.857 m s 2 toward the center of
the circle.
The length of the wire is unnecessary information. We
could, on the other hand, use it to find the radius of the
circle, the speed of the bob, and the period of the motion.
P6.10
(b)
v=
235 m
= 6.53 m s
36.0 s
The radius is given by
1
2πr = 235 m
4
r = 150 m
(a)
F v I toward center
GH r JK
b6.53 m sg at 35.0° north of west
=
2
ar =
2
150 m
e
je
e j
j
= 0.285 m s 2 cos 35.0° − i + sin 35.0° j
= −0.233 m s 2 i + 0.163 m s 2 j
(c)
a=
=
dv
f
− vi
i
t
e6.53 m s j − 6.53 m s ij
36.0 s
= −0.181 m s 2 i + 0.181 m s 2 j
FIG. P6.9
Chapter 6
*P6.11
b ge
j
Fg = mg = 4 kg 9.8 m s 2 = 39.2 N
Ta
1.5 m
2m
θ = 48.6°
r = 2 m cos 48.6° = 1.32 m
θ
sin θ =
39.2 N
a f
mv 2
∑ Fx = ma x = r
Ta cos 48.6°+Tb cos 48.6° =
Ta + Tb =
163
Tb
b gb
4 kg 6 m s
109 N
= 165 N
cos 48.6°
g
forces
ac
v
2
1.32 m
motion
FIG. P6.11
∑ Fy = ma y
+Ta sin 48.6°−Tb sin 48.6°−39.2 N = 0
Ta − Tb =
(a)
39. 2 N
= 52.3 N
sin 48.6°
To solve simultaneously, we add the equations in Ta and Tb :
Ta + Tb + Ta − Tb = 165 N + 52.3 N
Ta =
(b)
*P6.12
217 N
= 108 N
2
Tb = 165 N − Ta = 165 N − 108 N = 56.2 N
v2
. Let f represent the rotation rate. Each revolution carries each bit of metal through distance
r
2πr , so v = 2πrf and
ac =
ac =
v2
= 4π 2 rf 2 = 100 g .
r
A smaller radius implies smaller acceleration. To meet the criterion for each bit of metal we consider
the minimum radius:
F 100 g IJ
f =G
H 4π r K
2
12
F 100 ⋅ 9.8 m s I
=G
H 4π a0.021 mf JK
2
2
12
= 34.4
FG
H
IJ
K
1 60 s
= 2.06 × 10 3 rev min .
s 1 min
164
Circular Motion and Other Applications of Newton’s Laws
Section 6.2
P6.13
Nonuniform Circular Motion
M = 40.0 kg , R = 3.00 m, T = 350 N
(a)
T
Mv 2
R
R
M
∑ F = 2T − Mg =
gFGH IJK
F 3.00 IJ = 23.1 em s j
= 700 − a 40.0fa9.80f G
H 40.0 K
b
v 2 = 2T − Mg
v2
2
2
n − Mg = F =
n = Mg +
P6.14
(a)
Mv 2
R
child + seat
child alone
FIG. P6.13(a)
FIG. P6.13(b)
IJ
K
FG
H
Mv 2
23.1
= 40.0 9.80 +
= 700 N
R
3.00
Consider the forces acting on the system consisting of the child and the seat:
∑ Fy = ma y ⇒ 2T − mg = m
v2 = R
v=
(b)
FG 2T − gIJ
Hm K
F 2T − gIJ
RG
Hm K
v2
R
Consider the forces acting on the child alone:
F
∑ Fy = ma y ⇒ n = mGH g +
and from above, v 2 = R
v2
R
I
JK
FG 2T − gIJ , so
Hm K
FG
H
n=m g+
P6.15
n
Mg
v = 4.81 m s
(b)
Mg
T
IJ
K
2T
− g = 2T .
m
Let the tension at the lowest point be T.
∑ F = ma:
T − mg = ma c =
F
GH
mv 2
r
I
JK
L
b8.00 m sg
T = b85.0 kg gM9.80 m s +
10.0 m
MN
T=m g+
v2
r
2
2
OP
PQ = 1.38 kN > 1 000 N
He doesn’t make it across the river because the vine breaks.
FIG. P6.15
Chapter 6
P6.16
b
(a)
4.00 m s
v2
=
ac =
r
12.0 m
(b)
a = a c2 + a t2
a=
a1.33f + a1.20f
2
g
2
= 1.33 m s 2
2
at an angle θ = tan −1
= 1.79 m s 2
FG a IJ =
Ha K
c
FIG. P6.16
48.0° inward
t
P6.17
∑ Fy =
mv 2
= mg + n
r
But n = 0 at this minimum speed condition, so
mv 2
= mg ⇒ v = gr =
r
P6.18
e9.80 m s ja1.00 mf =
2
3.13 m s .
At the top of the vertical circle,
T=m
a
(a)
f a40..00500f − a0.400fa9.80f =
v2
− mg
R
2
or T = 0.400
P6.19
FIG. P6.17
8.88 N
v = 20.0 m s,
B
C
n = force of track on roller coaster, and
15 m
10 m
R = 10.0 m .
A
Mv 2
∑ F = R = n − Mg
FIG. P6.19
From this we find
n = Mg +
b
ge
500 kg 20.0 m s 2
Mv 2
= 500 kg 9.80 m s 2 +
R
10.0 m
b
ge
j
n = 4 900 N + 20 000 N = 2.49 × 10 4 N
(b)
At B, n − Mg = −
Mv 2
R
The max speed at B corresponds to
n=0
− Mg = −
2
Mv max
⇒ v max = Rg = 15.0 9.80 = 12.1 m s
R
a f
j
165
166
P6.20
Circular Motion and Other Applications of Newton’s Laws
b
e
g
2
(a)
v2
ac =
r
(b)
Let n be the force exerted by the rail.
13.0 m s
v2
=
= 8.62 m
r=
a c 2 9.80 m s 2
j
Newton’s law gives
Mv 2
Mg + n =
r
v2
n=M
− g = M 2 g − g = Mg , downward
r
I
JK
F
GH
(c)
ac =
v2
r
ac =
b13.0 m sg
b
FIG. P6.20
g
2
20.0 m
= 8.45 m s 2
If the force exerted by the rail is n1
then
Mv 2
= Ma c
r
n1 = M a c − g which is < 0 since a c = 8.45 m s 2
n1 + Mg =
b
g
Thus, the normal force would have to point away from the center of the curve. Unless they
have belts, the riders will fall from the cars. To be safe we must require n1 to be positive.
Then a c > g . We need
v2
> g or v > rg =
r
Section 6.3
P6.21
(a)
a20.0 mfe9.80 m s j , v > 14.0 m s .
2
Motion in Accelerated Frames
T
18.0 N
∑ Fx = Ma , a = M = 5.00 kg =
3.60 m s 2
to the right.
(b)
If v = const, a = 0, so T = 0 (This is also
an equilibrium situation.)
(c)
Someone in the car (noninertial observer)
claims that the forces on the mass along x
are T and a fictitious force (–Ma). Someone
at rest outside the car (inertial observer)
claims that T is the only force on M in the
x-direction.
5.00 kg
FIG. P6.21
167
Chapter 6
*P6.22
We adopt the view of an inertial observer. If it is on the verge of sliding, the
cup is moving on a circle with its centripetal acceleration caused by friction.
∑ Fy = ma y :
+n − mg = 0
∑ Fx = ma x :
f=
n
f
mv 2
= µ sn = µ s mg
r
ja
e
mg
FIG. P6.22
f
v = µ s gr = 0.8 9.8 m s 2 30 m = 15.3 m s
If you go too fast the cup will begin sliding straight across the dashboard to the left.
P6.23
The only forces acting on the suspended object are the force of gravity mg
and the force of tension T, as shown in the free-body diagram. Applying
Newton’s second law in the x and y directions,
or
(a)
∑ Fx = T sin θ = ma
∑ Fy = T cos θ − mg = 0
(1)
T cos θ = mg
(2)
T cos θ
T sin θ
mg
FIG. P6.23
Dividing equation (1) by (2) gives
tan θ =
a 3.00 m s 2
=
= 0.306 .
g 9.80 m s 2
Solving for θ, θ = 17.0°
(b)
From Equation (1),
T=
*P6.24
a
fc
a f
h
0.500 kg 3.00 m s 2
ma
=
= 5.12 N .
sin θ
sin 17.0°
The water moves at speed
v=
a
f
2πr 2π 0.12 m
=
= 0.104 m s .
T
7.25 s
The top layer of water feels a downward force of gravity mg and an outward fictitious force in the
turntable frame of reference,
b
mv 2 m 0.104 m s
=
r
0.12 m
g
2
= m9.01 × 10 −2 m s 2 .
It behaves as if it were stationary in a gravity field pointing downward and outward at
tan −1
0.090 1 m s 2
9.8 m s 2
= 0.527° .
Its surface slopes upward toward the outside, making this angle with the horizontal.
168
P6.25
Circular Motion and Other Applications of Newton’s Laws
Fmax = Fg + ma = 591 N
Fmin = Fg − ma = 391 N
(a)
Adding, 2 Fg = 982 N, Fg = 491 N
(b)
Since Fg = mg , m =
(c)
Subtracting the above equations,
491 N
= 50.1 kg
9.80 m s 2
∴ a = 2.00 m s 2
2ma = 200 N
P6.26
(a)
∑ Fr = mar
mg =
g=
T=
(b)
*P6.27
FG
H
mv 2 m 2πR
=
R
R T
IJ
K
2
4π 2 R
T2
4π 2 R
6.37 × 10 6 m
= 2π
= 5.07 × 10 3 s = 1.41 h
g
9.80 m s 2
speed increase factor =
FG
H
IJ
K
v new
T
2πR Tcurrent
24.0 h
=
= current =
= 17.1
2πR
1.41 h
v current Tnew
Tnew
The car moves to the right with acceleration a. We find the acceleration of a b of the block relative to
the Earth. The block moves to the right also.
∑ Fy = ma y :
∑ Fx = ma x :
+n − mg = 0 , n = mg , f = µ k mg
+ µ k mg = ma b , a b = µ k g
The acceleration of the block relative to the car is a b − a = µ k g − a . In this frame the block starts from
rest and undergoes displacement −A and gains speed according to
d
2
+ 2 a x x f − xi
v xf2 = v xi
v xf2
(a)
d b
v = 2A a − µ k g
continued on next page
gi
12
b
ga
i
f b
g
= 0 + 2 µ k g − a − A − 0 = 2A a − µ k g .
to the left
Chapter 6
(b)
169
The time for which the box slides is given by
1
v xi + v xf t
2
1
−A =
0 − 2A a − µ k g
2
d
i
LM d b
gi OPQt
N
F 2A I .
t=G
H a − µ g JK
∆x =
12
12
k
The car in the Earth frame acquires finals speed v xf = v xi
F 2A I
+ at = 0 + aG
H a − µ g JK
12
. The speed
k
of the box in the Earth frame is then
b
g
− a 2 Af b a − µ g g + a 2 A f
=
ba − µ g g
v be = v bc + v ce = − 2A a − µ k g
12
12
12
k
+a
a
12
*P6.28
b
µ k g 2A
2A a − µ k g
g
12
=
12
k
12
k
2 µ k gA
.
v
Consider forces on the backpack as it slides in the Earth frame of reference.
∑ Fy = ma y :
∑ Fx = ma x :
We solve for µ k : vt − L =
b
g
b
+n − mg = ma , n = m g + a , f k = µ k m g + a
b
g
− µ k m g + a = ma x
The motion across the floor is described by L = vt +
P6.29
=
12
k
k
=
F 2A I
GH a − µ g JK
µ g a 2 Af
ba − µ g g
a
f
b
g
g
1
1
a x t 2 = vt − µ k g + a t 2 .
2
2
2 vt − L
1
= µk .
µk g + a t2 ,
2
g + a t2
b
g
b
g
In an inertial reference frame, the girl is accelerating horizontally inward at
b
5.70 m s
v2
=
r
2.40 m
g
2
= 13.5 m s 2
In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass
times this acceleration. Together this force and the weight of her head add to have a magnitude
equal to the mass of her head times an acceleration of
g2 +
Fv I
GH r JK
2
2
=
a9.80f + a13.5f
2
2
m s 2 = 16.7 m s 2
16.7
= 1.71 .
9.80
Thus, the force required to lift her head is larger by this factor, or the required force is
This is larger than g by a factor of
a
f
F = 1.71 55.0 N = 93.8 N .
170
*P6.30
Circular Motion and Other Applications of Newton’s Laws
(a)
The chunk is at radius r =
0.137 m + 0.080 m
= 0.054 2 m . Its speed is
4
v=
20 000
2πr
= 2π 0.054 2 m
= 114 m s
60 s
T
b
g
and its acceleration
ac =
b
g
2
114 m s
v2
=
= 2.38 × 10 5 m s 2 horizontally inward
r
0.054 2 m
= 2.38 × 10 5 m s 2
(b)
F g I=
GH 9.8 m s JK
2
2.43 × 10 4 g .
In the frame of the turning cone, the chunk feels a
mv 2
. In this frame its
horizontally outward force of
r
3.3 cm
acceleration is up along the cone, at tan −1 a13 .7 − 8 f cm = 49.2° .
e
+n −
je
mv 2
r
f
49.2°
2
Take the y axis perpendicular to the cone:
∑ Fy = ma y :
a
n
FIG. P6.30(b)
mv 2
sin 49.2° = 0
r
j
n = 2 × 10 −3 kg 2.38 × 10 5 m s 2 sin 49.2° = 360 N
(c)
a
f
f = µ k n = 0.6 360 N = 216 N
∑ Fx = ma x :
e2 × 10
−3
je
2
mv
cos 49. 2°− f = ma x
r
j
e
j
kg 2.38 × 10 5 m s 2 cos 49.2°−216 N = 2 × 10 −3 kg a x
a x = 47.5 × 10 4 m s 2 radially up the wall of the cone
P6.31
F 4π R I cos 35.0° = 0.027 6 m s
GH T JK
We take the y axis along the local vertical.
ba g = 9.80 − ba g = 9.78 m s
ba g = 0.015 8 m s
ar =
2
2
net y
net x
θ = arctan
2
e
N
2
r y
2
ax
= 0.092 8°
ay
35.0° a r
(exaggerated size)
θ
g0
a net
35.0°
Equator
FIG. P6.31
Chapter 6
Section 6.4
P6.32
Motion in the Presence of Resistive Forces
m = 80.0 kg , vT = 50.0 m s , mg =
(a)
DρAvT2 DρA mg
∴
= 2 = 0.314 kg m
2
2
vT
v = 30.0 m s
At
a=g−
(b)
171
DρAv 2
2
m
= 9.80 −
a0.314fa30.0f
80.0
2
= 6. 27 m s 2 downward
At v = 50.0 m s , terminal velocity has been reached.
∑ Fy = 0 = mg − R
b
ge
j
⇒ R = mg = 80.0 kg 9.80 m s 2 = 784 N directed up
(c)
v = 30.0 m s
DρAv 2
= 0.314 30.0
2
At
a
P6.33
(a)
fa f
2
= 283 N upward
a = g − bv
When v = vT , a = 0 and g = bvT
b=
g
vT
The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.
P6.34
y 1.50 m
=
= 0.300 m s
5.00 s
t
Thus,
vT =
Then
b=
(b)
At t = 0 , v = 0 and
a = g = 9.80 m s 2 down
(c)
When v = 0.150 m s, a = g − bv = 9.80 m s 2 − 32.7 s −1 0.150 m s = 4.90 m s 2
(a)
ρ=
9.80 m s 2
= 32.7 s −1
0.300 m s
jb
e
g
down
m
1
, A = 0.020 1 m 2 , R = ρ air ADvT2 = mg
V
2
m = ρ beadV = 0.830 g cm 3
LM 4 π a8.00 cmf OP = 1.78 kg
N3
Q
3
Assuming a drag coefficient of D = 0.500 for this spherical object, and taking the density of
air at 20°C from the endpapers, we have
vT =
(b)
v 2f
=
vi2
b
ge
j =
0.500e1.20 kg m je0.020 1 m j
2 1.78 kg 9.80 m s 2
3
b53.8 m sg =
=
+ 2 gh = 0 + 2 gh : h =
2 g 2e9.80 m s j
v 2f
2
2
2
148 m
53.8 m s
172
P6.35
Circular Motion and Other Applications of Newton’s Laws
Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward
applied force equals the sum of the gravitational and drag forces (both downward):
F = mg + bv .
The mass of the copper ball is
FG IJ e
HK
4πρr 3
4
=
π 8.92 × 10 3 kg m3 2.00 × 10 −2 m
3
3
m=
je
j
3
= 0.299 kg .
The applied force is then
a
fa f a
fe
j
F = mg + bv = 0.299 9.80 + 0.950 9.00 × 10 −2 = 3.01 N .
P6.36
∑ Fy = ma y
+T cos 40.0°− mg = 0
b620 kgge9.80 m s j = 7.93 × 10
2
T=
3
cos 40.0°
F
ma
=
∑ x
x
N
− R + T sin 40.0° = 0
e
j
R = 7.93 × 10 3 N sin 40.0° = 5.10 × 10 3 N =
D=
P6.37
(a)
e
2 5.10 × 10 3 N
jFH
kg m s 2
N
IK
1
DρAv 2
2
2R
=
ρAv 2
1. 20 kg m 2 3.80 m 2 40.0 m s
e
je
jb
e
P6.38
= 1. 40
e
je
j
3.00 × 10 −3 kg 9.80 m s 2
mg
=
= 1.47 N ⋅ s m
vT
2.00 × 10 −2 m s
In the equation describing the time variation of the velocity, we have
v = vT 1 − e − bt m
(c)
2
R = vT b = mg
At terminal velocity,
∴b =
(b)
g
FIG. P6.36
j
v = 0.632 vT when e − bt m = 0.368
FG m IJ lna0.368f =
HbK
2.04 × 10 −3 s
or at time
t=−
At terminal velocity,
R = vT b = mg = 2.94 × 10 −2 N
The resistive force is
a
fe
jb
1
1
DρAv 2 = 0.250 1.20 kg m3 2.20 m 2 27.8 m s
2
2
R = 255 N
R=
a=−
255 N
R
=−
= −0.212 m s 2
m
1200 kg
je
g
2
Chapter 6
P6.39
af
a
v t = vi e − ct
(a)
f
v 20.0 s = 5.00 = vi e −20 .0 c , vi = 10.0 m s .
(b)
At t = 40.0 s
FG 1 IJ
H 2K
v = b10.0 m sge
(c)
v = vi e − ct
s=
So 5.00 = 10.0 e −20 .0 c and −20.0 c = ln
P6.40
∑ F = ma
− kmv 2 = m
− kdt =
dv
v2
c=−
−40 .0 c
c h=
ln
1
2
20.0
b
3.47 × 10 −2 s −1
f
ga
= 10.0 m s 0.250 = 2.50 m s
dv
= − cvi e − ct = − cv
dt
dv
dt
z z
t
v
0
v0
− k dt = v −2 dv
a f
−k t − 0 =
v −1
−1
v
=−
v0
1 1
+
v v0
1 + v 0 kt
1 1
=
+ kt =
v v0
v0
v0
v=
1 + v 0 kt
*P6.41
(a)
From Problem 40,
v=
v0
dx
=
dt 1 + v 0 kt
z z
x
t
dx = v 0
0
0
z
t
dt
1 v 0 kdt
=
1 + v 0 kt k 0 1 + v 0 kt
b
b
g
g
t
1
ln 1 + v 0 kt
0
k
1
x − 0 = ln 1 + v 0 kt − ln 1
k
1
x = ln 1 + v 0 kt
k
x
x0 =
b
(b)
*P6.42
b
g
g
We have ln 1 + v 0 kt = kx
v0
v
= kx0 = v 0 e − kx = v
1 + v 0 kt = e kx so v =
1 + v 0 kt e
We write − kmv 2 = −
1
DρAv 2 so
2
k=
e
je
j
−3
3
2
DρA 0.305 1.20 kg m 4.2 × 10 m
=
= 5.3 × 10 −3 m
2m
2 0.145 kg
b
b
g
v = v 0 e − kx = 40.2 m s e
e
− 5.3 ×10
−3
g
ja
m 18 .3 m
f=
36.5 m s
173
174
P6.43
Circular Motion and Other Applications of Newton’s Laws
a
fa
f
1
DρAv 2 , we estimate that D = 1.00 , ρ = 1.20 kg m3 , A = 0.100 m 0.160 m = 1.60 × 10 −2 m 2
2
and v = 27.0 m s. The resistance force is then
In R =
R=
a fe
jb
1
1.00 1.20 kg m3 1.60 × 10 −2 m 2 27.0 m s
2
je
g
2
= 7.00 N
or
R ~ 10 1 N
Section 6.5
Numerical Modeling in Particle Dynamics
a
f af a
f
Note: In some problems we compute each new position as x t + ∆t = x t + v t + ∆t ∆t , rather than
x t + ∆t = x t + v t ∆t as quoted in the text. This method has the same theoretical validity as that presented in
the text, and in practice can give quicker convergence.
a
P6.44
f af af
(a)
(b)
At v = vT , a = 0, − mg + bvT = 0
af
ts
0
0.005
0.01
0.015
vT =
xm
a f
vms
2
2
1.999 755
1.999 3
0
–0.049
–0.095 55
–0.139 77
e
je
j
3.00 × 10 −3 kg 9.80 m s 2
mg
=
= 0.980 m s
b
3.00 × 10 −2 kg s
a f
b g
F mN
–29.4
–27.93
–26.534
–25.2
e
a m s2
j
–9.8
–9.31
–8.844 5
–8.40
. . . we list the result after each tenth iteration
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
1.990
1.965
1.930
1.889
1.845
1.799
1.752
1.704
1.65
1.61
1.56
1.51
1.46
–0.393
–0.629
–0.770
–0.854
–0.904
–0.935
–0.953
–0.964
–0.970
–0.974
–0.977
–0.978
–0.979
–17.6
–10.5
–6.31
–3.78
–2.26
–1.35
–0.811
–0.486
–0.291
–0.174
–0.110
–0.062 4
–0.037 4
–5.87
–3.51
–2.10
–1.26
–0.754
–0.451
–0.270
–0.162
–0.096 9
–0.058 0
–0.034 7
–0.020 8
–0.012 5
Terminal velocity is never reached. The leaf is at 99.9% of vT after 0.67 s. The fall to the
ground takes about 2.14 s. Repeating with ∆t = 0.001 s , we find the fall takes 2.14 s.
Chapter 6
P6.45
(a)
When v = vT , a = 0,
∑ F = − mg + CvT2 = 0
mg
=−
vT = −
C
af
(b)
ts
a f
e4.80 × 10
−4
je
kg 9.80 m s 2
2.50 × 10
−5
kg m
a f
b g
xm
0
0
–0.392
–1.168
–2.30
–3.77
–5.51
–7.48
–9.65
–11.96
–14.4
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2
175
j=
−13.7 m s
e
vms
F mN
a m s2
0
–1.96
–3.88
–5.683 2
–7.306 8
–8.710 7
–9.880 3
–10.823
–11.563
–12.13
–12.56
– 4.704
– 4.608
– 4.327 6
–3.896 5
–3.369 3
–2.807 1
–2.263 5
–1.775 3
–1.361 6
–1.03
–0.762
–9.8
–9.599 9
–9.015 9
–8.117 8
–7.019 3
–5.848 1
–4.715 6
–3.698 6
–2.836 6
–2.14
–1.59
–0.154
–0.029 1
–0.005 42
–0.321
–0.060 6
–0.011 3
j
. . . listing results after each fifth step
–13.49
–13.67
–13.71
–27.4
–41.0
–54.7
3
4
5
The hailstone reaches 99% of vT after 3.3 s, 99.95% of vT after 5.0 s, 99.99% of vT after 6.0 s,
99.999% of vT after 7.4 s.
P6.46
(a)
At terminal velocity,
∑ F = 0 = − mg + CvT2
C=
(b)
(c)
e
mg
vT2
b0.142 kgge9.80 m s j =
b42.5 m sg
2
=
jb
2
Cv 2 = 7.70 × 10 −4 kg m 36.0 m s
g
2
7.70 × 10 −4 kg m
= 0.998 N
Elapsed
Time (s)
Altitude
(m)
Speed
(m/s)
Resistance
Force (N)
Net
Force (N)
0.000 00
0.050 00
…
2.950 00
3.000 00
3.050 00
…
6.250 00
6.300 00
0.000 00
1.757 92
Acceleration
m s2
36.000 00
35.158 42
–0.998 49
–0.952 35
–2.390 09
–2.343 95
–16.831 58
–16.506 67
48.623 27
48.640 00
48.632 24
0.824 94
0.334 76
–0.155 27
–0.000 52
–0.000 09
0.000 02
–1.392 12
–1.391 69
–1.391 58
–9.803 69
–9.800 61
–9.799 87
1.250 85
–0.106 52
–26.852 97
–27.147 36
0.555 55
0.567 80
–0.836 05
–0.823 80
–5.887 69
–5.801 44
e
j
Maximum height is about 49 m . It returns to the ground after about 6.3 s with a speed
of approximately 27 m s .
176
P6.47
Circular Motion and Other Applications of Newton’s Laws
(a)
At constant velocity
mg
=−
vT = −
C
vT = −
(b)
(a)
b50.0 kg ge9.80 m s j =
2
0.200 kg m
b50.0 kg gb9.80 m sg =
20.0 kg m
−49.5 m s with chute closed and
−4.95 m s with chute open.
We use time increments of 0.1 s for 0 < t < 10 s , then 0.01 s for 10 s < t < 12 s , and then 0.1 s
again.
time(s)
0
1
2
4
7
10
10.1
10.3
11
12
50
100
145
6.48
∑ F = 0 = − mg + CvT2
height(m)
1000
995
980
929
812
674
671
669
665
659
471
224
0
velocity(m/s)
0
–9.7
–18.6
–32.7
–43.7
–47.7
–16.7
–8.02
–5.09
–4.95
–4.95
–4.95
–4.95
We use a time increment of 0.01 s.
time(s)
0
0.100
0.200
0.400
1.00
1.92
2.00
4.00
5.00
6.85
x(m)
0
7.81
14.9
27.1
51.9
70.0
70.9
80.4
81.4
81.8
y(m)
0
5.43
10.2
18.3
32.7
38.5
38.5
26.7
17.7
0
(b)
range = 81.8 m
(c)
So we have maximum range at θ = 15.9°
with θ
30.0°
35.0°
25.0°
20.0°
15.0°
10.0°
17.0°
16.0°
15.5°
15.8°
16.1°
15.9°
we find range
86.410 m
81.8 m
90.181 m
92.874 m
93.812 m
90.965 m
93.732 m
93.839 8 m
93.829 m
93.839 m
93.838 m
93.840 2 m
Chapter 6
P6.49
(a)
∑ F = − mg + Cv 2 = 0 . Thus,
At terminal speed,
C=
(b)
(c)
177
mg
v2
b0.046 0 kgge9.80 m s j =
b44.0 m sg
2
=
2
2.33 × 10 −4 kg m
We set up a spreadsheet to calculate the motion, try different initial speeds, and home in on
53 m s as that required for horizontal range of 155 m, thus:
Time
t (s)
x
(m)
0.000 0
0.002 7
…
2.501 6
2.504 3
2.506 9
…
3.423 8
3.426 5
3.429 1
…
5.151 6
5.154 3
0.000 0
0.121 1
vx
(m/s)
e
ax
m s2
j
y
(m)
vy
ay
em s j
v=
2
v x2
+
v y2
tan −1
Fv I
GH v JK
y
x
(m/s)
(deg)
45.687 0 –10.565 9 0.000 0 27.451 5 –13.614 6
45.659 0 –10.552 9 0.072 7 27.415 5 –13.604 6
53.300 0
53.257 4
31.000 0
30.982 2
90.194 6 28.937 5 –4.238 8 32.502 4 0.023 5 –9.800 0
90.271 3 28.926 3 –4.235 5 32.502 4 –0.002 4 –9.800 0
90.348 0 28.915 0 –4.232 2 32.502 4 –0.028 4 –9.800 0
28.937 5
28.926 3
28.915 1
0.046 6
–0.004 8
–0.056 3
115.229 8 25.492 6 –3.289 6 28.397 2 –8.890 5 –9.399 9
115.297 4 25.483 9 –3.287 4 28.373 6 –8.915 4 –9.397 7
115.364 9 25.475 1 –3.285 1 28.350 0 –8.940 3 –9.395 4
26.998 4
26.998 4
26.998 4
–19.226 2
–19.282 2
–19.338 2
154.996 8 20.843 8 –2.199 2 0.005 9 –23.308 7 –7.049 8
155.052 0 20.838 0 –2.198 0 –0.055 9 –23.327 4 –7.045 4
31.269 2
31.279 2
–48.195 4
–48.226 2
(m/s)
Similarly, the initial speed is 42 m s . The motion proceeds thus:
Time
t (s)
x
(m)
0.000 0 0.000 0
0.003 5 0.100 6
…
2.740 5 66.307 8
2.744 0 66.379 7
2.747 5 66.451 6
…
3.146 5 74.480 5
3.150 0 74.549 5
3.153 5 74.618 5
…
5.677 0 118.969 7
5.680 5 119.024 8
vx
(m/s)
e
ax
m s2
j
y
(m)
vy
ay
em s j
2
v=
v x2
+
v y2
tan −1
Fv I
GH v JK
y
x
(m/s)
(deg)
0.000 0 30.826 6 –14.610 3
0.107 9 30.775 4 –14.594 3
42.150 0
42.102 6
47.000 0
46.967 1
20.548 4 –2.137 4 39.485 4 0.026 0 –9.800 0
20.541 0 –2.135 8 39.485 5 –0.008 3 –9.800 0
20.533 5 –2.134 3 39.485 5 –0.042 6 –9.800 0
20.548 5
20.541 0
20.533 5
0.072 5
–0.023 1
–0.118 8
19.715 6 –1.967 6 38.696 3 –3.942 3 –9.721 3
19.708 7 –1.966 2 38.682 5 –3.976 4 –9.720 0
19.701 8 –1.964 9 38.668 6 –4.010 4 –9.718 6
20.105 8
20.105 8
20.105 8
–11.307 7
–11.406 7
–11.505 6
15.739 4 –1.254 0 0.046 5 –25.260 0 –6.570 1
15.735 0 –1.253 3 –0.041 9 –25.283 0 –6.564 2
29.762 3
29.779 5
–58.073 1
–58.103 7
28.746 2 –4.182 9
28.731 6 –4.178 7
(m/s)
The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, the
ball reaches maximum height when it has covered about 57% of its range. Its speed is a
minimum somewhat later. The impact speeds are both about 30 m/s.
178
Circular Motion and Other Applications of Newton’s Laws
Additional Problems
*P6.50
When the cloth is at a lower angle θ, the radial
component of ∑ F = ma reads
68°
R
mv 2
n + mg sin θ =
.
r
mg
At θ = 68.0° , the normal force drops to zero and
v2
.
g sin 68° =
r
v = rg sin 68° =
p
p
mg sin68°
mg cos68°
FIG. P6.50
a0.33 mfe9.8 m s j sin 68° = 1.73 m s
2
The rate of revolution is
b
angular speed = 1.73 m s
*P6.51
(a)
b
v = 30 km h
IJ FG 2πr IJ =
gFGH 12rev
πr K H 2π a0.33 mf K
0.835 rev s = 50.1 rev min .
1 h I F 1 000 m I
gFGH 3 600
G
J = 8.33 m s
s JK H 1 km K
∑ Fy = ma y : +n − mg = −
mv 2
r
L
F vI
b8.33 m sg
n = mG g − J = 1 800 kg M9.8 m s −
20. 4 m
MN
H rK
2
2
n
2
OP
PQ
mg
FIG. P6.51
= 1.15 × 10 4 N up
(b)
Take n = 0 . Then mg =
mv 2
.
r
v = gr =
P6.52
(a)
∑ Fy = ma y =
mg − n =
(b)
mv 2
R
e9.8 m s ja20.4 mf =
2
mv 2
R
n = mg −
When n = 0 ,
mg =
Then,
v=
mv 2
R
mv 2
R
gR .
14.1 m s = 50.9 km h
Chapter 6
*P6.53
(a)
slope =
(b)
slope =
(c)
(d)
0.160 N − 0
= 0.016 2 kg m
9.9 m 2 s 2
R
v
2
=
1
2
DρAv 2
v
1
Dρ A
2
=
2
1
DρA = 0.016 2 kg m
2
2 0.016 2 kg m
D=
1.20 kg m3 π 0.105 m
e
179
b
g
ja
f
2
= 0.778
e
je
j
From the table, the eighth point is at force mg = 8 1.64 × 10 −3 kg 9.8 m s 2 = 0.129 N and
b
g
2
horizontal coordinate 2.80 m s . The vertical coordinate of the line is here
0.129 N − 0.127 N
2
= 1.5%.
0.016 2 kg m 2.8 m s = 0.127 N . The scatter percentage is
0.127 N
b
P6.54
gb
g
(e)
The interpretation of the graph can be stated thus: For stacked coffee filters falling at
terminal speed, a graph of air resistance force as a function of squared speed demonstrates
that the force is proportional to the speed squared within the experimental uncertainty
estimated as 2%. This proportionality agrees with that described by the theoretical equation
1
R = DρAv 2 . The value of the constant slope of the graph implies that the drag coefficient
2
for coffee filters is D = 0.78 ± 2% .
(a)
While the car negotiates the curve, the accelerometer is at the angle θ.
mv 2
r
Horizontally:
T sin θ =
Vertically:
T cos θ = mg
where r is the radius of the curve, and v is the speed of the car.
tan θ =
By division,
v2
= g tan θ :
Then a c =
r
v2
rg
e
a c = 9.80 m s
2
j tan 15.0°
a c = 2.63 m s 2
b23.0 m sg
r=
(b)
v2
r=
ac
(c)
v 2 = rg tan θ = 201 m 9.80 m s 2 tan 9.00°
2
2.63 m s 2
a
fe
j
v = 17.7 m s
= 201 m
FIG. P6.54
180
P6.55
Circular Motion and Other Applications of Newton’s Laws
Take x-axis up the hill
∑ Fx = ma x :
+T sin θ − mg sin φ = ma
T
a = sin θ − g sin φ
m
∑ Fy = ma y : +T cos θ − mg cos φ = 0
mg cos φ
T=
cos θ
g cos φ sin θ
− g sin φ
a=
cos θ
a = g cos φ tan θ − sin φ
b
*P6.56
(a)
a
f
2π 7.46 m
= 1.23 m s . The
38 s
total force on it must add to
The speed of the bag is
ma c =
b30 kg gb1.23 m sg
fs
y
n
ac
2
mg
= 6.12 N
7.46 m
g
FIG. P6.56
∑ Fx = ma x :
∑ Fy = ma y :
fs cos 20 − n sin 20 = 6.12 N
b
ge
j
fs sin 20 + n cos 20 − 30 kg 9.8 m s 2 = 0
f cos 20 − 6.12 N
n= s
sin 20
Substitute:
fs sin 20 + fs
cos 2 20
cos 20
− 6.12 N
= 294 N
sin 20
sin 20
f s 2.92 = 294 N + 16.8 N
a
f
a f
fs = 106 N
(b)
a
b
f
gb
2π 7.94 m
= 1.47 m s
34 s
2
30 kg 1.47 m s
= 8.13 N
ma c =
7.94 m
fs cos 20 − n sin 20 = 8.13 N
v=
g
fs sin 20 + n cos 20 = 294 N
fs cos 20 − 8.13 N
sin 20
cos 2 20
cos 20
− 8.13 N
= 294 N
fs sin 20 + fs
sin 20
sin 20
fs 2.92 = 294 N + 22.4 N
n=
a
a f
fs = 108 N
n=
f
a108 Nf cos 20 − 8.13 N = 273 N
sin 20
fs 108 N
= 0.396
µs = =
n 273 N
x
Chapter 6
P6.57
(a)
Since the centripetal acceleration of a person is downward (toward
the axis of the earth), it is equivalent to the effect of a falling
elevator. Therefore,
Fg′ = Fg −
(b)
181
mv 2
or Fg > Fg′
r
a f
At the poles v = 0 and Fg′ = Fg = mg = 75.0 9.80 = 735 N down.
b
FIG. P6.57
g
At the equator, Fg′ = Fg − ma c = 735 N − 75.0 0.033 7 N = 732 N down.
P6.58
(a)
(b)
(c)
P6.59
(a)
Since the object of mass m 2 is in equilibrium,
∑ Fy = T − m 2 g = 0
or
T = m2 g .
The tension in the string provides the required centripetal acceleration of the puck.
Thus,
Fc = T = m 2 g .
From
Fc =
we have
v=
b
v = 300 mi h
m1 v 2
R
RFc
=
m1
FG m IJ gR
Hm K
2
.
1
gFGH 6088.0.0 mift sh IJK = 440 ft s
At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase.
His apparent weight is
Fg′ = mg + m
(b)
FG
H
v2
160
= 160 +
r
32.0
IJ a440f
K 1 200
2
= 967 lb .
At the highest point, the force of the seat on the pilot is directed down and
Fg′ = mg − m
v2
= −647 lb .
r
Since the plane is upside down, the seat exerts this downward force.
(c)
mv 2
. If we vary the aircraft’s R and v such that the above is true,
R
then the pilot feels weightless.
When Fg′ = 0 , then mg =
182
P6.60
Circular Motion and Other Applications of Newton’s Laws
For the block to remain stationary,
e
∑ Fy = 0 and ∑ Fx = mar .
j
e
j
n1 = m p + m b g so f ≤ µ s1n1 = µ s1 m p + m b g .
mb g
At the point of slipping, the required centripetal force equals the
maximum friction force:
e
2
v max
n1
j r = µ em + m j g
a0.750fa0.120fa9.80f = 0.939 m s .
∴ mp + mb
or v max = µ s1 rg =
s1
p
mp g
fp
b
f
For the penny to remain stationary on the block:
mb g
∑ Fy = 0 ⇒ n 2 − m p g = 0 or n 2 = m p g
∑ Fx = ma r ⇒ f p = m p
and
n2
2
v
.
r
fp
When the penny is about to slip on the block, f p = f p , max = µ s 2 n 2
or µ s 2 m p g = m p
mp g
2
v max
r
mp g
v max = µ s 2 rg =
a0.520fa0.120fa9.80f = 0.782 m s
FIG. P6.60
This is less than the maximum speed for the block, so the penny slips before the block starts to slip.
The maximum rotation frequency is
Max rpm =
P6.61
P6.62
v=
a
gLM a
N
OPFG 60 s IJ =
f QH 1 min K
v max
1 rev
= 0.782 m s
2πr
2π 0.120 m
b
62.2 rev min .
f
2πr 2π 9.00 m
=
= 3.77 m s
T
15.0 s
a
f
v2
= 1.58 m s 2
r
(a)
ar =
(b)
Flow = m g + a r = 455 N
(c)
Fhigh
(d)
Fmid = m g 2 + a r2 = 397 N upward and at θ = tan −1
b g
= mb g − a g =
r
328 N
ar
1.58
= tan −1
= 9.15° inward .
g
9.8
Standing on the inner surface of the rim, and moving with it, each person will feel a normal force
exerted by the rim. This inward force causes the 3.00 m s 2 centripetal acceleration:
ac =
v2
:
r
The period of rotation comes from v =
so the frequency of rotation is
e3.00 m s ja60.0 mf = 13.4 m s
2πr 2π a60.0 mf
=
= 28.1 s
T=
v
13. 4 m s
1
1
1 F 60 s I
=
f= =
G J = 2.14 rev min
T 28.1 s 28.1 s H 1 min K
v = ac r =
2πr
:
T
2
.
Chapter 6
P6.63
(a)
The mass at the end of the chain is in vertical equilibrium.
T
Thus T cos θ = mg .
l
=
2.50
m
mv 2
Horizontally T sin θ = ma r =
θ
r
R = 4.00 m
a
f
r = a 2.50 sin 28.0°+4.00f m = 5.17 m
r
r = 2.50 sin θ + 4.00 m
v2
.
Then a r =
5.17 m
By division tan θ =
183
mg
FIG. P6.63
ar
v2
=
g 5.17 g
a fa fa
f
v 2 = 5.17 g tan θ = 5.17 9.80 tan 28.0° m 2 s 2
v = 5.19 m s
(b)
T cos θ = mg
b
ge
j
50.0 kg 9.80 m s 2
mg
T=
=
= 555 N
cos θ
cos 28.0°
P6.64
(a)
The putty, when dislodged, rises and returns to the original level in time t. To find t, we use
2v
where v is the speed of a point on the rim of the wheel.
v f = vi + at : i.e., − v = + v − gt or t =
g
2πR
2 v 2πR
If R is the radius of the wheel, v =
=
, so t =
.
t
g
v
Thus, v 2 = πRg and v = πRg .
(b)
The putty is dislodged when F, the force holding it to the wheel is
F=
P6.65
(a)
mv 2
R
n=
f − mg = 0
f = µ sn
T=
(b)
v=
f
2πR
T
4π 2 Rµ s
g
T = 2.54 s
#
mv 2
= mπ g .
R
n
FG
H
IJ
K
rev 1 rev 60 s
rev
=
= 23.6
min 2.54 s min
min
mg
FIG. P6.65
184
P6.66
Circular Motion and Other Applications of Newton’s Laws
Let the x–axis point eastward, the y-axis upward, and the z-axis point southward.
(a)
vi2 sin 2θ i
g
The initial speed of the ball is therefore
The range is Z =
vi =
gZ
=
sin 2θ i
a9.80fa285f = 53.0 m s
sin 96.0°
The time the ball is in the air is found from ∆y = viy t +
b
1
a y t 2 as
2
f e
ga
j
0 = 53.0 m s sin 48.0° t − 4.90 m s 2 t 2
giving t = 8.04 s .
e
j
6
2πR e cos φ i 2π 6.37 × 10 m cos 35.0°
=
= 379 m s
86 400 s
86 400 s
(b)
vix =
(c)
360° of latitude corresponds to a distance of 2πR e , so 285 m is a change in latitude of
∆φ =
FG S IJ a360°f = FG 285 m IJ a360°f = 2.56 × 10
GH 2π e6.37 × 10 mj JK
H 2πR K
6
e
−3
degrees
The final latitude is then φ f = φ i − ∆φ = 35.0°−0.002 56° = 34.997 4° .
The cup is moving eastward at a speed v fx =
2πR e cos φ f
86 400 s
, which is larger than the eastward
velocity of the tee by
∆v x = v fx − v fi =
=
2πR e
2πR e
cos φ f − cos φ i =
cos φ i − ∆φ − cos φ i
86 400 s
86 400 s
b
2πR e
cos φ i cos ∆φ + sin φ i sin ∆φ − cos φ i
86 400 s
Since ∆φ is such a small angle, cos ∆φ ≈ 1 and ∆v x ≈
∆v x ≈
(d)
b g e
g
e
2πR e
sin φ i sin ∆φ .
86 400 s
j sin 35.0° sin 0.002 56° =
2π 6.37 × 10 6 m
86 400 s
ja
f
∆x = ∆v x t = 1.19 × 10 −2 m s 8.04 s = 0.095 5 m = 9.55 cm
1.19 × 10 −2 m s
Chapter 6
P6.67
(a)
If the car is about to slip down the incline, f is directed up
the incline.
∑ Fy = n cos θ + f sin θ − mg = 0 where
n=
b
g
b
g
θt
2
v min
yields
R
b
Rg tan θ − µ s
v min =
n
f
mg
µ s mg
and f =
.
cos θ 1 + µ s tan θ
cos θ 1 + µ s tan θ
∑ Fx = n sin θ − f cos θ = m
Then,
f = µ sn gives
θt
mg
n cos θ
g
f sin θ
.
1 + µ s tan θ
f cos θ
n sin θ
When the car is about to slip up the incline, f is directed
down the incline. Then, ∑ Fy = n cos θ − f sin θ − mg = 0
with f = µ sn yields
mg
µ s mg
mg
n=
and f =
.
cos θ 1 − µ s tan θ
cos θ 1 − µ s tan θ
b
In this case,
g
∑ Fx = n sin θ + f cos θ = m
If v min =
(c)
v min =
b
Rg tan θ − µ s
b
1 + µ s tan θ
1 − µ s tan θ
g = 0 , then
g
θt
n
2
v max
, which gives
R
Rg tan θ + µ s
v max =
(b)
b
g
f
.
θ
t
mg
µ s = tan θ .
a100 mfe9.80 m s jatan 10.0°−0.100f =
1 + a0.100 f tan 10.0°
a100 mfe9.80 m s jatan 10.0°+0.100f =
1 − a0.100f tan 10.0°
n cos θ
2
n sin θ
8.57 m s
f cos θ
2
v max =
16.6 m s
f sin θ
mg
FIG. P6.67
185
186
P6.68
Circular Motion and Other Applications of Newton’s Laws
(a)
The bead moves in a circle with radius v = R sin θ at a speed
of
v=
2πr 2πR sin θ
=
T
T
The normal force has
an inward radial component of n sin θ
and an upward component of n cos θ
∑ Fy = ma y :
n cos θ − mg = 0
or
n=
Then
∑ Fx = n sin θ = m
mg
cos θ
FIG. P6.68(a)
FG mg IJ sinθ = m FG 2πR sinθ IJ
K
H cosθ K
R sin θ H
T
v2
becomes
r
which reduces to
g sin θ 4π 2 R sin θ
=
cos θ
T2
This has two solutions:
sin θ = 0 ⇒ θ = 0°
and
cos θ =
2
gT 2
4π 2 R
(1)
(2)
If R = 15.0 cm and T = 0.450 s, the second solution yields
e9.80 m s ja0.450 sf
cos θ =
4π a0.150 mf
2
2
2
= 0.335 and θ = 70.4°
Thus, in this case, the bead can ride at two positions θ = 70. 4° and θ = 0° .
(b)
At this slower rotation, solution (2) above becomes
e9.80 m s ja0.850 sf
cos θ =
4π a0.150 mf
2
2
2
= 1.20 , which is impossible.
In this case, the bead can ride only at the bottom of the loop, θ = 0° . The loop’s rotation
must be faster than a certain threshold value in order for the bead to move away from the
lowest position.
Chapter 6
P6.69
At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg = arv + br 2 v 2
(a)
e
j e
j
mg = 3.10 × 10 −9 v + 0.870 × 10 −10 v 2
LM 4 π e10 mj OP
N3
Q
= e3.10 × 10 jv + e0.870 × 10 jv
4.11 × 10 −11
3
−5
m = ρV = 1 000 kg m3
For water,
−9
−10
2
Assuming v is small, ignore the second term on the right hand side: v = 0.013 2 m s .
(b)
e
j e
j
mg = 3.10 × 10 −8 v + 0.870 × 10 −8 v 2
Here we cannot ignore the second term because the coefficients are of nearly equal
magnitude.
e
j e
j
a3.10f + 4a0.870fa4.11f = 1.03 m s
2a0.870f
4.11 × 10 −8 = 3.10 × 10 −8 v + 0.870 × 10 −8 v 2
v=
(c)
e
2
−3.10 ±
j e
j
mg = 3.10 × 10 −7 v + 0.870 × 10 −6 v 2
Assuming v > 1 m s , and ignoring the first term:
e
j
4.11 × 10 −5 = 0.870 × 10 −6 v 2
P6.70
187
v=
FG mg IJ LM1 − expFG −bt IJ OP where expaxf = e
H b KN H m KQ
v = 6.87 m s
x
is the exponential function.
mg
b
At t → ∞ ,
v → vT =
At t = 5.54 s
0.500 vT = vT 1 − exp
LM
MN
F −ba5.54 sf I OP
GH 9.00 kg JK PQ
F −ba5.54 sf I = 0.500 ;
GH 9.00 kg JK
− b a5.54 sf
= ln 0.500 = −0.693 ;
exp
9.00 kg
b=
b9.00 kg ga0.693f = 1.13 m s
5.54 s
b9.00 kg ge9.80 m s j =
2
mg
b
(a)
vT =
vT =
(b)
0.750 vT = vT 1 − exp
LM
N
FG −1.13t IJ OP
H 9.00 s K Q
1.13 kg s
FG −1.13t IJ = 0.250
H 9.00 s K
9.00aln 0.250 f
t=
s=
exp
−1.13
continued on next page
11.1 s
78.3 m s
188
Circular Motion and Other Applications of Newton’s Laws
(c)
FG IJ LM1 − expFG − bt IJ OP ;
H K N H mKQ
mg
dx
=
dt
b
z z FGH
x
dx =
x0
t
0
mg
b
IJ LM1 − expFG −bt IJ OPdt
KN H m KQ
F
GH
I FG IJ = mgt + F m g I LexpFG −bt IJ − 1O
JK H K b GH b JK MN H m K PQ
F b9.00 kg g e9.80 m s j I
5.54 s
J expa−0.693f − 1
x = 9.00 kg e9.80 m s j
+G
1.13 kg s G
H b1.13 m sg JK
x = 434 m + 626 ma −0.500f = 121 m
mgt
m2 g
− bt
+
x − x0 =
exp
2
b
m
b
t
2
2
0
2
2
At t = 5.54 s ,
P6.71
2
2
∑ Fy = L y − Ty − mg = L cos 20.0°−T sin 20.0°−7.35 N = ma y = 0
∑ Fx = L x + Tx = L sin 20.0°+T cos 20.0° = m
b
g
v2
r
2
35.0 m s
v2
= 0.750 kg
= 16.3 N
m
r
60.0 m cos 20.0°
a
f
∴ L sin 20.0°+T cos 20.0° = 16.3 N
L cos 20.0°−T sin 20.0° = 7.35 N
cos 20.0°
16.3 N
=
sin 20.0° sin 20.0°
sin 20.0°
7.35 N
=
L−T
cos 20.0° cos20.0°
16.3 N
7.35 N
−
T cot 20.0°+ tan 20.0° =
sin 20.0° cos 20.0°
T 3.11 = 39.8 N
L+T
a
a f
T = 12.8 N
f
FIG. P6.71
Chapter 6
P6.72
(a)
af
a f
2.00
18.9
3.00
42.1
4.00
73.8
(b)
ts
dm
4.88
1.00
700
600
6.00 154
7.00 199
500
8.00 246
9.00 296
(c)
347
11.0
399
12.0
452
13.0
505
14.0
558
15.0
611
16.0
664
17.0
717
18.0
770
19.0
823
20.0
876
d (m)
900
800
5.00 112
10.0
400
300
200
100
0
0
2
4
6
8
10 12 14 16 18 20
t (s)
A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminal
speed.
vT = slope =
*P6.73
189
876 m − 399 m
= 53.0 m s
20.0 s − 11.0 s
dv
dx
=0−k
= − kv
dt
dt
v = vi − kx implies the acceleration is
a=
Then the total force is
∑ F = ma = ma− kvf
∑ F = −kmv
The resistive force is opposite to the velocity:
.
ANSWERS TO EVEN PROBLEMS
P6.2
215 N horizontally inward
P6.4
6.22 × 10 −12 N
P6.6
(a) 1.65 km s ; (b) 6.84 × 10 3 s
P6.8
0.966 g
P6.10
e
j
(c) e −0.181 i + 0.181 jj m s
(a) −0.233 i + 0.163 j m s 2 ; (b) 6.53 m s ;
2
P6.12
2.06 × 10 3 rev min
P6.14
(a)
P6.16
(a) 1.33 m s 2 ; (b) 1.79 m s 2 forward and
48.0° inward
P6.18
8.88 N
R
FG 2T − gIJ ; (b) 2T upward
Hm K
190
Circular Motion and Other Applications of Newton’s Laws
P6.46
(a) 7.70 × 10 −4 kg m; (b) 0.998 N;
(c) The ball reaches maximum height 49 m.
Its flight lasts 6.3 s and its impact speed is
27 m s .
15.3 m s Straight across the dashboard to
the left
P6.48
(a) see the solution; (b) 81.8 m; (c) 15.9°
P6.24
0.527°
P6.50
0.835 rev s
P6.26
(a) 1.41 h; (b) 17.1
P6.52
(a) mg −
P6.28
µk =
P6.54
(a) 2.63 m s 2 ; (b) 201 m; (c) 17.7 m s
P6.30
(a) 2.38 × 10 5 m s 2 horizontally inward
= 2.43 × 10 4 g ; (b) 360 N inward
perpendicular to the cone;
(c) 47.5 × 10 4 m s 2
P6.56
(a) 106 N; (b) 0.396
P6.58
(a) m 2 g ; (b) m 2 g ; (c)
P6.32
(a) 6.27 m s 2 downward ; (b) 784 N up;
(c) 283 N up
P6.60
62.2 rev min
P6.62
2.14 rev min
P6.34
(a) 53.8 m s ; (b) 148 m
P6.64
P6.36
1.40
(a) v = πRg ; (b) mπ g
P6.38
−0. 212 m s 2
P6.66
(a) 8.04 s; (b) 379 m s; (c) 1.19 cm s ;
(d) 9.55 cm
P6.40
see the solution
P6.68
(a) either 70.4° or 0°; (b) 0°
P6.42
36.5 m s
P6.70
(a) 78.3 m s ; (b) 11.1 s; (c) 121 m
P6.44
(a) 0.980 m s ; (b) see the solution
P6.72
(a) and (b) see the solution; (c) 53.0 m s
P6.20
(a) 8.62 m; (b) Mg downward;
(c) 8.45 m s 2 , Unless they are belted in,
the riders will fall from the cars.
P6.22
a
2 vt − L
b g + agt
f
2
mv 2
; (b) v = gR
R
FG m IJ gR
Hm K
2
1
7
Energy and Energy Transfer
CHAPTER OUTLINE
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
Systems and Environments
Work Done by a Constant
Force
The Scalar Product of Two
Vectors
Work Done by a Varying
Force
Kinetic Energy and the
Work-Kinetic Energy
Theorem
The Non-Isolated
System—Conservation of
Energy
Situations Involving Kinetic
Friction
Power
Energy and the Automobile
ANSWERS TO QUESTIONS
Q7.1
The force is perpendicular to every increment of displacement.
Therefore, F ⋅ ∆r = 0 .
Q7.2
(a)
Positive work is done by the chicken on the dirt.
(b)
No work is done, although it may seem like there is.
(c)
Positive work is done on the bucket.
(d)
Negative work is done on the bucket.
(e)
Negative work is done on the person’s torso.
Q7.3
Yes. Force times distance over which the toe is in contact with
the ball. No, he is no longer applying a force. Yes, both air
friction and gravity do work.
Q7.4
Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on an
object at rest or moving across a level floor.
Q7.5
(a)
Tension
(c)
Positive in increasing velocity on the downswing.
Negative in decreasing velocity on the upswing.
(b)
Air resistance
Q7.6
No. The vectors might be in the third and fourth quadrants, but if the angle between them is less
than 90° their dot product is positive.
Q7.7
The scalar product of two vectors is positive if the angle between them is between 0 and 90°. The
scalar product is negative when 90° < θ < 180° .
Q7.8
If the coils of the spring are initially in contact with one another, as the load increases from zero, the
graph would be an upwardly curved arc. After the load increases sufficiently, the graph will be
linear, described by Hooke’s Law. This linear region will be quite large compared to the first region.
The graph will then be a downward curved arc as the coiled spring becomes a completely straight
wire. As the load increases with a straight wire, the graph will become a straight line again, with a
significantly smaller slope. Eventually, the wire would break.
Q7.9
k ′ = 2 k . To stretch the smaller piece one meter, each coil would have to stretch twice as much as one
coil in the original long spring, since there would be half as many coils. Assuming that the spring is
ideal, twice the stretch requires twice the force.
191
192
Energy and Energy Transfer
Q7.10
Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can
always do positive work in striking another object and causing it to move along the same direction
of motion.
Q7.11
Work is only done in accelerating the ball from rest. The work is done over the effective length of the
pitcher’s arm—the distance his hand moves through windup and until release.
Q7.12
Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy.
Q7.13
The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longer
distance, the change in kinetic energy will be larger.
Q7.14
(a)
Kinetic energy is proportional to squared speed. Doubling the speed makes an object's
kinetic energy four times larger.
(b)
If the total work on an object is zero in some process, its speed must be the same at the final
point as it was at the initial point.
Q7.15
The larger engine is unnecessary. Consider a 30 minute commute. If you travel the same speed in
each car, it will take the same amount of time, expending the same amount of energy. The extra
power available from the larger engine isn’t used.
Q7.16
If the instantaneous power output by some agent changes continuously, its average power in a
process must be equal to its instantaneous power at least one instant. If its power output is constant,
its instantaneous power is always equal to its average power.
Q7.17
It decreases, as the force required to lift the car decreases.
Q7.18
As you ride an express subway train, a backpack at your feet has no kinetic energy as measured by
you since, according to you, the backpack is not moving. In the frame of reference of someone on the
side of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic
energy.
Q7.19
The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at the
bottom; but it might instead make a deeper crater with an equal-size average force. The farther it
falls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that the
gravitational force does on it.
Q7.20
The normal force does no work because the angle between the normal force and the direction of
motion is usually 90°. Static friction usually does no work because there is no distance through
which the force is applied.
Q7.21
An argument for: As a glider moves along an airtrack, the only force that the track applies on the
glider is the normal force. Since the angle between the direction of motion and the normal force is
90°, the work done must be zero, even if the track is not level.
Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of the
airtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work.
Q7.22
Gaspard de Coriolis first stated the work-kinetic energy theorem. Jean Victor Poncelet, an engineer
who invaded Russia with Napoleon, is most responsible for demonstrating its wide practical
applicability, in his 1829 book Industrial Mechanics. Their work came remarkably late compared to the
elucidation of momentum conservation in collisions by Descartes and to Newton’s Mathematical
Principles of the Philosophy of Nature, both in the 1600’s.
Chapter 7
SOLUTIONS TO PROBLEMS
Section 7.1
Systems and Environments
Section 7.2
Work Done by a Constant Force
P7.1
W = F∆r cos θ = 16.0 N 2.20 m cos 25.0° = 31.9 J
a
(a)
fa
f
(b), (c) The normal force and the weight are both at 90° to the displacement in any time interval.
Both do 0 work.
∑ W = 31.9 J + 0 + 0 =
(d)
P7.2
31.9 J
The component of force along the direction of motion is
a
f
F cos θ = 35.0 N cos 25.0° = 31.7 N .
The work done by this force is
a
f
a
fa
f
W = F cos θ ∆r = 31.7 N 50.0 m = 1.59 × 10 3 J .
P7.3
Method One.
Let φ represent the instantaneous angle the rope makes with the vertical as
it is swinging up from φ i = 0 to φ f = 60° . In an incremental bit of motion
from angle φ to φ + dφ , the definition of radian measure implies that
a
f
∆r = 12 m dφ . The angle θ between the incremental displacement and the
b
g
force of gravity is θ = 90°+ φ . Then cos θ = cos 90°+φ = − sin φ .
The work done by the gravitational force on Batman is
z
z
f
φ = 60 °
i
φ =0
W = F cos θdr =
b
ga
FIG. P7.3
f
mg − sin φ 12 m dφ
f z sin φ dφ = b−80 kg ge9.8 m s ja12 mfb− cos φ g
= a −784 N fa12 mfa− cos 60°+1f = −4.70 × 10 J
a
= − mg 12 m
60 °
2
0
60 °
0
3
Method Two.
b
ge
j
The force of gravity on Batman is mg = 80 kg 9.8 m s 2 = 784 N down. Only his vertical
displacement contributes to the work gravity does. His original y-coordinate below the tree limb is
–12 m. His final y-coordinate is −12 m cos 60° = −6 m . His change in elevation is
a
f
a
f
−6 m − −12 m = 6 m . The work done by gravity is
a
fa f
W = F∆r cos θ = 784 N 6 m cos 180° = −4.70 kJ .
193
194
P7.4
Energy and Energy Transfer
ja fa f
e
(a)
W = mgh = 3.35 × 10 −5 9.80 100 J = 3.28 × 10 −2 J
(b)
Since R = mg , Wair resistance = −3.28 × 10 −2 J
Section 7.3
The Scalar Product of Two Vectors
P7.5
A = 5.00 ; B = 9.00 ; θ = 50.0°
A ⋅ B = AB cos θ = 5.00 9.00 cos 50.0° = 28.9
P7.6
A ⋅ B = A x i + A y j + A z k ⋅ B x i + B y j + Bz k
a fa f
e
je
j
A ⋅ B = A B e i ⋅ i j + A B e i ⋅ jj + A B e i ⋅ k j
+ A B e j ⋅ i j + A B e j ⋅ jj + A B e j ⋅ k j
+ A B ek ⋅ i j + A B ek ⋅ jj + A B ek ⋅ k j
x
x
x
y
x z
y x
y
y
y
z x
z y
z
z z
A ⋅ B = A x Bx + A y B y + A z Bz
P7.7
P7.8
a fa f
a
fa f
(a)
W = F ⋅ ∆r = Fx x + Fy y = 6.00 3.00 N ⋅ m + −2.00 1.00 N ⋅ m = 16.0 J
(b)
θ = cos −1
16
FG F ⋅ ∆r IJ = cos
=
H F∆r K
ae 6.00f + a−2.00f jea3.00f + a1.00f j
−1
2
2
2
2
36.9°
We must first find the angle between the two vectors. It is:
θ = 360°−118°−90.0°−132° = 20.0°
Then
a
fb
g
F ⋅ v = Fv cos θ = 32.8 N 0.173 m s cos 20.0°
or F ⋅ v = 5.33
P7.9
(a)
N ⋅m
J
= 5.33 = 5.33 W
s
s
A = 3.00 i − 2.00 j
B = 4.00 i − 4.00 j
(b)
θ = cos −1
B = 3.00 i − 4.00 j + 2.00k
A = −2.00 i + 4.00 j
(c)
FIG. P7.8
cos θ =
a fa f
A⋅B
−6.00 − 16.0
=
AB
20.0 29.0
A = i − 2.00 j + 2.00k
B = 3.00 j + 4.00k
A ⋅B
12.0 + 8.00
= cos −1
= 11.3°
AB
13.0 32.0
θ = cos −1
a fa f
FG A ⋅ B IJ = cos FG −6.00 + 8.00 IJ =
H AB K
H 9.00 ⋅ 25.0 K
−1
θ = 156°
82.3°
Chapter 7
P7.10
e
j e
A − B = 3.00 i + j − k − − i + 2.00 j + 5.00k
j
A − B = 4.00 i − j − 6.00k
a
f e
je
a
j
f a
f
C ⋅ A − B = 2.00 j − 3.00k ⋅ 4.00 i − j − 6.00k = 0 + −2.00 + +18.0 = 16.0
Section 7.4
Work Done by a Varying Force
z
f
P7.11
W = Fdx = area under curve from xi to x f
i
x f = 8.00 m
xi = 0
(a)
W = area of triangle ABC =
W0 → 8 =
FG 1 IJ × 8.00 m × 6.00 N =
H 2K
24.0 J
FIG. P7.11
x f = 10.0 m
xi = 8.00 m
(b)
FG 1 IJ AC × altitude,
H 2K
FG 1 IJ CE × altitude,
H 2K
F 1I
= G J × a 2.00 mf × a −3.00 N f = −3.00 J
H 2K
W = area of ∆CDE =
W8 →10
a
P7.12
f
W0 →10 = W0→ 8 + W8→10 = 24.0 + −3.00 = 21.0 J
(c)
a
f
Fx = 8 x − 16 N
(a)
See figure to the right
(b)
Wnet =
a
fa
f a
fa
f
− 2.00 m 16.0 N
1.00 m 8.00 N
+
= −12.0 J
2
2
FIG. P7.12
195
196
P7.13
Energy and Energy Transfer
z
W = Fx dx
and W equals the area under the Force-Displacement curve
For the region 0 ≤ x ≤ 5.00 m ,
(a)
W=
a3.00 Nfa5.00 mf =
2
7.50 J
For the region 5.00 ≤ x ≤ 10.0 ,
(b)
a
FIG. P7.13
fa
f
W = 3.00 N 5.00 m = 15.0 J
For the region 10.0 ≤ x ≤ 15.0 ,
(c)
W=
a3.00 Nfa5.00 mf =
2
7.50 J
For the region 0 ≤ x ≤ 15.0
(d)
a
f
W = 7.50 + 7.50 + 15.0 J = 30.0 J
z
f
P7.14
W = F ⋅ dr =
i
zb
5m
0
P7.15
k=
ze
5m
j
4x i + 3 y j N ⋅ dx i
0
x2
4 N m xdx + 0 = 4 N m
2
g
b
g
5m
= 50.0 J
0
a fa f
4.00 9.80 N
F Mg
=
=
= 1.57 × 10 3 N m
y
y
2.50 × 10 −2 m
(a)
For 1.50 kg mass y =
(b)
Work =
a fa f
1.50 9.80
mg
=
= 0.938 cm
k
1.57 × 10 3
1 2
ky
2
1
Work = 1.57 × 10 3 N ⋅ m 4.00 × 10 −2 m
2
e
P7.16
(a)
je
j
= 1.25 J
Spring constant is given by F = kx
k=
(b)
2
Work = Favg x =
a
fa
f
a
f
230 N
F
=
= 575 N m
x
0.400 m
a
1
230 N 0.400 m = 46.0 J
2
f
Chapter 7
*P7.17
(a)
b
Fapplied = k leaf x + k helper x h = k x + k h x − y 0
197
g
b
g
N
N
x + 3.60 × 10 5
x − 0.5 m
m
m
6.8 × 10 5 N
x =
= 0.768 m
8.85 × 10 5 N m
5 × 10 5 N = 5.25 × 10 5
(b)
W=
FG
H
IJ a
K
f
N
1
1
1
k x 2 + k h x h2 =
5.25 × 10 5
0.768 m
m
2
2
2
2
+
a
f
N
1
3.60 × 10 5
0.268 m
m
2
2
= 1.68 × 10 5 J
z
f
P7.18
(a)
W = F ⋅ dr
W=
i
0.600 m
ze
j
15 000 N + 10 000 x N m − 25 000 x 2 N m 2 dx cos 0°
0
10 000 x 2 25 000 x 3
−
W = 15 000 x +
2
3
0.600 m
0
W = 9.00 kJ + 1.80 kJ − 1.80 kJ = 9.00 kJ
(b)
Similarly,
b10.0 kN mga1.00 mf − e25.0 kN m ja1.00 mf
W = a15.0 kN fa1.00 mf +
2
3
2
2
3
W = 11.7 kJ , larger by 29.6%
P7.19
a
f
1
2
k 0.100 m
2
∴ k = 800 N m and to stretch the spring to 0.200 m requires
4.00 J =
∆W =
P7.20
(a)
a fa
1
800 0.200
2
f
2
− 4.00 J = 12.0 J
The radius to the object makes angle θ with the horizontal, so
its weight makes angle θ with the negative side of the x-axis,
when we take the x–axis in the direction of motion tangent to
the cylinder.
∑ Fx = ma x
F − mg cos θ = 0
F = mg cos θ
FIG. P7.20
z
f
(b)
W = F ⋅ dr
i
We use radian measure to express the next bit of displacement as dr = Rdθ in terms of the
next bit of angle moved through:
z
π 2
W=
0
mg cos θRdθ = mgR sin θ
a f
W = mgR 1 − 0 = mgR
π 2
0
198
*P7.21
Energy and Energy Transfer
The same force makes both light springs stretch.
(a)
The hanging mass moves down by
x = x1 + x 2 =
FG
H
mg mg
1
1
+
= mg
+
k1
k2
k1 k 2
= 1.5 kg 9.8 m s 2
(b)
IJ
K
F 1m + 1m I=
GH 1 200 N 1 800 N JK
2.04 × 10 −2 m
We define the effective spring constant as
FG
b
g H
F 1m + 1m I =
=G
H 1 200 N 1 800 N JK
k=
mg
F
1
1
=
=
+
k1 k 2
x mg 1 k1 + 1 k 2
IJ
K
−1
−1
*P7.22
720 N m
See the solution to problem 7.21.
FG 1 + 1 IJ
Hk k K
F1 1I
(b)
k=G + J
Hk k K
L F O N kg ⋅ m s =
k =M P= =
NxQ m m
(a)
x = mg
1
2
−1
1
P7.23
2
2
kg
s2
Section 7.5
Kinetic Energy and the Work-Kinetic Energy Theorem
Section 7.6
The Non-Isolated System—Conservation of Energy
P7.24
(a)
KA =
(b)
1
mv B2 = K B : v B =
2
(c)
∑ W = ∆K = K B − K A = 2 mevB2 − v 2A j = 7.50 J − 1.20 J =
(a)
K=
1
1
mv 2 = 0.300 kg 15.0 m s
2
2
(b)
K=
1
0.300 30.0
2
P7.25
b
gb
1
0.600 kg 2.00 m s
2
2K B
=
m
g
2
= 1.20 J
a2fa7.50f =
0.600
5.00 m s
1
b
a
fa f
gb
2
=
a
g
2
= 33.8 J
fa f a4f = 4a33.8f =
1
0.300 15.0
2
6.30 J
2
135 J
Chapter 7
P7.26
e
199
j
v i = 6.00 i − 2.00 j = m s
(a)
vi = vix2 + viy2 = 40.0 m s
Ki =
(b)
b
ge
1
1
mvi2 = 3.00 kg 40.0 m 2 s 2 = 60.0 J
2
2
j
v f = 8.00 i + 4.00 j
v 2f = v f ⋅ v f = 64.0 + 16.0 = 80.0 m 2 s 2
1
3.00
∆K = K f − K i = m v 2f − vi2 =
80.0 − 60.0 = 60.0 J
2
2
e
P7.27
a f
j
Consider the work done on the pile driver from the time it starts from rest until it comes to rest at
the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and
h = 0.12 m the distance it moves the piling.
1
1
mv 2f − mvi2
2
2
mg h + d cos 0°+ F d cos 180° = 0 − 0 .
∑ W = ∆K :
Wgravity + Wbeam =
b ga f
d ia f
bmg gah + df = b2 100 kgge9.80 m s ja5.12 mf =
F=
so
2
Thus,
driver is upward .
P7.28
(a)
a f
a f
(a)
a f
Ki + ∑ W = K f =
0 + ∑W =
(b)
(c)
F=
f
a
f
1
mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 15.0 m)
2
2 area
2 30.0 J
=
= 3.87 m s
m
4.00 kg
∆K = K f − K i =
vf =
P7.29
a
1
mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 10.0 m)
2
2 area
2 22.5 J
=
= 3.35 m s
m
4.00 kg
∆K = K f − K i =
vf =
(c)
a
f
1
mv 2f
2
jb
1
15.0 × 10 −3 kg 780 m s
2
e
g
2
= 4.56 kJ
4.56 × 10 3 J
W
=
= 6.34 kN
∆r cos θ
0.720 m cos 0°
a
f
v −v
b780 m sg − 0 = 422 km s
=
a=
2x
2a0.720 mf
∑ F = ma = e15 × 10 kg je422 × 10 m s j =
2
f
2
2
i
2
f
(d)
8.78 × 10 5 N . The force on the pile
1
mv 2f − 0 = ∑ W = (area under curve from x = 0 to x = 5.00 m)
2
2 area
2 7.50 J
=
= 1.94 m s
m
4.00 kg
∆K = K f − K i =
vf =
(b)
0.120 m
d
−3
3
2
6.34 kN
200
P7.30
Energy and Energy Transfer
(a)
e
Kf =
(b)
j
v f = 0.096 3 × 10 8 m s = 2.88 × 10 7 m s
1
1
mv 2f = 9.11 × 10 −31 kg 2.88 × 10 7 m s
2
2
e
Ki + W = K f :
je
j
2
= 3.78 × 10 −16 J
0 + F∆r cos θ = K f
a
f
F 0.028 m cos 0° = 3.78 × 10 −16 J
F = 1.35 × 10 −14 N
∑F =
1.35 × 10 −14 N
= 1. 48 × 10 +16 m s 2
9.11 × 10 −31 kg
(c)
∑ F = ma ;
a=
(d)
v xf = v xi + a x t
2.88 × 10 7 m s = 0 + 1.48 × 10 16 m s 2 t
m
e
j
t = 1.94 × 10 −9 s
1
v xi + v xf t
2
1
0.028 m = 0 + 0 + 2.88 × 10 7 m s t
2
x f = xi +
Check:
d
i
e
j
t = 1.94 × 10 −9 s
Section 7.7
P7.31
Situations Involving Kinetic Friction
∑ Fy = ma y :
n − 392 N = 0
n = 392 N
a fa f
= F∆r cos θ = a130 fa5.00f cos 0° =
f k = µ k n = 0.300 392 N = 118 N
(a)
WF
(b)
∆Eint = f k ∆x = 118 5.00 = 588 J
650 J
a fa f
a fa f
(c)
Wn = n∆r cos θ = 392 5.00 cos 90° = 0
(d)
W g = mg∆r cos θ = 392 5.00 cos −90° = 0
(e)
∆K = K f − K i = ∑ Wother − ∆Eint
1
mv 2f − 0 = 650 J − 588 J + 0 + 0 = 62.0 J
2
(f)
vf =
a fa f a f
2K f
m
=
a
f
2 62.0 J
= 1.76 m s
40.0 kg
FIG. P7.31
Chapter 7
P7.32
(a)
vf =
so
(b)
a fe
1 2 1 2 1
kx i − kx f = 500 5.00 × 10 −2
2
2
2
1
1
1
Ws = mv 2f − mvi2 = mv 2f − 0
2
2
2
Ws =
2
j
2
c∑ W h = 2a0.625f m s =
2.00
m
1
1
mvi2 − f k ∆x + Ws = mv 2f
2
2
a
fa fa fb g
1
0.282 J = b 2.00 kg gv
2
2a0.282 f
v =
m s = 0.531 m s
0 − 0.350 2.00 9.80 0.050 0 J + 0.625 J =
− 0 = 0.625 J
0.791 m s
1
mv 2f
2
2
f
f
P7.33
(a)
2.00
a
f
= b10.0 kg gd9.80 m s ia5.00 mf cos 110° =
W g = mg cos 90.0°+θ
Wg
(b)
FIG. P7.32
2
−168 J
f k = µ k n = µ k mg cos θ
∆Eint = f k = µ k mg cos θ
a
fa
fa fa f
∆Eint = 5.00 m 0.400 10.0 9.80 cos 20.0° = 184 J
(c)
a fa f
WF = F = 100 5.00 = 500 J
(d)
∆K = ∑ Wother − ∆Eint = WF + W g − ∆Eint = 148 J
(e)
∆K =
1
1
mv 2f − mvi2
2
2
2 ∆K
2 148
+ vi2 =
+ 1.50
vf =
10.0
m
a f
P7.34
∑ Fy = ma y :
a f a f
a
2
FIG. P7.33
= 5.65 m s
f
n + 70.0 N sin 20.0°−147 N = 0
n = 123 N
f k = µ k n = 0.300 123 N = 36.9 N
a
f
a
fa
f
a
fa
f
a
fa
f
(a)
W = F∆r cos θ = 70.0 N 5.00 m cos 20.0° = 329 J
(b)
W = F∆r cos θ = 123 N 5.00 m cos 90.0° = 0 J
(c)
W = F∆r cos θ = 147 N 5.00 m cos 90.0° = 0
(d)
∆Eint = F∆x = 36.9 N 5.00 m = 185 J
(e)
∆K = K f − K i = ∑ W − ∆Eint = 329 J − 185 J = +144 J
a
fa
f
FIG. P7.34
201
202
P7.35
Energy and Energy Transfer
µ k = 0.100
1
K i − f k ∆x + Wother = K f :
mvi2 − f k ∆x = 0
2
2
2.00 m s
v2
1
∆x = i =
= 2.04 m
mvi2 = µ k mg∆x
2
2 µ k g 2 0.100 9.80
vi = 2.00 m s
b
a
Section 7.8
*P7.36
Pav =
g
fa f
Power
b
W K f mv 2 0.875 kg 0.620 m s
=
=
=
2 ∆t
∆t ∆t
2 21 × 10 −3 s
e
W
t
P=
j
a
g
2
= 8.01 W
fa
f
mgh 700 N 10.0 m
=
= 875 W
t
8.00 s
P7.37
Power =
P7.38
A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine is
equal to its final kinetic energy,
b
gb
1
1 300 kg 24.6 m s
2
with power P =
P7.39
(a)
g
2
= 390 kJ
390 000 J
~ 10 4 W around 30 horsepower.
15.0 s
∑ W = ∆K , but ∆K = 0 because he moves at constant speed. The skier rises a vertical
a
f
distance of 60.0 m sin 30.0° = 30.0 m . Thus,
b
ja
ge
f
Win = −Wg = 70.0 kg 9.8 m s 2 30.0 m = 2.06 × 10 4 J = 20.6 kJ .
(b)
The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,
Pinput =
P7.40
(a)
W 2.06 × 10 4 J
=
= 686 W = 0.919 hp .
∆t
30.0 s
The distance moved upward in the first 3.00 s is
∆y = vt =
LM 0 + 1.75 m s OPa3.00 sf = 2.63 m .
N 2 Q
The motor and the earth’s gravity do work on the elevator car:
1
1
mvi2 + Wmotor + mg∆y cos 180° = mv 2f
2
2
1
2
Wmotor = 650 kg 1.75 m s − 0 + 650 kg g 2.63 m = 1.77 × 10 4 J
2
b
Also, W = P t so P =
(b)
gb
g
ga
b
f
W 1.77 × 10 4 J
=
= 5.91 × 10 3 W = 7.92 hp.
t
3.00 s
b
g
When moving upward at constant speed v = 1.75 m s the applied force equals the
b
ge
j = 6.37 × 10 N . Therefore,
P = Fv = e6.37 × 10 N jb1.75 m sg = 1.11 × 10
weight = 650 kg 9.80 m s
2
3
3
4
W = 14.9 hp .
Chapter 7
P7.41
203
energy = power × time
For the 28.0 W bulb:
a
fe
j
Energy used = 28.0 W 1.00 × 10 4 h = 280 kilowatt ⋅ hrs
a
fb
g
total cost = $17.00 + 280 kWh $0.080 kWh = $39.40
For the 100 W bulb:
a
fe
j
Energy used = 100 W 1.00 × 10 4 h = 1.00 × 10 3 kilowatt ⋅ hrs
1.00 × 10 4 h
= 13.3
# bulb used =
750 h bulb
b
g e
jb
g
total cost = 13.3 $0.420 + 1.00 × 10 3 kWh $0.080 kWh = $85.60
Savings with energy-efficient bulb = $85.60 − $39.40 = $46.20
*P7.42
(a)
FG 454 g IJ FG 9 kcal IJ FG 4 186 J IJ = 1.71 × 10
H 1 lb K H 1 g K H 1 kcal K
Burning 1 lb of fat releases energy
1 lb
The mechanical energy output is
e1.71 × 10 Jja0.20f = nF∆r cos θ .
Then
3.42 × 10 6 J = nmg∆y cos 0°
7
J.
7
b ge
jb
J = ne5.88 × 10 Jj
ga
f
3.42 × 10 6 J = n 50 kg 9.8 m s 2 80 steps 0.150 m
3.42 × 10 6
3
3.42 × 10 6 J
= 582 .
5.88 × 10 3 J
This method is impractical compared to limiting food intake.
where the number of times she must climb the steps is n =
(b)
Her mechanical power output is
P=
*P7.43
(a)
(b)
FG
H
IJ
K
1 hp
W 5.88 × 10 3 J
=
= 90.5 W = 90.5 W
= 0.121 hp .
t
65 s
746 W
IJ FG 1 kcal IJ FG 1.30 × 10 J IJ =
K H 4 186 J K H 1 gal K
1 h F 10 mi I F 1 kcal I F 1.30 × 10 J I
= 776 mi gal .
For bicycling
G J
400 kcal H h K GH 4 186 J JK GH 1 gal JK
The fuel economy for walking is
FG
H
8
1h
3 mi
220 kcal
h
8
423 mi gal .
204
Energy and Energy Transfer
Section 7.9
P7.44
Energy and the Automobile
At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW to
the wheels. If an additional load of 350 kg is added to the car, a larger output power of
P2 = P1 + (power input to move 350 kg at speed v)
will be required. The additional power output needed to move 350 kg at speed v is:
b g b
g
∆Pout = ∆f v = µ r mg v .
Assuming a coefficient of rolling friction of µ r = 0.016 0 , the power output now needed from the
engine is
b
gb
ge
jb
g
P2 = P1 + 0.016 0 350 kg 9.80 m s 2 26.8 m s = 18.3 kW + 1.47 kW .
With the assumption of constant efficiency of the engine, the input power must increase by the
same factor as the output power. Thus, the fuel economy must decrease by this factor:
bfuel economyg = FGH PP IJK bfuel economyg = FGH 18.318+.31.47 IJK b6.40 km Lg
or bfuel economy g = 5.92 km L .
1
2
1
2
2
P7.45
(a)
fuel needed =
=
(b)
(c)
1
2
mv 2f − 12 mvi2
useful energy per gallon
b900 kggb24.6 m sg =
a0.150fe1.34 × 10 J galj
=
b
mv 2f − 0
eff.× energy content of fuel
2
1
2
8
1.35 × 10 −2 gal
73.8
power =
FG 1 gal IJ FG 55.0 mi IJ FG 1.00 h IJ FG 1.34 × 10
H 38.0 mi K H 1.00 h K H 3 600 s K H 1 gal
Additional Problems
P7.46
1
2
b
g
b
g
At apex, v = b 40.0 m sg cos 30.0° i + 0 j = b34.6 m sgi
1
1
And K = mv = b0.150 kg gb34.6 m sg = 90.0 J
2
2
At start, v = 40.0 m s cos 30.0° i + 40.0 m s sin 30.0° j
2
2
8
J
I a0.150f =
JK
8.08 kW
g
Chapter 7
P7.47
b
gb
Concentration of Energy output = 0.600 J kg ⋅ step 60.0 kg
b
gb
1 step I
gFGH 1.50
J = 24.0 J m
mK
g
F = 24.0 J m 1 N ⋅ m J = 24.0 N
P = Fv
a
f
70.0 W = 24.0 N v
v = 2.92 m s
P7.48
(a)
a fa f
A ⋅ i = A 1 cos α . But also, A ⋅ i = A x .
a Afa1f cos α = A
Thus,
Similarly,
cos β =
and
cos γ =
Ax
.
A
Ay
A
Az
A
A = A x2 + A y2 + A z2 .
where
P7.49
or cos α =
x
(b)
cos 2 α + cos 2 β + cos 2 γ =
(a)
x = t + 2.00t 3
FG A IJ + FG A IJ + FG A IJ
H AK H AK H AK
x
2
y
2
z
2
=
A2
=1
A2
Therefore,
dx
= 1 + 6.00t 2
dt
1
1
K = mv 2 = 4.00 1 + 6.00t 2
2
2
v=
a fe
a12.0tf m s
F = ma = 4.00a12.0t f = a 48.0t f N
dv
=
dt
2
(b)
a=
(c)
P = Fv = 48.0t 1 + 6.00t 2 =
(d)
W=
a
z
fe
ze
2.00
2 .00
0
0
Pdt =
j e48.0t + 288t j W
3
j
48.0t + 288 t 3 dt = 1 250 J
j = e2.00 + 24.0t
2
2
j
+ 72.0t 4 J
205
206
*P7.50
Energy and Energy Transfer
(a)
We write
F = ax b
a
f
5 000 N = aa0.315 mf
F 0.315 IJ = 2.44
5=G
H 0.129 K
b
1 000 N = a 0.129 m
b
b
b
ln 5 = b ln 2.44
ln 5
= 1.80 = b
ln 2.44
1 000 N
a=
= 4.01 × 10 4 N m1.8 = a
1.80
0.129 m
b=
a
(b)
W=
z
z
0. 25 m
0. 25 m
0
0
Fdx =
= 4.01 × 10
4
4.01 × 10 4
N x 2 .8
m1.8 2.8
f
N 1.8
x dx
m1.8
0. 25 m
= 4.01 × 10
4
0
a
f
N 0.25 m
2.8
m1.8
2.8
= 294 J
*P7.51
The work done by the applied force is
z
f
W = Fapplied dx =
z
x max
e
0
i
z
z
x max
j
− − k1 x + k 2 x 2 dx
x max
x2
= k 1 x dx + k 2 x dx = k1
2
0
0
= k1
P7.52
(a)
2
x max
2
+ k2
2
x max
0
x3
+ k2
3
xmax
0
3
x max
3
The work done by the traveler is mghs N where N is the number of steps he climbs during
the ride.
N = (time on escalator)(n)
where
h
atime on escalatorf = vertical velocity
of person
and
vertical velocity of person = v + nhs
Then,
N=
nh
v + nhs
and the work done by the person becomes Wperson =
continued on next page
mgnhhs
v + nhs
Chapter 7
(b)
207
The work done by the escalator is
b
ga f a
fb
ga f
We = power time = force exerted speed time = mgvt
h
as above.
v + nhs
where
t=
Thus,
We =
mgvh
.
v + nhs
As a check, the total work done on the person’s body must add up to mgh, the work an
elevator would do in lifting him.
∑ W = Wperson + We =
It does add up as follows:
P7.53
(a)
(b)
*P7.54
∆K =
1
mv 2 − 0 = ∑ W , so
2
v2 =
2W
and v =
m
b
2W
m
W = F ⋅ d = Fx d ⇒ Fx =
W
d
During its whole motion from y = 10.0 m to y = −3.20 mm, the force of gravity and the force of the
plate do work on the ball. It starts and ends at rest
Ki + ∑ W = K f
0 + Fg ∆y cos 0°+ Fp ∆x cos 180° = 0
b
g b
g
5 kg e9.8 m s ja10 mf
=
= 1.53 × 10
mg 10.003 2 m − Fp 0.003 20 m = 0
2
Fp
P7.55
g
mgnhhs
mgvh mgh nhs + v
+
=
= mgh
v + nhs v + nhs
v + nhs
b
F
t =
m
OPa3.00 sf =
PQ
240 W
P = Fv = F vi + at = F 0 +
(b)
P=
2
m
IJ FG F IJ t
K HmK
g FGH
(a)
LM a20.0 Nf
MN 5.00 kg
3.2 × 10
−3
2
5
N upward
208
Energy and Energy Transfer
z
f
*P7.56
(a)
W1 = F1 dx =
z
xi 1 + x a
i
(b)
W2 =
k1 x dx =
xi 1
z
− xi 2 + x a
k 2 x dx =
− xi 2
(c)
1
k 1 x i1 + x a
2
b
1
k 2 − xi 2 + x a
2
b
g
2
g
2
− xi21 =
− xi22 =
e
(a)
j
j
Before the horizontal force is applied, the springs exert equal forces: k 1 xi1 = k 2 xi 2
k1 xi1
k2
1
1
k1 x a2 + k1 x a xi1 + k 2 x a2 − k 2 x a xi 2
2
2
k x
1
1
2
2
= k 1 x a + k 2 x a + k1 x a xi1 − k 2 x a 1 i1
2
2
k2
1
= k1 + k 2 x a2
2
W1 + W2 =
b
*P7.57
e
1
k 2 x a2 − 2 x a xi 2
2
xi 2 =
(d)
1
k1 x a2 + 2 x a xi1
2
g
z ze
t
t
0
0
v = a dt =
j
1.16t − 0.21t 2 + 0.24t 3 dt
t2
t3
t4
= 1.16 − 0.21 + 0. 24
2
3
4
t
= 0.58t 2 − 0.07t 3 + 0.06t 4
0
At t = 0 , vi = 0. At t = 2.5 s ,
ja f − e0.07 m s ja2.5 sf + e0.06 m s ja2.5 sf
e
v f = 0.58 m s 3 2.5 s
2
3
4
5
4
= 4.88 m s
Ki + W = K f
0+W =
(b)
b
1
1
mv 2f = 1 160 kg 4.88 m s
2
2
g
2
= 1.38 × 10 4 J
At t = 2.5 s ,
e
j
e
ja f + e0.240 m s ja2.5 sf
a = 1.16 m s 3 2.5 s − 0.210 m s 4 2.5 s
2
5
3
Through the axles the wheels exert on the chassis force
∑ F = ma = 1 160 kg 5.34
m s 2 = 6.19 × 10 3 N
and inject power
b
g
P = Fv = 6.19 × 10 3 N 4.88 m s = 3.02 × 10 4 W .
= 5.34 m s 2 .
Chapter 7
P7.58
(a)
The new length of each spring is
x 2 + L2 , so its extension is
x 2 + L2 − L and the force it exerts is k
FH
IK
x 2 + L2 − L toward its
fixed end. The y components of the two spring forces add to
zero. Their x components add to
F = −2 ik
FH
x 2 + L2 − L
IK
x
x 2 + L2
F
GH
= −2 kx i 1 −
z
L
x 2 + L2
z
f
(b)
0
W = Fx dx
FIG. P7.58
.
F
GH
I dx
J
x +L K
ex + L j
+ kL
b1 2g
W = −2 kx 1 −
i
A
z
0
W = −2 k x dx + kL
A
ze
0
A
j
2 −1 2
x2 + L
x2
W = −2 k
2
2 x dx
0
A
L
2
2
2
0
2 12
A
W = 2 kL2 + kA 2 − 2 kL A 2 + L2
W = −0 + kA 2 + 2 kL2 − 2 kL A 2 + L2
*P7.59
I
JK
For the rocket falling at terminal speed we have
∑ F = ma
+ R − Mg = 0
Mg =
(a)
1
DρAvT2
2
For the rocket with engine exerting thrust T and flying up at the same speed,
∑ F = ma
+T − Mg − R = 0
T = 2 Mg
The engine power is P = Fv = TvT = 2 MgvT .
(b)
For the rocket with engine exerting thrust Tb and flying down steadily at 3vT ,
1
2
Rb = DρA 3 vT = 9 Mg
2
b g
∑ F = ma
−Tb − Mg + 9 Mg = 0
Tb = 8 Mg
The engine power is P = Tv = 8 Mg 3 vT = 24MgvT .
209
210
P7.60
Energy and Energy Transfer
(a)
a fe
j e20.5i + 14.3 jj N
= a 42.0 N fecos 150° i + sin 150° jj = e−36.4i + 21.0 jj N
F1 = 25.0 N cos 35.0° i + sin 35.0° j =
F2
e−15.9 i + 35.3 jj N
(b)
∑ F = F1 + F2 =
(c)
a=
(d)
v f = v i + at = 4.00 i + 2.50 j m s + −3.18 i + 7.07 j m s 2 3.00 s
∑F =
e−3.18 i + 7.07 jj m s
e
vf =
(e)
m
2
j
e
ja
je
e−5.54i + 23.7 jj m s
r f = ri + v i t +
e
1 2
at
2
f 12 e−3.18i + 7.07 jjem s ja3.00 sf
jb ga
e−2.30i + 39.3 jj m
2
r f = 0 + 4.00 i + 2.50 j m s 3.00 s +
∆r = r f =
(f)
Kf =
(g)
Kf =
b
1
1
mv 2f = 5.00 kg
2
2
(a)
g a5.54f + a23.7f em s j =
2
1
mvi2 + ∑ F ⋅ ∆r
2
1
2
K f = 5.00 kg 4.00 + 2.50
2
K f = 55.6 J + 1 426 J = 1.48 kJ
b
P7.61
f
2
2
2
1.48 kJ
g a f a f bm sg + a−15.9 Nfa−2.30 mf + a35.3 Nfa39.3 mf
∑ W = ∆K :
2
2
Ws + W g = 0
a
f
j a
f a
1 2
kxi − 0 + mg∆x cos 90°+60° = 0
2
1
2
1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x = 0
2
∆x = 4.12 m
e
(b)
∑ W = ∆K + ∆Eint :
fa fa
f
Ws + W g − ∆Eint = 0
1 2
kxi + mg∆x cos 150°− µ k mg cos 60° ∆x = 0
2
1
2
1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x − 0.200 9.80 0.400 cos 60.0° ∆x = 0
2
∆x = 3.35 m
e
j a
f a
fa fa
f
a
fa fa
fa
f
Chapter 7
P7.62
(a)
211
a f Lammf FaNf Lammf
FN
2.00
15.0
14.0
112
4.00
32.0
16.0
126
6.00
49.0
18.0
149
8.00
64.0
20.0
175
10.0
79.0
22.0
190
12.0
98.0
FIG. P7.62
(b)
A straight line fits the first eight points, together with the origin. By least-square fitting, its
slope is
0.125 N mm ± 2% = 125 N m ± 2%
In F = kx , the spring constant is k =
(c)
P7.63
b
F
, the same as the slope of the F-versus-x graph.
x
f
ga
F = kx = 125 N m 0.105 m = 13.1 N
K i + Ws + W g = K f
1
1
1
1
mvi2 + kx i2 − kx 2f + mg∆x cos θ = mv 2f
2
2
2
2
1 2
1
FIG. P7.63
0 + kxi − 0 + mgxi cos 100° = mv 2f
2
2
1
1
1.20 N cm 5.00 cm 0.050 0 m − 0.100 kg 9.80 m s 2 0.050 0 m sin 10.0° = 0.100 kg v 2
2
2
−3
2
0.150 J − 8.51 × 10 J = 0.050 0 kg v
b
v=
P7.64
ga
fb
b
g b
g
ge
jb
g
b
0.141
= 1.68 m s
0.050 0
b
gea6.00f − a8.00f jbm sg
b
ge
1
1
m v 2f − vi2 : ∆Eint = − 0.400 kg
2
2
e
j
2
2
(a)
∆Eint = − ∆K = −
(b)
∆Eint = f∆r = µ k mg 2πr :
5.60 J = µ k 0.400 kg 9.80 m s 2 2π 1.50 m
Thus,
µ k = 0.152 .
(c)
g
a f
j a
2
= 5.60 J
f
After N revolutions, the object comes to rest and K f = 0 .
1
mvi2
2
Thus,
∆Eint = − ∆K = −0 + K i =
or
µ k mg N 2πr =
This gives
b8.00 m sg
=
=
N=
µ mg a 2πr f a0.152fe9.80 m s j2π a1.50 mf
a f
1
2
k
mvi2
1
mvi2 .
2
2
1
2
2
2.28 rev .
212
P7.65
Energy and Energy Transfer
If positive F represents an outward force, (same as direction as r), then
z
ze
rf
f
W = F ⋅ dr =
i
j
2 F0σ 13 r −13 − F0σ 7 r −7 dr
ri
2 F σ 13 r −12 F0σ 7 r −6
−
W= 0
−12
−6
W=
− F0σ
13
e
r f−12
− ri−12
6
W = 1.03 × 10
−77
r f−6
rf
ri
j + F σ er
− ri−6
0
7
−6
f
− ri−6
6
j= Fσ
0
6
− 1.89 × 10
−134
r f−12
7
r f−6 − ri−6 −
F0 σ 13 −12
r f − ri−12
6
− ri−12
W = 1.03 × 10 −77 1.88 × 10 −6 − 2.44 × 10 −6 10 60 − 1.89 × 10 −134 3.54 × 10 −12 − 5.96 × 10 −8 10 120
W = −2.49 × 10 −21 J + 1.12 × 10 −21 J = −1.37 × 10 −21 J
P7.66
P∆t = W = ∆K =
a ∆m f v
2
2
ρ=
The density is
∆m ∆m
=
.
vol A∆x
Substituting this into the first equation and solving for P , since
3
for a constant speed, we get
P=
ρAv
2
Also, since P = Fv,
F=
ρAv 2
.
2
∆x
= v,
∆t
FIG. P7.66
.
Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the
drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it.
For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a
streamlined object and can be greater than one if the airflow around the object is complicated.
z
23 .7
P7.67
We evaluate
375dx
by calculating
12.8 x + 3.75 x
3
a
f + 375a0.100f + … 375a0.100f = 0.806
a12.8f + 3.75a12.8f a12.9f + 3.75a12.9f a23.6f + 3.75a23.6f
375 0.100
3
3
3
and
a
f + 375a0.100f + … 375a0.100f = 0.791 .
a12.9f + 3.75a12.9f a13.0f + 3.75a13.0f a23.7f + 3.75a23.7f
375 0.100
3
3
3
The answer must be between these two values. We may find it more precisely by using a value for
∆x smaller than 0.100. Thus, we find the integral to be 0.799 N ⋅ m .
Chapter 7
*P7.68
P=
1
Dρπr 2 v 3
2
ja
f b8 m sg
1
1 1.20 kg m3 π 1.5 m
2
e
(a)
Pa =
(b)
24 m s
Pb v b3
= 3 =
Pa v a
8 ms
e
F
GH
I
JK
2
3
= 2.17 × 10 3 W
3
= 3 3 = 27
j
Pb = 27 2.17 × 10 3 W = 5.86 × 10 4 W
P7.69
(a)
The suggested equation P∆t = bwd implies all of the
following cases:
(1)
(3)
FG w IJ a2df
H 2K
F ∆t I F d I
P G J = bwG J
H 2 K H 2K
P∆t = b
(2)
and
(4)
FG ∆t IJ = bFG w IJ d
H 2 K H 2K
FG P IJ ∆t = bFG w IJ d
H 2K H 2K
P
v = constant
n
d
fk = µ k n
F
w
These are all of the proportionalities Aristotle lists.
FIG. P7.69
(b)
For one example, consider a horizontal force F pushing an object of weight w at constant
velocity across a horizontal floor with which the object has coefficient of friction µ k .
∑ F = ma implies that:
+n − w = 0 and F − µ k n = 0
so that F = µ k w
As the object moves a distance d, the agent exerting the force does work
W = Fd cos θ = Fd cos 0° = µ k wd and puts out power P =
W
∆t
This yields the equation P∆t = µ k wd which represents Aristotle’s theory with b = µ k .
Our theory is more general than Aristotle’s. Ours can also describe accelerated motion.
*P7.70
(a)
So long as the spring force is greater than the friction force,
the block will be gaining speed. The block slows down when
the friction force becomes the greater. It has maximum
speed when kx a − f k = ma = 0.
e1.0 × 10
(b)
3
j
N m x a − 4.0 N = 0
0
x = −4.0 × 10 −3 m
0
By the same logic,
e1.0 × 10
3
j
N m x b − 10.0 N = 0
x = −1.0 × 10 −2 m
FIG. P7.70
213
214
Energy and Energy Transfer
ANSWERS TO EVEN PROBLEMS
P7.2
1.59 × 10 3 J
P7.44
5.92 km L
P7.4
(a) 3.28 × 10 −2 J ; (b) −3.28 × 10 −2 J
P7.46
90.0 J
P7.6
see the solution
P7.8
5.33 W
P7.10
16.0
P7.12
(a) see the solution; (b) −12.0 J
P7.48
Ay
Ax
A
; cos β =
; cos γ = z ;
A
A
A
(b) see the solution
(a) cos α =
P7.50
(a) a =
P7.52
(a)
40.1 kN
; b = 1.80 ; (b) 294 J
m 1.8
mgnhhs
mgvh
; (b)
v + nhs
v + nhs
P7.14
50.0 J
P7.16
(a) 575 N m ; (b) 46.0 J
P7.54
1.53 × 10 5 N upward
P7.18
(a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6%
P7.56
see the solution
P7.20
(a) see the solution; (b) mgR
P7.58
(a) see the solution;
P7.22
mg mg
1
1
+
+
; (b)
(a)
k1 k 2
k1
k2
FG
H
IJ
K
(b) 2 kL2 + kA 2 − 2 kL A 2 + L2
−1
P7.24
(a) 1.20 J; (b) 5.00 m s ; (c) 6.30 J
P7.26
(a) 60.0 J; (b) 60.0 J
P7.60
e
j
(a) F1 = 20.5 i + 14.3 j N ;
e
j
F2 = −36.4i + 21.0 j N ;
e
j
(c) e −3.18 i + 7.07 jj m s ;
(d) e −5.54i + 23.7 jj m s ;
(e) e −2.30 i + 39.3 jj m ; (f) 1.48 kJ; (g) 1.48 kJ
(b) −15.9 i + 35.3 j N ;
2
P7.28
(a) 1.94 m s ; (b) 3.35 m s ; (c) 3.87 m s
P7.30
(a) 3.78 × 10 −16 J ; (b) 1.35 × 10 −14 N ;
(c) 1.48 × 10 +16 m s 2 ; (d) 1.94 ns
P7.32
(a) 0.791 m s; (b) 0.531 m s
P7.34
(a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 J
P7.36
8.01 W
P7.38
~ 10 4 W
P7.40
(a) 5.91 kW; (b) 11.1 kW
P7.42
No. (a) 582; (b) 90.5 W = 0.121 hp
P7.62
(a) see the solution; (b) 125 N m ± 2% ;
(c) 13.1 N
P7.64
(a) 5.60 J; (b) 0.152; (c) 2.28 rev
P7.66
see the solution
P7.68
(a) 2.17 kW; (b) 58.6 kW
P7.70
(a) x = −4.0 mm ; (b) −1.0 cm
8
Potential Energy
CHAPTER OUTLINE
8.1
8.2
8.3
8.4
8.5
8.6
Q8.4
Potential Energy of a System
The Isolated
System—Conservation of
Mechanical Energy
Conservative and
Nonconservative Forces
Changes in Mechanical
Energy for Nonconservative
Forces
Relationship Between
Conservative Forces and
Potential Energy
Energy Diagrams and the
Equilibrium of a System
ANSWERS TO QUESTIONS
Q8.1
The final speed of the children will not depend on the slide
length or the presence of bumps if there is no friction. If there is
friction, a longer slide will result in a lower final speed. Bumps
will have the same effect as they effectively lengthen the
distance over which friction can do work, to decrease the total
mechanical energy of the children.
Q8.2
Total energy is the sum of kinetic and potential energies.
Potential energy can be negative, so the sum of kinetic plus
potential can also be negative.
Q8.3
Both agree on the change in potential energy, and the kinetic
energy. They may disagree on the value of gravitational
potential energy, depending on their choice of a zero point.
(a)
mgh is provided by the muscles.
(b)
No further energy is supplied to the object-Earth system, but some chemical energy must be
supplied to the muscles as they keep the weight aloft.
(c)
The object loses energy mgh, giving it back to the muscles, where most of it becomes internal
energy.
Q8.5
Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero,
but the book-Earth system increases in gravitational potential energy. Stretch a rubber band to
encompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearl
drift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo)
increases in internal energy.
Q8.6
Three potential energy terms will appear in the expression of total mechanical energy, one for each
conservative force. If you write an equation with initial energy on one side and final energy on the
other, the equation contains six potential-energy terms.
215
216
Q8.7
Potential Energy
(a)
It does if it makes the object’s speed change, but not if it only makes the direction of the
velocity change.
(b)
Yes, according to Newton’s second law.
Q8.8
The original kinetic energy of the skidding can be degraded into kinetic energy of random molecular
motion in the tires and the road: it is internal energy. If the brakes are used properly, the same
energy appears as internal energy in the brake shoes and drums.
Q8.9
All the energy is supplied by foodstuffs that gained their energy from the sun.
Q8.10
Elastic potential energy of plates under stress plus gravitational energy is released when the plates
“slip”. It is carried away by mechanical waves.
Q8.11
The total energy of the ball-Earth system is conserved. Since the system initially has gravitational
energy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to its
original position. Air resistance will cause the ball to come back to a point slightly below its initial
position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the
demonstrator will have to duck.
Q8.12
Using switchbacks requires no less work, as it does not change the change in potential energy from
top to bottom. It does, however, require less force (of static friction on the rolling drive wheels of a
car) to propel the car up the gentler slope. Less power is required if the work can be done over a
longer period of time.
Q8.13
There is no work done since there is no change in kinetic energy. In this case, air resistance must be
negligible since the acceleration is zero.
Q8.14
There is no violation. Choose the book as the system. You did work and the earth did work on the
book. The average force you exerted just counterbalanced the weight of the book. The total work on
the book is zero, and is equal to its overall change in kinetic energy.
Q8.15
Kinetic energy is greatest at the starting point. Gravitational energy is a maximum at the top of the
flight of the ball.
Q8.16
Gravitational energy is proportional to mass, so it doubles.
Q8.17
In stirring cake batter and in weightlifting, your body returns to the same conformation after each
stroke. During each stroke chemical energy is irreversibly converted into output work (and internal
energy). This observation proves that muscular forces are nonconservative.
Chapter 8
Q8.18
Let the gravitational energy be zero at the lowest point in the
motion. If you start the vibration by pushing down on the block (2),
its kinetic energy becomes extra elastic potential energy in the
spring ( Us ). After the block starts moving up at its lower turning
point (3), this energy becomes both kinetic energy (K) and
gravitational potential energy ( U g ), and then just gravitational
energy when the block is at its greatest height (1). The energy then
turns back into kinetic and elastic potential energy, and the cycle
repeats.
Q8.19
217
FIG. Q8.18
(a)
Kinetic energy of the running athlete is transformed into elastic potential energy of the bent
pole. This potential energy is transformed to a combination of kinetic energy and
gravitational potential energy of the athlete and pole as the athlete approaches the bar. The
energy is then all gravitational potential of the pole and the athlete as the athlete hopefully
clears the bar. This potential energy then turns to kinetic energy as the athlete and pole fall
to the ground. It immediately becomes internal energy as their macroscopic motion stops.
(b)
Rotational kinetic energy of the athlete and shot is transformed into translational kinetic
energy of the shot. As the shot goes through its trajectory as a projectile, the kinetic energy
turns to a mix of kinetic and gravitational potential. The energy becomes internal energy as
the shot comes to rest.
(c)
Kinetic energy of the running athlete is transformed to a mix of kinetic and gravitational
potential as the athlete becomes projectile going over a bar. This energy turns back into
kinetic as the athlete falls down, and becomes internal energy as he stops on the ground.
The ultimate source of energy for all of these sports is the sun. See question 9.
Q8.20
Chemical energy in the fuel turns into internal energy as the fuel burns. Most of this leaves the car
by heat through the walls of the engine and by matter transfer in the exhaust gases. Some leaves the
system of fuel by work done to push down the piston. Of this work, a little results in internal energy
in the bearings and gears, but most becomes work done on the air to push it aside. The work on the
air immediately turns into internal energy in the air. If you use the windshield wipers, you take
energy from the crankshaft and turn it into extra internal energy in the glass and wiper blades and
wiper-motor coils. If you turn on the air conditioner, your end effect is to put extra energy out into
the surroundings. You must apply the brakes at the end of your trip. As soon as the sound of the
engine has died away, all you have to show for it is thermal pollution.
Q8.21
A graph of potential energy versus position is a straight horizontal line for a particle in neutral
equilibrium. The graph represents a constant function.
Q8.22
The ball is in neutral equilibrium.
Q8.23
The ball is in stable equilibrium when it is directly below the pivot point. The ball is in unstable
equilibrium when it is vertically above the pivot.
218
Potential Energy
SOLUTIONS TO PROBLEMS
Section 8.1
Potential Energy of a System
P8.1
With our choice for the zero level for potential energy when the car
is at point B,
(a)
UB = 0 .
When the car is at point A, the potential energy of the car-Earth
system is given by
FIG. P8.1
U A = mgy
where y is the vertical height above zero level. With 135 ft = 41.1 m , this height is found as:
a
f
y = 41.1 m sin 40.0° = 26.4 m .
Thus,
b
ge
ja
f
U A = 1 000 kg 9.80 m s 2 26.4 m = 2.59 × 10 5 J .
The change in potential energy as the car moves from A to B is
U B − U A = 0 − 2.59 × 10 5 J = −2.59 × 10 5 J .
(b)
With our choice of the zero level when the car is at point A, we have U A = 0 . The potential
energy when the car is at point B is given by U B = mgy where y is the vertical distance of
point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m.
Because this distance is now below the zero reference level, it is a negative number.
Thus,
b
ge
ja
f
U B = 1 000 kg 9.80 m s 2 −26.5 m = −2.59 × 10 5 J .
The change in potential energy when the car moves from A to B is
U B − U A = −2.59 × 10 5 J − 0 = −2.59 × 10 5 J .
Chapter 8
P8.2
(a)
219
We take the zero configuration of system
potential energy with the child at the
lowest point of the arc. When the string
is held horizontal initially, the initial
position is 2.00 m above the zero level.
Thus,
a
fa
f
U g = mgy = 400 N 2.00 m = 800 J .
(b)
From the sketch, we see that at an angle
of 30.0° the child is at a vertical height of
2.00 m 1 − cos 30.0° above the lowest
point of the arc. Thus,
a
fa
f
a
FIG. P8.2
fa
fa
f
U g = mgy = 400 N 2.00 m 1 − cos 30.0° = 107 J .
(c)
The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at this
location.
*P8.3
The volume flow rate is the volume of water going over the falls each second:
a
fb
g
3 m 0.5 m 1.2 m s = 1.8 m3 s
The mass flow rate is
m
V
= ρ = 1 000 kg m3 1.8 m3 s = 1 800 kg s
t
t
e
je
j
If the stream has uniform width and depth, the speed of the water below the falls is the same as the
speed above the falls. Then no kinetic energy, but only gravitational energy is available for
conversion into internal and electric energy.
ja f
energy mgy m
=
= gy = 1 800 kg s 9.8 m s 2 5 m = 8.82 × 10 4 J s
t
t
t
The output power is Puseful = efficiency Pin = 0.25 8.82 × 10 4 W = 2.20 × 10 4 W
The input power is Pin =
b
g
b
e
ge
j
The efficiency of electric generation at Hoover Dam is about 85%, with a head of water (vertical
drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators.
Section 8.2
*P8.4
(a)
The Isolated System—Conservation of Mechanical Energy
One child in one jump converts chemical energy into mechanical energy in the amount that
her body has as gravitational energy at the top of her jump:
mgy = 36 kg 9.81 m s 2 0.25 m = 88.3 J . For all of the jumps of the children the energy is
e
ja f
12e1.05 × 10 j88.3 J = 1.11 × 10 J .
6
(b)
9
0.01
1.11 × 10 9 J = 1.11 × 10 5 J , making the Richter
100
log E − 4.8 log 1.11 × 10 5 − 4.8 5.05 − 4.8
magnitude
=
=
= 0.2 .
1.5
1.5
1.5
The seismic energy is modeled as E =
220
P8.5
Potential Energy
a f 12 mv
1
g a3.50 Rf = 2 g a Rf + v
2
Ui + K i = U f + K f :
mgh + 0 = mg 2 R +
2
2
v = 3.00 gR
∑F = m
2
v
:
R
n + mg = m
v2
R
LM v − g OP = m L 3.00 gR − g O = 2.00mg
N R Q MN R PQ
n = 2.00e5.00 × 10 kg je9.80 m s j
2
n=m
−3
2
= 0.098 0 N downward
P8.6
FIG. P8.5
K i + Ui = K f + U f
1
2
m 6.00 m s + 0 = 0 + m 9.80 m s 2 y
2
From leaving ground to the highest point,
b
g
e
b6.00 m sg =
∴y =
a2fe9.80 m s j
j
2
The mass makes no difference:
*P8.7
(a)
2
1.84 m
1
1
1
1
mvi2 + kx i2 = mv 2f + kx 2f
2
2
2
2
1
1
2
0 + 10 N m −0.18 m = 0.15 kg v 2f + 0
2
2
f b
ga
g
F 10 N I FG 1 kg ⋅ m IJ =
= a0.18 mf G
H 0.15 kg ⋅ m JK H 1 N ⋅ s K
b
vf
(b)
2
K i + U si = K f + U sf
1
0 + 10 N m −0.18 m
2
b
0.162 J =
vf =
f
ga
a
b
g
2
b
1
0.15 kg v 2f
2
1
+ 10 N m 0. 25 m − 0.18 m
2
=
b
1
0.15 kg v 2f + 0.024 5 J
2
f
g
ga
1.47 m s
2 0.138 J
= 1.35 m s
0.15 kg
f
2
FIG. P8.7
Chapter 8
*P8.8
The energy of the car is E =
221
1
mv 2 + mgy
2
1
mv 2 + mgd sin θ where d is the distance it has moved along the track.
2
dE
dv
P=
= mv
+ mgv sin θ
dt
dt
E=
(a)
When speed is constant,
jb
e
g
P = mgv sin θ = 950 kg 9.80 m s 2 2.20 m s sin 30° = 1.02 × 10 4 W
(b)
2. 2 m s − 0
dv
=a=
= 0.183 m s 2
dt
12 s
Maximum power is injected just before maximum speed is attained:
b
ge
j
P = mva + mgv sin θ = 950 kg 2.2 m s 0.183 m s 2 + 1.02 × 10 4 W = 1.06 × 10 4 W
(c)
*P8.9
(a)
At the top end,
1
1
2.20 m s
mv 2 + mgd sin θ = 950 kg
2
2
FG b
H
g + e9.80 m s j1 250 m sin 30°IJK =
2
2
5.82 × 10 6 J
Energy of the object-Earth system is conserved as the object moves between the release
point and the lowest point. We choose to measure heights from y = 0 at the top end of the
string.
eK + U j = eK + U j :
g
g
i
0 + mgyi =
f
1
mv 2f + mgy f
2
e9.8 m s ja−2 m cos 30°f = 12 v + e9.8 m s ja−2 mf
v = 2e9.8 m s ja 2 mfa1 − cos 30°f = 2.29 m s
2
2
f
2
f
(b)
Choose the initial point at θ = 30° and the final point at θ = 15° :
a
f
f
2 gLacos 15°− cos 30°f = 2e9.8 m s ja 2 mfacos 15°− cos 30°f =
0 + mg − L cos 30° =
vf =
P8.10
2
a
1
mv 2f + mg − L cos 15°
2
2
1.98 m s
Choose the zero point of gravitational potential energy of the object-spring-Earth system as the
configuration in which the object comes to rest. Then because the incline is frictionless, we have
EB = E A :
K B + U gB + U sB = K A + U gA + U sA
a
f
or
0 + mg d + x sin θ + 0 = 0 + 0 +
Solving for d gives
d=
kx 2
−x .
2mg sin θ
1 2
kx .
2
222
P8.11
Potential Energy
From conservation of energy for the block-spring-Earth system,
U gt = U si ,
or
b0.250 kg ge9.80 m s jh = FGH 12 IJK b5 000 N mga0.100 mf
2
2
This gives a maximum height h = 10.2 m .
P8.12
(a)
FIG. P8.11
The force needed to hang on is equal to the force F the
trapeze bar exerts on the performer.
From the free-body diagram for the performer’s body, as
shown,
F − mg cos θ = m
v2
A
F = mg cos θ + m
v2
A
or
FIG. P8.12
Apply conservation of mechanical energy of the performer-Earth system as the performer
moves between the starting point and any later point:
b
g
a
f
mg A − A cos θ i = mg A − A cos θ +
Solve for
(b)
1
mv 2
2
mv 2
and substitute into the force equation to obtain F = mg 3 cos θ − 2 cos θ i
A
b
At the bottom of the swing, θ = 0° so
b
F = mg 3 − 2 cos θ i
b
g
F = 2mg = mg 3 − 2 cos θ i
which gives
θ i = 60.0° .
g
g
.
223
Chapter 8
P8.13
Using conservation of energy for the system of the Earth and the two objects
(a)
b5.00 kg gga4.00 mf = b3.00 kg gga4.00 mf + 12 a5.00 + 3.00fv
2
v = 19.6 = 4.43 m s
(b)
Now we apply conservation of energy for the system of the 3.00 kg
object and the Earth during the time interval between the instant
when the string goes slack and the instant at which the 3.00 kg
object reaches its highest position in its free fall.
FIG. P8.13
a f
1
3.00 v 2 = mg ∆y = 3.00 g∆y
2
∆y = 1.00 m
y max = 4.00 m + ∆y = 5.00 m
P8.14
m1 > m 2
(a)
m1 gh =
v=
(b)
b
g
1
m1 + m 2 v 2 + m 2 gh
2
b
bm
g
g
2 m1 − m 2 gh
1 + m2
Since m 2 has kinetic energy
1
m 2 v 2 , it will rise an additional height ∆h determined from
2
m 2 g ∆h =
1
m2 v 2
2
or from (a),
∆h =
The total height m 2 reaches is h + ∆h =
P8.15
b
b
g
g
m1 − m 2 h
v2
=
m1 + m 2
2g
2m1 h
.
m1 + m 2
The force of tension and subsequent force of compression in the
rod do no work on the ball, since they are perpendicular to each
step of displacement. Consider energy conservation of the ballEarth system between the instant just after you strike the ball and
the instant when it reaches the top. The speed at the top is zero if
you hit it just hard enough to get it there.
K i + U gi = K f + U gf :
a f
a fa f
1
mvi2 + 0 = 0 + mg 2L
2
vi = 4 gL = 4 9.80 0.770
vi = 5.49 m s
initial
final
L
vi
FIG. P8.15
L
224
*P8.16
Potential Energy
efficiency =
e=
useful output energy useful output power
=
total input energy
total input power
m water gy t
b1 2gm ev tj
air
2
=
b
g
e Av t j
2 ρ water v water t gy
ρ air πr
2
b
g
2 ρ w v w t gy
=
2
2 3
ρ a πr v
where A is the length of a cylinder of air passing through the mill and v w is the volume of water
pumped in time t. We need inject negligible kinetic energy into the water because it starts and ends
at rest.
e
ja fb g
e
je
j
F 1 000 L IJ FG 60 s IJ = 160 L min
sG
H 1 m K H 1 min K
2
3
v w eρ a πr 2 v 3 0.275 1.20 kg m π 1.15 m 11 m s
=
=
2 ρ w gy
t
2 1 000 kg m 3 9.80 m s 2 35 m
= 2.66 × 10 −3 m3
P8.17
(a)
3
K i + U gi = K f + U gf
1
1
mvi2 + 0 = mv 2f + mgy f
2
2
1
1
1
2
2
2
mv xi + mv yi
= mv xf
+ mgy f
2
2
2
But v xi = v xf , so for the first ball
yf =
2
v yi
2g
=
b1 000 sin 37.0°g
2a9.80f
2
= 1.85 × 10 4 m
and for the second
b1 000g =
=
2a9.80f
2
yf
(b)
5.10 × 10 4 m
The total energy of each is constant with value
b
gb
1
20.0 kg 1 000 m s
2
g
2
= 1.00 × 10 7 J .
3
Chapter 8
P8.18
In the swing down to the breaking point, energy is conserved:
mgr cos θ =
1
mv 2
2
at the breaking point consider radial forces
∑ Fr = mar
+Tmax − mg cos θ = m
Eliminate
v2
r
v2
= 2 g cos θ
r
Tmax − mg cos θ = 2mg cos θ
Tmax = 3mg cos θ
θ = cos −1
F T I = cos
GH 3mg JK
max
−1
F
I
GG 3 2.00 kg44.59.N80 m s JJ
ge
jK
Hb
2
θ = 40.8°
*P8.19
(a)
For a 5-m cord the spring constant is described by F = kx ,
mg = k 1.5 m . For a longer cord of length L the stretch distance
is longer so the spring constant is smaller in inverse proportion:
a
f
k=
5 m mg
= 3.33 mg L
L 1.5 m
eK + U
g
+ Us
j = eK + U
i
g
+ Us
j
initial
f
1
0 + mgyi + 0 = 0 + mgy f + kx 2f
2
mg 2
1 2 1
mg yi − y f = kx f = 3.33
xf
2
2
L
d
FIG. P8.19(a)
i
here yi − y f = 55 m = L + x f
1
2
55.0 mL = 3.33 55.0 m − L
2
55.0 mL = 5.04 × 10 3 m 2 − 183 mL + 1.67 L2
a
f
0 = 1.67L2 − 238 L + 5.04 × 10 3 = 0
L=
a fe
2a1.67f
238 ± 238 2 − 4 1.67 5.04 × 10 3
j = 238 ± 152 =
3.33
only the value of L less than 55 m is physical.
(b)
mg
25.8 m
∑ F = ma
k = 3.33
x max = x f = 55.0 m − 25.8 m = 29.2 m
+ kx max − mg = ma
mg
3.33
29. 2 m − mg = ma
25.8 m
a = 2.77 g = 27.1 m s 2
final
25.8 m
225
226
*P8.20
Potential Energy
When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm.
v
h
We then choose the final point to be when B has moved up by and has speed A . Then A has
3
2
2h
moved down
and has speed v A :
3
eK
A
+ KB + Ug
0+0+0=
j = eK
i
A
+ KB + Ug
FG IJ
H K
v
1
1
mv A2 + m A
2
2
2
2
+
j
f
mgh mg 2 h
−
3
3
mgh 5
= mv A2
3
8
8 gh
15
vA =
Section 8.3
P8.21
Conservative and Nonconservative Forces
b
ge
j
y
Fg = mg = 4.00 kg 9.80 m s 2 = 39.2 N
(a)
Work along OAC = work along OA + work along AC
= Fg OA cos 90.0°+ Fg AC cos 180°
a f
a f
= a39.2 N fa5.00 mf + a39.2 N fa5.00 mfa −1f
B
C
(5.00, 5.00) m
O
A
= −196 J
(b)
W along OBC = W along OB + W along BC
= 39.2 N 5.00 m cos 180°+ 39.2 N 5.00 m cos 90.0°
a
fa
f
a
fa
f
FIG. P8.21
= −196 J
(c)
a f
F 1 IJ =
2 mjG −
H 2K
Work along OC = Fg OC cos135°
a
fe
= 39.2 N 5.00 ×
−196 J
The results should all be the same, since gravitational forces are conservative.
P8.22
(a)
z
W = F ⋅ dr and if the force is constant, this can be written as
z
d
i
W = F ⋅ dr = F ⋅ r f − ri , which depends only on end points, not path.
(b)
z
W = F ⋅ dr =
a
f
ze
je
j a f z dx + a4.00 Nf z dy
+ a 4.00 N fy
= 15.0 J + 20.0 J = 35.0 J
3 i + 4j ⋅ dx i + dyj = 3.00 N
5.00 m
W = 3.00 N x 0
5.00 m
0
The same calculation applies for all paths.
5.00 m
5.00 m
0
0
x
Chapter 8
P8.23
z
dx i ⋅ 2 y i + x 2 j =
z
dyj ⋅ 2 y i + x 2 j =
5.00 m
WOA =
(a)
0
z
2 ydx
z
x dy
z
x dy
z
2 ydx
j
5.00 m
j
5 .00 m
2
e
j
5.00 m
2
e
j
5.00 m
e
227
0
WOA = 0
and since along this path, y = 0
5 .00 m
W AC =
0
e
For x = 5.00 m,
W AC = 125 J
and
WOAC = 0 + 125 = 125 J
WOB =
(b)
z
dyj ⋅ 2 y i + x 2 j =
z
dx i ⋅ 2 yi + x 2 j =
5 .00 m
0
0
0
WOB = 0
since along this path, x = 0 ,
WBC =
5.00 m
0
0
WBC = 50.0 J
since y = 5.00 m,
WOBC = 0 + 50.0 = 50.0 J
WOC =
(c)
WOC =
Since x = y along OC,
ze
ze
je
5.00 m
j
(d)
F is nonconservative since the work done is path dependent.
(a)
a ∆K f
a
= ∑ W = W g = mg∆h = mg 5.00 − 3.20
1
1
mv B2 − mv 2A = m 9.80 1.80
2
2
v B = 5.94 m s
A→B
f
a
(b)
Wg
A →C
A
a fa f
f
Similarly, vC = v A2 + 2 g 5.00 − 2.00 = 7.67 m s
a
f
= mg 3.00 m = 147 J
2
2 x + x 2 dx = 66.7 J
0
P8.24
j z e2ydx + x dyj
dx i + dyj ⋅ 2 y i + x 2 j =
B
5.00 m
3.20 m
FIG. P8.24
C
2.00 m
228
P8.25
Potential Energy
(a)
e
j
F = 3.00 i + 5.00 j N
m = 4.00 kg
e
j
r = 2.00 i − 3.00 j m
a f
a
f
W = 3.00 2.00 + 5.00 −3.00 = −9.00 J
The result does not depend on the path since the force is conservative.
(b)
W = ∆K
Fa f I
GH
JK
4.00
4.00 v 2
−9.00 =
− 4.00
2
2
so v =
(c)
2
32.0 − 9.00
= 3.39 m s
2.00
∆U = −W = 9.00 J
Section 8.4
Changes in Mechanical Energy for Nonconservative Forces
P8.26
U f = K i − K f + Ui
(a)
U f = 30.0 − 18.0 + 10.0 = 22.0 J
E = 40.0 J
(b)
P8.27
Yes, ∆Emech = ∆K + ∆U is not equal to zero. For conservative forces ∆K + ∆U = 0 .
The distance traveled by the ball from the top of the arc to the bottom is πR . The work done by the
non-conservative force, the force exerted by the pitcher,
is
a f
∆E = F∆r cos 0° = F πR .
We shall assign the gravitational energy of the ball-Earth system to be zero with the ball at the
bottom of the arc.
Then
becomes
or
1
1
mv 2f − mvi2 + mgy f − mgyi
2
2
1
1
mv 2f = mvi2 + mgyi + F πR
2
2
∆Emech =
a f
v f = vi2 + 2 gyi +
a f a15.0f + 2a9.80fa1.20f + 2a30.0fπ a0.600f
0.250
2 F πR
=
m
2
v f = 26.5 m s
*P8.28
The useful output energy is
a
f
d
i
120 Wh 1 − 0.60 = mg y f − yi = Fg ∆y
∆y =
b
g FG J IJ FG N ⋅ m IJ =
H W ⋅sKH J K
120 W 3 600 s 0.40
890 N
194 m
Chapter 8
*P8.29
229
As the locomotive moves up the hill at constant speed, its output power goes into internal energy
plus gravitational energy of the locomotive-Earth system:
Pt = mgy + f∆r = mg∆r sin θ + f∆r
P = mgv f sin θ + fv f
As the locomotive moves on level track,
F 746 W I = f b27 m sg
f = 2.76 × 10 N
GH 1 hp JK
F 5 m IJ + e2.76 × 10 Njv
Then also 746 000 W = b160 000 kg ge9.8 m s jv G
H 100 m K
P = fvi
4
1 000 hp
2
vf =
P8.30
746 000 W
1.06 × 10 5 N
4
f
f
= 7.04 m s
We shall take the zero level of gravitational potential energy to be at the lowest level reached by the
diver under the water, and consider the energy change from when the diver started to fall until he
came to rest.
1
1
mv 2f − mvi2 + mgy f − mgyi = f k d cos 180°
2
2
∆E =
d
i
0 − 0 − mg yi − y f = − f k d
fk =
P8.31
d
mg yi − y f
i = b70.0 kg ge9.80 m s ja10.0 m + 5.00 mf =
2
5.00 m
d
m 2 gh − fh =
Ui + K i + ∆Emech = U f + K f :
2.06 kN
1
1
m1 v 2 + m 2 v 2
2
2
f = µn = µm1 g
m 2 gh − µm1 gh =
v2 =
v=
P8.32
d
i e
∆Emech = K f − K i + U gf − U gi
b
b
g
1
m1 + m 2 v 2
2
gb g
2 m 2 − µm1 hg
m1 + m 2
e
FIG. P8.31
ja
f
b
2 9.80 m s 2 1.50 m 5.00 kg − 0.400 3.00 kg
8.00 kg
g
= 3.74 m s
j
But ∆Emech = Wapp − f∆x , where Wapp is the work the boy
did pushing forward on the wheels.
d
i e
j
1
= me v − v j + mg a − hf + f∆x
2
1
= a 47.0 f a6. 20f − a1.40f − a 47.0fa9.80fa 2.60 f + a 41.0 fa12.4f
2
Thus,
Wapp = K f − K i + U gf − U gi + f∆x
or
Wapp
Wapp
Wapp = 168 J
2
f
2
i
2
2
FIG. P8.32
230
P8.33
Potential Energy
1
1
m v 2f − vi2 = − mvi2 = −160 J
2
2
e
∆K =
(b)
∆U = mg 3.00 m sin 30.0° = 73.5 J
(c)
The mechanical energy converted due to friction is 86.5 J
a
f=
(d)
f
FIG. P8.33
86.5 J
= 28.8 N
3.00 m
f = µ k n = µ k mg cos 30.0° = 28.8 N
28.8 N
µk =
= 0.679
5.00 kg 9.80 m s 2 cos 30.0°
b
P8.34
j
(a)
ge
j
Consider the whole motion: K i + U i + ∆Emech = K f + U f
(a)
0 + mgyi − f1 ∆x1 − f 2 ∆x 2 =
1
mv 2f + 0
2
b80.0 kg ge9.80 m s j1 000 m − a50.0 Nfa800 mf − b3 600 Nga200 mf = 12 b80.0 kg gv
1
784 000 J − 40 000 J − 720 000 J = b80.0 kg gv
2
2b 24 000 Jg
= 24.5 m s
v =
2
2
f
2
f
f
(b)
(c)
80.0 kg
Yes this is too fast for safety.
Now in the same energy equation as in part (a), ∆x 2 is unknown, and ∆x1 = 1 000 m − ∆x 2 :
a
fb
g b
g
784 000 J − 50.0 N 1 000 m − ∆x 2 − 3 600 N ∆x 2 =
b
g
b
gb
1
80.0 kg 5.00 m s
2
g
2
784 000 J − 50 000 J − 3 550 N ∆x 2 = 1 000 J
∆x 2 =
(d)
733 000 J
= 206 m
3 550 N
Really the air drag will depend on the skydiver’s speed. It will be larger than her 784 N
weight only after the chute is opened. It will be nearly equal to 784 N before she opens the
chute and again before she touches down, whenever she moves near terminal speed.
P8.35
aK + Uf + ∆E
(a)
a
0+
b
f
= K +U f :
mech
i
1 2
1
kx − f∆x = mv 2 + 0
2
2
ge
1
8.00 N m 5.00 × 10 −2 m
2
(b)
j − e3.20 × 10
e
2 5. 20 × 10 −3 J
v=
231
Chapter 8
5.30 × 10
−3
j=
kg
2
−2
ja
f 12 e5.30 × 10
N 0.150 m =
−3
j
kg v 2
1.40 m s
When the spring force just equals the friction force, the ball will stop speeding up. Here
Fs = kx ; the spring is compressed by
3. 20 × 10 −2 N
= 0.400 cm
8.00 N m
and the ball has moved
5.00 cm − 0.400 cm = 4.60 cm from the start.
(c)
Between start and maximum speed points,
1 2
1
1
kxi − f∆x = mv 2 + kx 2f
2
2
2
1
1
1
−2 2
− 3.20 × 10 −2 4.60 × 10 −2 = 5.30 × 10 −3 v 2 + 8.00 4.00 × 10 −3
8.00 5.00 × 10
2
2
2
v = 1.79 m s
e
P8.36
j e
je
j e
j
e
∑ Fy = n − mg cos 37.0° = 0
∴ n = mg cos 37.0° = 400 N
a
f
f = µn = 0.250 400 N = 100 N
− f∆x = ∆Emech
a−100fa20.0f = ∆U + ∆U + ∆K + ∆K
∆U = m g d h − h i = a50.0fa9.80fa 20.0 sin 37.0°f = 5.90 × 10
∆U = m g d h − h i = a100 fa9.80fa −20.0f = −1.96 × 10
1
∆K = m e v − v j
2
m
1
∆K = m e v − v j =
∆K = 2 ∆K
2
m
A
A
A
B
B
f
A
B
B
A
B
i
f
A
B
4
i
2
f
2
f
3
2
i
2
i
B
A
A
A
Adding and solving, ∆K A = 3.92 kJ .
FIG. P8.36
j
2
232
P8.37
Potential Energy
(a)
The object moved down distance 1.20 m + x. Choose y = 0 at its lower point.
K i + U gi + U si + ∆Emech = K f + U gf + U sf
0 + mgyi + 0 + 0 = 0 + 0 +
1 2
kx
2
b1.50 kg ge9.80 m s ja1.20 m + xf = 12 b320 N mgx
0 = b160 N mgx − a14.7 N fx − 17.6 J
14.7 N ± a −14.7 N f − 4b160 N mga−17.6 N ⋅ mf
x=
2b160 N mg
2
2
2
2
x=
14.7 N ± 107 N
320 N m
The negative root tells how high the object will rebound if it is instantly glued to the spring.
We want
x = 0.381 m
(b)
From the same equation,
b1.50 kg ge1.63 m s ja1.20 m + xf = 12 b320 N mgx
2
2
0 = 160 x 2 − 2.44x − 2.93
The positive root is x = 0.143 m .
(c)
The equation expressing the energy version of the nonisolated system model has one more
term:
mgyi − f∆x =
1 2
kx
2
b1.50 kg ge9.80 m s ja1.20 m + xf − 0.700 Na1.20 m + xf = 12 b320 N mgx
2
17.6 J + 14.7 Nx − 0.840 J − 0.700 Nx = 160 N m x 2
160 x 2 − 14.0 x − 16.8 = 0
x=
14.0 ±
a14.0f − 4a160fa−16.8f
x = 0.371 m
2
320
2
Chapter 8
P8.38
233
The total mechanical energy of the skysurfer-Earth system is
1
mv 2 + mgh .
2
Emech = K + U g =
Since the skysurfer has constant speed,
a f
dEmech
dv
dh
= mv
+ mg
= 0 + mg − v = − mgv .
dt
dt
dt
The rate the system is losing mechanical energy is then
dEmech
= mgv = 75.0 kg 9.80 m s 2 60.0 m s = 44.1 kW .
dt
b
*P8.39
(a)
jb
g
Let m be the mass of the whole board. The portion on the rough surface has mass
normal force supporting it is
a=
(b)
ge
µ mgx
mxg
and the frictional force is k
= ma . Then
L
L
µ k gx
opposite to the motion.
L
In an incremental bit of forward motion dx, the kinetic energy converted into internal
µ mgx
energy is f k dx = k
dx . The whole energy converted is
L
z
L
µ k mgx
µ mg x 2
1
mv 2 =
dx = k
2
2
L
L
0
L
=
0
µ k mgL
2
v = µ k gL
Section 8.5
Relationship Between Conservative Forces and Potential Energy
P8.40
U = − − Ax + Bx 2 dx =
ze
x
(a)
j
0
(b)
∆U = −
z
3.00 m
Fdx =
∆K =
(a)
Ax 2 Bx 3
−
2
3
j a f
e
A 3.00 2 − 2.00
2.00 m
P8.41
mx
. The
L
FG − 5.00 A + 19.0 BIJ
H 2
3 K
z
W = Fx dx =
2
z a2x + 4fdx = FGH 2x2
5 .00 m
2
1
(b)
∆K + ∆U = 0
(c)
∆K = K f −
mv12
2
2
−
+ 4x
a f − a2.00f
B 3.00
3
3
I
JK
=
5.00
19.0
A−
B
2
3
5 .00 m
= 25.0 + 20.0 − 1.00 − 4.00 = 40.0 J
1
∆U = − ∆K = −W = −40.0 J
K f = ∆K +
3
mv12
= 62.5 J
2
234
P8.42
Potential Energy
e
e
j e
j e
∂ 3x 3 y − 7x
∂U
=−
= − 9x 2 y − 7 = 7 − 9x 2 y
Fx = −
∂x
∂x
∂ 3x 3 y − 7x
∂U
=−
= − 3 x 3 − 0 = −3 x 3
Fy = −
∂y
∂y
j
j
b g
Thus, the force acting at the point x , y is F = Fx i + Fy j =
P8.43
af
e7 − 9x yji − 3x j .
2
3
A
r
d A
A
∂U
Fr = −
=−
= 2 . The positive value indicates a force of repulsion.
dr r
∂r
r
Ur =
Section 8.6
FG IJ
H K
Energy Diagrams and the Equilibrium of a System
P8.44
stable
unstable
neutral
FIG. P8.44
P8.45
(a)
Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.
(b)
A and E are unstable, and C is stable.
(c)
Fx
B
A
C
E
D
FIG. P8.45
x (m)
Chapter 8
P8.46
(a)
235
There is an equilibrium point wherever the graph of potential energy is horizontal:
At r = 1.5 mm and 3.2 mm, the equilibrium is stable.
At r = 2.3 mm , the equilibrium is unstable.
A particle moving out toward r → ∞ approaches neutral equilibrium.
P8.47
(b)
The system energy E cannot be less than –5.6 J. The particle is bound if −5.6 J ≤ E < 1 J .
(c)
If the system energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, the
particle’s position is limited to 0.6 mm ≤ r ≤ 3.6 mm .
(d)
K + U = E . Thus, K max = E − U min = −3.0 J − −5.6 J = 2.6 J .
(e)
Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm .
(f)
−3 J + W = 1 J . Hence, the binding energy is W = 4 J .
(a)
When the mass moves distance x, the length of each spring
a
changes from L to
k
FH
IK
f
x 2 + L2 , so each exerts force
x 2 + L2 − L towards its fixed end. The y-components
cancel out and the x components add to:
Fx = −2 k
FH
x 2 + L2 − L
IK FG
H
x
2
2
x +L
I = −2kx +
JK
2 kLx
x 2 + L2
FIG. P8.47(a)
Choose U = 0 at x = 0 . Then at any point the potential energy of the system is
z z FGH −2kx + x2kLx+ L IJK dx = 2kz xdx − 2kLz
U a x f = kx + 2 kLFH L − x + L IK
U a x f = 40.0 x + 96.0FH 1.20 − x + 1.44 IK
af
x
U x = − Fx dx = −
0
x
2
0
2
(b)
x
2
2
2
0
x
0
x
2
x + L2
dx
2
2
af
For negative x, U x has the same value as for
positive x. The only equilibrium point (i.e., where
Fx = 0) is x = 0 .
(c)
K i + U i + ∆Emech = K f + U f
1
0 + 0.400 J + 0 = 1.18 kg v 2f + 0
2
v f = 0.823 m s
b
g
FIG. P8.47(b)
236
Potential Energy
Additional Problems
P8.48
The potential energy of the block-Earth system is mgh.
An amount of energy µ k mgd cos θ is converted into internal energy due to friction on the incline.
Therefore the final height y max is found from
mgy max = mgh − µ k mgd cos θ
where
y max
sin θ
∴ mgy max = mgh − µ k mgy max cot θ
d=
h
y max
θ
Solving,
y max =
P8.49
FIG. P8.48
h
1 + µ k cot θ
.
At a pace I could keep up for a half-hour exercise period, I climb two stories up, traversing forty
steps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the system
of the Earth and me,
b
ja
ge
f
mgy = 85 kg 9.80 m s 2 40 × 0.18 m = 6 000 J
6 000 J
= ~ 10 2 W .
20 s
making my sustainable power
P8.50
v = 100 km h = 27.8 m s
The retarding force due to air resistance is
R=
a
fe
jb
1
1
DρAv 2 = 0.330 1.20 kg m 3 2.50 m 2 27.8 m s
2
2
je
g
2
= 382 N
Comparing the energy of the car at two points along the hill,
K i + U gi + ∆E = K f + U gf
or
a f
K i + U gi + ∆We − R ∆s = K f + U gf
where ∆We is the work input from the engine. Thus,
a f d
i e
∆We = R ∆s + K f − K i + U gf − U gi
j
Recognizing that K f = K i and dividing by the travel time ∆t gives the required power input from
the engine as
FG ∆W IJ = RFG ∆s IJ + mgFG ∆y IJ = Rv + mgv sinθ
H ∆t K H ∆t K H ∆t K
P = a382 N fb 27.8 m sg + b1 500 kg ge9.80 m s jb 27.8 m sg sin 3.20°
P=
e
2
P = 33.4 kW = 44.8 hp
Chapter 8
P8.51
m = mass of pumpkin
R = radius of silo top
∑ Fr = mar ⇒ n − mg cos θ = −m
237
v2
R
When the pumpkin first loses contact with the surface, n = 0 .
Thus, at the point where it leaves the surface: v 2 = Rg cos θ .
FIG. P8.51
Choose U g = 0 in the θ = 90.0° plane. Then applying conservation of energy for the pumpkin-Earth
system between the starting point and the point where the pumpkin leaves the surface gives
K f + U gf = K i + U gi
1
mv 2 + mgR cos θ = 0 + mgR
2
Using the result from the force analysis, this becomes
1
mRg cos θ + mgR cos θ = mgR , which reduces to
2
cos θ =
b g
2
, and gives θ = cos −1 2 3 = 48.2°
3
as the angle at which the pumpkin will lose contact with the surface.
P8.52
b
ja
ge
f
(a)
U A = mgR = 0. 200 kg 9.80 m s 2 0.300 m = 0.588 J
(b)
K A + U A = KB + UB
K B = K A + U A − U B = mgR = 0.588 J
(c)
(d)
2K B
=
m
vB =
a
f
2 0.588 J
= 2.42 m s
0.200 kg
b
ge
ja
f
FIG. P8.52
UC = mghC = 0.200 kg 9.80 m s 2 0.200 m = 0.392 J
b
K C = K A + U A − U C = mg h A − hC
b
ge
ja
g
f
K C = 0.200 kg 9.80 m s 2 0.300 − 0.200 m = 0.196 J
P8.53
b
gb
1
1
mv B2 = 0.200 kg 1.50 m s
2
2
g
2
(a)
KB =
= 0.225 J
(b)
∆Emech = ∆K + ∆U = K B − K A + U B − U A
b
g
= 0.225 J + b0.200 kg ge9.80 m s ja0 − 0.300 mf
= K B + mg hB − hA
2
= 0.225 J − 0.588 J = −0.363 J
(c)
It’s possible to find an effective coefficient of friction, but not the actual value of µ since n
and f vary with position.
238
P8.54
Potential Energy
The gain in internal energy due to friction represents a loss in mechanical energy that must be equal
to the change in the kinetic energy plus the change in the potential energy.
Therefore,
− µ k mgx cos θ = ∆K +
1 2
kx − mgx sin θ
2
and since vi = v f = 0 , ∆K = 0.
Thus,
a fa fa
fa
f a100fa02.200f − a2.00fa9.80fasin 37.0°fa0.200f
2
− µ k 2.00 9.80 cos 37.0° 0.200 =
and we find µ k = 0.115 . Note that in the above we had a gain in elastic potential energy for the
spring and a loss in gravitational potential energy.
P8.55
(a)
Since no nonconservative work is done, ∆E = 0
k = 100 N/m
Also ∆K = 0
2.00 kg
therefore, Ui = U f
b
g
where Ui = mg sin θ x
and U f =
1 2
kx
2
FIG. P8.55
a fa f
a f 2x and solving we find
Substituting values yields 2.00 9.80 sin 37.0° = 100
x = 0.236 m
(b)
∑ F = ma . Only gravity and the spring force act on the block, so
− kx + mg sin θ = ma
For x = 0.236 m ,
a = −5.90 m s 2 . The negative sign indicates a is up the incline.
The acceleration depends on position .
(c)
U(gravity) decreases monotonically as the height decreases.
U(spring) increases monotonically as the spring is stretched.
K initially increases, but then goes back to zero.
Chapter 8
P8.56
k = 2.50 × 10 4 N m,
m = 25.0 kg
x A = −0.100 m,
Ug
(a)
Emech = K A + U gA + U sA
x =0
= Us
=0
Emech
1 2
kx A
2
= 25.0 kg 9.80 m s 2 −0.100 m
Emech
1
2.50 × 10 4
2
= −24.5 J + 125 J = 100 J
Emech = 0 + mgx A +
b
f
ja
N mja −0.100 mf
ge
e
+
(b)
x=0
2
Since only conservative forces are involved, the total energy of the child-pogo-stick-Earth
system at point C is the same as that at point A.
b
ge
j
0 + 25.0 kg 9.80 m s 2 xC + 0 = 0 − 24.5 J + 125 J
K C + U gC + U sC = K A + U gA + U sA :
x C = 0.410 m
b
a
g
f
1
25.0 kg v B2 + 0 + 0 = 0 + −24.5 J + 125 J
2
v B = 2.84 m s
(c)
K B + U gB + U sB = K A + U gA + U sA :
(d)
K and v are at a maximum when a = ∑ F m = 0 (i.e., when the magnitude of the upward
spring force equals the magnitude of the downward gravitational force).
This occurs at x < 0 where
k x = mg
or
x=
Thus,
K = K max at x = −9.80 mm
b25.0 kg ge9.8 m s j = 9.80 × 10
2
(e)
e
2.50 × 10 4 N m
x =−9.80 mm
sA
m
s x =−9.80 mm
2
max
2
4
yielding
−3
j + eU − U
j
1
b25.0 kg gv = b25.0 kgge9.80 m s j a−0.100 mf − b−0.009 8 mg
2
1
+ e 2.50 × 10 N mj a −0.100 mf − b−0.009 8 mg
2
K max = K A + U gA − U g
or
P8.57
239
2
v max = 2.85 m s
∆Emech = − f∆x
E f − Ei = − f ⋅ d BC
1 2
kx − mgh = − µmgd BC
2
mgh − 12 kx 2
= 0.328
µ=
mgd BC
FIG. P8.57
2
240
P8.58
Potential Energy
(a)
F=−
(b)
F=0
d
− x 3 + 2 x 2 + 3 x i =
dx
e
j e3 x
2
j
− 4x − 3 i
when x = 1.87 and − 0.535
(c)
The stable point is at
af
x = −0.535 point of minimum U x .
The unstable point is at
FIG. P8.58
af
x = 1.87 maximum in U x .
P8.59
aK + U f = aK + U f
1
0 + b30.0 kg ge9.80 m s ja0.200 mf + b 250 N mga0.200 mf
2
1
= b50.0 kg gv + b 20.0 kg ge9.80 m s ja0.200 mf sin 40.0°
2
58.8 J + 5.00 J = b 25.0 kg gv + 25.2 J
i
f
2
2
2
2
2
v = 1.24 m s
P8.60
(a)
FIG. P8.59
Between the second and the third picture, ∆Emech = ∆K + ∆U
− µmgd = −
b
1
1
mvi2 + kd 2
2
2
g
b
ge
b
ge
1
1
50.0 N m d 2 + 0.250 1.00 kg 9.80 m s 2 d − 1.00 kg 3.00 m s 2 = 0
2
2
−2.45 ± 21.25 N
d=
= 0.378 m
50.0 N m
(b)
j
j
Between picture two and picture four, ∆Emech = ∆K + ∆U
a f 12 mv − 12 mv
2
v = b3.00 m sg −
b1.00 kg g a2.45 Nfa2fa0.378 mf
2
− f 2d =
2
i
2
= 2.30 m s
(c)
For the motion from picture two to picture five,
∆Emech = ∆K + ∆U
a
f
b
gb
g
1
2
1.00 kg 3.00 m s
2
9.00 J
D=
− 2 0.378 m = 1.08 m
2 0.250 1.00 kg 9.80 m s 2
− f D + 2d = −
a
fb
ge
j
a
f
FIG. P8.60
Chapter 8
P8.61
(a)
Initial compression of spring:
ga f
b
1
450 N m ∆x
2
∴ ∆x = 0. 400 m
(b)
2
1 2 1
kx = mv 2
2
2
b
gb
1
0.500 kg 12.0 m s
2
=
g
2
Speed of block at top of track: ∆Emech = − f∆x
FIG. P8.61
FG mgh + 1 mv IJ − FG mgh + 1 mv IJ = − f aπRf
H
K H
K
2
2
b0.500 kgge9.80 m s ja2.00 mf + 12 b0.500 kg gv − 12 b0.500 kggb12.0 m sg
= −a7.00 N faπ fa1.00 mf
T
2
T
B
2
B
2
T
2
2
0.250 vT2 = 4.21
∴ vT = 4.10 m s
(c)
Does block fall off at or before top of track? Block falls if a c < g
ac =
a f
2
4.10
vT2
=
= 16.8 m s 2
R
1.00
Therefore a c > g and the block stays on the track .
P8.62
Let λ represent the mass of each one meter of the chain and T
represent the tension in the chain at the table edge. We imagine the
edge to act like a frictionless and massless pulley.
(a)
For the five meters on the table with motion impending,
∑ Fy = 0 :
+n − 5 λg = 0
n = 5 λg
b g
fs ≤ µ sn = 0.6 5λg = 3λg
∑ Fx = 0 :
+T − f s = 0
T = fs
T ≤ 3 λg
FIG. P8.62
The maximum value is barely enough to support the hanging segment according to
∑ Fy = 0 :
+T − 3 λg = 0
T = 3 λg
so it is at this point that the chain starts to slide.
continued on next page
241
242
Potential Energy
(b)
Let x represent the variable distance the chain has slipped since the start.
a f
+n − a5 − x fλg = 0
n = a5 − xfλg
f = µ n = 0.4a5 − xfλg = 2λg − 0.4xλg
Then length 5 − x remains on the table, with now
∑ Fy = 0 :
k
k
Consider energies of the chain-Earth system at the initial moment when the chain starts to
slip, and a final moment when x = 5 , when the last link goes over the brink. Measure
heights above the final position of the leading end of the chain. At the moment the final link
slips off, the center of the chain is at y f = 4 meters.
Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at
3
height 8 − = 6.5 m .
2
K i + U i + ∆Emech = K f + U f :
FG 1 mv + mgyIJ
H2
K
b5λg g8 + b3λg g6.5 − z b2λg − 0.4xλg gdx = 12 b8λ gv + b8λg g4
b
z
f
g
2
0 + m1 gy1 + m 2 gy 2 i − f k dx =
f
i
5
0
2
z
z
5
5
40.0 g + 19.5 g − 2.00 g dx + 0.400 g x dx = 4.00 v 2 + 32.0 g
0
0
5
27.5 g − 2.00 gx 0 + 0.400 g
a f
2 5
x
2
= 4.00 v 2
0
a f
27.5 g − 2.00 g 5.00 + 0.400 g 12.5 = 4.00 v 2
22.5 g = 4.00 v 2
a22.5 mfe9.80 m s j =
2
v=
P8.63
4.00
7.42 m s
Launch speed is found from
mg
FG 4 hIJ = 1 mv :
H5 K 2
2
v = 2g
FG 4 IJ h
H 5K
v y = v sin θ
The height y above the water (by conservation of energy
for the child-Earth system) is found from
mgy =
1
h
mv y2 + mg
2
5
FIG. P8.63
1
mv x2 is constant in projectile motion)
2
1 2 h 1 2
h
y=
vy + =
v sin 2 θ +
2g
5 2g
5
(since
y=
LM FG IJ OP sin θ + h =
5
N H KQ
1
4
2g h
2g
5
2
4
h
h sin 2 θ +
5
5
Chapter 8
*P8.64
(a)
The length of string between glider and pulley is given by A 2 = x 2 + h02 . Then 2 A
Now
(b)
eK
A
a
243
dA
dx
= 2x
+ 0.
dt
dt
f
dA
dA
x
= v y = v x = cos θ v x .
is the rate at which string goes over the pulley:
A
dt
dt
+ KB + U g
j = eK
i
b
0 + 0 + m B g y 30 − y 45
A
g
+ KB + U g
j
f
1
1
= m A v x2 + m B v y2
2
2
Now y 30 − y 45 is the amount of string that has gone over the pulley, A 30 − A 45 . We have
h
h
h0
h0
sin 30° = 0 and sin 45° = 0 , so A 30 − A 45 =
−
= 0. 40 m 2 − 2 = 0.234 m .
A 30
A 45
sin 30° sin 45°
From the energy equation
e
0.5 kg 9.8 m s 2 0.234 m =
1
1
1.00 kg v x2 + 0.500 kg v x2 cos 2 45°
2
2
1.15 J
= 1.35 m s
0.625 kg
vx =
P8.65
j
b
g
(c)
v y = v x cos θ = 1.35 m s cos 45° = 0.958 m s
(d)
The acceleration of neither glider is constant, so knowing distance and acceleration at one
point is not sufficient to find speed at another point.
b
g
The geometry reveals D = L sin θ + L sin φ , 50.0 m = 40.0 m sin 50°+ sin φ , φ = 28.9°
(a)
From takeoff to alighting for the Jane-Earth system
eK + U j + W = eK + U j
1
mv + mg a − L cos θ f + FDa −1f = 0 + mg b− L cos φ g
2
1
50 kg v + 50 kg e9.8 m s ja −40 m cos 50°f − 110 Na50 mf = 50 kg e9.8 m s ja −40 m cos 28.9°f
2
g
wind
i
g
f
2
i
2
i
2
2
1
50 kg vi2 − 1.26 × 10 4 J − 5.5 × 10 3 J = −1.72 × 10 4 J
2
vi =
(b)
a
f
2 947 J
= 6.15 m s
50 kg
For the swing back
b
g
e
a f
a
f
f
1
mvi2 + mg − L cos φ + FD +1 = 0 + mg − L cos θ
2
1
130 kg vi2 + 130 kg 9.8 m s 2 −40 m cos 28.9° + 110 N 50 m
2
e
ja
ja
= 130 kg 9.8 m s 2 −40 m cos 50°
f
1
130 kg vi2 − 4.46 × 10 4 J + 5 500 J = −3.28 × 10 4 J
2
vi =
b
g=
2 6 340 J
130 kg
9.87 m s
a
f
244
P8.66
Potential Energy
1
1
mv 2 = kx 2
2
2
2
5.00 kg 1.20 m s
mv
k= 2 =
x
10 −2 m 2
Case I: Surface is frictionless
b
e
je
eK + U j = eK + U j
g
g
A
= 7. 20 × 10 2 N m
(b)
11.1 m s
v2
=
ac =
r
6.3 m
(c)
∑ Fy = ma y
g
2
2
B
1
0 + mgy A = mv B2 + 0
2
b
2
j − a0.300fb5.00 kg ge9.80 m s je10
5.00 kg 2 1
v = 7.20 × 10 2 N m 10 −1 m
2
2
v = 0.923 m s
(a)
g
µ k = 0.300
1
1
mv 2 = kx 2 − µ k mgx
2
2
Case II: Surface is rough,
*P8.67
gb
e
−1
j
m
j
v B = 2 gy A = 2 9.8 m s 2 6.3 m = 11.1 m s
2
= 19.6 m s 2 up
+n B − mg = ma c
e
j
n B = 76 kg 9.8 m s 2 + 19.6 m s 2 = 2.23 × 10 3 N up
a
f
(d)
W = F∆r cos θ = 2.23 × 10 3 N 0.450 m cos 0° = 1.01 × 10 3 J
(e)
eK + U j
g
B
e
+ W = K +Ug
j
D
b
g
1
1
mv B2 + 0 + 1.01 × 10 3 J = mv D2 + mg y D − y B
2
2
1
1
2
76 kg 11.1 m s + 1.01 × 10 3 J = 76 kg v D2 + 76 kg 9.8 m s 2 6.3 m
2
2
b
g
e5.70 × 10
3
e
j
J − 4.69 × 10 3 J 2
76 kg
(f)
eK + U j = eK + U j
1
mv + 0 = 0 + mg b y
2
g
g
D
2
D
(g)
E
j
= v D = 5.14 m s
where E is the apex of his motion
E − yD
g
y E − yD =
b
g
2
5.14 m s
v D2
=
= 1.35 m
2 g 2 9.8 m s 2
e
j
Consider the motion with constant acceleration between takeoff and touchdown. The time
is the positive root of
1
y f = yi + v yi t + a y t 2
2
1
−2.34 m = 0 + 5.14 m s t + −9.8 m s 2 t 2
2
2
4.9t − 5.14t − 2.34 = 0
e
t=
a fa
j
f=
5.14 ± 5.14 2 − 4 4.9 −2.34
9.8
1.39 s
Chapter 8
*P8.68
If the spring is just barely able to lift the lower block from the table, the spring lifts it through no
noticeable distance, but exerts on the block a force equal to its weight Mg. The extension of the
spring, from Fs = kx , must be Mg k . Between an initial point at release and a final point when the
moving block first comes to rest, we have
FG
H
IJ
K
FG
H
IJ
K
FG IJ
H K
2
0 + mg −
c he−4m j = −m ±
M=
2c h
Only a positive mass is physical, so we take M = ma3 − 1f = 2m .
−m ± m 2 − 4
2
2
1
2
1
2
(a)
FG IJ
H K
4mg
Mg
1 Mg
1 4mg
+ k
= 0 + mg
+ k
2
2
k
k
k
k
2 2
2 2
2
2 2
mMg
M g
4m g
8m g
−
+
=
+
k
k
k
2k
2
M
4m 2 = mM +
2
M2
+ mM − 4m 2 = 0
2
K i + U gi + U si = K f + U gf + U sf :
P8.69
245
9m 2
Take the original point where the ball is
released and the final point where its
upward swing stops at height H and
horizontal displacement
a
x = L2 − L − H
f
2
= 2LH − H 2
Since the wind force is purely horizontal, it
does work
z
z
Wwind = F ⋅ ds = F dx = F 2LH − H 2
FIG. P8.69
The work-energy theorem can be written:
K i + U gi + Wwind = K f + U gf , or
0 + 0 + F 2LH − H 2 = 0 + mgH giving F 2 2LH − F 2 H 2 = m 2 g 2 H 2
Here H = 0 represents the lower turning point of the ball’s oscillation, and the upper limit is
at F 2 2L = F 2 + m 2 g 2 H . Solving for H yields
a f e
j
H=
2LF 2
2L
=
F + m2 g 2
1 + mg F
b
2
g
2
As F → 0 , H → 0 as is reasonable.
As F → ∞ , H → 2L , which would be hard to approach experimentally.
(b)
H=
a
f
2 2.00 m
b
ge
j
1 + 2.00 kg 9.80 m s 2 14.7 N
continued on next page
2
= 1.44 m
246
Potential Energy
(c)
Call θ the equilibrium angle with the vertical.
∑ Fx = 0 ⇒ T sin θ = F , and
∑ Fy = 0 ⇒ T cos θ = mg
Dividing: tan θ =
F
14.7 N
=
= 0.750 , or θ = 36.9°
mg 19.6 N
a
f a
fa
f
Therefore, H eq = L 1 − cos θ = 2.00 m 1 − cos 36.9° = 0.400 m
(d)
As F → ∞ , tan θ → ∞ , θ → 90.0° and H eq → L
A very strong wind pulls the string out horizontal, parallel to the ground. Thus,
eH j
eq
P8.70
max
=L .
Call φ = 180°−θ the angle between the upward vertical and
the radius to the release point. Call v r the speed here. By
conservation of energy
vi = Rg
The path
after string
is cut
K i + U i + ∆E = K r + U r
1
1
mvi2 + mgR + 0 = mv r2 + mgR cos φ
2
2
gR + 2 gR = v r2 + 2 gR cos φ
C
R
θ
v r = 3 gR − 2 gR cos φ
FIG. P8.70
The components of velocity at release are v x = v r cos φ and
v y = v r sin φ so for the projectile motion we have
x = vxt
1
y = v y t − gt 2
2
R sin φ = v r cos φt
− R cos φ = v r sin φt −
1 2
gt
2
By substitution
− R cos φ = v r sin φ
R sin φ g R 2 sin 2 φ
−
v r cos φ 2 v r2 cos 2 φ
with sin 2 φ + cos 2 φ = 1 ,
b
gR sin 2 φ = 2 v r2 cos φ = 2 cos φ 3 gR − 2 gR cos φ
g
sin 2 φ = 6 cos φ − 4 cos 2 φ = 1 − cos 2 φ
3 cos 2 φ − 6 cos φ + 1 = 0
cos φ =
6 ± 36 − 12
6
Only the – sign gives a value for cos φ that is less than one:
cos φ = 0.183 5
φ = 79.43°
so θ = 100.6°
247
Chapter 8
P8.71
Applying Newton’s second law at the bottom (b) and top (t) of the
circle gives
Tb − mg =
Adding these gives
mv b2
and −Tt − mg = −
R
Tb = Tt + 2mg +
m
e
v b2
−
vt
mg
mv t2
R
v t2
Tb
j
R
mg
Also, energy must be conserved and ∆U + ∆K = 0
So,
e
m v b2 − v t2
2
j + b0 − 2mgRg = 0 and mev
Tt
2
b
− v t2
R
j = 4mg
vb
FIG. P8.71
Substituting into the above equation gives Tb = Tt + 6mg .
P8.72
(a)
(b)
Energy is conserved in the swing of the pendulum, and the
stationary peg does no work. So the ball’s speed does not
change when the string hits or leaves the peg, and the ball
swings equally high on both sides.
Relative to the point of suspension,
a
Ui = 0, U f = − mg d − L − d
θ
L
d
Peg
f
From this we find that
a
f
1
− mg 2d − L + mv 2 = 0
2
Also for centripetal motion,
mg =
mv 2
where R = L − d .
R
Upon solving, we get d =
3L
.
5
FIG. P8.72
248
*P8.73
Potential Energy
(a)
At the top of the loop the car and riders are in free
fall:
∑ Fy = ma y :
mg down =
v = Rg
mv 2
down
R
Energy of the car-riders-Earth system is conserved
between release and top of loop:
K i + U gi = K f + U gf :
0 + mgh =
a f
1
mv 2 + mg 2 R
2
a f
1
Rg + g 2 R
2
h = 2.50 R
gh =
(b)
Let h now represent the height ≥ 2.5 R of the release
point. At the bottom of the loop we have
mgh =
1
mv b2
2
∑ Fy = ma y :
or
v b2 = 2 gh
n b − mg =
n b = mg +
mv b2
up
R
m 2 gh
b g
b g
R
a f
1
mv t2 + mg 2 R
2
v t2 = 2 gh − 4 gR
At the top of the loop, mgh =
∑ Fy = ma y :
−n t − mg = −
FIG. P8.73
mv t2
R
m
2 gh − 4 gR
n t = − mg +
R
m 2 gh
− 5mg
nt =
R
b
g
b g
Then the normal force at the bottom is larger by
n b − n t = mg +
b g − mb2 ghg + 5mg =
m 2 gh
R
R
6mg .
Chapter 8
*P8.74
(a)
249
Conservation of energy for the sled-rider-Earth system,
between A and C:
K i + U gi = K f + U gf
b
1
m 2.5 m s
2
vC =
(b)
g
2
ja
e
1
mvC2 + 0
2
b2.5 m sg + 2e9.80 m s ja9.76 mf =
2
2
b
gb
g b
ja
ge
f
1
2
80 kg 2.5 m s + 80 kg 9.80 m s 2 9.76 m − f k ∆x = 0 + 0
2
− f k ∆x = −7.90 × 10 3 J
7.90 × 10 3 J 7.90 × 10 3 N ⋅ m
=
= 158 N
∆x
50 m
The water exerts a frictional force
fk =
and also a normal force of
n = mg = 80 kg 9.80 m s 2 = 784 N
b
ge
a158 Nf + a784 Nf
2
The magnitude of the water force is
(d)
FIG. P8.74(a)
14.1 m s
Incorporating the loss of mechanical energy during the portion of the motion in the water,
we have, for the entire motion between A and D (the rider’s stopping point),
K i + U gi − f k ∆x = K f + U gf :
(c)
f
+ m 9.80 m s 2 9.76 m =
j
2
= 800 N
The angle of the slide is
θ = sin −1
9.76 m
= 10.4°
54.3 m
For forces perpendicular to the track at B,
∑ Fy = ma y :
FIG. P8.74(d)
n B − mg cos θ = 0
b
ge
j
n B = 80.0 kg 9.80 m s 2 cos 10.4° = 771 N
(e)
∑ Fy = ma y :
mvC2
r
nC = 80.0 kg 9.80 m s 2
+nC − mg =
b
ge
j
b80.0 kggb14.1 m sg
+
2
20 m
nC = 1.57 × 10 3 N up
FIG. P8.74(e)
The rider pays for the thrills of a giddy height at A, and a high speed and tremendous splash
at C. As a bonus, he gets the quick change in direction and magnitude among the forces we
found in parts (d), (e), and (c).
250
Potential Energy
ANSWERS TO EVEN PROBLEMS
P8.2
(a) 800 J; (b) 107 J; (c) 0
P8.4
(a) 1.11 × 10 9 J ; (b) 0.2
P8.6
1.84 m
P8.42
e7 − 9x yji − 3x j
P8.44
see the solution
P8.46
(a) r = 1.5 mm and 3.2 mm, stable; 2.3 mm
and unstable; r → ∞ neutral;
(b) −5.6 J ≤ E < 1 J ; (c) 0.6 mm ≤ r ≤ 3.6 mm ;
(d) 2.6 J; (e) 1.5 mm; (f) 4 J
P8.48
see the solution
P8.50
33.4 kW
P8.52
(a) 0.588 J; (b) 0.588 J; (c) 2.42 m s;
(d) 0.196 J; 0.392 J
6
P8.8
(a) 10.2 kW; (b) 10.6 kW; (c) 5.82 × 10 J
P8.10
d=
P8.12
(a) see the solution; (b) 60.0°
P8.14
(a)
kx 2
−x
2mg sin θ
b
bm
g
+m g
2 m1 − m 2 gh
1
2
; (b)
2m1 h
m1 + m 2
2
3
P8.16
160 L min
P8.54
0.115
P8.18
40.8°
P8.56
P8.20
FG 8 gh IJ
H 15 K
(a) 100 J; (b) 0.410 m; (c) 2.84 m s ;
(d) −9.80 mm ; (e) 2.85 m s
P8.58
(a) 3 x 2 − 4x − 3 i ; (b) 1.87; –-0.535;
P8.22
(a) see the solution; (b) 35.0 J
P8.24
(a) v B = 5.94 m s; vC = 7.67 m s ; (b) 147 J
P8.26
(a) U f = 22.0 J ; E = 40.0 J ; (b) Yes. The total
12
mechanical energy changes.
e
j
(c) see the solution
P8.60
(a) 0.378 m; (b) 2.30 m s ; (c) 1.08 m
P8.62
(a) see the solution; (b) 7.42 m s
P8.64
(a) see the solution; (b) 1.35 m s ;
(c) 0.958 m s ; (d) see the solution
P8.28
194 m
P8.30
2.06 kN up
P8.66
0.923 m s
P8.32
168 J
P8.68
2m
P8.34
(a) 24.5 m s ; (b) yes; (c) 206 m; (d) Air drag
depends strongly on speed.
P8.70
100.6°
P8.36
3.92 kJ
P8.72
see the solution
P8.38
44.1 kW
P8.74
(a) 14.1 m s; (b) −7.90 J ; (c) 800 N;
(d) 771 N; (e) 1.57 kN up
P8.40
(a)
Ax 2 Bx 3
−
;
2
3
5 A 19B
19B 5 A
(b) ∆U =
−
; ∆K =
−
2
3
3
2
9
Linear Momentum and Collisions
CHAPTER OUTLINE
9.1
9.2
9.3
9.4
9.5
9.6
9.7
Linear Momentum and Its
Conservation
Impulse and Momentum
Collisions in One Dimension
Two-Dimensional Collisions
The Center of Mass
Motion of a System of
Particles
Rocket Propulsion
(c)
ANSWERS TO QUESTIONS
Q9.1
No. Impulse, F∆t , depends on the force and the time for which
it is applied.
Q9.2
The momentum doubles since it is proportional to the speed.
The kinetic energy quadruples, since it is proportional to the
speed-squared.
Q9.3
The momenta of two particles will only be the same if the
masses of the particles of the same.
Q9.4
(a)
It does not carry force, for if it did, it could accelerate
itself.
(b)
It cannot deliver more kinetic energy than it possesses.
This would violate the law of energy conservation.
It can deliver more momentum in a collision than it possesses in its flight, by bouncing from
the object it strikes.
Q9.5
Provided there is some form of potential energy in the system, the parts of an isolated system can
move if the system is initially at rest. Consider two air-track gliders on a horizontal track. If you
compress a spring between them and then tie them together with a string, it is possible for the
system to start out at rest. If you then burn the string, the potential energy stored in the spring will
be converted into kinetic energy of the gliders.
Q9.6
No. Only in a precise head-on collision with momenta with equal magnitudes and opposite
directions can both objects wind up at rest. Yes. Assume that ball 2, originally at rest, is struck
squarely by an equal-mass ball 1. Then ball 2 will take off with the velocity of ball 1, leaving ball 1 at
rest.
Q9.7
Interestingly, mutual gravitation brings the ball and the Earth together. As the ball moves
downward, the Earth moves upward, although with an acceleration 10 25 times smaller than that of
the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is
conserved.
Q9.8
(a)
Linear momentum is conserved since there are no external forces acting on the system.
(b)
Kinetic energy is not conserved because the chemical potential energy initially in the
explosive is converted into kinetic energy of the pieces of the bomb.
251
252
Linear Momentum and Collisions
Q9.9
Momentum conservation is not violated if we make our system include the Earth along with the
clay. When the clay receives an impulse backwards, the Earth receives the same size impulse
forwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than the
acceleration of the clay, but the planet absorbs all of the momentum that the clay loses.
Q9.10
Momentum conservation is not violated if we choose as our system the planet along with you.
When you receive an impulse forward, the Earth receives the same size impulse backwards. The
resulting acceleration of the Earth due to this impulse is significantly smaller than your acceleration
forward, but the planet’s backward momentum is equal in magnitude to your forward momentum.
Q9.11
As a ball rolls down an incline, the Earth receives an impulse of the same size and in the opposite
direction as that of the ball. If you consider the Earth-ball system, momentum conservation is not
violated.
Q9.12
Suppose car and truck move along the same line. If one vehicle overtakes the other, the fastermoving one loses more energy than the slower one gains. In a head-on collision, if the speed of the
m + 3m c
times the speed of the car, the car will lose more energy.
truck is less than T
3mT + m c
Q9.13
The rifle has a much lower speed than the bullet and much less kinetic energy. The butt distributes
the recoil force over an area much larger than that of the bullet.
Q9.14
His impact speed is determined by the acceleration of gravity and the distance of fall, in
v 2f = vi2 − 2 g 0 − yi . The force exerted by the pad depends also on the unknown stiffness of the pad.
Q9.15
The product of the mass flow rate and velocity of the water determines the force the firefighters
must exert.
Q9.16
The sheet stretches and pulls the two students toward each other. These effects are larger for a
faster-moving egg. The time over which the egg stops is extended so that the force stopping it is
never too large.
Q9.17
(c) In this case, the impulse on the Frisbee is largest. According to Newton’s third law, the impulse
on the skater and thus the final speed of the skater will also be largest.
Q9.18
Usually but not necessarily. In a one-dimensional collision between two identical particles with the
same initial speed, the kinetic energy of the particles will not change.
Q9.19
g downward.
Q9.20
As one finger slides towards the center, the normal force exerted by the sliding finger on the ruler
increases. At some point, this normal force will increase enough so that static friction between the
sliding finger and the ruler will stop their relative motion. At this moment the other finger starts
sliding along the ruler towards the center. This process repeats until the fingers meet at the center of
the ruler.
Q9.21
The planet is in motion around the sun, and thus has momentum and kinetic energy of its own. The
spacecraft is directed to cross the planet’s orbit behind it, so that the planet’s gravity has a
component pulling forward on the spacecraft. Since this is an elastic collision, and the velocity of the
planet remains nearly unchanged, the probe must both increase speed and change direction for both
momentum and kinetic energy to be conserved.
b
g
Chapter 9
253
Q9.22
No—an external force of gravity acts on the moon. Yes, because its speed is constant.
Q9.23
The impulse given to the egg is the same regardless of how it stops. If you increase the impact time
by dropping the egg onto foam, you will decrease the impact force.
Q9.24
Yes. A boomerang, a kitchen stool.
Q9.25
The center of mass of the balls is in free fall, moving up and then down with the acceleration due to
gravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the time
when the juggler is engaged in catching and tossing, the center of mass must accelerate up with a
somewhat smaller average acceleration. The center of mass moves around in a little circle, making
three revolutions for every one revolution that one ball makes. Letting T represent the time for one
cycle and Fg the weight of one ball, we have FJ 0.60T = 3 Fg T and FJ = 5 Fg . The average force exerted
by the juggler is five times the weight of one ball.
Q9.26
In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn,
because no outside force acts on this system. According to the text’s ‘basic expression for rocket
propulsion,’ the change in speed of the rocket body will be larger than the speed of the exhaust
relative to the rocket, if the final mass is less than 37% of the original mass.
Q9.27
The gun recoiled.
Q9.28
Inflate a balloon and release it. The air escaping from the balloon gives the balloon an impulse.
Q9.29
There was a time when the English favored position (a), the Germans position (b), and the French
position (c). A Frenchman, Jean D’Alembert, is most responsible for showing that each theory is
consistent with the others. All are equally correct. Each is useful for giving a mathematically simple
solution for some problems.
SOLUTIONS TO PROBLEMS
Section 9.1
P9.1
Linear Momentum and Its Conservation
e
(a)
(b)
j
v = 3.00 i − 4.00 j m s
m = 3.00 kg ,
e
j
p = mv = 9.00 i − 12.0 j kg ⋅ m s
Thus,
p x = 9.00 kg ⋅ m s
and
p y = −12.0 kg ⋅ m s
a9.00f + a12.0f = 15.0 kg ⋅ m s
F p I = tan a−1.33f = 307°
GH p JK
2
p = p x2 + p y2 =
θ = tan −1
y
x
−1
2
254
P9.2
Linear Momentum and Collisions
(a)
At maximum height v = 0 , so p = 0 .
(b)
Its original kinetic energy is its constant total energy,
Ki =
a
f b
1
1
mvi2 = 0.100 kg 15.0 m s
2
2
g
2
= 11.2 J .
At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and
the other half is kinetic:
v=
b
gb
b
g
1
0.100 kg v 2
2
2 × 5.62 J
= 10.6 m s
0.100 kg
K = 5.62 J =
g
Then p = mv = 0.100 kg 10.6 m s j
p = 1.06 kg ⋅ m s j .
P9.3
I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with
speed given by
d
i
v 2f − vi2 = 2 a x f − x i :
ja
e
0 − vi2 = 2 −9.80 m s 2 0.250 m
f
vi = 2.20 m s
Total momentum of the system of the Earth and me is conserved as I push the earth down and
myself up:
j b
e
gb
0 = 5.98 × 10 24 kg v e + 85.0 kg 2.20 m s
g
v e ~ 10 −23 m s
P9.4
(a)
For the system of two blocks ∆p = 0 ,
or
pi = p f
Therefore,
0 = Mv m + 3 M 2.00 m s
Solving gives
v m = −6.00 m s (motion toward the
a fb
g
left).
(b)
a f
1 2 1
1
2
+ 3 M v 32M = 8. 40 J
kx = Mv M
2
2
2
FIG. P9.4
Chapter 9
P9.5
(a)
The momentum is p = mv , so v =
(b)
K=
Section 9.2
*P9.6
1
mv 2 implies v =
2
FG IJ
H K
p
p
1
1
and the kinetic energy is K = mv 2 = m
m
m
2
2
2K
2K
, so p = mv = m
=
m
m
2
=
Impulse and Momentum
a f
From the impulse-momentum theorem, F ∆t = ∆p = mv f − mvi , the average force required to hold
F=
d
i = b12 kg gb0 − 60 mi hg F 1 m s I = −6.44 × 10
GH 2.237 mi h JK
0.050 s − 0
a ∆t f
m v f − vi
3
N.
Therefore, the magnitude of the needed retarding force is 6.44 × 10 3 N , or 1 400 lb. A person
cannot exert a force of this magnitude and a safety device should be used.
*P9.8
p2
.
2m
2mK .
onto the child is
P9.7
255
(a)
z
I = Fdt = area under curve
jb
g
I=
1
1.50 × 10 −3 s 18 000 N = 13.5 N ⋅ s
2
(b)
F=
13.5 N ⋅ s
= 9.00 kN
1.50 × 10 −3 s
(c)
From the graph, we see that Fmax = 18.0 kN
e
FIG. P9.7
1
1
mv12 = mgy1 . The rebound speed is given by mgy 2 = mv 22 . The
2
2
impulse of the floor is the change in momentum,
The impact speed is given by
b
= me
g
mv 2 up − mv1 down = m v 2 + v1 up
j
2 gh2 + 2 gh1 up
e
= 0.15 kg 2 9.8 m s 2
je
= 1.39 kg ⋅ m s upward
j
0.960 m + 1.25 m up
256
P9.9
Linear Momentum and Collisions
∆p = F∆t
f
e
j a
= ma − v sin 60.0°− v sin 60.0°f = −2mv sin 60.0°
= −2b3.00 kg gb10.0 m sga0.866f
∆p y = m v fy − viy = m v cos 60.0° − mv cos 60.0° = 0
∆p x
= −52.0 kg ⋅ m s
Fave =
P9.10
P9.11
∆p x −52.0 kg ⋅ m s
=
= −260 N
0.200 s
∆t
FIG. P9.9
Assume the initial direction of the ball in the –x direction.
ga f b
b
ga fe j
(a)
Impulse, I = ∆p = p f − pi = 0.060 0 40.0 i − 0.060 0 50.0 − i = 5.40 i N ⋅ s
(b)
Work = K f − K i =
b
1
0.060 0
2
g a40.0f − a50.0f
2
2
= −27.0 J
Take x-axis toward the pitcher
(a)
b0.200 kg gb15.0 m sga− cos 45.0°f + I = b0.200 kggb40.0 m sg cos 30.0°
pix + I x = p fx :
x
I x = 9.05 N ⋅ s
b0.200 kg gb15.0 m sga− sin 45.0°f + I = b0.200 kg gb40.0 m sg sin 30.0°
I = e9.05 i + 6.12 jj N ⋅ s
piy + I y = p fy :
(b)
b
y
f
ga
a
e
Fm =
P9.12
f
a
1
1
0 + Fm 4.00 ms + Fm 20.0 ms + Fm 4.00 ms
2
2
Fm × 24.0 × 10 −3 s = 9.05 i + 6.12 j N ⋅ s
I=
f
j
e377 i + 255jj N
If the diver starts from rest and drops vertically into the water, the velocity just before impact is
found from
K f + U gf = K i + U gi
1
2
+ 0 = 0 + mgh ⇒ v impact = 2 gh
mv impact
2
With the diver at rest after an impact time of ∆t , the average force during impact is given by
F=
e
m 0 − v impact
∆t
j = −m
2 gh
∆t
or F =
m 2 gh
∆t
(directed upward).
Assuming a mass of 55 kg and an impact time of ≈ 1.0 s , the magnitude of this average force is
b55 kg g 2e9.8 m s ja10 mf = 770 N , or
F =
2
1.0 s
~ 10 3 N .
Chapter 9
P9.13
The force exerted on the water by the hose is
F=
b
gb
g
0.600 kg 25.0 m s − 0
∆p water mv f − mvi
=
=
= 15.0 N .
1.00 s
∆t
∆t
According to Newton's third law, the water exerts a force of equal magnitude back on the hose.
Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting water
stream) to hold the hose stationary.
*P9.14
(a)
Energy is conserved for the spring-mass system:
K i + U si = K f + U sf :
1 2 1
kx = mv 2 + 0
2
2
k
v=x
m
0+
k
larger.
m
(b)
From the equation, a smaller value of m makes v = x
(c)
I = p f − p i = mv f = 0 = mx
(d)
From the equation, a larger value of m makes I = x km larger.
(e)
For the glider, W = K f − K i =
k
= x km
m
1
1
mv 2 − 0 = kx 2
2
2
The mass makes no difference to the work.
Section 9.3
P9.15
Collisions in One Dimension
b200 g gb55.0 m sg = b46.0 g gv + b200 g gb40.0 m sg
v = 65.2 m s
*P9.16
bm v
g = bm v + m v g
22.5 g b35 m sg + 300 g b−2.5 m sg = 22.5 gv
1 1
v1 f =
+ m2 v2
i
1 1
2 2 f
1f
+0
37.5 g ⋅ m s
= 1.67 m s
22.5 g
FIG. P9.16
257
258
P9.17
Linear Momentum and Collisions
Momentum is conserved
10.0 × 10 −3 kg v = 5.01 kg 0.600 m s
j b
e
gb
g
v = 301 m s
P9.18
(a)
mv1i + 3mv 2 i = 4mv f where m = 2.50 × 10 4 kg
vf =
P9.19
a f
4.00 + 3 2.00
= 2.50 m s
4
a f
LM
N
a f OPQ e
ja
f
1
1
1
4m v 2f − mv12i + 3m v 22i = 2.50 × 10 4 12.5 − 8.00 − 6.00 = −3.75 × 10 4 J
2
2
2
(b)
K f − Ki =
(a)
The internal forces exerted by the actor do
not change the total momentum of the
system of the four cars and the movie actor
a4mfv = a3mfb2.00 m sg + mb4.00 m sg
i
6.00 m s + 4.00 m s
= 2.50 m s
vi =
4
(b)
(c)
g
b
g
fb g
a fb
1
1
2
2
3m 2.00 m s + m 4.00 m s − 4 m 2.50 m s
2
2
2.50 × 10 4 kg
2
=
12.0 + 16.0 − 25.0 m s = 37.5 kJ
2
Wactor = K f − K i =
Wactor
a fb
ja
FIG. P9.19
e
g
2
The event considered here is the time reversal of the perfectly inelastic collision in the
previous problem. The same momentum conservation equation describes both processes.
P9.20
v1 , speed of m1 at B before collision.
1
m1 v12 = m1 gh
2
a fa f
v1 = 2 9.80 5.00 = 9.90 m s
v1 f , speed of m1 at B just after collision.
m − m2
1
v1 f = 1
v1 = − 9.90 m s = −3.30 m s
m1 + m 2
3
At the highest point (after collision)
a f
m1 ghmax =
a
1
m1 −3.30
2
f
FIG. P9.20
b−3.30 m sg =
2e9.80 m s j
2
2
hmax =
2
0.556 m
Chapter 9
P9.21
259
(a), (b) Let v g and v p be the velocity of the girl and the plank
relative to the ice surface. Then we may say that v g − v p is
the velocity of the girl relative to the plank, so that
v g − v p = 1.50
(1)
But also we must have m g v g + m p v p = 0 , since total
momentum of the girl-plank system is zero relative to the
ice surface. Therefore
45.0 v g + 150 v p = 0 , or v g = −3.33 v p
Putting this into the equation (1) above gives
FIG. P9.21
−3.33 v p − v p = 1.50 or v p = −0.346 m s
a
f
Then v g = −3.33 −0.346 = 1.15 m s
*P9.22
For the car-truck-driver-driver system, momentum is conserved:
p 1i + p 2 i = p 1 f + p 2 f :
b4 000 kg gb8 m sgi + b800 kg gb8 m sge− ij = b4 800 kg gv i
f
vf =
25 600 kg ⋅ m s
= 5.33 m s
4 800 kg
For the driver of the truck, the impulse-momentum theorem is
F∆t = p f − pi :
a
f b
gb
g b
gb
g
F 0.120 s = 80 kg 5.33 m s i − 80 kg 8 m s i
e j
F = 1.78 × 10 3 N − i on the truck driver
For the driver of the car,
a
f b
gb
g b
gb
ge j
F 0.120 s = 80 kg 5.33 m s i − 80 kg 8 m s − i
F = 8.89 × 10 3 Ni on the car driver , 5 times larger.
P9.23
(a)
According to the Example in the chapter text, the fraction of total kinetic energy transferred
to the moderator is
f2 =
4m1 m 2
bm
1
+ m2
g
2
where m 2 is the moderator nucleus and in this case, m 2 = 12m1
f2 =
b
4m1 12m1
b13m g
1
2
g = 48 =
169
0. 284 or 28.4%
of the neutron energy is transferred to the carbon nucleus.
(b)
a fe
j
Jj =
= a0.716 fe1.6 × 10
K C = 0.284 1.6 × 10 −13 J = 4.54 × 10 −14 J
Kn
−13
1.15 × 10 −13 J
260
P9.24
Linear Momentum and Collisions
Energy is conserved for the bob-Earth system between bottom and
top of swing. At the top the stiff rod is in compression and the bob
nearly at rest.
1
Mv b2 + 0 = 0 + Mg 2 A
2
v b2 = g 4A so v b = 2 gA
K i + Ui = K f + U f :
FIG. P9.24
Momentum of the bob-bullet system is conserved in the collision:
mv = m
P9.25
v
+ M 2 gA
2
e
j
v=
4M
m
gA
At impact, momentum of the clay-block system is conserved, so:
b
g
mv1 = m1 + m 2 v 2
After impact, the change in kinetic energy of the clay-block-surface
system is equal to the increase in internal energy:
b
b
g
g
b
g
ge
1
m1 + m 2 v 22 = f f d = µ m1 + m 2 gd
2
1
0.112 kg v 22 = 0.650 0.112 kg 9.80 m s 2 7.50 m
2
v 22 = 95.6 m 2 s 2
v 2 = 9.77 m s
e12.0 × 10
P9.26
−3
b
j b
gb
kg v1 = 0.112 kg 9.77 m s
ja
g
f
FIG. P9.25
v1 = 91.2 m s
We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers
apart. These equal forces act backward on the two bullets.
For the first,
K i + ∆Emech = K f
For the second,
pi = p f
1
7.00 × 10 −3 kg v 2 − F 8.00 × 10 −2 m = 0
2
7.00 × 10 −3 kg v = 1.014 kg v f
e
e
j
j b
e7.00 × 10 jv
=
e
j
g
−3
vf
1.014
b
e
j
K i + ∆Emech = K f :
Substituting for v f ,
1
1
7.00 × 10 −3 v
7.00 × 10 −3 kg v 2 − Fd = 1.014 kg
2
2
1.014
e
b
j
e
−3
1
1 7.00 × 10
−3
2
Fd = 7.00 × 10 v −
2
2
1.014
e
Substituting for v,
e
g
1
1
7.00 × 10 −3 kg v 2 − Fd = 1.014 kg v 2f
2
2
Again,
j
jFGH
Fd = F 8.00 × 10 −2 m 1 −
7.00 × 10 −3
1.014
I
JK
j
gFGH
I
JK
2
2
v2
d = 7.94 cm
Chapter 9
*P9.27
(a)
261
c∑ ph = c∑ ph , gives
a4.0 + 10 + 3.0f kg v = b4.0 kggb5.0 m sg + b10 kg gb3.0 m sg + b3.0 kggb−4.0 m sg .
Using conservation of momentum,
after
before
Therefore, v = +2.24 m s , or 2. 24 m s toward the right .
(b)
No . For example, if the 10-kg and 3.0-kg mass were to stick together first, they would
move with a speed given by solving
b13 kggv = b10 kg gb3.0 m sg + b3.0 kg gb−4.0 m sg , or v
1
1
= +1.38 m s .
Then when this 13 kg combined mass collides with the 4.0 kg mass, we have
b17 kggv = b13 kggb1.38 m sg + b4.0 kg gb5.0 m sg , and v = +2.24 m s
just as in part (a). Coupling order makes no difference.
Section 9.4
P9.28
(a)
Two-Dimensional Collisions
First, we conserve momentum for the system of two football players in the x direction (the
direction of travel of the fullback).
b90.0 kg gb5.00 m sg + 0 = b185 kg gV cosθ
where θ is the angle between the direction of the final velocity V and the x axis. We find
V cos θ = 2.43 m s
(1)
Now consider conservation of momentum of the system in the y direction (the direction of
travel of the opponent).
b95.0 kg gb3.00 m sg + 0 = b185 kggaV sinθ f
which gives,
V sin θ = 1.54 m s
Divide equation (2) by (1)
tan θ =
V = 2.88 m s
Then, either (1) or (2) gives
b
b
gb
gb
1
90.0 kg 5.00 m s
2
1
K f = 185 kg 2.88 m s
2
Ki =
1.54
= 0.633
2.43
θ = 32.3°
From which
(b)
(2)
g
g
b
gb
1
95.0 kg 3.00 m s
2
2
+
2
= 7.67 × 10 2 J
g
2
= 1.55 × 10 3 J
Thus, the kinetic energy lost is 783 J into internal energy.
262
P9.29
Linear Momentum and Collisions
p xf = p xi
b
mvO cos 37.0°+ mv Y cos 53.0° = m 5.00 m s
0.799 vO + 0.602 v Y = 5.00 m s
g
(1)
p yf = p yi
mvO sin 37.0°− mv Y sin 53.0° = 0
0.602 vO = 0.799 v Y
(2)
Solving (1) and (2) simultaneously,
vO = 3.99 m s and v Y = 3.01 m s .
P9.30
p xf = p xi :
a
f
mvO cos θ + mv Y cos 90.0°−θ = mvi
vO cos θ + v Y sin θ = vi
p yf = p yi :
FIG. P9.29
a
(1)
f
mvO sin θ − mv Y sin 90.0°−θ = 0
vO sin θ = v Y cos θ
(2)
From equation (2),
vO = v Y
FG cosθ IJ
H sinθ K
(3)
FIG. P9.30
Substituting into equation (1),
vY
so
F cos θ I + v
GH sinθ JK
2
e
Y
sin θ = vi
j
v Y cos 2 θ + sin 2 θ = vi sin θ , and v Y = vi sin θ .
Then, from equation (3), vO = vi cos θ .
We did not need to write down an equation expressing conservation of mechanical energy. In the
problem situation, the requirement of perpendicular final velocities is equivalent to the condition of
elasticity.
Chapter 9
P9.31
The initial momentum of the system is 0. Thus,
a1.20mfv
and
Bi
b
= m 10.0 m s
g
v Bi = 8.33 m s
b
b g
a
g
a
fb
g
FG e
H
1
1
1
2
2
m 10.0 m s + 1. 20m 8.33 m s = m 183 m 2 s 2
2
2
2
1
1
1 1
2
2
K f = m vG + 1.20m v B =
m 183 m 2 s 2
2
2
2 2
Ki =
or
fb g
vG2 + 1.20 v B2 = 91.7 m 2 s 2
e
jIJK
j
(1)
From conservation of momentum,
a
f
mvG = 1.20m v B
or
vG = 1.20 v B
(2)
Solving (1) and (2) simultaneously, we find
vG = 7.07 m s (speed of green puck after collision)
and
P9.32
v B = 5.89 m s (speed of blue puck after collision)
We use conservation of momentum for the system of two vehicles
for both northward and eastward components.
For the eastward direction:
b
g
M 13.0 m s = 2 MV f cos 55.0°
For the northward direction:
Mv 2i = 2 MV f sin 55.0°
Divide the northward equation by the eastward equation to find:
b
g
v 2 i = 13.0 m s tan 55.0° = 18.6 m s = 41.5 mi h
Thus, the driver of the north bound car was untruthful.
FIG. P9.32
263
264
P9.33
Linear Momentum and Collisions
By conservation of momentum for the system of the two billiard
balls (with all masses equal),
b
g
5.00 m s + 0 = 4.33 m s cos 30.0°+ v 2 fx
v 2 fx = 1.25 m s
b
g
0 = 4.33 m s sin 30.0°+ v 2 fy
v 2 fy = −2.16 m s
v 2 f = 2.50 m s at − 60.0°
FIG. P9.33
Note that we did not need to use the fact that the collision is perfectly elastic.
P9.34
(a)
pi = p f
so
p xi = p xf
and
p yi = p yf
From (2),
mvi = mv cos θ + mv cos φ
(1)
0 = mv sin θ + mv sin φ
(2)
sin θ = − sin φ
θ = −φ
Furthermore, energy conservation for the system
of two protons requires
1
1
1
mvi2 = mv 2 + mv 2
2
2
2
vi
v=
so
2
so
(b)
Hence, (1) gives vi =
b
2 vi cos θ
g
FIG. P9.34
θ = 45.0°
2
φ = −45.0°
a f
v = e3.00 i − 1.20 jj m s
3.00 5.00 i − 6.00 j = 5.00 v
P9.35
m1 v 1i + m 2 v 2i = m1 + m 2 v f :
P9.36
x-component of momentum for the system of the two objects:
p1ix + p 2ix = p1 fx + p 2 fx :
− mvi + 3mvi = 0 + 3mv 2 x
y-component of momentum of the system:
0 + 0 = − mv1 y + 3 mv 2 y
by conservation of energy of the system:
+
also
1
1
1
1
mvi2 + 3mvi2 = mv12y + 3m v 22 x + v 22 y
2
2
2
2
2 vi
v2x =
3
v1 y = 3 v 2 y
So the energy equation becomes
4vi2 = 9 v 22 y +
we have
or
continued on next page
8 vi2
= 12 v 22 y
3
2 vi
v2y =
3
e
4vi2
+ 3 v 22 y
3
j
Chapter 9
(a)
The object of mass m has final speed
v1 y = 3 v 2 y =
and the object of mass 3 m moves at
v 22 x + v 22 y =
2 vi
v 22 x + v 22 y =
(b)
θ = tan −1
Fv I
GH v JK
2y
θ = tan −1
2x
P9.37
F
GH
m 0 = 17.0 × 10 −27 kg
v i = 0 (the parent nucleus)
m1 = 5.00 × 10 −27 kg
v 1 = 6.00 × 10 6 j m s
m 2 = 8.40 × 10 −27 kg
v 2 = 4.00 × 10 6 i m s
(a)
4vi2 2 vi2
+
9
9
2
vi
3
I
JK
2 vi 3
= 35.3°
3 2 vi
m1 v 1 + m 2 v 2 + m 3 v 3 = 0
where m 3 = m 0 − m1 − m 2 = 3.60 × 10 −27 kg
FIG. P9.37
−27
−27
−27
6
6
×
×
+
×
×
+
×
5.00 10
6.00 10 j
8.40 10
4.00 10 i
3.60 10
v3 = 0
e
v3 =
(b)
je
j e
je
j e
j
e−9.33 × 10 i − 8.33 × 10 jj m s
6
1
1
1
m1 v12 + m 2 v 22 + m 3 v 32
2
2
2
1
−27
5.00 × 10
6.00 × 10 6
E=
2
6
E=
LMe
N
je
j + e8.40 × 10 je4.00 × 10 j + e3.60 × 10 je12.5 × 10 j OQP
2
−27
6 2
−27
6 2
E = 4.39 × 10 −13 J
Section 9.5
P9.38
The Center of Mass
The x-coordinate of the center of mass is
x CM =
∑ m i xi
∑ mi
=
b
0+0+0+0
2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg
g
xCM = 0
and the y-coordinate of the center of mass is
yCM =
∑ m i yi
∑ mi
yCM = 1.00 m
=
b2.00 kgga3.00 mf + b3.00 kg ga2.50 mf + b2.50 kg ga0f + b4.00 kg ga−0.500 mf
2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg
265
266
P9.39
Linear Momentum and Collisions
Take x-axis starting from the oxygen nucleus and pointing toward the
middle of the V.
yCM = 0
Then
x CM =
and
x CM =
∑ mi x i
∑ mi
a
=
f
a
f
0 + 1.008 u 0.100 nm cos 53.0°+1.008 u 0.100 nm cos 53.0°
15.999 u + 1.008 u + 1.008 u
FIG. P9.39
xCM = 0.006 73 nm from the oxygen nucleus
*P9.40
Let the x axis start at the Earth’s center and point toward the Moon.
e
j
24
22
8
m1 x1 + m 2 x 2 5.98 × 10 kg 0 + 7.36 × 10 kg 3.84 × 10 m
=
m1 + m 2
6.05 × 10 24 kg
x CM =
= 4.67 × 10 6 m from the Earth’s center
The center of mass is within the Earth, which has radius 6.37 × 10 6 m.
P9.41
Let A1 represent the area of the bottom row of squares, A 2
the middle square, and A3 the top pair.
A = A1 + A 2 + A 3
M = M1 + M 2 + M 3
M1 M
=
A1
A
A1 = 300 cm 2 , A 2 = 100 cm 2 , A3 = 200 cm 2 , A = 600 cm 2
A
300 cm 2
M
M1 = M 1 =
M=
2
A
2
600 cm
M2
M3
x CM
x CM
FG IJ
H K
F A IJ = 100 cm
= MG
H A K 600 cm
F A IJ = 200 cm
= MG
H A K 600 cm
2
2
2
2
M=
M
6
FIG. P9.41
M
3
x M + x 2 M 2 + x 3 M 3 15.0 cm
= 1 1
=
M
= 11.7 cm
yCM =
3
1
2
a
f
2
M=
c M h + 5.00 cmc Mh + 10.0 cmc Mh
a
yCM = 13.3 cm
1
6
M
f c Mha25.0 cmf = 13.3 cm
M 5.00 cm + 16 M 15.0 cm +
M
1
2
1
3
1
3
Chapter 9
*P9.42
(a)
267
Represent the height of a particle of mass dm within the object as y. Its contribution to the
gravitational energy of the object-Earth system is dm gy . The total gravitational energy is
1
Ug =
gy dm = g y dm . For the center of mass we have yCM =
y dm , so U g = gMyCM .
M
all mass
z
(b)
a f
z
a
e
fa
f
1
3.6 m 15.7 m 64.8 m = 1.83 × 10 3 m 3 . Its mass is
2
1.83 × 10 3 m 3 = 6.96 × 10 6 kg . Its center of mass is above its base by one-
The volume of the ramp is
ρV = 3 800 kg m3
fa
z
je
j
1
third of its height, yCM = 15.7 m = 5.23 m . Then
3
U g = MgyCM = 6.96 × 10 6 kg 9.8 m s 2 5.23 m = 3.57 × 10 8 J .
e
P9.43
M=
(a)
z
z
0.300 m
0 .300 m
0
0
λdx =
j
50.0 g m + 20.0 x g m 2 dx
M = 50.0 x g m + 10.0 x 2 g m 2
x CM =
(b)
x CM
*P9.44
z
xdm
all mass
M
=
LM
MN
1
M
z
0.300 m
λ xdx =
0
0.300 m
0
1
M
z
= 15.9 g
0.300 m
50.0 x g m + 20.0 x 2 g m 2 dx
0
3
20 x g m 2
1
25.0 x 2 g m +
=
15.9 g
3
OP
PQ
0.300 m
= 0.153 m
0
Take the origin at the center of curvature. We have L =
r=
2L
π
1
2πr ,
4
. An incremental bit of the rod at angle θ from the x axis has
dm M
Mr
=
, dm =
dθ where we have used the
rdθ L
L
definition of radian measure. Now
mass given by
yCM =
=
z
z
135 °
1
1
Mr
r2
y dm =
r sin θ
dθ =
M all mass
M θ = 45 °
L
L
FG 2L IJ 1 a− cosθ f
HπK L
2
135°
45 °
=
4L
π
2
2L
π
−
4 2L
π
2
=
F1 − 2 2 I L =
π GH
π JK
2
z
θ
x
135 °
sin θ dθ
FIG. P9.44
45 °
FG 1 + 1 IJ = 4 2L
H 2 2K π
The top of the bar is above the origin by r =
by
y
0.063 5L .
2
2L
π
, so the center of mass is below the middle of the bar
268
Linear Momentum and Collisions
Section 9.6
P9.45
Motion of a System of Particles
(a)
v CM =
=
v CM =
P9.46
b
m1 v 1 + m 2 v 2
M
2.00 kg 2.00 i m s − 3.00 j m s + 3.00 kg 1.00 i m s + 6.00 j m s
M
=
ge
j b
ge
e1.40i + 2.40 jj m s
b
ge
j
e7.00i + 12.0jj kg ⋅ m s
p = Mv CM = 5.00 kg 1. 40 i + 2.40 j m s =
(a)
See figure to the right.
(b)
Using the definition of the position vector at the center of mass,
rCM =
2.00 kg + 3.00 kg
e−2.00i − 1.00jj m
FIG. P9.46
The velocity of the center of mass is
v CM =
v CM =
(d)
m1 r1 + m 2 r2
m1 + m 2
b2.00 kg ga1.00 m, 2.00 mf + b3.00 kg ga−4.00 m, − 3.00 mf
rCM =
(c)
j
5.00 kg
(b)
rCM =
P9.47
∑ mi v i
b
gb
g b
gb
g
2.00 kg 3.00 m s , 0.50 m s + 3.00 kg 3.00 m s , − 2.00 m s
P m1 v 1 + m 2 v 2
=
=
M
m1 + m 2
2.00 kg + 3.00 kg
b
e3.00i − 1.00jj m s
The total linear momentum of the system can be calculated as P = Mv CM
or as
P = m1 v 1 + m 2 v 2
Either gives
P=
e15.0 i − 5.00jj kg ⋅ m s
Let x = distance from shore to center of boat
A = length of boat
x ′ = distance boat moves as Juliet moves toward Romeo
The center of mass stays fixed.
Before:
h + M cx + h
dM + M + M i
M ax − x ′f + M cx + − x ′h + M c x + − x ′h
=
x
dM + M + M i
F 55.0 + 77.0 IJ = x ′a−80.0 − 55.0 − 77.0f + A a55.0 + 77.0f
AG −
H 2 2K
2
55.0 A 55.0a 2.70f
x′ =
=
= 0.700 m
x CM =
c
Mb x + M J x −
B
After:
B
A
2
R
J
R
J
A
2
A
2
R
CM
B
212
212
J
R
A
2
FIG. P9.47
g
Chapter 9
P9.48
(a)
269
Conservation of momentum for the two-ball system gives us:
b
g
b
g
0.200 kg 1.50 m s + 0.300 kg −0.400 m s = 0.200 kg v1 f + 0.300 kg v 2 f
Relative velocity equation:
v 2 f − v1 f = 1.90 m s
d
0.300 − 0.120 = 0.200 v1 f + 0.300 1.90 + v1 f
Then
v1 f = −0.780 m s
v 2 f = 1.12 m s
v 1 f = −0.780 i m s
(b)
v CM =
Before,
i
v 2 f = 1.12 i m s
b0.200 kg gb1.50 m sgi + b0.300 kg gb−0.400 m sgi
b
g
0.500 kg
v CM = 0.360 m s i
Afterwards, the center of mass must move at the same velocity, as momentum of the system
is conserved.
Section 9.7
P9.49
Rocket Propulsion
dM
dt
e
je
j
Thrust = 2.60 × 10 3 m s 1.50 × 10 4 kg s = 3.90 × 10 7 N
(a)
Thrust = v e
(b)
∑ Fy = Thrust − Mg = Ma :
ja f e
e
j
3.90 × 10 7 − 3.00 × 10 6 9.80 = 3.00 × 10 6 a
a = 3.20 m s 2
*P9.50
(a)
dM 12.7 g
=
= 6.68 × 10 −3 kg s
dt
1.90 s
The fuel burns at a rate
Thrust = v e
dM
:
dt
e
5.26 N = v e 6.68 × 10 −3 kg s
j
v e = 787 m s
(b)
F M I:
GH M JK
v f − vi = v e ln
g FGH 53.553g .+5 25g .+5 25g .−512g .7 g IJK
b
i
v f − 0 = 797 m s ln
f
v f = 138 m s
P9.51
v = v e ln
Mi
Mf
(a)
Mi = e v v e M f
Mi = e 5 3.00 × 10 3 kg = 4.45 × 10 5 kg
The mass of fuel and oxidizer is
∆M = Mi − M f = 445 − 3.00 × 10 3 kg = 442 metric tons
(b)
a
e
a
j
f
f
∆M = e 2 3.00 metric tons − 3.00 metric tons = 19.2 metric tons
Because of the exponential, a relatively small increase in fuel and/or engine efficiency causes
a large change in the amount of fuel and oxidizer required.
F M I = − v lnF M I
GH M JK
GH M JK
F M − kt IJ = − v lnFG 1 − k tIJ
= M − kt , so v = − v lnG
H M K
H M K
From Equation 9.41, v − 0 = v e ln
i
Now, M f
i
i
f
e
f
e
i
e
i
i
M
With the definition, Tp ≡ i , this becomes
k
F
GH
I
JK
F t IJ
= 144 s , v = −b1 500 m sg lnG 1 −
H 144 s K
af
v t = − v e ln 1 −
With v e = 1 500 m s, and Tp
a f vbm sg
v (m/s)
ts
1220
1780
120
2690
132
3730
FH
dt
p
IK OP F 1
Q = −v G
GH 1 −
e
t
Tp
IF 1 I F v IF 1
JJ GH − T JK = GH T JK GG 1 −
K
H
e
p
p
t
Tp
I
JJ , or
K
ve
Tp − t
1 500 m s
144 s − t
a f aem s j
2
0
10.4
20
12.1
40
14.4
60
17.9
80
23.4
100
34.1
120
62.5
132
125
continued on next page
2
a (m/s )
140
120
100
80
60
40
20
0
t (s)
FIG. P9.52(d)
140
ts
120
With v e = 1 500 m s, and Tp = 144 s , a =
100
(d)
LM
N
d − v e ln 1 − Tt
80
af
at =
FIG. P9.52(b)
60
af
dv
at =
=
dt
t (s)
140
80
100
120
808
100
60
2500
2000
1500
1000
500
0
80
488
60
40
40
0
224
40
(c)
0
20
4000
3500
3000
20
(b)
t
Tp
20
(a)
0
P9.52
Linear Momentum and Collisions
0
270
Chapter 9
(e)
LM F t I OP
LM1 − t OPF − dt I
−
−
=
v
ln
1
dt
v
T
ln
G
J
z z MN H T K PQ z MN T PQGH T JK
LF t I F t I F t I O
xat f = v T MG 1 − J lnG 1 − J − G 1 − J P
MNH T K H T K H T K PQ
F tI
xat f = v eT − t j lnG 1 − J + v t
H TK
t
af
x t = 0 + vdt =
0
t
t
e
e p
p
0
0
p
p
t
e p
e
p
p
p
0
e
p
With v e = 1 500 m s = 1.50 km s , and Tp = 144 s ,
f FGH
IJ
K
t
+ 1.50t
144
a f xakmf
ts
60
22.1
80
42.2
100
71.7
120
115
132
153
100
80
60
40
20
0
t (s)
FIG. P9.52(f)
*P9.53
The thrust acting on the spacecraft is
∑ F = b3 500 kg ge2.50 × 10 −6 je9.80
∑ F = ma :
thrust =
FG dM IJ v :
H dt K
e
8.58 × 10 −2 N =
∆M = 4.41 kg
F ∆M I b70 m sg
GH 3 600 s JK
j
m s 2 = 8.58 × 10 −2 N
140
9.23
120
40
140
120
100
2.19
80
20
160
60
0
0
0
x (km)
40
a
x = 1.50 144 − t ln 1 −
20
(f)
p
271
272
Linear Momentum and Collisions
Additional Problems
P9.54
(a)
When the spring is fully compressed, each cart moves with same velocity v. Apply
conservation of momentum for the system of two gliders
b
pi = p f :
(b)
g
v=
m1 v 1 + m 2 v 2 = m1 + m 2 v
m1 v 1 + m 2 v 2
m1 + m 2
b
g
1
1
1
1
m1 v12 + m 2 v 22 = m1 + m 2 v 2 + kx m2
2
2
2
2
Only conservative forces act, therefore ∆E = 0 .
Substitute for v from (a) and solve for x m .
x m2
bm
=
xm =
(c)
1
g
b
g
b
+ m 2 m1 v12 + m1 + m 2 m 2 v 22 − m1 v1
b
k m1 + m 2
e
m1 m 2 v12 + v 22 − 2 v1 v 2
b
k m1 + m 2
g
j = bv
1
− v2
g
g − bm v g
2
2 2
2
− 2m 1 m 2 v1 v 2
g kbmm m+ m g
1
2
1
2
m1 v 1 + m 2 v 2 = m1 v 1 f + m 2 v 2 f
d
i
d
i
Conservation of momentum:
m1 v 1 − v 1 f = m 2 v 2 f − v 2
Conservation of energy:
1
1
1
1
m1 v12 + m 2 v 22 = m1 v12 f + m 2 v 22 f
2
2
2
2
which simplifies to:
Factoring gives
e
m dv
j e
j
i ⋅ dv + v i = m dv
(1)
m1 v12 − v12 f = m 2 v 22 f − v 22
1
1
− v1 f
1
1f
2
2f
id
− v2 ⋅ v2 f + v2
i
and with the use of the momentum equation (equation (1)),
this reduces to
dv
or
v1 f = v 2 f + v 2 − v1
1
i d
+ v1 f = v 2 f + v 2
i
(2)
Substituting equation (2) into equation (1) and simplifying yields:
v2 f =
FG 2m IJ v + FG m
Hm +m K Hm
1
1
1
2
IJ
K
− m1
v2
+
1 m2
2
Upon substitution of this expression for v 2 f into equation 2, one finds
v1 f =
FG m
Hm
IJ
K
FG
H
IJ
K
− m2
2m 2
v1 +
v2
m1 + m 2
1 + m2
1
Observe that these results are the same as Equations 9.20 and 9.21, which should have been
expected since this is a perfectly elastic collision in one dimension.
P9.55
(a)
b60.0 kg g4.00 m s = a120 + 60.0f kgv
Chapter 9
273
f
v f = 1.33 m s i
(b)
b
∑ Fy = 0 :
g
n − 60.0 kg 9.80 m s 2 = 0
a
f
f k = µ k n = 0.400 588 N = 235 N
f = −235 N i
FIG. P9.55
k
(c)
For the person, pi + I = p f
mvi + Ft = mv f
b60.0 kg g4.00 m s − a235 Nft = b60.0 kg g1.33 m s
t = 0.680 s
(d)
a
f
person:
mv f − mv i = 60.0 kg 1.33 − 4.00 m s = −160 N ⋅ si
cart:
120 kg 1.33 m s − 0 = +160 N ⋅ si
b
g
a
f
(e)
x f − xi =
1
1
4.00 + 1.33 m s 0.680 s = 1.81 m
vi + v f t =
2
2
(f)
x f − xi =
1
1
vi + v f t = 0 + 1.33 m s 0.680 s = 0.454 m
2
2
(g)
1
1
1
mv 2f − mvi2 = 60.0 kg 1.33 m s
2
2
2
(h)
1
1
1
mv 2f − mvi2 = 120.0 kg 1.33 m s
2
2
2
(i)
d
d
i
b
i
g
b
b
g
2
−
g
2
b
1
60.0 kg 4.00 m s
2
g
2
= −427 J
− 0 = 107 J
The force exerted by the person on the cart must equal in magnitude and opposite in
direction to the force exerted by the cart on the person. The changes in momentum of
the two objects must be equal in magnitude and must add to zero. Their changes in
kinetic energy are different in magnitude and do not add to zero. The following
represent two ways of thinking about ’ why. ’ The distance the cart moves is different
from the distance moved by the point of application of the friction force to the cart.
The total change in mechanical energy for both objects together, − 320 J, becomes
+320 J of additional internal energy in this perfectly inelastic collision.
P9.56
The equation for the horizontal range of a projectile is R =
vi2 sin 2θ
. Thus, with θ = 45.0° , the initial
g
velocity is
vi = Rg =
a f
a200 mfe9.80 m s j = 44.3 m s
2
I = F ∆t = ∆p = mvi − 0
Therefore, the magnitude of the average force acting on the ball during the impact is:
F=
e
jb
g
46.0 × 10 −3 kg 44.3 m s
mvi
=
= 291 N .
∆t
7.00 × 10 −3 s
274
P9.57
Linear Momentum and Collisions
We hope the momentum of the wrench provides enough recoil so that the astronaut can reach the
ship before he loses life support! We might expect the elapsed time to be on the order of several
minutes based on the description of the situation.
No external force acts on the system (astronaut plus wrench), so the total momentum is constant.
Since the final momentum (wrench plus astronaut) must be zero, we have final momentum = initial
momentum = 0.
m wrench v wrench + m astronaut v astronaut = 0
b
gb
g
0.500 kg 20.0 m s
m wrench v wrench
=−
= −0.125 m s
m astronaut
80.0 kg
At this speed, the time to travel to the ship is
Thus v astronaut = −
t=
30.0 m
= 240 s = 4.00 minutes
0.125 m s
The astronaut is fortunate that the wrench gave him sufficient momentum to return to the ship in a
reasonable amount of time! In this problem, we were told that the astronaut was not drifting away
from the ship when he threw the wrench. However, this is not quite possible since he did not
encounter an external force that would reduce his velocity away from the ship (there is no air
friction beyond earth’s atmosphere). If this were a real-life situation, the astronaut would have to
throw the wrench hard enough to overcome his momentum caused by his original push away from
the ship.
P9.58
Using conservation of momentum from just before to just
after the impact of the bullet with the block:
a
f
m
mvi = M + m v f
or
vi =
FG M + m IJ v
H m K
f
vi
(1)
M
h
The speed of the block and embedded bullet just after
impact may be found using kinematic equations:
d = v f t and h =
Thus, t =
d
1 2
gt
2
g
2h
d
=
and v f = = d
g
t
2h
FIG. P9.58
gd 2
2h
Substituting into (1) from above gives vi =
FG M + m IJ
H m K
gd 2
.
2h
Chapter 9
*P9.59
(a)
275
Conservation of momentum:
e
j
e
j
0.5 kg 2 i − 3 j + 1k m s + 1.5 kg −1i + 2 j − 3k m s
e
j
= 0.5 kg −1i + 3 j − 8k m s + 1.5 kg v 2 f
v2 f =
e−0.5 i + 1.5j − 4k j kg ⋅ m s + e0.5i − 1.5j + 4k j kg ⋅ m s =
1.5 kg
0
The original kinetic energy is
1
1
0.5 kg 2 2 + 3 2 + 1 2 m 2 s 2 + 1.5 kg 1 2 + 2 2 + 3 2 m 2 s 2 = 14.0 J
2
2
e
j
e
j
1
0.5 kg 1 2 + 3 2 + 8 2 m 2 s 2 + 0 = 18.5 J different from the original
2
energy so the collision is inelastic .
e
The final kinetic energy is
(b)
j
We follow the same steps as in part (a):
e−0.5 i + 1.5j − 4k j kg ⋅ m s = 0.5 kge−0.25i + 0.75j − 2k j m s + 1.5 kg v
e−0.5i + 1.5j − 4k j kg ⋅ m s + e0.125 i − 0.375 j + 1k j kg ⋅ m s
v =
2f
2f
1.5 kg
e−0.250i + 0.750 j − 2.00k j m s
=
We see v 2 f = v 1 f , so the collision is perfectly inelastic .
(c)
Conservation of momentum:
e−0.5 i + 1.5j − 4k j kg ⋅ m s = 0.5 kge−1i + 3 j + ak j m s + 1.5 kg v
−0.5 i + 1.5 j − 4k j kg ⋅ m s + e0.5 i − 1.5 j − 0.5 ak j kg ⋅ m s
e
v =
2f
2f
1.5 kg
=
a−2.67 − 0.333 afk m s
Conservation of energy:
a
1
1
0.5 kg 1 2 + 3 2 + a 2 m 2 s 2 + 1.5 kg 2.67 + 0.333 a
2
2
2
= 2.5 J + 0. 25 a + 5.33 J + 1.33 a + 0.083 3 a 2
14.0 J =
e
j
f
2
m2 s 2
0 = 0.333 a 2 + 1.33 a − 6.167
a=
a
fa
−1.33 ± 1.33 2 − 4 0.333 −6.167
f
0.667
a = 2.74 or − 6.74. Either value is possible.
∴ a = 2.74 ,
a fh
c
= c−2.67 − 0.333a −6.74fhk m s =
v 2 f = −2.67 − 0.333 2.74 k m s = −3.58k m s
∴ a = −6.74 , v 2 f
−0.419 k m s
276
P9.60
Linear Momentum and Collisions
(a)
The initial momentum of the system is zero, which
remains constant throughout the motion.
Therefore, when m1 leaves the wedge, we must
have
m 2 v wedge + m1 v block = 0
(b)
or
b3.00 kg gv
so
v wedge = −0.667 m s
wedge
b
gb
g
+ 0.500 kg +4.00 m s = 0
v wedge
+x
Using conservation of energy for the block-wedgeEarth system as the block slides down the smooth
(frictionless) wedge, we have
FIG. P9.60
K block + U system + K wedge = K block + U system
i
or
*P9.61
(a)
0 + m1 gh + 0 =
i
f
LM 1 m a4.00f + 0OP + 1 m a−0.667f
N2
Q 2
1
2
2
2
+ K wedge
f
which gives h = 0.952 m .
Conservation of the x component of momentum for the cart-bucket-water system:
b
g
mvi + 0 = m + ρV v
(b)
v block = 4.00 m/s
vi =
m + ρV
v
m
Raindrops with zero x-component of momentum stop in the bucket and slow its horizontal
motion. When they drip out, they carry with them horizontal momentum. Thus the cart
slows with constant acceleration.
Chapter 9
P9.62
Consider the motion of the firefighter during the three
intervals:
(1) before, (2) during, and (3) after collision with the
platform.
(a)
v1
While falling a height of 4.00 m, his speed changes
from vi = 0 to v1 as found from
i b
d
v2
g
∆E = K f + U f − K i − U i , or
K f = ∆E − U f + K i + U i
When the initial position of the platform is taken as
the zero level of gravitational potential, we have
a f
1
mv12 = fh cos 180° − 0 + 0 + mgh
2
FIG. P9.62
Solving for v1 gives
v1 =
(b)
b
2 − fh + mgh
m
g = 2c−300a4.00f + 75.0a9.80f4.00h =
6.81 m s
75.0
During the inelastic collision, momentum is conserved; and if v 2 is the speed of the
firefighter and platform just after collision, we have mv1 = m + M v 2 or
a
v2 =
f
a f
75.0 6.81
m 1 v1
=
= 5.38 m s
m + M 75.0 + 20.0
Following the collision and again solving for the work done by non-conservative forces,
using the distances as labeled in the figure, we have (with the zero level of gravitational
potential at the initial position of the platform):
∆E = K f + U fg + U fs − K i − U ig − U is , or
1
1
− fs = 0 + m + M g − s + ks 2 − m + M v 2 − 0 − 0
2
2
a
fa f
a
f
This results in a quadratic equation in s:
a f
2 000s 2 − 931 s + 300s − 1 375 = 0 or s = 1.00 m
277
278
*P9.63
Linear Momentum and Collisions
(a)
Each object swings down according to
mgR =
1
mv12
2
MgR =
1
Mv12
2
a
v1 = 2 gR
f
The collision: −mv1 + Mv1 = + m + M v 2
M−m
v2 =
v1
M+m
Swinging up:
a
f a
f
a
f
2 gRa1 − cos 35°fa M + mf = a M − m f 2 gR
f
a
1
M + m v 22 = M + m gR 1 − cos 35°
2
v 2 = 2 gR 1 − cos 35°
0.425 M + 0.425m = M − m
1.425m = 0.575 M
m
= 0.403
M
P9.64
(b)
No change is required if the force is different. The nature of the forces within the system of
colliding objects does not affect the total momentum of the system. With strong magnetic
attraction, the heavier object will be moving somewhat faster and the lighter object faster
still. Their extra kinetic energy will all be immediately converted into extra internal energy
when the objects latch together. Momentum conservation guarantees that none of the extra
kinetic energy remains after the objects join to make them swing higher.
(a)
Use conservation of the horizontal component of
momentum for the system of the shell, the cannon,
and the carriage, from just before to just after the
cannon firing.
p xf = p xi :
or
(b)
m shell v shell cos 45.0°+ m cannon v recoil = 0
a200fa125f cos 45.0°+b5 000gv
recoil
=0
v recoil = −3.54 m s
FIG. P9.64
Use conservation of energy for the system of the cannon, the carriage, and the spring from
right after the cannon is fired to the instant when the cannon comes to rest.
K f + U gf + U sf = K i + U gi + U si :
0+0+
x max =
1 2
1
2
kx max = mv recoil
+0+0
2
2
2
mv recoil
=
k
e
b5 000ga−3.54f
2
2.00 × 10 4
ja
m = 1.77 m
f
Fs, max = 2.00 × 10 4 N m 1.77 m = 3.54 × 10 4 N
(c)
Fs, max = kx max
(d)
No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it
from recoiling vertically. Momentum is not conserved in the vertical direction. The spring
does not have time to stretch during the cannon firing. Thus, no external horizontal force is
exerted on the system (cannon, carriage, and shell) from just before to just after firing.
Momentum of this system is conserved in the horizontal direction during this interval.
Chapter 9
P9.65
(a)
Utilizing conservation of
momentum,
b
g
m + m2
= 1
m1
2 gh
279
v1i
m 1 v1 A = m 1 + m 2 v B
v1 A
y
v1 A ≅ 6.29 m s
(b)
x
Utilizing the two equations,
FIG. P9.65
1 2
gt = y and x = v1 A t
2
we combine them to find
v1 A =
x
2y
g
From the data, v1 A = 6.16 m s
Most of the 2% difference between the values for speed is accounted for by the uncertainty
0.01 0.1
1
1
0.1
in the data, estimated as
+
+
+
+
= 1.1% .
8.68 68.8 263 257 85.3
*P9.66
The ice cubes leave the track with speed determined by mgyi =
e
1
mv 2 ;
2
j
v = 2 9.8 m s 2 1.5 m = 5.42 m s .
Its speed at the apex of its trajectory is 5.42 m s cos 40° = 4.15 m s . For its collision with the wall we
have
mvi + F∆t = mv f
FG
H
0.005 kg 4.15 m s + F∆t = 0.005 kg −
IJ
K
1
4.15 m s
2
F∆t = −3.12 × 10 −2 kg ⋅ m s
The impulse exerted by the cube on the wall is to the right, +3.12 × 10 −2 kg ⋅ m s. Here F could refer
to a large force over a short contact time. It can also refer to the average force if we interpret ∆t as
1
s, the time between one cube’s tap and the next’s.
10
Fav =
3.12 × 10 −2 kg ⋅ m s
= 0.312 N to the right
0.1 s
280
P9.67
Linear Momentum and Collisions
(a)
Find the speed when the bullet emerges from the
block by using momentum conservation:
400 m/s
mvi = MVi + mv
The block moves a distance of 5.00 cm. Assume for
an approximation that the block quickly reaches its
maximum velocity, Vi , and the bullet kept going
with a constant velocity, v. The block then
compresses the spring and stops.
FIG. P9.67
1
1
MVi2 = kx 2
2
2
Vi =
v=
v
5.00 cm
b900 N mge5.00 × 10
j
−2
m
2
= 1.50 m s
1.00 kg
jb
e
g b
gb
5.00 × 10 −3 kg 400 m s − 1.00 kg 1.50 m s
mvi − MVi
=
m
5.00 × 10 −3 kg
g
v = 100 m s
(b)
jb
mj
1
5.00 × 10 −3 kg 100 m s
2
e
1
+ b900 N mge5.00 × 10
2
∆ E = ∆K + ∆ U =
−2
g
2
−
jb
1
5.00 × 10 −3 kg 400 m s
2
e
g
2
2
∆E = −374 J , or there is an energy loss of 374 J .
*P9.68
The orbital speed of the Earth is
vE =
S
CM
2πr 2π 1.496 × 10 11 m
=
= 2.98 × 10 4 m s
T
3.156 × 10 7 s
E
In six months the Earth reverses its direction, to undergo
momentum change
e
je
FIG. P9.68
j
m E ∆v E = 2m E v E = 2 5.98 × 10 24 kg 2.98 × 10 4 m s = 3.56 × 10 25 kg ⋅ m s .
Relative to the center of mass, the sun always has momentum of the same magnitude in the
opposite direction. Its 6-month momentum change is the same size, mS ∆v S = 3.56 × 10 25 kg ⋅ m s .
Then ∆v S =
3.56 × 10 25 kg ⋅ m s
1.991 × 10 30 kg
= 0.179 m s .
Chapter 9
P9.69
(a)
b3.00 kg gb7.00 m sgj + e12.0 Nija5.00 sf = b3.00 kg gv
v = e 20.0 i + 7.00 jj m s
p i + Ft = p f :
f
f
(b)
a=
(c)
a=
(d)
∆r = v i t +
(e)
(f)
(g)
P9.70
v f − vi
t
20.0 i + 7.00 j − 7.00 jj m s
e
a=
=
:
5.00 s
∑F :
a=
m
4.00 i m s 2
12.0 N i
= 4.00 i m s 2
3.00 kg
ja
f e
ja
1
∆r = 7.00 m s j 5.00 s + 4.00 m s 2 i 5.00 s
2
∆r = 50.0 i + 35.0 j m
1 2
at :
2
e
e
f
2
j
e
je
j
W = 12.0 N i ⋅ 50.0 m i + 35.0 mj = 600 J
W = F ⋅ ∆r :
b
ge
1
1
mv 2f = 3.00 kg 20.0 i + 7.00 j ⋅ 20.0 i + 7.00 j m 2 s 2
2
2
1
mv 2f = 1.50 kg 449 m 2 s 2 = 674 J
2
b
je
b
gb
1
1
mvi2 + W = 3.00 kg 7.00 m s
2
2
g
2
j
ge
j
+ 600 J = 674 J
b
g
M = 360 kg − 2.50 kg s t .
We find the mass from
b
gb
g
1 500 m s 2.50 kg s
3 750 N
Thrust v e dM dt
=
=
=
M
M
M
M
We find the velocity and position according to Euler,
v new = v old + a ∆t
from
x new = x old + v ∆t
and
If we take ∆t = 0.132 s , a portion of the output looks like this:
We find the acceleration from
a=
a f
a f
Time
t(s)
Total mass
(kg)
Acceleration
a m s2
Speed, v
(m/s)
Position
x(m)
0.000
0.132
0.264
...
65.868
66.000
66.132
...
131.736
131.868
132.000
360.00
359.67
359.34
10.4167
10.4262
10.4358
0.0000
1.3750
2.7513
0.0000
0.1815
0.54467
195.330
195.000
194.670
19.1983
19.2308
19.2634
916.54
919.08
921.61
27191
27312
27433
30.660
30.330
30.000
122.3092
123.6400
125.0000
3687.3
3703.5
3719.8
152382
152871
153362
(a)
The final speed is
(b)
The rocket travels
e
j
v f = 3.7 km s
153 km
281
282
P9.71
Linear Momentum and Collisions
The force exerted by the table is equal to the change in momentum
of each of the links in the chain.
By the calculus chain rule of derivatives,
F1 =
a f
dp d mv
dm
dv
=
=v
+m .
dt
dt
dt
dt
We choose to account for the change in momentum of each link by
having it pass from our area of interest just before it hits the table,
so that
v
FIG. P9.71
dm
dv
≠ 0 and m
= 0.
dt
dt
Since the mass per unit length is uniform, we can express each link of length dx as having a mass dm:
dm =
M
dx .
L
The magnitude of the force on the falling chain is the force that will be necessary to stop each of the
elements dm.
F1 = v
FG IJ
H K
FG IJ
H K
dm
M dx
M 2
=v
=
v
dt
L dt
L
After falling a distance x, the square of the velocity of each link v 2 = 2 gx (from kinematics), hence
F1 =
2 Mgx
.
L
The links already on the table have a total length x, and their weight is supported by a force F2 :
F2 =
Mgx
.
L
Hence, the total force on the chain is
Ftotal = F1 + F2 =
3 Mgx
.
L
That is, the total force is three times the weight of the chain on the table at that instant.
Chapter 9
P9.72
A picture one second later differs by showing five extra kilograms of sand moving on the belt.
b
gb
g
(a)
5.00 kg 0.750 m s
∆p x
=
= 3.75 N
∆t
1.00 s
(b)
The only horizontal force on the sand is belt friction,
p xi + f∆t = p xf
so from
(c)
∆p x
= 3.75 N
∆t
this is
f=
and
Fext = 3.75 N
The belt is in equilibrium:
∑ Fx = ma x :
+ Fext − f = 0
a
f
(d)
W = F∆r cos θ = 3.75 N 0.750 m cos 0° = 2.81 J
(e)
1
1
∆m v 2 = 5.00 kg 0.750 m s
2
2
a f
(f)
*P9.73
283
x CM =
b
g
2
= 1.41 J
Friction between sand and belt converts half of the input work into extra internal energy.
∑ m i xi
∑ mi
=
c
m1 R +
A
2
h + m a0 f =
2
c
m1 R +
m1 + m 2
A
2
h
y
m1 + m 2
x
R
A
2
FIG. P9.73
ANSWERS TO EVEN PROBLEMS
P9.2
(a) 0; (b) 1.06 kg ⋅ m s ; upward
P9.20
0.556 m
P9.4
(a) 6.00 m s to the left; (b) 8.40 J
P9.22
1.78 kN on the truck driver; 8.89 kN in the
opposite direction on the car driver
P9.6
The force is 6.44 kN
P9.24
v=
4M
m
gA
P9.8
1.39 kg ⋅ m s upward
P9.10
(a) 5.40 N ⋅ s toward the net; (b) −27.0 J
P9.26
7.94 cm
P9.12
~ 10 3 N upward
P9.28
(a) 2.88 m s at 32.3°; (b) 783 J becomes
internal energy
P9.14
(a) and (c) see the solution; (b) small;
(d) large; (e) no difference
P9.30
v Y = vi sin θ ; vO = vi cos θ
P9.16
1.67 m s
P9.32
No; his speed was 41.5 mi h
P9.18
(a) 2.50 m s ; (b) 3.75 × 10 4 J
P9.34
(a) v =
vi
2
; (b) 45.0° and –45.0°
284
Linear Momentum and Collisions
2
vi ; (b) 35.3°
3
P9.36
(a)
P9.38
a0, 1.00 mf
P9.40
4.67 × 10 6 m from the Earth’s center
P9.42
(a) see the solution; (b) 3.57 × 10 8 J
2vi ;
v2 f =
P9.44
0.063 5L
P9.46
(a) see the solution;
(b) −2.00 m, − 1.00 m ;
(c) 3.00 i − 1.00 j m s ;
a
f
e
j
(d) e15.0 i − 5.00 jj kg ⋅ m s
P9.48
(a) −0.780 i m s ; 1.12 i m s; (b) 0.360 i m s
P9.50
(a) 787 m s; (b) 138 m s
P9.52
see the solution
P9.54
(a)
m1 v 1 + m 2 v 2
;
m1 + m 2
b
(b) v1 − v 2
(c) v 1 f =
g kbmm m+ m g ;
1
1
2
2
FG m
Hm
IJ
K
FG
H
1
− m1
v2
1 + m2
FG 2m IJ v + FG m
Hm +m K Hm
1
1
IJ
K
− m2
2m 2
v1 +
v2 ;
m1 + m 2
1 + m2
1
2
2
IJ
K
P9.56
291 N
P9.58
FG M + m IJ
H m K
P9.60
(a) −0.667 m s; (b) 0.952 m
P9.62
(a) 6.81 m s; (b) 1.00 m
P9.64
(a) −3.54 m s ; (b) 1.77 m; (c) 35.4 kN;
(d) No. The rails exert a vertical force to
change the momentum
P9.66
0.312 N to the right
P9.68
0.179 m s
P9.70
(a) 3.7 km s ; (b) 153 km
P9.72
(a) 3.75 N to the right; (b) 3.75 N to the
right; (c) 3.75 N; (d) 2.81 J; (e) 1.41 J;
(f) Friction between sand and belt converts
half of the input work into extra internal
energy.
gd 2
2h
10
Rotation of a Rigid Object
About a Fixed Axis
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
Angular Position, Velocity,
and Acceleration
Rotational Kinematics:
Rotational Motion with
Constant Angular
Acceleration
Angular and Linear
Quantities
Rotational Energy
Calculation of Moments of
Inertia
Torque
Relationship Between
Torque and Angular
Acceleration
Work, Power, and Energy
in Rotational Motion
Rolling Motion of a Rigid
Object
Q10.1
1 rev/min, or
π
rad/s. Into the wall (clockwise rotation). α = 0.
30
FIG. Q10.1
Q10.2
+ k , − k
Q10.3
Yes, they are valid provided that ω is measured in degrees per
second and α is measured in degrees per second-squared.
Q10.4
The speedometer will be inaccurate. The speedometer measures the number of revolutions per
second of the tires. A larger tire will travel more distance in one full revolution as 2πr .
Q10.5
Smallest I is about x axis and largest I is about y axis.
Q10.6
ML2
if the mass was nonuniformly distributed, nor
12
could it be calculated if the mass distribution was not known.
The moment of inertia would no longer be
Q10.7
The object will start to rotate if the two forces act along different lines. Then the torques of the forces
will not be equal in magnitude and opposite in direction.
Q10.8
No horizontal force acts on the pencil, so its center of mass moves straight down.
Q10.9
You could measure the time that it takes the hanging object, m, to fall a measured distance after
being released from rest. Using this information, the linear acceleration of the mass can be
calculated, and then the torque on the rotating object and its angular acceleration.
Q10.10
You could use ω = αt and v = at . The equation v = Rω is valid in this situation since a = Rα .
Q10.11
The angular speed ω would decrease. The center of mass is farther from the pivot, but the moment
of inertia increases also.
285
286
Rotation of a Rigid Object About a Fixed Axis
Q10.12
The moment of inertia depends on the distribution of mass with respect to a given axis. If the axis is
changed, then each bit of mass that makes up the object is a different distance from the axis. In
example 10.6 in the text, the moment of inertia of a uniform rigid rod about an axis perpendicular to
the rod and passing through the center of mass is derived. If you spin a pencil back and forth about
this axis, you will get a feeling for its stubbornness against changing rotation. Now change the axis
about which you rotate it by spinning it back and forth about the axis that goes down the middle of
the graphite. Easier, isn’t it? The moment of inertia about the graphite is much smaller, as the mass
of the pencil is concentrated near this axis.
Q10.13
Compared to an axis through the center of mass, any other parallel axis will have larger average
squared distance from the axis to the particles of which the object is composed.
Q10.14
A quick flip will set the hard–boiled egg spinning faster and more smoothly. The raw egg loses
mechanical energy to internal fluid friction.
Q10.15
I CM = MR 2 , I CM = MR 2 , I CM =
Q10.16
Yes. If you drop an object, it will gain translational kinetic energy from decreasing gravitational
potential energy.
Q10.17
No, just as an object need not be moving to have mass.
Q10.18
No, only if its angular momentum changes.
Q10.19
Yes. Consider a pendulum at its greatest excursion from equilibrium. It is momentarily at rest, but
must have an angular acceleration or it would not oscillate.
Q10.20
Since the source reel stops almost instantly when the tape stops playing, the friction on the source
reel axle must be fairly large. Since the source reel appears to us to rotate at almost constant angular
velocity, the angular acceleration must be very small. Therefore, the torque on the source reel due to
the tension in the tape must almost exactly balance the frictional torque. In turn, the frictional torque
is nearly constant because kinetic friction forces don’t depend on velocity, and the radius of the axle
where the friction is applied is constant. Thus we conclude that the torque exerted by the tape on
the source reel is essentially constant in time as the tape plays.
v
must increase to keep the
As the source reel radius R shrinks, the reel’s angular speed ω =
R
tape speed v constant. But the biggest change is to the reel’s moment of inertia. We model the reel as
a roll of tape, ignoring any spool or platter carrying the tape. If we think of the roll of tape as a
1
uniform disk, then its moment of inertia is I = MR 2 . But the roll’s mass is proportional to its base
2
area π R 2 . Thus, on the whole the moment of inertia is proportional to R 4 . The moment of inertia
decreases very rapidly as the reel shrinks!
The tension in the tape coming into the read-and-write heads is normally dominated by
balancing frictional torque on the source reel, according to TR ≈ τ friction . Therefore, as the tape plays
the tension is largest when the reel is smallest. However, in the case of a sudden jerk on the tape, the
rotational dynamics of the source reel becomes important. If the source reel is full, then the moment
of inertia, proportional to R 4 , will be so large that higher tension in the tape will be required to give
the source reel its angular acceleration. If the reel is nearly empty, then the same tape acceleration
will require a smaller tension. Thus, the tape will be more likely to break when the source reel is
nearly full. One sees the same effect in the case of paper towels; it is easier to snap a towel free when
the roll is new than when it is nearly empty.
1
1
MR 2 , I CM = MR 2
3
2
Chapter 10
287
Q10.21
The moment of inertia would decrease. This would result in a higher angular speed of the earth,
shorter days, and more days in the year!
Q10.22
There is very little resistance to motion that can reduce the kinetic energy of the rolling ball. Even
though there is static friction between the ball and the floor (if there were none, then no rotation
would occur and the ball would slide), there is no relative motion of the two surfaces—by the
definition of “rolling”—and so no force of kinetic friction acts to reduce K. Air resistance and friction
associated with deformation of the ball eventually stop the ball.
Q10.23
In the frame of reference of the ground, no. Every point
moves perpendicular to the line joining it to the
instantaneous contact point. The contact point is not
moving at all. The leading and trailing edges of the
cylinder have velocities at 45° to the vertical as shown.
v
vCM
CM
v
P
FIG. Q10.23
Q10.24
The sphere would reach the bottom first; the hoop would reach the bottom last. If each object has
the same mass and the same radius, they all have the same torque due to gravity acting on them.
The one with the smallest moment of inertia will thus have the largest angular acceleration and
reach the bottom of the plane first.
Q10.25
To win the race, you want to decrease the moment of inertia of the wheels as much as possible.
Small, light, solid disk-like wheels would be best!
SOLUTIONS TO PROBLEMS
Section 10.1
P10.1
(a)
(b)
Angular Position, Velocity, and Acceleration
θ t= 0 = 5.00 rad
ω t =0 =
dθ
dt
α t=0 =
dω
dt
t=0
= 10.0 + 4.00t t = 0 = 10.0 rad s
= 4.00 rad s 2
t=0
θ t= 3.00 s = 5.00 + 30.0 + 18.0 = 53.0 rad
ω t = 3.00 s =
dθ
dt
α t = 3.00 s =
dω
dt
t = 3 .00 s
= 10.0 + 4.00t t = 3.00 s = 22.0 rad s
= 4.00 rad s 2
t = 3 .00 s
288
Rotation of a Rigid Object About a Fixed Axis
Section 10.2
*P10.2
P10.3
P10.4
Rotational Kinematics: Rotational Motion with Constant Angular Acceleration
ω f = 2.51 × 10 4 rev min = 2.63 × 10 3 rad s
P10.6
2.63 × 10 3 rad s − 0
= 8.22 × 10 2 rad s 2
3.2 s
α=
(b)
1
1
θ f = ω i t + αt 2 = 0 + 8.22 × 10 2 rad s 2 3.2 s
2
2
(a)
α=
(b)
1
1
θ = ω i t + αt 2 = 4.00 rad s 2 3.00 s
2
2
=
t
ja f
e
2
= 4.21 × 10 3 rad
ω − ω i 12.0 rad s
=
= 4.00 rad s 2
t
3.00 s
ja
e
f
2
= 18.0 rad
ω i = 2 000 rad s , α = −80.0 rad s 2
a fa f
(a)
ω f = ω i + αt = 2 000 − 80.0 10.0 = 1 200 rad s
(b)
0 = ω i + αt
t=
P10.5
ω f −ωi
(a)
ωi =
ωi
=
−α
2 000
= 25.0 s
80.0
FG
H
IJ FG 2π rad IJ = 10π
K H 1.00 rev K 3
100 rev 1 min
1.00 min 60.0 s
ω f −ωi
(a)
t=
(b)
θ f = ωt =
α
=
FG ω
H
f
0 − 103π
−2.00
+ωi
2
rad s , ω f = 0
s = 5.24 s
IJ t = FG 10π
K H6
rad s
IJ FG 10π sIJ =
KH 6 K
27.4 rad
ω i = 3 600 rev min = 3.77 × 10 2 rad s
θ = 50.0 rev = 3.14 × 10 2 rad and ω f = 0
ω 2f = ω i2 + 2αθ
e
0 = 3.77 × 10 2 rad s
j
2
e
+ 2α 3.14 × 10 2 rad
j
α = −2.26 × 10 2 rad s 2
P10.7
ω = 5.00 rev s = 10.0π rad s . We will break the motion into two stages: (1) a period during which the
tub speeds up and (2) a period during which it slows down.
0 + 10.0π rad s
8.00 s = 40.0π rad
2
a f
10.0π rad s + 0
= ωt =
a12.0 sf = 60.0π rad
2
While speeding up,
θ 1 = ωt =
While slowing down,
θ2
So,
θ total = θ 1 + θ 2 = 100π rad = 50.0 rev
Chapter 10
P10.8
1
θ f − θ i = ω i t + αt 2 and ω f = ω i + αt are two equations in two unknowns ω i and α
2
1
1
θ f − θ i = ω f − αt t + αt 2 = ω f t − αt 2
2
2
2π rad
1
37.0 rev
= 98.0 rad s 3.00 s − α 3.00 s
1 rev
2
d
ω i = ω f − αt :
i
FG
H
e
j
232 rad = 294 rad − 4.50 s 2 α :
P10.9
*P10.10
289
(a)
ω=
(b)
∆t =
α=
IJ
K
a
f
a
f
2
61.5 rad
= 13.7 rad s 2
2
4.50 s
∆θ 1 rev
2π rad
=
=
= 7.27 × 10 −5 rad s
∆t 1 day 86 400 s
∆θ
ω
=
FG
H
IJ
K
= b0.750 rad sgt . For the bone,
107°
2π rad
= 2.57 × 10 4 s or 428 min
−5
7.27 × 10 rad s 360°
The location of the dog is described by θ d
θb =
1
1
2π rad + 0.015 rad s 2 t 2 .
3
2
We look for a solution to
2π
+ 0.007 5t 2
3
0 = 0.007 5t 2 − 0.75t + 2.09 = 0
0.75t =
t=
b
g
0.75 ± 0.75 2 − 4 0.007 5 2.09
0.015
= 2.88 s or 97.1 s
2π
2π
− 2π + 0.007 5t 2 or if 0.75t =
+ 2π + 0.007 5t 2 that is, if
3
3
either the dog or the turntable gains a lap on the other. The first equation has
The dog and bone will also pass if 0.75t =
t=
b
ga
0.75 ± 0.75 2 − 4 0.007 5 −4.19
0.015
f = 105 s or − 5.30 s
only one positive root representing a physical answer. The second equation has
t=
b
g
0.75 ± 0.75 2 − 4 0.007 5 8.38
0.015
= 12.8 s or 87.2 s .
In order, the dog passes the bone at 2.88 s after the merry-go-round starts to turn, and again at
12.8 s and 26.6 s, after gaining laps on the bone. The bone passes the dog at 73.4 s, 87.2 s, 97.1 s,
105 s, and so on, after the start.
290
Rotation of a Rigid Object About a Fixed Axis
Section 10.3
P10.11
Angular and Linear Quantities
Estimate the tire’s radius at 0.250 m and miles driven as 10 000 per year.
θ=
θ = 6.44 × 10 7
P10.12
P10.13
FG
IJ
H
K
F 1 rev IJ = 1.02 × 10
rad yr G
H 2π rad K
s 1.00 × 10 4 mi 1 609 m
=
= 6.44 × 10 7 rad yr
0.250 m
1 mi
r
7
rev yr or ~ 10 7 rev yr
v 45.0 m s
=
= 0.180 rad s
r
250 m
(a)
v = rω ; ω =
(b)
45.0 m s
v2
=
ar =
r
250 m
b
g
2
= 8.10 m s 2 toward the center of track
Given r = 1.00 m, α = 4.00 rad s 2 , ω i = 0 and θ i = 57.3° = 1.00 rad
(a)
ω f = ω i + αt = 0 + αt
a
f
At t = 2.00 s , ω f = 4.00 rad s 2 2.00 s = 8.00 rad s
(b)
b
g
v = rω = 1.00 m 8.00 rad s = 8.00 m s
b
a r = a c = rω 2 = 1.00 m 8.00 rad s
e
g
2
= 64.0 m s 2
j
a t = rα = 1.00 m 4.00 rad s 2 = 4.00 m s 2
The magnitude of the total acceleration is:
a = a r2 + a t2 =
e64.0 m s j + e4.00 m s j
2 2
2 2
= 64.1 m s 2
The direction of the total acceleration vector makes an angle φ with respect to the radius to
point P:
φ = tan −1
FG a IJ = tan FG 4.00 IJ =
H 64.0 K
Ha K
t
−1
3.58°
c
(c)
a
f e
ja
1
1
θ f = θ i + ω i t + αt 2 = 1.00 rad + 4.00 rad s 2 2.00 s
2
2
f
2
= 9.00 rad
Chapter 10
*P10.14
(a)
Consider a tooth on the front sprocket. It gives this speed, relative to the frame, to the link of
the chain it engages:
v = rω =
(b)
FG 0.152 m IJ 76 rev min FG 2π rad IJ FG 1 min IJ =
H 2 K
H 1 rev K H 60 s K
v 0.605 m s
= 0.07 m = 17.3 rad s
r
2
c
h
Consider the wheel tread and the road. A thread could be unwinding from the tire with this
speed relative to the frame:
v = rω =
(d)
FG 0.673 m IJ 17.3 rad s =
H 2 K
ω=
(b)
ω 2f = ω i2 + 2α ∆θ
a f
b25.0 rad sg − 0 =
=
2a ∆θ f
2 a1.25 revfb 2π rad rev g
2
ω 2f − ω i2
∆ω
∆t =
(a)
s = vt = 11.0 m s 9.00 s = 99.0 m
α
=
25.0 rad s
(c)
b
θ=
(b)
FG 1 IJ = 1.39 m s
H 1 rad K
v 25.0 m s
=
= 25.0 rad s
r
1.00 m
(a)
α=
P10.16
5.82 m s
We did not need to know the length of the pedal cranks, but we could use that information
to find the linear speed of the pedals:
v = rω = 0.175 m 7.96 rad s
P10.15
0.605 m s
Consider the chain link engaging a tooth on the rear sprocket:
ω=
(c)
291
39.8 rad s 2
ga
39.8 rad s 2
= 0.628 s
f
s 99.0 m
=
= 341 rad = 54.3 rev
r 0.290 m
ωf =
vf
r
=
22.0 m s
= 75.9 rad s = 12.1 rev s
0.290 m
292
P10.17
P10.18
Rotation of a Rigid Object About a Fixed Axis
FG
H
IJ
K
2π rad 1 200 rev
= 126 rad s
1 rev
60.0 s
(a)
ω = 2πf =
(b)
v = ωr = 126 rad s 3.00 × 10 −2 m = 3.77 m s
(c)
a c = ω 2 r = 126
(d)
s = rθ = ωrt = 126 rad s 8.00 × 10 −2 m 2.00 s = 20.1 m
b
ge
j
a f e8.00 × 10 j = 1 260 m s
2
−2
b
2
so a r = 1.26 km s 2 toward the center
ja
ge
f
The force of static friction must act forward and then more and more inward on the tires, to produce
both tangential and centripetal acceleration. Its tangential component is m 1.70 m s 2 . Its radially
e
inward component is
mv
. This takes the maximum value
r
e
j
FG
H
mω 2f r = mr ω i2 + 2α∆θ = mr 0 + 2α
With skidding impending we have
IJ
K
π
= mπrα = mπa t = mπ 1.70 m s 2 .
2
e
µs =
(a)
e
j
∑ Fy = ma y , + n − mg = 0, n = mg
fs = µ sn = µ s mg = m 2 1.70 m s 2
*P10.19
j
2
1.70 m s 2
g
j
2
e
+ m 2π 2 1.70 m s 2
j
2
1 + π 2 = 0.572
Let RE represent the radius of the Earth. The base of the building moves east at v1 = ω RE
where ω is one revolution per day. The top of the building moves east at v 2 = ω RE + h . Its
eastward speed relative to the ground is v 2 − v1 = ω h . The object’s time of fall is given by
b
∆y = 0 +
1 2
gt , t =
2
g
2h
. During its fall the object’s eastward motion is unimpeded so its
g
b
g
deflection distance is ∆x = v 2 − v1 t = ω h
f FGH 92.8sm IJK
2
F I
GH JK
2h
2
= ω h3 2
g
g
12
.
12
(b)
2π rad
50 m
86 400 s
(c)
The deflection is only 0.02% of the original height, so it is negligible in many practical cases.
a
32
= 1.16 cm
Chapter 10
Section 10.4
P10.20
293
Rotational Energy
m1 = 4.00 kg , r1 = y1 = 3.00 m ;
m 2 = 2.00 kg , r2 = y 2 = 2.00 m;
m 3 = 3.00 kg , r3 = y 3 = 4.00 m ;
ω = 2.00 rad s about the x-axis
(a)
I x = m1 r12 + m 2 r22 + m 3 r32
a f
a f + 3.00a4.00f
1
= a92.0fa 2.00f = 184 J
2
I x = 4.00 3.00
KR =
(b)
1
I xω 2
2
2
+ 2.00 2.00
2
2
= 92.0 kg ⋅ m 2
2
a f
= r ω = 2.00a 2.00f =
= r ω = 4.00a 2.00f =
FIG. P10.20
v2
v3
K1 =
2
4.00 m s
3
8.00 m s
K = K 1 + K 2 + K 3 = 72.0 + 16.0 + 96.0 = 184 J =
P10.21
(a)
a fa f
a fa f
a fa f
1
1
2
m1 v12 = 4.00 6.00 = 72.0 J
2
2
1
1
2
K 2 = m 2 v 22 = 2.00 4.00 = 16.0 J
2
2
1
1
2
K 3 = m 3 v 32 = 3.00 8.00 = 96.0 J
2
2
v1 = r1ω = 3.00 2.00 = 6.00 m s
1
I xω 2
2
I = ∑ m j r j2
y (m)
j
4
In this case,
3.00 kg
r1 = r2 = r3 = r4
r=
I=
2
13.0 m
2.00 kg
2
a3.00 mf + a2.00 mf
2
3
2
= 13.0 m
1
3.00 + 2.00 + 2.00 + 4.00 kg
x (m)
0
1
2
= 143 kg ⋅ m 2
(b)
KR =
jb
1 2 1
Iω = 143 kg ⋅ m 2 6.00 rad s
2
2
e
g
2
2.00 kg
4.00 kg
= 2.57 × 10 3 J
FIG. P10.21
3
294
P10.22
Rotation of a Rigid Object About a Fixed Axis
a
I = Mx 2 + m L − x
f
2
a
x
f
dI
= 2 Mx − 2m L − x = 0 (for an extremum)
dx
mL
∴x =
M+m
d2I
= 2m + 2 M ; therefore I is minimum when the axis of
dx 2
mL
which is also the center
rotation passes through x =
M+m
of mass of the system. The moment of inertia about an axis
passing through x is
I CM = M
LM mL OP
NM + mQ
2
LM
N
+m 1−
m
M+m
OP L
Q
2
2
=
L
m
M
x
Mm 2
L = µL2
M+m
L−x
FIG. P10.22
Mm
.
where µ =
M+m
Section 10.5
P10.23
Calculation of Moments of Inertia
We assume the rods are thin, with radius much less than L.
Call the junction of the rods the origin of coordinates, and
the axis of rotation the z-axis.
For the rod along the y-axis, I =
y
1
mL2 from the table.
3
x
For the rod parallel to the z-axis, the parallel-axis theorem
gives
I=
FG IJ
H K
1
L
mr 2 + m
2
2
2
≅
axis of rotation
z
1
mL2
4
FIG. P10.23
In the rod along the x-axis, the bit of material between x and x + dx has mass
distance r = x 2 +
FG L IJ
H 2K
FG m IJ dx and is at
H LK
2
from the axis of rotation. The total rotational inertia is:
z FGH
L2
I total =
1
1
L2
mL2 + mL2 +
x2 +
3
4
4
−L 2
FG IJ
H K
7
m x3
=
mL2 +
12
L 3
=
L2
+
−L 2
I FG m IJ dx
JK H L K
L2
mL
x
4 −L 2
2
7
11mL2
mL
mL2
+
=
mL2 +
12
12
4
12
Note: The moment of inertia of the rod along the x axis can also be calculated from the parallel-axis
2
1
L
theorem as
.
mL2 + m
12
2
FG IJ
H K
Chapter 10
P10.24
295
Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of
inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a
hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.
Use I =
1
m R12 + R 22 for the moment of inertia of a hollow cylinder.
2
e
j
Sidewall:
a
f − a0.165 mf e6.35 × 10 mje1.10 × 10
1
= b1.44 kg g a0.165 mf + a0.305 mf = 8.68 × 10
2
m = π 0.305 m
I side
2
2
−3
2
2
3
−2
j
kg m 3 = 1.44 kg
kg ⋅ m 2
Tread:
a
f − a0.305 mf a0.200 mfe1.10 × 10 kg m j = 11.0 kg
1
= b11.0 kg g a0.330 mf + a0.305 mf = 1.11 kg ⋅ m
2
m = π 0.330 m
I tread
2
2
3
3
2
2
2
Entire Tire:
e
j
I total = 2 I side + I tread = 2 8.68 × 10 −2 kg ⋅ m 2 + 1.11 kg ⋅ m 2 = 1.28 kg ⋅ m 2
P10.25
Every particle in the door could be slid straight down into a high-density rod across its bottom,
without changing the particle’s distance from the rotation axis of the door. Thus, a rod 0.870 m long
with mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door:
I=
b
ga
f
1
1
ML2 = 23.0 kg 0.870 m
3
3
2
= 5.80 kg ⋅ m 2 .
The height of the door is unnecessary data.
P10.26
Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is
0.750 m
= 0.120 m
2π
and its moment of inertia is
b
ga
f
1
1
MR 2 = 60.0 kg 0.120 m
2
2
2
= 0.432 kg ⋅ m 2 ~ 10 0 kg ⋅ m 2 = 1 kg ⋅ m 2 .
296
P10.27
Rotation of a Rigid Object About a Fixed Axis
For a spherical shell dI =
2
2
4πr 2 dr ρ r 2
dmr 2 =
3
3
e
z z
ze
j
e j af
2
rI
F
4πr jG 14. 2 − 11.6 J e10 kg m jdr
I=
H
3
RK
R
R
F 2I
F 2I
= G J 4π e14. 2 × 10 j
− G J 4π e11.6 × 10 j
H 3K
H
K
5
3
6
8π
14
2
11
6
.
.
I=
e10 jR FGH 5 − 6 IJK
3
rI
F
M = z dm = z 4πr G 14.2 − 11.6 J 10 dr
H
RK
F 14.2 − 11.6 IJ R
= 4π × 10 G
H3 4K
b8π 3ge10 jR b14.2 5 − 11.6 6g = 2 FG .907 IJ = 0.330
I
=
4π × 10 R R b14.2 3 − 11.6 4g 3 H 1.83 K
MR
2
4πr 2 r 2 ρ r dr
3
I = dI =
R
4
3
3
0
3
3
5
3
5
5
R
2
3
0
3
3
3
2
3
5
3
2
∴ I = 0.330 MR 2
*P10.28
(a)
y h
hx
. The area of the front face
= , y=
L
x L
1
1
is hL. The volume of the plate is hLw . Its density is
2
2
2M
M
M
. The mass of the ribbon is
ρ=
= 1
=
V
hLw
2 hLw
By similar triangles,
dm = ρdV = ρywdx =
y
h
x
L
2 Mywdx 2 Mhx
2 Mxdx
.
=
dx =
hLw
hLL
L2
FIG. P10.28
The moment of inertia is
I=
z
r 2 dm =
x2
x=0
all mass
(b)
z
L
z
L
2 Mxdx 2 M 3
2 M L4
ML2
= 2 x dx = 2
=
.
2
2
L
L 0
L 4
From the parallel axis theorem I = I CM + M
I h = I CM + M
FG L IJ
H 3K
inertia I CM +
2
FG 2L IJ
H3K
2
= I CM +
4ML2
and
9
ML2
. The two triangles constitute a rectangle with moment of
9
1
ML2 1
+
= 2 M L2 . Then 2 I CM = ML2
9
3
9
= I CM +
4ML2
+ I CM
9
I = I CM +
a f
4ML2
1
8
1
=
ML2 +
ML2 =
ML2 .
9
18
18
2
Chapter 10
*P10.29
297
We consider the cam as the superposition of the original solid disk and a disk of negative mass cut
from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the
volume and one-quarter the mass M 0 of the original solid cylinder:
M0 −
1
M0 = M
4
M0 =
4
M.
3
By the parallel-axis theorem, the original cylinder had moment of inertia
FG R IJ = 1 M R + M R = 3 M R .
H 2K 2
4
4
1F 1
I F R I M R . The whole cam has
The negative-mass portion has I = G − M J G J = −
H
KH 2 K
2
4
32
2
I CM + M 0
2
0
2
0
I=
0
0
0
2
2
M R 2 23
3
23 4
23
1
1 23
23
M0 R 2 − 0
=
M0 R 2 =
MR 2 =
MR 2 and K = Iω 2 =
MR 2 ω 2 =
MR 2ω 2 .
4
32
32
32 3
24
2
2 24
48
Section 10.6
P10.30
2
Torque
Resolve the 100 N force into components perpendicular
to and parallel to the rod, as
a
f
Fpar = 100 N cos 57.0° = 54.5 N
and
a
f
Fperp = 100 N sin 57.0° = 83.9 N
The torque of Fpar is zero since its line of action passes
through the pivot point.
a
FIG. P10.30
f
The torque of Fperp is τ = 83.9 N 2.00 m = 168 N ⋅ m (clockwise)
P10.31
∑ τ = 0.100 ma12.0 N f − 0.250 ma9.00 N f − 0.250 ma10.0 N f =
−3.55 N ⋅ m
The thirty-degree angle is unnecessary information.
FIG. P10.31
P10.32
The normal force exerted by the ground on each wheel is
b
ge
j
1 500 kg 9.80 m s 2
mg
n=
=
= 3 680 N
4
4
The torque of friction can be as large as
τ max = f max r = µ sn r = 0.800 3 680 N 0.300 m = 882 N ⋅ m
b g a
fb
ga
f
The torque of the axle on the wheel can be equally as large as the light wheel starts to turn without
slipping.
298
P10.33
Rotation of a Rigid Object About a Fixed Axis
In the previous problem we calculated the maximum torque that can be applied without skidding to
be 882 N · m. This same torque is to be applied by the frictional force, f, between the brake pad and
the rotor for this wheel. Since the wheel is slipping against the brake pad, we use the coefficient of
kinetic friction to calculate the normal force.
b g
τ = fr = µ k n r , so n =
Section 10.7
P10.34
(a)
882 N ⋅ m
τ
=
= 8.02 × 10 3 N = 8.02 kN
0.500 0.220 m
µ kr
a
fa
f
Relationship Between Torque and Angular Acceleration
b
ge
1
1
MR 2 = 2.00 kg 7.00 × 10 −2 m
2
2
0.600
τ
α= =
= 122 rad s 2
I 4.90 × 10 −3
∆ω
α=
∆t
2π
∆ω 1 200 60
∆t =
=
= 1.03 s
122
α
I=
j
2
= 4.90 × 10 −3 kg ⋅ m 2
c h
(b)
P10.35
P10.36
∆θ =
b
ga
1 2 1
αt = 122 rad s 1.03 s
2
2
f
2
= 64.7 rad = 10.3 rev
m = 0.750 kg , F = 0.800 N
a
f
(a)
τ = rF = 30.0 m 0.800 N = 24.0 N ⋅ m
(b)
α=
(c)
at
24.0
τ
rF
=
=
I mr 2 0.750 30.0
= 0.035 6 rad s 2
a f
= αr = 0.035 6a30.0 f = 1.07 m s
ω f = ω i + αt :
2
FIG. P10.35
2
a
f
10.0 rad s = 0 + α 6.00 s
10.00
rad s 2 = 1.67 rad s 2
α=
6.00
∑τ
36.0 N ⋅ m
= 21.6 kg ⋅ m 2
2
1.67 rad s
(a)
∑ τ = 36.0 N ⋅ m = Iα :
I=
(b)
ω f = ω i + αt :
0 = 10.0 + α 60.0
α
=
a f
α = −0.167 rad s 2
e
je
j
τ = Iα = 21.6 kg ⋅ m 2 0.167 rad s 2 = 3.60 N ⋅ m
(c)
Number of revolutions θ f = θ i + ω i t +
During first 6.00 s
During next 60.0 s
1 2
αt
2
a fa f
a f a fa f
FG
IJ
H
K
1
2
1.67 6.00 = 30.1 rad
2
1
2
θ f = 10.0 60.0 − 0.167 60.0 = 299 rad
2
1 rev
θ total = 329 rad
= 52.4 rev
2π rad
θf =
Chapter 10
P10.37
For m1 ,
∑ Fy = ma y :
+n − m 1 g = 0
n1 = m1 g = 19.6 N
f k 1 = µ k n1 = 7.06 N
b
∑ Fx = ma x :
g
−7.06 N + T1 = 2.00 kg a
For the pulley,
∑ τ = Iα :
FG IJ
H K
1
= b10.0 kg ga
2
= b5.00 kg ga
−T1 R + T2 R =
−T1 + T2
−T1 + T2
1
a
MR 2
2
R
+n 2 − m 2 g cos θ = 0
For m 2 ,
(1)
(2)
ja
e
f
n 2 = 6.00 kg 9.80 m s 2 cos 30.0°
= 50.9 N
FIG. P10.37
fk 2 = µ kn 2
= 18.3 N : −18.3 N − T2 + m 2 sin θ = m 2 a
−18.3 N − T2 + 29.4 N = 6.00 kg a (3)
b
(a)
g
Add equations (1), (2), and (3):
b
g
−7.06 N − 18.3 N + 29.4 N = 13.0 kg a
a=
e
j
4.01 N
= 0.309 m s 2
13.0 kg
T1 = 2.00 kg 0.309 m s 2 + 7.06 N = 7.67 N
(b)
e
j
T2 = 7.67 N + 5.00 kg 0.309 m s 2 = 9.22 N
P10.38
b
ga
f
1
1
mR 2 = 100 kg 0.500 m
2
2
ω i = 50.0 rev min = 5.24 rad s
I=
α=
ω f −ωi
t
=
2
= 12.5 kg ⋅ m 2
0 − 5.24 rad s
= −0.873 rad s 2
6.00 s
e
j
τ = Iα = 12.5 kg ⋅ m 2 −0.873 rad s 2 = −10.9 N ⋅ m
The magnitude of the torque is given by fR = 10.9 N ⋅ m, where f is
the force of friction.
10.9 N ⋅ m
0.500 m
Therefore,
f=
yields
µk =
f 21.8 N
=
= 0.312
n 70.0 N
and
f = µ kn
FIG. P10.38
299
300
*P10.39
Rotation of a Rigid Object About a Fixed Axis
1
∑ τ = Iα = 2 MR 2α
a
f 12 b80 kggFGH 1.225 mIJK e−1.67 rad s j
f a
2
−135 N 0.230 m + T 0.230 m =
2
T = 21.5 N
Section 10.8
P10.40
Work, Power, and Energy in Rotational Motion
The moment of inertia of a thin rod about an axis through one end is I =
kinetic energy is given as
KR =
with
Ih =
a
f
m h L2h 60.0 kg 2.70 m
=
3
3
2
1
ML2 . The total rotational
3
1
1
2
I hω 2h + I mω m
2
2
= 146 kg ⋅ m 2
a f
F
I
GH
JK
2π rad F 1 h I
=
= 1.75 × 10 rad s
1 h GH 3 600 s JK
1
1
= a146 fe1.45 × 10 j + a675 fe1.75 × 10 j
2
2
2
and
In addition,
*P10.41
100 kg 4.50 m
m L2
= 675 kg ⋅ m 2
Im = m m =
3
3
2π rad
1h
ωh =
= 1.45 × 10 −4 rad s
12 h 3 600 s
while
ωm
Therefore,
KR
−3
−4 2
−3 2
= 1.04 × 10 −3 J
1
11
E
where E = Iω 2 =
MR 2ω 2 is the stored energy and
2
22
∆t
1
P∆x
∆x
is the time it can roll. Then MR 2ω 2 = P∆t =
and
∆t =
4
v
v
The power output of the bus is P =
a
fc
2
h
2
2π
MR 2ω 2 v 1 600 kg 0.65 m 4 000 ⋅ 60 s 11.1 m s
∆x =
=
= 24.5 km .
4P
4 18 ⋅ 746 W
P10.42
a
fa
a
f
f
Work done = F∆r = 5.57 N 0.800 m = 4.46 J
1
1
and Work = ∆K = Iω 2f − Iω i2
2
2
(The last term is zero because the top starts from rest.)
Thus, 4.46 J =
1
4.00 × 10 −4 kg ⋅ m 2 ω 2f
2
e
A′
F
j
A
and from this, ω f = 149 rad s .
FIG. P10.42
Chapter 10
*P10.43
(a)
g a f a f = 2.28 × 10 kg ⋅ m
j b
b
g
e
j
1
1
1
F 0.82 m s IJ
0.850 kg gb0.82 m sg + b0.42 kg gb0.82 m sg + e 2.28 × 10 kg ⋅ m jG
b
H 0.03 m K
2
2
2
+0.42 kg e9.8 m s ja0.7 mf − 0.25b0.85 kg ge9.8 m s ja0.7 mf
F v IJ
1
1
1
= b0.85 kg gv + b0.42 kg gv + e 2.28 × 10 kg ⋅ m jG
2
2
2
H 0.03 m K
0.512 J + 2.88 J − 1.46 J = b0.761 kg gv
1
1
2
M R12 + R 22 = 0.35 kg 0.02 m + 0.03 m
2
2
K 1 + K 2 + K rot + U g 2 − f k ∆x = K 1 + K 2 + K rot f
e
I=
2
−4
2
i
2
2
−4
2
2
2
2
2
f
−4
2
f
f
2
2
2
f
vf =
(b)
P10.44
ω=
1.94 J
= 1.59 m s
0.761 kg
v 1.59 m s
=
= 53.1 rad s
0.03 m
r
We assume the rod is thin. For the compound object
LM
N
OP
Q
1
2
M rod L2 + m ball R 2 + M ball D 2
3
5
1
2
2
I = 1.20 kg 0.240 m + 2.00 kg 4.00 × 10 −2 m
3
5
I = 0.181 kg ⋅ m 2
I=
a
(a)
f
e
j
2
a
f
+ 2.00 kg 0.280 m
2
K f + U f = K i + U i + ∆E
FG IJ
H K
a
f
ja
1 2
L
+ M ball g L + R + 0
Iω + 0 = 0 + M rod g
2
2
1
0.181 kg ⋅ m 2 ω 2 = 1.20 kg 9.80 m s 2 0.120 m + 2.00 kg 9.80 m s 2 0.280 m
2
1
0.181 kg ⋅ m 2 ω 2 = 6.90 J
2
e
e
j
j
e
(b)
ω = 8.73 rad s
(c)
v = rω = 0.280 m 8.73 rad s = 2.44 m s
(d)
v 2f = vi2 + 2 a y f − yi
a
f
e
f
d
e
i
ja
f
v f = 0 + 2 9.80 m s 2 0.280 m = 2.34 m s
The speed it attains in swinging is greater by
2.44
= 1.043 2 times
2.34
ja
f
301
302
P10.45
Rotation of a Rigid Object About a Fixed Axis
(a)
For the counterweight,
∑ Fy = ma y becomes:
For the reel
∑ τ = Iα
50.0 − T =
FG 50.0 IJ a
H 9.80 K
reads TR = Iα = I
I=
where
a
R
1
MR 2 = 0.093 8 kg ⋅ m 2
2
We substitute to eliminate the acceleration:
50.0 − T = 5.10
F TR I
GH I JK
2
T = 11.4 N
a=
d
v f = 2 7.57 6.00 = 9.53 m s
Use conservation of energy for the system of the object, the reel, and the Earth:
aK + U f = aK + U f :
i
1
1
mv 2 + Iω 2
2
2
v2
I
2mgh = mv 2 + I 2 = v 2 m + 2
R
R
mgh =
f
2mgh
m + I2
v=
F I FG
IJ
GH JK H
K
2a50.0 N fa6.00 mf
=
=
5.10 kg +
R
P10.46
FIG. P10.45
50.0 − 11.4
= 7.57 m s 2
5.10
a f
i
v 2f = vi2 + 2 a x f − xi :
(b)
and
0.093 8
a0. 250 f
9.53 m s
2
Choose the zero gravitational potential energy at the level where the masses pass.
K f + U gf = K i + U gi + ∆E
1
1
1
m1 v 2 + m 2 v 2 + Iω 2 = 0 + m1 gh1i + m 2 gh2 i + 0
2
2
2
a
b
LM a f OPFG IJ
N
QH K
f
2
a fa f
a fa
1
1 1
v
15.0 + 10.0 v 2 +
3.00 R 2
= 15.0 9.80 1.50 + 10.0 9.80 −1.50
2
2 2
R
1
26.5 kg v 2 = 73.5 J ⇒ v = 2.36 m s
2
P10.47
g
f
From conservation of energy for the object-turntable-cylinder-Earth
system,
FG IJ
H K
1 v
I
2 r
2
+
1
mv 2 = mgh
2
v2
= 2mgh − mv 2
r2
2 gh
I = mr 2
−1
v2
I
FG
H
IJ
K
FIG. P10.47
Chapter 10
P10.48
The moment of inertia of the cylinder is
I=
b
f
ga
1
1
mr 2 = 81.6 kg 1.50 m
2
2
2
= 91.8 kg ⋅ m 2
and the angular acceleration of the merry-go-round is found as
α=
a f a
fa
f
50.0 N 1.50 m
Fr
τ
=
=
= 0.817 rad s 2 .
2
I
I
91.8 kg ⋅ m
e
j
At t = 3.00 s, we find the angular velocity
ω = ω i + αt
ja
e
f
ω = 0 + 0.817 rad s 2 3.00 s = 2.45 rad s
P10.49
jb
and K =
1 2 1
Iω = 91.8 kg ⋅ m 2 2. 45 rad s
2
2
(a)
Find the velocity of the CM
e
g
aK + U f = aK + U f
i
2
= 276 J .
Pivot
f
R
1
0 + mgR = Iω 2
2
2mgR
2mgR
= 3
ω=
2
I
2 mR
4g
Rg
= 2
3R
3
vCM = R
*P10.50
g
FIG. P10.49
Rg
3
(b)
v L = 2 vCM = 4
(c)
vCM =
(a)
The moment of inertia of the cord on the spool is
2mgR
=
2m
Rg
ea
f + a0.09 mf j = 4.16 × 10
1
1
M R12 + R 22 = 0.1 kg 0.015 m
2
2
e
j
e
2
2
−4
kg ⋅ m 2 .
j
The protruding strand has mass 10 −2 kg m 0.16 m = 1.6 × 10 −3 kg and
I = I CM + Md 2 =
FG a
H
f + a0.09 m + 0.08 mf IJK
1
1
0.16 m
ML2 + Md 2 = 1.6 × 10 −3 kg
12
12
2
2
= 4.97 × 10 −5 kg ⋅ m 2
For the whole cord, I = 4.66 × 10 −4 kg ⋅ m 2 . In speeding up, the average power is
P=
(b)
a
fa
E
=
∆t
1
2
Iω 2
∆t
=
a
f
⋅ 2π I
fFGH 2 000
J=
60 s K
P = τω = 7.65 N 0.16 m + 0.09 m
FG
H
4.66 × 10 −4 kg ⋅ m 2 2 500 ⋅ 2π
2 0.215 s
60 s
401 W
IJ
K
2
= 74.3 W
303
304
Rotation of a Rigid Object About a Fixed Axis
Section 10.9
P10.51
P10.52
b
gb
1
1
mv 2 = 10.0 kg 10.0 m s
2
2
(a)
K trans =
(b)
K rot =
(c)
K total = K trans + K rot = 750 J
FG
H
1 2 1 1
Iω =
mr 2
2
2 2
b
W = K f − K i = K trans + K rot
W=
or
P10.53
Rolling Motion of a Rigid Object
g
2
IJ FG v IJ = 1 b10.0 kg gb10.0 m sg
KH r K 4
g − bK
f
2
2
2
trans
+ K rot
FG
H
g
i
1
1
1
1 2
Mv 2 + Iω 2 − 0 − 0 = Mv 2 +
MR 2
2
2 5
2
2
7
W=
Mv 2
10
(a)
= 500 J
FG IJ
H K
IJ FG v IJ
KH RK
= 250 J
2
τ = Iα
mgR sin θ = I CM + mR 2 α
e
a=
j
mgR 2 sin θ
I CM + mR
R
2
mgR 2 sin θ
a hoop =
2mR 2
f
1
=
g sin θ
2
mgR 2 sin θ
(b)
θ
FIG. P10.53
Rf = Iα
f = µn = µmg cos θ
Iα
f
R
µ=
=
=
mg cos θ mg cos θ
P10.54
mg
n
2
=
a disk = 3
g sin θ
2
3
mR
2
4
The disk moves with the acceleration of the hoop.
3
LM
N
OP
Q
c
1
1
1
I
mv 2 + Iω 2 =
m + 2 v2
2
2
2
R
Also,
Ui = mgh , U f = 0 ,
K=
2
3
g sin θ
he
1
2
mR 2
2
R mg cos θ
where
and
j=
1
tan θ
3
v
R
vi = 0
ω=
since no slipping.
LM
N
OP
Q
1
I
m + 2 v 2 = mgh
2
R
2 gh
v2 =
1+ I 2
Therefore,
Thus,
e j
mR
1
I = mR 2
2
For a disk,
So
For a ring,
Since v disk
v2 =
2 gh
1 + 12
or
v disk =
4 gh
3
2 gh
or
v ring = gh
2
> v ring , the disk reaches the bottom first.
I = mR 2 so v 2 =
Chapter 10
P10.55
305
1
∆x 3.00 m
=
= 2.00 m s = 0 + v f
1.50 s
2
∆t
vf
4.00 m s
8.00
rad s
v f = 4.00 m s and ω f =
=
=
−2
r
6
.
38
× 10 −2
6.38 × 10 m 2
d
v=
i
e
j
We ignore internal friction and suppose the can rolls without slipping.
e K + K + U j + ∆E
b0 + 0 + mgy g + 0 = FGH 12 mv
trans
rot
g
mech
i
i
2
f
+
e
= K trans + K rot + U g
1 2
Iω f + 0
2
IJ
K
j
f
e
j a3.00 mf sin 25.0° = 12 b0.215 kg gb4.00 m sg
2.67 J = 1.72 J + e7 860 s jt
0.215 kg 9.80 m s 2
2
+
FG
H
IJ
K
1
8.00
rad s
I
2 6.38 × 10 −2
2
−2
I=
P10.56
(a)
0.951 kg ⋅ m 2 s 2
= 1.21 × 10 −4 kg ⋅ m 2
7 860 s −2
The height of the can is unnecessary data.
Energy conservation for the system of the ball and the
Earth between the horizontal section and top of loop:
1
1
1
1
mv 22 + Iω 22 + mgy 2 = mv12 + Iω 12
2
2
2
2
FG
H
IJ FG v IJ + mgy
KH r K
1F2
IF v I
+ G mr J G J
KH r K
2H3
1
1 2
mv 22 +
mr 2
2
2 3
1
mv12
2
5 2
5
v 2 + gy 2 = v12
6
6
=
v 2 = v12 −
2
2
2
2
6
gy 2 =
5
1
b4.03 m sg
2
−
2
FIG. P10.56
f
e
ja
b2.38 m sg = 12.6 m s
=
6
9.80 m s 2 0.900 m = 2.38 m s
5
2
v2
2
The centripetal acceleration is 2
>g
0.450 m
r
Thus, the ball must be in contact with the track, with the track pushing downward on it.
(b)
FG
H
v 3 = v12 −
(c)
IJ FG v IJ + mgy = 1 mv + 1 FG 2 mr IJ FG v IJ
KH r K
KH r K
2
2H3
6
= b 4.03 m sg − e9.80 m s ja −0.200 mf =
5
1
1 2
mv 32 +
mr 2
2
2 3
6
gy 3
5
3
2
3
2
1
2
2
1
2
2
4.31 m s
1
1
mv 22 + mgy 2 = mv12
2
2
v 2 = v12 − 2 gy 2 =
b4.03 m sg − 2e9.80 m s ja0.900 mf =
2
2
−1.40 m 2 s 2
This result is imaginary. In the case where the ball does not roll, the ball starts with less
energy than in part (a) and never makes it to the top of the loop.
306
Rotation of a Rigid Object About a Fixed Axis
Additional Problems
P10.57
A
1
sin θ = mA 2α
2
3
3 g
sin θ
α=
2 A
3 g
sin θ r
at =
2 A
3 g
Then
r > g sin θ
2 A
2
for r > A
3
1
∴ About
the length of the chimney will have a
3
tangential acceleration greater than g sin θ .
mg
FG
H
P10.58
FG IJ
H K
IJ
K
θt
θt
g sin θ
g
FIG. P10.57
The resistive force on each ball is R = DρAv 2 . Here v = rω , where r is the radius of each ball’s path.
The resistive torque on each ball is τ = rR , so the total resistive torque on the three ball system is
τ total = 3rR .
The power required to maintain a constant rotation rate is P = τ totalω = 3rRω . This required power
may be written as
a f ω = e3r DAω jρ
2π rad F 10 rev I F 1 min I 1 000π
ω=
G J = 30.0 rad s
1 rev GH 1 min JK H 60.0 s K
F 1 000π IJ ρ
P = 3a0.100 mf a0.600 fe 4.00 × 10 m jG
H 30.0 s K
P = e0.827 m s jρ , where ρ is the density of the resisting medium.
P = τ totalω = 3r DρA rω
With
2
3
or
(a)
3
3
3
5
−4
2
3
3
In air, ρ = 1.20 kg m3 ,
e
j
and P = 0.827 m 5 s 3 1. 20 kg m 3 = 0.992 N ⋅ m s = 0.992 W
P10.59
at
(b)
In water, ρ = 1 000 kg m3 and P = 827 W .
(a)
W = ∆K =
(b)
(c)
1 2 1 2 1
1
Iω f − Iω i = I ω 2f − ω i2 where I = mR 2
2
2
2
2
1
2
2
1.00 kg 0.500 m 8.00 rad s − 0 = 4.00 J
2
e
j
F 1IF I
= G JG Jb
f b
ga
g
H 2KH K
ω − 0 ωr b8.00 rad sga0.500 mf
=
=
= 1.60 s
t=
f
α
2.50 m s 2
a
1
θ f = θ i + ω i t + αt 2 ; θ i = 0 ; ω i = 0
2
1 2 1 2.50 m s 2
2
1.60 s = 6.40 rad
θ f = αt =
2
2 0.500 m
F
GH
Ia
JK
s = rθ = a0.500 mfa6. 40 radf =
f
3.20 m < 4.00 m Yes
Chapter 10
*P10.60
Start
The quantity of tape is constant. Then the area of the rings you
see it fill is constant. This is expressed by
rt
π rt2 − π rs2 = π r 2 − π rs2 + π r22 − π rs2 or r2 = rt2 + rs2 − r 2 is the
outer radius of spool 2.
(a)
P10.61
(a)
j
Later
r2
r
v
v
and ω 2 = . The
rt
rs
takeup reel must spin at maximum speed. At the end,
v
v
and ω 1 = . The angular
r = rs and r2 = rt so ω 2 =
rt
rs
speeds are just reversed.
At the start, r = rt and r2 = rs so ω 1 =
v
FIG. P10.60
Since only conservative forces act within the system of the
rod and the Earth,
∆E = 0
where I =
K f + U f = K i + Ui
so
FG IJ
H K
1 2
L
Iω + 0 = 0 + Mg
2
2
1
ML2
3
FIG. P10.61
3g
L
ω=
Therefore,
(b)
v
v
Where the tape comes off spool 1, ω 1 = . Where the
r
−1 2
v
2
2
.
tape joins spool 2, ω 2 = = v rs + rt − r 2
r2
e
(b)
rs
∑ τ = Iα , so that in the horizontal orientation,
Mg
FG L IJ = ML
H 2K 3
α=
FG L IJ ω
H 2K
a x = a r = − rω 2 = −
(d)
Using Newton’s second law, we have
R x = Ma x = −
2
3g
2
(c)
= −
2
3g
2L
a y = − a t = − rα = −α
3 Mg
2
R y − Mg = Ma y = −
3 Mg
4
α
Ry =
Mg
4
FG L IJ =
H 2K
−
3g
4
307
308
P10.62
Rotation of a Rigid Object About a Fixed Axis
e
j
α = −10.0 rad s 2 − 5.00 rad s 3 t =
z z
ω
dω =
65.0
t
dω
dt
−10.0 − 5.00t dt = −10.0t − 2.50t 2 = ω − 65.0 rad s
0
dθ
ω=
= 65.0 rad s − 10.0 rad s 2 t − 2.50 rad s 3 t 2
dt
e
(a)
j e
j
At t = 3.00 s,
ja
e
f e
je
j
ω = 65.0 rad s − 10.0 rad s 2 3.00 s − 2.50 rad s 3 9.00 s 2 = 12.5 rad s
z z
θ
(b)
0
t
dθ = ω dt =
b
0
z
t
e
j e
j
65.0 rad s − 10.0 rad s 2 t − 2.50 rad s 3 t 2 dt
0
g e
j e
j
θ = 65.0 rad s t − 5.00 rad s 2 t 2 − 0.833 rad s 3 t 3
At t = 3.00 s,
b
ga
f e
j
e
j
θ = 65.0 rad s 3.00 s − 5.00 rad s 2 9.00 s 2 − 0.833 rad s 3 27.0 s 3
θ = 128 rad
P10.63
The first drop has a velocity leaving the wheel given by
1
mvi2 = mgh1 , so
2
e
ja
f
e
ja
f
v1 = 2 gh1 = 2 9.80 m s 2 0.540 m = 3.25 m s
The second drop has a velocity given by
v 2 = 2 gh2 = 2 9.80 m s 2 0.510 m = 3.16 m s
From ω =
v
, we find
r
ω1 =
3.16 m s
v1 3.25 m s
v
=
= 8.53 rad s and ω 2 = 2 =
= 8.29 rad s
0.381 m
0.381 m
r
r
or
α=
b
g b
2
8.29 rad s − 8.53 rad s
ω 22 − ω 12
=
2θ
4π
g
2
= −0.322 rad s 2
Chapter 10
P10.64
At the instant it comes off the wheel, the first drop has a velocity v1 , directed upward. The
magnitude of this velocity is found from
K i + U gi = K f + U gf
1
mv12 + 0 = 0 + mgh1 or v1 = 2 gh1
2
and the angular velocity of the wheel at the instant the first drop leaves is
ω1 =
2 gh1
v1
=
R
R2
Similarly for the second drop: v 2 = 2 gh2 and ω 2 =
v2
=
R
.
2 gh2
R2
.
The angular acceleration of the wheel is then
2 gh2
ω 2 − ω 12
=
a= 2
2θ
P10.65
R2
−
2 gh1
R2
a f
2 2π
=
b
g h2 − h1
2πR
1
1
1
1
Mv 2f + Iω 2f : U f = Mgh f = 0 ; K i = Mvi2 + Iω i2 = 0
2
2
2
2
1
v
Ui = Mgh i : f = µN = µMg cos θ ; ω = ; h = d sin θ and I = mr 2
2
r
Kf =
b g
(a)
∆E = E f − Ei or − fd = K f + U f − K i − U i
1
1
− fd = Mv 2f + Iω 2f − Mgh
2
2
F
GH
I − Mgd sinθ
b
g
JK 2
1L
mO
M + P v = Mgd sin θ − b µMg cos θ gd or
M
2N
2Q
bsinθ − µ cosθ g
v = 2 Mgd
1
mr 2
− µMg cos θ d = Mv 2 +
2
2
v2
r2
2
2
m
2
vd
(b)
+M
L
O
M
= M4 gd
sin θ − µ cos θ gP
b
N am + 2 M f
Q
12
v 2f = vi2 + 2 a∆x , v d2 = 2 ad
a=
FG
H
IJ b
K
v d2
M
sin θ − µ cos θ
= 2g
2d
m + 2M
g
2
g
.
309
310
P10.66
Rotation of a Rigid Object About a Fixed Axis
(a)
E=
FG
H
IJ e j
K
1 2
MR 2 ω 2
2 5
1 2
E = ⋅ 5.98 × 10 24 6.37 × 10 6
2 5
e
(b)
je
LM FG
MN H
π I
j FGH 862400
JK
2
IJ FG 2π IJ OP
K H T K PQ
1
dT
= MR a 2π f e −2T j
5
dt
1
F 2π I F −2 I dT
= MR G J G J
H T K H T K dt
5
F −2 I F 10 × 10
= e 2.57 × 10 JjG
H 86 400 s JK GH 3.16 × 10
= 2.57 × 10 29 J
2
dE d 1 2
MR 2
=
dt dt 2 5
2
2
2
−3
2
2
−6
29
7
I b86 400 s dayg
J
sK
s
dE
= −1.63 × 10 17 J day
dt
*P10.67
(a)
ω f = ω i + αt
α=
ω f −ωi
t
=
2π
Tf
−
t
e
2π
Ti
=
d
2π Ti − T f
i
Ti T f t
j F 1 d I FG 1 yr IJ =
~
1 d 1 d 100 yr GH 86 400 s JK H 3.156 × 10 s K
2π −10 −3 s
2
7
(b)
−10 −22 s −2
The Earth, assumed uniform, has moment of inertia
I=
2
2
MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m
5
5
e
∑ τ = Iα ~ 9.71 × 10 37
je
j = 9.71 × 10
kg ⋅ m e−2.67 × 10
s j = −10
2
−22
2
−2
37
16
kg ⋅ m 2
N⋅m
The negative sign indicates clockwise, to slow the planet’s counterclockwise rotation.
(c)
τ = Fd . Suppose the person can exert a 900-N force.
d=
τ 2.59 × 10 16 N ⋅ m
~ 10 13 m
=
900 N
F
This is the order of magnitude of the size of the planetary system.
Chapter 10
P10.68
∆ θ = ωt
t=
∆θ
=
∆θ = 31°
c h rev = 0.005 74 s
v
31.0 °
360°
900 rev
60 s
ω
0.800 m
= 139 m s
v=
0.005 74 s
ω
d
FIG. P10.68
P10.69
τ f will oppose the torque due to the hanging object:
∑ τ = Iα = TR − τ f :
τ f = TR − Iα
(1)
Now find T, I and α in given or known terms and substitute into
equation (1).
b
∑ Fy = T − mg = − ma :
T=m g−a
at 2
2
also ∆y = vi t +
a=
and
g
(2)
2y
(3)
t2
α=
2y
a
= 2:
R Rt
I=
1
R
M R2 +
2
2
LM
MN
FIG. P10.69
(4)
FG IJ OP = 5 MR
H K PQ 8
2
2
(5)
Substituting (2), (3), (4), and (5) into (1),
FG
H
τ f =m g−
we find
P10.70
(a)
2y
t2
IJ R − 5 MR b2yg = RLMmFG g − 2 y IJ − 5 My OP
K 8 Rt
N H t K 4t Q
2
2
2
2
W = ∆K + ∆ U
W = K f − K i + U f − Ui
1
1
1
mv 2 + Iω 2 − mgd sin θ − kd 2
2
2
2
1 2
1
ω I + mR 2 = mgd sin θ + kd 2
2
2
0=
e
ω=
j
2mgd sin θ + kd 2
I + mR 2
FIG. P10.70
(b)
ω=
ω=
b
ge
ja
fa
f
a
f
2 0.500 kg 9.80 m s 2 0.200 m sin 37.0° + 50.0 N m 0.200 m
a
f
1.00 kg ⋅ m 2 + 0.500 kg 0.300 m
1.18 + 2.00
= 3.04 = 1.74 rad s
1.05
2
2
311
312
P10.71
Rotation of a Rigid Object About a Fixed Axis
(a)
m 2 g − T2 = m 2 a
b
g
e
j
T2 = m 2 g − a = 20.0 kg 9.80 m s 2 − 2.00 m s 2 = 156 N
T1 − m1 g sin 37.0° = m1 a
b
f
ga
T1 = 15.0 kg 9.80 sin 37.0°+2.00 m s 2 = 118 N
(b)
FG a IJ
H RK
bT − T gR = a156 N − 118 Nfa0.250 mf
I=
bT
2
g
− T1 R = Iα = I
2
2
1
2.00 m s 2
a
P10.72
2
FIG. P10.71
= 1.17 kg ⋅ m 2
For the board just starting to move,
FG A IJ cosθ = FG 1 mA IJ α
H 2K
H3 K
3 F gI
α = G J cos θ
2 H AK
∑ τ = Iα :
2
mg
3
g cos θ
2
3
The vertical component is
a y = a t cos θ = g cos 2 θ
2
If this is greater than g, the board will pull ahead of the ball falling:
The tangential acceleration of the end is
a t = Aα =
2
3
FIG. P10.72
(a)
3
2
g cos 2 θ ≥ g gives cos 2 θ ≥ so
2
3
(b)
When θ = 35.3° , the cup will land underneath the release-point of the ball if
cos θ ≥
When A = 1.00 m, and θ = 35.3°
rc = 1.00 m
a
f
θ ≤ 35.3°
and
2
= 0.816 m
3
so the cup should be 1.00 m − 0.816 m = 0.184 m from the moving end
P10.73
At t = 0 , ω = 3.50 rad s = ω 0 e 0 . Thus, ω 0 = 3.50 rad s
At t = 9.30 s, ω = 2.00 rad s = ω 0 e −σ a 9.30 s f , yielding σ = 6.02 × 10 −2 s −1
(a)
e
dω d ω 0 e
=
dt
dt
At t = 3.00 s,
α=
b
− σt
j = ω a−σ fe
0
ge
−σt
j
α = 3.50 rad s −6.02 × 10 −2 s −1 e
z
t
(b)
θ = ω 0 e −σt dt =
0
e
−3.00 6 .02 × 10 −2
j=
−0.176 rad s 2
ω 0 − σt
ω
e − 1 = 0 1 − e −σt
−σ
σ
At t = 2.50 s ,
3.50 rad s
− 6 .02 × 10 −2 ja 2.50 f
1−e e
θ=
= 8.12 rad = 1.29 rev
−2
6.02 × 10 1 s
e
(c)
As t → ∞ , θ →
LM
j N
OP
Q
3.50 rad s
ω0
1 − e −∞ =
= 58.2 rad = 9.26 rev
σ
6.02 × 10 −2 s −1
e
j
rc = A cos θ
Chapter 10
P10.74
313
Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Then
the total torque (taking CCW as positive) of these hands about the center of the clock is given by
τ = −m h g
FG L IJ sinθ
H 2K
h
h
− mm g
FG L IJ sinθ
H2K
m
m
=−
g
m h L h sin θ h + m m Lm sin θ m
2
b
g
If we take t = 0 at 12 o’clock, then the angular positions of the hands at time t are
θ h = ω ht ,
π
rad h
6
where
ωh =
and
θ m = ω mt ,
where
ω m = 2π rad h
Therefore,
τ = −4.90 m s 2 60.0 kg 2.70 m sin
f FGH π6t IJK + 100 kga4.50 mf sin 2πtOPQ
L F πt I
O
τ = −794 N ⋅ mMsinG J + 2.78 sin 2πt P , where t is in hours.
H
K
N 6
Q
or
(a)
(i)
At 3:00, t = 3.00 h ,
so
(ii)
LM
N
a
LM FG π IJ + 2.78 sin 6π OP =
N H 2K
Q
τ = −794 N ⋅ m sin
At 5:15, t = 5 h +
−794 N ⋅ m
15
h = 5.25 h , and substitution gives:
60
τ = −2 510 N ⋅ m
(b)
(iii)
At 6:00,
τ = 0 N ⋅m
(iv)
At 8:20,
τ = −1 160 N ⋅ m
(v)
At 9:45,
τ = −2 940 N ⋅ m
The total torque is zero at those times when
FG πt IJ + 2.78 sin 2πt = 0
H6K
sin
We proceed numerically, to find 0, 0.515 295 5, ..., corresponding to the times
12:00:00
2:33:25
4:58:14
7:27:36
10:02:59
12:30:55
2:56:29
5:30:52
8:03:05
10:27:29
12:58:19
3:33:22
6:00:00
8:26:38
11:01:41
1:32:31
3:56:55
6:29:08
9:03:31
11:29:05
1:57:01
4:32:24
7:01:46
9:26:35
314
*P10.75
Rotation of a Rigid Object About a Fixed Axis
(a)
As the bicycle frame moves forward at speed v, the center of each wheel moves forward at
v
the same speed and the wheels turn at angular speed ω = . The total kinetic energy of the
R
bicycle is
K = K trans + K rot
or
K=
b
IJ b
K
FG
H
g
g FGH
1
1
1
1
m frame + 2m wheel v 2 + 2 I wheel ω 2 = m frame + 2m wheel v 2 +
m wheel R 2
2
2
2
2
IJ FG v IJ .
KH R K
2
2
This yields
K=
(b)
b
g
b
1
1
m frame + 3m wheel v 2 = 8.44 kg + 3 0.820 kg
2
2
g b3.35 m sg
2
= 61.2 J .
As the block moves forward with speed v, the top of each trunk moves forward at the same
v
speed and the center of each trunk moves forward at speed . The angular speed of each
2
v
roller is ω =
. As in part (a), we have one object undergoing pure translation and two
2R
identical objects rolling without slipping. The total kinetic energy of the system of the stone
and the trees is
K = K trans + K rot
or
K=
FG IJ
H K
1
1
v
m stone v 2 + 2 m tree
2
2
2
2
+2
FG 1 I
H2
treeω
2
IJ = 1 FG m
K 2H
stone
+
IJ
K
FG
H
1
1
m tree v 2 +
m tree R 2
2
2
IJ FG v IJ .
K H 4R K
This gives
K=
P10.76
FG
H
IJ
K
b
1
3
1
m stone + m tree v 2 = 844 kg + 0.75 82.0 kg
2
4
2
g b0.335 m sg
2
= 50.8 J .
Energy is conserved so ∆U + ∆K rot + ∆K trans = 0
a
f LMN 12 mv
fa
mg R − r cos θ − 1 +
2
OP 1 LM 2 mr OPω
Q 2 N5 Q
−0 +
2
2
=0
θ
R
Since rω = v , this gives
ω=
or
ω=
a
fa
f
10 R − r 1 − cos θ g
7
r2
a
10 Rg 1 − cos θ
7r
2
f
since R >> r .
FIG. P10.76
2
2
Chapter 10
P10.77
F aI
1
∑ F = T − Mg = − Ma: ∑ τ = TR = Iα = 2 MR 2 GH R JK
(a)
Combining the above two equations we find
b
T =M g−a
g
and
a=
2T
M
FIG. P10.77
T=
thus
Mg
3
FG IJ
H K
2T
2 Mg
2
=
=
g
3
M M 3
(b)
a=
(c)
v 2f = vi2 + 2 a x f − xi
d
i
v 2f = 0 + 2
vf =
FG 2 gIJ ah − 0f
H3 K
4 gh
3
For comparison, from conservation of energy for the system of the disk and the Earth we
have
U gi + K rot i + K trans i = U gf + K rot f + K trans f :
Mgh + 0 + 0 = 0 +
vf =
P10.78
(a)
4 gh
3
∑ Fx = F − f = Ma: ∑ τ = fR = Iα
Using I =
(b)
FG
H
1 1
MR 2
2 2
1
a
2F
MR 2 and α = , we find a =
2
3M
R
When there is no slipping, f = µ Mg .
Substituting this into the torque equation of part (a), we have
µ MgR =
1
F
.
MRa and µ =
2
3 Mg
IJ FG v IJ
KH R K
f
2
+
1
Mv f 2
2
315
316
P10.79
Rotation of a Rigid Object About a Fixed Axis
(a)
∆K rot + ∆K trans + ∆U = 0
m
Note that initially the center of mass of the sphere is a
distance h + r above the bottom of the loop; and as the
mass reaches the top of the loop, this distance above
the reference level is 2R − r . The conservation of
energy requirement gives
h
a f
a
r
f
1
1
mg h + r = mg 2 R − r + mv 2 + Iω 2
2
2
For the sphere I =
gh + 2 gr = 2 gR +
R
P
FIG. P10.79
2
mr 2 and v = rω so that the expression becomes
5
7 2
v
10
(1)
Note that h = hmin when the speed of the sphere at the top of the loop satisfies the condition
mv 2
∑ F = mg = aR − r f
a
or v 2 = g R − r
f
Substituting this into Equation (1) gives
a
f
a
f
a
f
hmin = 2 R − r + 0.700 R − r or hmin = 2.70 R − r = 2.70 R
(b)
When the sphere is initially at h = 3 R and finally at point P, the conservation of energy
equation gives
a
f
1
1
mg 3 R + r = mgR + mv 2 + mv 2 , or
2
5
10
2
2R + r g
v =
7
a
f
Turning clockwise as it rolls without slipping past point P, the sphere is slowing down with
counterclockwise angular acceleration caused by the torque of an upward force f of static
2
friction. We have ∑ Fy = ma y and ∑ τ = Iα becoming f − mg = − mα r and fr =
mr 2 α .
5
FG IJ
H K
Eliminating f by substitution yields α =
mv 2
∑ Fx = −n = − R − r = −
c h( 2R + r ) mg =
10
7
R−r
5g
so that
7r
−20mg
7
∑ Fy =
5
− mg
7
(since R >> r )
317
Chapter 10
P10.80
Consider the free-body diagram shown. The sum of torques
about the chosen pivot is
F 1 I F a IJ = FG 2 mlIJ a
H K H3 K
∑ τ = Iα ⇒ FA = GH 3 ml 2 JK G
(a)
CM
l
2
b
f
Hx
A
CM
mg
l
g
∑ Fx = maCM ⇒ F + H x = maCM or H x = maCM − F
b
Hy
(1)
CM
A = l = 1.24 m : In this case, Equation (1) becomes
3 14.7 N
3F
=
= 35.0 m s 2
aCM =
2m 2 0.630 kg
a
pivot
ge
F = 14.7 N
j
Thus, H x = 0.630 kg 35.0 m s 2 − 14.7 N = +7.35 N or
FIG. P10.80
H x = 7.35 i N .
(b)
A=
1
= 0.620 m : For this situation, Equation (1) yields
2
aCM =
Again,
a
f
3 14.7 N
3F
=
= 17.5 m s 2 .
4m 4 0.630 kg
b
g
∑ Fx = maCM ⇒ H x = maCM − F , so
b
ge
j
H x = 0.630 kg 17.5 m s 2 − 14.7 N = −3.68 N or H x = −3.68 i N .
(c)
If H x = 0, then
F
∑ Fx = maCM ⇒ F = maCM , or aCM = m .
Thus, Equation (1) becomes
FA=
P10.81
FG 2 mlIJ FG F IJ so A= 2 l = 2 a1.24 mf =
H 3 KH mK
3
3
b
g
0.827 m from the top .
2
Let the ball have mass m and radius r. Then I = mr 2 . If the ball takes four seconds to go down
5
twenty-meter alley, then v = 5 m s . The translational speed of the ball will decrease somewhat as
the ball loses energy to sliding friction and some translational kinetic energy is converted to
rotational kinetic energy; but its speed will always be on the order of 5.00 m s , including at the
starting point.
As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed.
v
When ω = , the ball has achieved pure rolling motion, and kinetic friction ceases. To determine the
r
elapsed time before pure rolling motion is achieved, consider:
F 2 I L b5.00 m sg r OP which gives
MN t PQ
∑ τ = Iα ⇒ b µ k mg gr = GH 5 mr 2 JK M
t=
2( 5.00 m s) 2.00 m s
=
5µ k g
µkg
Note that the mass and radius of the ball have canceled. If µ k = 0.100 for the polished alley, the
sliding distance will be given by
b
∆x = vt = 5.00 m s
O
L
gMM a0.1002f.009.80m ms s PP = 10.2 m or ∆x ~
jQ
N e
2
10 1 m .
318
P10.82
Rotation of a Rigid Object About a Fixed Axis
Conservation of energy between apex and the point where
the grape leaves the surface:
mg∆y =
1
1
mv 2f + Iω 2f
2
2
a
f
FG
H
IJ FG v IJ
KH R K
f
R
2
f
n
F I
GH JK
2
7 vf
which gives g 1 − cos θ =
10 R
f
∆y = R—R cosθ
θ
1
1 2
mgR 1 − cos θ = mv 2f +
mR 2
2
2 5
a
i
(1)
mg cosθ
mg sinθ
Consider the radial forces acting on the grape:
mg cos θ − n =
mv 2f
R
FIG. P10.82
.
At the point where the grape leaves the surface, n → 0 .
Thus, mg cos θ =
mv 2f
R
or
v 2f
R
= g cos θ .
Substituting this into Equation (1) gives
g − g cos θ =
P10.83
(a)
10
7
and θ = 54.0° .
g cos θ or cos θ =
17
10
There are not any horizontal forces acting on the rod, so the center of mass will not move
horizontally. Rather, the center of mass drops straight downward (distance h/2) with the rod
rotating about the center of mass as it falls. From conservation of energy:
K f + U gf = K i + U gi
FG IJ
H K
I = MgFG h IJ which reduces to
JK
H 2K
1
1
h
2
or
+ Iω 2 + 0 = 0 + Mg
MvCM
2
2
2
FG
H
1
1 1
2
+
MvCM
Mh 2
2
2 12
vCM =
(b)
IJ FG v
KH
2
CM
h
2
3 gh
4
In this case, the motion is a pure rotation about a fixed pivot point (the lower end of the rod)
with the center of mass moving in a circular path of radius h/2. From conservation of energy:
K f + U gf = K i + U gi
FG IJ
H K
1 2
h
or
Iω + 0 = 0 + Mg
2
2
1 1
Mh 2
2 3
IJ FG v IJ
KH K
vCM =
3 gh
4
FG
H
CM
h
2
2
= Mg
FG h IJ which reduces to
H 2K
Chapter 10
P10.84
(a)
319
Mr 2
where M is the initial mass of
R2
mr 2
,
+ ∆K rot = 0 . Thus, when I =
2
The mass of the roll decreases as it unrolls. We have m =
the roll. Since ∆E = 0 , we then have ∆U g + ∆K trans
bmgr − MgRg + mv2 + LM mr2
N
2
Since ω r = v , this becomes v =
e
4g R3 − r 3
3r
2
OP
Q
ω2
=0
2
j
2
(b)
Using the given data, we find v = 5.31 × 10 4 m s
(c)
We have assumed that ∆E = 0 . When the roll gets to the end, we will have an inelastic
collision with the surface. The energy goes into internal energy . With the assumption we
made, there are problems with this question. It would take an infinite time to unwrap the
tissue since dr → 0 . Also, as r approaches zero, the velocity of the center of mass approaches
infinity, which is physically impossible.
P10.85
(a)
∑ Fx = F + f = MaCM
∑ τ = FR − fR = Iα
b
g
FR − MaCM − F R =
f = MaCM − F = M
(c)
v 2f = vi2 + 2 a x f − xi
vf =
8 Fd
3M
Mg
IaCM
R
FG 4F IJ − F =
H 3M K
(b)
d
F
i
aCM =
1
F
3
4F
3M
n
f
FIG. P10.85
320
P10.86
Rotation of a Rigid Object About a Fixed Axis
Call ft the frictional force exerted by each roller
backward on the plank. Name as fb the rolling
resistance exerted backward by the ground on
each roller. Suppose the rollers are equally far
from the ends of the plank.
M
m
R
m
For the plank,
b
∑ Fx = ma x
F
R
FIG. P10.86
g
6.00 N − 2 f t = 6.00 kg a p
The center of each roller moves forward only half as far as the plank. Each roller has acceleration
and angular acceleration
ap 2
ap
2
ap
=
a5.00 cmf a0.100 mf
Then for each,
b
g a2
1
f a5.00 cmf + f a5.00 cmf = b 2.00 kg ga5.00 cmf
∑ τ = Iα
2
F1 I
f + f = G kg J a
H2 K
∑ Fx = ma x
t
So
t
p
+ f t − f b = 2.00 kg
b
b
ap
2
10.0 cm
p
Add to eliminate fb :
b
g
2 f t = 1.50 kg a p
(a)
b
g b
g
And 6.00 N − 1.50 kg a p = 6.00 kg a p
ap =
For each roller, a =
(b)
ap
2
a6.00 Nf =
b7.50 kg g
0.800 m s 2
= 0.400 m s 2
b
g
Substituting back, 2 f t = 1.50 kg 0.800 m s 2
Mg
6.00 N
ft = 0.600 N
0.600 N + f b =
fb = −0.200 N
1
kg 0.800 m s 2
2
e
j
ft
ft
nt
nt
nt
nt
ft
ft
The negative sign means that the horizontal force
of ground on each roller is 0.200 N forward
mg
mg
rather than backward as we assumed.
fb
nb
fb
nb
FIG. P10.86(b)
Chapter 10
P10.87
Rolling is instantaneous rotation about the contact point P. The
weight and normal force produce no torque about this point.
Now F1 produces a clockwise torque about P and makes the
321
F3
F2
F4
spool roll forward.
Counterclockwise torques result from F3 and F4 , making the
θc
spool roll to the left.
The force F2 produces zero torque about point P and does
P
not cause the spool to roll. If F2 were strong enough, it would
cause the spool to slide to the right, but not roll.
P10.88
FIG. P10.87
F2
The force applied at the critical angle exerts zero torque about
the spool’s contact point with the ground and so will not make
the spool roll.
From the right triangle shown in the sketch, observe that
θ c = 90°−φ = 90°− 90°−γ = γ .
b
g
γ
R
r
Thus, cos θ c = cos γ =
.
R
r
φ
θc
P
FIG. P10.88
P10.89
(a)
Consider motion starting from rest over distance x along the incline:
bK
trans
g
b
+ K rot + U i + ∆E = K trans + K rot + U
0 + 0 + Mgx sin θ + 0 =
a
FG
H
1
1
Mv 2 + 2 mR 2
2
2
f
g
f
IJ FG v IJ
KH RK
2
+0
2 Mgx sin θ = M + 2m v 2
Since acceleration is constant,
v 2 = vi2 + 2 ax = 0 + 2 ax , so
a
f
2 Mgx sin θ = M + 2m 2 ax
a=
Mg sin θ
M + 2m
a
f
y
x
∆x
θ
FIG. P10.88
continued on next page
F1
322
Rotation of a Rigid Object About a Fixed Axis
(c)
Suppose the ball is fired from a cart at rest. It moves with acceleration g sin θ = a x down the
incline and a y = − g cos θ perpendicular to the incline. For its range along the ramp, we have
y − yi = v yi t −
t=
2 v yi
g cos θ
x − xi = v xi t +
d=0+
d=
(b)
1
g cos θt 2 = 0 − 0
2
1
axt 2
2
F
GH
I
JK
2
4v yi
1
g sin θ 2
2
g cos 2 θ
2
2 v yi
sin θ
g cos 2 θ
In the same time the cart moves
x − xi = v xi t +
1
2
dc = 0 +
dc =
a
1
axt 2
2
F g sinθM I FG 4v IJ
GH aM + 2mf JK H g cos θ K
2
yi
2
2
2
2 v yi
sin θM
f
g M + 2m cos 2 θ
So the ball overshoots the cart by
∆x = d − d c =
∆x =
∆x =
2
2 v yi
2
sin θ
2 v yi
g cos 2 θ
sin θM +
2
4v yi
2
−
2
sin θM
2 v yi
sin θm −
a
a
g cos 2 θ M + 2m
2
2 v yi
g cos θ M + 2m
2
4mv yi
sin θ
aM + 2mfg cos
2
θ
f
f
sin θM
Chapter 10
P10.90
∑ Fx = ma x reads − f + T = ma . If we take torques around the center of mass,
we can use ∑ τ = Iα , which reads + fR 2 − TR1 = Iα . For rolling without
slipping, α =
a
. By substitution,
R2
b
Ia
I
=
T− f
R2 R 2 m
fR 22 m − TR1 R 2 m = IT − If
e
j b
f I + mR 22 = T I + mR1 R 2
f=
mg
T
f
fR 2 − TR1 =
F I + mR R I T
GH I + mR JK
1
2
2
323
g
n
FIG. P10.90
g
2
Since the answer is positive, the friction force is confirmed to be to the left.
ANSWERS TO EVEN PROBLEMS
P10.28
1
ML2
2
P10.30
168 N ⋅ m clockwise
−226 rad s 2
P10.32
882 N ⋅ m
P10.8
13.7 rad s 2
P10.34
(a) 1.03 s; (b) 10.3 rev
P10.10
(a) 2.88 s; (b) 12.8 s
P10.36
(a) 21.6 kg ⋅ m 2 ; (b) 3.60 N ⋅ m ; (c) 52.4 rev
P10.12
(a) 0.180 rad s;
(b) 8.10 m s 2 toward the center of the
track
P10.38
0.312
P10.40
1.04 × 10 −3 J
P10.14
(a) 0.605 m s ; (b) 17.3 rad s ; (c) 5.82 m s ;
(d) The crank length is unnecessary
P10.42
149 rad s
P10.16
(a) 54.3 rev; (b) 12.1 rev s
P10.44
(a) 6.90 J; (b) 8.73 rad s ; (c) 2.44 m s ;
(d) 1.043 2 times larger
P10.18
0.572
P10.46
2.36 m s
P10.20
(a) 92.0 kg ⋅ m 2 ; 184 J ;
(b) 6.00 m s ; 4.00 m s ; 8.00 m s ; 184 J
P10.48
276 J
P10.50
(a) 74.3 W; (b) 401 W
P10.22
see the solution
P10.52
7 Mv 2
10
P10.54
The disk;
P10.2
(a) 822 rad s 2 ; (b) 4.21 × 10 3 rad
P10.4
(a) 1.20 × 10 2 rad s ; (b) 25.0 s
P10.6
2
P10.24
1.28 kg ⋅ m
P10.26
~ 10 0 kg ⋅ m 2
4 gh
versus
3
gh
324
P10.56
Rotation of a Rigid Object About a Fixed Axis
a
f
(a) 2.38 m s ; (b) 4.31 m s;
(c) It will not reach the top of the loop.
P10.76
P10.58
(a) 0.992 W; (b) 827 W
P10.78
see the solution
P10.60
see the solution
P10.80
P10.62
(a) 12.5 rad s ; (b) 128 rad
(a) 35.0 m s 2 ; 7.35 i N ;
(b) 17.5 m s 2 ; −3.68 i N ;
P10.64
P10.66
b
g h2 − h1
2πR
g
(a) 2.57 × 10 29 J ; (b) −1.63 × 10 17 J day
P10.68
139 m s
P10.70
(a)
P10.72
see the solution
P10.74
(a) −794 N ⋅ m ; −2 510 N ⋅ m; 0;
−1 160 N ⋅ m; −2 940 N ⋅ m;
(b) see the solution
2mgd sin θ + kd 2
I + mR 2
; (b) 1.74 rad s
7r 2
(c) At 0.827 m from the top.
P10.82
2
10 Rg 1 − cos θ
54.0°
e
4g R3 − r 3
j ; (b) 5.31 × 10
4
m s;
P10.84
(a)
P10.86
(a) 0.800 m s 2 ; 0.400 m s 2 ;
(b) 0.600 N between each cylinder and the
plank; 0.200 N forward on each cylinder
by the ground
P10.88
see the solution
P10.90
see the solution; to the left
2
3r
(c) It becomes internal energy.
11
Angular Momentum
CHAPTER OUTLINE
11.1
11.2
11.3
11.4
11.5
11.6
The Vector Product and
Torque
Angular Momentum
Angular Momentum of a
Rotating Rigid Object
Conservation of Angular
Momentum
The Motion of Gyroscopes
and Tops
Angular Momentum as a
Fundamental Quantity
ANSWERS TO QUESTIONS
Q11.1
No to both questions. An axis of rotation must be defined to
calculate the torque acting on an object. The moment arm of
each force is measured from the axis.
Q11.2
A ⋅ B × C is a scalar quantity, since B × C is a vector. Since
A ⋅ B is a scalar, and the cross product between a scalar and a
vector is not defined, A ⋅ B × C is undefined.
a
f
a
f
a f
Q11.3
e j
(a)
Down–cross–left is away from you: − j × − i = − k
(b)
Left–cross–down is toward you: − i × − j = k
e j
FIG. Q11.3
Q11.4
The torque about the point of application of the force is zero.
Q11.5
You cannot conclude anything about the magnitude of the angular momentum vector without first
defining your axis of rotation. Its direction will be perpendicular to its velocity, but you cannot tell
its direction in three-dimensional space until an axis is specified.
Q11.6
Yes. If the particles are moving in a straight line, then the angular momentum of the particles about
any point on the path is zero.
Q11.7
Its angular momentum about that axis is constant in time. You cannot conclude anything about the
magnitude of the angular momentum.
Q11.8
No. The angular momentum about any axis that does not lie along the instantaneous line of motion
of the ball is nonzero.
325
326
Angular Momentum
Q11.9
There must be two rotors to balance the torques on the body of the helicopter. If it had only one
rotor, the engine would cause the body of the helicopter to swing around rapidly with angular
momentum opposite to the rotor.
Q11.10
The angular momentum of the particle about the center of rotation is constant. The angular
momentum about any point that does not lie along the axis through the center of rotation and
perpendicular to the plane of motion of the particle is not constant in time.
Q11.11
The long pole has a large moment of inertia about an axis along the rope. An unbalanced torque will
then produce only a small angular acceleration of the performer-pole system, to extend the time
available for getting back in balance. To keep the center of mass above the rope, the performer can
shift the pole left or right, instead of having to bend his body around. The pole sags down at the
ends to lower the system center of gravity.
Q11.12
The diver leaves the platform with some angular momentum about a horizontal axis through her
center of mass. When she draws up her legs, her moment of inertia decreases and her angular speed
increases for conservation of angular momentum. Straightening out again slows her rotation.
Q11.13
Suppose we look at the motorcycle moving to the right. Its drive wheel is turning clockwise. The
wheel speeds up when it leaves the ground. No outside torque about its center of mass acts on the
airborne cycle, so its angular momentum is conserved. As the drive wheel’s clockwise angular
momentum increases, the frame of the cycle acquires counterclockwise angular momentum. The
cycle’s front end moves up and its back end moves down.
Q11.14
The angular speed must increase. Since gravity does not exert a torque on the system, its angular
momentum remains constant as the gas contracts.
Q11.15
Mass moves away from axis of rotation, so moment of inertia increases, angular speed decreases,
and period increases.
Q11.16
The turntable will rotate counterclockwise. Since the angular momentum of the mouse-turntable
system is initially zero, as both are at rest, the turntable must rotate in the direction opposite to the
motion of the mouse, for the angular momentum of the system to remain zero.
Q11.17
Since the cat cannot apply an external torque to itself while falling, its angular momentum cannot
change. Twisting in this manner changes the orientation of the cat to feet-down without changing
the total angular momentum of the cat. Unfortunately, humans aren’t flexible enough to accomplish
this feat.
Q11.18
The angular speed of the ball must increase. Since the angular momentum of the ball is constant, as
the radius decreases, the angular speed must increase.
Q11.19
Rotating the book about the axis that runs across the middle pages perpendicular to the
binding—most likely where you put the rubber band—is the one that has the intermediate moment
of inertia and gives unstable rotation.
Q11.20
The suitcase might contain a spinning gyroscope. If the gyroscope is spinning about an axis that is
oriented horizontally passing through the bellhop, the force he applies to turn the corner results in a
torque that could make the suitcase swing away. If the bellhop turns quickly enough, anything at all
could be in the suitcase and need not be rotating. Since the suitcase is massive, it will want to follow
an inertial path. This could be perceived as the suitcase swinging away by the bellhop.
Chapter 11
SOLUTIONS TO PROBLEMS
Section 11.1
P11.1
The Vector Product and Torque
i
j
M×N= 6
2
k
−1 = −7.00 i + 16.0 j − 10.0k
2 −1 −3
P11.2
a
fa
f
area = A × B = AB sin θ = 42.0 cm 23.0 cm sin 65.0°−15.0° = 740 cm 2
(b)
A + B = 42.0 cm cos 15.0°+ 23.0 cm cos 65.0° i + 42.0 cm sin 15.0°+ 23.0 cm sin 65.0° j
A + B = 50.3 cm i + 31.7 cm j
a
a
f
a
f
f a
f
length = A + B = a50.3 cmf + a31.7 cmf
2
P11.3
f a
(a)
(a)
(b)
i
j k
2
3
a
2
f
a
f
= 59.5 cm
A × B = −3 4 0 = −17.0k
0
A × B = A B sin θ
17 = 5 13 sin θ
FG 17 IJ = 70.6°
H 5 13 K
A ⋅ B = −3.00a6.00f + 7.00a −10.0f + a −4.00fa9.00 f = −124
AB = a−3.00f + a7.00f + a −4.00f ⋅ a6.00f + a −10.0 f + a9.00 f
F A ⋅ B IJ = cos a−0.979f = 168°
(a)
cos G
H AB K
θ = arcsin
P11.4
2
2
2
−1
j
6.00 −10.0
= 127
k
9.00
a23.0f + a3.00f + a−12.0f = 26.1
FG A × B IJ = sin a0.206f = 11.9° or 168°
H AB K
A×B =
(c)
2
7.00 −4.00 = 23.0 i + 3.00 j − 12.0k
A × B = −3.00
sin −1
2
−1
i
(b)
2
2
2
2
−1
Only the first method gives the angle between the vectors unambiguously.
327
328
*P11.5
Angular Momentum
a
f a
f
τ = r × F = 0.450 m 0.785 N sin 90°−14° up × east
= 0.343 N ⋅ m north
FIG. P11.5
P11.6
The cross-product vector must be perpendicular to both of the factors, so its dot product with either
factor must be zero:
e
je
j
Does 2 i − 3 j + 4k ⋅ 4i + 3 j − k = 0 ?
8 − 9 − 4 = −5 ≠ 0
No . The cross product could not work out that way.
P11.7
A × B = A ⋅ B ⇒ AB sin θ = AB cos θ ⇒ tan θ = 1 or
θ = 45.0°
i
P11.8
j k
a f a f a f a−7.00 N ⋅ mfk
(a)
τ = r × F = 1 3 0 = i 0 − 0 − j 0 − 0 + k 2 − 9 =
3 2 0
(b)
The particle’s position vector relative to the new axis is 1 i + 3 j − 6 j = 1 i − 3 j .
i j k
a
f
τ = 1 −3 0 = 11.0 N ⋅ m k
3
2 0
P11.9
B
F3 = F1 + F2
The torque produced by F3 depends on the
perpendicular distance OD, therefore translating the
point of application of F3 to any other point along
F3
D
O
BC will not change the net torque .
A
C
F1
FIG. P11.9
F2
Chapter 11
*P11.10
i × i = 1 ⋅ 1 ⋅ sin 0° = 0
j
i
j × j and k × k are zero similarly since the
vectors being multiplied are parallel.
k
329
i × j = k
j × i = − k
j × k = i
k × j = − i
k × i = j
i × k = − j
i × j = 1 ⋅ 1 ⋅ sin 90° = 1
FIG. P11.10
Section 11.2
P11.11
Angular Momentum
L = ∑ mi vi ri
b
y
gb
f b
ga
gb
ga
f
= 4.00 kg 5.00 m s 0.500 m + 3.00 kg 5.00 m s 0.500 m
3.00 kg
2
L = 17.5 kg ⋅ m s , and
e
j
L = 17.5 kg ⋅ m 2 s k
x
1.00 m
4.00 kg
FIG. P11.11
P11.12
L=r×p
ge
e
j b
j
L = e −8.10k − 13.9k j kg ⋅ m s = e −22.0 kg ⋅ m sjk
L = 1.50 i + 2.20 j m × 1.50 kg 4.20 i − 3.60 j m s
2
P11.13
e
j
r = 6.00 i + 5.00tj m
so
v=
e
dr
= 5.00 j m s
dt
j
p = mv = 2.00 kg 5.00 j m s = 10.0 j kg ⋅ m s
i
and
2
L = r × p = 6.00
0
j
5.00t
10.0
k
0 =
0
e60.0 kg ⋅ m sjk
2
330
P11.14
Angular Momentum
mv 2
r
∑ Fx = ma x
T sin θ =
∑ Fy = ma y
T cos θ = mg
So
2
sin θ v
=
cos θ rg
v = rg
θ
l
sin θ
cos θ
L = rmv sin 90.0°
L = rm rg
m
sin θ
cos θ
FIG. P11.14
sin θ
cos θ
r = A sin θ , so
L = m 2 gr 3
L=
P11.15
m 2 gA 3
sin 4 θ
cos θ
The angular displacement of the particle around the circle is θ = ωt =
vt
.
R
y
v
The vector from the center of the circle to the mass is then
R cos θ i + R sin θ j .
R
m
θ
Q
P
The vector from point P to the mass is
r = R i + R cos θ i + R sin θ j
LMFG
NH
r = R 1 + cos
FG vt IJ IJ i + sinFG vt IJ jOP
H R KK H R K Q
FIG. P11.15
The velocity is
v=
So
FG IJ
H K
FG IJ
H K
dr
vt vt = − v sin
i + v cos
j
dt
R
R
L = r × mv
a
f
L F vt I O
mvRk McosG J + 1P
N H RK Q
L = mvR 1 + cos ωt i + sin ωtj × − sin ωt i + cos ωtj
L=
P11.16
(a)
The net torque on the counterweight-cord-spool system is:
b
ge
j
τ = r × F = 8.00 × 10 −2 m 4.00 kg 9.80 m s 2 = 3.14 N ⋅ m .
(b)
L = r × mv + Iω
(c)
τ=
b
L = Rmv +
g
dL
= 0.400 kg ⋅ m a
dt
a=
FG IJ FG
H K H
IJ
K
1
v
M
MR 2
=R m+
v=
2
R
2
3.14 N ⋅ m
= 7.85 m s 2
0.400 kg ⋅ m
b0.400 kg ⋅ mgv
x
Chapter 11
P11.17
(a)
(b)
vi = vxi i
zero
At the highest point of the trajectory,
vi
v 2 sin 2θ
1
x= R= i
and
2
2g
y = hmax =
b
vi sin θ
Lv
=M
MN
=
2
i
g
b
v2
R
2
FIG. P11.17
b
sin 2θ vi sin θ
i+
2g
2g
− m vi sin θ
θ
O
2g
L 1 = r1 × mv 1
(c)
331
g jOP × mv
PQ
2
xi i
g v cos θ k
2
i
2g
v 2 sin 2θ
L 2 = R i × mv 2 , where R = i
g
e
j
= mR i × vi cos θ i − vi sin θ j
− mvi3 sin 2θ sin θ = − mRvi sin θ k =
k
g
(d)
P11.18
The downward force of gravity exerts a torque in the –z direction.
Whether we think of the Earth’s surface as curved or flat, we interpret the problem to mean that the
plane’s line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel to
the wheat field. Let the positive x direction be eastward, positive y be northward, and positive z be
vertically upward.
(a)
a
f e
j
p = mv = 12 000 kg e −175 i m sj = −2.10 × 10 i kg ⋅ m s
L = r × p = e 4.30 × 10 k mj × e −2.10 × 10 i kg ⋅ m sj = e −9.03 × 10
r = 4.30 km k = 4.30 × 10 3 m k
6
3
(b)
6
a
9
j
kg ⋅ m 2 s j
f
No . L = r p sin θ = mv r sin θ , and r sin θ is the altitude of the plane. Therefore, L =
constant as the plane moves in level flight with constant velocity.
(c)
Zero . The position vector from Pike’s Peak to the plane is anti-parallel to the velocity of
the plane. That is, it is directed along the same line and opposite in direction.
Thus, L = mvr sin180° = 0 .
332
P11.19
Angular Momentum
The vector from P to the falling ball is
1
r = ri + v i t + at 2
2
FG
H
m
IJ
K
1 2 r = A cos θ i + A sin θ j + 0 −
gt j
2
e
j
l
θ
The velocity of the ball is
P
v = v i + at = 0 − gtj
L = r × mv
So
LMe
N
IJ OP e j
KQ
FG
H
1 2 L = m A cos θ i + A sin θ j + 0 −
gt j × − gtj
2
j
FIG. P11.19
L = − mAgt cos θ k
P11.20
In the vertical section of the hose, the water has zero angular
momentum about our origin (point O between the fireman’s feet).
As it leaves the nozzle, a parcel of mass m has angular momentum:
a
fb
L = r × mv = mrv sin 90.0° = m 1.30 m 12.5 m s
e
2
j
vf
g
L = 16.3 m s m
1.30 m
The torque on the hose is the rate of change in angular momentum.
Thus,
τ=
jb
O
g
dL
dm
= 16.3 m 2 s
= 16.3 m 2 s 6.31 kg s = 103 N ⋅ m
dt
dt
e
j
e
vi
FIG. P11.20
Section 11.3
*P11.21
P11.22
K=
Angular Momentum of a Rotating Rigid Object
1 2 1 I 2ω 2 L2
Iω =
=
2
2 I
2I
The moment of inertia of the sphere about an axis through its center is
I=
b
ga
f
2
2
MR 2 = 15.0 kg 0.500 m
5
5
2
= 1.50 kg ⋅ m 2
Therefore, the magnitude of the angular momentum is
e
jb
g
L = Iω = 1.50 kg ⋅ m 2 3.00 rad s = 4.50 kg ⋅ m 2 s
Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is
directed upward in the +z direction.
e
j
Thus, L = 4.50 kg ⋅ m 2 s k .
Chapter 11
P11.23
(a)
(b)
FG 1 MR IJω = 1 b3.00 kg ga0.200 mf b6.00 rad sg =
H2 K 2
L1
F RI O
L = Iω = M MR + M G J Pω
H 2 K PQ
MN 2
3
= b3.00 kg ga0.200 mf b6.00 rad sg = 0.540 kg ⋅ m s
4
2
2
L = Iω =
2
2
2
P11.24
2
The total angular momentum about the center point is given by L = I hω h + I mω m
Ih =
and
I m3 =
In addition,
ωh =
f
2
a
= 146 kg ⋅ m 2
f
m m L2m 100 kg 4.50 m
=
3
3
2
= 675 kg ⋅ m 2
F
I
GH
JK
2π rad F 1 h I
ω =
= 1.75 × 10 rad s
1 h GH 3 600 s JK
L = 146 kg ⋅ m e1.45 × 10 rad sj + 675 kg ⋅ m e1.75 × 10
2π rad
1h
= 1.45 × 10 −4 rad s
12 h 3 600 s
−3
m
−4
2
Thus,
2
−3
rad s
j
L = 1.20 kg ⋅ m 2 s
or
(a)
a
m h L2h 60.0 kg 2.70 m
=
3
3
with
while
P11.25
0.360 kg ⋅ m 2 s
I=
a
1
m1 L2 + m 2 0.500
12
f
a f
2
=
a
fa f
1
0.100 1.00
12
2
a
+ 0.400 0.500
f
2
= 0.108 3 kg ⋅ m 2
L = Iω = 0.108 3 4.00 = 0.433 kg ⋅ m 2 s
(b)
I=
a
fa f
1
1
m1 L2 + m 2 R 2 = 0.100 1.00
3
3
2
a f
+ 0.400 1.00
2
= 0.433
a f
L = Iω = 0.433 4.00 = 1.73 kg ⋅ m 2 s
*P11.26
∑ Fx = ma x :
+ fs = ma x
We must use the center of mass as the axis in
af a
f a
n
f
∑ τ = Iα :
Fg 0 − n 77.5 cm + fs 88 cm = 0
∑ Fy = ma y :
+n − Fg = 0
Fg
88 cm
fs
155 cm
2
FIG. P11.26
We combine the equations by substitution:
a
f a f
e9.80 m s j77.5 cm = 8.63 m s
=
− mg 77.5 cm + ma x 88 cm = 0
2
ax
88 cm
2
333
334
*P11.27
Angular Momentum
v2
= ω 2r
r
We require a c = g =
e9.80 m s j = 0.313 rad s
2
g
=
r
ω=
100 m
a
2
f
I = Mr = 5 × 10 4 kg 100 m
2
= 5 × 10 8 kg ⋅ m 2
(a)
L = Iω = 5 × 10 8 kg ⋅ m 2 0.313 s = 1.57 × 10 8 kg ⋅ m 2 s
(c)
∑ τ = Iα =
d
I ω f −ωi
i
∆t
∑ τ∆t = Iω f − Iω i = L f − Li
This is the angular impulse-angular momentum theorem.
(b)
Section 11.4
P11.28
(a)
∆t =
1.57 × 10 8 kg ⋅ m 2 s
= 6.26 × 10 3 s = 1.74 h
2 125 N 100 m
a
fa
f
From conservation of angular momentum for the system of two cylinders:
1
g
+ I 2 ω f = I 1ω i
Kf =
so
P11.29
∑τ
=
Conservation of Angular Momentum
bI
(b)
Lf −0
Iiω i = I f ω f :
b
g
1
I 1 + I 2 ω 2f
2
Kf
Ki
=
1
2
bI
1
+ I2
2
1
2 I 1ω i
or
ωf =
and
Ki =
g FG I ω IJ
HI +I K
1
1
2
i
2
=
ω 2 = 7.14 rev min
1
I 1ω i2
2
I1
which is less than 1 .
I1 + I 2
e250 kg ⋅ m jb10.0 rev ming = 250 kg ⋅ m
2
I1
ωi
I1 + I 2
2
a
f
+ 25.0 kg 2.00 m
2
ω2
Chapter 11
P11.30
(a)
335
The total angular momentum of the system of the student, the stool, and the weights about
the axis of rotation is given by
e j
I total = I weights + I student = 2 mr 2 + 3.00 kg ⋅ m 2
Before:
r = 1.00 m .
Thus,
I i = 2 3.00 kg 1.00 m
After:
r = 0.300 m
Thus,
I f = 2 3.00 kg 0.300 m
b
f
ga
b
2
+ 3.00 kg ⋅ m 2 = 9.00 kg ⋅ m 2
f
ga
2
+ 3.00 kg ⋅ m 2 = 3.54 kg ⋅ m 2
We now use conservation of angular momentum.
I f ω f = I iω i
ωf =
or
(b)
P11.31
F I I ω = FG 9.00 IJ b0.750 rad sg =
GH I JK H 3.54 K
i
f
i
jb
g
jb
g
Ki =
1
1
I iω i2 = 9.00 kg ⋅ m 2 0.750 rad s
2
2
Kf =
1
1
I f ω 2f = 3.54 kg ⋅ m 2 1.91 rad s
2
2
e
e
1.91 rad s
2
= 2.53 J
2
= 6.44 J
Let M = mass of rod and m = mass of each bead. From Iiω i = I f ω f , we have
(a)
LM 1 MA
N 12
2
OP
Q
+ 2mr12 ω i =
LM 1 MA
N 12
2
OP
Q
+ 2mr22 ω f
When A = 0.500 m , r1 = 0.100 m , r2 = 0.250 m , and with other values as stated in the
problem, we find
ω f = 9.20 rad s .
(b)
Since there is no external torque on the rod,
L = constant and ω is unchanged .
*P11.32
Let M represent the mass of all the ribs together and L the length of each. The original moment of
1
inertia is ML2 . The final effective length of each rib is L sin 22.5° and the final moment of inertia is
3
1
2
M L sin 22.5° angular momentum of the umbrella is conserved:
3
a
f
1
1
ML2ω i = ML2 sin 2 22.5° ω f
3
3
1.25 rad s
ωf =
= 8.54 rad s
sin 2 22.5°
336
P11.33
Angular Momentum
(a)
The table turns opposite to the way the woman walks, so its angular momentum cancels
that of the woman. From conservation of angular momentum for the system of the woman
and the turntable, we have L f = Li = 0
so,
L f = I womanω woman + I tableω table = 0
and
ω table = −
ω table
woman
woman rv woman
I table
500 kg ⋅ m 2
a
f
1
1 2
2
m woman v woman
+ Iω table
2
2
work done = ∆K = K f − 0 =
W=
P11.34
I woman
m
r2
ω woman = − woman
I table
I table
ω table = 0.360 rad s counterclockwise
or
(b)
F
I FG v IJ = − m
I
JK
GH
JK H r K
60.0 kg a 2.00 mfb1.50 m sg
=−
= −0.360 rad s
F
GH
b
gb
1
60 kg 1.50 m s
2
g
2
+
jb
1
500 kg ⋅ m 2 0.360 rad s
2
e
g
2
= 99.9 J
When they touch, the center of mass is distant from the center of the larger puck by
yCM =
a
f
0 + 80.0 g 4.00 cm + 6.00 cm
= 4.00 cm
120 g + 80.0 g
e
jb
je
g
(a)
L = r1 m1 v1 + r2 m 2 v 2 = 0 + 6.00 × 10 −2 m 80.0 × 10 −3 kg 1.50 m s = 7.20 × 10 −3 kg ⋅ m 2 s
(b)
The moment of inertia about the CM is
FG 1 m r + m d IJ + FG 1 m r + m d IJ
H2
K H2
K
1
I = b0.120 kg ge6.00 × 10 mj + b0.120 kg ge 4.00 × 10 j
2
1
+ e80.0 × 10 kg je 4.00 × 10 mj + e80.0 × 10 kg je6.00 × 10
2
I=
2
1 1
2
1 1
2
2 2
−2
−3
2
2 2
2
−2
−2 2
2
−3
I = 7.60 × 10 −4 kg ⋅ m 2
Angular momentum of the two-puck system is conserved: L = Iω
ω=
L 7.20 × 10 −3 kg ⋅ m 2 s
=
= 9. 47 rad s
I
7.60 × 10 −4 kg ⋅ m 2
−2
m
j
2
Chapter 11
P11.35
(a)
Li = mvA
∑ τ ext = 0 , so L f
337
= Li = mvA
a f
F m IJ v
=G
H m + MK
Lf = m + M vfA
vf
(b)
f
FG
H
IJ
K
Fraction of K lost =
1
2
m2
M +m
2
mv 2 − 12
1
2
v2
mv
=
FIG. P11.35
M
M+m
For one of the crew,
mv 2
= mω i2 r
r
∑ Fr = mar :
n=
We require
n = mg , so ω i =
Now,
Iiω i = I f ω f
g
r
a
f
5.00 × 10 8 kg ⋅ m 2 + 150 × 65.0 kg × 100 m
F 5.98 × 10 I
GH 5.32 × 10 JK
8
8
(a)
(b)
2
a
f
g
= 5.00 × 10 8 kg ⋅ m 2 + 50 × 65.0 kg 100 m
r
2
ωf
g
g
= ω f = 1.12
r
r
a r = ω 2f r = 1.26 g = 12.3 m s 2
Now,
P11.37
v
1
mv 2
2
1
K f = M + m v 2f
2
m
vf =
v ⇒ velocity of the bullet
M+m
and block
Ki =
a
P11.36
l
M
Consider the system to consist of the wad of clay
and the cylinder. No external forces acting on this
system have a torque about the center of the
cylinder. Thus, angular momentum of the system
is conserved about the axis of the cylinder.
L f = Li :
Iω = mvi d
or
LM 1 MR
N2
Thus,
ω=
2
OP
Q
+ mR 2 ω = mvi d
2mvi d
a M + 2 m fR
2
FIG. P11.37
.
No . Some mechanical energy changes to internal energy in this perfectly inelastic collision.
338
*P11.38
Angular Momentum
(a)
Let ω be the angular speed of the signboard when it is vertical.
1 2
Iω = Mgh
2
1 1
1
ML2 ω 2 = Mg L 1 − cos θ
∴
2 3
2
FG
H
a
IJ
K
a
3 g 1 − cos θ
L
∴ω =
f
f
Mg
m
ja
e
θ
v
f
3 9.80 m s 2 1 − cos 25.0°
=
FIG. P11.38
0.50 m
= 2.35 rad s
(b)
I f ω f = Iiω i − mvL represents angular momentum conservation
∴
FG 1 ML
H3
2
∴ω f =
=
(c)
1
3
1
3
IJ
K
+ mL2 ω f =
1
ML2ω i − mvL
3
MLω i − mv
c
1
3
h
M+m L
b2.40 kg ga0.5 mfb2.347 rad sg − b0.4 kg gb1.6 m sg =
b2.40 kgg + 0.4 kg a0.5 mf
1
3
0. 498 rad s
Let hCM = distance of center of mass from the axis of rotation.
hCM =
b2.40 kg ga0.25 mf + b0.4 kgga0.50 mf = 0.285 7 m .
2.40 kg + 0.4 kg
Apply conservation of mechanical energy:
aM + mfgh a1 − cos θ f = 12 FGH 13 ML + mL IJKω
L c M + mhL ω OP
∴θ = cos M1 −
MN 2aM + mfgh PQ
R| b2.40 kg g + 0.4 kg a0.50 mf b0.498 rad sg
= cos S1 −
|T 2b2.40 kg + 0.4 kg ge9.80 m s jb0.285 7 mg
2
CM
1
3
−1
2
2
2
2
CM
−1
1
3
2
2
= 5.58°
1
L
2
2
U|
V|
W
Chapter 11
P11.39
339
The meteor will slow the rotation of the Earth by the largest amount if its line of motion passes
farthest from the Earth’s axis. The meteor should be headed west and strike a point on the equator
tangentially.
Let the z axis coincide with the axis of the Earth with +z pointing northward. Then, conserving
angular momentum about this axis,
∑ L f = ∑ Li ⇒ Iω f = Iω i + mv × r
or
Thus,
2
2
MR 2ω f k = MR 2ω i k − mvRk
5
5
mvR
5mv
or
=
ωi −ω f = 2
2
2 MR
5 MR
ωi −ω f =
e
je
j = 5.91 × 10
kg je6.37 × 10 mj
5 3.00 × 10 13 kg 30.0 × 10 3 m s
e
2 5.98 × 10
24
−14
6
rad s
∆ω max ~ 10 −13 rad s
Section 11.5
*P11.40
The Motion of Gyroscopes and Tops
Angular momentum of the system of the spacecraft and the gyroscope is conserved. The gyroscope
and spacecraft turn in opposite directions.
0 = I 1ω 1 + I 2 ω 2 :
θ
t
− I 1ω 1 = I 2
b
g
−20 kg ⋅ m 2 −100 rad s = 5 × 10 5 kg ⋅ m 2
t=
*P11.41
I=
2.62 × 10 5 s
= 131 s
2 000
2
2
MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m
5
5
e
je
j
2
= 9.71 × 10 37 kg ⋅ m 2
F 2π rad I = 7.06 × 10 kg ⋅ m s
GH 86 400 s JK
F 2π rad I FG 1 yr IJ F 1 d I =
kg ⋅ m sjG
H 2.58 × 10 yr JK H 365.25 d K GH 86 400 s JK
L = Iω = 9.71 × 10 37 kg ⋅ m 2
e
τ = Lω p = 7.06 × 10 33
Section 11.6
P11.42
FG 30° IJ FG π rad IJ
H t K H 180° K
33
2
2
2
4
5.45 × 10 22 N ⋅ m
Angular Momentum as a Fundamental Quantity
6.626 1 × 10 −34 J ⋅ s
(a)
L=
h
h
= mvr so v =
2πmr
2π
(b)
K=
1
1
mv 2 = 9.11 × 10 −31 kg 2.19 × 10 6 m s
2
2
(c)
ω=
=
L
1.055 × 10 −34 J ⋅ s
=
=
I mr 2
9.11 × 10 −31 kg 0.529 × 10 −10 m
e
e
v=
je
je
e
je
j
2π 9.11 × 10 -31 kg 0.529 × 10 −10 m
j
2
j
= 2.18 × 10 −18 J
2
= 4.13 × 10 16 rad s
= 2.19 × 10 6 m s
340
Angular Momentum
Additional Problems
*P11.43
First, we define the following symbols:
I P = moment of inertia due to mass of people on the equator
I E = moment of inertia of the Earth alone (without people)
ω = angular velocity of the Earth (due to rotation on its axis)
2π
T=
= rotational period of the Earth (length of the day)
ω
R = radius of the Earth
The initial angular momentum of the system (before people start running) is
b
g
Li = I P ω i + I E ω i = I P + I E ω i
When the Earth has angular speed ω, the tangential speed of a point on the equator is v t = Rω .
Thus, when the people run eastward along the equator at speed v relative to the surface of the Earth,
vp
v
=ω + .
their tangential speed is v p = v t + v = Rω + v and their angular speed is ω P =
R
R
The angular momentum of the system after the people begin to run is
FG
H
L f = I Pω p + I Eω = I P ω +
IJ
K
b
g
I v
v
+ I Eω = I P + I E ω + P .
R
R
d
i
Since no external torques have acted on the system, angular momentum is conserved L f = Li ,
b
g
b
g
IPv
= I P + I E ω i . Thus, the final angular velocity of the Earth is
R
IPv
IPv
ω = ωi −
= ω i 1 − x = , where x ≡
.
I P + I E Rω i
IP + IE R
giving I P + I E ω +
b
a f
g
b
The new length of the day is T =
day is ∆T = T − Ti ≈ Ti x = Ti
LM
MN bI
P
2π
ω
=
g
T
2π
= i ≈ Ti 1 + x , so the increase in the length of the
ωi 1− x 1− x
OP
g PQ
a f
a f
IPv
Ti2 I P v
2π
, this may be written as ∆T ≈
. Since ω i =
.
Ti
+ I E Rω i
2π I P + I E R
b
To obtain a numeric answer, we compute
jb
e
I P = m p R 2 = 5.5 × 10 9 70 kg
g e6.37 × 10 mj
6
2
= 1.56 × 10 25 kg ⋅ m 2
2
= 9.71 × 10 37 kg ⋅ m 2 .
and
IE =
2
2
m E R 2 = 5.98 × 10 24 kg 6.37 × 10 6 m
5
5
e
je
j
e8.64 × 10 sj e1.56 × 10 kg ⋅ m jb2.5 m sg =
Thus, ∆T ≈
2π e1.56 × 10 + 9.71 × 10 j kg ⋅ m e6.37 × 10 mj
4
25
2
25
37
2
2
6
7.50 × 10 −11 s .
g
Chapter 11
*P11.44
(a)
341
bK + U g = bK + U g
s A
s B
1
0 + mgy A = mv B2 + 0
2
e
j
v B = 2 gy A = 2 9.8 m s 2 6.30 m = 11.1 m s
(b)
L = mvr = 76 kg 11.1 m s 6.3 m = 5.32 × 10 3 kg ⋅ m 2 s toward you along the axis of the
channel.
(c)
The wheels on his skateboard prevent any tangential force from acting on him. Then no
torque about the axis of the channel acts on him and his angular momentum is constant. His
legs convert chemical into mechanical energy. They do work to increase his kinetic energy.
The normal force acts forward on his body on its rising trajectory, to increase his linear
momentum.
(d)
L = mvr
(e)
v=
5.32 × 10 3 kg ⋅ m 2 s
= 12.0 m s
76 kg 5.85 m
eK + U j + W = e K + U j
1
1
76 kg b11.1 m sg + 0 + W = 76 kg b12.0 m sg
2
2
g
g
B
C
2
2
+ 76 kg 9.8 m s 2 0.45 m
W = 5.44 kJ − 4.69 kJ + 335 J = 1.08 kJ
(f)
eK + U j = e K + U j
1
1
76 kg b12.0 m sg + 0 = 76 kgv
2
2
g
C
g
D
2
2
D
+ 76 kg 9.8 m s 2 5.85 m
v D = 5.34 m s
(g)
Let point E be the apex of his flight:
eK + U j = e K + U j
1
76 kg b5.34 m sg + 0 = 0 + 76 kg e9.8 m s jb y
2
by − y g = 1.46 m
g
D
g
E
2
E
(h)
E
− yD
g
D
For the motion between takeoff and touchdown
1
y f = yi + v yi t + a y t 2
2
−2.34 m = 0 + 5.34 m s t − 4.9 m s 2 t 2
t=
(i)
2
a fa f =
−5.34 ± 5.34 2 + 4 4.9 2.34
−9.8
1. 43 s
This solution is more accurate. In chapter 8 we modeled the normal force as constant while
the skateboarder stands up. Really it increases as the process goes on.
342
P11.45
Angular Momentum
(a)
I = ∑ mi ri2
=m
FG 4d IJ
H3K
2
+m
FG d IJ
H 3K
2
+m
FG 2d IJ
H3K
2
m
m
1
d2
= 7m
3
2d
3
2
d
Think of the whole weight, 3mg, acting at the center of gravity.
τ =r×F=
FG d IJ e− ij × 3mge− jj = bmgdgk
H 3K
3g
τ 3mgd
=
=
counterclockwise
2
I 7md
7d
(c)
α=
(d)
a = αr =
FG 3 g IJ FG 2d IJ =
H 7d K H 3 K
2g
up
7
The angular acceleration is not constant, but energy is.
a K + U f + ∆E = a K + U f
F dI 1
0 + a3m f g G J + 0 = Iω
H 3K 2
i
(e)
maximum kinetic energy = mgd
(f)
ωf =
(g)
L f = Iω f =
(h)
vf =ω fr =
6g
7d
7md 2
3
6g
=
7d
6g d
=
7d 3
FG 14 g IJ
H3K
2 gd
21
12
md 3 2
f
2
f
3
d
FIG. P11.45
(b)
m
P
+0
Chapter 11
P11.46
(a)
af b
343
g
The radial coordinate of the sliding mass is r t = 0.012 5 m s t . Its angular momentum is
b
gb
gb
gb
g
2
L = mr 2ω = 1.20 kg 2.50 rev s 2π rad rev 0.012 5 m s t 2
e
j
L = 2.95 × 10 −3 kg ⋅ m 2 s 3 t 2
or
The drive motor must supply torque equal to the rate of change of this angular momentum:
τ=
b
ja f b0.005 89 Wgt
dL
= 2.95 × 10 −3 kg ⋅ m 2 s 3 2t =
dt
e
f
ga
(b)
τ f = 0.005 89 W 440 s = 2.59 N ⋅ m
(c)
P = τω = 0.005 89 W t 5π rad s =
(d)
P f = 0.092 5 W s 440 s = 40.7 W
(e)
T =m
(f)
W=
(g)
b
gb
b
g b0.092 5 W sgt
f
ga
v2
= mrω 2 = 1.20 kg 0.012 5 m s t 5π rad s
r
b
z
zb
440 s
440 s
0
0
Pdt =
gb
gb
g
0.092 5 W s tdt =
ja
g
2
1
0.092 5 J s 2 440 s
2
e
b3.70 N sgt
=
f
2
= 8.96 kJ
The power the brake injects into the sliding block through the string is
b
gb
g b
g
Pb = F ⋅ v = Tv cos 180° = − 3.70 N s t 0.012 5 m s = − 0.046 3 W s t =
Wb =
z
440 s
0
0
Pb dt = −
g
0.046 3 W s tdt
b
ga
1
0.046 3 W s 440 s
2
=−
(h)
zb
440 s
dWb
dt
∑ W = W + Wb = 8.96 kJ − 4.48 kJ =
f
2
= −4.48 kJ
4.48 kJ
Just half of the work required to increase the angular momentum goes into rotational kinetic
energy. The other half becomes internal energy in the brake.
P11.47
Using conservation of angular momentum, we have
e j
e j
Laphelion = Lperihelion or mra2 ω a = mrp2 ω p .
vp
e j vr = emr j r
Thus, mra2
a
a
ra v a = rp v p or v a =
2
p
rp
ra
giving
p
vp =
b
g
0.590 AU
54.0 km s = 0.910 km s .
35.0 AU
344
P11.48
Angular Momentum
(a)
∑ τ = MgR − MgR =
(b)
∑τ =
0
dL
, and since
dt
∑τ = 0 , L =
constant.
Since the total angular momentum of the system is zero, the
monkey and bananas move upward with the same speed
at any instant, and he will not reach the bananas (until they
get tangled in the pulley). Also, since the tension in the rope is
the same on both sides, Newton’s second law applied to the
monkey and bananas give the same acceleration upwards.
FIG. P11.48
P11.49
(a)
τ = r × F = r F sin180° = 0
Angular momentum is conserved.
L f = Li
mrv = mri vi
v=
b g
m ri vi
mv 2
=
r
r3
ri vi
r
2
(b)
T=
(c)
The work is done by the centripetal force in the
negative direction.
Method 1:
b g
W = z F ⋅ dA = − z Tdr ′ = − z
dr ′ =
2a r ′ f
Fr I
mbr v g F 1
1I
1
mv G − 1J
=
− J=
G
2
Hr r K 2 Hr K
r
ri
i i
b g
ar ′ f
FIG. P11.49
m ri vi
2
i
Method 2:
(d)
m ri vi
3
2
2
W = ∆K =
2
2
i
2 r
2
ri
2
i
2
F
GH
I
JK
r2
1
1
1
mv 2 − mvi2 =
mvi2 i2 − 1
2
2
2
r
Using the data given, we find
v = 4.50 m s
T = 10.1 N
W = 0.450 J
Chapter 11
P11.50
(a)
Angular momentum is conserved:
F
GH
FG
H
mvi d
1
d
=
Md 2 + m
2
12
2
ω=
(b)
m
IJ IJω
KK
2
ω
vi
6mvi
Md + 3md
d
O
O
(a)
1
The original energy is mvi2 .
2
(b)
FIG. P11.50
The final energy is
F
GH
I 36m v
JK aMd + 3mdf
2 2
i
1 2 1 1
md 2
Iω =
Md 2 +
2
2 12
4
2
=
3m 2 vi2 d
.
2 Md + 3md
a
f
The loss of energy is
3m 2 vi2 d
mMvi2 d
1
=
mvi2 −
2
2 Md + 3md
2 Md + 3md
a
f a
f
and the fractional loss of energy is
a
mMvi2 d 2
2 Md + 3md
P11.51
(a)
f
mvi2
=
M
.
M + 3m
FG d IJ
H 2K
L = 2b75.0 kg gb5.00 m sga5.00 mf
Li = m1 v1i r1i + m 2 v 2i r2i = 2mv
i
Li = 3 750 kg ⋅ m 2 s
(b)
345
1
1
m1 v12i + m 2 v 22i
2
2
1
75.0 kg 5.00 m s
Ki = 2
2
Ki =
FG IJ b
H K
gb
g
2
= 1.88 kJ
(c)
Angular momentum is conserved: L f = Li = 3 750 kg ⋅ m 2 s
(d)
vf =
(e)
Kf = 2
(f)
W = K f − K i = 5.62 kJ
Lf
=
3 750 kg ⋅ m 2 s
= 10.0 m s
2 75.0 kg 2.50 m
d i b
2 mr f
f
ga
FG 1 IJ b75.0 kggb10.0 m sg
H 2K
2
= 7.50 kJ
FIG. P11.51
346
P11.52
Angular Momentum
(a)
(b)
(c)
*P11.53
LM FG d IJ OP = Mvd
N H 2K Q
F1 I
K = 2G Mv J = Mv
H2 K
Li = 2 Mv
2
2
L f = Li = Mvd
Lf
(d)
Mvd
=
= 2v
vf =
2 Mr f 2 M d4
(e)
Kf = 2
(f)
W = K f − K i = 3 Mv 2
FIG. P11.52
ch
FG 1 Mv IJ = Ma2 vf
H2 K
2
f
2
= 4Mv 2
The moment of inertia of the rest of the Earth is
I=
2
2
MR 2 = 5.98 × 10 24 kg 6.37 × 10 6 m
5
5
e
j
2
= 9.71 × 10 37 kg ⋅ m 2 .
For the original ice disks,
I=
1
1
Mr 2 = 2.30 × 10 19 kg 6 × 10 5 m
2
2
e
j
2
= 4.14 × 10 30 kg ⋅ m 2 .
For the final thin shell of water,
I=
2
2
Mr 2 = 2.30 × 10 19 kg 6.37 × 10 6 m
3
3
e
j
2
= 6.22 × 10 32 kg ⋅ m 2 .
Conservation of angular momentum for the spinning planet is expressed by Iiω i = I f ω f
π
2π
= e6. 22 × 10 + 9.71 × 10 j
e4.14 × 10 + 9.71 × 10 j 86 2400
s
b86 400 s + δ g
F1 + δ I F 1 + 4.14 × 10 I = F 1 + 6.22 × 10 I
GH 86 400 s JK GH 9.71 × 10 JK GH 9.71 × 10 JK
30
37
32
30
32
37
37
δ
6.22 × 10 32 4.14 × 10 30
=
−
86 400 s 9.71 × 10 37 9.71 × 10 37
δ = 0.550 s
37
Chapter 11
P11.54
For the cube to tip over, the center of mass (CM) must rise so that it
is over the axis of rotation AB. To do this, the CM must be raised a
e
distance of a
j
CM
2 −1 .
e
1
I cubeω 2
2
j
∴ Mga
2 −1 =
D
A
From conservation of angular momentum,
F
GH
4a
8 Ma
mv =
3
3
mv
ω=
2 Ma
F
GH
2
P11.55
C
Iω
JK
D
B
4a/3
A
I
JK
1 8 Ma 2 m 2 v 2
= Mga
2
3
4M 2 a 2
M
3 ga
m
v=
347
e
2 −1
e
2 −1
j
FIG. P11.54
j
Angular momentum is conserved during the
inelastic collision.
Mva = Iω
ω=
Mva 3 v
=
I
8a
The condition, that the box falls off the table, is that
the center of mass must reach its maximum height
as the box rotates, hmax = a 2 . Using conservation
of energy:
1 2
Iω = Mg a 2 − a
2
e
F
GH
j
I FG 3v IJ = Mgea
JK H 8 a K
16
v =
gae 2 − 1j
3
O
L ga
v = 4M e 2 − 1jP
3
Q
N
1 8 Ma 2
2
3
2
2 −a
j
FIG. P11.55
2
12
P11.56
(a)
The net torque is zero at the point of contact, so the angular momentum before and after the
collision must be equal.
FG 1 MR IJ ω = FG 1 MR IJω + eMR jω
H2 K H2 K
2
(b)
∆E
=
E
e
1 1
2 2
MR 2
ωi 2
3
je j
Rω i 2
3
e j
e MR jω
+ 12 M
1 1
2 2
2
i
2
2
i
−
e
1 1
2 2
2
j
MR 2 ω i2
= −
ω=
2
3
ωi
3
348
P11.57
Angular Momentum
∆p
Rω i
Mv
MRω
=
=
=
µMg µMg
3 µg
f
(a)
∆t =
(b)
W = ∆K =
µMgx =
1 2 1
Iω =
MR 2ω i2
2
18
1
MR 2ω i2
18
x=
(See Problem 11.56)
R 2ω i2
18 µg
ANSWERS TO EVEN PROBLEMS
P11.2
(a) 740 cm 2 ; (b) 59.5 cm
P11.32
8.54 rad s
P11.4
(a) 168°; (b) 11.9° principal value;
(c) Only the first is unambiguous.
P11.34
(a) 7.20 × 10 −3 kg ⋅ m 2 s ; (b) 9.47 rad s
P11.36
12.3 m s 2
P11.38
(a) 2.35 rad s; (b) 0.498 rad s ; (c) 5.58°
P11.40
131 s
P11.42
(a) 2.19 × 10 6 m s ; (b) 2.18 × 10 −18 J ;
P11.6
No; see the solution
P11.8
(a) −7.00 N ⋅ m k ; (b) 11.0 N ⋅ m k
P11.10
see the solution
P11.12
e−22.0 kg ⋅ m sjk
P11.14
see the solution
P11.16
(a) 3.14 N ⋅ m ; (b) 0. 400 kg ⋅ m v ;
a
f
(c) 4.13 × 10 16 rad s
2
P11.44
b
(c) 7.85 m s
P11.18
a
f
g
2
e
j
(a) +9.03 × 10 9 kg ⋅ m 2 s south; (b) No;
P11.46
b
g
b
g
(e) b3.70 N sgt ; (f) 8.96 kJ; (g) −4.48 kJ
(a) 0.005 89 W t ; (b) 2.59 N ⋅ m ;
(c) 0.092 5 W s t ; (d) 40.7 W ;
(c) 0
(h) +4.48 kJ
P11.20
103 N ⋅ m
P11.22
e4.50 kg ⋅ m sj up
P11.24
(a) 11.1 m s; (b) 5.32 × 10 3 kg ⋅ m 2 s ;
(c) see the solution; (d) 12.0 m s;
(e) 1.08 kJ ; (f) 5.34 m s; (g) 1.46 m;
(h) 1.43 s; (i) see the solution
2
2
1.20 kg ⋅ m s perpendicularly into the
clock face
P11.26
8.63 m s 2
P11.28
Kf
I1
Iω
=
(a) 1 i ; (b)
Ki I1 + I 2
I1 + I 2
P11.30
(a) 1.91 rad s ; (b) 2.53 J; 6.44 J
P11.48
(a) 0; (b) 0; no
P11.50
(a)
P11.52
(a) Mvd ; (b) Mv 2 ; (c) Mvd ; (d) 2v;
(e) 4Mv 2 ; (f) 3 Mv 2
6mvi
M
; (b)
M + 3m
Md + 3md
P11.54
M
3 ga
m
P11.56
(a)
e
2 −1
j
ωi
2
∆E
=−
; (b)
E
3
3
12
Static Equilibrium and Elasticity
CHAPTER OUTLINE
12.1
12.2
12.3
12.4
The Conditions for
Equilibrium
More on the Center of
Gravity
Examples of Rigid Objects
in Static Equilibrium
Elastic Properties of Solids
ANSWERS TO QUESTIONS
Q12.1
When you bend over, your center of gravity shifts forward.
Once your CG is no longer over your feet, gravity contributes
to a nonzero net torque on your body and you begin to rotate.
Q12.2
Yes, it can. Consider an object on a spring oscillating back and
forth. In the center of the motion both the sum of the torques
and the sum of the forces acting on the object are (separately)
zero. Again, a meteoroid flying freely through interstellar space
feels essentially no forces and keeps moving with constant
velocity.
Q12.3
No—one condition for equilibrium is that
∑ F = 0 . For this to
be true with only a single force acting on an object, that force
would have to be of zero magnitude; so really no forces act on
that object.
Q12.4
(a)
Consider pushing up with one hand on one side of a steering wheel and pulling down
equally hard with the other hand on the other side. A pair of equal-magnitude oppositelydirected forces applied at different points is called a couple.
(b)
An object in free fall has a non-zero net force acting on it, but a net torque of zero about its
center of mass.
Q12.5
No. If the torques are all in the same direction, then the net torque cannot be zero.
Q12.6
(a)
Yes, provided that its angular momentum is constant.
(b)
Yes, provided that its linear momentum is constant.
Q12.7
A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide
at the edge of a swimming pool each have a center of mass that is not within the bulk of the object.
Q12.8
Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywood
along the string of the plumb bob. Now suspend the plywood with the nail through a different point
on the plywood, not along the first line you drew. Again hang the plumb bob from the nail and trace
along the string. The center of gravity is located halfway through the thickness of the plywood
under the intersection of the two lines you drew.
349
350
Static Equilibrium and Elasticity
Q12.9
The center of gravity must be directly over the point where the chair leg contacts the floor. That
way, no torque is applied to the chair by gravity. The equilibrium is unstable.
Q12.10
She can be correct. If the dog stands on a relatively thick scale, the dog’s legs on the ground might
support more of its weight than its legs on the scale. She can check for and if necessary correct for
this error by having the dog stand like a bridge with two legs on the scale and two on a book of
equal thickness—a physics textbook is a good choice.
Q12.11
If their base areas are equal, the tall crate will topple first. Its center of gravity is higher off the incline
than that of the shorter crate. The taller crate can be rotated only through a smaller angle before its
center of gravity is no longer over its base.
Q12.12
The free body diagram demonstrates that it is necessary to have
friction on the ground to counterbalance the normal force of the
wall and to keep the base of the ladder from sliding. Interestingly
enough, if there is friction on the floor and on the wall, it is not
possible to determine whether the ladder will slip from the
equilibrium conditions alone.
FIG. Q12.12
Q12.13
When you lift a load with your back, your back muscles must supply the torque not only to rotate
your upper body to a vertical position, but also to lift the load. Since the distance from the
pivot—your hips—to the load—essentially your shoulders—is great, the force required to supply
the lifting torque is very large. When lifting from your knees, your back muscles need only keep
your back straight. The force required to do that is much smaller than when lifting with your back,
as the torque required is small, because the moment arm of the load is small—the line of action of
the load passes close to your hips. When you lift from your knees, your much stronger leg and hip
muscles do the work.
Q12.14
Shear deformation.
Q12.15
The vertical columns experience simple compression due to gravity acting upon their mass. The
horizontal slabs, however, suffer significant shear stress due to gravity. The bottom surface of a
sagging lintel is under tension. Stone is much stronger under compression than under tension, so
horizontal slabs are more likely to fail.
351
Chapter 12
SOLUTIONS TO PROBLEMS
Section 12.1
P12.1
The Conditions for Equilibrium
To hold the bat in equilibrium, the player must exert both a
force and a torque on the bat to make
∑ Fx = ∑ Fy = 0
and
F
0.600 m
∑τ = 0
∑ Fy = 0 ⇒ F − 10.0 N = 0 , or the player must exert a net
O
upward force of F = 10.0 N
10.0 N
To satisfy the second condition of equilibrium, the player must
exert an applied torque τ a to make
FIG. P12.1
∑ τ = τ a − a0.600 mfa10.0 N f = 0 . Thus, the required torque is
τ a = +6.00 N ⋅ m or 6.00 N ⋅ m counterclockwise
P12.2
Use distances, angles, and forces as shown. The conditions of
equilibrium are:
∑ Fy = 0 ⇒
∑ Fx = 0 ⇒
∑τ = 0 ⇒
Fy
Fx − R x = 0
Fy A cos θ − Fg
Fx
l
Fy + R y − Fg = 0
FG A IJ cos θ − F A sinθ = 0
H 2K
x
Ry
Fg
θ
Rx
O
FIG. P12.2
P12.3
Take torques about P.
L
∑ τ p = −n0 MN
OP
Q
LM
N
OP
Q
A
A
+ d + m1 g + d + m b gd − m 2 gx = 0
2
2
m1
We want to find x for which n 0 = 0 .
x=
bm g + m ggd + m g
1
b
m2 g
1
A
2
=
bm
1
g
+ mb d
m2
mb g
m1 g
O
m2 g
d
A
2
m2
P
CG
+ m1 2A
x
nO
nP
A
FIG. P12.3
352
Static Equilibrium and Elasticity
Section 12.2
P12.4
More on the Center of Gravity
The hole we can count as negative mass
xCG =
m 1 x1 − m 2 x 2
m1 − m 2
Call σ the mass of each unit of pizza area.
xCG =
xCG =
P12.5
c h c− h
− σπ c h
σπR 2 0 − σπ
σπR 2
R
8
3
4
=
R 2
2
R
2
R 2
2
R
6
The coordinates of the center of gravity of piece 1 are
4.00 cm
x1 = 2.00 cm and y1 = 9.00 cm .
The coordinates for piece 2 are
18.0 cm
1
x 2 = 8.00 cm and y 2 = 2.00 cm .
2
The area of each piece is
12.0 cm
A1 = 72.0 cm 2 and A 2 = 32.0 cm 2 .
FIG. P12.5
And the mass of each piece is proportional to the area. Thus,
∑ m i xi
xCG =
∑ mi
e72.0 cm ja2.00 cmf + e32.0 cm ja8.00 cmf =
=
2
2
72.0 cm 2 + 32.0 cm 2
3.85 cm
and
yCG =
∑ m i yi
∑ mi
e72.0 cm ja9.00 cmf + e32.0 cm ja2.00 cmf =
2
=
2
104 cm 2
6.85 cm .
4.00 cm
Chapter 12
P12.6
Let σ represent the mass-per-face area. A
vertical strip at position x, with width dx and
height
ax − 3.00f
y
2
1.00 m
has mass
9
a
f
y = (x — 3.00)2/9
2
σ x − 3.00 dx
dm =
.
9
The total mass is
x
z
a f
z
3.00
2
σ x − 3 dx
M = dm =
9
x =0
FG σ IJ z ex − 6 x + 9jdx
H 9K
F σ I L x 6x + 9xOP
M = G JM −
H 9 KN 3 2 Q
3.00
M=
353
x
0
2
FIG. P12.6
0
3
3.00 m
dx
3.00
2
=σ
0
The x-coordinate of the center of gravity is
xCG =
P12.7
z
xdm
M
=
1
9σ
z
3.00
a f
2
σx x − 3 dx =
0
σ
9σ
ze
3.00
j
x 3 − 6 x 2 + 9 x dx =
0
LM
N
1 x 4 6x3 9x 2
−
+
9 4
3
2
OP
Q
3.00
=
0
6.75 m
= 0.750 m
9.00
Let the fourth mass (8.00 kg) be placed at (x, y), then
xCG = 0 =
x=−
Similarly,
yCG = 0 =
a3.00fa4.00f + m axf
4
12.0 + m 4
12.0
= −1.50 m
8.00
a3.00fa4.00f + 8.00byg
12.0 + 8.00
y = −1.50 m
P12.8
In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus,
the center of gravity of the triangle is located at x = 6.67 m , y = 2.33 m (see the Example on the
center of mass of a triangle in Chapter 9).
The coordinates of the center of gravity of the three-object system are then:
b6.00 kg ga5.50 mf + b3.00 kgga6.67 mf + b5.00 kg ga−3.50 mf
a6.00 + 3.00 + 5.00f kg
xCG =
∑ m i xi
∑ mi
xCG =
35.5 kg ⋅ m
= 2.54 m and
14.0 kg
yCG =
∑ m i yi
∑ mi
yCG =
66.5 kg ⋅ m
= 4.75 m
14.0 kg
=
=
b6.00 kg ga7.00 mf + b3.00 kgga2.33 mf + b5.00 kgga+3.50 mf
14.0 kg
354
Static Equilibrium and Elasticity
Section 12.3
P12.9
Examples of Rigid Objects in Static Equilibrium
∑ τ = 0 = mga3r f − Tr
2T − Mg sin 45.0° = 0
3r
bg
Mg sin 45.0° 1 500 kg g sin 45.0°
=
2
2
= 530 9.80 N
T=
a fa f
m=
T 530 g
=
= 177 kg
3g
3g
m
1 500 kg
θ = 45°
FIG. P12.9
*P12.10
(a)
For rotational equilibrium of the lowest rod about its point of support,
m1 = 9.00 g
+12.0 g g 3 cm − m1 g 4 cm
(b)
For the middle rod,
+ m 2 2 cm − 12.0 g + 9.0 g 5 cm = 0
b
(c)
P12.11
g
m 2 = 52.5 g
For the top rod,
52.5 g + 12.0 g + 9.0 g 4 cm − m3 6 cm = 0
b
∑τ = 0 .
g
m3 = 49.0 g
Fg → standard weight
24.0 cm
26.0 cm
Fg′ → weight of goods sold
a
f a f
F 13 I
F = F′ G J
H 12 K
F F − F ′ I 100 = F 13 − 1I × 100 =
GH F ′ JK GH 12 JK
Fg 0.240 = Fg′ 0.260
g
Fg
F′g
g
g
g
FIG. P12.11
8.33%
g
*P12.12
(a)
Consider the torques about an axis perpendicular
to the page and through the left end of the
horizontal beam.
T
∑ τ = +aT sin 30.0°fd − a196 N fd = 0 ,
V
giving T = 392 N .
(b)
30.0°
H
196 N
d
FIG. P12.12
a
f
From
∑ Fx = 0 ,
From
∑ Fy = 0 , V + T sin 30.0°−200 N = 0 , or V = 196 N − a392 N f sin 30.0° =
H − T cos 30.0° = 0 , or H = 392 N cos 30.0° = 339 N to the right .
0 .
355
Chapter 12
P12.13
(a)
∑ Fx = f − n w = 0
∑ Fy = n g − 800 N − 500 N = 0
nw
Taking torques about an axis at the foot of the ladder,
a800 Nfa4.00 mf sin 30.0°+a500 Nfa7.50 mf sin 30.0°
−n a15.0 cmf cos 30.0° = 0
500 N
ng
w
Solving the torque equation,
nw
a4.00 mfa800 Nf + a7.50 mfa500 Nf tan 30.0° = 268 N .
=
800 N
f
A
15.0 m
Next substitute this value into the Fx equation to find
f = n w = 268 N
Solving the equation
FIG. P12.13
in the positive x direction.
∑ Fy = 0 ,
n g = 1 300 N in the positive y direction.
(b)
In this case, the torque equation
∑τ A = 0
gives:
a9.00 mfa800 Nf sin 30.0°+a7.50 mfa500 Nf sin 30.0°−a15.0 mfbn g sin 60.0° = 0
w
or
n w = 421 N .
Since f = n w = 421 N and f = fmax = µn g , we find
fmax
421 N
=
= 0.324 .
1 300 N
ng
µ=
P12.14
(a)
∑ Fx = f − n w = 0
∑ Fy = n g − m1 g − m 2 g = 0
nw
(1)
m2 g
(2)
F LI
∑ τ A = −m1 g GH 2 JK cos θ − m 2 gx cos θ + n w L sin θ = 0
From the torque equation,
nw =
LM 1 m g + FG x IJ m g OP cot θ
N2 H LK Q
1
Then, from equation (1):
and from equation (2):
(b)
m1 g
2
L1 F xI O
f = n = M m g + G J m g P cot θ
N2 H LK Q
n = bm + m g g
w
g
1
1
2
2
If the ladder is on the verge of slipping when x = d ,
then
µ=
f
x=d
ng
=
e
m1
2
+
m2d
L
j cot θ
m1 + m 2
.
θ
f
A
ng
FIG. P12.14
356
P12.15
Static Equilibrium and Elasticity
(a)
Taking moments about P,
aR sin 30.0°f0 + aR cos 30.0°fa5.00 cmf − a150 Nfa30.0 cmf = 0
R = 1 039.2 N = 1.04 kN
The force exerted by the hammer on the nail is equal in magnitude
and opposite in direction:
1.04 kN at 60° upward and to the right.
(b)
FIG. P12.15
f = R sin 30.0°−150 N = 370 N
n = R cos 30.0° = 900 N
a
f a
f
Fsurface = 370 N i + 900 N j
P12.16
See the free-body diagram at the right.
When the plank is on the verge of tipping about point P, the
normal force n1 goes to zero. Then, summing torques about
point P gives
∑ τ p = −mgd + Mgx = 0
or
F mI
x = G Jd .
H MK
From the dimensions given on the free-body diagram, observe
that d = 1.50 m Thus, when the plank is about to tip,
x=
P12.17
F 30.0 kg I a1.50 mf =
GH 70.0 kg JK
Mg
3.00 m
x
P
n1
mg
d
6.00 m
FIG. P12.16
0.643 m .
Torque about the front wheel is zero.
a
fb g a
fb g
0 = 1.20 m mg − 3.00 m 2 Fr
Thus, the force at each rear wheel is
Fr = 0.200mg = 2.94 kN .
The force at each front wheel is then
Ff =
mg − 2 Fr
= 4. 41 kN .
2
FIG. P12.17
n2
1.50 m
Chapter 12
P12.18
∑ Fx = Fb − Ft + 5.50 N = 0
(1)
357
5.50 N
∑ Fy = n − mg = 0
Summing torques about point O,
10.0 m
∑ τ O = Ft a1.50 mf − a5.50 mfa10.0 mf = 0
mg
Ft
which yields Ft = 36.7 N to the left
1.50 m
Then, from Equation (1),
Fb
O
Fb = 36.7 N − 5.50 N = 31.2 N to the right
n
FIG. P12.18
P12.19
P12.20
(a)
Te sin 42.0° = 20.0 N
Te = 29.9 N
(b)
Te cos 42.0° = Tm
Tm = 22.2 N
Relative to the hinge end of the bridge, the cable is attached
horizontally out a distance x = 5.00 m cos 20.0° = 4.70 m and
a f
vertically down a distance y = a5.00 mf sin 20.0° = 1.71 m . The
cable then makes the following angle with the horizontal:
θ = tan −1
(a)
LM a12.0 + 1.71f m OP = 71.1° .
N 4.70 m Q
af
a
f
f
f
a
which yields T = 35.5 kN
∑ Fx = 0 ⇒ Rx − T cos 71.1° = 0
or
(c)
a
f
b
R x = 35.5 kN cos 71.1° = 11.5 kN right
g
∑ Fy = 0 ⇒ R y − 19.6 kN + T sin 71.1°−9.80 kN = 0
Thus,
a
f
R y = 29. 4 kN − 35.5 kN sin 71.1° = −4.19 kN
= 4.19 kN down
y
19.6 kN
9.80 kN
f
−9.80 kN 7.00 m cos 20.0° = 0
(b)
20.0°
Rx
7.00 m
−T cos 71.1° 1.71 m + T sin 71.1° 4.70 m
a
x
5.00 m
R x 0 + R y 0 − 19.6 kN 4.00 m cos 20.0°
a
T
4.00 m
Take torques about the hinge end of the bridge:
af
Ry
FIG. P12.20
358
*P12.21
Static Equilibrium and Elasticity
(a)
We model the horse as a particle. The drawbridge will fall
out from under the horse.
α = mg
=
(b)
e
1
2
A cos θ 0
1
3
mA
j
3 9.80 m s 2 cos 20.0°
a
2 8.00 m
f
1 2
Iω = mgh
2
1 1
1
∴ ⋅ mA 2ω 2 = mg ⋅ A 1 − sin θ 0
2 3
2
b
∴ω =
(c)
Rx
θ0
A
3g
cos θ 0
2A
=
2
Ry
= 1.73 rad s
mg
2
FIG. P12.21(a)
g
e
ja
3 9.80 m s 2
3g
1 − sin θ 0 =
1 − sin 20° = 1.56 rad s
A
8.00 m
b
g
f
The linear acceleration of the bridge is:
a
fe
Ry
θ0
1
1
a = Aα = 8.0 m 1.73 rad s 2 = 6.907 m s 2
2
2
j
The force at the hinge + the force of gravity produce the
acceleration a = 6.907 m s 2 at right angles to the bridge.
b
ge
a
Rx
mg
FIG. P12.21(c)
j
R x = ma x = 2 000 kg 6.907 m s 2 cos 250° = −4.72 kN
Ry − mg = ma y
j b
e
g
e
j
∴ R y = m g + a y = 2 000 kg 9.80 m s 2 + 6.907 m s 2 sin 250° = 6.62 kN
e
j
Thus: R = −4.72 i + 6.62 j kN .
(d)
Rx = 0
a=ω
2
FG 1 AIJ = b1.56 rad sg a4.0 mf = 9.67 m s
H2 K
2
R y − mg = ma
b
ge
j
Ry
2
∴ R y = 2 000 kg 9.8 m s 2 + 9.67 m s 2 = 38.9 kN
Thus: R y = 38.9 j kN
a
mg
FIG. P12.21(d)
Rx
Chapter 12
P12.22
Call the required force F, with
components Fx = F cos 15.0° and
Fx
center of the wheel by the handles.
R
F cos 15.0°−n x
b
b
nx
8.00 cm
Just as the wheel leaves the ground, the
ground exerts no force on it.
∑ Fx = 0 :
∑ Fy = 0 :
400 N
Fy
Fy = − F sin 15.0° , transmitted to the
ny
a
a
distances
(1)
forces
− F sin 15.0°−400 N + n y = 0 (2)
FIG. P12.22
Take torques about its contact point with the brick. The needed distances are seen to be:
a
f
b = R − 8.00 cm = 20.0 − 8.00 cm = 12.0 cm
2
2
a = R − b = 16.0 cm
(a)
a
∑τ = 0 :
f
− Fx b + Fy a + 400 N a = 0 , or
a
f
a
f
a
fa
f
F − 12.0 cm cos 15.0°+ 16.0 cm sin 15.0° + 400 N 16.0 cm = 0
F=
so
(b)
6 400 N ⋅ cm
= 859 N
7.45 cm
Then, using Equations (1) and (2),
a
f
n x = 859 N cos 15.0° = 830 N and
a
f
n y = 400 N + 859 N sin 15.0° = 622 N
n = n x2 + n y2 = 1.04 kN
θ = tan −1
F n I = tan a0.749f =
GH n JK
y
−1
36.9° to the left and upward
x
*P12.23
When x = x min , the rod is on the verge of slipping, so
b g
f = fs
From
max
= µ sn = 0.50n .
∑ Fx = 0 , n − T cos 37° = 0 , or n = 0.799T .
a
f
∑ Fy = 0 ,
Using
∑τ = 0
a
2.0 m
37°
n
f
x
Fg
Fg
2.0 m
FIG. P12.23
Thus, f = 0.50 0.799T = 0.399T
From
f + T sin 37°−2 Fg = 0 , or 0.399T − 0.602T − 2 Fg = 0 , giving T = 2.00 Fg .
for an axis perpendicular to the page and through the left end of the beam gives
f e j
359
a
f
− Fg ⋅ x min − Fg 2.0 m + 2 Fg sin 37° 4.0 m = 0 , which reduces to x min = 2.82 m .
360
P12.24
Static Equilibrium and Elasticity
x=
3L
4
L
If the CM of the two bricks does not lie over the edge, then
the bricks balance.
If the lower brick is placed
L
over the edge, then the
4
second brick may be placed so that its end protrudes
over the edge.
P12.25
x
3L
4
FIG. P12.24
To find U, measure distances and forces from point A. Then, balancing torques,
a0.750fU = 29.4a2.25f
U = 88.2 N
To find D, measure distances and forces from point B. Then, balancing torques,
a0.750fD = a1.50fa29.4f
Also, notice that U = D + Fg , so
*P12.26
D = 58.8 N
∑ Fy = 0 .
Consider forces and torques on the beam.
∑ Fx = 0 :
∑ Fy = 0 :
∑τ = 0 :
R cos θ − T cos 53° = 0
R sin θ + T sin 53°−800 N = 0
aT sin 53°f8 m − a600 Nfx − a200 Nf4 m = 0
600 Nx + 800 N ⋅ m
= 93.9 N m x + 125 N . As x increases from 2 m, this expression
8 m sin 53°
grows larger.
b
(a)
Then T =
(b)
From substituting back,
g
R cos θ = 93.9 x + 125 cos 53°
R sin θ = 800 N − 93.9 x + 125 sin 53°
Dividing, tan θ =
800 N
R sin θ
= − tan 53°+
93.9 x +125 cos 53°
R cos θ
a
f
F 32 − 1IJ
tan θ = tan 53° G
H 3x + 4 K
As x increases the fraction decreases and θ decreases .
continued on next page
Chapter 12
(c)
To find R we can work out R 2 cos 2 θ + R 2 sin 2 θ = R 2 . From the expressions above for
R cos θ and R sin θ ,
a
R 2 = T 2 cos 2 53°+T 2 sin 2 53°−1 600 NT sin 53°+ 800 N
R 2 = T 2 − 1 600T sin 53°+640 000
a
R 2 = 93.9 x + 125
f
2
a
f
2
f
− 1 278 93.9 x + 125 + 640 000
e
R = 8 819 x 2 − 96 482 x + 495 678
j
12
At x = 0 this gives R = 704 N . At x = 2 m , R = 581 N . At x = 8 m , R = 537 N . Over the
range of possible values for x, the negative term −96 482x dominates the positive term
8 819 x 2 , and R decreases as x increases.
Section 12.4
P12.27
∆L
F
=Y
A
Li
∆L =
P12.28
Elastic Properties of Solids
(a)
a fa fa f
e
je
j
200 9.80 4.00
FLi
=
= 4.90 mm
AY
0.200 × 10 −4 8.00 × 10 10
stress =
F
F
= 2
A πr
a
f FGH d2 IJK
e
8
F = stress π
F = 1.50 × 10
2
F 2.50 × 10 m I
N m jπ G
H 2 JK
-2
2
2
F = 73.6 kN
(b)
a
∆L =
*P12.29
f
stress = Y strain =
Y∆L
Li
astressfL = e1.50 × 10
i
Y
The definition of Y =
8
ja
f=
N m 2 0.250 m
1.50 × 10
10
N m
2
2.50 mm
stress
means that Y is the slope of the graph:
strain
Y=
300 × 10 6 N m 2
= 1.0 × 10 11 N m 2 .
0.003
361
362
P12.30
Static Equilibrium and Elasticity
Count the wires. If they are wrapped together so that all support nearly equal stress, the number
should be
20.0 kN
= 100 .
0.200 kN
Since cross-sectional area is proportional to diameter squared, the diameter of the cable will be
a1 mmf
P12.31
100 ~ 1 cm .
From the defining equation for the shear modulus, we find ∆x as
e
ja
f
5.00 × 10 −3 m 20.0 N
hf
∆x =
=
= 2.38 × 10 −5 m
−4
6
2
2
SA
3.0 × 10 N m 14.0 × 10 m
e
je
j
or ∆x = 2.38 × 10 −2 mm .
P12.32
The force acting on the hammer changes its momentum according to
a f
mvi + F ∆t = mv f so F =
Hence, F =
30.0 kg −10.0 m s − 20.0 m s
0.110 s
m v f − vi
∆t
.
= 8.18 × 10 3 N .
By Newton’s third law, this is also the magnitude of the average force exerted on the spike by the
hammer during the blow. Thus, the stress in the spike is:
stress =
and the strain is: strain =
P12.33
(a)
a fa f
= π e5.00 × 10
F 8.18 × 10 3 N
=
= 1.97 × 10 7 N m 2
b0.023 0 mg2
A
π
4
stress 1.97 × 10 7 N m 2
=
= 9.85 × 10 −5 .
Y
20.0 × 10 10 N m 2
F = A stress
F
−3
m
j e4.00 × 10
2
8
N m2
j
3.0 ft
= 3.14 × 10 4 N
(b)
t
A
The area over which the shear occurs is equal to
the circumference of the hole times its thickness.
Thus,
a f
e
je
j
A = 2πr t = 2π 5.00 × 10 −3 m 5.00 × 10 −3 m
= 1.57 × 10
af
−4
e
m
FIG. P12.33
2
je
j
So, F = A Stress = 1.57 × 10 −4 m 2 4.00 × 10 8 N m 2 = 6.28 × 10 4 N .
Chapter 12
P12.34
363
Let the 3.00 kg mass be mass #1, with the 5.00 kg mass, mass # 2. Applying Newton’s second law to
each mass gives:
m1 a = T − m1 g
(1)
m2 a = m2 g − T
and
(2)
where T is the tension in the wire.
Solving equation (1) for the acceleration gives: a =
and substituting this into equation (2) yields:
T
− g,
m1
m2
T − m2 g = m2 g − T .
m1
Solving for the tension T gives
b
gb
ge
j
2
2m1 m 2 g 2 3.00 kg 5.00 kg 9.80 m s
=
= 36.8 N .
8.00 kg
m 2 + m1
T=
From the definition of Young’s modulus, Y =
FLi
, the elongation of the wire is:
A ∆L
a f
a36.8 Nfa2.00 mf
TL
∆L =
=
YA
e2.00 × 10 N m jπ e2.00 × 10
i
11
P12.35
2
−3
j
m
2
= 0.029 3 mm .
Consider recompressing the ice, which has a volume 1.09V0 .
F ∆V IJ = −e2.00 × 10 N m ja−0.090f =
∆P = − BG
1.09
HV K
9
2
1.65 × 10 8 N m 2
i
*P12.36
B=−
∆P
∆V
Vi
=−
∆PVi
∆V
e
j
1.13 × 10 8 N m 2 1 m3
∆PVi
=−
= −0.053 8 m 3
B
0.21 × 10 10 N m 2
(a)
∆V = −
(b)
The quantity of water with mass 1.03 × 10 3 kg occupies volume at the bottom
1 m 3 − 0.053 8 m 3 = 0.946 m 3 . So its density is
(c)
*P12.37
1.03 × 10 3 kg
0.946 m
3
= 1.09 × 10 3 kg m3 .
With only a 5% volume change in this extreme case, liquid water is indeed nearly
incompressible.
Part of the load force extends the cable and part compresses the column by the same distance ∆A :
F=
∆A =
YA A A ∆A Ys As ∆A
+
AA
As
F
YA A A
AA
+
Ys As
As
= 8.60 × 10
−4
=
e
8 500 N
7 ×10 10 π 0.162 4 2 − 0 .161 4 2
a f
4 3. 25
m
j + 20 ×10 π b0.012 7 g
4a 5.75 f
10
2
364
Static Equilibrium and Elasticity
Additional Problems
*P12.38
(a)
The beam is perpendicular to the wall, since 3 2 + 4 2 = 5 2 . Then sin θ =
(b)
∑ τ hinge = 0 :
a f
f
+T sin θ 3 m − 250 N 10 m = 0
T=
(c)
a
4m
; θ = 53.1° .
5m
2 500 Nm
= 1.04 × 10 3 N
3 m sin 53.1°
1.04 × 10 3 N
T
=
= 0.126 m
k 8.25 × 10 3 N m
The cable is 5.126 m long. From the law of cosines,
x=
a fa
5.126 m
4m
α
f
4 2 = 5.126 2 + 3 2 − 2 3 5.126 cos θ
θ = cos −1
(d)
2
2
3m
2
3 + 5.126 − 4
= 51.2°
2 3 5.126
a fa
f
θ
FIG. P12.38
From the law of sines, the angle the hinge makes with the wall satisfies
sin α
sin 51.2°
=
5.126 m
4m
sin α = 0.998 58
∑ τ hinge = 0
a f
a
fa
f
+T 3 m sin 51.2°−250 N 10 m 0.998 58 = 0
T = 1.07 × 10 3 N
x=
(e)
1.07 × 10 3 N
= 0.129 m
8.25 × 10 3 N m
θ = cos −1
(f)
3 2 + 5.129 2 − 4 2
= 51.1°
2 3 5.129
a fa
f
Now the answers are self-consistent:
sin 51.1°
= 0.998 51
4m
T 3 m sin 51.1°−250 N 10 m 0.998 51 = 0
sin α = 5.129 m
a f
a
fa
f
3
T = 1.07 × 10 N
x = 0.129 5 m
θ = 51.1°
P12.39
Let n A and n B be the normal forces at the points of
support.
Choosing the origin at point A with
∑ Fy = 0 and ∑ τ = 0,
A
B
we find:
e
j e
j
−e3.00 × 10 jb g g15.0 − e8.00 × 10 jb g g25.0 + n a50.0 f = 0
n A + n B − 8.00 × 10 4 g − 3.00 × 10 4 g = 0 and
4
4
B
15.0 m
50.0 m
FIG. P12.39
The equations combine to give n A = 5.98 × 10 5 N and b B = 4.80 × 10 5 N .
365
Chapter 12
P12.40
When the concrete has cured and the pre-stressing tension has been released, the rod presses in on
the concrete and with equal force, T2 , the concrete produces tension in the rod.
(a)
Thus, ∆L =
or
(b)
f FGH ∆LL IJK
a
In the concrete: stress = 8.00 × 10 6 N m 2 = Y ⋅ strain = Y
astressfL = e
i
ja
8.00 × 10 6 N m 2 1.50 m
9
Y
30.0 × 10 N m
i
f
2
∆L = 4.00 × 10 −4 m = 0.400 mm .
In the concrete: stress =
T2
= 8.00 × 10 6 N m 2 , so
Ac
e
je
j
T2 = 8.00 × 10 6 N m 2 50.0 × 10 −4 m 2 = 40.0 kN
(c)
FG IJ
H K
T2 Li
T2
∆L
=
Ysteel so ∆L =
AR
Li
A R Ysteel
For the rod:
e4.00 × 10 Nja1.50 mf
∆L =
= 2.00 × 10
e1.50 × 10 m je20.0 × 10 N m j
4
−4
10
2
m = 2.00 mm
(d)
The rod in the finished concrete is 2.00 mm longer than its unstretched length. To remove
stress from the concrete, one must stretch the rod 0.400 mm farther, by a total of 2. 40 mm .
(e)
For the stretched rod around which the concrete is poured:
FG
H
IJ
K
FG ∆L IJ A Y
H L K
F 2.40 × 10 m I e1.50 × 10 m je20.0 × 10
T =G
H 1.50 m JK
∆Ltotal
T1
=
Ysteel
AR
Li
or T1 =
−3
−4
1
*P12.41
−3
2
total
i
2
With A as large as possible, n1 and n 2 will both be large. The
equality sign in f 2 ≤ µ sn 2 will be true, but the less-than sign
in f1 < µ sn1 . Take torques about the lower end of the pole.
n 2 A cos θ + Fg
R steel
FG 1 AIJ cos θ − f A sinθ = 0
H2 K
f
1
Fg = 0
2
Since n 2 > 0 , it is necessary that
1 − 0.576 tan θ < 0
1
= 1.736
0.576
∴θ > 60.1°
∴ tan θ >
∴A =
f2
A
n2
Setting f 2 = 0.576n 2 , the torque equation becomes
a
j
N m 2 = 48.0 kN
θ
2
n 2 1 − 0.576 tan θ +
10
7.80 ft
d
<
= 9.00 ft
sin θ sin 60.1°
n1
Fg
θ
f1
FIG. P12.41
d
366
P12.42
Static Equilibrium and Elasticity
Call the normal forces A and B. They make angles α and β
with the vertical.
∑ Fx = 0:
∑ Fy = 0:
Mg
A sin α − B sin β = 0
A
A cos α − Mg + B cos β = 0
β
α
A sin α
sin β
Substitute B =
A cos α + A cos β
b
sin α
= Mg
sin β
g
Mg
A cos α sin β + sin α cos β = Mg sin β
sin β
A = Mg
sin α + β
b
sin α
B = Mg
sin α + β
b
P12.43
B
(a)
See the diagram.
(b)
If x = 1.00 m , then
B sin α
A sin α
g
B cos α
A cos α
g
FIG. P12.42
T
Ry
60.0°
Rx
O
∑ τ O = a−700 N fa1.00 mf − a200 N fa3.00 mf
a fa f
+aT sin 60.0°fa6.00 mf = 0
x
700 N
− 80.0 N 6.00 m
3.00 m
200 N
80.0 N
3.00 m
FIG. P12.43
Solving for the tension gives: T = 343 N .
(c)
From
∑ Fx = 0 , R x = T cos 60.0° =
From
∑ Fy = 0 , R y = 980 N − T sin 60.0° =
171 N .
683 N .
If T = 900 N :
∑ τ O = a−700 N fx − a 200 N fa3.00 mf − a80.0 N fa6.00 mf +
Solving for x gives: x = 5.13 m .
a900 Nf sin 60.0° a6.00 mf = 0 .
367
Chapter 12
P12.44
(a)
Sum the torques about top hinge:
∑ τ = 0:
T cos 30.0°
C
af af
af
C 0 + D 0 + 200 N cos 30.0° 0
a
f
+200 N sin 30.0° 3.00 m
a
f a
1.80 m
f
−392 N 1.50 m + A 1.80 m
af
160 N bright g
Giving A =
392 N
A
+B 0 = 0
1.50 m
1.50 m
B
.
FIG. P12.44
∑ Fx = 0 :
(b)
T sin 30.0°
D
−C − 200 N cos 30.0°+ A = 0
C = 160 N − 173 N = −13.2 N
In our diagram, this means 13.2 N to the right .
∑ Fy = 0 : +B + D − 392 N + 200 N sin 30.0° = 0
(c)
b g
B + D = 392 N − 100 N = 292 N up
Given C = 0: Take torques about bottom hinge to obtain
(d)
af af a
f af
a
f
a
f
a
f
A 0 + B 0 + 0 1.80 m + D 0 − 392 N 1.50 m + T sin 30.0° 3.00 m + T cos 30.0° 1.80 m = 0
so T =
P12.45
Using
588 N ⋅ m
= 192 N .
1.50 m + 1.56 m
a
f
∑ Fx = ∑ Fy = ∑ τ = 0, choosing the origin at the left end
of the beam, we have (neglecting the weight of the beam)
∑ Fx = Rx − T cos θ = 0 ,
∑ Fy = Ry + T sin θ − Fg = 0 ,
and
∑ τ = − Fg aL + d f + T sin θ a2L + d f = 0.
Solving these equations, we find:
(a)
T=
(b)
Rx
a f
sin θ a 2L + d f
F aL + d f cot θ
=
Fg L + d
g
2L + d
Ry =
Fg L
2L + d
FIG. P12.45
368
P12.46
Static Equilibrium and Elasticity
∑ τ point 0 = 0 gives
T sin 25.0°
aT cos 25.0°fFGH 34A sin 65.0°IJK + aT sin 25.0°fFGH 34A cos 65.0°IJK
FA
I
= b 2 000 N gaA cos 65.0°f + b1 200 N gG cos 65.0°J
H2
K
l
3l
4
1 200 N
From which, T = 1 465 N = 1.46 kN
From
65.0°
H
∑ Fx = 0 ,
b
g
H = T cos 25.0° = 1 328 N toward right = 1.33 kN
From
2 000 N
T cos 25.0°
V
FIG. P12.46
∑ Fy = 0 ,
b
g
V = 3 200 N − T sin 25.0° = 2 581 N upward = 2.58 kN
P12.47
We interpret the problem to mean that the support at
point B is frictionless. Then the support exerts a force in
the x direction and
FBy = 0
∑ Fx = FBx − FAx = 0
b
g
and ∑ τ = −b3 000 g ga 2.00f − b10 000 g ga6.00f + F a1.00 f = 0 .
FAy − 3 000 + 10 000 g = 0
Bx
These equations combine to give
FIG. P12.47
5
FAx = FBx = 6.47 × 10 N
FAy = 1.27 × 10 5 N
P12.48
a
f
n= M+m g H = f
a
H
f
H max = f max = µ s m + M g
mgL
cos 60.0°+ Mgx cos 60.0°− HL sin 60.0°
2
µ m + M tan 60.0° m
x H tan 60.0° m
=
−
= s
−
2M
2M
L
Mg
M
∑τ A = 0 =
a
3
1
= µ s tan 60.0°− = 0.789
2
4
x
f
Mg
mg
n
60.0°
A
f
FIG. P12.48
Chapter 12
P12.49
From the free-body diagram, the angle T makes with the rod is
T
20°
θ = 60.0°+20.0° = 80.0°
and the perpendicular component of T is T sin 80.0°.
Summing torques around the base of the rod,
a
∑ τ = 0:
fb
a
g
f
10 000 N
− 4.00 m 10 000 N cos 60°+T 4.00 m sin 80° = 0
T=
∑ Fx = 0 :
b10 000 Ng cos 60.0° =
sin 80.0°
FV
5.08 × 10 3 N
60°
FH − T cos 20.0° = 0
FH
FH = T cos 20.0° = 4.77 × 10 3 N
∑ Fy = 0 :
FIG. P12.49
FV + T sin 20.0°−10 000 N = 0
b
g
and FV = 10 000 N − T sin 20.0° = 8.26 × 10 3 N
P12.50
Choosing the origin at R,
(1)
(2)
(3)
R
∑ Fx = + R sin 15.0°−T sin θ = 0
∑ Fy = 700 − R cos 15.0°+T cos θ = 0
∑ τ = −700 cos θ a0.180f + T b0.070 0g = 0
Solve the equations for θ
from (3), T = 1 800 cos θ from (1), R =
T
90°
15.0°
1 800 sin θ cos θ
sin 15.0°
1 800 sin θ cos θ cos 15.0°
Then (2) gives 700 −
+ 1 800 cos 2 θ = 0
sin 15.0°
or
θ
cos 2 θ + 0.388 9 − 3.732 sin θ cos θ = 0
θ
18.0 cm
25.0 cm
n
FIG. P12.50
Squaring, cos 4 θ − 0.880 9 cos 2 θ + 0.010 13 = 0
Let
u = cos 2 θ then using the quadratic equation,
u = 0.011 65 or 0.869 3
Only the second root is physically possible,
∴θ = cos −1 0.869 3 = 21.2°
∴ T = 1.68 × 10 3 N
P12.51
and
Choosing torques about R, with
−
a
f a
R = 2.34 × 10 3 N
∑τ = 0
fFGH IJK a
f
2L
L
350 N + T sin 12.0°
− 200 N L = 0 .
2
3
From which, T = 2.71 kN .
Let R x = compression force along spine, and from
R x = Tx = T cos 12.0° = 2.65 kN .
∑ Fx = 0
FIG. P12.51
369
370
P12.52
Static Equilibrium and Elasticity
(a)
(b)
Just three forces act on the rod: forces perpendicular to the
sides of the trough at A and B, and its weight. The lines of
action of A and B will intersect at a point above the rod.
They will have no torque about this point. The rod’s weight
will cause a torque about the point of intersection as in
Figure 12.52(a), and the rod will not be in equilibrium
unless the center of the rod lies vertically below the
intersection point, as in Figure 12.52(b). All three forces
must be concurrent. Then the line of action of the weight is
a diagonal of the rectangle formed by the trough and the
normal forces, and the rod’s center of gravity is vertically
above the bottom of the trough.
2
AO =
So cos θ =
(a)
2
L
1+
2
cos 2 30.0 °
cos 2 60.0 °
2
F cos
GH cos
2
30.0°
2
60.0°
Fg
O
FIG. P12.52(a)
I
JK
B
Fg
θ
A
30.0°
L
=
2
60.0°
O
FIG. P12.52(b)
AO 1
= and θ = 60.0° .
2
L
Locate the origin at the bottom left corner of the cabinet
and let x = distance between the resultant normal force and
the front of the cabinet. Then we have
∑ Fx = 200 cos 37.0°− µn = 0
∑ Fy = 200 sin 37.0°+n − 400 = 0
∑ τ = na0.600 − xf − 400a0.300f + 200 sin 37.0° a0.600 f
(2)
−200 cos 37.0° 0.400 = 0
(3)
a
From (2),
From (3),
From (1),
(b)
A
In Figure (b), AO cos 30.0° = BO cos 60.0° and
L2 = AO + BO = AO + AO
P12.53
B
(1)
f
n = 400 − 200 sin 37.0° = 280 N
a
f
72.2 − 120 + 280 0.600 − 64.0
280
x = 20.1 cm to the left of the front edge
x=
µk =
200 cos 37.0°
= 0.571
280
In this case, locate the origin x = 0 at the bottom right
corner of the cabinet. Since the cabinet is about to tip, we
can use ∑ τ = 0 to find h:
∑ τ = 400a0.300f − a300 cos 37.0°fh = 0
h=
FIG. P12.53
120
= 0.501 m
300 cos 37.0°
Chapter 12
P12.54
(a), (b) Use the first diagram and sum the torques about the lower
front corner of the cabinet.
∑ τ = 0 ⇒ − F 1.00 m + 400 N 0.300 m = 0
a
f a fa
a400 Nfa0.300 mf = 120 N
yielding F =
1.00 m
=
0
⇒
−
F
f + 120 N = 0 ,
∑ x
∑ Fy = 0 ⇒ −400 N + n = 0 ,
0.300 m
F
f
400 N
1.00 m
f = 120 N
n = 400 N
or
so
f
f 120 N
Thus, µ s = =
= 0.300 .
n 400 N
(c)
n
Apply F ′ at the upper rear corner and directed so
θ + φ = 90.0° to obtain the largest possible lever arm.
θ = tan −1
F’
θ
FG 1.00 m IJ = 59.0°
H 0.600 m K
400 N
θ
f
a1.00 mf + a0.600 mf + a400 Nfa0.300 mf = 0
2
φ
1.00 m
Thus, φ = 90.0°−59.0° = 31.0° .
Sum the torques about the lower front corner of the
cabinet:
−F′
371
2
n
0.600 m
120 N ⋅ m
= 103 N .
so
F′ =
1.17 m
Therefore, the minimum force required to tip the cabinet is
FIG. P12.54
103 N applied at 31.0° above the horizontal at the upper left corner .
P12.55
(a)
We can use
∑ Fx = ∑ Fy = 0 and ∑ τ = 0 with pivot point at
the contact on the floor.
Then
P
T
∑ Fx = T − µ sn = 0 ,
L/2
∑ Fy = n − Mg − mg = 0, and
FL
Mg
I
∑ τ = MgaL cos θ f + mg GH 2 cos θ JK − T aL sin θ f = 0
L/2
mg
Solving the above equations gives
M=
FG
H
m 2 µ s sin θ − cos θ
2 cos θ − µ s sin θ
n
θ
IJ
K
f
FIG. P12.55
This answer is the maximum vaue for M if µ s < cot θ . If µ s ≥ cot θ , the mass M can increase
without limit. It has no maximum value, and part (b) cannot be answered as stated either. In
the case µ s < cot θ , we proceed.
(b)
At the floor, we have the normal force in the y-direction and frictional force in the xdirection. The reaction force then is
b g
R = n 2 + µ sn
2
=
a M + mf g
1 + µ s2 .
At point P, the force of the beam on the rope is
b g
F = T 2 + Mg
2
a
= g M 2 + µ s2 M + m
f
2
.
372
P12.56
Static Equilibrium and Elasticity
(a)
The height of pin B is
1000 N
a10.0 mf sin 30.0° = 5.00 m .
B
10.0 m
The length of bar BC is then
nA
nC
45.0°
30.0°
BC =
5.00 m
= 7.07 m.
sin 45.0°
C
A
FIG. P12.56(a)
Consider the entire truss:
∑ Fy = n A − 1 000 N + nC = 0
∑ τ A = −b1 000 N g10.0 cos 30.0°+nC 10.0 cos 30.0°+7.07 cos 45.0°
=0
Which gives nC = 634 N .
Then, n A = 1 000 N − nC = 366 N .
(b)
(c)
Suppose that a bar exerts on a pin a force not along the
length of the bar. Then, the pin exerts on the bar a
force with a component perpendicular to the bar. The
only other force on the bar is the pin force on the other
end. For ∑ F = 0 , this force must also have a
component perpendicular to the bar. Then, the total
torque on the bar is not zero. The contradiction proves
that the bar can only exert forces along its length.
FIG. P12.56(b)
Joint A:
CAB
∑ Fy = 0 : −C AB sin 30.0°+366 N = 0 ,
so
C AB = 732 N
A
TAC
nA = 366 N
∑ Fx = 0 : −C AB cos 30.0°+TAC = 0
a
1000 N
f
TAC = 732 N cos 30.0° = 634 N
30.0°
Joint B:
a
f
∑ Fx = 0 : 732 N cos 30.0°−CBC cos 45.0° = 0
C BC =
a732 Nf cos 30.0° =
cos 45.0°
897 N
CAB = 732 N
B
45.0°
CBC
FIG. P12.56(c)
Chapter 12
P12.57
From geometry, observe that
cos θ =
1
4
θ = 75.5°
and
For the left half of the ladder, we have
∑ Fx = T − R x = 0
∑ Fy = R y + n A − 686 N = 0
∑ τ top = 686 Na1.00 cos 75.5°f + T a2.00 sin 75.5°f
(1)
−n A 4.00 cos 75.5° = 0
(3)
a
f
(2)
FIG. P12.57
For the right half of the ladder we have
∑ Fx = Rx − T = 0
∑ Fy = nB − Ry = 0
∑ τ top = nB a 4.00 cos 75.5°f − T a 2.00 sin 75.5°f = 0
(4)
(5)
Solving equations 1 through 5 simultaneously yields:
(a)
T = 133 N
(b)
n A = 429 N
and
n B = 257 N
(c)
R x = 133 N
and
R y = 257 N
The force exerted by the left half of the ladder on the right half is to the right and
downward.
P12.58 (a)
x CG =
=
yCG =
∑ m i xi
∑ mi
b1 000 kg g10.0 m + b125 kgg0 + b125 kg g0 + b125 kg g20.0 m =
1 375 kg
9.09 m
b1 000 kg g10.0 m + b125 kgg20.0 m + b125 kgg20.0 m + b125 kg g0
1 375 kg
= 10.9 m
(b)
By symmetry, x CG = 10.0 m
There is no change in yCG = 10.9 m
(c)
P12.59
vCG =
FG 10.0 m − 9.09 m IJ =
H 8.00 s K
0.114 m s
Considering the torques about the point at the bottom of the bracket yields:
b0.050 0 mga80.0 Nf − Fb0.060 0 mg = 0 so
F = 66.7 N .
373
374
P12.60
Static Equilibrium and Elasticity
When it is on the verge of slipping, the cylinder is in equilibrium.
and
f1 = n 2 = µ sn1
f 2 = µ sn 2
∑ Fx = 0 :
∑ Fy = 0 :
∑ τ = 0:
P + n1 + f 2 = Fg
P = f1 + f 2
As P grows so do f1 and f 2
n
1
Therefore, since µ s = , f1 = 1
2
2
n
then
(1)
P + n1 + 1 = Fg
4
5
So
P + n1 = Fg
4
3
Therefore, P =
Fg
8
P12.61
(a)
(b)
P12.62
(a)
and
a f
F
A
∆L
Li
=
FIG. P12.60
(2)
FG IJ
H K
becomes
F = k ∆L , Young’s modulus is Y =
Thus, Y =
n 2 n1
=
2
4
n
n
3
P = 1 + 1 = n1
2
4 4
5 4
P+
P = Fg
4 3
f2 =
and
or
8
P = Fg
3
FLi
A ∆L
a f
kLi
YA
and k =
A
Li
z
za
∆L
∆L
0
0
W = − Fdx = −
f
− kx dx =
YA
Li
z
∆L
xdx = YA
a ∆L f
0
2
2 Li
Take both balls together. Their weight is 3.33 N
and their CG is at their contact point.
P1
∑ Fx = 0 : + P3 − P1 = 0
P2 = 3.33 N
∑ Fy = 0 : + P2 − 3.33 N = 0
∑ τ A = 0: − P3 R + P2 R − 3.33 NaR + R cos 45.0°f
a
f
+ P1 R + 2 R cos 45.0° = 0
Substituting,
a
f a
fa
+ P Ra1 + 2 cos 45.0°f = 0
− P1 R + 3.33 N R − 3.33 N R 1 + cos 45.0°
3.33 N
P3
Fg
f
1
P2
a3.33 Nf cos 45.0° = 2 P cos 45.0°
1
P1 = 1.67 N so P3 = 1.67 N
(b)
FIG. P12.62(a)
Take the upper ball. The lines of action of its weight, of P1 ,
and of the normal force n exerted by the lower ball all go
through its center, so for rotational equilibrium there can be
no frictional force.
∑ Fx = 0 : n cos 45.0°− P1 = 0
1.67 N
= 2.36 N
cos 45.0°
∑ Fy = 0 : n sin 45.0°−1.67 N = 0 gives the same result
n=
1.67 N
n cos 45.0°
n sin 45.0°
FIG. P12.62(b)
P1
Chapter 12
P12.63
∑ Fy = 0 :
375
+380 N − Fg + 320 N = 0
Fg = 700 N
Take torques about her feet:
a
∑ τ = 0:
f a
f a
f
−380 N 2.00 m + 700 N x + 320 N 0 = 0
x = 1.09 m
P12.64
FIG. P12.63
The tension in this cable is not uniform, so this becomes a fairly difficult problem.
dL
F
=
L YA
At any point in the cable, F is the weight of cable below that point. Thus, F = µgy where µ is the mass
per unit length of the cable.
Then, ∆y =
z FGH
Li
0
IJ
K
z
L
2
µg i
1 µgLi
dL
dy =
ydy =
2 YA
L
YA 0
a fa fa f
e
je
j
a10.0 − 1.00f m s =
F ∆v I
F = mG J = b1.00 kg g
H ∆t K
0.002 s
2
∆y =
P12.65
(a)
(b)
(c)
2. 40 9.80 500
1
= 0.049 0 m = 4.90 cm
2 2.00 × 10 11 3.00 × 10 −4
stress =
4 500 N
4 500 N
F
=
= 4.50 × 10 6 N m 2
0.010 m 0.100 m
A
a
fa
f
Yes . This is more than sufficient to break the board.
376
P12.66
Static Equilibrium and Elasticity
The CG lies above the center of the bottom. Consider a disk of water at height y above the bottom.
Its radius is
fFGH 30.0y cm IJK = 25.0 cm + 3y
yI
yI
yI
F
F
F
Its area is π G 25.0 cm + J . Its volume is π G 25.0 cm + J dy and its mass is πρ G 25.0 cm + J
H
K3
H
K3
H
3K
a
25.0 cm + 35.0 − 25.0 cm
2
2
whole mass of the water is
M=
z
z
30 .0 cm
30 .0 cm
y =0
0
dm =
F
GH
I
JK
50.0 y y 2
+
dy
3
9
πρ 625 +
L
50.0 y
y O
+ P
M = πρ M625 y +
6
27 PQ
MN
L
50.0a30.0f
a30.0f OP
+
M = πρ M625a30.0f +
6
27 PQ
MN
M = π e10 kg cm je 27 250 cm j = 85.6 kg
2
3
30.0
0
2
−3
3
3
3
The height of the center of gravity is
yCG =
z
30 .0 cm
y =0
= πρ
ydm
M
z
30 .0 cm
0
F 625y + 50.0 y
GH
3
2
+
I
JK
y 3 dy
9 M
LM
OP
MN
PQ
50.0a30.0 f
a30.0f
πρ L 625a30.0f
=
+
+
M
2
9
36
M MN
π e10 kg cm j
453 750 cm
=
πρ 625 y 2 50.0 y 3 y 4
=
+
+
2
9
36
M
2
−3
30 .0 cm
0
3
3
4
yCG
M
1.43 × 10 3 kg ⋅ cm
=
= 16.7 cm
85.6 kg
4
OP
PQ
2
dy . The
Chapter 12
P12.67
Let θ represent the angle of the wire with the vertical. The radius of
the circle of motion is r = 0.850 m sin θ .
For the mass:
a
f
v2
= mrω 2
r
T sin θ = m 0.850 m sin θ ω 2
a
a f
f
a f a
f
π e3.90 × 10 mj e7.00 × 10 N m je1.00 × 10 j
AY ⋅ astrainf
ω=
=
ma0.850 mf
b1.20 kg ga0.850 mf
a
2
P12.68
10
mg
FIG. P12.67
T
Further, = Y ⋅ strain or T = AY ⋅ strain
A
Thus, AY ⋅ strain = m 0.850 m ω 2 , giving
or
θ
r
f
−4
T
θ
∑ Fr = mar = m
377
−3
2
ω = 5.73 rad s .
For the bridge as a whole:
D
B
∑ τ A = n A a0f − a13.3 kNfa100 mf + nE a 200 mf = 0
so
nE =
a13.3 kNfa100 mf =
200 m
A
6.66 kN
E
C
nA
∑ Fy = n A − 13.3 kN + n E = 0 gives
100 m
100 m
13.3 kN
n A = 13.3 kN − n E = 6.66 kN
At Pin A:
∑ Fy = − FAB sin 40.0°+6.66 kN = 0 or
b
6.66 kN
= 10.4 kN compression
sin 40.0°
∑ Fx = FAC − 10.4 kN cos 40.0° = 0 so
FAB =
a
f
a
f
FAB
a
FAC = 10.4 kN cos 40.0° = 7.94 kN tension
At Pin B:
40.0°
g
FAC
f
∑ Fy = a10.4 kN f sin 40.0°− FBC sin 40.0° = 0
a
Thus, FBC = 10. 4 kN tension
a
40.0°
f
b
b
g
= 10.4 kN atensionf
= 7.94 kN atensionf
By symmetry: FDE = FAB = 10.4 kN compression
FEC = FAC
We can check by analyzing Pin C:
∑ Fx = +7.94 kN − 7.94 kN = 0 or 0 = 0
∑ Fy = 2a10.4 kNf sin 40.0°−13.3 kN = 0
which yields 0 = 0 .
40.0°
FBC
FBD = 2 10.4 kN cos 40.0° = 15.9 kN compression
FDC = FBC
FBD
f
∑ Fx = FAB cos 40.0°+ FBC cos 40.0°− FBD = 0
and
nA = 6.66 kN
FAB = 10.4 kN
g
10.4 kN
10.4 kN
40.0°
40.0°
7.94 kN
7.94 kN
13.3 kN
FIG. P12.68
nE
378
P12.69
Static Equilibrium and Elasticity
Member AC is not in pure compression or tension. It
also has shear forces present. It exerts a downward
force S AC and a tension force FAC on Pin A and on
Pin C. Still, this member is in equilibrium.
SAC
25.0 m
FAC
FAC
C
A
′ = 0 ⇒ FAC = FAC
′
∑ Fx = FAC − FAC
∑ τ A = 0: −a14.7 kNfa25.0 mf + S ′AC a50.0 mf = 0
or
14.7 kN
S ′AC = 7.35 kN
∑ Fy = S AC − 14.7 kN + 7.35 kN = 0 ⇒ S AC = 7.35 kN
D
B
Then S AC = S ′AC and we have proved that the loading by the car A
is equivalent to one-half the weight of the car pulling down on n
A
each of pins A and C, so far as the rest of the truss is concerned.
For the Bridge as a whole:
SAC
25.0 m
E
C
75.0 m
nE
25.0 m 14.7 kN
∑ τ A = 0:
a
fa
f a
f
7.35 kN
− 14.7 kN 25.0 m + n E 100 m = 0
n E = 3.67 kN
FAB
30.0°
∑ Fy = n A − 14.7 kN + 3.67 kN = 0
FAC
n A = 11.0 kN
At Pin A:
nA = 11.0 kN
∑ Fy = −7.35 kN + 11.0 kN − FAB sin 30.0° = 0
b
FAB = 7.35 kN compression
FBD
g
30.0°
∑ Fx = FAC − a7.35 kN f cos 30.0° = 0
a
FAC = 6.37 kN tension
At Pin B:
7.35 kN
60.0°
FBC
f
4.24 kN
∑ Fy = −a7.35 kN f sin 30.0°− FBC sin 60.0° = 0
60.0°
a
f
∑ F = a7.35 kNf cos 30.0°+a 4.24 kN f cos 60.0°− F
F = 8.49 kN bcompressiong
FBC = 4.24 kN tension
x
BD
FCD
60.0°
6.37 kN
FCE
=0
7.35 kN
BD
At Pin C:
∑ Fy = a 4.24 kNf sin 60.0°+ FCD sin 60.0°−7.35 kN = 0
a
FCD = 4.24 kN tension
f
30.0°
∑ Fx = −6.37 kN − a4.24 kNf cos 60.0°+a 4.24 kNf cos 60.0°+ FCE = 0
a
f
FCE = 6.37 kN tension
At Pin E:
6.37 kN
3.67 kN
FIG. P12.69
∑ Fy = − FDE sin 30.0°+3.67 kN = 0
b
FDE = 7.35 kN compression
or ∑ Fx = −6.37 kN − FDE cos 30.0° = 0
which gives FDE = 7.35 kN as before.
FDE
g
379
Chapter 12
P12.70
(1)
(2)
ph = Iω
ω
p
p = MvCM
h
vCM
If the ball rolls without slipping, Rω = vCM
So, h =
P12.71
(a)
2
Iω
Iω
I
=
=
=
R
5
p
MvCM MR
FIG. P12.70
If the acceleration is a, we have Px = ma and
Py + n − Fg = 0 . Taking the origin at the center of
H
L
d
gravity, the torque equation gives
a
f
CG
Py L − d + Px h − nd = 0 .
P
h
Solving these equations, we find
(b)
F d − ah I .
L GH
g JK
ah e 2.00 m s ja1.50 mf
=
=
If P = 0 , then d =
(c)
Using the given data, Px = −306 N and Py = 553 N .
Py =
n
Fg
Fgy
FIG. P12.71
2
y
9.80 m s 2
g
e
0.306 m .
j
Thus, P = −306 i + 553 j N .
*P12.72
When the cyclist is on the point of tipping over forward,
the normal force on the rear wheel is zero. Parallel to the
plane we have f1 − mg sin θ = ma . Perpendicular to the
plane, n1 − mg cos θ = 0 . Torque about the center of mass:
af a
f a
mg
f
mg 0 − f1 1.05 m + n1 0.65 m = 0 .
f1
Combining by substitution,
FIG. P12.72
ma = f1 − mg sin θ =
FG
H
n1 0.65 m
0.65 m
− mg sin θ = mg cos θ
− mg sin θ
1.05 m
1.05 m
IJ
K
0.65
− sin 20° = 2.35 m s 2
1.05
When the car is on the point of rolling over, the normal
force on its inside wheels is zero.
a = g cos 20°
*P12.73
∑ Fy = ma y :
n − mg = 0
∑ Fx = ma x :
f=
n1
mg
h
mv 2
R
Take torque about the center of mass: fh − n
Then by substitution
2
mv max
R
h−
mgd
=0
2
f
d
= 0.
2
v max =
mg
d
gdR
2h
FIG. P12.73
A wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover.
380
Static Equilibrium and Elasticity
ANSWERS TO EVEN PROBLEMS
P12.2
Fy + R y − Fg = 0 ; Fx − R x = 0 ;
Fy A cos θ − Fg
FG A IJ cos θ − F A sinθ = 0
H 2K
P12.40
(a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm;
(d) 2.40 mm; (e) 48.0 kN
P12.42
at A: Mg
P12.44
(a) 160 N to the right;
(b) 13.2 N to the right; (c) 292 N up;
(d) 192 N
x
sin β
sin α
; at B: Mg
sin α + β
sin α + β
b
g
b
g
P12.4
see the solution
P12.6
0.750 m
P12.8
a2.54 m, 4.75 mf
P12.10
(a) 9.00 g; (b) 52.5 g; (c) 49.0 g
P12.46
1.46 kN ; 1.33 i + 2.58 j kN
P12.12
(a) 392 N; (b) 339 i + 0 j N
e
P12.48
0.789
P12.14
(a) f =
P12.50
T = 1.68 kN ; R = 2.34 kN; θ = 21.2°
P12.52
(a) see the solution; (b) 60.0°
P12.54
(a) 120 N; (b) 0.300; (c) 103 N at 31.0° above
the horizontal to the right
P12.56
(a), (b) see the solution;
(c) C AB = 732 N ; TAC = 634 N ; C BC = 897 N
P12.58
(a) 9.09 m, 10.9 m ; (b) 10.0 m, 10.9 m ;
(c) 0.114 m s to the right
ng
j
LM m g + m gx OP cot θ ;
N2 L Q
e + j cot θ
= bm + m g g ; (b) µ =
m +m
1
2
m1
2
1
2
m2d
L
1
2
P12.16
see the solution; 0.643 m
P12.18
36.7 N to the left ; 31.2 N to the right
P12.20
(a) 35.5 kN; (b) 11.5 kN to the right;
(c) 4.19 kN down
P12.22
(a) 859 N; (b) 104 kN at 36.9° above the
horizontal to the left
P12.24
3L
4
P12.60
e
a
j
f
a
f
3
Fg
8
P12.62
(a) P1 = 1.67 N ; P2 = 3.33 N ; P3 = 1.67 N ;
(b) 2.36 N
P12.26
(a) see the solution; (b) θ decreases ;
(c) R decreases
P12.64
4.90 cm
P12.28
(a) 73.6 kN; (b) 2.50 mm
P12.66
16.7 cm above the center of the bottom
P12.30
~ 1 cm
P12.68
P12.32
9.85 × 10 −5
P12.34
0.029 3 mm
C AB = 10.4 kN ; TAC = 7.94 kN ;
TBC = 10.4 kN ; C BD = 15.9 kN ;
C DE = 10.4 kN ; TDC = 10. 4 kN ;
TEC = 7.94 kN
P12.36
(a) −0.053 8 m3 ; (b) 1.09 × 10 3 kg m3 ;
(c) Yes, in most practical circumstances
P12.38
(a) 53.1°; (b) 1.04 kN; (c) 0.126 m, 51.2°;
(d) 1.07 kN; (e) 0.129 m, 51.1°; (f) 51.1°
P12.70
2
R
5
P12.72
2.35 m s 2
13
Universal Gravitation
CHAPTER OUTLINE
13.1
13.2
13.3
13.4
13.5
13.6
13.7
Newton’s Law of Universal
Gravitation
Measuring the Gravitational
Constant
Free-Fall Acceleration and
the Gravitational Force
Kepler’s Laws and the
Motion of Planets
The Gravitational Field
Gravitational Potential
Energy
Energy Considerations in
Planetary and Satellite
Motion
ANSWERS TO QUESTIONS
Q13.1
Because g is the same for all objects near the Earth’s surface.
The larger mass needs a larger force to give it just the same
acceleration.
Q13.2
To a good first approximation, your bathroom scale reading is
unaffected because you, the Earth, and the scale are all in free
fall in the Sun’s gravitational field, in orbit around the Sun. To
a precise second approximation, you weigh slightly less at
noon and at midnight than you do at sunrise or sunset. The
Sun’s gravitational field is a little weaker at the center of the
Earth than at the surface subsolar point, and a little weaker still
on the far side of the planet. When the Sun is high in your sky,
its gravity pulls up on you a little more strongly than on the
Earth as a whole. At midnight the Sun pulls down on you a
little less strongly than it does on the Earth below you. So you
can have another doughnut with lunch, and your bedsprings
will still last a little longer.
Q13.3
Kepler’s second law states that the angular momentum of the Earth is constant as the Earth orbits
the sun. Since L = mωr , as the orbital radius decreases from June to December, then the orbital speed
must increase accordingly.
Q13.4
Because both the Earth and Moon are moving in orbit about the Sun. As described by
Fgravitational = ma centripetal , the gravitational force of the Sun merely keeps the Moon (and Earth) in a
nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in its
orbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, at
new moon or full moon.
Q13.5
Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameter
of the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit,
however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!
Q13.6
Kepler’s third law, which applies to all planets, tells us that the period of a planet is proportional to
r 3 2 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. The
Sun’s gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiences
much smaller centripetal acceleration than Earth and has a correspondingly longer period.
381
382
Q13.7
Universal Gravitation
Ten terms are needed in the potential energy:
U = U 12 + U 13 + U 14 + U 15 + U 23 + U 24 + U 25 + U 34 + U 35 + U 45 .
With N particles, you need
N
∑ ai − 1f =
i =1
Q13.8
N2 − N
terms.
2
No, the escape speed does not depend on the mass of the rocket. If a rocket is launched at escape
speed, then the total energy of the rocket-Earth system will be zero. When the separation distance
GM E m
1
= 0 , the mass
becomes infinite U = 0 the rocket will stop K = 0 . In the expression mv 2 −
2
r
m of the rocket divides out.
a
f
a
f
Q13.9
It takes 100 times more energy for the 10 5 kg spacecraft to reach the moon than the 10 3 kg
spacecraft. Ideally, each spacecraft can reach the moon with zero velocity, so the only term that need
be analyzed is the change in gravitational potential energy. U is proportional to the mass of the
spacecraft.
Q13.10
The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factor
of 5. The energy required—and fuel—would be proportional to v 2 , or 25 times more fuel is required
to leave the Earth versus leaving the Moon.
Q13.11
The satellites used for TV broadcast are in geosynchronous orbits. The centers of their orbits are the
center of the Earth, and their orbital planes are the Earth’s equatorial plane extended. This is the
plane of the celestial equator. The communication satellites are so far away that they appear quite
close to the celestial equator, from any location on the Earth’s surface.
Q13.12
For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be located
at the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it must
be over the southern hemisphere for the other half. We could share with Easter Island a satellite that
would look straight down on Arizona each morning and vertically down on Easter Island each
evening.
Q13.13
The absolute value of the gravitational potential energy of the Earth-Moon system is twice the
kinetic energy of the moon relative to the Earth.
Q13.14
In a circular orbit each increment of displacement is perpendicular to the force applied. The dot
product of force and displacement is zero. The work done by the gravitational force on a planet in an
elliptical orbit speeds up the planet at closest approach, but negative work is done by gravity and
the planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in one
complete orbit is zero.
Q13.15
Every point q on the sphere that does not lie
along the axis connecting the center of the
sphere and the particle will have companion
point q’ for which the components of the
gravitational force perpendicular to the axis
will cancel. Point q’ can be found by rotating
the sphere through 180° about the axis. The
forces will not necessarily cancel if the mass is
not uniformly distributed, unless the center of
mass of the non-uniform sphere still lies along
the axis.
q
Fpq
Fpq
q’ (behind the sphere)
FIG. Q13.15
p
Chapter 13
Q13.16
Speed is maximum at closest approach. Speed is minimum at farthest distance.
Q13.17
Set the universal description of the gravitational force, Fg =
Fg = ma gravitational , where M X and R X
383
GM X m
, equal to the local description,
R X2
are the mass and radius of planet X, respectively, and m is the
mass of a “test particle.” Divide both sides by m.
Q13.18
The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as one
interpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pull
in different outward directions on the extra particle.
Q13.19
Cavendish determined G. Then from g =
Q13.20
The gravitational force is conservative. An encounter with a stationary mass cannot permanently
speed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planet
will gain kinetic energy as the planet’s gravity does net positive work on it.
Q13.21
Method one: Take measurements from an old kinescope of Apollo astronauts on the moon. From the
motion of a freely falling object or from the period of a swinging pendulum you can find the
acceleration of gravity on the moon’s surface and calculate its mass. Method two: One could
determine the approximate mass of the moon using an object hanging from an extremely sensitive
balance, with knowledge of the position and distance of the moon and the radius of the Earth. First
weigh the object when the moon is directly overhead. Then weigh of the object when the moon is
just rising or setting. The slight difference between the measured weights reveals the cause of tides
in the Earth’s oceans, which is a difference in the strength of the moon’s gravity between different
points on the Earth. Method three: Much more precisely, from the motion of a spacecraft in orbit
around the moon, its mass can be determined from Kepler’s third law.
Q13.22
The spacecraft did not have enough fuel to stop dead in its high-speed course for the Moon.
GM
, one may determine the mass of the Earth.
R2
SOLUTIONS TO PROBLEMS
Section 13.1
P13.1
Newton’s Law of Universal Gravitation
For two 70-kg persons, modeled as spheres,
Fg =
P13.2
F = m1 g =
g=
Gm 2
r
2
Gm1 m 2
r
2
e6.67 × 10
=
−11
jb
gb
N ⋅ m 2 kg 2 70 kg 70 kg
a2 mf
2
g
~ 10 −7 N .
Gm1 m 2
=
r2
e6.67 × 10
−11
je
a100 mf
N ⋅ m 2 kg 2 4.00 × 10 4 × 10 3 kg
2
j=
2.67 × 10 −7 m s 2
384
P13.3
Universal Gravitation
(a)
At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objects
are oppositely directed,
(b)
Gm1 m 2
and from
Fg =
we have
∑F =
r2
b
gb
G 50.0 kg 500 kg − 200 kg
a0.200 mf
2
g=
2.50 × 10 −5 N toward the 500-kg object.
At a point between the two objects at a distance d from the 500-kg objects, the net force on
the 50.0-kg object will be zero when
b
gb g = Gb50.0 kggb500 kg g
d
a0.400 m − df
G 50.0 kg 200 kg
2
d = 0.245 m
or
P13.4
2
m1 + m 2 = 5.00 kg
F =G
m 2 = 5.00 kg − m1
m1 m 2
r
2
b5.00 kg gm
g
bm − 3.00 kg gbm
e
⇒ 1.00 × 10 −8 N = 6.67 × 10 −11 N ⋅ m 2 kg 2
2
1 − m1 =
b
e1.00 × 10
−8
6.67 × 10
je
N 0.040 0 m 2
−11
2
N ⋅ m kg
2
j m ba50..00200kgm−f m g
1
1
2
j = 6.00 kg
2
Thus, m12 − 5.00 kg m1 + 6.00 kg = 0
or
1
1
g
− 2.00 kg = 0
giving m1 = 3.00 kg, so m 2 = 2.00 kg . The answer m1 = 2.00 kg and m 2 = 3.00 kg is physically
equivalent.
P13.5
The force exerted on the 4.00-kg mass by the 2.00-kg mass is
directed upward and given by
F24 = G
m4m2 j = 6.67 × 10 −11 N ⋅ m 2 kg 2
2
r24
e
j b4.00a3kg.00gbm2.00f kg g j
2
= 5.93 × 10 −11 j N
The force exerted on the 4.00-kg mass by the 6.00-kg mass is
directed to the left
F64 = G
m 4 m6
2
r64
e− ij = e−6.67 × 10
−11
N ⋅ m 2 kg 2
j b4.00a4kg.00gbm6.00f kg g i
2
FIG. P13.5
= −10.0 × 10 −11 iN
Therefore, the resultant force on the 4.00-kg mass is F4 = F24 + F64 =
e−10.0i + 5.93 jj × 10
−11
N .
Chapter 13
P13.6
(a)
385
The Sun-Earth distance is 1.496 × 10 11 m and the Earth-Moon distance is 3.84 × 10 8 m , so the
distance from the Sun to the Moon during a solar eclipse is
1.496 × 10 11 m − 3.84 × 10 8 m = 1.492 × 10 11 m
M S = 1.99 × 10 30 kg
The mass of the Sun, Earth, and Moon are
M E = 5.98 × 10 24 kg
M M = 7.36 × 10 22 kg
and
We have FSM =
(b)
(c)
FEM =
FSE
e6.67 × 10
e6.67 × 10
=
Gm1 m 2
r2
−11
e6.67 × 10 je1.99 × 10 je7.36 × 10 j =
=
e1.492 × 10 j
−11
30
22
11 2
je
je
j=
je
j=
N ⋅ m 2 kg 2 5.98 × 10 24 7.36 × 10 22
e3.84 × 10 j
8 2
−11
je
N ⋅ m 2 kg 2 1.99 × 10 30 5.98 × 10 24
e1.496 × 10 j
11 2
4.39 × 10 20 N
1.99 × 10 20 N
3.55 × 10 22 N
Note that the force exerted by the Sun on the Moon is much stronger than the force of the
Earth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. The
Moon’s path is everywhere concave toward the Sun. Only by subtracting out the solar
orbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass of
this system.
Section 13.2
P13.7
Measuring the Gravitational Constant
b
ge
j
1.50 kg 15.0 × 10 −3 kg
GMm
−11
2
2
F=
= 6.67 × 10
N ⋅ m kg
= 7.41 × 10 −10 N
2
−2
r2
4.50 × 10 m
e
j
e
j
386
P13.8
Universal Gravitation
Let θ represent the angle each cable makes with the vertical, L the
cable length, x the distance each ball scrunches in, and d = 1 m the
original distance between them. Then r = d − 2 x is the separation of
the balls. We have
∑ Fy = 0 :
T cos θ − mg = 0
∑ Fx = 0 :
T sin θ −
Then
tan θ =
Gmm
=0
r2
FIG. P13.8
Gmm
r 2 mg
x
2
L −x
2
=
a
Gm
g d − 2x
f
a
x d − 2x
2
f
2
=
Gm 2
L − x2 .
g
Gm
is numerically small. There are two possibilities: either x is small or else d − 2 x is
g
The factor
small.
Possibility one: We can ignore x in comparison to d and L, obtaining
e6.67 × 10
xa1 mf =
2
−11
jb
N ⋅ m 2 kg 2 100 kg
e9.8 m s j
2
e
g 45 m
x = 3.06 × 10 −8 m.
j
The separation distance is r = 1 m − 2 3.06 × 10 −8 m = 1.000 m − 61.3 nm .
Possibility two: If d − 2 x is small, x ≈ 0.5 m and the equation becomes
a0.5 mfr = e6.67 × 10 b9N.8 ⋅Nm kgkgg jb100 kgg a45 mf − a0.5 mf
−11
2
2
2
2
2
r = 2.74 × 10 −4 m .
For this answer to apply, the spheres would have to be compressed to a density like that of the
nucleus of atom.
Section 13.3
P13.9
a=
Free-Fall Acceleration and the Gravitational Force
MG
b 4R g
E
2
=
9.80 m s 2
= 0.613 m s 2
16.0
toward the Earth.
e j = 4 πGρR
4 πR 3
GM Gρ 3
g= 2 =
R
R2
P13.10
If
gM 1
= =
gE 6
then
g
ρM
= M
gE
ρE
3
4πGρ M R M
3
4πGρ E RE
3
F I F R I = FG 1 IJ a4f =
GH JK GH R JK H 6 K
E
M
2
.
3
Chapter 13
P13.11
(a)
At the zero-total field point,
so
(b)
GmM E
rE2
387
GmM M
=
rM2
MM
r
7.36 × 10 22
= rE
= E
24
ME
9
.01
5.98 × 10
r
rE + rM = 3.84 × 10 8 m = rE + E
9.01
3.84 × 10 8 m
rE =
= 3.46 × 10 8 m
1.11
rM = rE
At this distance the acceleration due to the Earth’s gravity is
gE =
GM E
rE2
e6.67 × 10
=
−11
je
mj
N ⋅ m 2 kg 2 5.98 × 10 24 kg
e3.46 × 10
8
j
2
g E = 3.34 × 10 −3 m s 2 directed toward the Earth
Section 13.4
P13.12
Kepler’s Laws and the Motion of Planets
b
g
3
2πr 2π 384 400 × 10 m
=
= 1.02 × 10 3 m s .
T
27.3 × 86 400 s
(a)
v=
(b)
In one second, the Moon falls a distance
b
g
e
e
j a f
j
2
3
1
1 v 2 2 1 1.02 × 10
x = at 2 =
t =
× 1.00
2
2 r
2 3.844 × 10 8
2
= 1.35 × 10 −3 m = 1.35 mm .
The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straight-line
path.
P13.13
Applying Newton’s 2nd Law,
GMM
a 2r f
2
=
∑ F = ma yields Fg = ma c for each star:
Mv 2
r
M=
or
4v 2 r
.
G
We can write r in terms of the period, T, by considering the time and
distance of one complete cycle. The distance traveled in one orbit is the
circumference of the stars’ common orbit, so 2πr = vT . Therefore
M=
e
FG IJ
H K
4v 2 r 4v 2 vT
=
G
G 2π
ja
3
fb
FIG. P13.13
g
3
2 v 3 T 2 220 × 10 m s 14.4 d 86 400 s d
=
= 1.26 × 10 32 kg = 63.3 solar masses
so, M =
−11
2
2
πG
π 6.67 × 10
N ⋅ m kg
e
j
388
P13.14
Universal Gravitation
Since speed is constant, the distance traveled between t1 and t 2 is equal to the distance traveled
between t 3 and t 4 . The area of a triangle is equal to one-half its (base) width across one side times its
(height) dimension perpendicular to that side.
So
b
g
b
1
1
bv t 2 − t1 = bv t 4 − t 3
2
2
g
states that the particle’s radius vector sweeps out equal areas in equal times.
P13.15
T2 =
M=
4π 2 a 3
GM
4π 2 a 3
GT 2
(Kepler’s third law with m << M )
=
e
j
4π 2 4.22 × 10 8 m
e
jb
3
6.67 × 10 −11 N ⋅ m 2 kg 2 1.77 × 86 400 s
g
2
= 1.90 × 10 27 kg
(Approximately 316 Earth masses)
P13.16
By conservation of angular momentum for the satellite,
vp
rp v p = ra v a
va
=
ra 2 289 km + 6.37 × 10 3 km 8 659 km
=
=
= 1.27 .
6 829 km
rp
459 km + 6.37 × 10 3 km
We do not need to know the period.
P13.17
By Kepler’s Third Law, T 2 = ka 3
(a = semi-major axis)
For any object orbiting the Sun, with T in years and a in A.U.,
k = 1.00 . Therefore, for Comet Halley
a75.6f = a1.00fFGH 0.5702 + y IJK
2
3
The farthest distance the comet gets from the Sun is
a f
y = 2 75.6
P13.18
∑ F = ma :
23
FIG. P13.17
− 0.570 = 35.2 A. U. (out around the orbit of Pluto)
Gm planet M star
r2
=
m planet v 2
r
GM star
= v 2 = r 2ω 2
r
GM star = r 3ω 3 = rx3ω 2x = ry3ω 2y
ωy =ωx
Fr I
GH r JK
x
y
3 2
ωy =
F 90.0° I 3
GH 5.00 yr JK
3 2
=
468°
5.00 yr
So planet Y has turned through 1.30 revolutions .
FIG. P13.18
Chapter 13
P13.19
GM J
dR + di
=
2
d
4π 2 R J + d
T
J
d
−11
i
2
GM J T 2 = 4π 2 R J + d
e6.67 × 10
389
i
3
jb
je
N ⋅ m 2 kg 2 1.90 × 10 27 kg 9.84 × 3 600
g
2
e
= 4π 2 6.99 × 10 7 + d
j
3
d = 8.92 × 10 7 m = 89 200 km above the planet
P13.20
The gravitational force on a small parcel of material at the star’s equator supplies the necessary
centripetal force:
GM s m
Rs2
so
ω=
=
mv 2
= mRsω 2
Rs
GM s
Rs3
=
e6.67 × 10
−11
j e
mj
N ⋅ m 2 kg 2 2 1.99 × 10 30 kg
e10.0 × 10
3
j
3
ω = 1.63 × 10 4 rad s
*P13.21
The speed of a planet in a circular orbit is given by
GM sun m
∑ F = ma :
r2
=
mv 2
r
GM sun
.
r
v=
For Mercury the speed is
vM =
e6.67 × 10 je1.99 × 10 j m
e5.79 × 10 j s
and for Pluto,
vP =
e6.67 × 10 je1.99 × 10 j m
e5.91 × 10 j s
−11
30
10
2
−11
30
12
2
2
= 4.79 × 10 4 m s
2
= 4.74 × 10 3 m s .
With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original
distances rP and rM perpendicular to their lines of motion, they will be equally far from the Sun
after time t where
2 2
rP2 + v P2 t 2 = rM2 + v M
t
e
j
2
rP2 − rM2 = v M
− v P2 t 2
t=
e5.91 × 10
e4.79 × 10
4
j − e5.79 × 10
m sj − e 4.74 × 10
12
m
2
2
10
3
j
m
2
ms
j
2
=
3.49 × 10 25 m 2
= 1.24 × 10 8 s = 393 yr .
2.27 × 10 9 m 2 s 2
390
*P13.22
Universal Gravitation
GM s m
∑ F = ma :
For the Earth,
r2
Also the angular momentum
We eliminate
FG LT IJ
H 2πm K
FG IJ
H K
mv 2 m 2πr
=
r
r T
2
.
GM s T 2 = 4π 2 r 3 .
2πr
L = mvr = m
r is a constant for the Earth.
T
LT
r=
between the equations:
2πm
Then
GM s T 2 = 4π 2
=
32
GM sT 1 2 = 4π 2
FG L IJ
H 2πm K
3 2
.
Now the rate of change is described by
GM s
FG 1 T
H2
∆T ≈ − ∆t
−1 2
dM F T I ∆T
dT
IJ
IJ FG
=−
G2 J ≈
K H
K
dt
dt H M K T
FG 2 T IJ = −5 000 yrF 3.16 × 10 s I e−3.64 × 10 kg sjF 2 1 yr
GH 1 yr JK
GH 1.991 × 10
H MK
dM s 1 2
dT
+G 1
=0
T
dt
dt
dM s
dt
s
s
7
9
s
30
I
J
kg K
∆T = 1.82 × 10 −2 s
Section 13.5
P13.23
g=
so
g=
The Gravitational Field
Gm Gm Gm
i + 2 j + 2 cos 45.0° i + sin 45.0 j
2l
l2
l
e
g=
Gm
l2
FG
H
FG
H
j
GM
1
1+
2
l
2 2
2+
y
m
IJ ei + jj or
K
l
m
l
IJ
K
1
toward the opposite corner
2
m
O
FIG. P13.23
P13.24
(a)
(b)
e
j e
je
j
6.67 × 10 −11 N ⋅ m 2 kg 2 100 1.99 × 10 30 kg 10 3 kg
GMm
F=
=
= 1.31 × 10 17 N
2
4
r2
1.00 × 10 m + 50.0 m
e
∆F =
GMm GMm
− 2
2
rfront
rback
∆g =
2
2
∆F GM rback − rfront
=
2
2
m
rfront
rback
e
j
j
e6.67 × 10 j 100e1.99 × 10
∆g =
e1.00 × 10
−11
∆g = 2.62 × 10 12 N kg
j LNMe1.01 × 10 mj − e1.00 × 10 mj OPQ
mj e1.01 × 10 mj
30
4
2
4
2
4
4
2
2
FIG. P13.24
x
Chapter 13
P13.25
g1 = g 2 =
MG
r 2 + a2
g1y = − g 2y
g y = g1y + g 2 y
r
cos θ =
2
a + r2
g 1 x = g 2 x = g 2 cos θ
e
e j
j
391
12
g = 2 g 2 x − i
or
Section 13.6
P13.26
g=
2 MGr
e
2
r + a2
j
3 2
toward the center of mass
FIG. P13.25
Gravitational Potential Energy
(a)
e
jb
je
f
g
6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg
GM E m
=−
= −4.77 × 10 9 J .
U=−
6
r
6.37 + 2.00 × 10 m
a
(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward on
the satellite and upward on the planet.
F=
P13.27
U = −G
so that
Mm
r
GM E m
r
2
and
=
e6.67 × 10
−11
jb
je
N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg
e8.37 × 10 mj
6
g=
2
569 N
GM E
RE2
FG 1 − 1 IJ = 2 mgR
H 3R R K 3
2
∆U = b1 000 kg ge9.80 m s je6.37 × 10 mj =
3
∆U = −GMm
E
E
E
2
P13.28
g=
6
4.17 × 10 10 J .
The height attained is not small compared to the radius of the Earth, so U = mgy does not apply;
GM 1 M 2
U=−
does. From launch to apogee at height h,
r
GM E M p
GM E M p
1
+0=0−
K i + Ui + ∆Emch = K f + U f :
M p vi2 −
2
RE
RE + h
1
10.0 × 10 3 m s
2
e
j e
2
e
= − 6.67 × 10 −11 N ⋅ m 2 kg 2
e5.00 × 10
7
.98 × 10
jFGH 56.37
× 10
5.98 × 10 kg I
jFGH 6.37
× 10 m + h JK
− 6.67 × 10 −11 N ⋅ m 2 kg 2
j e
h = 2.52 × 10 7 m
3.99 × 10 14 m3 s 2
1. 26 × 10 7 m 2 s 2
6
I
J
mK
kg
24
6
j
m 2 s 2 − 6.26 × 10 7 m 2 s 2 =
6.37 × 10 6 m + h =
24
−3.99 × 10 14 m3 s 2
6.37 × 10 6 m + h
= 3.16 × 10 7 m
392
P13.29
P13.30
Universal Gravitation
MS
(a)
ρ=
(b)
g=
(c)
Ug = −
GM S
W = − ∆U = −
(a)
j
mj
e
4π 6.37 × 10
e6.67 × 10
3
6
−11
= 1.84 × 10 9 kg m 3
je
mj
N ⋅ m 2 kg 2 1.99 × 10 30 kg
e6.37 × 10
6
2
e
j=
3.27 × 10 6 m s 2
jb
je
FG −Gm m
H r
1
−11
2
−0
IJ
K
je
je
N ⋅ m 2 kg 2 7.36 × 10 22 kg 1.00 × 10 3 kg
6
1.74 × 10 m
FG
H
U Tot = U 12 + U 13 + U 23 = 3U 12 = 3 −
(b)
(a)
e
g
6.67 × 10 −11 N ⋅ m 2 kg 2 1.99 × 10 30 kg 1.00 kg
GM S m
=−
= −2.08 × 10 13 J
rE
6.37 × 10 6 m
U Tot = −
*P13.32
=
rE2
e+6.67 × 10
W=
P13.31
=
2
4
3 πrE
3 1.99 × 10 30 kg
e
Gm 1 m 2
r12
je
2.82 × 10 9 J
IJ
K
3 6.67 × 10 −11 N ⋅ m 2 kg 2 5.00 × 10 −3 kg
0.300 m
j=
j
2
= −1.67 × 10 −14 J
At the center of the equilateral triangle
Energy conservation of the object-Earth system from release to radius r:
eK + U j
g
altitude h
e
= K +Ug
j
radius r
GM E m
GM E m 1
= mv 2 −
0−
RE + h 2
r
F
GH
v = 2GM E
f
dt = −
i
∆t =
i
RE + h
=−
dr
dt
dr i dr
=
. The time of fall is
v f v
2GM E
RE
∆t =
12
E
z z z
z FGH FGH
z LMN
f
(b)
FG 1 − 1 IJ I
H r R + h K JK
6.87 × 10 6 m
1
1
−
r RE + h
IJ I
K JK
−1 2
dr
2 × 6.67 × 10 −11 × 5.98 × 10 24
6.37 ×10 6 m
FG 1 − 1 IJ OP
H r 6.87 × 10 m K Q
6
−1 2
dr
We can enter this expression directly into a mathematical calculation program.
r
Alternatively, to save typing we can change variables to u = 6 . Then
10
−1 2
6. 87
−
1
2
10 6 6.87 1
1
1
1
−8
6
10
3
541
10
∆t = 7.977 × 10 14
−
du
=
×
−
.
−
6
6
1
2
u 6.87
6.87 × 10
6. 37 10 u
6. 37
10 6
e
j
z FGH
IJ
K
e j
z FGH
IJ
K
−1 2
du
A mathematics program returns the value 9.596 for this integral, giving for the time of
fall ∆t = 3.541 × 10 −8 × 10 9 × 9.596 = 339.8 = 340 s .
Chapter 13
Section 13.7
P13.33
Energy Considerations in Planetary and Satellite Motion
F
GH
I
JK
FG
H
1 2
1
1
= v 2f
vi + GM E 0 −
2
2
RE
or
v 2f = v12 −
and
vf
vf
P13.34
IJ
K
1
1 1
1
−
= mv 2f
mvi2 + GM E m
2
2
r f ri
2GM E
RE
F 2GM IJ
= Gv −
H R K
= Le 2.00 × 10 j − 1.25 × 10 O
MN
PQ
2
1
12
E
E
4 2
8
12
= 1.66 × 10 4 m s
2 M SunG
= 42.1 km s
RE⋅Sun
(a)
v solar escape =
(b)
Let r = RE⋅S x represent variable distance from the Sun, with x in astronomical units.
v=
2 M SunG 42.1
=
RE⋅S x
x
If v =
125 000 km
, then x = 1.47 A.U. = 2.20 × 10 11 m
3 600 s
(at or beyond the orbit of Mars, 125 000 km/h is sufficient for escape).
P13.35
To obtain the orbital velocity, we use
∑F =
or
v=
or
vi2
GM E
=
RE + h
RE + h
b
g
MG
R
1
mMG
2
=
mv esc
2
R
2 MG
=
v esc =
R
We can obtain the escape velocity from
P13.36
mMG mv 2
=
R
R2
2v
2
FG
H
IJ
K
1
1 GM E m
1
=
K i = mvi2 =
+
2
2 RE h
2
LM e6.67 × 10 N ⋅ m kg je5.98 × 10 kg jb500 kgg OP
MN
PQ = 1.45 × 10
e6.37 × 10 mj + e0.500 × 10 mj
−11
2
24
2
6
6
10
J
The change in gravitational potential energy of the satellite-Earth system is
F
I
GH
JK
= e6.67 × 10
N ⋅ m kg je5.98 × 10 kg jb500 kg ge −1.14 × 10
1
1
= mv = b500 kg ge 2.00 × 10 m sj = 1.00 × 10 J .
2
2
∆U =
GM E m GM E m
1
1
−
= GM E m
−
Ri
Rf
Ri R f
−11
Also,
Kf
2
2
2
f
24
3
2
9
The energy transformed due to friction is
a
f
∆Eint = K i − K f − ∆U = 14.5 − 1.00 + 2.27 × 10 9 J = 1.58 × 10 10 J .
−8
j
m −1 = −2.27 × 10 9 J
393
394
P13.37
Universal Gravitation
Fc = FG gives
mv 2 GmM E
=
r
r2
which reduces to v =
2πr
r
= 2πr
.
v
GM E
and period =
(a)
GM E
r
r = RE + 200 km = 6 370 km + 200 km = 6 570 km
Thus,
e
6
period = 2π 6.57 × 10 m
j e6.67 × 10
e6.57 × 10 mj
N ⋅ m kg je5.98 × 10
6
−11
2
2
3
T = 5.30 × 10 s = 88.3 min = 1.47 h
GM E
=
r
e6.67 × 10
−11
je
N ⋅ m 2 kg 2 5.98 × 10 24 kg
(b)
v=
(c)
K f + U f = K i + Ui + energy input, gives
input =
e6.57 × 10 mj
6
F
GH
I F
JK GH
−GM E m
−GM E m
1
1
−
mv 2f − mvi2 +
2
2
rf
ri
j=
7.79 km s
IJ
K
(1)
ri = RE = 6.37 × 10 6 m
vi =
2πRE
= 4.63 × 10 2 m s
86 400 s
Substituting the appropriate values into (1) yields the
minimum energy input = 6.43 × 10 9 J
24
kg
j
Chapter 13
P13.38
The gravitational force supplies the needed centripetal acceleration.
GM E m
Thus,
=
b R + hg b
E
2
b
mv 2
RE + h
2πr 2π RE + h
=
T=
GM E
v
(a)
v2 =
or
g
g
GM E
RE + h
bR + hg
E
T = 2π
3
GM E
b R + hg
E
GM E
RE + h
(b)
v=
(c)
Minimum energy input is
e
j e
j
∆Emin = K f + U gf − K i − U gi .
It is simplest to
launch the satellite from a location on the equator, and launch it toward the east.
1
mvi2
2
GM E m
.
U gi = −
RE
Ki =
This choice has the object starting with energy
vi =
Thus,
∆Emin =
Etot = −
∆E =
FG
H
GMm
2r
F
GH
and
LM
MN b
IJ
K
GM E
GM E m 1
4π 2 RE2
1
m
−
− m
2
RE + h
RE + h 2
86 400 s
∆Emin = GM E m
or
P13.39
2πRE
2πRE
=
1.00 day 86 400 s
with
I e
JK
LM R + 2 h OP − 2π R m
MN 2 R bR + hg PQ b86 400 sg
je
2
E
E
E
j
2
E
OP GM m
+
g PQ R
E
2
E
2
F
GH
6.67 × 10 −11 5.98 × 10 24 10 3 kg
1
1
GMm 1 1
−
=
−
ri r f
2
2
10 3 m 6 370 + 100 6 370 + 200
I
JK
∆E = 4.69 × 10 8 J = 469 MJ
gE =
P13.40
Gm E
GmU
gU =
rE2
FG IJ
H K
(a)
gU mU rE2
1
=
= 14.0
3.70
g E m E rU2
(b)
v esc ,E =
2
rU2
a fe
gU = 1.02 9.80 m s 2 = 10.0 m s 2
2GmU
:
rU
v esc ,E
2Gm E
; v esc ,U =
rE
v esc ,U
=
mU rE
14.0
=
= 1.95
3.70
m E rU
For the Earth, from the text’s table of escape speeds, v esc ,E = 11.2 km s
a fb
j
= 1.02
g
∴ v esc ,U = 1.95 11.2 km s = 21.8 km s
395
396
P13.41
Universal Gravitation
The rocket is in a potential well at Ganymede’s surface with energy
e
6.67 × 10 −11 N ⋅ m 2 m 2 1.495 × 10 23 kg
Gm 1 m 2
=−
U1 = −
r
kg 2 2.64 × 10 6 m
e
2
6
U1 = −3.78 × 10 m 2 m s
j
j
2
The potential well from Jupiter at the distance of Ganymede is
U2 = −
e
6.67 × 10 −11 N ⋅ m 2 m 2 1.90 × 10 27 kg
Gm1 m 2
=−
r
kg 2 1.071 × 10 9 m
e
2
8
U 2 = −1.18 × 10 m 2 m s
To escape from both requires
j
j
2
1
2
m 2 v esc
= + 3.78 × 10 6 + 1.18 × 10 8 m 2 m 2 s 2
2
e
j
v esc = 2 × 1.22 × 10 8 m 2 s 2 = 15.6 km s
P13.42
We interpret “lunar escape speed” to be the escape speed from the surface of a stationary moon
alone in the Universe:
GM m m
1
2
mv esc
=
2
Rm
v esc =
2GM m
Rm
2GM m
Rm
v launch = 2
Now for the flight from moon to Earth
aK + U f = a K + U f
i
f
GmM m GmM E 1
GmM m GmM E
1
2
2
−
−
= mv impact
−
−
mv launch
2
2
Rm
rel
rm 2
RE
4GM m GM m GM E 1 2
GM m GM E
−
−
= v impact −
−
2
Rm
Rm
rel
rm 2
RE
L F 3 M + M + M − M I OP
= M 2G G
MN H R r R r JK PQ
L F 3 × 7.36 × 10 kg + 7.36 × 10
= M 2G G
MN H 1.74 × 10 m 3.84 × 10
12
v impact
m
m
m
E
m2
E
E
el
22
6
22
kg
8
m
+
5.98 × 10 24 kg
6.37 × 10 6 m
e
j
−
3.84 × 10 8
= 2G 1.27 × 10 17 + 1.92 × 10 14 + 9.39 × 10 17 − 1.56 × 10 16 kg m
e
j
= 2 6.67 × 10 −11 N ⋅ m 2 kg 2 10.5 × 10 17 kg m
12
I OP
m JK PQ
5.98 × 10 24 kg
12
= 11.8 km s
12
Chapter 13
*P13.43
(a)
397
Energy conservation for the object-Earth system from firing to apex:
eK + U j = eK + U j
g
g
i
f
GmM E
GmM E
1
=0−
mvi2 −
RE
RE + h
2
where
GmM E
1
2
=
. Then
mv esc
RE
2
RE
1 2 1 2
1 2
vi − v esc = − v esc
2
2
2
RE + h
2
− vi2 =
v esc
1
2
v esc
h=
h=
a f
a11.2f − a8.76f
6.37 × 10 6 m 8.76
2
− vi2
=
2
v esc
RE
RE + h
RE + h
2
v esc
RE
2
v esc
RE
2
− vi2
v esc
− RE =
2
2
v esc
RE − v esc
RE + vi2 RE
2
− vi2
v esc
RE vi2
2
− vi2
v esc
= 1.00 × 10 7 m
(b)
h=
(c)
The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it is
described by the same energy equation
2
2
FG
H
2
vi2 = v esc
1−
FG
H
IJ
K
IJ e
K
RE
h
2
= v esc
= 11.2 × 10 3 m s
RE + h
RE + h
j FGH 6.37 × 102.51m×+102.51m× 10 m IJK
7
2
6
7
= 1.00 × 10 8 m 2 s 2
vi = 1.00 × 10 4 m s
(d)
With vi << v esc , h ≈
RE vi2
2
v esc
b ga f
0 2 = vi2 + 2 − g h − 0 .
P13.44
=
RE vi2 RE
GM E
vi2
=
. But g =
,
so
h
, in agreement with
2GM E
2g
RE2
For a satellite in an orbit of radius r around the Earth, the total energy of the satellite-Earth system is
GM E
E=−
. Thus, in changing from a circular orbit of radius r = 2 RE to one of radius r = 3 RE , the
2r
required work is
W = ∆E = −
LM
N
OP
Q
GM E m GM E m
GM E m
1
1
+
= GM E m
−
=
.
2r f
2ri
4 RE 6 R E
12 RE
398
*P13.45
Universal Gravitation
(a)
The major axis of the orbit is
Further, in Figure 13.5,
Then
(b)
2 a = 50.5 AU
so
a + c = 50 AU
so
c 24.75
= 0.980
e= =
a 25.25
a = 25.25 AU
c = 24.75 AU
In T 2 = K s a 3 for objects in solar orbit, the Earth gives us
b1 yrg
2
a
= K s 1 AU
f
b1 yrg
=
a1 AU f
2
3
Ks
3
b1 yrg a25.25 AU f T = 127 yr
T =
a1 AU f
N ⋅ m kg je1.991 × 10 kg je1.2 × 10 kg j
= −2.13 × 10
50e1.496 × 10 mj
2
Then
*P13.46
e
(c)
6.67 × 10 −11
GMm
U=−
=−
r
(a)
For the satellite
v0 =
(b)
3
2
FG GM IJ
H r K
E
∑ F = ma
2
3
2
30
17
11
GmM E
r2
=
J
mv 02
r
12
Conservation of momentum in the forward direction for the exploding satellite:
c∑ mvh = c∑ mvh
i
f
5mv 0 = 4mvi + m0
vi =
(c)
10
FG
H
5
5 GM E
v0 =
r
4
4
IJ
K
12
With velocity perpendicular to radius, the orbiting fragment is at perigee. Its apogee
distance and speed are related to r and vi by 4mrvi = 4mr f v f and
GM E 4m 1
vr
GM E 4m
1
= 4mv 2f −
. Substituting v f = i we have
4mvi2 −
2
2
r
rf
rf
1 2 GM E 1 vi2 r 2 GM E
25 GM E
gives
=
−
. Further, substituting vi2 =
vi −
2
2
2 rf
16 r
r
rf
25 GM E GM E 25 GM E r GM E
−
=
−
32 r
32 r f2
r
rf
−7
25r
1
=
−
32r 32r f2 r f
Clearing of fractions, −7r f2 = 25r 2 − 32rr f or 7
rf
=
14
f
2
− 32
FG r IJ + 25 = 0 giving
HrK
f
a fa f = 50 or 14 . The latter root describes the starting point. The outer
+32 ± 32 2 − 4 7 25
14
14
25
25r
=
; rf =
.
end of the orbit has
7
7
r
r
FG r IJ
HrK
rf
399
Chapter 13
Additional Problems
P13.47
Let m represent the mass of the spacecraft, rE the radius of the Earth’s orbit, and x the distance from
Earth to the spacecraft.
Fs =
The Sun exerts on the spacecraft a radial inward force of
GM s m
br − x g
E
FE =
while the Earth exerts on it a radial outward force of
2
GM E m
x2
The net force on the spacecraft must produce the correct centripetal acceleration for it to have an
orbital period of 1.000 year.
FS − FE =
Thus,
GM S m
br − x g
E
which reduces to
GM S
br − x g
E
2
−
GM E
x2
2
−
=
GM E m
b
br − x g .
x2
4π 2
mv 2
m
=
=
rE − x
rE − x
g b
E
T2
LM 2π br − xg OP
g MN T PQ
2
E
(1)
Cleared of fractions, this equation would contain powers of x ranging from the fifth to the zeroth.
We do not solve it algebraically. We may test the assertion that x is between 1.47 × 10 9 m and
1.48 × 10 9 m by substituting both of these as trial solutions, along with the following data:
M S = 1.991 × 10 30 kg , M E = 5.983 × 10 24 kg , rE = 1.496 × 10 11 m, and T = 1.000 yr = 3.156 × 10 7 s .
With x = 1. 47 × 10 9 m substituted into equation (1), we obtain
6.052 × 10 −3 m s 2 − 1.85 × 10 −3 m s 2 ≈ 5.871 × 10 −3 m s 2
or
5.868 × 10 −3 m s 2 ≈ 5.871 × 10 −3 m s 2
With x = 1. 48 × 10 9 m substituted into the same equation, the result is
6.053 × 10 −3 m s 2 − 1.82 × 10 −3 m s 2 ≈ 5.870 8 × 10 −3 m s 2
or
5.870 9 × 10 −3 m s 2 ≈ 5.870 8 × 10 −3 m s 2 .
Since the first trial solution makes the left-hand side of equation (1) slightly less than the right hand
side, and the second trial solution does the opposite, the true solution is determined as between the
trial values. To three-digit precision, it is 1.48 × 10 9 m .
As an equation of fifth degree, equation (1) has five roots. The Sun-Earth system has five Lagrange
points, all revolving around the Sun synchronously with the Earth. The SOHO and ACE satellites
are at one. Another is beyond the far side of the Sun. Another is beyond the night side of the Earth.
Two more are on the Earth’s orbit, ahead of the planet and behind it by 60°. Plans are under way to
gain perspective on the Sun by placing a spacecraft at one of these two co-orbital Lagrange points.
The Greek and Trojan asteroids are at the co-orbital Lagrange points of the Jupiter-Sun system.
400
P13.48
Universal Gravitation
The acceleration of an object at the center of the Earth due
to the gravitational force of the Moon is given by
M
a = G Moon
d2
MM
At the point A nearest the Moon,
a+ = G
2
d−r
a f
At the point B farthest from the Moon, a − = G
MM
ad + r f
FIG. P13.48
2
∆a = a + − a = GM M
*P13.49
2
−
1
d2
OP
PQ
= 1.11 × 10 −6 m s 2
For d >> r ,
∆a =
Across the planet,
∆g 2 ∆a 2. 22 × 10 −6 m s 2
=
=
= 2. 26 × 10 −7
g
g
9.80 m s 2
d
3
Energy conservation for the two-sphere system from release to contact:
−
Gmm
Gmm 1
1
=−
+ mv 2 + mv 2
R
2r
2
2
Gm
(a)
(b)
FG 1 − 1 IJ = v
H 2r R K
FG L 1 − 1 OIJ
H MN 2r R PQK
2
12
v = Gm
The injected impulse is the final momentum of each sphere,
FG L 1 − 1 OIJ
H MN 2r R PQK
mv = m 2 2 Gm
12
LM
N
= Gm 3
FG 1 − 1 IJ OP
H 2r R K Q
12
.
If they now collide elastically each sphere reverses its velocity to receive impulse
LM
N
a f
mv − − mv = 2mv = 2 Gm 3
P13.50
2GM M r
LM 1
MN ad − r f
FG 1 − 1 IJ OP
H 2r R K Q
12
Momentum is conserved:
m1 v 1i + m 2 v 2i = m1 v 1 f + m 2 v 2 f
0 = Mv 1 f + 2 Mv 2 f
v2 f = −
1
v1 f
2
Energy is conserved:
aK + U f + ∆E = aK + U f
i
0−
−
f
Gm 1 m 2
Gm1 m 2
1
1
+ 0 = m1 v12 f + m 2 v 22 f −
rf
ri
2
2
a f
a fFGH
GM 2 M
1
1
1
v1 f
= Mv12 f + 2 M
12 R
2
2
2
v1 f =
2 GM
R
3
v2 f =
IJ
K
2
−
a f
GM 2 M
4R
1
1 GM
v1 f =
R
2
3
Chapter 13
P13.51
e1.25 × 10
=
(a)
v2
ac =
r
(b)
diff = 10.2 − 9.90 = 0.312 m s 2 =
ac
6.67 × 10
P13.52
(a)
11
−11
2
ms
1.53 × 10
N ⋅ m kg
11
j
2
= 10.2 m s 2
m
GM
r2
e0.312 m s je1.53 × 10 mj
M=
2
6
401
2
= 1.10 × 10 32 kg
2
FIG. P13.51
GM E
The free-fall acceleration produced by the Earth is g =
r2
= GM E r −2 (directed downward)
a f
dg
= GM E −2 r −3 = −2GM E r −3 .
dr
Its rate of change is
The minus sign indicates that g decreases with increasing height.
dg
2GM E
=−
.
dr
RE3
At the Earth’s surface,
(b)
For small differences,
∆g
∆r
*P13.53
=
∆g
h
=
2GM E
∆g =
Thus,
RE3
e
ja
je
2GM E h
f=
2 6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 6.00 m
RE3
1.85 × 10 −5 m s 2
(c)
∆g =
(a)
Each bit of mass dm in the ring is at the same distance from the object at A. The separate
GmM ring
Gmdm
to the system energy add up to −
. When the object is at A,
contributions −
r
r
this is
e6.37 × 10 mj
6
3
−6.67 × 10 −11 N ⋅ m 2 1 000 kg 2.36 × 10 20 kg
kg 2
(b)
8
8
2
= −7.04 × 10 4 J .
When the object is at the center of the ring, the potential energy is
−
(c)
e1 × 10 mj + e2 × 10 mj
2
6.67 × 10 −11 N ⋅ m 2 1 000 kg 2.36 × 10 20 kg
kg 2 1 × 10 8 m
= −1.57 × 10 5 J .
Total energy of the object-ring system is conserved:
e K + U j = eK + U j
g
g
A
B
1
0 − 7.04 × 10 J = 1 000 kgv B2 − 1.57 × 10 5 J
2
4
vB =
F 2 × 8.70 × 10 J I
GH 1 000 kg JK
4
12
= 13.2 m s
402
P13.54
Universal Gravitation
To approximate the height of the sulfur, set
mv 2
= mg Io h
2
h = 70 000 m
GM
= 1.79 m s 2
r2
g Io =
a fb
g
b
v = 2 1.79 70 000 ≈ 500 m s over 1 000 mi h
v = 2 g Io h
g
A more precise answer is given by
1
GMm
GMm
=−
mv 2 −
2
r1
r2
1 2
v = 6.67 × 10 −11 8.90 × 10 22
2
e
P13.55
*P13.56
je
jFGH 1.82 1× 10
−
6
1
1.89 × 10
6
IJ
K
From the walk, 2πr = 25 000 m. Thus, the radius of the planet is r =
From the drop:
∆y =
so,
g=
a
1 2 1
gt = g 29. 2 s
2
2
a
f
f = 3.28 × 10
2 1.40 m
a
29.2 s
f
2
2
v = 492 m s
25 000 m
= 3.98 × 10 3 m
2π
= 1.40 m
−3
m s2 =
MG
r2
∴ M = 7.79 × 10 14 kg
The distance between the orbiting stars is d = 2r cos 30° = 3 r since
3
. The net inward force on one orbiting star is
cos 30° =
2
Gmm
GMm Gmm
mv 2
30
°+
+
30
°
=
cos
cos
r
d2
r2
d2
2 2
Gm 2 cos 30° GM 4π r
+ 2 =
r
rT 2
3r 2
m
4π 2 r 3
+M =
G
T2
3
FG
H
T2 =
r
d
60°
F
F
4π 2 r 3
G M+
30°
r
IJ
K
e
j
F r
T = 2π G
GH GeM +
m
3
3
m
3
P13.57
30°
F
I
JJ
jK
12
FIG. P13.56
For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 g = 9.80 m s 2
v2
= ω 2r
r
g
= 0.057 2 rad s
ω=
r
g=
The required rotation rate of the cylinder is
1 rev
110 s
(For a description of proposed cities in space, see Gerard K. O’Neill in Physics Today, Sept. 1974.)
Chapter 13
P13.58
(a)
G has units
N ⋅ m 2 kg ⋅ m ⋅ m 2
m3
=
=
kg 2
s 2 ⋅ kg 2
s 2 ⋅ kg
and dimensions G =
L3
.
T2 ⋅ M
The speed of light has dimensions of c =
as angular momentum or h =
M ⋅ L2
.
T
L
, and Planck’s constant has the same dimensions
T
We require G p c q h r = L , or L3 p T −2 p M − p Lq T − q M r L2 r T − r = L1 M 0 T 0 .
Thus, 3 p + q + 2r = 1
−2 p − q − r = 0
−p + r = 0
which reduces (using r = p ) to
3p + q + 2p = 1
−2 p − q − p = 0
These equations simplify to
Then, 5 p − 3 p = 1 , yielding p =
5 p + q = 1 and q = −3 p .
1
3
1
, q = − , and r = .
2
2
2
Therefore, Planck length = G 1 2 c − 3 2 h1 2 .
(b)
P13.59
e6.67 × 10 j e3 × 10 j e6.63 × 10 j
−11 1 2
8 −3 2
−34 1 2
e
= 1.64 × 10 −69
Gm p m 0
1
2
=
m 0 v esc
2
R
v esc =
2Gm p
R
4
With m p = ρ πR 3 , we have
3
v esc =
=
So, v esc ∝ R .
403
2Gρ 34 πR 3
R
8πGρ
R
3
j
12
= 4.05 × 10 −35 m ~ 10 −34 m
404
*P13.60
Universal Gravitation
For both circular orbits,
GM E m
∑ F = ma :
=
r2
v=
mv 2
r
GM E
r
FIG. P13.60
e6.67 × 10 N ⋅ m kg je5.98 × 10
e6.37 × 10 m + 2 × 10 mj
kg
j=
7.79 × 10 3 m s .
N ⋅ m 2 kg 2 5.98 × 10 24 kg
j=
7.85 × 10 3 m s .
−11
(a)
(b)
The original speed is vi =
The final speed is
vi =
2
2
6
e6.67 × 10
−11
e
24
5
je
6
j
6.47 × 10 m
The energy of the satellite-Earth system is
K +Ug =
GM E m
GM E m 1 GM E GM E
1
= m
−
=−
mv 2 −
2
2
2r
r
r
r
g=
−3.04 × 10 9 J .
jb
g=
−3.08 × 10 9 J .
Originally
(d)
Finally
Ef = −
(e)
Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is
so large that the total energy decreases by
−11
N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg
e
i
e6.67 × 10
je
jb
(c)
e6.67 × 10
E =−
6
j
2 6.57 × 10 m
−11
je
N ⋅ m 2 kg 2 5.98 × 10 24 kg 100 kg
e
6
j
2 6. 47 × 10 m
e
j
Ei − E f = −3.04 × 10 9 J − −3.08 × 10 9 J = 4.69 × 10 7 J .
(f)
The only forces on the object are the backward force of air resistance R, comparatively very
small in magnitude, and the force of gravity. Because the spiral path of the satellite is not
perpendicular to the gravitational force, one component of the gravitational force pulls
forward on the satellite to do positive work and make its speed increase.
405
Chapter 13
P13.61
(a)
At infinite separation U = 0 and at rest K = 0 . Since energy of the two-planet system is
conserved we have,
0=
Gm1 m 2
1
1
m1 v12 + m 2 v 22 −
d
2
2
(1)
The initial momentum of the system is zero and momentum is conserved.
0 = m1 v1 − m 2 v 2
Therefore,
b
Therefore,
(a)
K1 =
1
m1 v12 = 1.07 × 10 32 J
2
b
g
g
d
and
v 2 = 2.58 × 10 3 m s
and
K2 =
1
m 2 v 22 = 2.67 × 10 31 J
2
The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earth
is conserved;
mra v a = mrp v p and v a = v p
(b)
b
2G
d m1 + m 2
Substitute given numerical values into the equation found for v1 and v 2 in part (a) to find
v1 = 1.03 × 10 4 m s
P13.62
v 2 = m1
and
g
2G m 1 + m 2
b g
v r = v1 − − v 2 =
Relative velocity
(b)
2G
d m1 + m 2
v1 = m 2
Combine equations (1) and (2):
(2)
Kp =
F r I = e3.027 × 10
GH r JK
p
je
e
je
ms
a
1
1
mv p2 = 5.98 × 10 24 3.027 × 10 4
2
2
e
4
j
2
471 I
JK =
jFGH 11..521
2.93 × 10 4 m s
= 2.74 × 10 33 J
je
j
6.673 × 10 −11 5.98 × 10 24 1.99 × 10 30
GmM
=−
= −5.40 × 10 33 J
Up = −
11
rp
1.471 × 10
(c)
Using the same form as in part (b), K a = 2.57 × 10 33 J and U a = −5.22 × 10 33 J .
Compare to find that K p + U p = −2.66 × 10 33 J and K a + U a = −2.65 × 10 33 J . They agree.
406
P13.63
Universal Gravitation
(a)
The work must provide the increase in gravitational energy
W = ∆U g = U gf − U gi
GM E M p
=−
rf
GM E M p
=−
RE + y
= GM E M p
=
+
+
F1
GH R
F 6.67 × 10
GH
kg
GM E M p
ri
GM E M p
RE
1
RE + y
−
E
−11
N ⋅ m2
2
I
JK
I 5.98 × 10
JK e
24
jb
kg 100 kg
gFGH 6.37 ×110
6
m
−
W = 850 MJ
(b)
In a circular orbit, gravity supplies the centripetal force:
GM E M p
bR
Then,
E
+y
Mpv2
=
g bR
2
E
+y
g
1
1 GM E M p
Mp v2 =
2
2 RE + y
b
g
So, additional work = kinetic energy required
=
e
je
jb
−11
N ⋅ m 2 5.98 × 10 24 kg 100 kg
1 6.67 × 10
2
kg 2 7.37 × 10 6 m
e je
j
∆W = 2.71 × 10 9 J
P13.64
Centripetal acceleration comes from gravitational acceleration.
v 2 M c G 4π 2 r 2
= 2 =
r
r
T 2r
GM c T 2 = 4π 2 r 3
e6.67 × 10 ja20fe1.99 × 10 je5.00 × 10 j
−11
−3 2
30
= 4π 2 r 3
rorbit = 119 km
P13.65
e
j
15
2πr 2π 30 000 × 9.46 × 10 m
=
= 7 × 10 15 s = 2 × 10 8 yr
v
2.50 × 10 5 m s
(a)
T=
(b)
4π 2 30 000 × 9.46 × 10 15 m
4π 2 a 3
M=
=
GT 2
6.67 × 10 −11 N ⋅ m 2 kg 2 7.13 × 10 15 s
e
e
j
3
je
M = 1.34 × 10 11 solar masses ~ 10 11 solar masses
The number of stars is on the order of 10 11 .
j
2
= 2.66 × 10 41 kg
g
1
7.37 × 10 6 m
IJ
K
Chapter 13
P13.66
(a)
From the data about perigee, the energy of the satellite-Earth system is
E=
or
(b)
407
6.67 × 10 je5.98 × 10 ja1.60f
j e
7.02 × 10
a fe
GM E m 1
1
= 1.60 8.23 × 10 3
mv p2 −
rp
2
2
2
−11
24
−
6
E = −3.67 × 10 7 J
b
ge
je
j
L = mvr sin θ = mv p rp sin 90.0° = 1.60 kg 8.23 × 10 3 m s 7.02 × 10 6 m
= 9.24 × 10 10 kg ⋅ m 2 s
(c)
Since both the energy of the satellite-Earth system and the angular momentum of the Earth
are conserved,
at apogee we must have
1
GMm
=E
mv a2 −
2
ra
and
mv a ra sin 90.0° = L .
Thus,
6.67 × 10 −11 5.98 × 10 24 1.60
1
2
= −3.67 × 10 7 J
1.60 v a −
2
rs
and
b1.60 kg gv r
e
a f
a a
ja f
je
= 9.24 × 10 10 kg ⋅ m 2 s .
e
ja fa f
je
6.67 × 10 −11 5.98 × 10 24 1.60 1.60 v a
1
2
Solving simultaneously,
= −3.67 × 10 7
1.60 v a −
2
9.24 × 10 10
a f
0.800 v a2 − 11 046 v a + 3.672 3 × 10 7 = 0
which reduces to
va =
so
11 046 ±
b11 046g − 4a0.800fe3.672 3 × 10 j .
2a0.800 f
2
7
This gives v a = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at the
apogee while the larger refers to perigee.
ra =
Thus,
(d)
b
ge
j
The major axis is 2a = rp + ra , so the semi-major axis is
a=
(e)
9.24 × 10 10 kg ⋅ m 2 s
L
=
= 1.04 × 10 7 m .
mv a
1.60 kg 5.58 × 10 3 m s
T=
4π 2 a 3
=
GM E
1
7.02 × 10 6 m + 1.04 × 10 7 m = 8.69 × 10 6 m
2
e
j
e
j
4π 2 8.69 × 10 6 m
e6.67 × 10
T = 8 060 s = 134 min
−11
je
3
N ⋅ m 2 kg 2 5.98 × 10 24 kg
j
408
*P13.67
Universal Gravitation
Let m represent the mass of the meteoroid and vi its speed when far away.
No torque acts on the meteoroid, so its angular momentum is conserved as
it moves between the distant point and the point where it grazes the Earth,
moving perpendicular to the radius:
Li = L f :
FIG. P13.67
mri × v i = mr f × v f
b
g
m 3 RE vi = mRE v f
v f = 3 vi
Now energy of the meteoroid-Earth system is also conserved:
GM E m
1
1
mvi2 + 0 = mv 2f −
2
2
RE
eK + U j = eK + U j :
g
g
i
f
GM E
1 2 1
vi = 9 vi2 −
2
2
RE
e j
GM E
= 4vi2 :
RE
*P13.68
vi =
GM E
4 RE
From Kepler’s third law, minimum period means minimum orbit size. The “treetop satellite” in
Figure P13.35 has minimum period. The radius of the satellite’s circular orbit is essentially equal to
the radius R of the planet.
FG
H
R
R e 4π R j
GρV =
GMm
∑ F = ma :
2
=
mv 2 m 2πR
=
R
R T
2
2
IJ
K
2
2
RT 2
4
4π 2 R 3
G ρ πR 3 =
3
T2
FG
H
IJ
K
The radius divides out: T 2 Gρ = 3π
P13.69
T=
3π
Gρ
If we choose the coordinate of the center of mass at the origin, then
0=
bMr
2
− mr1
g
M+m
and
Mr2 = mr1
(Note: this is equivalent to saying that the net torque must be zero and
the two experience no angular acceleration.) For each mass F = ma so
mr1ω 12 =
MGm
d2
and
Mr2ω 22 =
MGm
d2
FIG. P13.69
b
g
Combining these two equations and using d = r1 + r2 gives r1 + r2 ω 2 =
with ω 1 = ω 2 = ω
2π
and T =
ω
we find T 2 =
4π 2 d 3
G M+m
a
f
.
aM + mfG
d2
409
Chapter 13
P13.70
The gravitational force exerted on m 2 by the Earth (mass m1 ) accelerates m 2 according to:
Gm1 m 2
. The equal magnitude force exerted on the Earth by m 2 produces negligible
m2 g 2 =
r2
acceleration of the Earth. The acceleration of relative approach is then
(a)
g2 =
Gm1
r
2
=
e6.67 × 10
−11
je
mj
N ⋅ m 2 kg 2 5.98 × 10 24 kg
e1.20 × 10
7
2
j=
2.77 m s 2 .
Again, m 2 accelerates toward the center of mass with g 2 = 2.77 m s 2 . Now the Earth
accelerates toward m 2 with an acceleration given as
(b)
m1 g 1 =
g1 =
Gm1 m 2
r2
Gm 2
r
2
e6.67 × 10
=
−11
je
mj
N ⋅ m 2 kg 2 2.00 × 10 24 kg
e1.20 × 10
7
2
j = 0.926 m s
2
The distance between the masses closes with relative acceleration of
g rel = g 1 + g 2 = 0.926 m s 2 + 2.77 m s 2 = 3.70 m s 2 .
P13.71
Initial Conditions and Constants:
Mass of planet:
5.98 × 10 24 kg
Radius of planet:
Initial x:
Initial y:
Initial v x :
6.37 × 10 6 m
0.0 planet radii
2.0 planet radii
+5 000 m/s
Time interval:
10.9 s
Initial
vy :
0.0 m/s
FIG. P13.71
t (s)
0.0
10.9
21.7
32.6
…
5 431.6
5 442.4
5 453.3
5 464.1
…
10 841.3
10 852.2
10 863.1
x (m)
y (m)
r (m)
vx
(m/s)
vy
ax
(m/s)
em s j
2
ay
em s j
2
0.0
54 315.3
108 629.4
162 941.1
12 740 000.0
12 740 000.0
12 739 710.0
12 739 130.0
12 740 000.0
12 740 115.8
12 740 173.1
12 740 172.1
5 000.0
4 999.9
4 999.7
4 999.3
0.0
–26.7
–53.4
–80.1
0.000 0
–0.010 0
–0.021 0
–0.031 0
–2.457 5
–2.457 4
–2.457 3
–2.457 2
112 843.8
31 121.4
–50 603.4
–132 324.3
–8 466 816.0
–8 467 249.7
–8 467 026.9
–8 466 147.7
8 467 567.9
8 467 306.9
8 467 178.2
8 467 181.7
–7 523.0
–7 523.2
–7 522.8
–7 521.9
–39.9
20.5
80.9
141.4
–0.074 0
–0.020 0
0.033 0
0.087 0
5.562 5
5.563 3
5.563 4
5.562 8
–108 629.0
–54 314.9
0.4
12 739 134.4
12 739 713.4
12 740 002.4
12 739 597.5
12 739 829.2
12 740 002.4
4 999.9
5 000.0
5 000.0
53.3
26.6
–0.1
0.021 0
0.010 0
0.000 0
–2.457 5
–2.457 5
–2.457 5
The object does not hit the Earth ; its minimum radius is 1.33 RE .
Its period is 1.09 × 10 4 s . A circular orbit would require a speed of 5.60 km s .
410
Universal Gravitation
ANSWERS TO EVEN PROBLEMS
P13.2
2.67 × 10 −7 m s 2
P13.40
(a) 10.0 m s 2 ; (b) 21.8 km s
P13.4
3.00 kg and 2.00 kg
P13.42
11.8 km s
P13.6
(a) 4.39 × 10 20 N toward the Sun;
(b) 1.99 × 10 20 N toward the Earth;
(c) 3.55 × 10 22 N toward the Sun
P13.44
GM E m
12 RE
P13.8
see the solution; either 1 m − 61.3 nm or
2.74 × 10 −4 m
P13.46
(a) v 0
F GM IJ
=G
H r K
(c) r f =
P13.10
2
3
P13.12
(a) 1.02 km s ; (b) 1.35 mm
P13.14
see the solution
P13.16
1.27
P13.18
Planet Y has turned through
1.30 revolutions
P13.20
1.63 × 10 4 rad s
P13.22
18.2 ms
P13.24
(a) 1.31 × 10 17 N toward the center;
(b) 2.62 × 10 12 N kg
E
12
; (b) vi =
5
GM E 1 2
r
e j
25r
7
P13.48
2.26 × 10 −7
P13.50
2 GM 1 GM
;
3
R
3
R
4
;
P13.52
(a), (b) see the solution;
(c) 1.85 × 10 −5 m s 2
P13.54
492 m s
P13.56
see the solution
P13.58
(a) G 1 2 c − 3 2 h1 2 ; (b) ~ 10 −34 m
P13.60
(a) 7.79 km s; (b) 7.85 km s;(c) −3.04 GJ ;
(d) −3.08 GJ ; (e) loss = 46.9 MJ ;
(f) A component of the Earth’s gravity
pulls forward on the satellite in its
downward banking trajectory.
P13.62
(a) 29.3 km s ; (b) K p = 2.74 × 10 33 J ;
P13.26
(a) −4.77 × 10 9 J ; (b) 569 N down;
(c) 569 N up
P13.28
2.52 × 10 7 m
P13.30
2.82 × 10 9 J
P13.32
(a) see the solution; (b) 340 s
P13.34
(a) 42.1 km s ; (b) 2.20 × 10 11 m
P13.64
119 km
P13.36
1.58 × 10 10 J
P13.66
(a) −36.7 MJ ; (b) 9.24 × 10 10 kg ⋅ m 2 s ;
(c) 5.58 km s; 10.4 Mm; (d) 8.69 Mm;
(e) 134 min
P13.68
see the solution
P13.70
(a) 2.77 m s 2 ; (b) 3.70 m s 2
P13.38
b
U p = −5.40 × 10 33 J ;(c) K a = 2.57 × 10 33 J;
U a = −5.22 × 10 33 J; yes
g bGM g ;
(b) bGM g b R + hg
;
L R + 2 h OP − 2π R m
(c) GM m M
MN 2 R bR + hg PQ b86 400 sg
(a) 2π RE + h
E
E
−1 2
E
−1 2
3 2
12
E
2
E
E
E
2
E
2
The satellite should be launched from the
Earth’s equator toward the east.
14
Fluid Mechanics
CHAPTER OUTLINE
14.1
14.2
14.3
14.4
14.5
14.6
14.7
Pressure
Variation of Pressure with
Depth
Pressure Measurements
Buoyant Forces and
Archimede’s Principle
Fluid Dynamics
Bernoulli’s Equation
Other Applications of Fluid
Dynamics
ANSWERS TO QUESTIONS
Q14.1
The weight depends upon the total volume of glass. The
pressure depends only on the depth.
Q14.2
Both must be built the same. The force on the back of each dam
is the average pressure of the water times the area of the dam.
If both reservoirs are equally deep, the force is the same.
FIG. Q14.2
Q14.3
If the tube were to fill up to the height of several stories of the building, the pressure at the bottom of
the depth of the tube of fluid would be very large according to Equation 14.4. This pressure is much
larger than that originally exerted by inward elastic forces of the rubber on the water. As a result,
water is pushed into the bottle from the tube. As more water is added to the tube, more water
continues to enter the bottle, stretching it thin. For a typical bottle, the pressure at the bottom of the
tube can become greater than the pressure at which the rubber material will rupture, so the bottle
will simply fill with water and expand until it bursts. Blaise Pascal splintered strong barrels by this
method.
Q14.4
About 1 000 N: that’s about 250 pounds.
Q14.5
The submarine would stop if the density of the surrounding water became the same as the average
density of the submarine. Unfortunately, because the water is almost incompressible, this will be
much deeper than the crush depth of the submarine.
Q14.6
Yes. The propulsive force of the fish on the water causes the scale reading to fluctuate. Its average
value will still be equal to the total weight of bucket, water, and fish.
Q14.7
The boat floats higher in the ocean than in the inland lake. According to Archimedes’s principle, the
magnitude of buoyant force on the ship is equal to the weight of the water displaced by the ship.
Because the density of salty ocean water is greater than fresh lake water, less ocean water needs to
be displaced to enable the ship to float.
411
412
Fluid Mechanics
Q14.8
In the ocean, the ship floats due to the buoyant force from salt water. Salt water is denser than fresh
water. As the ship is pulled up the river, the buoyant force from the fresh water in the river is not
sufficient to support the weight of the ship, and it sinks.
Q14.9
Exactly the same. Buoyancy equals density of water times volume displaced.
Q14.10
At lower elevation the water pressure is greater because pressure increases with increasing depth
below the water surface in the reservoir (or water tower). The penthouse apartment is not so far
below the water surface. The pressure behind a closed faucet is weaker there and the flow weaker
from an open faucet. Your fire department likely has a record of the precise elevation of every fire
hydrant.
Q14.11
As the wind blows over the chimney, it creates a lower pressure at the top of the chimney. The
smoke flows from the relatively higher pressure in front of the fireplace to the low pressure outside.
Science doesn’t suck; the smoke is pushed from below.
Q14.12
The rapidly moving air above the ball exerts less pressure than the atmospheric pressure below the
ball. This can give substantial lift to balance the weight of the ball.
Q14.13
The ski–jumper gives her body the shape of an airfoil. She
deflects downward the air stream as it rushes past and it
deflects her upward by Newton’s third law. The air exerts
on her a lift force, giving her a higher and longer trajectory.
To say it in different words, the pressure on her back is less
than the pressure on her front.
FIG. Q14.13
Q14.14
The horizontal force exerted by the outside fluid, on an area element of the object’s side wall, has
equal magnitude and opposite direction to the horizontal force the fluid exerts on another element
diametrically opposite the first.
Q14.15
The glass may have higher density than the liquid, but the air inside has lower density. The total
weight of the bottle can be less than the weight of an equal volume of the liquid.
Q14.16
Breathing in makes your volume greater and increases the buoyant force on you. You instinctively
take a deep breath if you fall into the lake.
Q14.17
No. The somewhat lighter barge will float higher in the water.
Q14.18
The level of the pond falls. This is because the anchor displaces more water while in the boat. A
floating object displaces a volume of water whose weight is equal to the weight of the object. A
submerged object displaces a volume of water equal to the volume of the object. Because the density
of the anchor is greater than that of water, a volume of water that weighs the same as the anchor will
be greater than the volume of the anchor.
Q14.19
The metal is more dense than water. If the metal is sufficiently thin, it can float like a ship, with the
lip of the dish above the water line. Most of the volume below the water line is filled with air. The
mass of the dish divided by the volume of the part below the water line is just equal to the density of
water. Placing a bar of soap into this space to replace the air raises the average density of the
compound object and the density can become greater than that of water. The dish sinks with its
cargo.
Chapter 14
413
Q14.20
The excess pressure is transmitted undiminished throughout the container. It will compress air
inside the wood. The water driven into the wood raises its average density and makes if float lower
in the water. Add some thumbtacks to reach neutral buoyancy and you can make the wood sink or
rise at will by subtly squeezing a large clear–plastic soft–drink bottle. Bored with graph paper and
proving his own existence, René Descartes invented this toy or trick.
Q14.21
The plate must be horizontal. Since the pressure of a fluid increases with increasing depth, other
orientations of the plate will give a non-uniform pressure on the flat faces.
Q14.22
The air in your lungs, the blood in your arteries and veins, and the protoplasm in each cell exert
nearly the same pressure, so that the wall of your chest can be in equilibrium.
Q14.23
Use a balance to determine its mass. Then partially fill a graduated cylinder with water. Immerse the
rock in the water and determine the volume of water displaced. Divide the mass by the volume and
you have the density.
Q14.24
When taking off into the wind, the increased airspeed over the wings gives a larger lifting force,
enabling the pilot to take off in a shorter length of runway.
Q14.25
Like the ball, the balloon will remain in front of you. It will not bob up to the ceiling. Air pressure
will be no higher at the floor of the sealed car than at the ceiling. The balloon will experience no
buoyant force. You might equally well switch off gravity.
Q14.26
Styrofoam is a little more dense than air, so the first ship floats lower in the water.
Q14.27
We suppose the compound object floats. In both orientations it displaces its own weight of water, so
it displaces equal volumes of water. The water level in the tub will be unchanged when the object is
turned over. Now the steel is underwater and the water exerts on the steel a buoyant force that was
not present when the steel was on top surrounded by air. Thus, slightly less wood will be below the
water line on the block. It will appear to float higher.
Q14.28
A breeze from any direction speeds up to go over the mound and the air pressure drops. Air then
flows through the burrow from the lower entrance to the upper entrance.
Q14.29
Regular cola contains a considerable mass of dissolved sugar. Its density is higher than that of water.
Diet cola contains a very small mass of artificial sweetener and has nearly the same density as water.
The low–density air in the can has a bigger effect than the thin aluminum shell, so the can of diet
cola floats.
Q14.30
(a)
Lowest density: oil; highest density: mercury
(b)
The density must increase from top to bottom.
(a)
Since the velocity of the air in the right-hand section of the pipe is lower than that in the
middle, the pressure is higher.
(b)
The equation that predicts the same pressure in the far right and left-hand sections of the
tube assumes laminar flow without viscosity. Internal friction will cause some loss of
mechanical energy and turbulence will also progressively reduce the pressure. If the
pressure at the left were not higher than at the right, the flow would stop.
Q14.31
414
Q14.32
Fluid Mechanics
Clap your shoe or wallet over the hole, or a seat cushion, or your hand. Anything that can sustain a
force on the order of 100 N is strong enough to cover the hole and greatly slow down the escape of
the cabin air. You need not worry about the air rushing out instantly, or about your body being
“sucked” through the hole, or about your blood boiling or your body exploding. If the cabin pressure
drops a lot, your ears will pop and the saliva in your mouth may boil—at body temperature—but
you will still have a couple of minutes to plug the hole and put on your emergency oxygen mask.
Passengers who have been drinking carbonated beverages may find that the carbon dioxide
suddenly comes out of solution in their stomachs, distending their vests, making them belch, and all
but frothing from their ears; so you might warn them of this effect.
SOLUTIONS TO PROBLEMS
Section 14.1
P14.1
Pressure
e
M = ρ ironV = 7 860 kg m3
M = 0.111 kg
P14.2
jLMN 34 π b0.015 0 mg OPQ
3
The density of the nucleus is of the same order of magnitude as that of one proton, according to the
assumption of close packing:
ρ=
m 1.67 × 10 −27 kg
~
~ 10 18 kg m3 .
3
−15
V
4π
10
m
3
e
j
With vastly smaller average density, a macroscopic chunk of matter or an atom must be mostly
empty space.
a f
50.0 9.80
F
=
A π 0.500 × 10 −2
= 6.24 × 10 6 N m 2
P14.3
P=
P14.4
Let Fg be its weight. Then each tire supports
so
yielding
P14.5
e
j
2
Fg
4
,
F Fg
=
A 4A
Fg = 4 AP = 4 0.024 0 m 2 200 × 10 3 N m 2 = 1.92 × 10 4 N
P=
e
je
j
The Earth’s surface area is 4πR 2 . The force pushing inward over this area amounts to
e
j
e
j
F = P0 A = P0 4πR 2 .
This force is the weight of the air:
Fg = mg = P0 4πR 2
so the mass of the air is
m=
e
P0 4πR 2
g
j = e1.013 × 10
5
jLMN e
j OPQ
N m 2 4π 6.37 × 10 6 m
9.80 m s 2
2
= 5.27 × 10 18 kg .
Chapter 14
Section 14.2
P14.6
415
Variation of Pressure with Depth
e
jb
je
g
P = P0 + ρgh = 1.013 × 10 5 Pa + 1 024 kg m3 9.80 m s 2 1 000 m
(a)
P = 1.01 × 10 7 Pa
(b)
The gauge pressure is the difference in pressure between the water outside and the air
inside the submarine, which we suppose is at 1.00 atmosphere.
Pgauge = P − P0 = ρgh = 1.00 × 10 7 Pa
The resultant inward force on the porthole is then
a
F = Pgauge A = 1.00 × 10 7 Pa π 0.150 m
P14.7
Fel = Ffluid
and
h=
= 7.09 × 10 5 N .
kx
ρgA
−3
2
3
3
2
−2
=
O
j PQ
m
2
15 000
F
= 2
200
3.00
e
1.62 m
FIG. P14.7
F1
F
= 2
A1 A 2
Since the pressure is the same on both sides,
In this case,
P14.9
2
kx = ρghA
or
e1 000 N m je5.00 × 10 mj
h=
e10 kg m je9.80 m s jLMNπ e1.00 × 10
P14.8
f
F2 = 225 N
or
j
Fg = 80.0 kg 9.80 m s 2 = 784 N
When the cup barely supports the student, the normal force of the
ceiling is zero and the cup is in equilibrium.
e
j
Fg = F = PA = 1.013 × 10 5 Pa A
A=
Fg
P
=
784
= 7.74 × 10 −3 m 2
5
1.013 × 10
FIG. P14.9
P14.10
(a)
Suppose the “vacuum cleaner” functions as a high–vacuum pump. The air below the brick
will exert on it a lifting force
LM e
N
j OPQ =
F = PA = 1.013 × 10 5 Pa π 1.43 × 10 −2 m
(b)
2
65.1 N .
The octopus can pull the bottom away from the top shell with a force that could be no larger
than
b
g
e
je
ja
f LNMπ e1.43 × 10
F = PA = P0 + ρgh A = 1.013 × 10 5 Pa + 1 030 kg m3 9.80 m s 2 32.3 m
F = 275 N
−2
j OQP
m
2
416
P14.11
Fluid Mechanics
The excess water pressure (over air pressure) halfway down is
e
ja
je
f
Pgauge = ρgh = 1 000 kg m3 9.80 m s 2 1.20 m = 1.18 × 10 4 Pa .
The force on the wall due to the water is
ja
e
fa
f
F = Pgauge A = 1.18 × 10 4 Pa 2.40 m 9.60 m = 2.71 × 10 5 N
horizontally toward the back of the hole.
P14.12
P14.13
The pressure on the bottom due to the water is Pb = ρgz = 1.96 × 10 4 Pa
So,
Fb = Pb A = 5.88 × 10 6 N
On each end,
F = PA = 9.80 × 10 3 Pa 20.0 m 2 = 196 kN
On the side,
F = PA = 9.80 × 10 3
2
588 kN
In the reference frame of the fluid, the cart’s acceleration causes a fictitious force to act backward, as if
a
the acceleration of gravity were g 2 + a 2 directed downward and backward at θ = tan −1
from the
g
d
vertical. The center of the spherical shell is at depth below the air bubble and the pressure there is
2
F I
GH JK
P = P0 + ρg eff h = P0 +
P14.14
e
j
Pae60.0 m j =
1
ρd g 2 + a 2 .
2
The air outside and water inside both exert atmospheric pressure,
so only the excess water pressure ρgh counts for the net force. Take
a strip of hatch between depth h and h + dh . It feels force
a
f
dF = PdA = ρgh 2.00 m dh .
(a)
1.00 m
The total force is
z
F = dF =
z
2.00 m
2.00 m
a
f
ρgh 2.00 m dh
h = 1.00 m
FIG. P14.14
a2.00 mf a2.00 mf − a1.00 mf
= e1 000 kg m je9.80 m s j
f
2
F = 29.4 kN bto the right g
The lever arm of dF is the distance a h − 1.00 mf from hinge to strip:
τ = z dτ = z ρgha 2.00 mfa h − 1.00 mfdh
Lh
h O
τ = ρg a 2.00 mfM − a1.00 mf P
2 Q
N3
F 7.00 m − 3.00 m I
τ = e1 000 kg m je9.80 m s ja 2.00 mfG
JK
2
H 3
h2
F = ρg 2.00 m
2
a
(b)
2.00 m
2.00 m
3
2
2
1.00 m
2.00 m
h = 1.00 m
3
2
2.00 m
1.00 m
3
2
τ = 16.3 kN ⋅ m counterclockwise
3
3
2
Chapter 14
P14.15
417
The bell is uniformly compressed, so we can model it with any shape. We choose a sphere of
diameter 3.00 m.
The pressure on the ball is given by: P = Patm + ρ w gh so the change in pressure on the ball from
when it is on the surface of the ocean to when it is at the bottom of the ocean is ∆P = ρ w gh .
In addition:
∆V =
ρ ghV
4πρ w ghr 3
−V∆P
=− w
=−
, where B is the Bulk Modulus.
3B
B
B
∆V = −
je
jb
a3fe14.0 × 10 Paj
e
ga
f
4π 1 030 kg m3 9.80 m s 2 10 000 m 1.50 m
10
3
= −0.010 2 m3
Therefore, the volume of the ball at the bottom of the ocean is
V − ∆V =
a
f
4
π 1.50 m
3
3
− 0.010 2 m 3 = 14.137 m3 − 0.010 2 m 3 = 14.127 m 3 .
This gives a radius of 1.499 64 m and a new diameter of 2.999 3 m. Therefore the diameter decreases
by 0.722 mm .
Section 14.3
P14.16
(a)
Pressure Measurements
We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the
water surface in the basin and point 2 at the water surface in the straw:
P1 + ρgy1 = P2 + ρgy 2
e
je
j
1.013 × 10 5 N m 2 + 0 = 0 + 1 000 kg m 3 9.80 m s 2 y 2
(b)
P14.17
y 2 = 10.3 m
No atmosphere can lift the water in the straw through zero height difference.
P0 = ρgh
h=
P0
10.13 × 10 5 Pa
=
= 10.5 m
ρg 0.984 × 10 3 kg m3 9.80 m s 2
e
je
j
No. Some alcohol and water will evaporate. The equilibrium
vapor pressures of alcohol and water are higher than the vapor
pressure of mercury.
FIG. P14.17
418
P14.18
Fluid Mechanics
(a)
Using the definition of density, we have
hw =
(b)
100 g
m water
=
= 20.0 cm
2
A 2 ρ water 5.00 cm 1.00 g cm 3
e
j
Sketch (b) at the right represents the situation after
the water is added. A volume A 2 h2 of mercury
has been displaced by water in the right tube. The
additional volume of mercury now in the left tube
is A1 h . Since the total volume of mercury has not
changed,
b
A 2 h2 = A1 h
g
FIG. P14.18
h2 =
or
A1
h
A2
(1)
At the level of the mercury–water interface in the right tube, we may write the absolute
pressure as:
P = P0 + ρ water ghw
The pressure at this same level in the left tube is given by
b
g
P = P0 + ρ Hg g h + h2 = P0 + ρ water ghw
which, using equation (1) above, reduces to
LM
N
ρ Hg h 1 +
or h =
ρ water hw
e
ρ Hg 1 +
A1
A2
j
OP
Q
A1
= ρ water hw
A2
.
e1.00 g cm ja20.0 cmf =
Thus, the level of mercury has risen a distance of h =
e13.6 g cm jc1 + h
3
3
10 .0
5.00
0.490 cm
above the original level.
b
g
P14.19
∆P0 = ρg∆h = −2.66 × 10 3 Pa :
P = P0 + ∆P0 = 1.013 − 0.026 6 × 10 5 Pa = 0.986 × 10 5 Pa
P14.20
Let h be the height of the water column added to the right
side of the U–tube. Then when equilibrium is reached, the
situation is as shown in the sketch at right. Now consider
two points, A and B shown in the sketch, at the level of the
water–mercury interface. By Pascal’s Principle, the absolute
pressure at B is the same as that at A. But,
PA = P0 + ρ w gh + ρ Hg gh2 and
b
h1
water
h
B
h2
Mercury
A
g
PB = P0 + ρ w g h1 + h + h2 .
Thus, from PA = PB , ρ w h1 + ρ w h + ρ w h2 = ρ w h + ρ Hg h2 , or
h1 =
LM ρ
Nρ
Hg
w
OP
Q
a
fa
f
− 1 h2 = 13.6 − 1 1.00 cm = 12.6 cm .
FIG. P14.20
Chapter 14
*P14.21
P = P0 + ρgh
The gauge pressure is
(a)
ja
e
f
P − P0 = ρgh = 1 000 kg 9.8 m s 2 0.160 m = 1.57 kPa = 1.57 × 10 3 Pa
419
FG 1 atm IJ
H 1.013 × 10 Pa K
5
= 0.015 5 atm .
It would lift a mercury column to height
1 568 Pa
P − P0
h=
=
= 11.8 mm .
ρg
13 600 kg m 3 9.8 m s 2
e
(b)
je
j
Increased pressure of the cerebrospinal fluid will raise the level of the fluid in the
spinal tap.
(c)
Blockage of the fluid within the spinal column or between the skull and the spinal
column would prevent the fluid level from rising.
Section 14.4
P14.22
Buoyant Forces and Archimede’s Principle
(a)
The balloon is nearly in equilibrium:
− Fg
∑ Fy = ma y ⇒ B − Fg
e j
helium
e j
payload
=0
ρ air gV − ρ helium gV − m payload g = 0
This reduces to
m payload = ρ air − ρ helium V = 1.29 kg m3 − 0.179 kg m 3 400 m 3
or
b
g e
je
j
m payload = 444 kg
(b)
Similarly,
m payload = ρ air − ρ hydrogen V = 1.29 kg m3 − 0.089 9 kg m3 400 m 3
e
j e
je
j
m payload = 480 kg
The air does the lifting, nearly the same for the two balloons.
P14.23
At equilibrium
∑ F = 0 or
where
The applied force,
B is the buoyant force.
Fapp = B − mg
b
g
B = Vol ρ water g
where
a f
So,
F = aVolf g b ρ
g
b
4
F = π e1.90 × 10 mj e9.80 m s je10 kg m
3
F = bm + ρ V g g must be equal to F = ρ Vg
and
app
water
−2
app
P14.24
Fapp + mg = B
g
3
m = Vol ρ ball .
4
− ρ ball = πr 3 g ρ water − ρ ball
3
s
2
3
b
w
3
g
FIG. P14.23
j
− 84.0 kg m 3 = 0.258 N
Since V = Ah , m + ρ s Ah = ρ w Ah
and A =
b
m
ρw − ρs h
g
FIG. P14.24
420
Fluid Mechanics
P14.25
(a)
Before the metal is immersed:
∑ Fy = T1 − Mg = 0 or
b
ge
T1 = Mg = 1.00 kg 9.80 m s 2
scale
j
= 9.80 N
B
T1
(b)
After the metal is immersed:
∑ Fy = T2 + B − Mg = 0
or
b
Mg
g
Mg
T2 = Mg − B = Mg − ρ w V g
V=
M
ρ
=
1.00 kg
2 700 kg m3
a
e
T2 = Mg − B = 9.80 N − 1 000 kg m3
(a)
b
FIG. P14.25
Thus,
*P14.26
T2
Fg
∑ Fy = 0 :
(b)
F
jGH 2 7001.00kgkgm
3
I 9.80 m s =
j
JK e
2
6.17 N .
−15 N − 10 N + B = 0
B = 25.0 N
T
B
FIG. P14.26(a)
(c)
The oil pushes horizontally inward on each side of the block.
(d)
String tension increases . The oil causes the water below to be
under greater pressure, and the water pushes up more strongly
on the bottom of the block.
(e)
Consider the equilibrium just before the string breaks:
15 N
−15 N − 60 N + 25 N+ Boil = 0
Boil = 50 N
60 N
For the buoyant force of the water we have
B = ρVg
jb
e
g
25 N = 1 000 kg m3 0.25Vblock 9.8 m s 2
Vblock = 1.02 × 10 −2 m3
25 N
Boil
FIG. P14.26(e)
For the buoyant force of the oil
e
j e
j
50 N = 800 kg m3 f e 1.02 × 10 −2 m 3 9.8 m s 2
f e = 0.625 = 62.5%
(f)
e
j e
j
−15 N + 800 kg m3 f f 1.02 × 10 −2 m 3 9.8 m s 2 = 0
15 N
f f = 0.187 = 18.7%
Boil
FIG. P14.26(f)
Chapter 14
P14.27
421
P = P0 + ρgh
Taking P0 = 1.013 × 10 5 N m 2 and h = 5.00 cm
(a)
we find
Ptop = 1.017 9 × 10 5 N m 2
For h = 17.0 cm, we get
Pbot = 1.029 7 × 10 5 N m 2
Since the areas of the top and bottom are
A = 0.100 m
we find
Ftop = Ptop A = 1.017 9 × 10 3 N
and
Fbot = 1.029 7 × 10 3 N
T + B − Mg = 0
(b)
e
a
f
je
2
= 10 −2 m 2
je
B = ρ w Vg = 10 3 kg m3 1.20 × 10 −3 m3 9.80 m s 2 = 11.8 N
and
Mg = 10.0 9.80 = 98.0 N
Therefore,
T = Mg − B = 98.0 − 11.8 = 86.2 N
a f
b
FIG. P14.27
j
where
g
Fbot − Ftop = 1.029 7 − 1.017 9 × 10 3 N = 11.8 N
(c)
which is equal to B found in part (b).
P14.28
Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at 0°C
and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is
B − Fg ,He − Fg , env = ρ air Vg − ρ HeVg − m env g
b
gFGH 34 πr IJK g − m g
L4
O
= a1. 29 − 0.179 f kg m M π a0.125 mf Pe9.80 m s j − 5.00 × 10
3
N
Q
3
Fup = ρ air − ρ He
Fup
env
3
3
2
−3
e
j
kg 9.80 m s 2 = 0.040 1 N
If your weight (including harness, strings, and submarine sandwich) is
e
j
70.0 kg 9.80 m s 2 = 686 N
686 N
= 17 000 ~ 10 4 .
0.040 1 N
you need this many balloons:
P14.29
(a)
e
a
j
e
ja
f
3
But B = Weight of block = mg = ρ woodVwood g = 0.650 g cm 3 20.0 cm g
a f
a fa fa
f
20.0 − h = 20.0a0.650 f so h = 20.0a1 − 0.650f = 7.00 cm
3
0.650 20.0 g = 1.00 20.0 20.0 20.0 − h g
(b)
f
According to Archimedes, B = ρ water Vwater g = 1.00 g cm 3 20.0 × 20.0 × 20.0 − h g
B = Fg + Mg where M = mass of lead
a f
a f
M = a1.00 − 0.650fa 20.0f = 0.350a 20.0 f
3
3
1.00 20.0 g = 0.650 20.0 g + Mg
3
3
= 2 800 g = 2.80 kg
422
*P14.30
Fluid Mechanics
(a)
The weight of the ball must be equal to the buoyant force of the water:
4 3
πrouter g
3
1.26 kgg = ρ water
router
(b)
F 3 × 1.26 kg I
=G
H 4π 1 000 kg m JK
13
3
= 6.70 cm
The mass of the ball is determined by the density of aluminum:
FG 4 πr
H3
IJ
K
FG IJ ea
H K
4
− πri3
3
4
1.26 kg = 2 700 kg m3
π 0.067 m
3
m = ρ Al V = ρ Al
3
0
f
3
− ri3
j
1.11 × 10 −4 m 3 = 3.01 × 10 −4 m 3 − ri3
e
ri = 1.89 × 10 −4 m 3
*P14.31
j
13
= 5.74 cm
Let A represent the horizontal cross-sectional area of the rod, which we presume to be constant. The
rod is in equilibrium:
∑ Fy = 0 :
− mg + B = 0 = − ρ 0 Vwhole rod g + ρ fluidVimmersed g
a
f
ρ 0 ALg = ρA L − h g
The density of the liquid is
*P14.32
ρ=
ρ 0L
.
L−h
We use the result of Problem 14.31. For the rod floating in a liquid of density 0.98 g cm 3 ,
ρ = ρ0
L
L−h
0.98 g cm3 =
a
3
ρ 0L
L − 0.2 cm
f
e
j
0.98 g cm L − 0.98 g cm 3 0.2 cm = ρ 0 L
For floating in the dense liquid,
ρ 0L
1.14 g cm 3 =
1.14 g cm 3
(a)
By substitution,
aL − 1.8 cmf
− e1.14 g cm j1.8 cm = ρ L
3
a
0
f
a f
1.14L − 1.14 1.8 cm = 0.98L − 0.2 0.98
0.16L = 1.856 cm
L = 11.6 cm
(b)
Substituting back,
a
ρ 0 = 0.963 g cm
(c)
f
0.98 g cm3 11.6 cm − 0.2 cm = ρ 0 11.6 cm
3
ρ 0L
is not of the form ρ = a + bh , equal-size
L−h
steps of ρ do not correspond to equal-size steps of h.
The marks are not equally spaced. Because ρ =
Chapter 14
P14.33
P14.34
The balloon stops rising when
bρ
Therefore,
V=
air
g
− ρ He gV = Mg
and
400
M
=
ρ air − ρ He 1.25 e −1 − 0.180
bρ
air
423
g
− ρ He V = M ,
V = 1 430 m3
Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced
water = weight of the frog, so
ρ oozeVg = m frog g
or
m frog = ρ oozeV = ρ ooze
FG
H
IJ e
K
1 4 3
2π
πr = 1.35 × 10 3 kg m3
6.00 × 10 −2 m
2 3
3
j e
j
3
Hence, m frog = 0.611 kg .
P14.35
B = Fg
V
= ρ sphere gV
2
1
ρ sphere = ρ H 2O = 500 kg m3
2
4
ρ glycerin g
V − ρ sphere gV = 0
10
10
ρ glycerin =
500 kg m3 = 1 250 kg m3
4
ρ H 2O g
FG
H
IJ
K
e
P14.36
FIG. P14.35
j
Constant velocity implies zero acceleration, which means that the submersible is in equilibrium
under the gravitational force, the upward buoyant force, and the upward resistance force:
∑ Fy = ma y = 0
e
j
− 1. 20 × 10 4 kg + m g + ρ w gV + 1 100 N = 0
where m is the mass of the added water and V is the sphere’s volume.
1.20 × 10 4 kg + m = 1.03 × 10 3
so
P14.37
LM 4 π a1.50f OP + 1 100 N
N3
Q 9.8 m s
3
2
m = 2.67 × 10 3 kg
By Archimedes’s principle, the weight of the fifty planes is equal to the weight of a horizontal slice of
water 11.0 cm thick and circumscribed by the water line:
a f
50e 2.90 × 10 kg j g = e1 030 kg m j g a0.110 mf A
∆B = ρ water g ∆V
4
3
giving A = 1.28 × 10 4 m 2 . The acceleration of gravity does not affect the answer.
424
Fluid Mechanics
Section 14.5
Fluid Dynamics
Section 14.6
Bernoulli’s Equation
P14.38
By Bernoulli’s equation,
b
b
g
g
dm
= ρAv = 1 000π 5.00 × 10 −2
dt
e
P14.39
b
g
1
1
1 000 v 2 = 6.00 × 10 4 N m 2 + 1 000 16 v 2
2
2
1
2.00 × 10 4 N m 2 = 1 000 15 v 2
2
v = 1.63 m s
8.00 × 10 4 N m 2 +
jb
2
FIG. P14.38
g
1.63 m s = 12.8 kg s
Assuming the top is open to the atmosphere, then
P1 = P0 .
Note
P2 = P0 .
Flow rate = 2.50 × 10 −3 m3 min = 4.17 × 10 −5 m3 s .
(a)
so
A1 >> A 2
Assuming v1 = 0 ,
v1 << v 2
P1 +
ρv12
ρv 2
+ ρgy1 = P2 + 2 + ρgy 2
2
2
b
v 2 = 2 gy1
(b)
Flow rate = A 2 v 2 =
F πd
GH 4
2
g
12
a fa f
= 2 9.80 16.0
I a17.7f = 4.17 × 10
JK
−5
12
= 17.7 m s
m3 s
d = 1.73 × 10 −3 m = 1.73 mm
*P14.40
Take point 1 at the free surface of the water in the tank and 2 inside the nozzle. F
Fair
water
1
1
(a)
With the cork in place P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22 becomes
2
2
f
P0 + 1 000 kg m3 9.8 m s 2 7.5 m + 0 = P2 + 0 + 0 ; P2 − P0 = 7.35 × 10 4 Pa .
For the stopper ∑ Fx = 0
FIG. P14.40
Fwater − Fair − f = 0
P2 A − P0 A = f
a
f
f = 7.35 × 10 4 Paπ 0.011 m
(b)
2
= 27.9 N
Now Bernoulli’s equation gives
P0 + 7.35 × 10 4 Pa + 0 = P0 + 0 +
v 2 = 12.1 m s
1
1 000 kg m 3 v 22
2
e
j
The quantity leaving the nozzle in 2 h is
e
ja
f b12.1 m sg7 200 s =
ρV = ρAv 2 t = 1 000 kg m3 π 0.011 m
continued on next page
2
3.32 × 10 4 kg .
Chapter 14
(c)
Take point 1 in the wide hose and 2 just outside the nozzle. Continuity:
A1 v 1 = A 2 v 2
π
FG 6.6 cm IJ
H 2 K
2
v1 = π
FG 2.2 cm IJ
H 2 K
2
12.1 m s
12.1 m s
= 1.35 m s
9
1
1
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22
2
2
1
1
2
P1 + 0 + 1 000 kg m3 1.35 m s = P0 + 0 + 1 000 kg m 3 12.1 m s
2
2
v1 =
jb
e
g
jb
e
P1 − P0 = 7.35 × 10 4 Pa − 9.07 × 10 2 Pa = 7.26 × 10 4 Pa
P14.41
Flow rate Q = 0.012 0 m3 s = v1 A1 = v 2 A 2
v2 =
*P14.42
*P14.43
(a)
P=
(b)
PEL
FG IJ
H K
= 0.85e8.5 × 10 ja9.8 fa87f =
Q 0.012 0
=
= 31.6 m s
A2
A2
∆E ∆mgh
∆m
=
=
gh = Rgh
∆t
∆t
∆t
5
616 MW
The volume flow rate is
FG
H
125 cm3
0.96 cm
= Av1 = π
16.3 s
2
IJ
K
2
v1 .
The speed at the top of the falling column is
v1 =
7.67 cm 3 s
0.724 cm 2
= 10.6 cm s .
Take point 2 at 13 cm below:
P1 + ρgy1 +
1 2
1
ρv1 = P2 + ρgy 2 + ρv 22
2
2
e
je
= P0 + 0 +
e
j
1
1 000 kg m3 v 22
2
e
e
j
b
v 2 = 2 9.8 m s 2 0.13 m + 0.106 m s
g
2
= 1.60 m s
The volume flow rate is constant:
7.67 cm 3 s = π
d = 0. 247 cm
FG d IJ
H 2K
jb
1
1 000 kg m 3 0.106 m s
2
j
P0 + 1 000 kg m 3 9.8 m s 2 0.13 m +
2
160 cm s
g
2
g
2
425
426
*P14.44
Fluid Mechanics
(a)
P1 +
Between sea surface and clogged hole:
e
ja f
je
1 atm + 0 + 1 030 kg m3 9.8 m s 2 2 m = P2 + 0 + 0
1 2
1
ρv1 + ρgy1 = P2 + ρv 22 + ρgy 2
2
2
P2 = 1 atm + 20.2 kPa
The air on the back of his hand pushes opposite the water, so the net force on his hand is
e
F = PA = 20. 2 × 10 3 N m 2
(b)
jFGH π4 IJK e1.2 × 10
2
F = 2.28 N
1
1 030 kg m 3 v 22 + 0
2
e
j
v 2 = 6. 26 m s
π
1.2 × 10 −2 m
4
j b6.26 m sg = 7.08 × 10
e
2
The volume rate of flow is
A2 v 2 =
One acre–foot is
4 047 m 2 × 0.304 8 m = 1 234 m3
1 234 m 3
Requiring
(a)
j
m
Now, Bernoulli’s theorem is
1 atm + 0 + 20. 2 kPa = 1 atm +
P14.45
−2
7.08 × 10
−4
3
m s
−4
= 1.74 × 10 6 s = 20.2 days
FG P + 1 ρv + ρgyIJ = FG P + 1 ρv + ρgyIJ
K H 2
K
H 2
P + 0 + ρg a564 mf = 1 atm + 0 + ρg b 2 096 mg
P = 1 atm + e1 000 kg m je9.8 m s jb1 532 mg = 1 atm + 15.0 MPa
2
Suppose the flow is very slow:
2
river
3
(b)
e
rim
2
4 500 m3 d = Av =
The volume flow rate is
v = 4 500 m3 d
(c)
m3 s
πd 2 v
4
I
F
jFGH 861400d s IJK GH π a0.1504 mf JK =
2
2.95 m s
Imagine the pressure as applied to stationary water at the bottom of the pipe:
FG P + 1 ρv
H 2
IJ = FG P + 1 ρv + ρgyIJ
K
H 2
K
1
P + 0 = 1 atm + e1 000 kg m jb 2.95 m sg + 1 000 kg e9.8 m s jb1 532 mg
2
2
2
+ ρgy
bottom
top
3
P = 1 atm + 15.0 MPa + 4.34 kPa
The additional pressure is 4.34 kPa .
2
2
Chapter 14
P14.46
(a)
427
For upward flight of a water-drop projectile from geyser vent to fountain–top,
2
v yf2 = v yi
+ 2 a y ∆y
ja
e
(b)
f
Then 0 = vi2 + 2 −9.80 m s 2 +40.0 m and
vi = 28.0 m s
Between geyser vent and fountain–top:
P1 +
Air is so low in density that very nearly
P1 = P2 = 1 atm
Then,
1 2
vi + 0 = 0 + 9.80 m s 2 40.0 m
2
1 2
1
ρv1 + ρgy1 = P2 + ρv 22 + ρgy 2
2
2
ja
e
f
v1 = 28.0 m s
(c)
P1 +
Between the chamber and the fountain-top:
f
je
ja
P − P = e1 000 kg m je9.80 m s ja 215 mf =
e
1 2
1
ρv1 + ρgy1 = P2 + ρv 22 + ρgy 2
2
2
e
ja
je
f
P1 + 0 + 1 000 kg m 3 9.80 m s 2 −175 m = P0 + 0 + 1 000 kg m 3 9.80 m s 2 +40.0 m
1
P14.47
P1 +
3
0
2
2.11 MPa
ρv12
ρ2
A
= P2 + 2 (Bernoulli equation), v1 A1 = v 2 A 2 where 1 = 4
2
2
A2
∆P = P1 − P2 =
F
GH
I
JK
A2
ρ 2
ρ
ρv 2
v 2 − v12 = v12 12 − 1 and ∆P = 1 15 = 21 000 Pa
2
2
2
A2
e
j
v1 = 2.00 m s ; v 2 = 4v1 = 8.00 m s :
The volume flow rate is
Section 14.7
P14.48
P14.49
v1 A1 = 2.51 × 10 −3 m3 s
Other Applications of Fluid Dynamics
a f
Mg = P1 − P2 A
for a balanced condition
16 000 9.80
= 7.00 × 10 4 − P2
A
where
A = 80.0 m 2
∴ P2 = 7.0 × 10 4 − 0.196 × 10 4 = 6.80 × 10 4 Pa
b
ρ air
v=
g
v2
= ∆P = ρ Hg g∆h
2
2 ρ Hg g∆h
ρ air
= 103 m s
v air
A
∆h
Mercury
FIG. P14.49
428
P14.50
Fluid Mechanics
The assumption of incompressibility is surely unrealistic, but allows an estimate of the speed:
1 2
1
ρv1 = P2 + ρgy 2 + ρv 22
2
2
1
1.00 atm + 0 + 0 = 0.287 atm + 0 + 1.20 kg m 3 v 22
2
P1 + ρgy1 +
e
v2 =
P14.51
(a)
P0 + ρgh + 0 = P0 + 0 +
a
fe
2 1.00 − 0.287 1.013 × 10 5 N m 2
1.20 kg m
1 2
ρv 3
2
P + ρgy +
j=
347 m s
v 3 = 4.43 m s
1 2
1
ρv 2 = P0 + 0 + ρv 32
2
2
P = P0 − ρgy
Since v 2 = v 3 ,
y≤
Since P ≥ 0
*P14.52
3
v 3 = 2 gh
If h = 1.00 m ,
(b)
j
FIG. P14.51
5
P0
1.013 × 10 Pa
=
= 10.3 m
3
ρg 10 kg m3 9.8 m s 2
e
je
j
Take points 1 and 2 in the air just inside and outside the window pane.
1 2
1
ρv1 + ρgy1 = P2 + ρv 22 + ρgy 2
2
2
1
P0 + 0 = P2 + 1.30 kg m3 11.2 m s
2
P1 +
jb
e
(a)
g
2
P2 = P0 − 81.5 Pa
The total force exerted by the air is outward,
e
ja fa
f
jb
g a4 mfa1.5 mf =
P1 A − P2 A = P0 A − P0 A + 81.5 N m 2 4 m 1.5 m = 489 N outward
(b)
P14.53
P1 A − P2 A =
1 2
1
ρv 2 A = 1.30 kg m 3 22.4 m s
2
2
e
∆P =
From the equation of continuity:
A1 v1 = A 2 v 2
−5
j e
j
m 2 v1 = 1.00 × 10 −8 m 2 v 2
1.96 kN outward
2.00 N
= 8.00 × 10 4 Pa
2.50 × 10 −5 m 2
In the reservoir, the gauge pressure is
e2.50 × 10
2
e
j
v1 = 4.00 × 10 −4 v 2
Thus, v12 is negligible in comparison to v 22 .
Then, from Bernoulli’s equation:
bP − P g + 12 ρv
1
2
2
1
+ ρgy1 =
8.00 × 10 4 Pa + 0 + 0 = 0 +
v2 =
e
2 8.00 × 10 4 Pa
1 000 kg m
3
j=
1 2
ρv 2 + ρgy 2
2
1
1 000 kg m3 v 22
2
e
12.6 m s
j
Chapter 14
Additional Problems
P14.54
Consider the diagram and apply Bernoulli’s
A
equation to points A and B, taking y = 0 at
the level of point B, and recognizing that v A
is approximately zero. This gives:
af
a
af
1
2
ρ w 0 + ρ w g h − L sin θ
2
1
= PB + ρ w v B2 + ρ w g 0
2
PA +
h
f
L
Valve
B
θ
Now, recognize that PA = PB = Patmosphere
since both points are open to the atmosphere
(neglecting variation of atmospheric
pressure with altitude). Thus, we obtain
a
f
FIG. P14.54
e
a
j
f
v B = 2 g h − L sin θ = 2 9.80 m s 2 10.0 m − 2.00 m sin 30.0°
v B = 13.3 m s
Now the problem reduces to one of projectile motion with v yi = v B sin 30.0° = 6.64 m s . Then,
b g
2
v yf2 = v yi
+ 2 a ∆y gives at the top of the arc (where y = y max and v yf = 0 )
b
g + 2e−9.80 m s jby
2.25 m babove the level where the water emergesg .
0 = 6.64 m s
or y max =
P14.55
2
2
max
−0
g
When the balloon comes into equilibrium, we must have
∑ Fy = B − Fg , balloon − Fg , He − Fg , string = 0
He
Fg , string is the weight of the string above the ground, and B
is the buoyant force. Now
Fg , balloon = m balloon g
h
Fg , He = ρ HeVg
B = ρ air Vg
and
Fg , string = m string
h
g
L
FIG. P14.55
Therefore, we have
ρ air Vg − m balloon g − ρ HeVg − m string
or
h=
bρ
air
g
− ρ He V − m balloon
m string
h
g=0
L
L
giving,
a1.29 − 0.179fekg m jFH
3
h=
a
4π 0. 400 m
3
0.050 0 kg
f
3
I − 0.250 kg
K
a2.00 mf =
1.91 m .
429
430
P14.56
Fluid Mechanics
Assume v inside ≈ 0
Pgauge
P14.57
a fa
ga f
b
f
1
2
1 000 30.0 + 1 000 9.80 0.500
2
= P − 1 atm = 4.50 × 10 5 + 4.90 × 10 3 = 455 kPa
P + 0 + 0 = 1 atm +
The “balanced” condition is one in which the apparent weight of the
body equals the apparent weight of the weights. This condition can be
written as:
Fg − B = Fg′ − B ′
where B and B ′ are the buoyant forces on the body and weights
respectively. The buoyant force experienced by an object of volume V
in air equals:
b
FIG. P14.57
g
Buoyant force = Volume of object ρ air g
P14.58
so we have
B = Vρ air g
Therefore,
Fg = Fg′ + V −
F
GH
B′ =
and
Iρ
ρg JK
Fg′
F F′ I ρ
GH ρg JK
g
air g .
air g .
The cross–sectional area above water is
a
f − a0.200 cmfa0.566 cmf = 0.330 cm
2.46 rad
π 0.600 cm
2π
a
Aall = π 0.600
0.400 cm
f
2
2
2
0.80 cm
= 1.13 cm 2
ρ water gAunder = ρ wood Aall g
1.13 − 0.330
= 0.709 g cm 3 = 709 kg m3
ρ wood =
1.13
P14.59
P14.60
At equilibrium,
∑ Fy = 0 :
FIG. P14.58
B − Fspring − Fg , He − Fg , balloon = 0
b
g
giving
Fspring = kL = B − m He + m balloon g .
But
B = weight of displaced air = ρ air Vg
and
m He = ρ HeV .
Therefore, we have:
kL = ρ air Vg − ρ HeVg − m balloon g
or
L=
From the data given,
L=
Thus, this gives
L = 0.604 m .
bρ
air
g
− ρ He V − m balloon
k
e1.29 kg m
3
FIG. P14.59
g.
j
− 0.180 kg m3 5.00 m3 − 2.00 × 10 −3 kg
90.0 N m
a f
P = ρgh
1.013 × 10 5 = 1.29 9.80 h
h = 8.01 km
For Mt. Everest,
29 300 ft = 8.88 km
Yes
e9.80 m s j .
2
Chapter 14
P14.61
z z
z b
The torque is
τ = dτ = rdF
From the figure
τ = y ρg H − y wdy =
H
g
0
431
1
ρgwH 3
6
1
ρgwH 2
The total force is given as
2
If this were applied at a height y eff such that the torque remains
unchanged, we have
LM
N
1
1
ρgwH 3 = y eff ρgwH 2
6
2
P14.62
(a)
(b)
P14.63
OP
Q
y eff =
and
1
H .
3
FIG. P14.61
The pressure on the surface of the two hemispheres is constant
at all points, and the force on each element of surface area is
directed along the radius of the hemispheres. The applied force
along the axis must balance the force on the “effective” area,
which is the projection of the actual surface onto a plane
perpendicular to the x axis,
A = πR 2
bP − PgπR
2
Therefore,
F=
For the values given
F = P0 − 0.100 P0 π 0.300 m
0
g a
b
f
2
FIG. P14.62
= 0.254P0 = 2.58 × 10 4 N
Looking first at the top scale and the iron block, we have:
T1 + B = Fg , iron
where T1 is the tension in the spring scale, B is the buoyant force, and Fg , iron is the weight of the iron
block. Now if m iron is the mass of the iron block, we have
m iron = ρ ironV
V=
so
m iron
ρ iron
= Vdisplaced oil
Then, B = ρ oil Viron g
Therefore, T1 = Fg , iron − ρ oilViron g = m iron g − ρ oil
or
FG
H
T1 = 1 −
IJ
K
F
GH
m iron
ρ iron
g
I a fa f
JK
ρ oil
916
m iron g = 1 −
2.00 9.80 = 17.3 N
7 860
ρ iron
Next, we look at the bottom scale which reads T2 (i.e., exerts an upward force T2 on the system).
Consider the external vertical forces acting on the beaker–oil–iron combination.
∑ Fy = 0 gives
T1 + T2 − Fg , beaker − Fg , oil − Fg , iron = 0
or
b
g
b
ge
j
T2 = m beaker + m oil + m iron g − T1 = 5.00 kg 9.80 m s 2 − 17.3 N
Thus, T2 = 31.7 N is the lower scale reading.
432
P14.64
Fluid Mechanics
Looking at the top scale and the iron block:
FG m IJ g
Hρ K
is the buoyant force exerted on the iron block by the oil.
F m IJ g
−B=m g−ρ G
Thus,
T =F
Hρ K
F ρ IJ m g is the reading on the top scale.
or
T = G1 −
H ρ K
T1 + B = Fg , Fe
1
Fe
B = ρ 0 VFe g = ρ 0
where
g , Fe
Fe
0
1
Fe
Fe
0
Fe
Fe
Fe
Now, consider the bottom scale, which exerts an upward force of T2 on the beaker–oil–iron
combination.
T1 + T2 − Fg , beaker − Fg , oil − Fg , Fe = 0
∑ Fy = 0 :
g FGH
b
T2 = Fg , beaker + Fg , oil + Fg , Fe − T1 = m b + m 0 + m Fe g − 1 −
P14.65
LMm
MN
T2 =
or
b
+ m0 +
FG ρ IJ m OP g
H ρ K PQ
0
is the reading on the bottom scale.
Fe
Fe
IJ
K
ρ0
m Fe g
ρ Fe
ρ CuV = 3.083 g
ρ Zn xV + ρ Cu 1 − x V = 2.517 g
a f a f
F 3.083 IJ x + 3.083a1 − xf = 2.517
ρ G
Hρ K
FG 1 − 7.133 IJ x = FG 1 − 2.517 IJ
H 8.960 K H 3.083 K
Zn
Cu
x = 0.900 4
%Zn = 90.04%
P14.66
(a)
From
∑ F = ma
b
g
B − m shell g − m He g = m total a = m shell + m He a
Where
B = ρ water Vg
and
πd 3
4
Also,
V = πr 3 =
3
6
Putting these into equation (1) above,
Fm
GH
shell
+ ρ He
I F
JK GH
bρ
a=
(b)
t=
m He = ρ HeV
I
JK
πd 3
πd 3
πd 3
− m shell − ρ He
a = ρ water
g
6
6
6
which gives
or
(1)
water
− ρ He
g
πd 3
6
− m shell
3
m shell + ρ He πd6
g
a
f − 4.00 kg
1 000 − 0.180 ge kg m j
b
a=
9.80 m s
a
f
4.00 kg + e0.180 kg m j
2a h − d f
2a 4.00 m − 0.200 mf
2x
=
=
= 4.06 s
3 π 0. 200 m
6
3
3 π 0. 200 m
6
a
a
0.461 m s 2
3
2
= 0.461 m s 2
Chapter 14
P14.67
Inertia of the disk: I =
b
f
ga
1
1
MR 2 = 10.0 kg 0.250 m
2
2
2
= 0.312 kg ⋅ m 2
Angular acceleration: ω f = ω i + αt
α=
Braking torque:
FG 0 − 300 rev min IJ FG 2π rad IJ FG 1 min IJ = −0.524 rad s
H 60.0 s K H 1 rev K H 60.0 s K
∑ τ = Iα ⇒ − fd = Iα , so
f=
2
− Iα
d
e0.312 kg ⋅ m je0.524 rad s j = 0.744 N
Friction force: f =
2
2
0.220 m
Normal force: f = µ k n ⇒ n =
gauge pressure: P =
P14.68
f
0.744 N
=
= 1.49 N
µk
0.500
1.49 N
n
=
A π 2.50 × 10 −2 m
e
j
2
= 758 Pa
The incremental version of P − P0 = ρgy is
dP = − ρgdy
We assume that the density of air is proportional to pressure, or
P
=
ρ
P0
ρ0
dP = − P
Combining these two equations we have
z
P
P0
ρ0
gdy
P0
z
h
ρ
dP
= − g 0 dy
P
P0 0
FG P IJ = − ρ gh
HP K P
and integrating gives
0
ln
0
so where α =
P14.69
ρ0 g
,
P0
P = P0 e −αh
Energy for the fluid-Earth system is conserved.
aK + Uf + ∆E
i
mech
a
f
= K +U f :
0+
mgL
1
+ 0 = mv 2 + 0
2
2
e
j
v = gL = 2.00 m 9.8 m s 2 = 4.43 m s
0
433
434
P14.70
Fluid Mechanics
Let s stand for the edge of the cube, h for the depth of immersion, ρ ice stand for the density of the
ice, ρ w stand for density of water, and ρ a stand for density of the alcohol.
(a)
According to Archimedes’s principle, at equilibrium we have
ρ ice gs 3 = ρ w ghs 2 ⇒ h = s
With
ρ ice
ρw
ρ ice = 0.917 × 10 3 kg m3
ρ w = 1.00 × 10 3 kg m3
(b)
and
s = 20.0 mm
we get
h = 20.0 0.917 = 18.34 mm ≈ 18.3 mm
a
f
We assume that the top of the cube is still above the alcohol surface. Letting ha stand for the
thickness of the alcohol layer, we have
ρ a gs 2 ha + ρ w gs 2 hw = ρ ice gs 3
(c)
so
hw =
With
ρ a = 0.806 × 10 3 kg m3
and
ha = 5.00 mm
we obtain
hw = 18.34 − 0.806 5.00 = 14.31 mm ≈ 14.3 mm
Here
hw′ = s − ha′ , so Archimedes’s principle gives
FG ρ IJ s − FG ρ IJ h
H ρ K Hρ K
ice
a
w
w
a f
b
g
b
g
ρ a gs 2 ha′ + ρ w gs 2 s − ha′ = ρ ice gs 3 ⇒ ρ a ha′ + ρ w s − ha′ = ρ ice s
ha′ = s
bρ
bρ
w
w
g = 20.0 a1.000 − 0.917f = 8.557 ≈
−ρ g
a1.000 − 0.806f
− ρ ice
a
8.56 mm
a
Chapter 14
P14.71
435
Note: Variation of atmospheric pressure with altitude is included in
this solution. Because of the small distances involved, this effect is
unimportant in the final answers.
(a)
Consider the pressure at points A and B in part (b) of the
figure:
a
f
Using the left tube: PA = Patm + ρ a gh + ρ w g L − h where the
second term is due to the variation of air pressure with
altitude.
Using the right tube: PB = Patm + ρ 0 gL
But Pascal’s principle says that PA = PB .
f
Patm + ρ 0 gL = Patm + ρ a gh + ρ w g L − h
or
bρ
h=
(b)
a
Therefore,
FG ρ
Hρ
w
g b
g
− ρ a h = ρ w − ρ 0 L , giving
IJ F
K GH
I
JK
1 000 − 750
− ρ0
L=
5.00 cm = 1.25 cm
−
1
000 − 1.29
ρ
w
a
w
Consider part (c) of the diagram showing the situation
when the air flow over the left tube equalizes the fluid
levels in the two tubes. First, apply Bernoulli’s equation to
points A and B y A = y B , v A = v , and v B = 0
b
This gives: PA +
1
ρ a v 2 + ρ a gy A
2
g
1
= P + ρ a0 f
2
B
a
and since y A = y B , this reduces to: PB − PA =
2
+ ρ a gy B
1
ρav2
2
FIG. P14.71
(1)
Now consider points C and D, both at the level of the
oil–water interface in the right tube. Using the variation of
pressure with depth in static fluids, we have:
PC = PA + ρ a gH + ρ w gL
and
PD = PB + ρ a gH + ρ 0 gL
But Pascal’s principle says that PC = PD . Equating these two gives:
PB + ρ a gH + ρ 0 gL = PA + ρ a gH + ρ w gL
or
Substitute equation (1) for PB − PA into (2) to obtain
or
v=
b
2 gL ρ w − ρ 0
ρa
v = 13.8 m s
g
b
g
1
ρ a v 2 = ρ w − ρ 0 gL
2
g = 2e9.80 m s jb0.050 0 mgFG 1 000 − 750 IJ
H 1.29 K
2
b
PB − PA = ρ w − ρ 0 gL
(2)
436
P14.72
Fluid Mechanics
(a)
The flow rate, Av, as given may be expressed as follows:
25.0 liters
= 0.833 liters s = 833 cm3 s .
30.0 s
The area of the faucet tap is π cm 2 , so we can find the velocity as
v=
(b)
flow rate 833 cm 3 s
=
= 265 cm s = 2.65 m s .
A
π cm 2
We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. A1 v1 = A 2 v 2
gives v1 = 0.295 m s . Bernoulli’s equation is:
P1 − P2 =
j b
1
ρ v 22 − v12 + ρg y 2 − y1
2
e
g
and gives
P1 − P2 =
or
P14.73
(a)
1
10 3 kg m3
2
e
j b2.65 m sg − b0.295 m sg + e10
2
2
3
je
ja
f
kg m3 9.80 m s 2 2.00 m
Pgauge = P1 − P2 = 2.31 × 10 4 Pa .
Since the upward buoyant force is balanced by the weight of the sphere,
m1 g = ρVg = ρ
FG 4 πR IJ g .
H3 K
3
In this problem, ρ = 0.789 45 g cm 3 at 20.0°C, and R = 1.00 cm so we find:
m1 = ρ
(b)
FG 4 πR IJ = e0.789 45 g cm jLM 4 π a1.00 cmf OP =
H3 K
N3
Q
3
3
3.307 g .
Following the same procedure as in part (a), with ρ ′ = 0.780 97 g cm 3 at 30.0°C, we find:
m2 = ρ ′
(c)
3
FG 4 πR IJ = e0.780 97 g cm jLM 4 π a1.00 cmf OP =
H3 K
N3
Q
3
3
3
3.271 g .
When the first sphere is resting on the bottom of the tube,
n + B = Fg 1 = m1 g , where n is the normal force.
Since B = ρ ′Vg
e
ja
f
n = m1 g − ρ ′Vg = 3.307 g − 0.780 97 g cm 3 1.00 cm
n = 34.8 g ⋅ cm s 2 = 3.48 × 10 −4 N
3
980 cm s 2
Chapter 14
*P14.74
(a)
437
Take point 1 at the free water surface in the tank and point 2 at the bottom end of the tube:
1 2
1
ρv1 = P2 + ρgy 2 + ρv 22
2
2
1 2
P0 + ρgd + 0 = P0 + 0 + ρv 2
2
v 2 = 2 gd
P1 + ρgy1 +
The volume flow rate is
V Ah
Ah
Ah
.
=
= v 2 A ′ . Then t =
=
t
t
v 2 A ′ A ′ 2 gd
a0.5 mf 0.5 m
=
m 2e9.8 m s j10 m
2
*P14.75
(b)
t=
(a)
For diverging stream lines that pass just above and just below the hydrofoil we have
2 × 10 −4
2
2
44.6 s
Pt + ρgy t +
1 2
1
ρv t = Pb + ρgy b + ρv b2 .
2
2
Ignoring the buoyant force means taking y t ≈ y b
b g
1
1
2
ρ nv b = Pb + ρv b2
2
2
1 2 2
Pb − Pt = ρv b n − 1
2
Pt +
e
b
g
The lift force is Pb − Pt A =
(b)
j
1 2 2
ρv b n − 1 A .
2
e
j
For liftoff,
1 2 2
ρv b n − 1 A = Mg
2
e
vb
j
F 2Mg I
=G
GH ρen − 1jA JJK
12
2
The speed of the boat relative to the shore must be nearly equal to this speed of the water
below the hydrofoil relative to the boat.
(c)
e
j
v 2 n 2 − 1 Aρ = 2 Mg
A=
b g
b9.5 m sg e1.05 − 1j1 000 kg m
2 800 kg 9.8 m s 2
2
2
3
= 1.70 m 2
438
Fluid Mechanics
ANSWERS TO EVEN PROBLEMS
P14.2
~ 10 18 kg m3 ; matter is mostly empty
space
P14.4
1.92 × 10 4 N
P14.6
(a) 1.01 × 10 7 Pa ;
(b)7.09 × 10 5 N outward
P14.8
255 N
P14.10
(a) 65.1 N; (b) 275 N
P14.12
P14.38
12.8 kg s
P14.40
(a) 27.9 N; (b) 3.32 × 10 4 kg ;
(c) 7.26 × 10 4 Pa
6
5.88 × 10 N down; 196 kN outward;
588 kN outward
P14.42
(a) see the solution; (b) 616 MW
P14.44
(a) 2.28 N toward Holland; (b) 1.74 × 10 6 s
P14.46
(a), (b) 28.0 m s ; (c) 2.11 MPa
P14.48
6.80 × 10 4 Pa
P14.50
347 m s
P14.52
(a) 489 N outward; (b) 1.96 kN outward
P14.54
2.25 m above the level where the water
emerges
P14.14
(a) 29.4 kN to the right;
(b) 16.3 kN ⋅ m counterclockwise
P14.16
(a) 10.3 m; (b) zero
P14.18
(a) 20.0 cm; (b) 0.490 cm
P14.56
455 kPa
P14.20
12.6 cm
P14.58
709 kg m3
P14.22
(a) 444 kg; (b) 480 kg
P14.60
8.01 km; yes
P14.62
(a) see the solution; (b) 2.58 × 10 4 N
P14.64
top scale: 1 −
m
ρw − ρs h
P14.24
b
P14.26
(a) see the solution; (b) 25.0 N up;
(c) horizontally inward;
(d) tension increases; see the solution;
(e) 62.5%; (f) 18.7%
g
P14.28
~ 10 4 balloons of 25-cm diameter
P14.30
(a) 6.70 cm; (b) 5.74 cm
FG
H
IJ
K
ρ0
m Fe g ;
ρ Fe
FG
H
bottom scale: m b + m 0 +
P14.66
(a) 0.461 m s 2 ; (b) 4.06 s
P14.68
see the solution
IJ
K
ρ 0 m Fe
g
ρ Fe
3
P14.32
(a) 11.6 cm; (b) 0.963 g cm ;
(c) no; see the solution
P14.70
(a) 18.3 mm; (b) 14.3 mm; (c) 8.56 mm
P14.34
0.611 kg
P14.72
(a) 2.65 m s ; (b) 2.31 × 10 4 Pa
P14.36
2.67 × 10 3 kg
P14.74
(a) see the solution; (b) 44.6 s
15
Oscillatory Motion
CHAPTER OUTLINE
15.1
15.2
15.3
15.4
15.5
15.6
15.7
Q15.4
Motion of an Object
Attached to a Spring
Mathematical Representation
of Simple Harmonic Motion
Energy of the Simple
Harmonic Oscillator
Comparing Simple Harmonic
Motion with Uniform Circular
Motion
The Pendulum
Damped Oscillations
Forced Oscillations
ANSWERS TO QUESTIONS
Q15.1
Neither are examples of simple harmonic motion, although
they are both periodic motion. In neither case is the acceleration
proportional to the position. Neither motion is so smooth as
SHM. The ball’s acceleration is very large when it is in contact
with the floor, and the student’s when the dismissal bell rings.
Q15.2
You can take φ = π , or equally well, φ = −π . At t = 0 , the particle
is at its turning point on the negative side of equilibrium, at
x = −A .
Q15.3
The two will be equal if and only if the position of the particle
at time zero is its equilibrium position, which we choose as the
origin of coordinates.
(a)
In simple harmonic motion, one-half of the time, the velocity is in the same direction as the
displacement away from equilibrium.
(b)
Velocity and acceleration are in the same direction half the time.
(c)
Acceleration is always opposite to the position vector, and never in the same direction.
Q15.5
No. It is necessary to know both the position and velocity at time zero.
Q15.6
The motion will still be simple harmonic motion, but the period of oscillation will be a bit larger. The
F kI
effective mass of the system in ω = G
H m JK
12
will need to include a certain fraction of the mass of the
eff
spring.
439
440
Q15.7
Oscillatory Motion
We assume that the coils of the spring do not hit one another. The frequency will be higher than f by
the factor 2 . When the spring with two blocks is set into oscillation in space, the coil in the center
of the spring does not move. We can imagine clamping the center coil in place without affecting the
motion. We can effectively duplicate the motion of each individual block in space by hanging a
single block on a half-spring here on Earth. The half-spring with its center coil clamped—or its other
half cut off—has twice the spring constant as the original uncut spring, because an applied force of
the same size would produce only one-half the extension distance. Thus the oscillation frequency in
space is
FG 1 IJ FG 2 k IJ
H 2π K H m K
12
= 2 f . The absence of a force required to support the vibrating system in
orbital free fall has no effect on the frequency of its vibration.
Q15.8
No; Kinetic, Yes; Potential, No. For constant amplitude, the total energy
1 2
kA stays constant. The
2
1
mv 2 would increase for larger mass if the speed were constant, but here the greater
2
mass causes a decrease in frequency and in the average and maximum speed, so that the kinetic and
potential energies at every point are unchanged.
kinetic energy
Q15.9
Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2
are valid.
Equation
x t = A cos ωt + φ
Information given by equation
position as a function of time
v t = −ωA sin ωt + φ
velocity as a function of time
af
b g
af
b g
va x f = ±ω e A − x j
aat f = −ω A cosbωt + φ g
aat f = −ω xat f
2
2 12
velocity as a function of position
2
acceleration as a function of time
2
acceleration as a function of position
The angular frequency ω appears in every equation. It is a good idea to figure out the value of angular
frequency early in the solution to a problem about vibration, and to store it in calculator memory.
Q15.10
Lf
Li
2 Li
and T f =
=
= 2Ti . The period gets larger by
g
g
g
mass has no effect on the period of a simple pendulum.
We have Ti =
2 times. Changing the
Q15.11
(a)
Q15.12
No, the equilibrium position of the pendulum will be shifted (angularly) towards the back of the car.
The period of oscillation will increase slightly, since the restoring force (in the reference frame of the
accelerating car) is reduced.
Q15.13
The motion will be periodic—that is, it will repeat. The period is nearly constant as the angular
amplitude increases through small values; then the period becomes noticeably larger as θ increases
farther.
Q15.14
Shorten the pendulum to decrease the period between ticks.
Q15.15
No. If the resistive force is greater than the restoring force of the spring (in particular, if b 2 > 4mk ),
the system will be overdamped and will not oscillate.
Period decreases.
(b)
Period increases.
(c)
No change.
Chapter 15
441
Q15.16
Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in
time. Without damping, the amplitude would increase without limit at resonance.
Q15.17
The phase constant must be π rad .
Q15.18
Higher frequency. When it supports your weight, the center of the diving board flexes down less
than the end does when it supports your weight. Thus the stiffness constant describing the center of
1
k
is greater
the board is greater than the stiffness constant describing the end. And then f =
2π m
for you bouncing on the center of the board.
FG IJ
H K
Q15.19
The release of air from one side of the parachute can make the parachute turn in the opposite
direction, causing it to release air from the opposite side. This behavior will result in a periodic driving
force that can set the parachute into side-to-side oscillation. If the amplitude becomes large enough,
the parachute will not supply the needed air resistance to slow the fall of the unfortunate skydiver.
Q15.20
An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the
wind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches the
natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven
into a large-amplitude resonance vibration. Note that it is not the size of the driving force that sets
the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf
will be set into resonant oscillation.
Q15.21
We assume the diameter of the bob is not very small compared to the length of the cord supporting
it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length
of the pendulum and slightly lowering its frequency. As the last drops of water dribble out, the
center of mass of the bob hops back up to the center of the sphere, and the pendulum frequency
quickly increases to its original value.
SOLUTIONS TO PROBLEMS
Section 15.1
P15.1
Motion of an Object Attached to a Spring
(a)
Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then
repeat the motion over and over again. Thus, the motion is periodic .
(b)
To determine the period, we use: x =
1 2
gt .
2
The time for the ball to hit the ground is t =
a
a
f
2 4.00 m
2x
=
= 0.909 s
g
9.80 m s 2
f
This equals one-half the period, so T = 2 0.909 s = 1.82 s .
(c)
No . The net force acting on the ball is a constant given by F = − mg (except when it is in
contact with the ground), which is not in the form of Hooke’s law.
442
Oscillatory Motion
Section 15.2
P15.2
P15.3
Mathematical Representation of Simple Harmonic Motion
π
6
IJ
K
x = 5.00 cm cos 2t +
(b)
v=
dx
π
= − 10.0 cm s sin 2t +
dt
6
(c)
a=
π
dv
= − 20.0 cm s 2 cos 2t +
dt
6
(d)
A = 5.00 cm
a
g FGH
b
IJ
K
j FGH
e
f a
IJ
K
f FGH π6 IJK =
a
At t = 0 ,
x = 5.00 cm cos
At t = 0 ,
v = −5.00 cm s
At t = 0 ,
a = −17.3 cm s 2
and
T=
f
b
2π
ω
=
4.33 cm
2π
= 3.14 s
2
g
x = 4.00 m cos 3.00πt + π Compare this with x = A cos ωt + φ to find
(a)
ω = 2π f = 3.00π
or
*P15.4
f FGH
a
(a)
T=
f = 1.50 Hz
1
= 0.667 s
f
(b)
A = 4.00 m
(c)
φ = π rad
(d)
x t = 0.250 s = 4.00 m cos 1.75π = 2.83 m
(a)
The spring constant of this spring is
a
f a
f a
f
k=
F 0.45 kg 9.8 m s 2
=
= 12.6 N m
x
0.35 m
we take the x-axis pointing downward, so φ = 0
x = A cos ωt = 18.0 cm cos
(d)
12.6 kg
0.45 kg ⋅ s 2
84. 4 s = 18.0 cm cos 446.6 rad = 15.8 cm
a f
Now 446.6 rad = 71 × 2π + 0.497 rad . In each cycle the object moves 4 18 = 72 cm , so it has
a
f a
f
moved 71 72 cm + 18 − 15.8 cm = 51.1 m .
(b)
By the same steps, k =
x = A cos
(e)
a f
0. 44 kg 9.8 m s 2
= 12.1 N m
0.355 m
k
12.1
t = 18.0 cm cos
84.4 = 18.0 cm cos 443.5 rad = −15.9 cm
m
0.44
443.5 rad = 70 2π + 3.62 rad
a
f
Distance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m
(c)
The answers to (d) and (e) are not very different given the difference in the data about the
two vibrating systems. But when we ask about details of the future, the imprecision in our
knowledge about the present makes it impossible to make precise predictions. The two
oscillations start out in phase but get completely out of phase.
Chapter 15
P15.5
(a)
At t = 0 , x = 0 and v is positive (to the right). Therefore, this situation corresponds to
x = A sin ωt
and
v = vi cos ωt
Since f = 1.50 Hz ,
ω = 2π f = 3.00π
a
a
f
v max = vi = Aω = 2.00 3.00π = 6.00π cm s = 18.8 cm s
The particle has this speed at t = 0 and next at
(c)
a
a max = Aω 2 = 2.00 3.00π
f
2
P15.6
t=
T
1
=
s
2
3
t=
3
T = 0.500 s
4
= 18.0π 2 cm s 2 = 178 cm s 2
This positive value of acceleration first occurs at
(d)
f
x = 2.00 cm sin 3.00π t
Also, A = 2.00 cm, so that
(b)
443
2
s and A = 2.00 cm, the particle will travel 8.00 cm in this time.
3
3
Hence, in 1.00 s = T , the particle will travel
8.00 cm + 4.00 cm = 12.0 cm .
2
Since T =
FG
H
IJ
K
af
FG v IJ sin ωt
HωK
i
The proposed solution
x t = xi cos ωt +
implies velocity
v=
dx
= − x iω sin ωt + vi cos ωt
dt
and acceleration
a=
v
dv
= − x iω 2 cos ωt − viω sin ωt = −ω 2 x i cos ωt + i sin ωt = −ω 2 x
ω
dt
FG
H
FG IJ
H K
IJ
K
(a)
The acceleration being a negative constant times position means we do have SHM, and its
angular frequency is ω. At t = 0 the equations reduce to x = xi and v = vi so they satisfy all
the requirements.
(b)
v 2 − ax = − x iω sin ωt + vi cos ωt
b
g − e− x ω
2
i
2
jFGH
cos ωt − vi sin ωt xi cos ωt +
FG v IJ sin ωtIJ
HωK K
i
v 2 − ax = xi2ω 2 sin 2 ωt − 2 xi viω sin ωt cos ωt + vi2 cos 2 ωt
+ x i2ω 2 cos 2 ωt + x i viω cos ωt sin ωt + xi viω sin ωt cos ωt + vi2 sin 2 ωt = x i2ω 2 + vi2
So this expression is constant in time. On one hand, it must keep its original value vi2 − ai xi .
On the other hand, if we evaluate it at a turning point where v = 0 and x = A , it is
A 2ω 2 + 0 2 = A 2ω 2 . Thus it is proved.
P15.7
(a)
T=
12.0 s
= 2.40 s
5
(b)
f=
1
1
=
= 0.417 Hz
T 2. 40
(c)
ω = 2π f = 2π 0.417 = 2.62 rad s
a
f
444
*P15.8
Oscillatory Motion
The mass of the cube is
ja
e
f
m = ρV = 2.7 × 10 3 kg m3 0.015 m
3
= 9.11 × 10 −3 kg
The spring constant of the strip of steel is
k=
f=
P15.9
f=
ω
1
=
π
2
2π
k
m
14.3 N
F
=
= 52.0 N m
x 0.027 5 m
ω
2π
k=
x = A cos ωt
52 kg
k
1
=
m 2π
1
2π
T=
or
Solving for k,
*P15.10
=
s 2 9.11 × 10 −3 kg
= 12.0 Hz
1
m
= 2π
f
k
4π 2 m
A = 0.05 m
T
2
=
b
4π 2 7.00 kg
a2.60 sf
2
g=
40.9 N m .
a = − Aω 2 cos ωt
v = − Aω sin ωt
If f = 3 600 rev min = 60 Hz , then ω = 120π s −1
a
f
v max = 0.05 120π m s = 18.8 m s
P15.11
(a)
ω=
k
=
m
8.00 N m
= 4.00 s −1
0.500 kg
From this we find that
(b)
t=
FG 1 IJ sin FG x IJ and when
H 4.00 K H 10.0 K
−1
Using t =
f
2
m s 2 = 7.11 km s 2
a
a f
a = −160 sina 4.00t f cm s
v = 40.0 cos 4.00t cm s
v max = 40.0 cm s
2
amax = 160 cm s 2 .
x = 6.00 cm, t = 0.161 s.
a f
a = −160 sin 4.00a0.161f =
FG 1 IJ sin FG x IJ
H 4.00 K H 10.0 K
−1
when x = 0 , t = 0 and when
x = 8.00 cm, t = 0.232 s.
Therefore,
∆t = 0.232 s .
f
x = 10.0 sin 4.00t cm .
so position is given by
v = 40.0 cos 4.00 0.161 = 32.0 cm s
We find
(c)
a
amax = 0.05 120π
−96.0 cm s 2 .
Chapter 15
P15.12
445
m = 1.00 kg , k = 25.0 N m, and A = 3.00 cm. At t = 0 , x = −3.00 cm
(a)
ω=
k
=
m
25.0
= 5.00 rad s
1.00
2π
2π
T=
=
= 1.26 s
ω 5.00
so that,
(b)
b
g
v max = Aω = 3.00 × 10 −2 m 5.00 rad s = 0.150 m s
b
amax = Aω 2 = 3.00 × 10 −2 m 5.00 rad s
(c)
g
2
= 0.750 m s 2
Because x = −3.00 cm and v = 0 at t = 0 , the required solution is x = − A cos ωt
a
f
x = −3.00 cos 5.00t cm
or
a
a
f
f
dx
= 15.0 sin 5.00t cm s
dt
dv
a=
= 75.0 cos 5.00t cm s 2
dt
v=
P15.13
The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.
ω=
b
2π
= 6. 28 s
T
f
ga
and v max = ωA = 6.28 s 0.100 m = 0.628 m s .
P15.14
(a)
v max = ωA
A=
(b)
Section 15.3
P15.15
(a)
v max
ω
=
v
ω
x = − A sin ωt = −
FG v IJ sin ωt
HωK
Energy of the Simple Harmonic Oscillator
Energy is conserved for the block-spring system between the maximum-displacement and
the half-maximum points:
aK + U f = aK + U f
1
b6.50 N mga0.100 mf
2
i
32.5 mJ =
k
=
m
0+
f
b
2
1
m 0.300 m s
2
b
=
1
m 0.300 m s
2
g
+ 8.12 mJ
2
6.50 N m
= 3.46 rad s
0.542 kg
(b)
ω=
(c)
amax = Aω 2 = 0.100 m 3.46 rad s
b
g
2
g
2
+
1 2 1
1
kA = mv 2 + kx 2
2
2
2
b
ge
1
6.50 N m 5.00 × 10 −2 m
2
m=
a
2 24.4 mJ
9.0 × 10
∴T =
= 1.20 m s 2
j
2π
ω
=
−2
f
2
m s2
2
= 0.542 kg
2π rad
= 1.81 s
3.46 rad s
446
P15.16
P15.17
Oscillatory Motion
b
P15.19
k = mω 2 = 0.200 kg 25.1 rad s
(b)
E=
kA 2
⇒A=
2
2E
=
k
2
126
Choose the car with its shock-absorbing bumper as the system; by conservation of energy,
k
= 3.16 × 10 −2 m
m
e
v=x
j
e
j
(a)
(b)
v max = Aω
(c)
a max = Aω 2 = 3.50 × 10 −2 m 22.4 s −1
(a)
E=
(b)
v = ω A2 − x2 =
e
b
j
2
ge
250
= 22.4 s −1
0.500
v max = 0.784 m s
= 17.5 m s 2
1 2 1
kA = 35.0 N m 4.00 × 10 −2 m
2
2
j
2
= 28.0 mJ
k
A2 − x2
m
35.0
e4.00 × 10 j − e1.00 × 10 j = 1.02 m s
50.0 × 10
1
1
1
1
mv = kA − kx = a35.0 fLe 4.00 × 10 j − e3.00 × 10 j O =
MN
PQ
2
2
2
2
v=
(c)
= 0.153 J
k
=
m
ω=
where
5.00 × 10 6
= 2.23 m s
10 3
2
−2
kA 2 250 N m 3.50 × 10 m
E=
=
2
2
.
P15.20
g = 126 N m
2a 2.00f
= 0.178 m
(a)
1
1
mv 2 = kx 2 :
2
2
P15.18
2π
2π
=
= 25.1 rad s
T
0. 250
m = 200 g , T = 0.250 s, E = 2.00 J ; ω =
−2 2
−3
2
2
−2 2
−2 2
2
(d)
1 2
1
kx = E − mv 2 = 15.8 mJ
2
2
(a)
k=
(b)
ω=
(c)
v max = ωA = 50.0 0. 200 = 1.41 m s at x = 0
(d)
amax = ω 2
(e)
E=
(f)
v = ω A 2 − x 2 = 50.0
(g)
a = ω 2 x = 50.0
−2 2
F
20.0 N
=
= 100 N m
x 0.200 m
k
= 50.0 rad s
m
1 2
kA
2
f=
so
a f
A = 50.0a0.200 f = 10.0 m s
1
= a100 fa0.200 f = 2.00 J
2
2
at x = ± A
2
FG 0.200 IJ =
H 3 K
a
8
0.200
9
f
2
3.33 m s 2
= 1.33 m s
ω
2π
= 1.13 Hz
12.2 mJ
Chapter 15
P15.21
(a)
E=
a f
1 2
1
kA , so if A ′ = 2 A , E ′ = k A ′
2
2
2
=
a f
1
k 2A
2
2
447
= 4E
Therefore E increases by factor of 4 .
*P15.22
(b)
v max =
k
A , so if A is doubled, v max is doubled .
m
(c)
a max =
k
A , so if A is doubled, a max also doubles .
m
(d)
T = 2π
(a)
y f = yi + v yi t +
m
is independent of A, so the period is unchanged .
k
1
ayt 2
2
1
−11 m = 0 + 0 + −9.8 m s 2 t 2
2
e
t=
(b)
j
22 m ⋅ s 2
= 1.50 s
9.8 m
Take the initial point where she steps off the bridge and the final point at the bottom of her
motion.
eK + U
g
+ Us
j = eK + U
i
g
+ Us
j
f
1
0 + mgy + 0 = 0 + 0 + kx 2
2
1
65 kg 9.8 m s 2 36 m = k 25 m
2
k = 73. 4 N m
a
(c)
The spring extension at equilibrium is x =
f
2
F 65 kg 9.8 m s 2
=
= 8.68 m , so this point is
k
73.4 N m
11 + 8.68 m = 19.7 m below the bridge and the amplitude of her oscillation is
36 − 19.7 = 16.3 m .
k
=
m
73.4 N m
= 1.06 rad s
65 kg
(d)
ω=
(e)
Take the phase as zero at maximum downward extension. We find what the phase was 25 m
higher when x = −8.68 m:
In x = A cos ωt ,
FG
H
t
−8.68 m = 16.3 m cos 1.06
s
t = −2.01 s
IJ
K
16.3 m = 16.3 m cos 0
t
1.06 = −122° = −2.13 rad
s
Then +2.01 s is the time over which the spring stretches.
(f)
total time = 1.50 s + 2.01 s = 3.50 s
448
P15.23
Oscillatory Motion
Model the oscillator as a block-spring system.
v2 + ω 2x2 = ω 2 A2
From energy considerations,
v max = ωA and v =
ωA
2
From this we find x 2 =
P15.24
3 2
A
4
so
FG ωA IJ
H2K
and
x=
2
+ ω 2x2 = ω 2 A2
3
A = ±2.60 cm where A = 3.00 cm
2
The potential energy is
a f
1 2 1 2
kx = kA cos 2 ωt .
2
2
Us =
The rate of change of potential energy is
a f
a f
dU s 1 2
1
= kA 2 cos ωt −ω sin ωt = − kA 2ω sin 2ωt .
2
2
dt
(a)
This rate of change is maximal and negative at
2ωt =
Then, t =
π
π
π
, 2ωt = 2π + , or in general, 2ωt = 2nπ + for integer n.
2
2
2
a
f
π 4n + 1
π
4n + 1 =
4ω
4 3.60 s −1
a
f
e
j
For n = 0 , this gives t = 0.218 s while n = 1 gives t = 1.09 s .
All other values of n yield times outside the specified range.
(b)
Section 15.4
P15.25
dU s
dt
=
max
1 2
1
kA ω = 3.24 N m 5.00 × 10 −2 m
2
2
b
ge
j e3.60 s j =
2
−1
14.6 mW
Comparing Simple Harmonic Motion with Uniform Circular Motion
(a)
The motion is simple harmonic because the tire is rotating with constant velocity and you
are looking at the motion of the bump projected in a plane perpendicular to the tire.
(b)
Since the car is moving with speed v = 3.00 m s , and its radius is 0.300 m, we have:
ω=
3.00 m s
= 10.0 rad s .
0.300 m
Therefore, the period of the motion is:
T=
2π
ω
=
2π
b10.0 rad sg =
0.628 s .
Chapter 15
P15.26
The angle of the crank pin is θ = ωt .
Its x-coordinate is
ω
Piston
x = A cos θ = A cos ωt
A
where A is the distance from the
center of the wheel to the crank pin.
This is of the form x = A cos ωt + φ ,
so the yoke and piston rod move
with simple harmonic motion.
b
Section 15.5
P15.27
(a)
P15.28
P15.29
g
FIG. P15.26
The Pendulum
T = 2π
L=
(b)
x = –A
gT 2
4π 2
L
g
e9.80 m s ja12.0 sf
2
=
Tmoon = 2π
4π 2
L
= 2π
g moon
2
= 35.7 m
35.7 m
1.67 m s 2
= 29.1 s
The period in Tokyo is
TT = 2π
LT
gT
and the period in Cambridge is
TC = 2π
LC
gC
We know
TT = TC = 2.00 s
For which, we see
LT LC
=
gT gC
or
g C LC 0.994 2
=
=
= 1.001 5
g T LT 0.992 7
The swinging box is a physical pendulum with period T = 2π
I
.
mgd
The moment of inertia is given approximately by
I=
1
mL2 (treating the box as a rod suspended from one end).
3
Then, with L ≈ 1.0 m and d ≈
T ≈ 2π
1
3
L
,
2
mL2
mg
ch
L
2
= 2π
a
f
2 1.0 m
2L
= 2π
= 1.6 s or T ~ 10 0 s .
2
3g
3 9.8 m s
e
j
x ( t)
449
450
P15.30
P15.31
Oscillatory Motion
ω=
2π
:
T
T=
ω=
g
:
L
L=
2π
=
ω
2π
= 1.42 s
4. 43
g
9.80
=
= 0.499 m
2
ω2
4.43
a f
Using the simple harmonic motion model:
A = rθ = 1 m 15°
π
= 0. 262 m
180°
g
9.8 m s 2
=
= 3.13 rad s
L
1m
ω=
(a)
v max = Aω = 0. 262 m 3.13 s = 0.820 m s
(b)
a max = Aω 2 = 0.262 m 3.13 s
b
a tan = rα
(c)
g
2
= 2.57 m s 2
a tan 2.57 m s 2
=
= 2.57 rad s 2
r
1m
α=
FIG. P15.31
F = ma = 0.25 kg 2.57 m s 2 = 0.641 N
More precisely,
(a)
1
mv 2
and
2
∴ v max = 2 gL 1 − cos θ = 0.817 m s
mgh =
a
(b)
f
f
Iα = mgL sin θ
α max =
P15.32
a
h = L 1 − cos θ
mgL sin θ
2
mL
=
g
sin θ i = 2.54 rad s 2
L
a fa
f
(c)
Fmax = mg sin θ i = 0.250 9.80 sin 15.0° = 0.634 N
(a)
The string tension must support the weight of the bob, accelerate it upward, and also provide
the restoring force, just as if the elevator were at rest in a gravity field 9.80 + 5.00 m s 2
a
T = 2π
L
5.00 m
= 2π
g
14.8 m s 2
T = 3.65 s
(b)
T = 2π
(c)
g eff =
5.00 m
e9.80 m s
2
− 5.00 m s 2
j
= 6.41 s
e9.80 m s j + e5.00 m s j
T = 2π =
2 2
5.00 m
11.0 m s 2
2 2
= 4.24 s
= 11.0 m s 2
f
Chapter 15
P15.33
Referring to the sketch we have
x
R
For small displacements,
tan θ ≈ sin θ
mg
F=−
and
x = − kx
R
Since the restoring force is proportional to the displacement from
equilibrium, the motion is simple harmonic motion.
F = − mg sin θ
tan θ =
and
Comparing toF = − mω 2 x shows ω =
P15.34
T=
(a)
g
.
R
k
=
m
total measured time
50
a f
Period, T asf
Length, L m
4
1.000 0.750 0.500
3
1.996 1.732 1.422
2
L
4π 2 L
T = 2π
so
g=
g
T2
The calculated values for g are:
(b)
af
0
1.996 1.732 1.422
g m s2
9.91
j
9.87
From T 2 =
Thus, g =
0.25
0.5
0.75
9.76
this agrees with the accepted value of g = 9.80 m s 2 within 0.5%.
F 4π I L , the slope of T
GH g JK
2
2
versus L graph =
4π 2
= 4.01 s 2 m .
g
4π 2
= 9.85 m s 2 . This is the same as the value in (b).
slope
f = 0. 450 Hz , d = 0.350 m, and m = 2.20 kg
T=
1
;
f
T = 2π
I =T2
I
;
mgd
mgd
4π
2
=
T2 =
FG 1 IJ
H fK
2
1.0
L, m
FIG. P15.34
Thus, g ave = 9.85 m s 2
(c)
1
Period, T s
e
FIG. P15.33
T2, s2
The measured periods are:
P15.35
451
4π 2 I
mgd
mgd
4π
2
=
a fa f =
e0.450 s j
2.20 9.80 0.350
4π
2
−1 2
0.944 kg ⋅ m 2
FIG. P15.35
452
P15.36
Oscillatory Motion
(a)
The parallel-axis theorem:
I = I CM + Md 2 =
=M
FG 13 m IJ
H 12 K
a
f
1
1
ML2 + Md 2 =
M 1.00 m
12
12
2
a
f
+ M 1.00 m
2
2
e
j
M 13 m 2
13 m
I
= 2π
= 2π
= 2.09 s
T = 2π
12 Mg 1.00 m
Mgd
12 9.80 m s 2
(b)
(a)
e
j
FIG. P15.36
1.00 m
9.80 m s
2
difference =
= 2.01 s
2.09 s − 2.01 s
= 4.08%
2.01 s
The parallel axis theorem says directly I = I CM + md 2
so
(b)
f
For the simple pendulum
T = 2π
P15.37
a
T = 2π
eI
I
= 2π
mgd
CM
+ md 2
j
mgd
When d is very large T → 2π
d
gets large.
g
When d is very small T → 2π
I CM
gets large.
mgd
So there must be a minimum, found by
j bmgdg
F 1I
FG 1 IJ eI
mg + 2π bmgd g
= 2π e I
+ md j G − J bmgd g
H 2K
H 2K
−π e I
+ md jmg
2π md mgd
=
+
=0
+
+
I
md
mgd
I
md
mgd
b
g
b
g
e
j
e
j
dT
d
=0=
2π I CM + md 2
dd
dd
e
12
−1 2
2 12
CM
−1 2
2
CM
2 12
CM
−3 2
3 2
CM
2 12
3 2
This requires
− I CM − md 2 + 2md 2 = 0
or
P15.38
ICM = md 2 .
We suppose the stick moves in a horizontal plane. Then,
b
f
ga
1
1
2.00 kg 1.00 m
mL2 =
12
12
I
T = 2π
I=
κ
κ=
4π 2 I
T
2
=
e
4π 2 0.167 kg ⋅ m 2
a180 sf
2
j=
2
= 0.167 kg ⋅ m 2
203 µN ⋅ m
CM
+ md 2
j
−1 2
2md
Chapter 15
P15.39
e
je
(a)
I = 5.00 × 10 −7 kg ⋅ m 2
(b)
I
d 2θ
= −κθ ;
dt 2
e
Section 15.6
θ
π I
JK
jFGH 0.2250
FIG. P15.39
2
= 3.16 × 10 −4
N⋅m
rad
Damped Oscillations
1
1
mv 2 + kx 2
2
2
2
dE
d x
= mv 2 + kxv
dt
dt
2
md x
= − kx − bv
dt 2
dE
= v − kx − bv + kvx
dt
dE
= − bv 2 < 0
dt
E=
The total energy is
Taking the time-derivative,
Use Equation 15.31:
a
Thus,
P15.41
2
κ
2π
=ω =
I
T
κ = Iω 2 = 5.00 × 10 −7
P15.40
j
T = 0.250 s, I = mr 2 = 20.0 × 10 −3 kg 5.00 × 10 −3 m
f
b
θ i = 15.0°
g
θ t = 1 000 = 5.50°
x1 000 Ae − bt 2 m 5.50
− b 1 000 g
=
=
=e b
15.0
xi
A
x = Ae − bt 2 m
2m
FG 5.50 IJ = −1.00 = −bb1 000g
H 15.0 K
2m
ln
∴
P15.42
b
= 1.00 × 10 −3 s −1
2m
b
x = Ae − bt 2 m cos ωt + φ
Show that
is a solution of
− kx − b
where
ω=
b
x = Ae − bt 2 m cos ωt + φ
FG
H
g
g
2
dx
d x
=m 2
dt
dt
FG IJ
H K
k
b
−
m
2m
(1)
2
.
(2)
IJ b g
b g
K
d x
b L
FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP
=−
Ae
H 2m K
2m MN
dt
Q
L
FG − b IJω sinbωt + φ g + Ae ω cosbωt + φ gOP
− M Ae
H 2m K
N
Q
dx
b
= Ae − bt 2 m −
cos ωt + φ − Ae − bt 2 mω sin ωt + φ
dt
2m
2
− bt 2 m
2
− bt 2 m
continued on next page
(3)
(4)
− bt 2 m
− bt 2 m
2
(5)
453
454
Oscillatory Motion
Substitute (3), (4) into the left side of (1) and (5) into the right side of (1);
2
b g 2bm Ae cosbωt + φ g + bωAe sinbωt + φ g
bL
FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP
= − M Ae
H 2m K
2N
Q
b
cosbωt + φ g
+ Ae
ω sinbωt + φ g − mω Ae
2
Compare the coefficients of Ae
cosbωt + φ g and Ae
sinbωt + φ g :
F k b I = −k + b
b
bF b I
b
= − G−
− mω =
− mG −
cosine-term: − k +
J
K
H
2m
2
2m
4m
2m
H m 4m JK
− kAe − bt 2 m cos ωt + φ +
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
2
− bt 2 m
− bt 2 m
2
sine-term:
bω = +
2
2
2
2
2
af af
b
b
ω + ω = bω
2
2
b
g
Since the coefficients are equal, x = Ae − bt 2 m cos ωt + φ is a solution of the equation.
*P15.43
The frequency if undamped would be ω 0 =
(a)
k
=
m
With damping
ω = ω 02 −
FG b IJ
H 2m K
2
=
2.05 × 10 4 N m
= 44.0 s.
10.6 kg
FG 44 1 IJ − FG 3 kg IJ
H s K H s 2 10.6 kg K
= 1 933.96 − 0.02 = 44.0
f=
(b)
b
ω 44.0
=
= 7.00 Hz
2π 2π s
2
1
s
g
In x = A 0 e − bt 2 m cos ωt + φ over one cycle, a time T =
A0 e − b 2π
2 mω
2
2π
ω
, the amplitude changes from A0 to
for a fractional decrease of
A 0 − A 0 e − πb mω
= 1 − e −π 3 a10.6⋅44.0 f = 1 − e −0 .020 2 = 1 − 0.979 98 = 0.020 0 = 2.00% .
A0
(c)
The energy is proportional to the square of the amplitude, so its fractional rate of decrease is
twice as fast:
E=
We specify
1 2 1 2 − 2 bt 2 m
= E0 e − bt m .
kA = kA 0 e
2
2
0.05E0 = E0 e − 3 t 10.6
0.05 = e − 3 t 10.6
e + 3 t 10 .6 = 20
3t
= ln 20 = 3.00
10.6
t = 10.6 s
Chapter 15
Section 15.7
P15.44
(a)
Forced Oscillations
For resonance, her frequency must match
f0 =
(b)
ω0
1
=
2π 2π
4.30 × 10 3 N m
= 2.95 Hz .
12.5 kg
k
1
=
m 2π
dx
dv
= − Aω sin ωt , and a =
= − Aω 2 cos ωt , the maximum acceleration
dt
dt
is Aω 2 . When this becomes equal to the acceleration due to gravity, the normal force
exerted on her by the mattress will drop to zero at one point in the cycle:
From x = A cos ωt , v =
2
Aω = g
P15.45
or
b g
F = 3.00 cos 2π t N
2π
= 2π rad s
T
(a)
ω=
(b)
In this case,
A=
g
ω2
=
g
k
m
e9.80 m s jb12.5 kgg =
A=
2
gm
=
k
4.30 × 10 3 N m
and
k = 20.0 N m
so
T = 1.00 s
ω0 =
k
=
m
2.85 cm
20.0
= 3.16 rad s
2.00
The equation for the amplitude of a driven oscillator,
P15.46
455
FG F IJ eω
H mK
0
2
− ω 02
j
−1
a f
3
4π 2 − 3.16
2
with b = 0, gives
A=
Thus
A = 0.050 9 m = 5.09 cm .
F0 cos ωt − kx = m
b
x = A cos ωt + φ
d2x
dt 2
ω0 =
=
2 −1
k
m
(1)
g
(2)
b
dx
= − Aω sin ωt + φ
dt
g
d2x
= − Aω 2 cos ωt + φ
dt 2
b
(3)
g
(4)
b g e j b
j cosbωt + φ g = F cos ωt
Substitute (2) and (4) into (1):
F0 cos ωt − kA cos ωt + φ = m − Aω 2 cos ωt + φ
Solve for the amplitude:
ekA − mAω
2
0
These will be equal, provided only that φ must be zero and kA − mAω 2 = F0
Thus, A =
F0
m
c h−ω
k
m
2
g
456
P15.47
Oscillatory Motion
From the equation for the amplitude of a driven oscillator with no damping,
F0 m
A=
eω
2
− ω 02
j
2
e
ω = 2π f = 20.0π s −1
e
F0 = mA ω
F0 =
P15.48
j
FG 40.0 IJ e2.00 × 10 jb3 950 − 49.0g =
H 9.80 K
−2
eω
2
− ω 02
j + b bω m g
2
k
200
=
= 49.0 s −2
40 .0
m
9.80
c h
318 N
A=
2
Fext m
eω
2
− ω 02
j
2
=
e
Fext m
±ω
2
− ω 02
j
=±
ω 2 = ω 02 ±
This yields
ω = 8.23 rad s or ω = 4.03 rad s
Then,
f=
ω
2π
Fext m
ω 2 − ω 02
Fext m k Fext 6.30 N m
1.70 N
= ±
=
±
0.150 kg
A
m mA
0.150 kg 0.440 m
Thus,
b
gives either f = 1.31 Hz
or
f
ga
f = 0.641 Hz
The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm:
f=
*P15.50
− ω 02
ω 02 =
Fext m
A=
With b = 0,
P15.49
2
j
1
2π
g
1
=
L 2π
9.80 m s 2
= 1.74 Hz .
0.082 1 m
For the resonance vibration with the occupants in the car, we have for the spring constant of the
suspension
f=
1
2π
k
m
e j d1 130 kg + 4b72.4 kg gi = 1.82 × 10
F 4b72. 4 kg ge9.8 m s j
x= =
= 1.56 × 10 m
k = 4π 2 f 2 m = 4π 2 1.8 s −1
2
2
Now as the occupants exit
k
1.82 × 10
5
kg s
2
−2
5
kg s 2
Chapter 15
457
Additional Problems
P15.51
Let F represent the tension in the rod.
pivot
(a)
At the pivot, F = Mg + Mg = 2 Mg
A fraction of the rod’s weight Mg
FG y IJ as well as the
H LK
P
L
weight of the ball pulls down on point P. Thus, the
tension in the rod at point P is
F = Mg
FG y IJ + Mg =
H LK
FG
H
Mg 1 +
y
L
IJ
K
y
.
M
FIG. P15.51
(b)
Relative to the pivot, I = I rod + I ball =
1
4
ML2 + ML2 = ML2
3
3
I
where m = 2 M and d is the distance from the
mgd
pivot to the center of mass of the rod and ball combination. Therefore,
For the physical pendulum, T = 2π
d=
For L = 2.00 m, T =
P15.52
(a)
Total energy =
M
4π
3
c h + ML = 3L and T = 2π
L
2
M+M
a
4
f=
2 2.00 m
9.80 m s
2
b
4
3
ML2
a 2 M f gc h
3L
4
4π
3
=
2L
.
g
2.68 s .
f
ga
1 2 1
kA = 100 N m 0.200 m
2
2
2
= 2.00 J
At equilibrium, the total energy is:
b
g
b
g
b
g
1
1
m1 + m 2 v 2 = 16.0 kg v 2 = 8.00 kg v 2 .
2
2
Therefore,
b8.00 kg gv
2
= 2.00 J , and v = 0.500 m s .
This is the speed of m1 and m 2 at the equilibrium point. Beyond this point, the mass m 2
moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to the
restoring force of the spring.
continued on next page
458
Oscillatory Motion
(b)
The energy of the m1 -spring system at equilibrium is:
b
gb
1
1
m1 v 2 = 9.00 kg 0.500 m s
2
2
This is also equal to
Therefore,
g
2
= 1.125 J .
a f
1
2
k A′ , where A ′ is the amplitude of the m1 -spring system.
2
a fa f
1
100 A ′
2
2
= 1.125 or A ′ = 0.150 m.
m1
= 1.885 s
k
The period of the m1 -spring system is T = 2π
1
T = 0.471 s after it passes the equilibrium point for the spring to become fully
4
stretched the first time. The distance separating m1 and m 2 at this time is:
and it takes
D=v
P15.53
F d xI
GH dt JK
2
FG T IJ − A ′ = 0.500 m s a0.471 sf − 0.150 m = 0.085 6 =
H 4K
µs
= Aω 2
2
B
max
P
fmax = µ sn = µ s mg = mAω 2
A=
8.56 cm .
µsg
= 6.62 cm
ω2
n
f
B
mg
FIG. P15.53
P15.54
The maximum acceleration of the oscillating system is a max = Aω 2 = 4π 2 Af 2 . The friction force
exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B
is about to slip,
e
f = fmax = µ sn = µ s mg = m 4π 2 Af 2
P15.55
j
or
µs g
.
4π 2 f 2
A=
Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one
proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D
and to the diatomic molecule of hydrogen-1 as H.
MD = 2MH
ωD
=
ωH
k
MD
k
MH
=
MH
=
MD
1
2
fD =
fH
2
= 0.919 × 10 14 Hz
Chapter 15
P15.56
1
1
mv 2 + IΩ 2 ,
2
2
where Ω is the rotation rate of the ball about its
center of mass. Since the center of the ball moves
along a circle of radius 4R, its displacement from
equilibrium is s = 4R θ and its speed is
ds
dθ
= 4R
v=
. Also, since the ball rolls without
dt
dt
slipping,
459
The kinetic energy of the ball is K =
5R
a f
FG IJ
H K
v=
θ
ds
= RΩ
dt
Ω=
so
R
h
FG IJ
H K
v
dθ
=4
R
dt
s
FIG. P15.56
The kinetic energy is then
K=
=
FG
H
1
dθ
m 4R
2
dt
IJ
K
2
FG IJ
H K
112mR 2 dθ
10
dt
+
FG
H
1 2
mR 2
2 5
IJ FG 4 dθ IJ
K H dt K
2
2
a
f
When the ball has an angular displacement θ, its center is distance h = 4R 1 − cos θ higher than
when at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR 1 − cos θ . For small
a
f
angles, 1 − cos θ ≈
a
f
2
θ
(see Appendix B). Hence, U g ≈ 2mgRθ 2 , and the total energy is
2
E = K +Ug =
FG IJ
H K
112mR 2 dθ
10
dt
2
+ 2mgRθ 2 .
FG IJ
H K
28 R d θ
d θ
F 5 g IJθ .
+ gθ = 0 , or
This reduces to
= −G
H 28R K
5 dt
dt
Since E = constant in time,
2
FG IJ
H K
112 mR 2 dθ d 2θ
dE
dθ
.
+ 4mgRθ
=0=
5
dt
dt dt 2
dt
2
2
2
With the angular acceleration equal to a negative constant times the angular position, this is in the
5g
.
defining form of a simple harmonic motion equation with ω =
28 R
The period of the simple harmonic motion is then T =
2π
ω
= 2π
28 R
.
5g
460
P15.57
Oscillatory Motion
(a)
Li
L
a
h
a
FIG. P15.57(a)
(b)
T = 2π
π 1 dL
dT
=
dt
g L dt
L
g
af
We need to find L t and
dL
. From the diagram in (a),
dt
FG IJ
H K
1 dh
a h dL
=−
− ;
.
2 dt
2 2 dt
L = Li +
But
(1)
dM
dV
dh
=ρ
= − ρA . Therefore,
dt
dt
dt
IJ
K
(2)
FG 1 IJ FG dM IJ t = L − L
H 2 ρA K H dt K
(3)
FG
H
1 dM dL
dh
1 dM
=−
=
;
ρA dt dt
dt
2 ρA dt
z
L
Also,
dL =
Li
i
Substituting Equation (2) and Equation (3) into Equation (1):
F
GH
π
1
g 2 ρa 2
dT
=
dt
(c)
I FG dM IJ
JK H dt K
1
Li +
1
2 ρa 2
c ht
dM
dt
.
Substitute Equation (3) into the equation for the period.
2π
T=
g
Li +
FG IJ
H K
1
dM
t
2
dt
2 ρa
Or one can obtain T by integrating (b):
F
GH
I FG dM IJ
z
JK H dt K z L + dt c ht
L 2 OPL
O
1 F dM I
π F 1 I F dM I M
T −T =
L +
t− L P
M
G
J
G
J
G
J
g H 2 ρa K H dt K M
PQ
N c h PQMN 2 ρa H dt K
L
2π
1 F dM I
, so T =
L +
G Jt .
g
2 ρa H dt K
g
Ti
dT =
T
i
But Ti = 2π
i
t
π
1
g 2 ρa 2
0
2
i
2
i
1
2 ρa 2
1
2 ρa 2
dM
dt
dM
dt
i
2
i
Chapter 15
P15.58
ω=
(a)
P15.59
k
2π
=
m
T
2
k =ω m =
4π 2 m
m′ =
(b)
T2
a f
k T′
= m
4π 2
FG T ′ IJ
HTK
2
Hy
We draw a free-body diagram of the pendulum.
The force H exerted by the hinge causes no torque
about the axis of rotation.
τ = Iα
Hx
h
d 2θ
= −α
dt 2
and
τ = MgL sin θ + kxh cos θ = − I
kx
Lθ
x
2
d θ
dt 2
k
m
mg
For small amplitude vibrations, use the
approximations: sin θ ≈ θ , cos θ ≈ 1, and x ≈ s = hθ .
Therefore,
F
GH
(a)
I
JK
MgL + kh 2
d 2θ
θ = −ω 2θ
=
−
I
dt 2
b
g
ω=
ML2
1
2π
we have at t = 0
v = −ωA sin φ = − v max
This requires φ = 90° , so
x = A cos ωt + 90°
And this is equivalent to
x = − A sin ωt
Numerically we have
ω=
and v max = ωA
20 m s = 10 s −1 A
In
a
MgL + kh 2
ML2
f
50 N m
= 10 s −1
0.5 kg
e
a
= 2π f
g
v = −ωA sin ωt + φ
j
f e
A=2m
j
x = −2 m sin 10 s −1 t
So
(b)
b
MgL + kh 2
In x = A cos ωt + φ ,
k
=
m
L sinθ
FIG. P15.59
f=
*P15.60
2
1
1
1
mv 2 + kx 2 = kA 2 ,
2
2
2
implies
FG
H
1 2
1
kx = 3 mv 2
2
2
11 2 1 2 1 2
kx + kx = kA
32
2
2
x=±
continued on next page
IJ
K
3
A = ±0.866 A = ±1.73 m
4
4 2
x = A2
3
hcosθ
461
462
Oscillatory Motion
(c)
ω=
(d)
In
g
L
g
L=
a
ω2
=
9.8 m s 2
e10 s j
−1 2
f e
= 0.098 0 m
j
x = −2 m sin 10 s −1 t
the particle is at x = 0 at t = 0 , at 10t = π s , and so on.
The particle is at
x=1 m
when
−
with solutions
e10 s jt = − π6
1
= sin 10 s −1 t
2
e
j
−1
e10 s jt = π + π6 , and so on.
FπI
The minimum time for the motion is ∆t in 10 ∆t = G J s
H 6K
FπI
∆t = G J s = 0.052 4 s
H 60 K
−1
P15.61
(a)
FIG. P15.60(d)
At equilibrium, we have
F LI
∑ τ = 0 − mg GH 2 JK + kx0 L
where x 0 is the equilibrium compression.
After displacement by a small angle,
FIG. P15.61
F LI
F LI
∑ τ = − mg GH 2 JK + kxL = −mg GH 2 JK + kbx0 − Lθ gL = − kθL2
But,
1
∑ τ = Iα = 3 mL2
d 2θ
.
dt 2
So
d 2θ
3k
= − θ.
2
m
dt
The angular acceleration is opposite in direction and proportional to the displacement, so
3k
we have simple harmonic motion with ω 2 =
.
m
(b)
f=
ω
1
=
2π 2π
3k
1
=
m 2π
b
g=
3 100 N m
5.00 kg
1.23 Hz
Chapter 15
*P15.62
463
As it passes through equilibrium, the 4-kg object has speed
v max = ωA =
100 N m
k
A=
2 m = 10.0 m s.
m
4 kg
In the completely inelastic collision momentum of the two-object system is conserved. So the new
10-kg object starts its oscillation with speed given by
b
g b g b
g
4 kg 10 m s + 6 kg 0 = 10 kg v max
v max = 4.00 m s
(a)
The new amplitude is given by
1
1
2
= kA 2
mv max
2
2
b
10 kg 4 m s
g = b100 N mgA
2
2
A = 1.26 m
(b)
(c)
Thus the amplitude has decreased by
2.00 m − 1.26 m = 0.735 m
The old period was
T = 2π
4 kg
m
= 2π
= 1.26 s
k
100 N m
The new period is
T = 2π
10 2
s = 1.99 s
100
The period has increased by
1.99 m − 1.26 m = 0.730 s
The old energy was
1
1
2
= 4 kg 10 m s
mv max
2
2
The new mechanical energy is
1
10 kg 4 m s
2
b gb
b
gb
g
2
g
2
= 200 J
= 80 J
The energy has decreased by 120 J .
P15.63
(d)
The missing mechanical energy has turned into internal energy in the completely inelastic
collision.
(a)
T=
(b)
E=
(c)
At maximum angular displacement
2π
ω
= 2π
L
= 3.00 s
g
a fa f
1
1
mv 2 = 6.74 2.06
2
2
a
h = L − L cos θ = L 1 − cos θ
2
f
= 14.3 J
mgh =
1
mv 2
2
cos θ = 1 −
h
L
h=
v2
= 0.217 m
2g
θ = 25.5°
464
P15.64
Oscillatory Motion
One can write the following equations of motion:
T − kx = 0
(describes the spring)
mg − T ′ = ma = m
a
f
R T′ − T = I
2
d x
dt 2
(for the hanging object)
d 2θ I d 2 x
=
dt 2 R dt 2
(for the pulley)
with I =
FIG. P15.64
1
MR 2
2
Combining these equations gives the equation of motion
FG m + 1 MIJ d x + kx = mg .
H 2 K dt
2
2
af
mg
mg
(where
arises because of the extension of the spring due to
k
k
the weight of the hanging object), with frequency
The solution is x t = A sin ωt +
f=
P15.65
ω
2π
=
k
1
=
m + 12 M 2π
1
2π
(a)
For M = 0
f = 3.56 Hz
(b)
For M = 0.250 kg
f = 2.79 Hz
(c)
For M = 0.750 kg
f = 2.10 Hz
100 N m
.
0.200 kg + 12 M
Suppose a 100-kg biker compresses the suspension 2.00 cm.
Then,
k=
F
980 N
=
= 4.90 × 10 4 N m
x 2.00 × 10 −2 m
If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is
f=
1
2π
1
k
=
m 2π
4.90 × 10 4 N m
= 1.58 Hz
500 kg
If he encounters washboard bumps at the same frequency, resonance will make the motorcycle
bounce a lot. Assuming a speed of 20.0 m/s, we find these ridges are separated by
20.0 m s
1.58 s −1
= 12.7 m ~ 10 1 m .
In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also
vibrate at higher frequencies by rocking back and forth between front and rear wheels, by having
just the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps will
excite all of these other resonances.
Chapter 15
P15.66
(a)
For each segment of the spring
dK =
Also,
vx =
x
v
A
a f
1
dm v x2 .
2
dm =
and
m
dx .
A
FIG. P15.66
Therefore, the total kinetic energy of the block-spring system is
K=
(b)
ω=
k
m eff
Therefore,
P15.67
(a)
∑ F = −2T sin θ j
z FGH
A
0
I
JK
FG
H
IJ
K
x2v2 m
m 2
1
dx =
M+
v .
2
2
3
A
A
FG
H
IJ
K
1
1
m 2
m eff v 2 =
M+
v
2
2
3
and
T=
1
1
Mv 2 +
2
2
2π
ω
= 2π
M + m3
k
.
where θ = tan −1
FG y IJ
H LK
Therefore, for a small displacement
y
sin θ ≈ tan θ =
L
(b)
and
FIG. P15.67
−2Ty
∑ F = L j
The total force exerted on the ball is opposite in direction and proportional to its
displacement from equilibrium, so the ball moves with simple harmonic motion. For a
spring system,
∑ F = −kx
becomes here
Therefore, the effective spring constant is
2T
and
L
∑F = −
ω=
2T
y.
L
k
=
m
2T
.
mL
465
466
P15.68
Oscillatory Motion
(a)
Assuming a Hooke’s Law type spring,
F = Mg = kx
and empirically
Mg = 1.74x − 0.113
k = 1.74 N m ± 6% .
so
(b)
M , kg
0.020 0
x, m
0.17
Mg , N
0.196
0.040 0
0.293
0.392
0.050 0
0.353
0.49
0.060 0
0.413
0.588
0.070 0
0.471
0.686
0.080 0
0.493
0.784
We may write the equation as theoretically
T2 =
4π 2
4π 2
M+
ms
3k
k
FIG. P15.68
and empirically
T 2 = 21.7 M + 0.058 9
k=
so
4π 2
= 1.82 N m ± 3%
21.7
Time, s T , s
7.03 0.703
M , kg
0.020 0
T 2 , s2
0.494
9.62
0.962
0.040 0
0.925
10.67
1.067
0.050 0
1.138
11.67
1.167
0.060 0
1.362
12.52
1.252
0.070 0
1.568
13.41
1.341
0.080 0
1.798
The k values 1.74 N m ± 6%
and
so
(c)
1.82 N m ± 3% differ by 4%
they agree.
Utilizing the axis-crossing point,
ms = 3
FG 0.058 9 IJ kg =
H 21.7 K
8 grams ± 12%
in agreement with 7.4 grams.
Chapter 15
P15.69
(a)
∆K + ∆U = 0
Thus, K top + U top = K bot + U bot
M
R
where K top = U bot = 0
1 2
Iω , but
2
h = R − R cos θ = R 1 − cos θ
Therefore, mgh =
a
θ
f
v
R
MR 2 mr 2
and I =
+
+ mR 2
2
2
Substituting we find
ω=
2
2
2
(b)
M
m
M
m
2
+
r2
R2
r2
R2
+2
1
2
dCM =
af
mR + M 0
m+M
MR 2 + 12 mr 2 + mR 2
mgR
We require Ae − bt 2 m =
A
2
bt
= ln 2
or
2m
The spring constant is irrelevant.
or
(b)
2
2
I
mT gd CM
T = 2π
T = 2π
(a)
2
2
mT = m + M
P15.70
e + bt 2 m = 2
0.100 kg s
t = 0.693
2 0.375 kg
b
∴ t = 5.20 s
g
We can evaluate the energy at successive turning points, where
1
1
1
1 1 2
cos ωt + φ = ±1 and the energy is kx 2 = kA 2 e − bt 2 m . We require kA 2 e − bt 2 m =
kA
2
2
2
2 2
m ln 2 0.375 kg 0.693
∴t =
=
= 2.60 s .
or
e + bt m = 2
b
0.100 kg s
b
g
a
(c)
m
FIG. P15.69
2
so
θ
v
f 12 FGH MR2 + mr2 + mR IJK Rv
L M mr + m OPv
mgRa1 − cos θ f = M +
N 4 4R 2 Q
a1 − cos θ f
v = 4 gR
e + + 2j
Rg a1 − cos θ f
v=2
a
mgR 1 − cos θ =
and
467
f
1 2
kA , the fractional rate of change of energy over time is
2
2
d 1
dE
dA
dA
1
dt 2 kA
dt
dt
2 k 2 A dt
= 1
=
=
2
2
1 kA 2
E
A
kA
2
2
From E =
e
j
a f
two times faster than the fractional rate of change in amplitude.
FG
H
IJ
K
468
P15.71
Oscillatory Motion
(a)
When the mass is displaced a distance x from
equilibrium, spring 1 is stretched a distance x1 and
spring 2 is stretched a distance x 2 .
By Newton’s third law, we expect
k1 x1 = k 2 x 2 .
When this is combined with the requirement that
x = x1 + x 2 ,
FIG. P15.71
LM k OPx
Nk + k Q
L k k OP x = ma
F =M
Nk + k Q
x1 =
we find
The force on either spring is given by
(b)
2
1
2
This is in the form
F = k eff x = ma
and
T = 2π
b
m k1 + k 2
m
= 2π
k eff
k1 k 2
g
In this case each spring is distorted by the distance x which the mass is displaced. Therefore,
the restoring force is
b
g
F = − k1 + k 2 x
so that
P15.72
1
1 2
1
where a is the acceleration of the mass m.
2
T = 2π
b
k eff = k1 + k 2
and
m
k1 + k 2
g
.
Let A represent the length below water at equilibrium and M the tube’s mass:
∑ Fy = 0 ⇒ − Mg + ρπ r 2 Ag = 0 .
Now with any excursion x from equilibrium
a f
− Mg + ρπ r 2 A − x g = Ma .
Subtracting the equilibrium equation gives
− ρπ r 2 gx = Ma
a=−
F ρπ r g I x = −ω x
GH M JK
2
2
The opposite direction and direct proportionality of a to x imply SHM with angular frequency
ω=
T=
ρπ r 2 g
M
2π
ω
=
FG 2 IJ
HrK
πM
ρg
Chapter 15
P15.73
For θ max = 5.00° , the motion calculated by the Euler method
agrees quite precisely with the prediction of θ max cos ωt . The
period is T = 2.20 s .
Time,
t (s)
0.000
0.004
0.008
…
0.544
0.548
0.552
…
1.092
1.096
1.100
1.104
…
1.644
1.648
1.652
…
2.192
2.196
2.200
2.204
Angle,
θ (°)
5.000 0
4.999 3
4.998 0
Ang. speed
(°/s)
0.000 0
–0.163 1
–0.326 2
Ang. Accel.
° s2
e j
θ max cos ωt
–40.781 5
–40.776 2
–40.765 6
5.000 0
4.999 7
4.998 7
0.056 0
–0.001 1
–0.058 2
–14.282 3
–14.284 2
–14.284 1
–0.457 6
0.009 0
0.475 6
0.081 0
0.023 9
–0.033 3
–4.999 4
–5.000 0
–5.000 0
–4.999 3
–0.319 9
–0.156 8
0.006 3
0.169 4
40.776 5
40.781 6
40.781 4
40.775 9
–4.998 9
–4.999 8
–5.000 0
–4.999 6
–0.063 8
0.003 3
0.060 4
14.282 4
14.284 2
14.284 1
0.439 7
–0.027 0
–0.493 6
–0.071 6
–0.014 5
0.042 7
4.999 4
5.000 0
5.000 0
4.999 3
0.313 7
0.150 6
–0.012 6
–0.175 7
–40.776 8
–40.781 7
–40.781 3
–40.775 6
4.999 1
4.999 9
5.000 0
4.999 4
For θ max = 100° , the simple harmonic motion approximation
θ max cos ωt diverges greatly from the Euler calculation. The
period is T = 2.71 s , larger than the small-angle period by 23%.
Time,
Angle,
t (s)
θ (°)
0.000 100.000 0
0.004
99.992 6
0.008
99.977 6
…
1.096 –84.744 9
1.100 –85.218 2
1.104 –85.684 0
…
1.348 –99.996 0
1.352 –100.000 8
1.356 –99.998 3
…
2.196
40.150 9
2.200
41.045 5
2.204
41.935 3
…
2.704
99.998 5
2.708 100.000 8
2.712
99.995 7
Ang. speed
(°/s)
0.000 0
–1.843 2
–3.686 5
Ang. Accel.
° s2
e j
θ max cos ωt
–460.606 6
–460.817 3
–460.838 2
100.000 0
99.993 5
99.973 9
–120.191 0
–118.327 2
–116.462 0
465.948 8
466.286 9
466.588 6
–99.995 4
–99.999 8
–99.991 1
–3.053 3
–1.210 0
0.633 2
460.812 5
460.805 7
460.809 3
–75.797 9
–75.047 4
–74.287 0
224.867 7
223.660 9
222.431 8
–301.713 2
–307.260 7
–312.703 5
99.997 1
99.999 3
99.988 5
2.420 0
0.576 8
–1.266 4
–460.809 0
–460.805 7
–460.812 9
12.642 2
11.507 5
10.371 2
FIG. P15.73
469
470
*P15.74
Oscillatory Motion
(a)
The block moves with the board in what we take as the positive x direction, stretching the
spring until the spring force −kx is equal in magnitude to the maximum force of static
µ mg
.
friction µ sn = µ s mg . This occurs at x = s
k
(b)
Since v is small, the block is nearly at the rest at this break point. It starts almost immediately
to move back to the left, the forces on it being −kx and + µ k mg . While it is sliding the net
force exerted on it can be written as
− kx + µ k mg = − kx +
FG
H
IJ
K
kµ k mg
µ mg
= −k x − k
= − kx rel
k
k
where x rel is the excursion of the block away from the point
µ k mg
.
k
Conclusion: the block goes into simple harmonic motion centered about the equilibrium
µ mg
position where the spring is stretched by k
.
k
(d)
The amplitude of its motion is its original displacement, A =
b
g
µ s mg µ k mg
−
. It first comes to
k
k
2 µ k − µ s mg
µ k mg
. Almost immediately at this point it
−A=
k
k
latches onto the slowly-moving board to move with the board. The board exerts a force of
static friction on the block, and the cycle continues.
rest at spring extension
(c)
The graph of
the motion
looks like this:
FIG. P15.74(c)
(e)
b
g
2 A 2 µ s − µ k mg
=
.
v
kv
The time for which the block is springing back is one half a cycle of simple harmonic motion,
The time during each cycle when the block is moving with the board is
F
GH
I
JK
1
m
m
=π
. We ignore the times at the end points of the motion when the speed of
2π
2
k
k
2A
the block changes from v to 0 and from 0 to v. Since v is small compared to
, these
π
times are negligible. Then the period is
T=
continued on next page
b
g
2 µ s − µ k mg
kv
+π
m
.
k
m
k
Chapter 15
(f)
T=
a
fb
b0.024 m sgb12 N mg
f=
Then
*P15.75
ge
2 0.4 − 0.25 0.3 kg 9.8 m s 2
b
g
2 µ s − µ k mg
j +π
0.3 kg
= 3.06 s + 0.497 s = 3.56 s
12 N m
1
= 0.281 Hz .
T
+π
m
increases as m increases, so the frequency decreases .
k
(g)
T=
(h)
As k increases, T decreases and f increases .
(i)
As v increases, T decreases and f increases .
(j)
As µ s − µ k increases, T increases and f decreases .
(a)
Newton’s law of universal gravitation is
F=−
Thus,
F=−
Which is of Hooke’s law form with
k=
(b)
kv
b
471
g
GMm
r
=−
2
FG
H
IJ
K
Gm 4 3
πr ρ
r2 3
FG 4 πρGmIJ r
H3 K
4
πρGm
3
FG 4 IJ πρGmr = ma
H 3K
F 4I
a = − G J πρGr = −ω r
H 3K
The sack of mail moves without friction according to
−
2
Since acceleration is a negative constant times excursion from equilibrium, it executes SHM
with
ω=
4πρG
3
and period
T=
The time for a one-way trip through the earth is
T
=
2
We have also
g=
so
g
4ρG
=
3
πR e
b g
and
2π
ω
3π
ρG
=
3π
4 ρG
GM e
R e2
=
G 4πR e3 ρ
3 R e2
(a) 4.33 cm; (b) −5.00 cm s ;
P15.6
see the solution
P15.8
12.0 Hz
P15.10
18.8 m s; 7.11 km s 2
2
(c) −17.3 cm s ; (d) 3.14 s; 5.00 cm
P15.4
4
πρGR e
3
Re
6.37 × 10 6 m
T
=π
=π
= 2.53 × 10 3 s = 42.2 min .
2
2
g
9.8 m s
ANSWERS TO EVEN PROBLEMS
P15.2
=
(a) 15.8 cm; (b) −15.9 cm;
(c) see the solution; (d) 51.1 m; (e) 50.7 m
472
Oscillatory Motion
(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ;
15 cm
sin 5t ;
(c) x = −3 cmcos 5t ; v =
s
75 cm
a=
cos 5t
s2
P15.42
see the solution
P15.44
(a) 2.95 Hz; (b) 2.85 cm
P15.46
see the solution
P15.48
either 1.31 Hz or 0.641 Hz
P15.14
F vI
(a) ; (b) x = − G J sin ωt
HωK
ω
P15.50
1.56 cm
P15.16
(a) 126 N m; (b) 0.178 m
P15.52
(a) 0.500 m s ; (b) 8.56 cm
P15.18
(a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2
P15.54
A=
P15.20
(a) 100 N m; (b) 1.13 Hz;
(c) 1.41 m s at x = 0 ;
P15.56
see the solution
(f) 1.33 m s ; (g) 3.33 m s 2
P15.58
(a) k =
P15.22
(a) 1.50 s; (b) 73.4 N m;
(c) 19.7 m below the bridge; (d) 1.06 rad s;
(e) 2.01 s; (f) 3.50 s
P15.60
(a) x = −2 m sin 10t ; (b) at x ± 1.73 m;
(c) 98.0 mm; (d) 52.4 ms
P15.24
(a) 0.218 s and 1.09 s; (b) 14.6 mW
P15.62
P15.26
The position of the piston is given by
x = A cos ωt .
(a) decreased by 0.735 m;
(b) increased by 0.730 s;
(c) decreased by 120 J; (d) see the solution
P15.64
(a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 Hz
P15.66
(a)
P15.68
see the solution; (a) k = 1.74 N m ± 6% ;
(b) 1.82 N m ± 3%; they agree;
(c) 8 g ± 12%; it agrees
P15.70
(a) 5.20 s; (b) 2.60 s; (c) see the solution
P15.72
see the solution; T =
P15.74
see the solution; (f) 0.281 Hz ;
(g) decreases; (h) increases; (i) increases;
(j) decreases
P15.12
FG
H
FG
H
IJ
K
IJ
K
v
µsg
4π 2 f 2
2
(d) 10.0 m s at x = ± A ; (e) 2.00 J;
P15.28
gC
= 1.001 5
gT
P15.30
1.42 s; 0.499 m
P15.32
(a) 3.65 s; (b) 6.41 s; (c) 4.24 s
P15.34
(a) see the solution;
(b), (c) 9.85 m s 2 ; agreeing with the
accepted value within 0.5%
P15.36
(a) 2.09 s; (b) 4.08%
P15.38
203 µN ⋅ m
P15.40
see the solution
FG IJ
H K
4π 2 m
T′
; (b) m ′ = m
2
T
T
a
FG
H
2
f a f
IJ
K
M + m3
1
m 2
M+
v ; (b) T = 2π
2
3
k
FG 2 IJ
HrK
πM
ρg
16
Wave Motion
CHAPTER OUTLINE
16.1
16.2
16.3
16.4
16.5
16.6
Propagation of a
Disturbance
Sinusoidal Waves
The Speed of Waves on
Strings
Reflection and Transmission
Rate of Energy Transfer by
Sinusoidal Waves on Strings
The Linear Wave Equation
ANSWERS TO QUESTIONS
Q16.1
As the pulse moves down the string, the particles of the string
itself move side to side. Since the medium—here, the
string—moves perpendicular to the direction of wave
propagation, the wave is transverse by definition.
Q16.2
To use a slinky to create a longitudinal wave, pull a few coils
back and release. For a transverse wave, jostle the end coil
side to side.
Q16.3
From v =
Q16.4
It depends on from what the wave reflects. If reflecting from a
less dense string, the reflected part of the wave will be right
side up.
T
µ
, we must increase the tension by a factor of 4.
2π vA
Q16.5
Yes, among other things it depends on. v max = ωA = 2π fA =
Q16.6
Since the frequency is 3 cycles per second, the period is
Q16.7
Amplitude is increased by a factor of
Q16.8
The section of rope moves up and down in SHM. Its speed is always changing. The wave continues on
with constant speed in one direction, setting further sections of the rope into up-and-down motion.
Q16.9
Each element of the rope must support the weight of the rope below it. The tension increases with
λ
. Here v is the speed of the wave.
1
second = 333 ms.
3
2 . The wave speed does not change.
height. (It increases linearly, if the rope does not stretch.) Then the wave speed v =
T
µ
increases
with height.
Q16.10
The difference is in the direction of motion of the elements of the medium. In longitudinal waves,
the medium moves back and forth parallel to the direction of wave motion. In transverse waves, the
medium moves perpendicular to the direction of wave motion.
473
474
Wave Motion
Q16.11
Slower. Wave speed is inversely proportional to the square root of linear density.
Q16.12
As the wave passes from the massive string to the less massive string, the wave speed will increase
according to v =
T
. The frequency will remain unchanged. Since v = fλ , the wavelength must
µ
increase.
Q16.13
Higher tension makes wave speed higher. Greater linear density makes the wave move more
slowly.
Q16.14
The wave speed is independent of the maximum particle speed. The source determines the
maximum particle speed, through its frequency and amplitude. The wave speed depends instead on
properties of the medium.
Q16.15
Longitudinal waves depend on the compressibility of the fluid for their propagation. Transverse
waves require a restoring force in response to sheer strain. Fluids do not have the underlying
structure to supply such a force. A fluid cannot support static sheer. A viscous fluid can
temporarily be put under sheer, but the higher its viscosity the more quickly it converts input
work into internal energy. A local vibration imposed on it is strongly damped, and not a source of
wave propagation.
Q16.16
Let ∆t = ts − t p represent the difference in arrival times of the two waves at a station at distance
d = v s ts = v p t p
F1 1I
from the hypocenter. Then d = ∆tG − J
Hv v K
s
−1
. Knowing the distance from the first
p
station places the hypocenter on a sphere around it. A measurement from a second station limits it
to another sphere, which intersects with the first in a circle. Data from a third non-collinear station
will generally limit the possibilities to a point.
Q16.17
The speed of a wave on a “massless” string would be infinite!
SOLUTIONS TO PROBLEMS
Section 16.1
P16.1
Propagation of a Disturbance
Replace x by x − vt = x − 4.5t
6
to get
y=
2
x − 4.5t + 3
a
f
Chapter 16
475
P16.2
FIG. P16.2
P16.3
a
5.00 e −a x + 5 t f is of the form f x + vt
2
f
so it describes a wave moving to the left at v = 5.00 m s .
P16.4
(a)
The longitudinal wave travels a shorter distance and is moving faster, so it will arrive at
point B first.
(b)
The wave that travels through the Earth must travel
e
j
a distance of
2 R sin 30.0° = 2 6.37 × 10 6 m sin 30.0° = 6.37 × 10 6 m
at a speed of
7 800 m/s
Therefore, it takes
6.37 × 10 6 m
= 817 s
7 800 m s
The wave that travels along the Earth’s surface must travel
FG π radIJ = 6.67 × 10
H3 K
a distance of
s = Rθ = R
at a speed of
4 500 m/s
Therefore, it takes
6.67 × 10 6
= 1 482 s
4 500
The time difference is
665 s = 11.1 min
6
m
476
P16.5
Wave Motion
b
g b
where t is the travel time for the faster wave.
a
fb g b
ga f
b4.50 km sga17.3 sf = 23.6 s
or t =
a7.80 − 4.50f km s
and the distance is d = b7.80 km sga 23.6 sf = 184 km
Then, 7.80 − 4.50 km s t = 4.50 km s 17.3 s
Section 16.2
P16.6
.
Sinusoidal Waves
Using data from the observations, we have λ = 1.20 m
and f =
8.00
12.0 s
fFGH 128..000 s IJK =
a
Therefore, v = λf = 1.20 m
P16.7
0.800 m s
f=
40.0 vibrations 4
= Hz
30.0 s
3
λ=
v 42.5 cm s
= 4
= 31.9 cm = 0.319 m
f
3 Hz
a
fa
v=
425 cm
= 42.5 cm s
10.0 s
f
P16.8
v = fλ = 4.00 Hz 60.0 cm = 240 cm s = 2.40 m s
P16.9
y = 0.020 0 m sin 2.11x − 3.62t in SI units
A = 2.00 cm
k = 2.11 rad m
λ=
2π
= 2.98 m
k
ω = 3.62 rad s
f=
ω
= 0.576 Hz
2π
b
v = fλ =
P16.10
ga
The distance the waves have traveled is d = 7.80 km s t = 4.50 km s t + 17.3 s
b
g a
f
ω 2π 3.62
=
= 1.72 m s
2π k
2.11
g a
f
y = 0.005 1 m sin 310 x − 9.30t SI units
v=
ω 9.30
=
= 0.030 0 m s
k 310
s = vt = 0.300 m in positive x - direction
f
Chapter 16
*P16.11
a
f db
g b
(a)
The transverse velocity is
∂y
= − Aω cos kx − ωt
∂t
Its maximum magnitude is
Aω = 12 cm 31.4 rad s = 3.77 m s
ay =
(b)
∂v y
∂t
=
a
b
a
c
fh
f
g
a
∂
− Aω cos kx − ωt = − Aω 2 sin kx − ωt
∂t
a
f
fe
Aω 2 = 0.12 m 31.4 s −1
The maximum value is
P16.12
gi
From y = 12.0 cm sin 1.57 rad m x − 31.4 rad s t
a
f a
j
2
= 118 m s 2
f
At time t, the phase of y = 15.0 cm cos 0.157 x − 50.3t at coordinate x is
b
g b
g
φ = 0.157 rad cm x − 50.3 rad s t . Since 60.0° =
φB = φ A ±
π
rad , or (since x A = 0 ),
3
π
rad , the requirement for point B is that
3
b0.157 rad cmgx − b50.3 rad sgt = 0 − b50.3 rad sgt ± π3 rad .
B
This reduces to x B =
P16.13
±π rad
= ±6.67 cm .
3 0.157 rad cm
b
a
g
f
y = 0.250 sin 0.300 x − 40.0t m
a
Compare this with the general expression y = A sin kx − ωt
(a)
A = 0.250 m
(b)
ω = 40.0 rad s
(c)
k = 0.300 rad m
(d)
λ=
(e)
v = fλ =
(f)
The wave moves to the right, in + x direction .
2π
2π
=
= 20.9 m
k
0.300 rad m
FG ω IJ λ = FG 40.0 rad s IJ a20.9 mf =
H 2π K H 2π K
133 m s
f
477
478
P16.14
Wave Motion
(a)
See figure at right.
(b)
T=
2π
=
ω
y (cm)
2π
= 0.125 s
50.3
10
0
This agrees with the period found in the example
in the text.
t (s)
0.1
—10
0.2
FIG. P16.14
P16.15
(a)
k=
φ = −0.785
In general,
Assuming
b
(a)
y (mm)
0.2
0.1
0.0
–0.1
–0.2
t=0
0.2
0.4
x (mm)
FIG. P16.16(a)
(b)
2π
= 18.0 rad m
λ 0.350 m
1
1
T= =
= 0.083 3 s
f 12.0 s
k=
2π
=
ω = 2π f = 2π 12.0 s = 75.4 rad s
f
b ga
y = A sinb kx + ωt + φ g specializes to
y = 0.200 m sinb18.0 x m + 75.4t s + φ g
v = fλ = 12.0 s 0.350 m = 4.20 m s
(c)
at x = 0 , t = 0 we require
b g
−3.00 × 10 −2 m = 0. 200 m sin +φ
φ = −8.63° = −0.151 rad
so
b g a0.200 mf sinb18.0 x m + 75.4t s − 0.151 radg
y x, t =
g
y = 0.080 0 sin 7.85 x + 6π t − 0.785 m
Therefore,
P16.16
−1
or
b g
Or (where y 0 , t = 0 at t = 0 )
λ
=
2π
then we require that
Therefore,
(b)
2π
a0.800 mf = 7.85 m
ω = 2π f = 2π a3.00f = 6.00π rad s
y = A sina kx + ωt f
y = b0.080 0g sinb7.85 x + 6π t g m
y = 0.080 0 sinb7.85 x + 6π t + φ g
yb x, 0g = 0 at x = 0.100 m
0 = 0.080 0 sinb0.785 + φ g
A = y max = 8.00 cm = 0.080 0 m
Chapter 16
P16.17
f FGH π8 x + 4π tIJK
a
y = 0.120 m sin
(a)
(b)
v=
dy
:
dt
a=
dv
:
dt
fa f FGH π8 x + 4π tIJK
va0.200 s, 1.60 mf = −1.51 m s
Fπ
I
a = a −0.120 mfa 4π f sinG x + 4π tJ
H8
K
aa0.200 s, 1.60 mf = 0
a
x = 0.120 4π cos
2
π 2π
=
:
8
λ
2π
ω = 4π =
:
T
k=
λ = 16.0 m
T = 0.500 s
v=
P16.18
479
(a)
λ 16.0 m
=
= 32.0 m s
T 0.500 s
b g
b
g
yb0 , 0g = A sin φ = 0.020 0 m
y x , t = A sin kx + ωt + φ
Let us write the wave function as
dy
dt
= Aω cos φ = −2.00 m s
0, 0
ω=
Also,
2π
2π
=
= 80.0 π s
T
0.025 0 s
A 2 = xi2 +
FG v IJ = b0.020 0 mg + FG 2.00 m s IJ
HωK
H 80.0 π s K
i
2
2
2
A = 0.021 5 m
(b)
A sin φ 0.020 0
= −2 = −2.51 = tan φ
A cos φ
80 .0π
a
f
Your calculator’s answer tan −1 −2.51 = −1.19 rad has a negative sine and positive cosine,
just the reverse of what is required. You must look beyond your calculator to find
φ = π − 1.19 rad = 1.95 rad
b
g
(c)
v y, max = Aω = 0.021 5 m 80.0π s = 5.41 m s
(d)
λ = v x T = 30.0 m s 0.025 0 s = 0.750 m
b
k=
2π
λ
=
ga
2π
= 8.38 m
0.750 m
b g b
g b
f
ω = 80.0π s
y x , t = 0.021 5 m sin 8.38 x rad m + 80.0π t rad s + 1.95 rad
g
480
P16.19
Wave Motion
(a)
f=
v
=
λ
b1.00 m sg =
2.00 m
b
0.500 Hz
g
ω = 2π f = 2π 0.500 s = 3.14 rad s
2π
2π
= 3.14 rad m
2.00 m
(b)
k=
(c)
y = A sin kx − ωt + φ becomes
λ
b
g
a0.100 mf sinb3.14x m − 3.14t s + 0g
y=
(d)
=
For x = 0 the wave function requires
a
f b
a
f b
y = 0.100 m sin −3.14t s
(e)
(f)
g
y = 0.100 m sin 4.71 rad − 3.14 t s
vy =
g
∂y
= 0.100 m − 3.14 s cos 3.14x m − 3.14t s
∂t
b
g b
g
The cosine varies between +1 and –1, so
b
v y ≤ 0.314 m s
P16.20
g
a0.100 mf sina1.00 rad − 20.0tf
(a)
at x = 2.00 m , y =
(b)
y = 0.100 m sin 0.500 x − 20.0t = A sin kx − ωt
a
f a
f
so ω = 20.0 rad s and f =
Section 16.3
P16.21
P16.22
ω
2π
a
f
= 3.18 Hz
The Speed of Waves on Strings
The down and back distance is 4.00 m + 4.00 m = 8.00 m .
a
f
The speed is then
v=
d total 4 8.00 m
T
=
= 40.0 m s =
t
0.800 s
µ
Now,
µ=
0.200 kg
= 5.00 × 10 −2 kg m
4.00 m
So
T = µv 2 = 5.00 × 10 −2 kg m 40.0 m s
The mass per unit length is: µ =
jb
e
g
2
= 80.0 N
0.060 0 kg
= 1.20 × 10 −2 kg m .
5.00 m
b
gb
The required tension is: T = µv 2 = 0.012 0 kg m 50.0 m s
g
2
= 30.0 N .
Chapter 16
P16.23
v=
P16.24
(a)
T
µ
1 350 kg ⋅ m s 2
=
5.00 × 10 −3 kg m
= 520 m s
a f
ω = 2π f = 2π 500 = 3 140 rad s , k =
j b
e
y = 2.00 × 10 −4 m sin 16.0 x − 3 140t
v = 196 m s =
(b)
481
ω 3 140
=
= 16.0 rad m
v
196
g
T
4.10 × 10 −3 kg m
T = 158 N
P16.25
P16.26
T
Mg
T = Mg is the tension;
v=
Then,
MgL L2
= 2
m
t
and
g=
=
µ
m
L
MgL L
= is the wave speed.
m
t
=
e
e
j
1.60 m 4.00 × 10 −3 kg
Lm
=
= 1.64 m s 2
Mt 2 3.00 kg 3.61 × 10 −3 s 2
j
T
v=
µ
T = µv 2 = ρAv 2 = ρπr 2 v 2
ja fe
e
j b200 m sg
T = 8 920 kg m3 π 7.50 × 10 −4 m
2
2
T = 631 N
P16.27
Since µ is constant, µ =
T2
v 22
=
T2
P16.28
T1
v12
and
Fv I
=G J
Hv K
2
1
2
F 30.0 m s I a6.00 Nf =
T =G
H 20.0 m s JK
The period of the pendulum is T = 2π
2
1
13.5 N .
L
g
Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is
vertical and stationary. The speed of waves in the string is then:
v=
Mg
MgL
F
= m =
µ
m
L
Since it might be difficult to measure L precisely, we eliminate
so v =
Mg T g
Tg
=
m 2π
2π
M
.
m
L=
T g
2π
482
P16.29
Wave Motion
If the tension in the wire is T, the tensile stress is
Stress =
T
A
a
f
T = A stress .
so
The speed of transverse waves in the wire is
v=
T
µ
=
a
A Stress
m
L
f=
Stress
m
AL
=
Stress
m
Volume
=
Stress
ρ
where ρ is the density. The maximum velocity occurs when the stress is a maximum:
v max =
P16.30
2.70 × 10 8 Pa
= 185 m s .
7 860 kg m 3
mg = 2T sin θ
From the free-body diagram
T=
The angle θ is found from
mg
2 sin θ
cos θ =
3L
8
L
2
=
3
4
FIG. P16.30
∴θ = 41.4°
(a)
v=
T
v=
µ
v=
or
(b)
P16.31
mg
=
2 µ sin 41. 4°
F 30.4
GH
I
J
kg K
ms
F
I
9.80 m s
GG
JJ
×
°
2
8
.
00
10
kg
m
sin
41
.
4
e
j
H
K
2
−3
m
m
m = 3.89 kg
v = 60.0 = 30.4 m and
The total time is the sum of the two times.
µ
L
=L
v
T
Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as
m = ρV = ρAL and also as m = µL .
In each wire
t=
Then we have
µ = ρA =
Thus,
t=L
For copper,
For steel,
The total time is
πρd 2
4
F πρd I
GH 4T JK
L aπ fb8 920ge1.00 × 10
t = a 20.0 fM
MM
a4fa150f
N
L aπ fb7 860ge1.00 × 10
t = a30.0fM
MM
a4fa150f
N
2
12
0.137 + 0.192 = 0.329 s
j OP
PP
Q
j OP
PP
Q
12
−3 2
= 0.137 s
12
−3 2
= 0.192 s
Chapter 16
P16.32
Refer to the diagrams. From the free-body diagram of point A:
∑ Fy = 0 ⇒ T1 sin θ = Mg
∑ Fx = 0 ⇒ T1 cos θ = T
and
Combining these equations to eliminate T1 gives the tension in the
Mg
.
string connecting points A and B as: T =
tan θ
v=
T
=
µ
Mg
tan θ
m
L
=
D
L/4
L/4
θ
A
d
B
MgL
m tan θ
θ
d
L/2
M
The speed of transverse waves in this segment of string is then
483
M
T1
θ
A
T
and the time for a pulse to travel from A to B is
t=
*P16.33
L
2
v
=
f has units Hz = 1 s , so T =
(a)
Mg
mL tan θ
.
4Mg
FIG. P16.32
1
has units of seconds, s . For the other T we have T = µv 2 ,
f
kg m 2 kg ⋅ m
=
= N .
m s2
s2
with units
The first T is period of time; the second is force of tension.
(b)
Section 16.4
Reflection and Transmission
Problem 7 in Chapter 18 can be assigned with this section.
Section 16.5
P16.34
f=
P=
P16.35
Rate of Energy Transfer by Sinusoidal Waves on Strings
v
λ
=
30.0
= 60.0 Hz
0.500
FG
H
ω = 2π f = 120π rad s
IJ a
K
1
1 0.180
µω 2 A 2 v =
120π
2
2 3.60
f a0.100f a30.0f =
2
2
1.07 kW
Suppose that no energy is absorbed or carried down into the water. Then a fixed amount of power is
spread thinner farther away from the source, spread over the circumference 2π r of an expanding
circle. The power-per-width across the wave front
P
2π r
is proportional to amplitude squared so amplitude is proportional to
P
.
2π r
484
P16.36
Wave Motion
T = constant; v =
T
µ
; P=
1
µω 2 A 2 v
2
(a)
If L is doubled, v remains constant and P is constant .
(b)
If A is doubled and ω is halved, P ∝ ω 2 A 2 remains constant .
(c)
If λ and A are doubled, the product ω 2 A 2 ∝
A2
remains constant, so
λ2
P remains constant .
(d)
If L and λ are halved, then ω 2 ∝
1
is quadrupled, so P is quadrupled .
λ2
(Changing L doesn’t affect P ).
P16.37
A = 5.00 × 10 −2 m
T
v=
Therefore,
P=
µ = 4.00 × 10 −2 kg m
1
µω 2 A 2 v :
2
µ
ω2 =
P = 300 W
T = 100 N
= 50.0 m s
a f
2 300
2P
=
µA 2 v 4.00 × 10 −2 5.00 × 10 −2
e
je
j a50.0f
2
ω = 346 rad s
ω
= 55.1 Hz
f=
2π
P16.38
µ = 30.0 g m = 30.0 × 10 −3 kg m
λ = 1.50 m
f = 50.0 Hz:
ω = 2π f = 314 s −1
2 A = 0.150 m:
A = 7.50 × 10 −2 m
(a)
FG 2π x − ωtIJ
Hλ K
y = e7.50 × 10 j sina 4.19 x − 314t f
y = A sin
−2
P16.39
IJ W
ja f e7.50 × 10 j FGH 4314
.19 K
1
1
µω 2 A 2 v = 30.0 × 10 −3 314
2
2
e
FIG. P16.38
(b)
P=
(a)
v = fλ =
(b)
λ=
2π
2π
=
m = 7.85 m
k
0.800
(c)
f=
50.0
= 7.96 Hz
2π
(d)
P=
1
1
µω 2 A 2 v = 12.0 × 10 −3 50.0
2
2
−2 2
2
ω 2π ω 50.0
m s = 62.5 m s
= =
k 0.800
2π k
e
ja f a0.150f a62.5f W =
2
2
21.1 W
P = 625 W
Chapter 16
*P16.40
FG
H
Comparing y = 0.35 sin 10πt − 3πx +
k=
(a)
The rate of energy transport is
jb
1
1
µω 2 A 2 v = 75 × 10 −3 kg m 10π s
2
2
e
g a0.35 mf 3.33 m s =
2
2
15.1 W .
The energy per cycle is
Eλ = P T =
P16.41
IJ with y = A sinbkx − ωt + φ g = A sinbωt − kx − φ + π g we have
K
3π
λ ω 10π s
= =
= 3.33 m s .
, ω = 10π s , A = 0.35 m . Then v = fλ = 2π f
m
2π k 3π m
P=
(b)
π
4
485
jb
1
1
µω 2 A 2 λ = 75 × 10 −3 kg m 10π s
2
2
e
g a0.35 mf
2
2
2π m
= 3.02 J .
3π
Originally,
1
µω 2 A 2 v
2
1
T
P0 = µω 2 A 2
µ
2
P0 =
1
P0 = ω 2 A 2 Tµ
2
The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, it
can carry power larger by 2 times.
2 P0 =
*P16.42
1 2 2
ω A T 2µ
2
As for a strong wave, the rate of energy transfer is proportional to the square of the amplitude and to
the speed. We write P = FvA 2 where F is some constant. With no absorption of energy,
2
2
= Fv mudfill A mudfill
Fv bedrock A bedrock
v bedrock
A
= mudfill =
v mudfill
A bedrock
The amplitude increases by 5.00 times.
25 v mudfill
=5
v mudfill
486
Wave Motion
Section 16.6
P16.43
The Linear Wave Equation
a
f
(a)
A = 7.00 + 3.00 4.00 yields A = 40.0
(b)
In order for two vectors to be equal, they must have the same magnitude and the same
direction in three-dimensional space. All of their components must be equal. Thus,
7.00 i + 0 j + 3.00k = A i + Bj + Ck requires A = 7.00 , B = 0 , and C = 3.00 .
(c)
In order for two functions to be identically equal, they must be equal for every value of
every variable. They must have the same graphs. In
a
f
a
f
A + B cos Cx + Dt + E = 0 + 7.00 mm cos 3.00 x + 4.00t + 2.00 ,
the equality of average values requires that A = 0 . The equality of maximum values
requires B = 7.00 mm . The equality for the wavelength or periodicity as a function of x
requires C = 3.00 rad m . The equality of period requires D = 4.00 rad s , and the
equality of zero-crossings requires E = 2.00 rad .
*P16.44
∂2y
The linear wave equation is
∂x
1 ∂2 y
v 2 ∂t 2
If
y = e b a x − vt f
then
∂y
∂y
= − bve b a x − vt f and
= be b a x − vt f
∂t
∂x
∂2 y
∂t 2
= b 2 v 2 e b a x − vt f and
∂2y
Therefore,
P16.45
=
2
∂t
1 ∂2y
The linear wave equation is
a
To show that y = ln b x − vt
2
∂2y
∂x
= b 2 e b a x − vt f
∂x 2
, demonstrating that e b a x − vt f is a solution
2
∂2y
=
v 2 ∂t 2
f
= v2
∂2y
∂x 2
is a solution, we find its first and second derivatives with respect to x
and t and substitute into the equation.
∂2y
∂y
1
=
− bv
∂t b x − vt
fa f
a
∂y
= ba x − vt f
∂x
−1
=
∂t 2
∂2y
b
∂x
e j
a f
2
−v
1 ∂2y 1
Then 2 2 = 2
v ∂t
v x − vt
2
=−
2
a f
−1 − bv
a
f
2
b
x − vt
b
f
b 2 x − vt
=−
1
ax − vtf
2
2
a
=
∂2y
∂x 2
=−
2
v2
ax − vtf
=−
2
1
ax − vtf
2
so the given wave function is a solution.
Chapter 16
P16.46
(a)
From y = x 2 + v 2 t 2 ,
∂2y
∂y
= 2x
∂x
evaluate
∂x 2
∂2y
∂y
= v 2 2t
∂t
Does
∂2y
∂t 2
equation.
Note
So
(c)
a
1
x + vt
2
a
∂t 2
=2
= 2v 2
1 ∂2y
?
v 2 ∂t 2
=
By substitution: 2 =
(b)
487
f
2
+
1
2 v 2 and this is true, so the wave function does satisfy the wave
v2
a
1
x − vt
2
f 12 ax + vtf
f x + vt =
f
2
2
1 2
1
1
1
x + xvt + v 2 t 2 + x 2 − xvt + v 2 t 2
2
2
2
2
2
2 2
= x + v t as required.
=
a
f 12 ax − vtf
and g x − vt =
2
.
y = sin x cos vt makes
∂2y
∂y
= cos x cos vt
∂x
∂x 2
∂2y
∂y
= − v sin x sin vt
∂t
Then
∂2y
∂x
2
=
∂t 2
= − sin x cos vt
= − v 2 sin x cos vt
1 ∂2 y
v 2 ∂t 2
−1 2
v sin x cos vt which is true as required.
v2
Note sin x + vt = sin x cos vt + cos x sin vt
becomes − sin x cos vt =
a
f
a
f
sin x − vt = sin x cos vt − cos x sin vt .
a f a f
1
f a x + vtf = sina x + vt f
2
So sin x cos vt = f x + vt + g x − vt with
and
a
f
g x − vt =
a
1
sin x − vt
2
f
.
Additional Problems
P16.47
Assume a typical distance between adjacent people ~ 1 m .
Then the wave speed is
v=
∆x 1 m
~
~ 10 m s
∆t 0.1 s
Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the
stadium is
T=
e j
2
2π r 2π 10
~
= 63 s ~ 1 min .
v
10 m s
488
P16.48
Wave Motion
a
f
Compare the given wave function y = 4.00 sin 2.00 x − 3.00t cm to the general form
a
f
y = A sin kx − ωt to find
P16.49
(a)
amplitude A = 4.00 cm = 0.040 0 m
(b)
k=
(c)
ω = 2π f = 3.00 s −1 and f = 0.477 Hz
(d)
T=
(e)
The minus sign indicates that the wave is traveling in the positive x -direction .
(a)
Let u = 10π t − 3π x +
2π
λ
= 2.00 cm −1 and λ = π cm = 0.031 4 m
1
= 2.09 s
f
π
4
du
dx
= 10π − 3π
= 0 at a point of constant phase
dt
dt
dx 10
=
= 3.33 m s
3
dt
The velocity is in the positive x -direction .
f FGH
y 0.100 , 0 = 0.350 m sin −0.300π +
(c)
k=
(d)
*P16.50
g a
b
(b)
(a)
2π
λ
IJ
K
π
= −0.054 8 m = −5.48 cm
4
= 3π : λ = 0.667 m
ω = 2π f = 10π : f = 5.00 Hz
fa f FGH
0.175 m = a0.350 mf sin b99.6 rad sgt
∴ sin b99.6 rad sgt = 0.5
vy =
∂y
π
= 0.350 10π cos 10π t − 3π x +
∂t
4
a
IJ
K
a fa
f
v y, max = 10π 0.350 = 11.0 m s
The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e., 0.523 6 rad
and 2.618 rad.
99.6 rad s t1 = 0.523 6 rad , thus t1 = 5.26 ms
b
g
b99.6 rad sgt
2
= 2.618 rad , thus t 2 = 26.3 ms
∆t ≡ t 2 − t1 = 26.3 ms − 5.26 ms = 21.0 ms
(b)
P16.51
Distance traveled by the wave =
FG ω IJ ∆t = FG 99.6 rad s IJ e21.0 × 10 sj =
H k K H 1.25 rad m K
−3
The equation v = λf is a special case of
speed = (cycle length)(repetition rate).
e
jb
g
Thus, v = 19.0 × 10 −3 m frame 24.0 frames s = 0.456 m s .
1.68 m .
Chapter 16
P16.52
Assuming the incline to be frictionless and taking the positive x-direction to be up the incline:
∑ Fx = T − Mg sinθ = 0
T = Mg sin θ
or the tension in the string is
v=
The speed of transverse waves in the string is then
The time interval for a pulse to travel the string’s length is ∆t =
P16.53
µ
=
g
+ Us
j
top
0 + Mgx + 0 + 0 = 0 + 0 +
x=
b
2 Mg
k
ge
j
e
+ ∆E = K + U g + U s
1 2
kx
2
(a)
T = kx = 2 Mg = 2 2.00 kg 9.80 m s 2 = 39.2 N
(b)
L = L0 + x = L0 +
(c)
v=
2 Mg
k
39.2 N
L = 0.500 m +
= 0.892 m
100 N m
v=
T
µ
=
TL
m
39.2 N × 0.892 m
5.0 × 10 −3 kg
v = 83.6 m s
Mgx =
1 2
kx
2
(a)
T = kx = 2 Mg
(b)
L = L0 + x = L0 +
(c)
v=
T
µ
=
TL
=
m
2 Mg
k
FG
H
2 Mg
2 Mg
L0 +
m
k
IJ
K
Mg sin θ
m
L
=
L
m
=L
=
v
MgL sin θ
Energy is conserved as the block moves down distance x:
eK + U
P16.54
T
j
bottom
MgL sin θ
m
mL
Mg sin θ
489
490
P16.55
Wave Motion
T
v=
(b)
From Equation 16.21, P =
µ
=
80.0 N
(a)
e5.00 × 10
−3
= 179 m s
j
kg 2.00 m
FG IJ
H K
1
v
µvω 2 A 2 and ω = 2π
2
λ
FG IJ
H K
1
2πv
µvA 2
λ
2
P=
2π 2
P=
FH
2
=
2π 2 µA 2 v 3
λ2
IK b0.040 0 mg b179 m sg
a0.160 mf
5.00 ×10 −3 kg
2.00 m
2
3
2
P = 1.77 × 10 4 W = 17.7 kW
P16.56
v=
T
µ
Now v = fλ implies v =
ω
so that
k
FG IJ
H K
µ ω
m=
g k
*P16.57
µv 2
.
g
and in this case T = mg ; therefore, m =
2
0.250 kg m
=
9.80 m s 2
LM 18π s OP
N 0.750π m Q
−1
−1
Let M = mass of block, m = mass of string. For the block,
2
= 14.7 kg .
∑ F = ma implies T =
speed of a wave on the string is then
T
v=
t=
µ
(a)
µ=
v=
M
m
m
M
m
=
M
0.003 2 kg
= 0.084 3 rad
0.450 kg
dm
dx
= ρA
= ρA
dL
dx
T
µ
=
T
=
ρA
T
ρ ax + b
a
With all SI units, v =
(b)
= rω
m
r
r 1
=
v ω
θ = ωt =
P16.58
Mω 2 r
=
v x= 0 =
v x=10 .0 =
e
f
=
T
j
ρ 10 x + 10 −2 10 −4
b2 700ge0 + 10 je10 j
−2
24.0
−2
−4
j
ρ 10 x + 10 −2 cm 2
−3
24.0
b2 700ge10
e
T
−3
ms
= 94.3 m s
je j
+ 10 −2 10 −4
= 66.7 m s
mv b2
= mω 2 r . The
r
Chapter 16
P16.59
v=
T
where
µ
Therefore,
dx
, so that
But v =
dt
T = µxg , the weight of a length x, of rope.
v = gx
dx
dt =
gx
t=
and
z
L
0
P16.60
dx
gx
=
1
x
1
2
g
L
= 2
0
L
g
FG mxg IJ + Mg , so the wave speed is:
H LK
T
TL
F MgL IJ = dx .
=
= xg + G
v=
H m K dt
µ
m
L F MgL IJ OP dx
1 xg + b MgL m g
t=
Then t = z dt = z M xg + G
g
N H m KQ
MgL I
2 LF
LF m+M − MI
F MgL IJ OP
−G
t = MG Lg +
t=2
J
JK
H m K PQ
g MNH
m K
g GH
m
L F m − 0I
When M = 0 , as in the previous problem,
t=2
J = 2 Lg
g GH
m K
F 1 m − 1 m + …I
F mI
As m → 0 we expand m + M = M G 1 + J = M G 1 +
H MK
H 2 M 8 M JK
F M + em M j − em M j + … − M I
L
G
JJ
to obtain
t=2
gG
m
H
K
LF1 m I
mL
=
t≈2
J
G
gH2 MK
Mg
At distance x from the bottom, the tension is T =
(a)
t
L
0
0
(b)
1 2 x=L
−1 2
1
2
12
2
2
1
2
(a)
x=0
12
12
(c)
P16.61
491
2
1
8
32
The speed in the lower half of a rope of length L is the same function of distance (from the
L
bottom end) as the speed along the entire length of a rope of length
.
2
L
L′
with L ′ =
Thus, the time required = 2
g
2
FG IJ
H K
and the time required = 2
F
GH
L
L
= 0.707 2
2g
g
I
JK
.
It takes the pulse more that 70% of the total time to cover 50% of the distance.
(b)
By the same reasoning applied in part (a), the distance climbed in τ is given by d =
L
L
, we find the distance climbed =
.
g
4
1
In half the total trip time, the pulse has climbed of the total length.
4
For τ =
t
=
2
gτ 2
.
4
492
P16.62
P16.63
Wave Motion
(a)
v=
ω 15.0
=
= 5.00 m s in positive x -direction
k 3.00
(b)
v=
15.0
= 5.00 m s in negative x -direction
3.00
(c)
v=
15.0
= 7.50 m s in negative x -direction
2.00
(d)
v=
12.0
1
2
= 24.0 m s in positive x -direction
T
A
∆L
L
Young’s modulus for the wire may be written as Y =
, where T is the tension maintained in the
wire and ∆L is the elongation produced by this tension. Also, the mass density of the wire may be
µ
.
A
The speed of transverse waves in the wire is then
expressed as ρ =
v=
T
µ
=
T
A
µ
=
Y
A
c h
∆L
L
ρ
∆L ρv 2
=
.
L
Y
If the wire is aluminum and v = 100 m s, the strain is
and the strain in the wire is
jb
e
2.70 × 10 3 kg m3 100 m s
∆L
=
L
7.00 × 10 10 N m 2
*P16.64
(a)
g
2
= 3.86 × 10 −4 .
Consider a short section of chain at the top of the loop. A freebody diagram is shown. Its length is s = R 2θ and its mass is
µR2θ . In the frame of reference of the center of the loop,
Newton’s second law is
mv 02
µR 2θv 02
2T sin θ down =
down =
∑ Fy = ma y
R
R
θ
θ
a f
2θ
T
R
FIG. P16.64(a)
For a very short section, sin θ = θ and T = µ v 02 .
T
(b)
The wave speed is v =
(c)
In the frame of reference of the center of the loop, each pulse moves with equal speed
clockwise and counterclockwise.
µ
= v0 .
v
v0
v0
FIG. P16.64(c-1)
continued on next page
v0
v
T
Chapter 16
493
In the frame of reference of the ground, once pulse moves backward at speed v 0 + v = 2 v 0
and the other forward at v 0 − v = 0 . The one pulse makes two revolutions while the loop
makes one revolution and the other pulse does not move around the loop. If it is generated
at the six-o’clock position, it will stay at the six-o’clock position.
v0
v0
v0
FIG. P16.64(c-2)
P16.65
(a)
Assume the spring is originally stationary throughout, extended to have a length L much
greater than its equilibrium length. We start moving one end forward with the speed v at
which a wave propagates on the spring. In this way we create a single pulse of compression
that moves down the length of the spring. For an increment of spring with length dx and
mass dm, just as the pulse swallows it up, ∑ F = ma
becomes kdx = adm or
But
k
dm
dx
= a.
dm
k
= µ so a = .
µ
dx
dv v
v2
= when vi = 0. But L = vt , so a =
.
dt t
L
Also, a =
Equating the two expressions for a, we have
(b)
Using the expression from part (a) v =
F T I F 2T IJ
v=G J =G
H µK H µ K
F T I F 2T IJ
v′ = G J = G
H µ ′ K H 3µ K
(a)
0
0
12
= v0
= v0
0
∆t left =
L
2
v
∆t right =
=
L
2
v′
L
2 v0 2
=
=
L
2 v0
2
3
2 where v 0 ≡
12
0
(b)
µ
12
12
P16.66
kL
∆t 0
2 2
=
∆t left + ∆t right = 0.966 ∆t 0
=
µ
2
3
v2
or v =
L
kL2
=
m
FG T IJ
Hµ K
12
0
0
2
3
= 0.354∆t 0 where ∆t 0 ≡
∆t 0
2
=
k
= 0.612 ∆t 0
L
v0
kL
µ
.
b100 N mga2.00 mf
0.400 kg
2
= 31.6 m s .
494
P16.67
P16.68
Wave Motion
af
FG IJ
H K
1
1
ω
µω 3 2 −2 bx
A0 e
µω 2 A 2 v = µω 2 A02 e −2 bx
=
2
2
2k
k
(a)
P x =
(b)
P 0 =
(c)
P x
= e −2 bx
P 0
v=
af
µω 3 2
A0
2k
af
af
4 450 km
= 468 km h = 130 m s
9.50 h
b
g
e
j
µ a x f is a linear function, so it is of the form
To have µ a0f = µ we require b = µ . Then
2
130 m s
v2
d=
=
= 1 730 m
g
9.80 m s 2
*P16.69
(a)
0
af
µ aL f = µ
µ x = mx + b
0
a f bµ
µ x =
Then
(b)
= mL + µ 0
µL − µ0
L
m=
so
L
L
g
− µ0 x
L
+ µ0
dx
dx
, the time required to move from x to x + dx is
. The time required to move
v
dt
from 0 to L is
From v =
∆t =
z z
z FGH b
L
dx L dx
1
=
=
T
v
T
0
0
µ
z
L
af
µ x dx
0
I FG µ − µ IJ dxF
∆t =
JK H L K GH µ
T
I 1
L I F b µ − µ gx
1 F
∆t =
+µ J
G
J
G
L
T H µ − µ KH
K
2L
∆t =
eµ − µ j
3 T bµ − µ g
2Le µ − µ je µ + µ µ + µ j
∆t =
3 T e µ − µ je µ + µ j
2L F µ + µ µ + µ I
∆t =
G µ + µ JK
3 TH
1
L
0
g
12
µL − µ0 x
+ µ0
L
0
L
32
L
L
0
0
0
L
0
L
0
L
32
L
32
0
L
L
L
0
L
L
L
0
0
0
0
3
2 0
0
0
L
L
L − µ0
IJ
K
Chapter 16
495
ANSWERS TO EVEN PROBLEMS
P16.2
see the solution
P16.4
(a) the P wave; (b) 665 s
P16.6
0.800 m s
P16.8
2.40 m s
P16.10
0.300 m in the positive x-direction
P16.12
±6.67 cm
P16.14
(a) see the solution; (b) 0.125 s; in
agreement with the example
P16.16
(a) see the solution; (b) 18.0 m ; 83.3 ms ;
75.4 rad s ; 4.20 m s ;
(c) 0.2 m sin 18 x + 75.4t − 0.151
a
P16.18
f a
b g
b0.021 5 mg sinb8.38x + 80.0π t + 1.95g
P16.20
(a) see the solution; (b) 3.18 Hz
P16.22
30.0 N
P16.24
(a) y = 0.2 mm sin 16 x − 3 140t ;
(b) 158 N
P16.26
631 N
P16.28
v=
P16.30
F
m I
(a) v = G 30.4
H s ⋅ kg JK
P16.32
a
Tg
2π
f b
P16.40
(a) 15.1 W ; (b) 3.02 J
P16.42
The amplitude increases by 5.00 times
P16.44
see the solution
P16.46
(a) see the solution;
1
1
2
2
x + vt + x − vt ;
(b)
2
2
1
1
(c) sin x + vt + sin x − vt
2
2
a
a
f
a
f
a
f
P16.48
(a) 0.040 0 m; (b) 0.031 4 m ;
(c) 0.477 Hz; (d) 2.09 s;
(e) positive x -direction
P16.50
(a) 21.0 ms ; (b) 1.68 m
P16.52
∆t =
P16.54
(a) 2Mg ; (b) L0 +
g
M
m
f
(a) y = 0.075 0 sin 4.19 x − 314t ;
(b) 625 W
f
(a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m s ;
(d) y x , t =
g a
b
P16.38
(c)
f
mL
Mg sin θ
2 Mg
;
k
2 Mg
2 Mg
L0 +
m
k
P16.56
14.7 kg
P16.58
(a) v =
FG
H
e
IJ
K
T
−7
ρ 10 x + 10 −6
j
in SI units;
(b) 94.3 m s; 66.7 m s
m ; (b) 3.89 kg
mL tan θ
4Mg
P16.34
1.07 kW
P16.36
(a), (b), (c) P is constant ;
(d) P is quadrupled
P16.60
see the solution
P16.62
(a) 5.00 i m s ; (b) −5.00 i m s ;
(c) −7.50 i m s ; (d) 24.0 i m s
P16.64
(a) µ v 02 ; (b) v 0 ;
(c) One travels 2 rev and the other does
not move around the loop.
496
Wave Motion
F 2T IJ = v 2 ;
(a) v = G
Hµ K
F 2T IJ = v 2 ; (b) 0.966∆t
v′ = G
3
H 3µ K
12
P16.66
0
P16.68
0
0
12
0
0
0
0
130 m s ; 1.73 km
17
Sound Waves
CHAPTER OUTLINE
17.1
17.2
17.3
17.4
17.5
17.6
Speed of Sound Waves
Periodic Sound Waves
Intensity of Periodic Sound
Waves
The Doppler Effect
Digital Sound Recording
Motion Picture Sound
ANSWERS TO QUESTIONS
Q17.1
Sound waves are longitudinal because elements of the
medium—parcels of air—move parallel and antiparallel to the
direction of wave motion.
Q17.2
We assume that a perfect vacuum surrounds the clock. The
sound waves require a medium for them to travel to your ear.
The hammer on the alarm will strike the bell, and the vibration
will spread as sound waves through the body of the clock. If a
bone of your skull were in contact with the clock, you would
hear the bell. However, in the absence of a surrounding
medium like air or water, no sound can be radiated away. A
larger-scale example of the same effect: Colossal storms raging
on the Sun are deathly still for us.
What happens to the sound energy within the clock?
Here is the answer: As the sound wave travels through the
steel and plastic, traversing joints and going around corners, its
energy is converted into additional internal energy, raising the
temperature of the materials. After the sound has died away,
the clock will glow very slightly brighter in the infrared portion
of the electromagnetic spectrum.
1
meter from the sonic ranger, then the sensor would have to measure how long it
2
would take for a sound pulse to travel one meter. Since sound of any frequency moves at about
343 m s, then the sonic ranger would have to be able to measure a time difference of under
0.003 seconds. This small time measurement is possible with modern electronics. But it would be
more expensive to outfit sonic rangers with the more sensitive equipment than it is to print “do not
1
use to measure distances less than meter” in the users’ manual.
2
Q17.3
If an object is
Q17.4
The speed of sound to two significant figures is 340 m s. Let’s assume that you can measure time to
1
second by using a stopwatch. To get a speed to two significant figures, you need to measure a
10
time of at least 1.0 seconds. Since d = vt , the minimum distance is 340 meters.
Q17.5
The frequency increases by a factor of 2 because the wave speed, which is dependent only on the
medium through which the wave travels, remains constant.
497
498
Sound Waves
Q17.6
When listening, you are approximately the same distance from all of the members of the group. If
different frequencies traveled at different speeds, then you might hear the higher pitched
frequencies before you heard the lower ones produced at the same time. Although it might be
interesting to think that each listener heard his or her own personal performance depending on
where they were seated, a time lag like this could make a Beethoven sonata sound as if it were
written by Charles Ives.
Q17.7
Since air is a viscous fluid, some of the energy of sound vibration is turned into internal energy. At
such great distances, the amplitude of the signal is so decreased by this effect you re unable to hear
it.
Q17.8
We suppose that a point source has no structure, and radiates sound equally in all directions
(isotropically). The sound wavefronts are expanding spheres, so the area over which the sound
energy spreads increases according to A = 4π r 2 . Thus, if the distance is tripled, the area increases by
a factor of nine, and the new intensity will be one-ninth of the old intensity. This answer according
to the inverse-square law applies if the medium is uniform and unbounded.
For contrast, suppose that the sound is confined to move in a horizontal layer. (Thermal
stratification in an ocean can have this effect on sonar “pings.”) Then the area over which the sound
energy is dispersed will only increase according to the circumference of an expanding circle:
A = 2π rh , and so three times the distance will result in one third the intensity.
In the case of an entirely enclosed speaking tube (such as a ship’s telephone), the area
perpendicular to the energy flow stays the same, and increasing the distance will not change the
intensity appreciably.
Q17.9
He saw the first wave he encountered, light traveling at 3.00 × 10 8 m s . At the same moment,
infrared as well as visible light began warming his skin, but some time was required to raise the
temperature of the outer skin layers before he noticed it. The meteor produced compressional waves
in the air and in the ground. The wave in the ground, which can be called either sound or a seismic
wave, traveled much faster than the wave in air, since the ground is much stiffer against
compression. Our witness received it next and noticed it as a little earthquake. He was no doubt
unable to distinguish the P and S waves. The first air-compression wave he received was a shock
wave with an amplitude on the order of meters. It transported him off his doorstep. Then he could
hear some additional direct sound, reflected sound, and perhaps the sound of the falling trees.
Q17.10
A microwave pulse is reflected from a moving object. The waves that are reflected back are Doppler
shifted in frequency according to the speed of the target. The receiver in the radar gun detects the
reflected wave and compares its frequency to that of the emitted pulse. Using the frequency shift,
the speed can be calculated to high precision. Be forewarned: this technique works if you are either
traveling toward or away from your local law enforcement agent!
Q17.11
As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency;
as you move away, the echo would be shifted down in frequency.
Q17.12
Normal conversation has an intensity level of about 60 dB.
Q17.13
A rock concert has an intensity level of about 120 dB.
A cheering crowd has an intensity level of about 90 dB.
Normal conversation has an intensity level of about 50–60 dB.
Turning a page in the textbook has an intensity level of about 10–20 dB.
Chapter 17
499
Q17.14
One would expect the spectra of the light to be Doppler shifted up in frequency (blue shift) as the
star approaches us. As the star recedes in its orbit, the frequency spectrum would be shifted down
(red shift). While the star is moving perpendicular to our line of sight, there will be no frequency
shift at all. Overall, the spectra would oscillate with a period equal to that of the orbiting stars.
Q17.15
For the sound from a source not to shift in frequency, the radial velocity of the source relative to the
observer must be zero; that is, the source must not be moving toward or away from the observer.
The source can be moving in a plane perpendicular to the line between it and the observer. Other
possibilities: The source and observer might both have zero velocity. They might have equal
velocities relative to the medium. The source might be moving around the observer on a sphere of
constant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequency
heard will be equal to the frequency emitted by the source.
Q17.16
Wind can change a Doppler shift but cannot cause one. Both v o and v s in our equations must be
interpreted as speeds of observer and source relative to the air. If source and observer are moving
relative to each other, the observer will hear one shifted frequency in still air and a different shifted
frequency if wind is blowing. If the distance between source and observer is constant, there will
never be a Doppler shift.
Q17.17
If the object being tracked is moving away from the observer, then the sonic pulse would never
reach the object, as the object is moving away faster than the wave speed. If the object being tracked
is moving towards the observer, then the object itself would reach the detector before reflected
pulse.
Q17.18
New-fallen snow is a wonderful acoustic absorber as it reflects very little of the sound that reaches it.
It is full of tiny intricate air channels and does not spring back when it is distorted. It acts very much
like acoustic tile in buildings. So where does the absorbed energy go? It turns into internal
energy—albeit a very small amount.
Q17.19
As a sound wave moves away from the source, its intensity decreases. With an echo, the sound must
move from the source to the reflector and then back to the observer, covering a significant distance.
Q17.20
The observer would most likely hear the sonic boom of the plane itself and then beep, baap, boop.
Since the plane is supersonic, the loudspeaker would pull ahead of the leading “boop” wavefront
before emitting the “baap”, and so forth.
“How are you?” would be heard as “?uoy era woH”
Q17.21
This system would be seen as a star moving in an elliptical path. Just like the light from a star in a
binary star system, described in the answer to question 14, the spectrum of light from the star would
undergo a series of Doppler shifts depending on the star’s speed and direction of motion relative to
the observer. The repetition rate of the Doppler shift pattern is the period of the orbit. Information
about the orbit size can be calculated from the size of the Doppler shifts.
SOLUTIONS TO PROBLEMS
Section 17.1
Speed of Sound Waves
b
ga
f
P17.1
Since v light >> v sound : d ≈ 343 m s 16. 2 s = 5.56 km
P17.2
v=
B
ρ
=
2.80 × 10 10
= 1.43 km s
13.6 × 10 3
500
P17.3
Sound Waves
a20.0 m − 1.75 mf = 5.32 × 10
Sound takes this time to reach the man:
−2
343 m s
so the warning should be shouted no later than
before the pot strikes.
Since the whole time of fall is given by y =
1 2
gt :
2
s
0.300 s + 5.32 × 10 −2 s = 0.353 s
18.25 m =
1
9.80 m s 2 t 2
2
e
j
t = 1.93 s
P17.4
the warning needs to come
1.93 s − 0.353 s = 1.58 s
into the fall, when the pot has fallen
1
9.80 m s 2 1.58 s
2
to be above the ground by
20.0 m − 12.2 m = 7.82 m
(a)
At 9 000 m, ∆T =
e
ja
f
2
= 12.2 m
FG 9 000 IJ a−1.00° Cf = −60.0° C so T = −30.0° C .
H 150 K
Using the chain rule:
a
fFGH IJK
a
f
dv dv dT dx
dv dT
v
dv
1
, so dt = 247 s
=
=v
= v 0.607
=
dt dT dx dt
dT dx
v
150
247
f z dvv
z
L 331.5 + 0.607a30.0f OP
Fv I
t = a 247 sf lnG J = a 247 sf ln M
Hv K
N 331.5 + 0.607a−30.0f Q
t
a
vf
dt = 247 s
0
vi
f
i
t = 27.2 s for sound to reach ground.
(b)
t=
9 000
h
=
= 25.7 s
v 331.5 + 0.607 30.0
a f
It takes longer when the air cools off than if it were at a uniform temperature.
*P17.5
Let x 1 represent the cowboy’s distance from the nearer canyon wall and x 2 his distance from the
farther cliff. The sound for the first echo travels distance 2 x1 . For the second, 2 x 2 . For the third,
2 x 2 − 2 x1
2 x1 + 2 x 2 − 2 x 2
2 x1 + 2 x 2 . For the fourth echo, 2 x1 + 2 x 2 + 2 x1 . Then
= 1.92 s and
= 1.47 s .
340 m s
340 m s
2x2
1
= 1.92 s + 1.47 s ; x 2 = 576 m.
Thus x 1 = 340 m s 1.47 s = 250 m and
2
340 m s
(a)
(b)
So x 1 + x 2 = 826 m
b
2 x 1 + 2 x 2 + 2 x 1 − 2 x1 + 2 x 2
340 m s
g=
1.47 s
Chapter 17
P17.6
It is easiest to solve part (b) first:
(b)
The distance the sound travels to the plane is d s = h 2 +
FG h IJ
H 2K
2
=
h 5
.
2
The sound travels this distance in 2.00 s, so
b
ga f
2a686 mf
= 614 m .
giving the altitude of the plane as h =
ds =
h 5
= 343 m s 2.00 s = 686 m
2
5
(a)
a
Thus, the speed of the plane is: v =
Section 17.2
P17.7
*P17.8
λ=
*P17.9
P17.10
26° C
= 346 m s
273° C
Let t represent the time for the echo to return. Then
1
1
vt = 346 m s 24 × 10 −3 s = 4.16 m .
2
2
Let ∆t represent the duration of the pulse:
∆t =
(c)
307 m
= 153 m s .
2.00 s
340 m s
v
=
= 5.67 mm
f 60.0 × 10 3 s −1
d=
(b)
h
= 307 m .
2
Periodic Sound Waves
The sound speed is v = 331 m s 1 +
(a)
f
The distance the plane has traveled in 2.00 s is v 2.00 s =
L = 10 λ =
b
10λ 10 λ 10
10
=
=
=
= 0.455 µs .
v
fλ
f
22 × 10 6 1 s
g
10 v 10 346 m s
=
= 0.157 mm
f
22 × 10 6 1 s
v 1 500 m s
=
= 1.50 mm
f
10 6 s
1 500 m s
= 75.0 µm
If f = 20 MHz , λ =
2 × 10 7 s
If f = 1 MHz , λ =
∆Pmax = ρvω smax
smax
e
j
4.00 × 10 −3 N m 2
∆Pmax
=
=
= 1.55 × 10 −10 m
ρvω
1.20 kg m3 343 m s 2π 10.0 × 10 3 s −1
e
jb
ga fe
j
501
502
P17.11
Sound Waves
A = 2.00 µm
(a)
2π
= 0.400 m = 40.0 cm
15.7
ω 858
= 54.6 m s
v= =
k 15.7
λ=
P17.12
a fb
g a fe
j
(b)
s = 2.00 cos 15.7 0.050 0 − 858 3.00 × 10 −3 = −0.433 µm
(c)
v max = Aω = 2.00 µm 858 s −1 = 1.72 mm s
b
ge
j
F π x − 340π t IJ (SI units)
∆P = a1.27 Paf sinG
Hm s K
(a)
The pressure amplitude is: ∆Pmax = 1.27 Pa .
P17.13
(b)
ω = 2π f = 340π s, so f = 170 Hz
(c)
k=
(d)
v = λf = 2.00 m 170 Hz = 340 m s
k=
ω=
2π
=
λ
2π v
λ
a
smax =
2π
fa
f
2π
= 62.8 m −1
0.100 m
=
ω = 2π f =
k=
= π m , giving λ = 2.00 m
λ
a
Therefore,
P17.14
2π
f
b
2π 343 m s
g = 2.16 × 10
4
−1
s
a0.100 mf
∆P = a0.200 Paf sin 62.8 x m − 2.16 × 10
2π v
λ
=
b
2π 343 m s
a0.100 mf
g = 2.16 × 10
a
jb
4
f
ge
0.200 Pa
∆Pmax
=
= 2. 25 × 10 −8 m
3
ρvω
1.20 kg m 343 m s 2.16 × 10 4 s −1
e
=
2π
j
−1
max
P17.15
ts .
rad s
a0.100 mf = 62.8 m
Therefore, s = s
cosb kx − ω t g = e 2.25 × 10
λ
4
FG 2π v IJ s
H λK
2π a1.20fa343 f e5.50 × 10 j
=
=
∆Pmax = ρvω smax = ρv
2πρv 2 smax
λ=
∆Pmax
−8
j e
max
2
0.840
j
m cos 62.8 x m − 2.16 × 10 4 t s .
−6
5.81 m
503
Chapter 17
P17.16
(a)
The sound “pressure” is extra tensile stress for one-half of each cycle. When it becomes
0.500% 13.0 × 10 10 Pa = 6.50 × 10 8 Pa , the rod will break. Then, ∆Pmax = ρvω smax
a
fe
j
smax =
(b)
6.50 × 10 8 N m 2
∆Pmax
=
= 4.63 mm .
ρvω
8.92 × 10 3 kg m3 5 010 m s 2π 500 s
a
jb
e
From s = smax cos kx − ωt
f
a
f
b
g
f
ga
jb
gb
1
1
1
2
2
ρv ωsmax = ρvvmax
= 8.92 × 10 3 kg m 3 5 010 m s 14.5 m s
2
2
2
= 4.73 × 10 9 W m 2
I=
(c)
*P17.17
g
∂s
= −ωsmax sin kx − ωt
∂t
v max = ωsmax = 2π 500 s 4.63 mm = 14.5 m s
v=
b
gb
af
e
g
2
Let P x represent absolute pressure as a function of x. The net force
to the right on the chunk of air is + P x A − P x + ∆x A . Atmospheric
∂∆P
∆xA .
pressure subtracts out, leaving − ∆P x + ∆x + ∆P x A = −
P x + ∆x A
Px A
∂x
∂2s
The mass of the air is ∆m = ρ∆V = ρA∆x and its acceleration is 2 . So
∂t
FIG. P17.17
Newton’s second law becomes
2
∂∆P
∂ s
−
∆xA = ρA∆x 2
∂x
∂t
∂
∂s
∂2s
−B
−
=ρ 2
∂x
∂x
∂t
af a f
a f af
FG
H
af
a
f
IJ
K
B ∂2s ∂2s
=
ρ ∂x 2 ∂t 2
Into this wave equation as a trial solution we substitute the wave function s x , t = smax cos kx − ωt
we find
∂s
= − ksmax sin kx − ωt
∂x
∂2s
= − k 2 smax cos kx − ωt
2
∂x
∂s
= +ωsmax sin kx − ωt
∂t
∂2s
= −ω 2 smax cos kx − ωt
∂t 2
B
B ∂2s ∂2s
=
becomes − k 2 smax cos kx − ωt = −ω 2 smax cos kx − ωt
ρ ∂x 2 ∂t 2
ρ
b g
a
f
a
f
a
This is true provided
B 4π 2
ρ λ2
f
f
a
B
ρ
.
f
= 4π 2 f 2 .
The sound wave can propagate provided it has λ2 f 2 = v 2 =
speed v =
f
f
a
a
a
B
ρ
; that is, provided it propagates with
504
Sound Waves
Section 17.3
*P17.18
Intensity of Periodic Sound Waves
The sound power incident on the eardrum is ℘ = IA where I is the intensity of the sound and
A = 5.0 × 10 −5 m 2 is the area of the eardrum.
(a)
At the threshold of hearing, I = 1.0 × 10 −12 W m 2 , and
e
je
j
℘ = 1.0 × 10 −12 W m 2 5.0 × 10 −5 m 2 = 5.00 × 10 −17 W .
(b)
At the threshold of pain, I = 1.0 W m 2 , and
e
I=
JK
je
j
℘ = 1.0 W m 2 5.0 × 10 −5 m 2 = 5.00 × 10 −5 W .
P17.19
FG I IJ = 10 logF 4.00 × 10
66.0 dB
GH 1.00 × 10
HI K
F
I
I
70.0 dB = 10 log G
H 1.00 × 10 W m JK
W m j10 b
Therefore, I = e1.00 × 10
−6
β = 10 log
−12
0
P17.20
(a)
−12
−12
(b)
I=
2
70 .0 10
2
g=
1.00 × 10 −5 W m 2 .
2
∆Pmax
, so
2ρ v
jb
e
ge
∆Pmax = 2 ρvI = 2 1.20 kg m3 343 m s 1.00 × 10 −5 W m 2
j
∆Pmax = 90.7 mPa
P17.21
I=
(a)
1
2
ρω 2 smax
v
2
At f = 2 500 Hz , the frequency is increased by a factor of 2.50, so the intensity (at constant
a f = 6.25 .
Therefore, 6.25a0.600f = 3.75 W m
smax ) increases by 2.50
(b)
P17.22
2
.
0.600 W m 2
The original intensity is I 1 =
(a)
2
1
2
2
ρω 2 smax
v = 2π 2 ρvf 2 smax
2
If the frequency is increased to f ′ while a constant displacement amplitude is maintained,
the new intensity is
b g
2
2
I 2 = 2π 2 ρv f ′ smax
so
continued on next page
b g
FG IJ
H K
2
2
I 2 2π ρv f ′ smax
f′
=
=
2
2 2
I1
f
2π ρvf smax
2
or I 2 =
FG f ′ IJ
H fK
2
I1 .
Chapter 17
(b)
If the frequency is reduced to f ′ =
intensity is
505
f
while the displacement amplitude is doubled, the new
2
I 2 = 2π 2 ρv
FG f IJ b2s g
H 2K
2
max
2
2
= 2π 2 ρvf 2 smax
= I1
or the intensity is unchanged .
*P17.23
(a)
For the low note the wavelength is λ =
For the high note λ =
v 343 m s
=
= 2.34 m .
f
146.8 s
343 m s
= 0.390 m .
880 s
880 Hz
= 5.99 nearly
146.8 Hz
equal to a small integer. This fact is associated with the consonance of the notes D and A.
We observe that the ratio of the frequencies of these two notes is
(b)
β = 10 dB log
I=
F
GH 10
I
−12
W m
2
I = 75 dB gives I = 3.16 × 10
JK
−5
W m2
2
∆Pmax
2 ρv
jb
e
g
∆Pmax = 3.16 × 10 −5 W m 2 2 1.20 kg m3 343 m s = 0.161 Pa
for both low and high notes.
(c)
b
g
1
1
2
2
ρv ωsmax = ρv 4π 2 f 2 smax
2
2
I
smax =
2
2π ρvf 2
for the low note,
I=
smax =
3.16 × 10 −5 W m 2
1
2π 1.20 kg m 343 m s 146.8 s
2
3
6.24 × 10 −5
m = 4.25 × 10 −7 m
146.8
for the high note,
6.24 × 10 −5
smax =
m = 7.09 × 10 −8 m
880
=
(d)
*P17.24
146.8
880
=
= 1.093 , the
134.3 804.9
wavelengths and displacement amplitudes are made 1.093 times larger, and the pressure
amplitudes are unchanged.
With both frequencies lower (numerically smaller) by the factor
The power necessarily supplied to the speaker is the power carried away by the sound wave:
P=
b
1
ρAv ωsmax
2
e
g
2
2
= 2π 2 ρAvf 2 smax
j FGH 0.082 m IJK b343 m sgb600 1 sg e0.12 × 10
= 2π 2 1.20 kg m3 π
2
2
−2
j
m
2
= 21.2 W
506
P17.25
Sound Waves
(a)
I 1 = 1.00 × 10 −12 W m 2 10 b
e
or
j
I 1 = 1.00 × 10
e
−4
I 2 = 1.00 × 10
W m
−12
β 1 10
g = e1.00 × 10 −12
j
W m 2 10 80.0 10
2
W m 2 10 b
j
−4.5
β 2 10
g = e1.00 × 10 −12
−5
2
j
W m 2 10 75.0 10
2
or
I 2 = 1.00 × 10
W m = 3.16 × 10 W m
When both sounds are present, the total intensity is
I = I 1 + I 2 = 1.00 × 10 −4 W m 2 + 3.16 × 10 −5 W m 2 = 1.32 × 10 −4 W m 2 .
(b)
The decibel level for the combined sounds is
β = 10 log
*P17.26
(a)
(b)
F 1.32 × 10
GH 1.00 × 10
−4
W m2
−12
2
W m
I = 10 log 1.32 × 10 =
e
j
JK
8
81.2 dB .
v
and f is the same for all three waves. Since the speed is smallest in air, λ is
f
1 493 m s
smallest in air. It is larger by
= 4.51 times in water and by
331 m s
5 950
= 18.0 times in iron .
331
We have λ =
From I =
1
2
ρvω 2 smax
; smax =
2
2I 0
ρvω 02
, smax is smallest in iron, larger in water by
7 860 ⋅ 5 950
7 860 ⋅ 5 950
ρ iron v iron
=
= 5.60 times , and larger in air by
= 331 times .
1 000 ⋅ 1 493
1.29 ⋅ 331
ρ water v water
(c)
From I =
2
∆Pmax
; ∆Pmax = 2 Iρv , ∆Pmax is smallest in air, larger in water by
2 ρv
1 000 ⋅ 1 493
= 59.1 times , and larger in iron by
1.29 ⋅ 331
(d)
b
7 860 ⋅ 5 950
= 331 times .
1.29 ⋅ 331
g
331 m s 2π
v v2π
=
=
= 0.331 m in air
f
ω
2 000 π s
1 493 m s
5 950 m s
λ=
= 1.49 m in water
λ=
= 5.95 m in iron
1 00 0 s
1 000 s
λ=
smax =
2I0
ρvω 02
=
2 × 10 −6 W m 2
e1.29 kg m jb331 m sgb6 283 1 sg
3
2
smax =
2 × 10 −6
1
= 1.84 × 10 −10 m in water
1 000 1 493 6 283
smax =
2 × 10 −6
1
= 3.29 × 10 −11 m in iron
7 860 5 950 6 283
b
b
∆Pmax = 2 Iρv =
g
g
2e10
−6
je
j
W m 2 1.29 kg m3 331 m s = 0.029 2 Pa in air
b g
b7 860gb5 950g =
∆Pmax = 2 × 10 −6 1 000 1 493 = 1.73 Pa in water
∆Pmax = 2 × 10 −6
= 1.09 × 10 −8 m in air
9.67 Pa in iron
Chapter 17
P17.27
(a)
120 dB = 10 dB log
I = 1.00 W m 2 =
r=
℘
=
4π I
LM
MN 10
I
−12
W m
2
℘
OP
PQ
4π r 2
6.00 W
e
4π 1.00 W m 2
= 0.691 m
j
We have assumed the speaker is an isotropic point source.
(b)
0 dB = 10 dB log
F
GH 10
−12
I
JK
I
W m2
I = 1.00 × 10 −12 W m 2
r=
℘
=
4π I
6.00 W
e
4π 1.00 × 10 -12 W m 2
j
= 691 km
We have assumed a uniform medium that absorbs no energy.
P17.28
We begin with β 2 = 10 log
FG I IJ , and β
HI K
2
0
1
I
℘
℘
, and I 1 =
, giving 2
I1
4π r22
4π r12
Then, β 2 − β 1 = 10 log
P17.29
FG r IJ
Hr K
1
2
2
= 20 log
1
0
β2
Also, I 2 =
FG I IJ , so
HI K
FI I
− β = 10 log G J .
HI K
Fr I
=G J .
Hr K
= 10 log
2
1
1
2
1
2
FG r IJ
Hr K
1
.
2
Since intensity is inversely proportional to the square of the distance,
a f
a fa f
2
I4 =
10.0
∆P 2
1
I 0. 4 and I 0. 4 = max =
= 0.121 W m 2 .
100
2 ρv
2 1.20 343
The difference in sound intensity level is
∆β = 10 log
FG I
HI
4 km
0.4 km
At 0.400 km,
β 0. 4 = 10 log
At 4.00 km,
IJ = 10a−2.00f = −20.0 dB .
K
F 0.121 W m I = 110.8 dB .
GH 10 W m JK
2
−12
2
a
f
β 4 = β 0.4 + ∆β = 110.8 − 20.0 dB = 90.8 dB .
Allowing for absorption of the wave over the distance traveled,
b
ga
f
β ′4 = β 4 − 7.00 dB km 3.60 km = 65.6 dB .
This is equivalent to the sound intensity level of heavy traffic.
507
508
P17.30
Sound Waves
Let r1 and r2 be the distance from the speaker to the observer that hears 60.0 dB and 80.0 dB,
respectively. Use the result of problem 28,
β 2 − β 1 = 20 log
Thus, log
FG r IJ , to obtain 80.0 − 60.0 = 20 logFG r IJ .
Hr K
Hr K
1
1
2
2
FG r IJ = 1 , so r
Hr K
1
= 10.0r2 . Also: r1 + r2 = 110 m , so
1
2
10.0r2 + r2 = 110 m giving r2 = 10.0 m , and r1 = 100 m .
P17.31
We presume the speakers broadcast equally in all directions.
rAC = 3.00 2 + 4.00 2 m = 5.00 m
1.00 × 10 −3 W
℘
I=
=
= 3.18 × 10 −6 W m 2
2
4π r 2
4π 5.00 m
(a)
a
β = 10 dB log
f
F 3.18 × 10 W m I
GH 10 W m JK
−6
2
−12
2
β = 10 dB 6.50 = 65.0 dB
rBC = 4. 47 m
(b)
I=
1.50 × 10 −3 W
a
f
4π 4.47 m
β = 10 dB log
2
= 5.97 × 10 −6 W m 2
F 5.97 × 10 I
GH 10 JK
−6
−12
β = 67.8 dB
I = 3.18 µW m 2 + 5.97 µW m 2
(c)
β = 10 dB log
P17.32
In I =
F 9.15 × 10 I =
GH 10 JK
−6
69.6 dB
−12
℘
1
, intensity I is proportional to 2 ,
2
r
4π r
so between locations 1 and 2:
I 2 r12
=
.
I 1 r22
2
2
2
1
b g , intensity is proportional to s , so II = ss .
F s I F r I F 1I F r I
Then, G J = G J or G J = G J , giving r = 2r = 2a50.0 mf = 100 m .
H s K H r K H 2K H r K
But, r = a50.0 mf + d yields d = 86.6 m .
In I =
1
ρv ωsmax
2
2
2
max
2
2
2
2
1
1
2
2
2
1
2
2
2
2
1
2
1
Chapter 17
P17.33
β = 10 log
FG I IJ
H 10 K
I = 10 b
−12
β 10
g e10 −12 j
509
W m2
I a120 dB f = 1.00 W m 2 ; I a100 dB f = 1.00 × 10 −2 W m 2 ; I a10 dB f = 1.00 × 10 −11 W m 2
(a)
℘ = 4π r 2 I so that r12 I 1 = r22 I 2
r2
(b)
P17.34
P17.35
r2
FI
=r G
HI
FI
=r G
HI
1
1
1
2
1
2
IJ
K
IJ
K
12
a
f
1.00
= 30.0 m
1.00 × 10 −2
a
f
1.00
= 9.49 × 10 5 m
1.00 × 10 −11
= 3.00 m
12
= 3.00 m
a
f e7.00 × 10
2
ja
−2
(a)
E =℘t = 4π r 2 It = 4π 100 m
(b)
β = 10 log
(a)
The sound intensity inside the church is given by
F 7.00 × 10 I =
GH 1.00 × 10 JK
−2
−12
f
W m 2 0.200 s = 1.76 kJ
108 dB
FG I IJ
HI K
F I I
101 dB = a10 dBf lnG
H 10 W m JK
I = 10 e10
W m j = 10
W m
β = 10 ln
0
−12
10.1
−12
2
−1.90
2
2
= 0.012 6 W m 2
We suppose that sound comes perpendicularly out through the windows and doors. Then,
the radiated power is
e
je
j
℘ = IA = 0.012 6 W m 2 22.0 m 2 = 0.277 W .
Are you surprised by how small this is? The energy radiated in 20.0 minutes is
b
ga
E =℘t = 0.277 J s 20.0 min
(b)
60.0 s I
fFGH 1.00
J=
min K
332 J .
If the ground reflects all sound energy headed downward, the sound power, ℘ = 0.277 W ,
covers the area of a hemisphere. One kilometer away, this area is
b
g
2
A = 2π r 2 = 2π 1 000 m = 2π × 10 6 m 2 .
The intensity at this distance is
I=
0. 277 W
℘
=
= 4.41 × 10 −8 W m 2
A 2π × 10 6 m 2
and the sound intensity level is
a
f FGH 14..0041××1010
β = 10 dB ln
−8
W m2
−12
2
W m
I=
JK
46.4 dB .
510
*P17.36
Sound Waves
Assume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensity
I
of this sound is given by 100 dB = 10 dB log −12
; I = 10 −2 W m 2 . If the lawnmower
10
W m2
℘
radiates as a point source, its sound power is given by I =
.
4π r 2
a f
2
℘ = 4π 1 m 10 −2 W m 2 = 0.126 W
Now let your neighbor have an identical lawnmower 20 m away. You receive from it sound with
0.126 W
intensity I =
= 2.5 × 10 −5 W m 2 . The total sound intensity impinging on you is
2
4π 20 m
a
10
−2
f
W m + 2.5 × 10 −5 W m 2 = 1.002 5 × 10 −2 W m 2 . So its level is
2
10 dB log
1.002 5 × 10 −2
10 −12
= 100.01 dB .
If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as a
change from 100 dB.
Section 17.4
P17.37
f′= f
The Doppler Effect
bv ± v g
bv ± v g
O
S
(a)
(b)
P17.38
(a)
a343 + 40.0f = 338 Hz
a343 + 20.0f
a343 + 20.0f = 483 Hz
f ′ = 510
a343 + 40.0f
F 115 min I = 12.0 rad s
ω = 2π f = 2π G
H 60.0 s min JK
v
= ωA = b12.0 rad sge1.80 × 10 mj =
f ′ = 320
−3
max
(b)
The heart wall is a moving observer.
f′= f
(c)
0.021 7 m s
FG v + v IJ = b2 000 000 HzgFG 1 500 + 0.021 7 IJ =
H v K
H 1 500 K
O
2 000 028.9 Hz
Now the heart wall is a moving source.
f ′′ = f ′
FG v IJ = b2 000 029 HzgF 1 500 I =
GH 1 500 − 0.021 7 JK
Hv−v K
s
2 000 057.8 Hz
Chapter 17
P17.39
f′=
Approaching ambulance:
511
f
b1 − v vg
S
f
f ′′ =
Departing ambulance:
d1 − b− v vgi
F vI F vI
560G 1 − J = 480G 1 + J
H vK H vK
S
S
Since f ′ = 560 Hz and f ′′ = 480 Hz
S
vS
= 80.0
v
80.0 343
m s = 26.4 m s
vS =
1 040
1 040
a f
P17.40
(a)
The maximum speed of the speaker is described by
1
1
2
= kA 2
mv max
2
2
v max =
a
f
20.0 N m
0.500 m = 1.00 m s
5.00 kg
k
A=
m
The frequencies heard by the stationary observer range from
′ =f
fmin
FG v IJ to f ′
Hv+v K
max
max
=f
FG v IJ
Hv−v K
max
where v is the speed of sound.
(b)
β = 10 dB log
F 343 m s I =
GH 343 m s + 1.00 m s JK
F 343 m s I =
= 440 HzG
H 343 m s − 1.00 m s JK
fmin
′ = 440 Hz
439 Hz
fmax
′
441 Hz
FG I IJ = 10 dB logF ℘ 4π r I
GH I JK
HI K
2
0
0
The maximum intensity level (of 60.0 dB) occurs at r = rmin = 1.00 m . The minimum intensity
level occurs when the speaker is farthest from the listener (i.e., when
r = rmax = rmin + 2 A = 2.00 m).
F ℘ I − 10 dB logF ℘ I
GH 4π I r JK
GH 4π I r JK
F ℘ 4π I r I = 10 dB logF r I .
β
or
−β
= 10 dB log G
GH r JK
JK
℘
H 4π I r
This gives: 60.0 dB − β
= 10 dB log a 4.00f = 6.02 dB , and β
= 54.0 dB
Thus, β max − β min = 10 dB log
max
min
2
0 min
2
0 min
min
2
0 max
2
0 max
2
max
2
min
min
.
512
P17.41
Sound Waves
f′= f
F 340 I
GH 340 − b−9.80t g JK
485a340f + a 485fd9.80t i = a512 fa340 f
F 512 − 485 IJ 340 = 1.93 s
t =G
H 485 K 9.80
FG v IJ
Hv−v K
485 = 512
s
fall
f
f
d1 =
1 2
gt f = 18.3 m :
2
t return =
18.3
= 0.053 8 s
340
The fork continues to fall while the sound returns.
t total fall = t f + treturn = 1.93 s + 0.053 8 s = 1.985 s
P17.42
b
g
d total =
1 2
gt total fall = 19.3 m
2
a
f
m
−10° C = 325 m s
s⋅° C
(a)
v = 331 m s + 0.6
(b)
Approaching the bell, the athlete hears a frequency of
After passing the bell, she hears a lower frequency of
*P17.43
(a)
FG v + v IJ
H v K
F v + b− v g I
f ′′ = f G
H v JK
f′= f
O
O
The ratio is
f ′′ v − vO 5
=
=
f ′ v + vO 6
which gives 6 v − 6 v o = 5 v + 5 v o or
vO =
a
v 325 m s
=
= 29.5 m s
11
11
f
Sound moves upwind with speed 343 − 15 m s . Crests pass a stationary upwind point at
frequency 900 Hz.
Then
λ=
v 328 m s
=
= 0.364 m
f
900 s
(b)
By similar logic,
λ=
343 + 15 m s
v
=
= 0.398 m
f
900 s
(c)
The source is moving through the air at 15 m/s toward the observer. The observer is
stationary relative to the air.
f′= f
a
f
FG v + v IJ = 900 HzFG 343 + 0 IJ =
H 343 − 15 K
H v−v K
o
941 Hz
s
(d)
The source is moving through the air at 15 m/s away from the downwind firefighter. Her
speed relative to the air is 30 m/s toward the source.
f′= f
FG v + v IJ = 900 HzF 343 + 30 I = 900 HzFG 373 IJ =
GH 343 − a−15f JK
H 358 K
H v−v K
o
s
938 Hz
Chapter 17
*P17.44
The half-angle of the cone of the shock wave is θ where
θ = sin −1
FG v
Hv
sound
source
IJ = sin FG 1 IJ = 41.8° .
H 1.5 K
K
−1
φ
As shown in the sketch, the angle between the direction of propagation
of the shock wave and the direction of the plane’s velocity is
The half angle of the shock wave cone is given by sin θ =
vS =
P17.46
θ = sin −1
P17.47
(b)
sin θ
=
v light
vS
.
2.25 × 10 8 m s
= 2.82 × 10 8 m s
sin 53.0°
a
f
v
1
= sin −1
= 46.4°
vS
1.38
sin θ =
v
1
=
; θ = 19.5°
vS 3.00
tan θ =
h
h
; x=
x
tan θ
x=
(a)
v light
v shock
FIG. P17.44
φ = 90°−θ = 90°−41.8° = 48.2° .
P17.45
20 000 m
= 5.66 × 10 4 m = 56.6 km
tan 19.5°
It takes the plane t =
5.66 × 10 4 m
x
=
= 56.3 s to travel this distance.
vS 3.00 335 m s
b
g
x
θ
t=0
θ
h
h
Observer
a.
v plane
θ
Observer hears the boom
b.
FIG. P17.47(a)
513
514
Sound Waves
Section 17.5
Digital Sound Recording
Section 17.6
Motion Picture Sound
*P17.48
For a 40-dB sound,
40 dB = 10 dB log
LM
MN 10
I = 10 −8 W m 2 =
2
∆Pmax
2 ρv
−12
OP
PQ
I
W m2
jb
e
g
∆Pmax = 2 ρvI = 2 1.20 kg m 2 343 m s 10 −8 W m 2 = 2.87 × 10 −3 N m 2
*P17.49
code =
(b)
For sounds of 40 dB or softer, too few digital words are available to represent the wave form
with good fidelity.
(c)
In a sound wave ∆P is negative half of the time but this coding scheme has no words
available for negative pressure variations.
65 536 = 7
28.7 N m 2
If the source is to the left at angle θ from the direction you are
facing, the sound must travel an extra distance d sin θ to reach your
right ear as shown, where d is the distance between your ears. The
d sin θ
. Then
delay time is ∆t in v =
∆t
θ = sin −1
*P17.50
2.87 × 10 −3 N m 2
(a)
103 dB = 10 dB log
(a)
b
g
343 m s 210 × 10
v∆ t
= sin −1
d
0.19 m
LM
MN 10
I
−12
W m
2
I = 2.00 × 10 −2 W m 2 =
−6
s
= 22.3° left of center .
θ
θ
ear
ear
FIG. P17.49
OP
PQ
℘
4π r 2
=
a
℘
f
4π 1.6 m
2
℘ = 0.642 W
(b)
efficiency =
sound output power 0.642 W
=
= 0.004 28
total input power
150 W
Additional Problems
P17.51
Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus
time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all
directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose
that, at the ambient temperature, sound moves at 340 m/s; and suppose that the horizontal width of
each row of seats is 60 cm. Then there is a time delay of
0.6 m
b340 m sg = 0.002 s
continued on next page
Chapter 17
515
between your sound impulse reaching each riser and the next. Whatever its material, each will
reflect much of the sound that reaches it. The reflected wave sounds very different from the sharp
pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty
crests, each separated from the next in time by
a
f = 0.004 s .
b340 m sg
2 0.6 m
This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once
again after its reflection. Thus, you hear a sound of definite pitch, with period about 0.004 s, frequency
1
~ 300 Hz
0.003 5 s
wavelength
b
g
340 m s
v
=
= 1.2 m ~ 10 0 m
f
300 s
λ=
b
and duration
g
a
f
20 0.004 s ~ 10 −1 s .
P17.52
v 343 m s
=
= 0.232 m
f 1 480 s −1
(a)
λ=
(b)
β = 81.0 dB = 10 dB log
e
j
LM
MN 10
I
−12
W m
2
OP
PQ
I = 10 −12 W m 2 10 8.10 = 10 −3.90 W m 2 = 1.26 × 10 −4 W m 2 =
smax =
(c)
P17.53
λ′ =
2I
=
ρ vω 2
e
2 1.26 × 10 −4 W m 2
e1.20 kg m jb343 m sg4π e1 480 s j
3
v 343 m s
=
= 0.246 m
f ′ 1 397 s −1
Since cos 2 θ + sin 2 θ = 1 ,
j
−1 2
2
1
2
ρvω 2 smax
2
= 8.41 × 10 −8 m
∆λ = λ ′ − λ = 13.8 mm
sin θ = ± 1 − cos 2 θ (each sign applying half the time)
a
f
a
∆P = ∆Pmax sin kx − ωt = ± ρvω smax 1 − cos 2 kx − ωt
f
2
2
2
∆P = ± ρvω smax
− smax
− s2
cos 2 kx − ωt = ± ρvω smax
Therefore
P17.54
a
f
The trucks form a train analogous to a wave train of crests with speed v = 19.7 m s
2
and unshifted frequency f =
= 0.667 min −1 .
3.00 min
(a)
(b)
The cyclist as observer measures a lower Doppler-shifted frequency:
19.7 + −4. 47
v + vo
= 0.667 min −1
= 0.515 min
f′= f
v
19.7
FG IJ e
H K
F v + v ′ IJ = e0.667 min
f ′′ = f G
H v K
o
jFGH a f IJK
jFGH 19.7 +19a.−71.56f IJK =
−1
0.614 min
The cyclist’s speed has decreased very significantly, but there is only a modest increase in
the frequency of trucks passing him.
516
Sound Waves
ja
f
P17.55
v=
2d
vt 1
:d=
= 6.50 × 10 3 m s 1.85 s = 6.01 km
t
2 2
P17.56
(a)
The speed of a compression wave in a bar is
e
Y
v=
(b)
ρ
20.0 × 10 10 N m 2
7 860 kg m
3
= 5.04 × 10 3 m s .
The signal to stop passes between layers of atoms as a sound wave, reaching the back end of
the bar in time
t=
(c)
=
0.800 m
L
=
= 1.59 × 10 −4 s .
v 5.04 × 10 3 m s
As described by Newton’s first law, the rearmost layer of steel has continued to move
forward with its original speed vi for this time, compressing the bar by
b
ge
j
∆L = vi t = 12.0 m s 1.59 × 10 −4 s = 1.90 × 10 −3 m = 1.90 mm .
(d)
The strain in the rod is:
(e)
The stress in the rod is:
σ =Y
∆L 1.90 × 10 −3 m
=
= 2.38 × 10 −3 .
L
0.800 m
FG ∆L IJ = e20.0 × 10
HLK
10
je
j
N m 2 2.38 × 10 −3 = 476 MPa .
Since σ > 400 MPa , the rod will be permanently distorted.
(f)
We go through the same steps as in parts (a) through (e), but use algebraic expressions
rather than numbers:
The speed of sound in the rod is v =
Y
ρ
.
The back end of the rod continues to move forward at speed vi for a time of t =
traveling distance ∆L = vi t after the front end hits the wall.
The strain in the rod is:
∆L vi t
ρ
=
= vi
.
L
L
Y
The stress is then: σ = Y
FG ∆L IJ = Yv
HLK
ρ
i
Y
ρ
L
=L
,
v
Y
= vi ρY .
For this to be less than the yield stress, σ y , it is necessary that
vi ρY < σ y or vi <
σy
ρY
.
With the given numbers, this speed is 10.1 m/s. The fact that the length of the rod divides
out means that the steel will start to bend right away at the front end of the rod. There it will
yield enough so that eventually the remainder of the rod will experience only stress within
the elastic range. You can see this effect when sledgehammer blows give a mushroom top to
a rod used as a tent stake.
Chapter 17
P17.57
(a)
f′= f
517
v
bv − v g
diver
so 1 −
v diver
f
=
v
f′
FG
H
⇒ v diver = v 1 −
f
f′
IJ
K
with v = 343 m s , f ′ = 1 800 Hz and f = 2 150 Hz
we find
F
GH
v diver = 343 1 −
(b)
I
JK
1 800
= 55.8 m s .
2 150
If the waves are reflected, and the skydiver is moving into them, we have
f ′′ = f ′
bv + v g ⇒ f ′′ = f LM v OP bv + v g
v
MN bv − v g QP v
diver
diver
diver
so f ′′ = 1 800
P17.58
(a)
f′=
a343 + 55.8f =
a343 − 55.8f
fv
v−u
2 500 Hz .
FG 1 − 1 IJ
H v − u v + uK
2bu v g
fva v + u − v + uf
2uvf
∆f =
=
=
f
v −u
1 − eu v j
v e1 − eu v jj
2a36.1fa 400f
∴ ∆f =
=
340 1 − a36.1f 340
f ′′ =
fv
v − −u
2
(b)
P17.59
130 km h = 36.1 m s
f ′ − f ′′ = fv
a f
2
2
2
2
2
2
2
2
85.9 Hz
v − vO
where vO
v − vS
and vS are measured relative to the medium in which the sound is propagated. In this case the
ocean current is opposite the direction of travel of the ships and
When observer is moving in front of and in the same direction as the source, f ′ = f
b
g
= 64.0 km h − b −10.0 km hg = 74.0 km h = 20.55 m s
vO = 45.0 km h − −10.0 km h = 55.0 km h = 15.3 m s , and
vS
b
Therefore, f ′ = 1 200.0 Hz
520 m s − 15.3 m s
=
g 11520
m s − 20.55 m s
1 204.2 Hz .
518
P17.60
Sound Waves
Use the Doppler formula, and remember that the bat is a moving source.
If the velocity of the insect is v x ,
40.4 = 40.0
a340 + 5.00fb340 − v g .
a340 − 5.00fb340 + v g
x
x
Solving,
v x = 3.31 m s .
Therefore, the bat is gaining on its prey at 1.69 m s .
P17.61
sin β =
a
v
1
=
vS N M
h = v 12.8 s
a
f
x = vS 10.0 s
shock front
β
f
h
h
v 1.28
tan β = = 1.28
=
x
vS N M
sin β
1
cos β =
=
tan β 1.28
β = 38.6°
NM =
P17.62
(a)
FIG. P17.62(a)
λ=
(c)
λ′ =
(d)
(e)
v 343 m s
=
= 0.343 m
f 1 000 s −1
f
IJ a
K
v
v F v + v I a343 + 40.0f m s
λ ′′ =
= G
J = 1 000 s = 0.383 m
f ′′ f H v K
F v − v IJ = b1 000 Hzg a343 − 30.0f m s = 1.03 kHz
f′ = fG
a343 − 40.0f m s
H v−v K
FG
H
343 − 40.0 m s
v v v − vS
=
=
= 0.303 m
f′ f
v
1 000 s −1
S
O
S
shock front
FIG. P17.61
1
= 1.60
sin β
(b)
vs
x
−1
Chapter 17
P17.63
∆t = L
FG 1
Hv
−
air
519
IJ
K
v − v air
1
= L cu
v cu
v air v cu
b
ge
je
331 m s 3.56 × 10 3 m s
v air v cu
∆t =
6.40 × 10 −3 s
L=
v cu − v air
3 560 − 331 m s
b
g
j
L = 2.34 m
P17.64
P
The shock wavefront connects all observers
first hearing the plane, including our observer
O and the plane P, so here it is vertical. The
angle φ that the shock wavefront makes with
the direction of the plane’s line of travel is
given by
sin φ =
φ
340 m s
v
=
= 0.173
vS 1 963 m s
C
so φ = 9.97° .
O
θ
Using the right triangle CPO, the angle θ is
seen to be
FIG. P17.64
θ = 90.0°−φ = 90.0°−9.97° = 80.0° .
P17.65
(a)
F v I = sin F 331 I =
GH 20.0 × 10 JK
GH v JK
F 1 533 IJ = 4.40°
θ ′ = sin G
H 20.0 × 10 K
θ = sin −1
−1
sound
3
obj
(b)
P17.66
−1
0.948°
3
℘
1
℘1 β 1 − β 2 = 10 log 1
20.0
℘2
80.0 − β 2 = 10 log 20.0 = +13.0
℘2 =
β 2 = 67.0 dB
FG Y IJ
H ρK
FTI
=G J .
H µK
12
P17.67
For the longitudinal wave v L =
.
12
For the transverse wave vT
If we require
vL
µY
m
= 8.00 , we have T =
and
where µ =
L
64.0 ρ
vT
ρ=
e
je
2
mass
m
=
.
volume π r 2 L
j
−3
10
2
π r 2 Y π 2.00 × 10 m 6.80 × 10 N m
=
= 1.34 × 10 4 N .
This gives T =
64.0
64.0
520
P17.68
Sound Waves
The total output sound energy is eE =℘∆t , where ℘ is the power radiated.
eE eE
eE
eE
=
=
=
Thus, ∆t =
.
2
℘ IA
π
d 2I
4
4π r I
e
But, β = 10 log
j
FG I IJ . Therefore, I = I e10 j and ∆t =
HI K
0
0
P17.69
β 10
eE
4π d 2 I 0 10 β
10
.
(a)
If the source and the observer are moving away from each other, we have: θ S − θ 0 = 180° ,
and since cos180° = −1 , we get Equation 17.12 with negative values for both vO and vS .
(b)
If vO = 0 m s then f ′ =
v
f
v − vS cos θ S
Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the
intersection,
cos θ S =
so f ′ =
ga
343 m s
500 Hz
343 m s − 0.800 25.0 m s
b
4
5
f
or f ′ = 531 Hz .
Note that as the train approaches, passes, and departs from the intersection, θ S varies from
0° to 180° and the frequency heard by the observer varies from:
P17.70
a
f
fmax
′ =
343 m s
v
f=
500 Hz = 539 Hz
v − vS cos 0°
343 m s − 25.0 m s
fmin
′ =
343 m s
v
f=
500 Hz = 466 Hz
v − vS cos 180°
343 m s + 25.0 m s
a
f
Let T represent the period of the source vibration, and E be the energy put into each wavefront.
E
Then ℘av = . When the observer is at distance r in front of the source, he is receiving a spherical
T
wavefront of radius vt, where t is the time since this energy was radiated, given by vt − vS t = r . Then,
t=
a f
The area of the sphere is 4π vt
2
=
4π v 2 r 2
bv − v g
Tbv − v g
S
2
r
.
v − vS
. The energy per unit area over the spherical wavefront
2
E ℘av
S
=
.
A
4π v 2 r 2
The observer receives parcels of energy with the Doppler shifted frequency
v
v
f′= f
=
, so the observer receives a wave with intensity
v − vS
T v − vS
is uniform with the value
FG
H
IJ
K b
g
I=
FG E IJ f ′ = FG ℘ T bv − v g
H A K GH 4π v r
av
S
2 2
2
IF v I
JJK GH Tbv − v g JK =
S
℘av
4π r
2
FG v − v IJ
H v K
S
.
Chapter 17
P17.71
(a)
The time required for a sound pulse to travel distance L at
L
L
. Using this expression
speed v is given by t = =
v
Y ρ
521
L2
L1
L3
we find
FIG. P17.71
L1
t1 =
t2 =
=
Y1 ρ 1
e7.00 × 10
1.50 m − L1
Y2 ρ 2
=
L1
10
N m
e1.60 × 10
e
2
j e2 700 kg m j
3
e
1.50 m − L1
N m2
10
j e11.3 × 10
3
j
= 1.96 × 10 −4 L1 s
kg m3
j
j
t 2 = 1.26 × 10 −3 − 8.40 × 10 −4 L1 s
or
t3 =
1.50 m
e11.0 × 10
10
N m3
j e8 800 kg m j
3
t 3 = 4.24 × 10 −4 s
We require t1 + t 2 = t 3 , or
1.96 × 10 −4 L1 + 1.26 × 10 −3 − 8.40 × 10 −4 L1 = 4.24 × 10 −4 .
This gives L1 = 1.30 m and L 2 = 1.50 − 1.30 = 0.201 m .
The ratio of lengths is then
(b)
The ratio of lengths
L1
= 6.45 .
L2
L1
is adjusted in part (a) so that t1 + t 2 = t 3 . Sound travels the two paths
L2
in equal time and the phase difference, ∆φ = 0 .
P17.72
To find the separation of adjacent molecules, use a model where each molecule occupies a sphere of
radius r given by
ρ air =
3
or 1.20 kg m =
4.82 × 10 −26 kg
4
3π
r3
average mass per molecule
4
3π
r3
L 3e4.82 × 10 kg j OP
,r=M
MN 4π e1.20 kg m j PQ
−26
3
13
= 2.12 × 10 −9 m .
Intermolecular separation is 2r = 4. 25 × 10 −9 m, so the highest possible frequency sound wave is
fmax =
343 m s
v
v
=
=
= 8.03 × 10 10 Hz ~ 10 11 Hz .
λ min 2r 4.25 × 10 −9 m
522
Sound Waves
ANSWERS TO EVEN PROBLEMS
P17.2
1.43 km s
P17.36
no
P17.4
(a) 27.2 s; (b) longer than 25.7 s, because
the air is cooler
P17.38
(a) 2.17 cm s ; (b) 2 000 028.9 Hz ;
(c) 2 000 057.8 Hz
P17.6
(a) 153 m s ; (b) 614 m
P17.40
(a) 441 Hz; 439 Hz; (b) 54.0 dB
P17.8
(a) 4.16 m; (b) 0.455 µs ; (c) 0.157 mm
P17.42
(a) 325 m s; (b) 29.5 m s
P17.10
1.55 × 10 −10 m
P17.44
48. 2°
P17.12
(a) 1.27 Pa; (b) 170 Hz; (c) 2.00 m;
(d) 340 m s
P17.46
46. 4°
P17.48
(a) 7; (b) and (c) see the solution
P17.50
(a) 0.642 W ; (b) 0.004 28 = 0.428%
P17.52
(a) 0.232 m; (b) 84.1 nm; (c) 13.8 mm
P17.54
(a) 0.515 min ; (b) 0.614 min
P17.56
(a) 5.04 km s ; (b) 159 µs ; (c) 1.90 mm;
(d) 0.002 38 ; (e) 476 MPa ;
(f) see the solution
P17.58
(a) see the solution; (b) 85.9 Hz
P17.60
The gap between bat and insect is closing
at 1.69 m s .
P17.62
(a) see the solution; (b) 0.343 m;
(c) 0.303 m; (d) 0.383 m; (e) 1.03 kHz
P17.64
80.0°
P17.66
67.0 dB
P17.68
∆t =
e
4
j
P17.14
s = 22.5 nm cos 62.8 x − 2.16 × 10 t
P17.16
(a) 4.63 mm; (b) 14.5 m s ;
(c) 4.73 × 10 9 W m 2
P17.18
(a) 5.00 × 10 −17 W ; (b) 5.00 × 10 −5 W
P17.20
(a) 1.00 × 10
−5
F f ′I
=G J
HfK
2
W m ; (b) 90.7 mPa
2
I 1 ; (b) I 2 = I 1
P17.22
(a) I 2
P17.24
21.2 W
P17.26
(a) 4.51 times larger in water than in air
and 18.0 times larger in iron;
(b) 5.60 times larger in water than in iron
and 331 times larger in air;
(c) 59.1 times larger in water than in air
and 331 times larger in iron;
(d) 0.331 m; 1.49 m; 5.95 m; 10.9 nm;
184 pm; 32.9 pm; 29.2 mPa; 1.73 Pa; 9.67 Pa
eE
4π d I 0 10 β
2
P17.28
see the solution
P17.30
10.0 m; 100 m
P17.70
see the solution
P17.32
86.6 m
P17.72
~ 10 11 Hz
P17.34
(a) 1.76 kJ ; (b) 108 dB
10
18
Superposition and Standing Waves
CHAPTER OUTLINE
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
Superposition and
Interference
Standing Waves
Standing Waves in a String
Fixed at Both Ends
Resonance
Standing Waves in Air
Columns
Standing Waves in Rod and
Plates
Beats: Interference in Time
Non-Sinusoidal Wave
Patterns
ANSWERS TO QUESTIONS
Q18.1
No. Waves with other waveforms are also trains of disturbance
that add together when waves from different sources move
through the same medium at the same time.
Q18.2
The energy has not disappeared, but is still carried by the wave
pulses. Each particle of the string still has kinetic energy. This is
similar to the motion of a simple pendulum. The pendulum
does not stop at its equilibrium position during
oscillation—likewise the particles of the string do not stop at
the equilibrium position of the string when these two waves
superimpose.
Q18.3
No. A wave is not a solid object, but a chain of disturbance. As
described by the principle of superposition, the waves move
through each other.
Q18.4
They can, wherever the two waves are nearly enough in phase that their displacements will add to
create a total displacement greater than the amplitude of either of the two original waves.
When two one-dimensional sinusoidal waves of the same amplitude interfere, this
condition is satisfied whenever the absolute value of the phase difference between the two waves is
less than 120°.
Q18.5
When the two tubes together are not an efficient transmitter of sound from source to receiver, they
are an efficient reflector. The incoming sound is reflected back to the source. The waves reflected by
the two tubes separately at the junction interfere constructively.
Q18.6
No. The total energy of the pair of waves remains the same. Energy missing from zones of
destructive interference appears in zones of constructive interference.
Q18.7
Each of these standing wave patterns is made of two superimposed waves of identical frequencies
traveling, and hence transferring energy, in opposite directions. Since the energy transfer of the
waves are equal, then no net transfer of energy occurs.
Q18.8
Damping, and non–linear effects in the vibration turn the energy of vibration into internal energy.
Q18.9
The air in the shower stall can vibrate in standing wave patterns to intensify those frequencies in
your voice which correspond to its free vibrations. The hard walls of the bathroom reflect sound
very well to make your voice more intense at all frequencies, giving the room a longer reverberation
time. The reverberant sound may help you to stay on key.
523
524
Superposition and Standing Waves
Q18.10
The trombone slide and trumpet valves change the length of the air column inside the instrument,
to change its resonant frequencies.
Q18.11
The vibration of the air must have zero amplitude at the closed end. For air in a pipe closed at one
end, the diagrams show how resonance vibrations have NA distances that are odd integer
submultiples of the NA distance in the fundamental vibration. If the pipe is open, resonance
vibrations have NA distances that are all integer submultiples of the NA distance in the
fundamental.
FIG. Q18.11
Q18.12
What is needed is a tuning fork—or other pure-tone generator—of the desired frequency. Strike the
tuning fork and pluck the corresponding string on the piano at the same time. If they are precisely
in tune, you will hear a single pitch with no amplitude modulation. If the two pitches are a bit off,
you will hear beats. As they vibrate, retune the piano string until the beat frequency goes to zero.
Q18.13
Air blowing fast by a rim of the pipe creates a “shshshsh” sound called edgetone noise, a mixture of
all frequencies, as the air turbulently switches between flowing on one side of the edge and the
other. The air column inside the pipe finds one or more of its resonance frequencies in the noise. The
air column starts vibrating with large amplitude in a standing wave vibration mode. It radiates
sound into the surrounding air (and also locks the flapping airstream at the edge to its own
frequency, making the noise disappear after just a few cycles).
Q18.14
A typical standing–wave vibration possibility for a bell is similar to that for the glass shown in
Figure 18.17. Here six node-to-node distances fit around the circumference of the rim. The
circumference is equal to three times the wavelength of the transverse wave of in-and-out bending
of the material. In other states the circumference is two, four, five, or higher integers times the
wavelengths of the higher–frequency vibrations. (The circumference being equal to the wavelength
would describe the bell moving from side to side without bending, which it can do without
producing any sound.) A tuned bell is cast and shaped so that some of these vibrations will have
their frequencies constitute higher harmonics of a musical note, the strike tone. This tuning is lost if
a crack develops in the bell. The sides of the crack vibrate as antinodes. Energy of vibration may be
rapidly converted into internal energy at the end of the crack, so the bell may not ring for so long a
time.
Q18.15
The bow string is pulled away from equilibrium and released, similar to the way that a guitar string
is pulled and released when it is plucked. Thus, standing waves will be excited in the bow string. If
the arrow leaves from the exact center of the string, then a series of odd harmonics will be excited.
Even harmonies will not be excited because they have a node at the point where the string exhibits
its maximum displacement.
Chapter 18
525
Q18.16
Walking makes the person’s hand vibrate a little. If the frequency of this motion is equal to the
natural frequency of coffee sloshing from side to side in the cup, then a large–amplitude vibration of
the coffee will build up in resonance. To get off resonance and back to the normal case of a smallamplitude disturbance producing a small–amplitude result, the person can walk faster, walk slower,
or get a larger or smaller cup. Alternatively, even at resonance he can reduce the amplitude by
adding damping, as by stirring high–fiber quick–cooking oatmeal into the hot coffee.
Q18.17
Beats. The propellers are rotating at slightly different frequencies.
Q18.18
Instead of just radiating sound very softly into the surrounding air, the tuning fork makes the
chalkboard vibrate. With its large area this stiff sounding board radiates sound into the air with
higher power. So it drains away the fork’s energy of vibration faster and the fork stops vibrating
sooner. This process exemplifies conservation of energy, as the energy of vibration of the fork is
transferred through the blackboard into energy of vibration of the air.
Q18.19
The difference between static and kinetic friction makes your finger alternately slip and stick as it
slides over the glass. Your finger produces a noisy vibration, a mixture of different frequencies, like
new sneakers on a gymnasium floor. The glass finds one of its resonance frequencies in the noise.
The thin stiff wall of the cup starts vibrating with large amplitude in a standing wave vibration
mode. A typical possibility is shown in Figure 18.17. It radiates sound into the surrounding air, and
also can lock your squeaking finger to its own frequency, making the noise disappear after just a few
cycles. Get a lot of different thin–walled glasses of fine crystal and try them out. Each will generally
produce a different note. You can tune them by adding wine.
Q18.20
Helium is less dense than air. It carries sound at higher speed. Each cavity in your vocal apparatus
has a standing-wave resonance frequency, and each of these frequencies is shifted to a higher value.
Your vocal chords can vibrate at the same fundamental frequency, but your vocal tract amplifies by
resonance a different set of higher frequencies. Then your voice has a different quacky quality.
Warning: Inhaling any pressurized gas can cause a gas embolism which can lead to stroke or
death, regardless of your age or health status. If you plan to try this demonstration in class, inhale
your helium from a balloon, not directly from a pressurized tank.
Q18.21
Stick a bit of chewing gum to one tine of the second fork. If the beat frequency is then faster than 4
beats per second, the second has a lower frequency than the standard fork. If the beats have slowed
down, the second fork has a higher frequency than the standard. Remove the gum, clean the fork,
add or subtract 4 Hz according to what you found, and your answer will be the frequency of the
second fork.
SOLUTIONS TO PROBLEMS
Section 18.1
P18.1
Superposition and Interference
a
f
a
f
y = y1 + y 2 = 3.00 cos 4.00 x − 1.60t + 4.00 sin 5.0 x − 2.00t evaluated at the given x values.
a
a
f
f
(a)
x = 1.00 , t = 1.00
y = 3.00 cos 2.40 rad + 4.00 sin +3.00 rad = −1.65 cm
(b)
x = 1.00 , t = 0.500
y = 3.00 cos +3.20 rad + 4.00 sin +4.00 rad = −6.02 cm
(c)
x = 0.500 , t = 0
y = 3.00 cos +2.00 rad + 4.00 sin +2.50 rad = 1.15 cm
a
f
a
f
a
f
a
f
526
Superposition and Standing Waves
P18.2
FIG. P18.2
P18.3
(a)
y2
(b)
a f
= f a x + vtf , so wave 2 travels in the
y1 = f x − vt , so wave 1 travels in the +x direction
To cancel, y1 + y 2 = 0 :
−x direction
5
=
2
for the positive root,
+5
a 3 x − 4t f + 2 a 3 x + 4t − 6 f
a3 x − 4tf = a3x + 4t − 6f
3 x − 4t = ±a3 x + 4t − 6f
2
2
+2
2
8t = 6
t = 0.750 s
(at t = 0.750 s , the waves cancel everywhere)
(c)
for the negative root,
x = 1.00 m
6x = 6
(at x = 1.00 m , the waves cancel always)
P18.4
Suppose the waves are sinusoidal.
a4.00 cmf sinakx − ωtf + a4.00 cmf sinakx − ωt + 90.0°f
2a 4.00 cmf sina kx − ωt + 45.0°f cos 45.0°
So the amplitude is a8.00 cmf cos 45.0° = 5.66 cm .
The sum is
P18.5
The resultant wave function has the form
FG φ IJ sinFG kx − ωt + φ IJ
H 2K H
2K
L − π 4 OP =
FφI
A = 2 A cosG J = 2a5.00f cos M
H 2K
N 2 Q
y = 2 A0 cos
(a)
(b)
0
f=
ω 1 200π
=
= 600 Hz
2π
2π
9.24 m
Chapter 18
P18.6
FG φ IJ = A
H 2K
FG IJ
H K
1
φ
π
= cos −1
= 60.0° =
2
2
3
2π
Thus, the phase difference is
φ = 120° =
3
λ
T
1
=
=
This phase difference results if the time delay is
3 3 f 3v
3.00 m
= 0.500 s
Time delay =
3 2.00 m s
2 A 0 cos
0
so
b
P18.7
P18.8
g
(a)
If the end is fixed, there is inversion of the pulse upon reflection. Thus, when they meet,
they cancel and the amplitude is zero .
(b)
If the end is free, there is no inversion on reflection. When they meet, the amplitude is
2 A = 2 0.150 m = 0.300 m .
(a)
∆x = 9.00 + 4.00 − 3.00 = 13 − 3.00 = 0.606 m
a
f
v 343 m s
=
= 1.14 m
f
300 Hz
∆x 0.606
=
= 0.530 of a wave ,
1.14
λ
∆φ = 2π 0.530 = 3.33 rad
λ=
The wavelength is
Thus,
a
or
(b)
∆x
f
∆x
v
v
343
=
= 283 Hz
f=
2 ∆x 2 0.606
For destructive interference, we want
λ
where ∆x is a constant in this set up.
P18.9
527
= 0.500 = f
a
f
We suppose the man’s ears are at the same level as the lower speaker. Sound from the upper
speaker is delayed by traveling the extra distance L2 + d 2 − L .
2n − 1 λ
with n = 1, 2 , 3 , …
He hears a minimum when this is
2
n−1 2 v
L2 + d 2 − L =
Then,
f
n−1 2 v
L2 + d 2 =
+L
f
a
b
g
b
g
bn − 1 2g v
=
2
2
L +d
2
2
L=
f
f
2
2
b gv
2bn − 1 2gv f
d − n −1 2
2
2
+ L2 +
f
b
g
2 n − 1 2 vL
f
2
n = 1, 2 , 3 , …
This will give us the answer to (b). The path difference starts from nearly zero when the man is very
far away and increases to d when L = 0. The number of minima he hears is the greatest integer
n−1 2 v
solution to d ≥
f
df 1
n = greatest integer ≤ + .
v 2
b
continued on next page
g
528
Superposition and Standing Waves
(a)
a
fb
g
4.00 m 200 s 1
df 1
+ =
+ = 2.92
330 m s
2
v 2
He hears two minima.
(b)
With n = 1,
L=
b gv
2b1 2g v f
d2 − 1 2
2
2
f2
a4.00 mf − b330 m sg 4b200 sg
b330 m sg 200 s
2
2
=
2
L = 9.28 m
with n = 2
L=
P18.10
b gv
2b3 2 g v f
d2 − 3 2
2
2
f2
= 1.99 m .
Suppose the man’s ears are at the same level as the lower speaker. Sound from the upper speaker is
delayed by traveling the extra distance ∆r = L2 + d 2 − L .
a
He hears a minimum when ∆r = 2n − 1
Then,
fFGH λ2 IJK with n = 1, 2, 3, …
IJ FG v IJ
KH f K
F 1IF vI
L + d = Gn − J G J + L
H 2KH f K
F 1I F vI F 1IF vI
L + d = G n − J G J + 2G n − J G J L + L
H 2K H f K H 2KH f K
F 1I F vI F 1IF vI
d − G n − J G J = 2G n − J G J L
H 2K H f K H 2KH f K
FG
H
L2 + d 2 − L = n −
2
2
2
1
2
2
2
2
2
2
2
2
(1)
Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The
path difference ∆r starts from nearly zero when the man is very far away and increases to d when
L = 0.
(a)
The number of minima he hears is the greatest integer value for which L ≥ 0 . This is the
1 v
same as the greatest integer solution to d ≥ n −
, or
2 f
FG
H
IJ FG IJ
KH K
number of minima heard = n max = greatest integer ≤ d
(b)
FG f IJ + 1
H vK 2
.
From equation 1, the distances at which minima occur are given by
Ln =
b g bv f g
2bn − 1 2gb v f g
d2 − n − 1 2
2
2
where n = 1, 2 , … , n max .
P18.11
(a)
Chapter 18
f b
b
ga
ga f
= b 25.0 rad cmga5.00 cmf − b 40.0 rad sga 2.00 sf = 45.0 rad
529
φ 1 = 20.0 rad cm 5.00 cm − 32.0 rad s 2.00 s = 36.0 rad
φ1
∆φ = 9.00 radians = 516° = 156°
(b)
∆φ = 20.0 x − 32.0t − 25.0 x − 40.0t = −5.00 x + 8.00t
At t = 2.00 s , the requirement is
a f a
f
∆φ = −5.00 x + 8.00 2.00 = 2n + 1 π for any integer n.
For x < 3.20 , −5.00 x + 16.0 is positive, so we have
a
f
−5.00 x + 16.0 = 2n + 1 π , or
x = 3.20 −
a2n + 1fπ
5.00
The smallest positive value of x occurs for n = 2 and is
x = 3.20 −
P18.12
(a)
a4 + 1fπ = 3.20 − π =
5.00
0.058 4 cm .
v 344 m s
=
= 16.0 m
f 21.5 Hz
First we calculate the wavelength:
λ=
Then we note that the path difference equals
9.00 m − 1.00 m =
1
λ
2
Therefore, the receiver will record a minimum in sound intensity.
(b)
We choose the origin at the midpoint between the speakers. If the receiver is located at point
(x, y), then we must solve:
Then,
ax + 5.00f
ax + 5.00f
2
+ y2 −
2
+ y2 =
λ2
=λ
4
ax − 5.00f
Square both sides and simplify to get:
20.0 x −
Upon squaring again, this reduces to:
400 x 2 − 10.0 λ2 x +
Substituting λ = 16.0 m , and reducing,
or
ax − 5.00f
ax − 5.00f
(When plotted this yields a curve called a hyperbola.)
2
2
1
λ
2
1
+ y2 + λ
2
+ y2 =
+ y2
λ4
= λ2 x − 5.00
16.0
9.00 x 2 − 16.0 y 2 = 144
y2
x2
−
=1
16.0 9.00
2
a
f
2
+ λ2 y 2
530
Superposition and Standing Waves
Section 18.2
P18.13
Standing Waves
a
f a
f a f
y = 1.50 m sin 0.400 x cos 200t = 2 A0 sin kx cos ωt
Therefore, k =
2π
λ
λ=
= 0.400 rad m
and ω = 2π f so
f=
2π
= 15.7 m
0.400 rad m
ω
2π
The speed of waves in the medium is v = λf =
P18.14
y = 0.030 0 m cos
(a)
200 rad s
= 31.8 Hz
2π rad
=
200 rad s
λ
ω
= 500 m s
2π f = =
k 0.400 rad m
2π
FG x IJ cosa40tf
H 2K
nodes occur where y = 0 :
π
x
= 2n + 1
2
2
a
f
a2n + 1fπ = π , 3π , 5π , … .
F 0.400 IJ = 0.029 4 m
= 0.030 0 m cosG
H 2 K
so x =
(b)
P18.15
y max
The facing speakers produce a standing wave in the space between them, with the spacing between
nodes being
d NN =
343 m s
v
λ
=
=
= 0. 214 m
2 2 f 2 800 s −1
e
j
If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a
distance from either speaker of
1.25 m
= 0.625 .
2
Then there isa node at
0.625 −
0. 214
= 0.518 m
2
a node at
0.518 m − 0.214 m = 0.303 m
a node at
0.303 m − 0.214 m = 0.089 1 m
a node at
0.518 m + 0.214 m = 0.732 m
a node at
0.732 m + 0.214 m = 0.947 m
and a node at
0.947 m + 0.214 m = 1.16 m from either speaker.
531
Chapter 18
P18.16
y = 2 A 0 sin kx cos ωt
∂2y
∂x 2
P18.17
∂2 y
= −2 A0 k 2 sin kx cos ωt
∂t 2
= −2 A0ω 2 sin kx cos ωt
Substitution into the wave equation gives
−2 A 0 k 2 sin kx cos ωt =
This is satisfied, provided that
v=
ω
k
f
a
f
y = y + y = 3.00 sinbπ x g cosb0.600π t g + 3.00 sinbπ x g cosb0.600π t g cm
y = a6.00 cmf sinbπ x g cosb0.600π t g
(a)
We can take cosb0.600π t g = 1 to get the maximum y.
y
= a6.00 cmf sina0.250π f =
At x = 0.250 cm,
2
0
2
sin kx cos ωt
j
a
y1 = 3.00 sin π x + 0.600t cm; y 2 = 3.00 sin π x − 0.600t cm
1
2
max
a
At x = 0.500 cm ,
(c)
Now take cos 0.600π t = −1 to get y max :
b
f
g
a
f a
fa f
At x = 1.50 cm,
y max = 6.00 cm sin 1.50π −1 = 6.00 cm
The antinodes occur when
x=
But k =
(a)
f a
4.24 cm
y max = 6.00 cm sin 0.500π = 6.00 cm
(b)
(d)
P18.18
FG 1 IJ e−2 A ω
Hv K
2π
λ
= π , so
nλ
n = 1, 3 , 5 , …
4
b
g
λ = 2.00 cm
λ
= 0.500 cm as in (b)
4
3λ
= 1.50 cm as in (c)
x2 =
4
5λ
= 2.50 cm
x3 =
4
and
x1 =
The resultant wave is
y = 2 A sin kx +
FG
H
IJ FG
K H
φ
φ
cos ωt −
2
2
IJ
K
φ
= nπ
2
nπ φ
−
so
x=
k
2k
φ
which means that each node is shifted
to the left.
2k
(b)
The nodes are located at
kx +
The separation of nodes is
∆x = n + 1
LMa f π − φ OP − LM nπ − φ OP
N k 2k Q N k 2k Q
The nodes are still separated by half a wavelength.
∆x =
π λ
=
k 2
532
Superposition and Standing Waves
Section 18.3
P18.19
Standing Waves in a String Fixed at Both Ends
L = 30.0 m ; µ = 9.00 × 10 −3 kg m; T = 20.0 N ; f1 =
where
FTI
v=G J
H µK
so
f1 =
12
= 47.1 m s
47.1
= 0.786 Hz
60.0
f 2 = 2 f1 = 1.57 Hz
f3 = 3 f1 = 2.36 Hz
*P18.20
P18.21
v
2L
f 4 = 4 f1 = 3.14 Hz
b ge
j
The tension in the string is
T = 4 kg 9.8 m s 2 = 39.2 N
Its linear density is
µ=
and the wave speed on the string is
v=
In its fundamental mode of vibration, we have
λ = 2L = 2 5 m = 10 m
Thus,
f=
(a)
(b)
m 8 × 10 −3 kg
=
= 1.6 × 10 −3 kg m
L
5m
T
µ
39.2 N
=
1.6 × 10 −3 kg m
= 156.5 m s
a f
v
λ
=
156.5 m s
= 15.7 Hz
10 m
Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then
n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For
2L
v
, and the frequency is f = .
standing waves, λ =
n
λ
Thus,
f=
n Tn
2L µ
and also
f=
n + 1 Tn +1
µ
2L
Thus,
Tn
n +1
=
=
n
Tn +1
Therefore,
4n + 4 = 5n , or n = 4
Then,
f=
4
2 2.00 m
a
f
b25.0 kg gg = 5
b16.0 kggg 4
b25.0 kg ge9.80 m s j =
2
0.002 00 kg m
350 Hz
The largest mass will correspond to a standing wave of 1 loop
an = 1f so
1
350 Hz =
2 2.00 m
yielding
m = 400 kg
a
f
e
m 9.80 m s 2
j
0.002 00 kg m
Chapter 18
*P18.22
533
The first string has linear density
1.56 × 10 −3 kg
= 2.37 × 10 −3 kg m.
0.658 m
µ1 =
The second, µ 2 =
6.75 × 10 −3 kg
= 7.11 × 10 −3 kg m.
0.950 m
The tension in both is T = 6.93 kg 9.8 m s 2 = 67.9 N . The speed of waves in the first string is
67.9 N
T
=
= 169 m s
µ1
2.37 × 10 −3 kg m
v1 =
T
and in the second v 2 =
= 97.8 m s . The two strings vibrate at the same frequency, according to
µ2
n 1 v1 n 2 v 2
=
2 L1
2L 2
n1 169 m s n 2 97.8 m s
=
2 0.658 m
2 0.950 m
a
f a
f
n2
5
= 2.50 = . Thus n1 = 2 and n 2 = 5 are the number of antinodes on each string in the lowest
2
n1
resonance with a node at the junction.
(b)
The first string has 2 + 1 = 3 nodes and the second string 5
more nodes, for a total of 8, or 6 other than the vibrator
junction
and pulley.
(a)
*P18.23
The frequency is
b
g=
2a0.658 mf
2 169 m s
FIG. P18.22(b)
257 Hz .
af
For the E-string on a guitar vibrating as a whole, v = fλ = 330 Hz 2 64.0 cm . When it is stopped at
64.0 cm
the first fret we have 12 2 330 Hz 2 L F = v = 330 Hz 2 64.0 cm . So L F = 12
. Similarly for the
2
64.0 cm
. The spacing between the first
second fret, 2 2 12 330 Hz 2 L F# = v = 330 Hz 2 64.0 cm . L F# =
2 2 12
and second frets is
af
af
af
af
FG 1
H2
64.0 cm
1 12
−
1
2
2 12
IJ = 64.0 cmFG 1 − 1 IJ = 3.39 cm.
K
H 1.059 5 1.059 5 K
2
This is a more precise version of the answer to the example in the text.
Now the eighteenth fret is distant from the bridge by L18 =
much string vibrate: L19 =
64.0 cm
. And the nineteenth lets this
2 18 12
64.0 cm
. The distance between them is
2 19 12
FG 1
H2
64.0 cm
18 12
−
1
2
19 12
IJ = 64.0 cm 1 FG 1 − 1 IJ =
K
K
2 H
2
1.5
1 12
1. 27 cm .
534
*P18.24
Superposition and Standing Waves
For the whole string vibrating, d NN = 0.64 m =
λ
; λ = 1.28 m . The
2
1
speed of a pulse on the string is v = fλ = 330 1.28 m = 422 m s .
s
λ
2
(a)
When the string is stopped at the fret, d NN = 0.64 m = ;
3
2
λ = 0.853 m
v 422 m s
f= =
= 495 Hz .
λ 0.853 m
(b)
The light touch at a point one third of the way along the
string damps out vibration in the two lowest vibration
states of the string as a whole. The whole string vibrates in
its third resonance possibility: 3d NN = 0.64 m = 3
λ = 0.427 m
f=
P18.25
P18.26
f1 =
FIG. P18.24(a)
FG IJ
H K
v
T
, where v =
µ
2L
v
λ
=
λ
;
2
FIG. P18.24(b)
422 m s
= 990 Hz .
0.427 m
12
1
.
2
(a)
If L is doubled, then f1 ∝ L−1 will be reduced by a factor
(b)
If µ is doubled, then f1 ∝ µ −1 2 will be reduced by a factor
(c)
If T is doubled, then f1 ∝ T will increase by a factor of
1
2
.
2.
L = 60.0 cm = 0.600 m; T = 50.0 N ; µ = 0.100 g cm = 0.010 0 kg m
fn =
where
nv
2L
FG T IJ = 70.7 m s
H µK
F 70.7 IJ = 58.9n = 20 000 Hz
= nG
H 1.20 K
12
v=
fn
Largest n = 339 ⇒ f = 19.976 kHz .
P18.27
d NN = 0.700 m
λ = 1.40 m
fλ = v = 308 m s =
T
e1.20 × 10 j a0.700f
(a)
T = 163 N
(b)
f3 = 660 Hz
−3
FIG. P18.27
535
Chapter 18
P18.28
a
f fv ; λ = 2L = fv
Ff I
F fI
F 392 IJ = 0.038 2 m
− G J L = L G 1 − J = a0.350 mfG 1 −
H 440 K
Hf K
H fK
λ G = 2 0.350 m =
LG − L A = LG
A
G
G
G
A
A
A
G
G
A
Thus, L A = LG − 0.038 2 m = 0.350 m − 0.038 2 m = 0.312 m ,
or the finger should be placed 31.2 cm from the bridge .
LA =
v
1
=
2 fA 2 fA
T
µ
; dL A =
dT
4 f A Tµ
;
dL A 1 dT
=
2 T
LA
dL
0.600 cm
dT
=2 A =2
= 3.84%
35.0 − 3.82 cm
T
LA
a
P18.29
f
In the fundamental mode, the string above the rod has only
two nodes, at A and B, with an anti-node halfway between A
and B. Thus,
A
L
λ
2L
= AB =
or λ =
.
cos θ
cos θ
2
Since the fundamental frequency is f, the wave speed in this
segment of string is
v = λf =
θ
B
L
2Lf
.
cos θ
M
Also, v =
T
T
=
=
µ
m AB
TL
m cos θ
T
where T is the tension in this part of the string. Thus,
2Lf
=
cos θ
F
θ
4L2 f 2
TL
TL
or
=
2
m cos θ
m
cos θ
cos θ
and the mass of string above the rod is:
T cos θ
m=
4Lf 2
Mg
[Equation 1]
FIG. P18.29
Now, consider the tension in the string. The light rod would rotate about point P if the string exerted
any vertical force on it. Therefore, recalling Newton’s third law, the rod must exert only a horizontal
force on the string. Consider a free-body diagram of the string segment in contact with the end of
the rod.
Mg
∑ Fy = T sin θ − Mg = 0 ⇒ T = sin θ
Then, from Equation 1, the mass of string above the rod is
m=
FG Mg IJ cos θ =
H sinθ K 4Lf
2
Mg
4Lf 2 tan θ
.
536
*P18.30
Superposition and Standing Waves
Let m = ρV represent the mass of the copper cylinder. The original tension in the wire is
V
g on the cylinder, to reduce the tension to
T1 = mg = ρVg . The water exerts a buoyant force ρ water
2
T2 = ρVg − ρ water
FG IJ
H K
FG V IJ g = FG ρ − ρ IJ Vg .
H 2K H 2 K
The speed of a wave on the string changes from
f1 =
v1
λ
=
water
T1
µ
T1 1
µ λ
to
T2
µ
to f 2 =
. The frequency changes from
T2 1
µ λ
where we assume λ = 2L is constant.
Then
*P18.31
f2
T2
=
=
f1
T1
ρ − ρ water 2
8.92 − 1.00 2
=
ρ
8.92
f 2 = 300 Hz
8.42
= 291 Hz
8.92
a
f db
g i db
gi
Comparing
y = 0.002 m sin π rad m x cos 100π rad s t
with
y = 2 A sin kx cos ωt
2π
k=
= π m −1 , λ = 2.00 m , and ω = 2π f = 100π s −1 : f = 50.0 Hz
we find
(a)
λ
λ
= 1.00 m
2
L
3.00 m
=
= 3 loops
d NN 1.00 m
d NN =
Then the distance between adjacent nodes is
and on the string are
e
(b)
λb
, λ b = 6.00 m
2
100 m s
v
= 16.7 Hz
fb = a =
λb
6.00 m
In the simplest standing wave vibration, d NN = 3.00 m =
and
(c)
j
v = fλ = 50 s −1 2 m = 100 m s
For the speed we have
In v 0 =
T0
µ
increases to
, if the tension increases to Tc = 9T0 and the string does not stretch, the speed
vc =
Then
9T0
=3
T0
b
g
= 3 v 0 = 3 100 m s = 300 m s
µ
300 m s
λ
v
= 6.00 m
λc = c =
d NN = c = 3.00 m
−1
2
fa
50 s
µ
and one loop fits onto the string.
Chapter 18
Section 18.4
P18.32
537
Resonance
The natural frequency is
f=
9.80 m s 2
= 0.352 Hz .
2.00 m
g
1
=
L 2π
1
1
=
T 2π
The big brother must push at this same frequency of 0.352 Hz to produce resonance.
P18.33
v=
9.15 m
= 3.66 m s
2.50 s
(a)
The wave speed is
(b)
From the figure, there are antinodes at both ends of the pond, so the distance between
adjacent antinodes
λ
= 9.15 m ,
2
is
d AA =
and the wavelength is
λ = 18.3 m
The frequency is then
f=
v
λ
=
3.66 m s
= 0.200 Hz
18.3 m
We have assumed the wave speed is the same for all wavelengths.
P18.34
v = gd =
The wave speed is
e9.80 m s ja36.1 mf = 18.8 m s
2
The bay has one end open and one closed. Its simplest resonance is with a node of horizontal
velocity, which is also an antinode of vertical displacement, at the head of the bay and an antinode
of velocity, which is a node of displacement, at the mouth. The vibration of the water in the bay is
like that in one half of the pond shown in Figure P18.33.
Then,
d NA = 210 × 10 3 m =
and
λ = 840 × 10 3 m
Therefore, the period is
T=
λ
4
1 λ 840 × 10 3 m
= =
= 4. 47 × 10 4 s = 12 h 24 min
18.8 m s
f v
This agrees precisely with the period of the lunar excitation , so we identify the extra-high tides as
amplified by resonance.
P18.35
The distance between adjacent nodes is one-quarter of the circumference.
d NN = d AA =
so λ = 10.0 cm and f =
λ 20.0 cm
=
= 5.00 cm
2
4
v 900 m s
=
= 9 000 Hz = 9.00 kHz .
λ 0.100 m
The singer must match this frequency quite precisely for some interval of time to feed enough
energy into the glass to crack it.
538
Superposition and Standing Waves
Section 18.5
P18.36
P18.37
Standing Waves in Air Columns
d AA = 0.320 m ; λ = 0.640 m
v
(a)
f=
(b)
λ = 0.085 0 m; d AA = 42.5 mm
(a)
For the fundamental mode in a closed pipe, λ = 4L , as
in the diagram.
λ
= 531 Hz
But v = fλ , therefore L =
So, L =
(b)
e
4 240 s −1
j
= 0.357 m
A
λ=
A
FIG. P18.37
j
The wavelength is
N
λ /2
343 m s
v
=
= 0.715 m
2 f 2 240 s −1
e
N
λ /4
L
v
4f
For an open pipe, λ = 2L , as in the diagram.
So, L =
P18.38
343 m s
A
v 343 m s
=
= 1.31 m
f
261.6 s
so the length of the open pipe vibrating in its simplest (A-N-A) mode is
d A to A =
A closed pipe has
1
λ = 0.656 m
2
(N-A) for its simplest resonance,
(N-A-N-A) for the second,
*P18.39
and
(N-A-N-A-N-A) for the third.
Here, the pipe length is
5d N to A =
5λ 5
= 1.31 m = 1.64 m
4
4
a
f
Assuming an air temperature of T = 37° C = 310 K , the speed of sound inside the pipe is
b
v = 331 m s
g
310 K
= 353 m s .
273 K
In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is
λ = 4L . Thus, for the whooping crane
a
f
λ = 4 5.0 ft = 2.0 × 10 1 ft and f =
v
λ
=
b353 m sg FG 3.281 ft IJ =
2.0 × 10 ft H 1 m K
1
57.9 Hz .
Chapter 18
P18.40
539
The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and
antinode at the open end,
with
d N to A = 3 cm =
so
λ = 0.12 m
and
f=
λ
4
v 343 m s
=
≈ 3 kHz
λ
0.12 m
A small-amplitude external excitation at this frequency can, over time, feed energy
into a larger-amplitude resonance vibration of the air in the canal, making it audible.
P18.41
For a closed box, the resonant frequencies will have nodes at both sides, so the permitted
1
wavelengths will be L = nλ , n = 1, 2 , 3 , … .
2
b
i.e., L =
g
nλ nv
nv
=
.
and f =
2L
2
2f
Therefore, with L = 0.860 m and L ′ = 2.10 m, the resonant frequencies are
a
fn = n 206 Hz
a
f
for L = 0.860 m for each n from 1 to 9
and fn′ = n 84.5 Hz
P18.42
f
for L ′ = 2.10 m for each n from 2 to 23.
λ=
The wavelength of sound is
The distance between water levels at resonance is d =
t=
and
P18.43
v
2f
∴ Rt = π r 2 d =
π r 2v
2f
π r 2v
.
2 Rf
For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of
resonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz. These are
odd-integer multipliers of the fundamental frequency of 50.0 Hz . Then the pipe length is
d NA =
P18.44
v
f
λ v 340 m s
=
=
= 1.70 m .
4 4f
4 50 s
b g
L
λ
= d AA = or
n
2
Since λ =
v
f
With v = 343 m s and
L=
nλ
2
L=n
FG v IJ
H2fK
for n = 1, 2 , 3 , …
for n = 1, 2 , 3 , …
f = 680 Hz,
L=n
F 343 m s I = na0.252 mf
GH 2a680 Hzf JK
for n = 1, 2 , 3 , …
a
f
Possible lengths for resonance are: L = 0.252 m, 0.504 m, 0.757 m, … , n 0. 252 m .
540
P18.45
Superposition and Standing Waves
For resonance in a narrow tube open at one end,
f =n
(a)
b
f = 384 Hz
g
v
n = 1, 3 , 5 , … .
4L
warm
air
Assuming n = 1 and n = 3 ,
384 =
v
3v
and 384 =
.
4 0.228
4 0.683
a
a
f
f
22.8 cm
68.3 cm
In either case, v = 350 m s .
(b)
For the next resonance n = 5 , and L =
b
e
g
j
5 v 5 350 m s
=
= 1.14 m .
4f
4 384 s −1
FIG. P18.45
P18.46
The length corresponding to the fundamental satisfies f =
v
v
34
: L1 =
=
= 0.167 m .
4L
4 f 4 512
a f
Since L > 20.0 cm, the next two modes will be observed, corresponding to f =
or L 2 =
P18.47
3v
5v
= 0.502 m and L3 =
= 0.837 m .
4f
4f
We suppose these are the lowest resonances of the enclosed air columns.
λ
= 0.670 m
2
For one,
λ=
v 343 m s
=
= 1.34 m
f
256 s −1
length = d AA =
For the other,
λ=
v 343 m s
=
= 0.780 m
f
440 s −1
length = 0.390 m
So,
(b)
original length = 1.06 m
λ = 2d AA = 2.12 m
P18.48
3v
5v
and f =
.
4L 2
4L3
343 m s
= 162 Hz
2.12 m
(a)
f=
(a)
For the fundamental mode of an open tube,
L=
(b)
v = 331 m s 1 +
343 m s
λ v
=
=
= 0.195 m .
2 2 f 2 880 s −1
e
j
a−5.00f = 328 m s
273
We ignore the thermal expansion of the metal.
f=
The flute is flat by a semitone.
v
λ
=
328 m s
v
=
= 841 Hz
2L 2 0.195 m
a
f
Chapter 18
Section 18.6
P18.49
541
Standing Waves in Rod and Plates
5 100
v
=
= 1.59 kHz
2L
2 1.60
(a)
f=
(b)
Since it is held in the center, there must be a node in the center as well as antinodes at the
ends. The even harmonics have an antinode at the center so only the odd harmonics are
a fa f
present.
(c)
P18.50
f=
3 560
v′
=
= 1.11 kHz
2L
2 1.60
When the rod is clamped at one-quarter of its length, the vibration pattern reads ANANA and the
rod length is L = 2d AA = λ .
Therefore, L =
Section 18.7
P18.51
a fa f
v 5 100 m s
=
= 1.16 m
f
4 400 Hz
Beats: Interference in Time
f ∝v∝ T
f new = 110
540
= 104.4 Hz
600
∆f = 5.64 beats s
P18.52
(a)
The string could be tuned to either 521 Hz or 525 Hz from this evidence.
(b)
Tightening the string raises the wave speed and frequency. If the frequency were originally
521 Hz, the beats would slow down.
Instead, the frequency must have started at 525 Hz to become 526 Hz .
(c)
From f =
v
λ
=
T µ
2L
=
1 T
2L µ
FG IJ
H K
f2
T2
f
=
and T2 = 2
f1
T1
f1
2
T1 =
FG 523 Hz IJ
H 526 Hz K
2
T1 = 0.989T1 .
The fractional change that should be made in the tension is then
fractional change =
T1 − T2
= 1 − 0.989 = 0.011 4 = 1.14% lower.
T1
The tension should be reduced by 1.14% .
542
P18.53
Superposition and Standing Waves
For an echo f ′ = f
bv + v g the beat frequency is f
bv − v g
s
s
b
= f′− f .
Solving for fb .
gives fb = f
b2 v g
s
bv − v g when approaching wall.
2a1.33f
= a 256f
a343 − 1.33f = 1.99 Hz beat frequency
s
(a)
fb
(b)
When he is moving away from the wall, v s changes sign. Solving for v s gives
vs =
*P18.54
2
- foot pipes produces actual frequencies of 131 Hz and 196 Hz and a
3
combination tone at 196 − 131 Hz = 65.4 Hz , so this pair supplies the so-called missing fundamental.
The 4 and 2-foot pipes produce a combination tone 262 − 131 Hz = 131 Hz , so this does not work.
2
The 2 and 2 - foot pipes produce a combination tone at 262 − 196 Hz = 65.4 Hz , so this works.
3
2
Also, 4, 2 , and 2 - foot pipes all playing together produce the 65.4-Hz combination tone.
3
Using the 4 and 2
a
Section 18.8
P18.55
a fa f
a fa f
5 343
fb v
=
= 3.38 m s .
2 f − fb
2 256 − 5
f
a
a
f
f
Non-Sinusoidal Wave Patterns
We list the frequencies of the harmonics of each note in Hz:
Note
A
C#
E
1
440.00
554.37
659.26
2
880.00
1 108.7
1 318.5
Harmonic
3
1 320.0
1 663.1
1 977.8
4
1 760.0
2 217.5
2 637.0
The second harmonic of E is close the the third harmonic of A, and the fourth
harmonic of C# is close to the fifth harmonic of A.
P18.56
We evaluate
s = 100 sin θ + 157 sin 2θ + 62.9 sin 3θ + 105 sin 4θ
+51.9 sin 5θ + 29.5 sin 6θ + 25.3 sin 7θ
where s represents particle displacement in nanometers
and θ represents the phase of the wave in radians. As θ
advances by 2π , time advances by (1/523) s. Here is the
result:
FIG. P18.56
5
2 200.0
2 771.9
3 296.3
Chapter 18
543
Additional Problems
P18.57
f = 87.0 Hz
speed of sound in air: v a = 340 m s
(a)
ja
e
f
v = fλ b = 87.0 s −1 0.400 m
λb =
v = 34.8 m s
(b)
λ a = 4L
va = λ a f
UV
W
L=
340 m s
va
=
= 0.977 m
4 f 4 87.0 s −1
e
j
FIG. P18.57
*P18.58
(a)
Use the Doppler formula
f′= f
bv ± v g .
bv ∓ v g
0
s
With f1′ = frequency of the speaker in front of student and
f 2′ = frequency of the speaker behind the student.
m s + 1.50 m sg
= 458 Hz
f b343b343
m s − 0g
b343 m s − 1.50 m sg = 454 Hz
f ′ = a 456 Hzf
b343 m s + 0g
a
f1′ = 456 Hz
2
Therefore, fb = f1′ − f 2′ = 3.99 Hz .
(b)
The waves broadcast by both speakers have λ =
v 343 m s
=
= 0.752 m . The standing wave
f
456 s
λ
= 0.376 m . The student walks from one maximum to the next in
2
0.376 m
1
= 0.251 s , so the frequency at which she hears maxima is f = = 3.99 Hz .
time ∆t =
T
1.50 m s
between them has d AA =
P18.59
Moving away from station, frequency is depressed:
343
f ′ = 180 − 2.00 = 178 Hz :
178 = 180
343 − −v
Solving for v gives
Therefore,
v=
a2.00fa343f
a f
178
v = 3.85 m s away from station
Moving toward the station, the frequency is enhanced:
343
f ′ = 180 + 2.00 = 182 Hz :
182 = 180
343 − v
2.00 343
Solving for v gives
4=
182
Therefore,
v = 3.77 m s toward the station
a fa f
544
P18.60
Superposition and Standing Waves
v=
a48.0fa2.00f = 141 m s
4.80 × 10 −3
d NN = 1.00 m ; λ = 2.00 m ; f =
λa =
P18.61
v
λ
= 70.7 Hz
v a 343 m s
=
= 4.85 m
f
70.7 Hz
Call L the depth of the well and v the speed of sound.
f λ4 = a2n − 1f 4vf = a2n4− 511fb.5343s m sg
e
j
a2n + 1fb343 m sg
λ
v
= a 2n + 1f
=
L = 2an + 1f − 1
4
4f
4e60.0 s j
a2n − 1fb343 m sg = a2n + 1fb343 m sg
4e51.5 s j
4e60.0 s j
a
L = 2n − 1
Then for some integer n
1
−1
1
2
and for the next resonance
2
Thus,
−1
and we require an integer solution to
The equation gives n =
−1
−1
2n + 1 2n − 1
=
60.0
51.5
111.5
= 6.56 , so the best fitting integer is n = 7 .
17
Then
L=
and
L=
af b
g = 21.6 m
4e51.5 s j
2a7 f + 1 b343 m sg
= 21.4 m
4e60.0 s j
2 7 − 1 343 m s
−1
−1
suggest the best value for the depth of the well is 21.5 m .
P18.62
The second standing wave mode of the air in the pipe reads ANAN, with d NA =
so
λ = 2.33 m
and
f=
v
λ
=
343 m s
= 147 Hz
2.33 m
For the string, λ and v are different but f is the same.
λ
0.400 m
= d NN =
2
2
so
λ = 0.400 m
a
fa
f
T
v = λf = 0. 400 m 147 Hz = 58.8 m s =
e
jb
T = µv 2 = 9.00 × 10 −3 kg m 58.8 m s
g
µ
2
= 31.1 N
λ 1.75 m
=
3
4
Chapter 18
P18.63
(a)
Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The
frequency and tension are the same in both sections, so
f=
(b)
1
2L
T
µ
=
1
2f
thin wire.
so L ′ =
(a)
1
2 0.400
a
4.60
= 59.9 Hz .
2.00 × 10 −3
f
As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire.
µ ′ = 8.00 g m
P18.64
545
T
µ′
L′ =
LM 1 OP
N a2fa59.9f Q
4.60
= 20.0 cm half the length of the
8.00 × 10 −3
For the block:
∑ Fx = T − Mg sin 30.0° = 0
so T = Mg sin 30.0° =
(b)
1
Mg .
2
The length of the section of string parallel to the incline is
h
= 2 h . The total length of the string is then 3h .
sin 30.0°
FIG. P18.64
m
3h
(c)
The mass per unit length of the string is
µ=
(d)
The speed of waves in the string is
v=
(e)
In the fundamental mode, the segment of length h vibrates as one loop. The distance
between adjacent nodes is then d NN =
f=
3 Mgh
2m
v
λ
=
1 3 Mgh
=
2h
2m
3 Mg
8mh
λ= h .
The period of the standing wave of 3 nodes (or two loops) is
T=
(h)
FG Mg IJ FG 3 h IJ =
H 2 KH m K
When the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2
the wavelength is
(f)
µ
=
λ
= h , so the wavelength is λ = 2h .
2
The frequency is
(g)
T
e
j
fb = 1.02 f − f = 2.00 × 10 −2 f =
e2.00 × 10 j
−2
3 Mg
8mh
1 λ
2m
= =h
=
3 Mgh
f v
2mh
3 Mg
FG λ IJ and
H 2K
546
P18.65
Superposition and Standing Waves
(a)
f=
n
2L
T
so
f′ L
L 1
=
=
=
f L ′ 2L 2
µ
The frequency should be halved to get the same number of antinodes for twice the
length.
(b)
n′
T
=
n
T′
FG IJ = LM n OP
H K Nn + 1 Q
L n OP T
T′ = M
Nn + 1 Q
T ′ F nf ′L ′ I
=G
J
T H n ′fL K
T′
n
=
T
n′
so
2
2
2
The tension must be
(c)
FG IJ
H K
T′
3
=
T
2⋅2
P18.66
2
f ′ n ′L T ′
=
f nL ′ T
so
2
T′ 9
=
T 16
to get twice as many antinodes.
0.010 0 kg
= 5.00 × 10 −3 kg m :
For the wire, µ =
2.00 m
T
v=
µ
e200 kg ⋅ m s j
2
=
5.00 × 10 −3 kg m
v = 200 m s
If it vibrates in its simplest state, d NN = 2.00 m =
f=
v
λ
b200 m sg = 50.0 Hz
=
4.00 m
(a)
The tuning fork can have frequencies 45.0 Hz or 55.0 Hz .
(b)
If f = 45.0 Hz , v = fλ = 45.0 s 4.00 m = 180 m s .
b
b
g
Then, T = v 2 µ = 180 m s
g e5.00 × 10
2
or if f = 55.0 Hz , T = v 2 µ = f 2
P18.67
λ
:
2
j
λ µ = b55.0 sg a 4.00 mf e5.00 × 10
−3
kg m = 162 N
2
2
2
−3
j
kg m = 242 N .
We look for a solution of the form
a
f
a
f
b
g
= A sina 2.00 x − 10.0t f cos φ + A cosa 2.00 x − 10.0t f sin φ
5.00 sin 2.00 x − 10.0t + 10.0 cos 2.00 x − 10.0 t = A sin 2.00 x − 10.0 t + φ
This will be true if both
5.00 = A cos φ and 10.0 = A sin φ ,
requiring
a5.00f + a10.0f
2
2
= A2
A = 11.2 and φ = 63.4°
a
The resultant wave 11.2 sin 2.00 x − 10.0t + 63.4°
f
is sinusoidal.
Chapter 18
P18.68
P18.69
2π
and ω = 2π f =
2π v
FG 2π x IJ cosFG 2π vt IJ
H λK H λ K
b g
y x , t = 2 A sin kx cos ωt = 2 A sin
(a)
With k =
(b)
For the fundamental vibration,
λ 1 = 2L
so
y1 x , t = 2 A sin
(c)
For the second harmonic λ 2 = L and
y 2 x , t = 2 A sin
(d)
In general, λ n =
(a)
Let θ represent the angle each slanted rope
makes with the vertical.
λ
λ
:
2L
and
n
b g
FG π x IJ cosFG π vt IJ
HLK H L K
b g
FG 2π x IJ cosFG 2π vt IJ
H LK H L K
b g
FG nπ x IJ cosFG nπ vt IJ
H LK H L K
yn x , t = 2 A sin
In the diagram, observe that:
sin θ =
1.00 m 2
=
1.50 m 3
or θ = 41.8° .
Considering the mass,
∑ Fy = 0 : 2T cos θ = mg
b12.0 kg ge9.80 m s j =
or T =
FIG. P18.69
2
2 cos 41.8°
(b)
*P18.70
d AA =
78.9 N
T
v=
For the standing wave pattern shown (3 loops),
d=
3
λ
2
or
λ=
2 2.00 m
= 1.33 m
3
Thus, the required frequency is
f=
µ
a
v
λ
=
=
78.9 N
= 281 m s
0.001 00 kg m
The speed of transverse waves in the string is
f
281 m s
= 211 Hz
1.33 m
λ
= 7.05 × 10 −3 m is the distance between antinodes.
2
Then λ = 14.1 × 10 −3 m
and f =
v
λ
=
3.70 × 10 3 m s
14.1 × 10
−3
m
547
= 2.62 × 10 5 Hz .
The crystal can be tuned to vibrate at 2 18 Hz , so that binary counters can
derive from it a signal at precisely 1 Hz.
FIG. P18.70
548
Superposition and Standing Waves
ANSWERS TO EVEN PROBLEMS
P18.2
see the solution
P18.38
0.656 m; 1.64 m
P18.4
5.66 cm
P18.40
3 kHz; see the solution
P18.6
0.500 s
P18.42
∆t =
P18.8
(a) 3.33 rad; (b) 283 Hz
P18.10
(a) The number is the greatest
f
1
+ ;
integer ≤ d
v
2
P18.44
L = 0.252 m, 0.504 m, 0.757 m, … ,
n 0.252 m for n = 1, 2 , 3 , …
P18.46
0.502 m; 0.837 m
P18.48
(a) 0.195 m; (b) 841 m
P18.50
1.16 m
P18.52
(a) 521 Hz or 525 Hz; (b) 526 Hz;
(c) reduce by 1.14%
FG IJ
H K
d − bn − 1 2g b v f g
=
2bn − 1 2 gb v f g
2
(b) Ln
2
2
where
n = 1, 2 , … , n max
P18.12
λ
;
2
(b) along the hyperbola 9 x 2 − 16 y 2 = 144
(a) ∆x =
a
f
π r 2v
2 Rf
a
f
2
2
4-foot and 2 -foot ; 2 and 2 - foot; and
3
3
all three together
P18.14
(a) 2n + 1 π m for n = 0 , 1, 2 , 3 , …;
(b) 0.029 4 m
P18.54
P18.16
see the solution
P18.56
see the solution
P18.18
see the solution
P18.58
(a) and (b) 3.99 beats s
P18.20
15.7 Hz
P18.60
4.85 m
P18.22
(a) 257 Hz; (b) 6
P18.62
31.1 N
P18.24
(a) 495 Hz; (b) 990 Hz
P18.64
(a)
P18.26
19.976 kHz
P18.28
3.84%
3 Mgh
1
m
Mg ; (b) 3h; (c)
; (d)
;
2m
2
3h
3 Mg
2mh
(e)
; (g) h;
; (f)
8mh
3 Mg
e
(h) 2.00 × 10 −2
j
3 Mg
8mh
P18.30
291 Hz
P18.32
0.352 Hz
P18.66
(a) 45.0 Hz or 55.0 Hz; (b) 162 N or 242 N
P18.34
see the solution
P18.68
see the solution
P18.36
(a) 531 Hz; (b) 42.5 mm
P18.70
262 kHz
19
Temperature
CHAPTER OUTLINE
19.1
19.2
19.3
19.4
19.5
Temperature and the Zeroth
Law of Thermodynamics
Thermometers and the
Celsius Temperature Scale
The Constant-Volume Gas
Thermometer and the
Absolute Temperature Scale
Thermal Expansion of
Solids and Liquids
Macroscopic Description of
an Ideal Gas
ANSWERS TO QUESTIONS
Q19.1
Two objects in thermal equilibrium need not be in contact.
Consider the two objects that are in thermal equilibrium in
Figure 19.1(c). The act of separating them by a small distance
does not affect how the molecules are moving inside either
object, so they will still be in thermal equilibrium.
Q19.2
The copper’s temperature drops and the water temperature
rises until both temperatures are the same. Then the metal and
the water are in thermal equilibrium.
Q19.3
The astronaut is referring to the temperature of the lunar
surface, specifically a 400°F difference. A thermometer would
register the temperature of the thermometer liquid. Since there
is no atmosphere in the moon, the thermometer will not read a
realistic temperature unless it is placed into the lunar soil.
Q19.4
Rubber contracts when it is warmed.
Q19.5
Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact with
the hot water. Then the mercury heats up, and it expands.
Q19.6
If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than the
cavity in your tooth when you take a sip of your ever-beloved coffee, resulting in a broken or
cracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam would
contract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isn’t it nice
that your dentist knows thermodynamics?
Q19.7
The measurements made with the heated steel tape will be too short—but only by a factor of
5 × 10 −5 of the measured length.
Q19.8
(a)
One mole of H 2 has a mass of 2.016 0 g.
(b)
One mole of He has a mass of 4.002 6 g.
(c)
One mole of CO has a mass of 28.010 g.
Q19.9
The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because the
ideal gas law cannot work all the way down to or below the temperature at which gas turns to
liquid, or in the case of CO 2 , a solid.
549
550
Temperature
Q19.10
Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The air
inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall
moves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas to
start with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as the
water vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns to
slush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of the
original volume.
Q19.11
Cylinder A must be at lower pressure. If the gas is thin, it will be at one-third the absolute pressure
of B.
Q19.12
At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong
latches hold them together, but they would explode apart if you tried to open the hot cooker.
Q19.13
(a)
The water level in the cave rises by a smaller distance than the water outside, as the trapped
air is compressed. Air can escape from the cave if the rock is not completely airtight, and also
by dissolving in the water.
(b)
The ideal cave stays completely full of water at low tide. The water in the cave is supported
by atmospheric pressure on the free water surface outside.
(a)
(b)
FIG. Q19.13
Q19.14
Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies that
were mostly liquid water—or if they just liked its taste or its cleaning properties—it is conceivable
that they might place one hundred degrees between its freezing and boiling points. It is very
unlikely, on the other hand, that these would be our familiar “normal” ice and steam points, because
atmospheric pressure would surely be different where the aliens come from.
Q19.15
As the temperature increases, the brass expands. This would effectively increase the distance, d,
from the pivot point to the center of mass of the pendulum, and also increase the moment of inertia
of the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physical
I
pendulum is T = 2π
, the period would increase, and the clock would run slow.
mgd
Q19.16
As the water rises in temperature, it expands. The excess volume would spill out of the cooling
system. Modern cooling systems have an overflow reservoir to take up excess volume when the
coolant heats up and expands.
Q19.17
The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar,
both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a
certain temperature at which the inner diameter of the lid has expanded more than the top of the
jar, and the lid will be easier to remove.
Chapter 19
Q19.18
The sphere expands when heated, so that it no longer fits through the
ring. With the sphere still hot, you can separate the sphere and ring by
heating the ring. This more surprising result occurs because the thermal
expansion of the ring is not like the inflation of a blood-pressure cuff.
Rather, it is like a photographic enlargement; every linear dimension,
including the hole diameter, increases by the same factor. The reason
for this is that the atoms everywhere, including those around the inner
circumference, push away from each other. The only way that the
atoms can accommodate the greater distances is for the
circumference—and corresponding diameter—to grow. This property
was once used to fit metal rims to wooden wagon and horse-buggy
wheels. If the ring is heated and the sphere left at room temperature,
the sphere would pass through the ring with more space to spare.
551
FIG. Q19.18
SOLUTIONS TO PROBLEMS
Section 19.1
Temperature and the Zeroth Law of Thermodynamics
No problems in this section
Section 19.2
Thermometers and the Celsius Temperature Scale
Section 19.3
The Constant-Volume Gas Thermometer and the Absolute Temperature Scale
P19.1
Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A and
B are constants. To find A and B, we use the data
a
f
1.635 atm = A + a78.0° CfB
0.900 atm = A + −80.0° C B
(1)
(2)
Solving (1) and (2) simultaneously,
we find
A = 1.272 atm
and
B = 4.652 × 10 −3 atm ° C
Therefore,
P = 1.272 atm + 4.652 × 10 −3 atm ° C T
(a)
P = 0 = 1.272 atm + 4.652 × 10 −3 atm ° C T
At absolute zero
which gives
e
j
e
j
T = −274° C .
(b)
At the freezing point of water P = 1.272 atm + 0 = 1.27 atm .
(c)
And at the boiling point P = 1.272 atm + 4.652 × 10 −3 atm ° C 100° C = 1.74 atm .
e
ja
f
552
P19.2
Temperature
P1V = nRT1
and P2 V = nRT2
imply that
P19.3
P19.4
P2 T2
=
P1 T1
a
fa
f
(a)
P2 =
0.980 atm 273 K + 45.0 K
P1T2
=
= 1.06 atm
273 + 20.0 K
T1
(b)
T3 =
T1 P3
P1
a
f
a293 K fa0.500 atmf = 149 K =
=
0.980 atm
−124° C
FIG. P19.2
a
f
9
9
TC + 32.0° F = −195.81 + 32.0 = −320° F
5
5
(a)
TF =
(b)
T = TC + 273.15 = −195.81 + 273.15 = 77.3 K
(a)
To convert from Fahrenheit to Celsius, we use
TC =
and the Kelvin temperature is found as
T = TC + 273 = 310 K
(b)
In a fashion identical to that used in (a), we find TC = −20.6° C
T = 253 K
and
P19.5
P19.6
b
FG 212° F − 32.0° F IJ =
H 100° C − 0.00° C K
(a)
∆T = 450° C = 450° C
(b)
∆T = 450° C = 450 K
810° F
a
f
100° C = aa60.0° Sf + b
Subtracting, 100° C = aa75.0° Sf
Require
0.00° C = a −15.0° S + b
a
a = 1.33 C° S° .
f
Then 0.00° C = 1.33 −15.0° S C°+ b
b = 20.0° C .
b
g
So the conversion is TC = 1.33 C° S° TS + 20.0° C .
P19.7
(a)
T = 1 064 + 273 = 1 337 K melting point
T = 2 660 + 273 = 2 933 K boiling point
(b)
g a
f
5
5
TF − 32.0 = 98.6 − 32.0 = 37.0° C
9
9
∆T = 1 596° C = 1 596 K . The differences are the same.
Chapter 19
Section 19.4
P19.8
553
Thermal Expansion of Solids and Liquids
α = 1.10 × 10 −5 ° C −1 for steel
e
a
j
f
∆L = 518 m 1.10 × 10 −5 ° C −1 35.0° C − −20.0° C = 0.313 m
P19.9
The wire is 35.0 m long when TC = −20.0° C .
b
∆L = Liα T − Ti
g
a f
a f for Cu.
∆L = a35.0 mfe1.70 × 10 aC°f jc35.0° C − a −20.0° C fh =
−1
α = α 20.0° C = 1.70 × 10 −5 C°
−1
−5
a
fe
ja
+3.27 cm
f
P19.10
∆L = Liα ∆T = 25.0 m 12.0 × 10 −6 C° 40.0° C = 1.20 cm
P19.11
For the dimensions to increase, ∆L = αLi ∆T
a
fa
1.00 × 10 −2 cm = 1.30 × 10 −4 ° C −1 2.20 cm T − 20.0° C
f
T = 55.0° C
*P19.12
P19.13
*P19.14
ja
e
fa
f
∆L = αLi ∆T = 22 × 10 −6 C° 2.40 cm 30° C = 1.58 × 10 −3 cm
a
fa
f
a
fa
f
(a)
∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 30.0 cm 65.0° C = 0.176 mm
(b)
∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 1.50 cm 65.0° C = 8.78 × 10 −4 cm
(c)
∆V = 3αVi ∆T = 3 9.00 × 10 −6 ° C −1
e
F
jGH 30.0aπ fa4 1.50f
Ia
JK
2
f
cm3 65.0° C = 0.093 0 cm3
The horizontal section expands according to ∆L = αLi ∆T .
ja
e
fa
f
∆x = 17 × 10 −6 ° C −1 28.0 cm 46.5° C − 18.0° C = 1.36 × 10 −2 cm
The vertical section expands similarly by
ja
e
FIG. P19.14
fa
f
∆y = 17 × 10 −6 ° C −1 134 cm 28.5° C = 6.49 × 10 −2 cm .
The vector displacement of the pipe elbow has magnitude
∆r = ∆x 2 + ∆ y 2 =
a0.136 mmf + a0.649 mmf
2
2
= 0.663 mm
and is directed to the right below the horizontal at angle
θ = tan −1
FG ∆y IJ = tan FG 0.649 mm IJ = 78.2°
H 0.136 mm K
H ∆x K
−1
∆r = 0.663 mm to the right at 78.2° below the horizontal
554
P19.15
Temperature
b
g
b
L Al 1 + α Al ∆T = LBrass 1 + α Brass ∆T
L Al − LBrass
∆T =
LBrassα Brass − L Alα Al
(a)
∆T =
g
a10.01 − 10.00f
a10.00fe19.0 × 10 j − a10.01fe24.0 × 10 j
−6
−6
∆T = −199° C so T = −179° C. This is attainable.
∆T =
(b)
a10.02 − 10.00f
a10.00fe19.0 × 10 j − a10.02fe24.0 × 10 j
−6
−6
∆T = −396° C so T = −376° C which is below 0 K so it cannot be reached.
P19.16
g a50.0° Cf
jb
e
∆A = 2 17.0 × 10 −6 ° C −1 0.080 0 m
∆A = 2αAi ∆T :
(a)
2
∆A = 1.09 × 10 −5 m 2 = 0.109 cm 2
(b)
The length of each side of the hole has increased. Thus, this represents an increase in the
area of the hole.
b
g
e
e
P19.17
∆V = β − 3α Vi ∆T = 5.81 × 10 −4 − 3 11.0 × 10 −6
P19.18
(a)
a
f
jjb50.0 galga20.0f =
0.548 gal
a
5.050 cm = 5.000 cm 1 + 24.0 × 10 −6 ° C −1 T − 20.0° C
L = Li 1 + α∆T :
f
T = 437° C
(b)
L Al = LBrass for some ∆T , or
We must get
b
g
b
Li , Al 1 + α Al ∆T = Li , Brass 1 + α Brass ∆T
e
g
j
e
j
5.000 cm 1 + 24.0 × 10 −6 ° C −1 ∆T = 5.050 cm 1 + 19.0 × 10 −6 ° C −1 ∆T
Solving for ∆T , ∆T = 2 080° C ,
T = 3 000° C
so
This will not work because aluminum melts at 660° C .
P19.19
b
a
g
f
(a)
V f = Vi 1 + β∆T = 100 1 + 1.50 × 10 −4 −15.0 = 99.8 mL
(b)
∆Vacetone = βVi ∆T
b
b
∆Vflask = βVi ∆T
g
g
acetone
Pyrex
b
= 3αVi ∆T
g
Pyrex
for same Vi , ∆T ,
∆Vacetone β acetone
1.50 × 10 −4
1
=
=
=
−6
∆Vflask
β flask
×
6
.
40
10 −2
3 3.20 × 10
e
j
The volume change of flask is
about 6% of the change in the acetone’s volume .
Chapter 19
P19.20
(a),(b)
555
The material would expand by ∆L = αLi ∆T ,
∆L
= α∆T , but instead feels stress
Li
F Y∆L
=
= Yα∆T = 7.00 × 10 9 N m 2 12.0 × 10 −6 C°
A
Li
e
a f a30.0° Cf
j
−1
= 2.52 × 10 6 N m 2 . This will not break concrete.
P19.21
(a)
b
g
∆V = Vt β t ∆T − VAl β Al ∆T = β t − 3α Al Vi ∆T
e
j
e
ja
= 9.00 × 10 −4 − 0.720 × 10 −4 ° C −1 2 000 cm3 60.0° C
∆V = 99.4 cm3
(b)
f
overflows.
The whole new volume of turpentine is
ja
e
f
2 000 cm3 + 9.00 × 10 −4 ° C −1 2 000 cm3 60.0° C = 2 108 cm3
so the fraction lost is
99.4 cm3
= 4.71 × 10 −2
3
2 108 cm
and this fraction of the cylinder’s depth will be empty upon cooling:
a
f
4.71 × 10 −2 20.0 cm = 0.943 cm .
*P19.22
The volume of the sphere is
VPb =
a
f
4 3 4
π r = π 2 cm
3
3
3
= 33.5 cm3 .
The amount of mercury overflowing is
e
j
overflow = ∆VHg + ∆VPb − ∆Vglass = β Hg VHg + β PbVPb − β glassVglass ∆T
where Vglass = VHg + VPb is the initial volume. Then
e
j e
j
e
j e
1
1
L
O
= Ma182 − 27f10
118 cm + a87 − 27f10
33.5 cm P 40° C = 0.812 cm
C°
C°
N
Q
j
overflow = β Hg − β glass VHg + β Pb − β glass VPb ∆T = β Hg − 3α glass VHg + 3α Pb − 3α glass VPb ∆T
−6
P19.23
In
−6
3
3
F Y∆L
=
require ∆L = αLi ∆T
A
Li
F
= Yα∆T
A
F
500 N
=
∆T =
−4
2
AYα
2.00 × 10 m 20.0 × 10 10 N m 2 11.0 × 10 −6 C°
e
∆T = 1.14° C
je
je
j
3
556
*P19.24
Temperature
Model the wire as contracting according to ∆L = αLi ∆T and then stretching according to
∆L Y
F
stress = = Y
= αLi ∆T = Yα∆T .
A
Li Li
e
1
45° C = 396 N
C°
j
(a)
F = YAα∆T = 20 × 10 10 N m 2 4 × 10 −6 m 2 11 × 10 −6
(b)
∆T =
3 × 10 8 N m 2
stress
=
= 136° C
Yα
20 × 10 10 N m 2 11 × 10 −6 C°
e
j
To increase the stress the temperature must decrease to 35° C − 136° C = −101° C .
(c)
*P19.25
The original length divides out, so the answers would not change.
The area of the chip decreases according to
∆A = γA1 ∆T = A f − Ai
b
a
g
A f = Ai 1 + γ∆T = Ai 1 + 2α∆T
f
The star images are scattered uniformly, so the number N of stars that fit is proportional to the area.
a
f
ja
e
f
Then N f = N i 1 + 2α∆T = 5 342 1 + 2 4.68 × 10 −6 ° C −1 −100° C − 20° C = 5 336 star images .
Section 19.5
P19.26
P19.27
Macroscopic Description of an Ideal Gas
a
fe
n=
(b)
N = nN A
(a)
Initially, PV
i i = n i RTi
a
= a 2.99 molfe6.02 × 10
j
fa
f
j
a1.00 atmfV = n Ra10.0 + 273.15f K
P b0. 280V g = n Ra 40.0 + 273.15f K
23
molecules mol = 1.80 × 10 24 molecules
i
Finally, Pf V f = n f RT f
f
i
i
i
0.280 Pf
giving
313.15 K
=
1.00 atm 283.15 K
Pf = 3.95 atm
or
Pf = 4.00 × 10 5 Pa abs. .
Dividing these equations,
(b)
je
9.00 atm 1.013 × 10 5 Pa atm 8.00 × 10 −3 m 3
PV
=
= 2.99 mol
8.314 N ⋅ mol K 293 K
RT
(a)
a f
P a1.02fb0.280V g = n Ra85.0 + 273.15f K
After being driven
d
i
i
Pd = 1.121Pf = 4.49 × 10 5 Pa
P19.28
a fa f
a fa f
3 150 0.100
3 PV
=
= 884 balloons
3
4π r P ′ 4π 0.150 3 1.20
If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be
full of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced
0.100 m3 3
to to 884 −
= 877 .
3
4π 0.15 m
PV = NP ′V ′ =
e
a
4 3
π r NP ′ :
3
j
f
N=
Chapter 19
P19.29
The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N.
e
ja
fa
ga
fa
f
1.01 × 10 5 N m 2 10.0 m 20.0 m 30.0 m
PV
=
= 2. 49 × 10 5 mol
n=
RT
8.314 J mol ⋅ K 293 K
b
e
5
N = nN A = 2.49 × 10 mol 6.022 × 10
*P19.30
(a)
PV
i i = n i RTi =
23
f
j
molecules mol = 1.50 × 10 29 molecules
mi
RTi
M
e
j
3
4.00 × 10 −3 kg 1.013 × 10 5 N 4π 6.37 × 10 6 m mole ⋅ K
MPV
i i
=
mi =
RTi
8.314 Nm 50 K
mole
m2 3
= 1.06 × 10 21 kg
(b)
Pf V f
PV
i i
=
n f RT f
ni RTi
F 1.06 × 10 kg + 8.00 × 10 kg I T
GH
JK 50 K
1.06 × 10 kg
F 1 IJ = 56.9 K
T = 100 K G
H 1.76 K
nRT F 9.00 g I F 8.314 J I F
773 K
I
=G
P=
G
J
G
J
V
H 18.0 g mol K H mol K K H 2.00 × 10 m JK =
21
2 ⋅1 =
20
f
21
f
P19.31
P19.32
P19.33
557
−3
a
1.61 MPa = 15.9 atm
fa f
(a)
T2 = T1
P2
= 300 K 3 = 900 K
P1
(b)
T2 = T1
P2 V2
= 300 2 2 = 1 200 K
P1V1
∑ Fy = 0 :
3
a fa f
b
g
ρ out gV − ρ in gV − 200 kg g = 0
bρ
out
ge
− ρ in 400 m
3
j = 200 kg
The density of the air outside is 1.25 kg m3 .
n
P
=
From PV = nRT ,
V RT
The density is inversely proportional to the temperature, and the density
of the hot air is
jFGH 283T K IJK
e1.25 kg m jFGH 1 − 283T K IJK e400 m j = 200 kg
e
ρ in = 1.25 kg m3
Then
in
3
3
in
283 K
1−
= 0.400
Tin
283 K
0.600 =
Tin = 472 K
Tin
FIG. P19.33
558
*P19.34
P19.35
Temperature
Consider the air in the tank during one discharge process. We suppose that the process is slow
enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to
1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L
to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes
from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water
comes out. Were it not for male pattern dumbness, each person could more efficiently use his device
by starting with the tank half full of water.
(a)
PV = nRT
e
je
j
b
ga f
m = nM = a 41.6 molfb 28.9 g molg =
1.013 × 10 5 Pa 1.00 m3
PV
n=
=
= 41.6 mol
RT
8.314 J mol ⋅ K 293 K
(b)
1.20 kg , in agreement with the tabulated density of
3
1.20 kg m at 20.0°C.
*P19.36
e
j
2
The void volume is 0.765Vtotal = 0.765π r 2 = 0.765π 1.27 × 10 −2 m 0.2 m = 7.75 × 10 −5 m3 . Now for
the gas remaining PV = nRT
n=
P19.37
(a)
b
n=
PV = nRT
e
j
ga
−5
5
2
3
PV 12.5 1.013 × 10 N m 7.75 × 10 m
=
= 3.96 × 10 −2 mol
RT
8.314 Nm mole K 273 + 25 K
f
PV
RT
a
fe
3
−3
5
PVM 1.013 × 10 Pa 0.100 m 28.9 × 10 kg mol
=
m = nM =
RT
8.314 J mol ⋅ K 300 K
b
f
ga
j
m = 1.17 × 10 −3 kg
P19.38
e
j
(b)
Fg = mg = 1.17 × 10 −3 kg 9.80 m s 2 = 11.5 mN
(c)
F = PA = 1.013 × 10 5 N m 2 0.100 m
(d)
The molecules must be moving very fast to hit the walls hard.
ja
e
At depth,
P = P0 + ρgh
At the surface,
P0 V f = nRT f :
Therefore
and
2
= 1.01 kN
PVi = nRTi
P0 V f
bP + ρghgV
F T I FG P + ρgh IJ
GH T JK H P K
F 293 K IJ FG 1.013 × 10
= 1.00 cm G
H 278 K K GH
V f = Vi
Vf
f
f
0
i
=
Tf
Ti
0
i
0
3
V f = 3.67 cm3
5
e
je
ja
f IJ
JK
Pa + 1 025 kg m 3 9.80 m s 2 25.0 m
1.013 × 10 5 Pa
Chapter 19
P19.39
mf
PV = nRT :
mi
=
nf
m f = mi
so
Pf V f RTi Pf
=
RT f PV
Pi
i i
=
ni
FP I
GH P JK
f
i
∆m = mi − m f = mi
P19.40
559
F P − P I = 12.0 kgFG 41.0 atm − 26.0 atm IJ =
GH P JK
H 41.0 atm K
i
f
4.39 kg
i
My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20° C = 293 K . Think
of the air as 80.0% N 2 and 20.0% O 2 .
Avogadro’s number of molecules has mass
a0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg mol
F mI
PV = nRT = G J RT
H MK
PVM e1.00 × 10 N m je38.4 m jb0.028 8 kg molg
=
= 45.4 kg
m=
RT
b8.314 J mol ⋅ K ga293 K f
Then
5
gives
*P19.41
2
3
~ 10 2 kg
The CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar
mass is M = 12.0 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder is
m sample
6.50 g
=
= 0.148 mol
n=
M
44.0 g mol
b
Then
PV = nRT
gives
V=
g
b
ga
nRT 0.148 mol 8.314 J mol ⋅ K 273 K + 20 K
=
P
1.013 × 10 5 N m 2
e
je
je
j
f FG 1 N ⋅ m IJ F 10 L I =
H 1 J K GH 1 m JK
3
3
3.55 L
P19.42
10 −9 Pa 1.00 m 3 6.02 × 10 23 molecules mol
PVN A
=
= 2.41 × 10 11 molecules
N=
RT
8.314 J K ⋅ mol 300 K
P19.43
P0 V = n1 RT1 =
b
FG m IJ RT
H MK
F m IJ RT
P V = n RT = G
HMK
P VM F 1
1I
− J
m −m =
G
R HT T K
0
2
1
2
1
1
2
2
2
0
1
2
ga
f
560
P19.44
Temperature
(a)
Initially the air in the bell satisfies P0 Vbell = nRTi
or
a
f
P0 2.50 m A = nRTi
(1)
When the bell is lowered, the air in the bell satisfies
a
f
Pbell 2.50 m − x A = nRT f
(2)
where x is the height the water rises in the bell. Also, the pressure in the bell, once it is
lowered, is equal to the sea water pressure at the depth of the water level in the bell.
a
a
f
f
Pbell = P0 + ρg 82.3 m − x ≈ P0 + ρg 82.3 m
(3)
The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from
(1) into (2),
a
fa
f
P0 + ρg 82.3 m 2.50 m − x A = P0 Vbell
Tf
Ti
.
Using P0 = 1 atm = 1.013 × 10 5 Pa and ρ = 1.025 × 10 3 kg m3
L
O
fMM TT FGH1 + ρga82P.3 mf IJK PP
N
Q
LM 277.15 K F e1.025 × 10 kg m je9.80 m s ja82.3 mf I
G1 +
JJ
= a 2.50 mfM1 −
293.15 K G
1.013 × 10 N m
K
H
MN
a
x = 2.50 m 1 −
−1
f
0
0
3
3
−1
2
2
5
OP
PP
Q
x = 2.24 m
(b)
If the water in the bell is to be expelled, the air pressure in the bell must be raised to the
water pressure at the bottom of the bell. That is,
a
f
Pbell = P0 + ρg 82.3 m
5
e
ja
je
f
= 1.013 × 10 Pa + 1.025 × 10 3 kg m 3 9.80 m s 2 82.3 m
5
Pbell = 9.28 × 10 Pa = 9.16 atm
Additional Problems
P19.45
The excess expansion of the brass is
b
a f a
f
∆a ∆Lf = 2.66 × 10 m
∆ ∆L = 19.0 − 11.0 × 10 −6
−4
(a)
The rod contracts more than tape to
a length reading 0.950 0 m − 0.000 266 m = 0.949 7 m
(b)
g
a° Cf a0.950 mfa35.0° Cf
∆Lrod − ∆Ltape = α brass − α steel Li ∆T
0.950 0 m + 0.000 266 m = 0.950 3 m
−1
Chapter 19
P19.46
At 0°C, 10.0 gallons of gasoline has mass,
ρ=
from
m
V
jb
e
m = ρV = 730 kg m3 10.0 gal
80 m I
= 27.7 kg
gFGH 0.003
1.00 gal JK
3
The gasoline will expand in volume by
b
ga
f
∆V = βVi ∆T = 9.60 × 10 −4 ° C −1 10.0 gal 20.0° C − 0.0° C = 0.192 gal
At 20.0°C,
10.192 gal = 27.7 kg
10.0 gal = 27.7 kg
F 10.0 gal I = 27.2 kg
GH 10.192 gal JK
The extra mass contained in 10.0 gallons at 0.0°C is
27.7 kg − 27.2 kg = 0.523 kg .
P19.47
Neglecting the expansion of the glass,
∆h =
∆h =
V
β∆T
A
b0.250 cm 2g e1.82 × 10
π e 2.00 × 10 cmj
3
4
3π
−3
2
−4
ja
f
° C −1 30.0° C = 3.55 cm
FIG. P19.47
P19.48
(a)
The volume of the liquid increases as ∆V = Vi β∆T . The volume of the flask increases as
∆Vg = 3αVi ∆T . Therefore, the overflow in the capillary is Vc = Vi ∆T β − 3α ; and in the
b
capillary Vc = A∆h .
Therefore, ∆h =
(b)
Vi
β − 3α ∆T .
A
b
g
b g
For a mercury thermometer
β Hg = 1.82 × 10 −4 ° C −1
and for glass,
3α = 3 × 3.20 × 10 −6 ° C −1
Thus
β − 3α ≈ β
or
α << β .
g
561
562
P19.49
Temperature
The frequency played by the cold-walled flute is fi =
v
v
=
.
λ i 2 Li
When the instrument warms up
ff =
fi
v
v
v
=
=
=
.
λ f 2L f 2Li 1 + α∆T 1 + α∆T
a
f
The final frequency is lower. The change in frequency is
FG
H
IJ
K
1
1 + α∆T
α∆T
v
≈
α∆T
1 + α∆T
2 Li
∆f = f i − f f = f i 1 −
FG
H
IJ
K
a f
b343 m sge24.0 × 10 C°ja15.0° Cf =
∆f ≈
2a0.655 mf
∆f =
v
2 Li
−6
0.094 3 Hz
This change in frequency is imperceptibly small.
P19.50
(a)
P0 V P ′V ′
=
T
T′
V ′ = V + Ah
P ′ = P0 +
k
kh
A
FG P + kh IJ aV + Ahf = P V FG T ′ IJ
H AK
HTK
e1.013 × 10 N m + 2.00 × 10 N m hj
e5.00 × 10 m + e0.010 0 m jhj
= e1.013 × 10 N m je5.00 × 10
0
5
2
−3
3
5
3
FIG. P19.50
2
−3
2
m3
KI
J
jFGH 523
293 K K
2 000 h 2 + 2 013 h − 397 = 0
(b)
20°C
0
5
h=
250°C
h
−2 013 ± 2 689
= 0.169 m
4 000
e
ja
2.00 × 10 3 N m 0.169
kh
5
= 1.013 × 10 Pa +
P′ = P +
A
0.010 0 m 2
P ′ = 1.35 × 10 5 Pa
f
Chapter 19
P19.51
(a)
ρ=
m
m
and dρ = − 2 dV
V
V
For very small changes in V and ρ, this can be expressed as
∆ρ = −
m ∆V
= − ρβ∆T .
V V
The negative sign means that any increase in temperature causes the density to decrease
and vice versa.
(b)
*P19.52
For water we have β =
1.000 0 g cm 3 − 0.999 7 g cm 3
∆ρ
=
= 5 × 10 −5 ° C −1 .
3
ρ∆T
1.000 0 g cm 10.0° C − 4.0° C
ja
e
f
The astronauts exhale this much CO 2 :
n=
m sample
M
=
F
GH
Ia
JK
gFGH
I
JK
1.09 kg
1 000 g
1 mol
= 520 mol .
3 astronauts 7 days
astronaut ⋅ day 1 kg
44.0 g
fb
Then 520 mol of methane is generated. It is far from liquefaction and behaves as an ideal gas.
P=
P19.53
(a)
b
f
ga
nRT 520 mol 8.314 J mol ⋅ K 273 K − 45 K
= 6.57 × 10 6 Pa
=
V
150 × 10 −3 m3
We assume that air at atmospheric pressure is above the
piston.
mg
+ P0
A
In equilibrium
Pgas =
Therefore,
nRT mg
=
+ P0
hA
A
h=
or
nRT
mg + P0 A
where we have used V = hA as the volume of the gas.
(b)
From the data given,
h=
e
20.0 kg 9.80 m s
= 0.661 m
b
ga
0.200 mol 8.314 J K ⋅ mol 400 K
2
j e
5
+ 1.013 × 10 N m
2
je
f
0.008 00 m 2
j
FIG. P19.53
563
564
P19.54
Temperature
The angle of bending θ, between tangents to the two ends of the strip, is
equal to the angle the strip subtends at its center of curvature. (The angles
are equal because their sides are perpendicular, right side to the right side
and left side to left side.)
(a)
Li + ∆L1 = θ r1
The definition of radian measure gives
FIG. P19.54
Li + ∆L 2 = θ r2
and
b
∆L 2 − ∆L1 = θ r2 − r1
By subtraction,
g
α 2 Li ∆T − α 1 Li ∆T = θ ∆r
θ=
2
g
− α 1 Li ∆T
∆r
b
g
(b)
In the expression from part (a), θ is directly proportional to ∆T and also to α 2 − α 1 .
Therefore θ is zero when either of these quantities becomes zero.
(c)
The material that expands more when heated contracts more when cooled, so the bimetallic
strip bends the other way. It is fun to demonstrate this with liquid nitrogen.
(d)
θ=
b
g
2 α 2 − α 1 Li ∆T
2 ∆r
=
ee
j
ja
fa f
2 19 × 10 −6 − 0.9 × 10 −6 ° C −1 200 mm 1° C
0.500 mm
= 1.45 × 10 −2 = 1. 45 × 10 −2 rad
P19.55
bα
FG 180° IJ =
H π rad K
0.830°
From the diagram we see that the change in area is
∆A = ∆w + w∆ + ∆w∆ .
Since ∆ and ∆w are each small quantities, the product ∆w∆ will
be very small. Therefore, we assume ∆w∆ ≈ 0.
Since
∆w = wα∆T
we then have
∆A = wα∆T + wα∆T
and
∆ = α∆T ,
FIG. P19.55
and since A = w , ∆A = 2αA∆T .
The approximation assumes ∆w∆ ≈ 0, or α∆T ≈ 0 . Another way of stating this is α∆T << 1 .
Chapter 19
P19.56
L
Ti = 2π i
g
(a)
Li =
so
Ti2 g
4π 2
565
a1.000 sf e9.80 m s j = 0.248 2 m
=
2
2
4π 2
b
f
ga
∆L = αLi ∆T = 19.0 × 10 −6 ° C −1 0.284 2 m 10.0° C = 4.72 × 10 −5 m
0.248 3 m
Li + ∆L
= 2π
= 1.000 095 0 s
g
9.80 m s 2
T f = 2π
∆T = 9.50 × 10 −5 s
e
In one week, the time lost is time lost = 1 week 9.50 × 10 −5 s lost per second
(b)
b
time lost = 7.00 d week
gFGH 861.00400ds IJK FGH 9.50 × 10
−5
time lost = 57.5 s lost
P19.57
z
I = r 2 dm
a f b ga
f
I aT f
= a1 + α∆T f
I bT g
I aT f − I bT g
≈ 2α∆T
I bT g
r T = r Ti 1 + α∆T
and since
2
for α∆T << 1 we find
i
i
thus
i
(a)
(b)
P19.58
(a)
With α = 17.0 × 10 −6 ° C −1 and
∆T = 100° C
we find for Cu:
∆I
= 2 17.0 × 10 −6 ° C −1 100° C = 0.340%
I
With
α = 24.0 × 10 −6 ° C −1
and
∆T = 100° C
we find for Al:
∆I
= 2 24.0 × 10 −6 ° C −1 100° C = 0.480%
I
e
B = ρgV ′
B=
(b)
e
P ′ = P0 + ρgd
ρgP0 Vi
=
P′
ja
f
ja
f
P ′V ′ = P0 Vi
ρgP0 Vi
bP + ρgdg
0
Since d is in the denominator, B must decrease as the depth increases.
(The volume of the balloon becomes smaller with increasing pressure.)
(c)
b
af
af
g
ρgP0 Vi P0 + ρgd
P0
1 Bd
=
=
=
ρgP0Vi P0
2 B0
P0 + ρgd
P0 + ρgd = 2 P0
d=
1.013 × 10 5 N m 2
P0
=
= 10.3 m
ρg 1.00 × 10 3 kg m3 9.80 m s 2
e
je
j
j
s lost
s
IJ
K
566
*P19.59
Temperature
The effective coefficient is defined by ∆Ltotal = α effective Ltotal ∆T where ∆Ltotal = ∆LCu + ∆LPb and
Ltotal = LCu + LPb = xLtotal + 1 − x Ltotal . Then by substitution
a f
b
g
α Cu LCu ∆T + α Pb LPb ∆T = α eff LCu + LPb ∆T
a f
gx = α
α Cu x + α Pb 1 − x = α eff
bα
Cu
x=
*P19.60
eff
20 × 10
−6
17 × 10
−6
− α Pb
1 C° − 29 × 10 −6 1 C°
1 C° − 29 × 10
−6
1 C°
9
= 0.750
12
=
(a)
No torque acts on the disk so its angular momentum is constant. Its moment of inertia
decreases as it contracts so its angular speed must increase .
(b)
I iω i = I f ω f =
1
1
1
1
2
MRi2ω i = MR 2f ω f = M Ri + Riα∆T ω f = MRi2 1 − α ∆T
2
2
2
2
−2
ω f = ω i 1 − α ∆T
P19.61
− α Pb
=
25.0 rad s
e1 − e17 × 10
−6
j
1 C° 830° C
j
2
=
2
ωf
25.0 rad s
= 25.7 rad s
0.972
After expansion, the length of one of the spans is
a
f
a
f
L f = Li 1 + α∆T = 125 m 1 + 12 × 10 −6 ° C −1 20.0° C = 125.03 m .
L f , y, and the original 125 m length of this span form a right triangle with y as the altitude. Using the
Pythagorean theorem gives:
a125.03 mf
2
a
f
= y 2 + 125 m
2
yielding y = 2.74 m .
P19.62
a
f
After expansion, the length of one of the spans is L f = L 1 + α∆T . L f , y, and the original length L of
this span form a right triangle with y as the altitude. Using the Pythagorean theorem gives
L2f = L2 + y 2 ,
Since
P19.63
(a)
or
y = L2f − L2 = L
2
Let m represent the sample mass. The number of moles is n =
Then, ρ =
m
m
RT or PM = RT .
M
V
m
PM
.
=
V
RT
e
a f
− 1 = L 2α∆T + α∆T
2
y ≈ L 2α∆T .
α∆T << 1 ,
So PV = nRT becomes PV =
(b)
a1 + α∆T f
jb
g
1.013 × 10 5 N m 2 0.032 0 kg mol
PM
ρ=
=
= 1.33 kg m 3
RT
8.314 J mol ⋅ K 293 K
b
ga
f
m
m
and the density is ρ = .
V
M
Chapter 19
P19.64
(a)
(b)
FG nR IJ T
H PK
From PV = nRT , the volume is:
V=
Therefore, when pressure is held constant,
dV nR V
=
=
dT
P T
Thus,
β≡
At T = 0° C = 273 K , this predicts
β=
FG 1 IJ dV = FG 1 IJ V , or β =
H V K dT H V K T ′
1
T
1
= 3.66 × 10 −3 K −1
273 K
β He = 3.665 × 10 −3 K −1 and β air = 3.67 × 10 −3 K −1
Experimental values are:
They agree within 0.06% and 0.2%, respectively.
P19.65
For each gas alone, P1 =
N 1 kT
N kT
N kT
and P2 = 2
and P3 = 3 , etc.
V
V
V
For all gases
b
g
P1V1 + P2 V2 + P3 V3 … = N 1 + N 2 + N 3 … kT and
bN
1
g
+ N 2 + N 3 … kT = PV
Also, V1 = V2 = V3 = … = V , therefore P = P1 + P2 + P3 … .
P19.66
(a)
Using the Periodic Table, we find the molecular masses of the air components to be
b g
a f
b g
M N 2 = 28.01 u , M O 2 = 32.00 u , M Ar = 39.95 u
b
g
and M CO 2 = 44.01 u .
Thus, the number of moles of each gas in the sample is
75.52 g
= 2.696 mol
b g 28.01
g mol
23.15 g
nbO g =
= 0.723 4 mol
32.00 g mol
1.28 g
na Ar f =
= 0.032 0 mol
39.95 g mol
n N2 =
2
b g
n CO 2 =
0.05 g
= 0.001 1 mol
44.01 g mol
The total number of moles is n 0 = ∑ ni = 3.453 mol . Then, the partial pressure of N 2 is
b g
P N2 =
2.696 mol
1.013 × 10 5 Pa = 79.1 kPa .
3.453 mol
e
j
Similarly,
b g
P O 2 = 21.2 kPa
continued on next page
567
a f
P Ar = 940 Pa
b
g
P CO 2 = 33.3 Pa
568
Temperature
(b)
Solving the ideal gas law equation for V and using T = 273.15 + 15.00 = 288.15 K , we find
V=
Then, ρ =
(c)
a
fb
100 × 10 −3 kg
m
=
= 1.22 kg m3 .
V 8.166 × 10 −2 m3
The 100 g sample must have an appropriate molar mass to yield n 0 moles of gas: that is
a f
M air =
*P19.67
f
ga
3.453 mol 8.314 J mol ⋅ K 288.15 K
n 0 RT
= 8.166 × 10 −2 m3 .
=
P
1.013 × 10 5 Pa
100 g
= 29.0 g mol .
3.453 mol
Consider a spherical steel shell of inner radius r and much smaller thickness t, containing helium at
pressure P. When it contains so much helium that it is on the point of bursting into two
hemispheres, we have Pπ r 2 = 5 × 10 8 N m 2 2π rt . The mass of the steel is
e
ρ sV = ρ s 4π r 2 t = ρ s 4π r 2
Pr
10 9 Pa
j
. For the helium in the tank, PV = nRT becomes
m
4
P π r 3 = nRT = He RT = 1 atmVballoon .
3
M He
The buoyant force on the balloon is the weight of the air it displaces, which is described by
m
4
1 atmVballoon = air RT = P π r 3 . The net upward force on the balloon with the steel tank hanging
M air
3
from it is
+ m air g − m He g − m s g =
M air P 4π r 3 g M He P 4π r 3 g ρ s P 4π r 3 g
−
−
3 RT
3 RT
10 9 Pa
The balloon will or will not lift the tank depending on whether this quantity is positive or negative,
M air − M He
ρ
− 9 s . At 20°C this quantity is
which depends on the sign of
3 RT
10 Pa
b
=
g
a28.9 − 4.00f × 10 kg mol − 7 860 kg m
3b8.314 J mol ⋅ K g293 K
10 N m
−3
9
3
2
= 3.41 × 10 −6 s 2 m 2 − 7.86 × 10 −6 s 2 m 2
where we have used the density of iron. The net force on the balloon is downward so the helium
balloon is not able to lift its tank.
Chapter 19
P19.68
With piston alone:
T = constant, so PV = P0 V0
or
P Ahi = P0 Ah0
569
b g b g
Fh I
P=P G J
Hh K
With A = constant,
0
0
i
P = P0 +
But,
mp g
A
FIG. P19.68
where m p is the mass of the piston.
P0 +
Thus,
hi =
which reduces to
mp g
= P0
A
h0
1+
mp g
P0 A
FG h IJ
Hh K
0
i
50.0 cm
=
1+
e
20.0 kg 9.80 m s 2
a
= 49.81 cm
j
1.013 × 10 5 Pa π 0.400 m
f
2
With the man of mass M on the piston, a very similar calculation (replacing m p by m p + M ) gives:
h′ =
h0
em
1+
p +M
P0 A
jg
50.0 cm
=
1+
e
95.0 kg 9.80 m s 2
a
j
1.013 × 10 5 Pa π 0.400 m
= 49.10 cm
f
2
Thus, when the man steps on the piston, it moves downward by
∆h = hi − h ′ = 49.81 cm − 49.10 cm = 0.706 cm = 7.06 mm .
(b)
P = const, so
(a)
(b)
z
Ti
dL
= αdT :
L
a
Ahi Ah ′
=
T
Ti
hi
49.81
= 293 K
= 297 K
T = Ti
49.10
h′
or
FG IJ
H K
giving
P19.69
V V′
=
T Ti
Ti
f
L f = 1.00 m e
αdT =
z
Li
Li
FG
H
IJ
K
FG IJ = α∆T ⇒
H K
Lf
dL
⇒ ln
L
Li
a
2.00 × 10 −5 °C −1 100 °C
f
L f = Li eα∆T
= 1.002 002 m
a
f
L ′f = 1.00 m 1 + 2.00 × 10 −5 ° C −1 100° C = 1.002 000 m :
a
f
L f = 1.00 m e
a
2.00 ×10 −2 °C −1 100 °C
a
f
(or 24°C)
L f − L ′f
Lf
= 2.00 × 10 −6 = 2.00 × 10 −4%
= 7.389 m
f
L ′f = 1.00 m 1 + 0.020 0° C −1 100° C = 3.000 m :
L f − L ′f
Lf
= 59. 4%
570
P19.70
Temperature
At 20.0°C, the unstretched lengths of the steel and copper wires are
a f a
f
a f a−20.0° Cf = 1.999 56 m
L a 20.0° C f = a 2.000 mf 1 + 17.0 × 10 aC°f a −20.0° C f = 1.999 32 m
Ls 20.0° C = 2.000 m 1 + 11.0 × 10 −6 C°
−1
−6
c
−1
Under a tension F, the length of the steel and copper wires are
LM
N
Ls′ = Ls 1 +
F
YA
OP
Q
LM
N
L c′ = L c 1 +
s
F
YA
OP
Q
where Ls′ + L c′ = 4.000 m .
c
Since the tension, F, must be the same in each wire, solve for F:
F=
b L ′ + L ′ g − bL
s
c
Ls
Ys As
+
s + Lc
Lc
Yc A c
g.
When the wires are stretched, their areas become
e
= π e1.000 × 10
j
mj
As = π 1.000 × 10 −3 m
Ac
−3
2
2
e
ja f
1 + e17.0 × 10 ja −20.0 f
1 + 11.0 × 10 −6 −20.0
2
−6
2
= 3.140 × 10 −6 m 2
= 3.139 × 10 −6 m 2
Recall Ys = 20.0 × 10 10 Pa and Yc = 11.0 × 10 10 Pa . Substituting into the equation for F, we obtain
F=
b
4.000 m − 1.999 56 m + 1.999 32 m
1.999 56 m
e20.0 × 10
10
je
j
Pa 3.140 × 10 −6 m 2 + 1.999 32 m
g
e11.0 × 10
10
je
F = 125 N
To find the x-coordinate of the junction,
b
L
gMM 20.0 × 10
N e
Ls′ = 1.999 56 m 1 +
125 N
10
je
N m 2 3.140 × 10 −6
Thus the x-coordinate is −2.000 + 1.999 958 = −4.20 × 10 −5 m .
OP
= 1.999 958 m
m jP
Q
2
j
Pa 3.139 × 10 −6 m 2
Chapter 19
P19.71
e
j e7.86 × 10
(a)
µ = π r 2 ρ = π 5.00 × 10 −4 m
(b)
f1 =
2
1
v
T
and v =
so f1 =
µ
2L
2L
3
j
kg m3 = 6.17 × 10 −3 kg m
T
µ
b g = e6.17 × 10 ja2 × 0.800 × 200f
2
Therefore, T = µ 2Lf1
(c)
−3
2
= 632 N
First find the unstressed length of the string at 0°C:
FG
H
L = L natural 1 +
e
A = π 5.00 × 10 −4
Therefore,
IJ
K
mj
T
L
so L natural =
+
AY
1 T AY
2
= 7.854 × 10 −7 m 2 and Y = 20.0 × 10 10 Pa
T
632
=
= 4.02 × 10 −3 , and
−7
10
AY
7.854 × 10
20.0 × 10
e
je
j
L natural =
a0.800 mf
e1 + 4.02 × 10 j
−3
= 0.796 8 m .
a
f
The unstressed length at 30.0°C is L30 °C = L natural 1 + α 30.0° C − 0.0° C ,
b
g e
ja f
LM1 + T ′ OP , where T ′ is the tension in the string at 30.0°C,
N AY Q
L L − 1OP = e7.854 × 10 je20.0 × 10 jLM 0.800 − 1OP = 580 N .
T ′ = AY M
NL
Q
N 0.797 06 Q
or L30 °C = 0.796 8 m 1 + 11.0 × 10 −6 30.0 = 0.797 06 m .
Since L = L30 °C
−7
10
30 °C
To find the frequency at 30.0°C, realize that
a
f1′
T′
=
so f1′ = 200 Hz
f1
T
*P19.72
f
580 N
= 192 Hz .
632 N
Some gas will pass through the porous plug from the reaction chamber 1 to the reservoir 2 as the
reaction chamber is heated, but the net quantity of gas stays constant according to
n i1 + n i 2 = n f 1 + n f 2 .
Assuming the gas is ideal, we apply n =
PV
to each term:
RT
b g
b g
a f a f a f a f
F 5 IJ = P FG 1 + 4 IJ
1 atmG
P = 1.12 atm
H 300 K K H 673 K 300 K K
Pf V0
Pf 4V0
Pi 4V0
PV
i 0
+
=
+
300 K R 300 K R
673 K R 300 K R
f
f
571
572
P19.73
Temperature
Let 2θ represent the angle the curved rail subtends. We have
a
Li + ∆L = 2θR = Li 1 + α∆T
and
sin θ =
Thus,
θ=
Li
2
R
=
a
f
Li
2R
f a
f
Li
1 + α∆T = 1 + α∆T sin θ
2R
FIG. P19.73
a
f
b
g
and we must solve the transcendental equation
θ = 1 + α∆T sin θ = 1.000 005 5 sin θ
Homing in on the non-zero solution gives, to four digits,
θ = 0.018 16 rad = 1.040 5°
Now,
h = R − R cos θ =
a
Li 1 − cos θ
2 sin θ
f
This yields h = 4.54 m , a remarkably large value compared to ∆L = 5.50 cm .
*P19.74
(a)
Let xL represent the distance of the stationary line below the
top edge of the plate. The normal force on the lower part of the
plate is mg 1 − x cos θ and the force of kinetic friction on it is
µ k mg 1 − x cos θ up the roof. Again, µ k mgx cos θ acts down the
roof on the upper part of the plate. The near-equilibrium of the
plate requires ∑ Fx = 0
a f
a f
a f
motion
f kt
f kb
xL
temperature rising
− µ k mgx cos θ + µ k mg 1 − x cos θ − mg sin θ = 0
FIG. P19.74(a)
−2 µ k mgx cos θ = mg sin θ − µ k mg cos θ
2 µ k x = µ k − tan θ
x=
1 tan θ
−
2 2µ k
and the stationary line is indeed below the top edge by xL =
(b)
With the temperature falling, the plate contracts faster than the
roof. The upper part slides down and feels an upward frictional
force µ k mg 1 − x cos θ . The lower part slides up and feels
downward frictional force µ k mgx cos θ . The equation ∑ Fx = 0
is then the same as in part (a) and the stationary line is above
L
tan θ
the bottom edge by xL =
.
1−
µk
2
a f
FG
H
(c)
IJ
K
Start thinking about the plate at dawn, as the temperature
starts to rise. As in part (a), a line at distance xL below the top
edge of the plate stays stationary relative to the roof as long as
the temperature rises. The point P on the plate at distance xL
above the bottom edge is destined to become the fixed point
when the temperature starts falling. As the temperature rises,
this point moves down the roof because of the expansion of the
central part of the plate. Its displacement for the day is
continued on next page
IJ
K
FG
H
tan θ
L
1−
.
2
µk
motion
f kt
f kb
xL
temperature falling
FIG. P19.74(b)
xL
xL
P
FIG. P19.74(c)
Chapter 19
f
ga
L L F tanθ I OPbT − T g
− α gML − 2 G 1 −
NM 2 H µ JK PQ
F L tanθ IJ bT − T g .
= bα − α gG
H µ K
b
= bα
573
∆L = α 2 − α 1 L − xL − xL ∆T
2
1
2
1
h
k
k
h
c
c
At dawn the next day the point P is farther down the roof by the distance ∆L . It represents
the displacement of every other point on the plate.
bα
(d)
−α1
2
θI
gFGH L tan
bT − T g = FGH 24 × 10
µ JK
k
h
c
−6
IJ
K
1
1 1.20 m tan 18.5°
− 15 × 10 −6
32° C = 0.275 mm
C°
C°
0. 42
If α 2 < α 1 , the diagram in part (a) applies to temperature falling and the diagram in part (b)
applies to temperature rising. The weight of the plate still pulls it step by step down the
roof. The same expression describes how far it moves each day.
(e)
ANSWERS TO EVEN PROBLEMS
P19.2
(a) 1.06 atm ; (b) −124° C
P19.32
(a) 900 K; (b) 1 200 K
P19.4
(a) 37.0° C = 310 K ; (b) −20.6° C = 253 K
P19.34
see the solution
P19.6
TC = 1.33 C° S° TS + 20.0° C
P19.36
3.96 × 10 −2 mol
P19.8
0.313 m
P19.38
3.67 cm3
P19.10
1.20 cm
P19.40
between 10 1 kg and 10 2 kg
P19.12
15.8 µ m
P19.42
2.41 × 10 11 molecules
P19.14
0.663 mm to the right at 78.2° below the
horizontal
P19.44
(a) 2.24 m; (b) 9.28 × 10 5 Pa
P19.46
0.523 kg
P19.48
(a) see the solution; (b) α << β
P19.50
(a) 0.169 m; (b) 1.35 × 10 5 Pa
P19.52
6.57 MPa
P19.54
(a) θ =
; (b) see the solution;
∆r
(c) it bends the other way; (d) 0.830°
b
g
2
P19.16
(a) 0.109 cm ; (b) increase
P19.18
(a) 437°C ; (b) 3 000°C ; no
P19.20
(a) 2.52 × 10 6 N m 2 ; (b) no
P19.22
0.812 cm3
P19.24
(a) 396 N; (b) −101° C ; (c) no change
P19.26
(a) 2.99 mol ; (b) 1.80 × 10 24 molecules
P19.56
(a) increase by 95.0 µs ; (b) loses 57.5 s
P19.28
884 balloons
P19.58
P19.30
(a) 1.06 × 10 21 kg ; (b) 56.9 K
(a) B = ρgP0 Vi P0 + ρgd
(c) 10.3 m
bα
2
g
− α 1 Li ∆T
b
g
−1
up; (b) decrease;
574
Temperature
P19.60
(a) yes; see the solution; (b) 25.7 rad s
P19.62
y ≈ L 2α∆T
P19.64
(a) see the solution;
(b) 3.66 × 10 −3 K −1 , within 0.06% and 0.2%
of the experimental values
P19.66
a
f
12
(a) 79.1 kPa for N 2 ; 21.2 kPa for O 2 ;
940 Pa for Ar; 33.3 Pa for CO 2 ;
(b) 81.7 L; 1.22 kg m3 ; (c) 29.0 g mol
P19.68
(a) 7.06 mm; (b) 297 K
P19.70
125 N ; −42.0 µm
P19.72
1.12 atm
P19.74
(a), (b), (c) see the solution; (d) 0.275 mm;
(e) see the solution
20
Heat and the First Law of
Thermodynamics
CHAPTER OUTLINE
20.1
20.2
20.3
20.4
20.5
20.6
20.7
ANSWERS TO QUESTIONS
Heat and Internal Energy
Specific Heat and
Calorimetry
Latent Heat
Work and Heat in
Thermodynamic Processes
The First Law of
Thermodynamics
Some Applications of
the First Law of
Thermodynamics
Energy Transfer
Mechanisms
Q20.1
Temperature is a measure of molecular motion. Heat is energy
in the process of being transferred between objects by random
molecular collisions. Internal energy is an object’s energy of
random molecular motion and molecular interaction.
Q20.2
The ∆T is twice as great in the ethyl alcohol.
Q20.3
The final equilibrium temperature will show no significant
increase over the initial temperature of the water.
Q20.4
Some water may boil away. You would have to very precisely
measure how much, and very quickly measure the
temperature of the steam; it is not necessarily 100°C .
Q20.5
The fingers are wetted to create a layer of steam between the fingers and the molten lead. The steam
acts as an insulator and can prevent or delay serious burns. The molten lead demonstration is
dangerous, and we do not recommend it.
Q20.6
Heat is energy being transferred, not energy contained in an object. Further, a large-mass object, or
an object made of a material with high specific heat, can contain more internal energy than a highertemperature object.
Q20.7
There are three properties to consider here: thermal conductivity, specific heat, and mass. With dry
aluminum, the thermal conductivity of aluminum is much greater than that of (dry) skin. This
means that the internal energy in the aluminum can more readily be transferred to the atmosphere
than to your fingers. In essence, your skin acts as a thermal insulator to some degree (pun intended).
If the aluminum is wet, it can wet the outer layer of your skin to make it into a good conductor of
heat; then more internal energy from the aluminum can get into you. Further, the water itself, with
additional mass and with a relatively large specific heat compared to aluminum, can be a significant
source of extra energy to burn you. In practical terms, when you let go of a hot, dry piece of
aluminum foil, the heat transfer immediately ends. When you let go of a hot and wet piece of
aluminum foil, the hot water sticks to your skin, continuing the heat transfer, and resulting in more
energy transfer to you!
Q20.8
Write 1 000 kg 4 186 J kg ⋅° C 1° C = V 1.3 kg m3 1 000 J kg ⋅° C 1° C to find V = 3. 2 × 10 3 m 3 .
b
ga f e
jb
575
ga f
576
Heat and the First Law of Thermodynamics
Q20.9
The large amount of energy stored in concrete during the day as the sun falls on it is released at
night, resulting in an higher average evening temperature than the countryside. The cool air in the
surrounding countryside exerts a buoyant force on the warmer air in the city, pushing it upward
and moving into the city in the process. Thus, evening breezes tend to blow from country to city.
Q20.10
If the system is isolated, no energy enters or leaves the system by heat, work, or other transfer
processes. Within the system energy can change from one form to another, but since energy is
conserved these transformations cannot affect the total amount of energy. The total energy is
constant.
Q20.11
(a) and (b) both increase by minuscule amounts.
Q20.12
The steam locomotive engine is a perfect example of turning internal energy into mechanical
energy. Liquid water is heated past the point of vaporization. Through a controlled mechanical
process, the expanding water vapor is allowed to push a piston. The translational kinetic energy of
the piston is usually turned into rotational kinetic energy of the drive wheel.
Q20.13
Yes. If you know the different specific heats of zinc and copper, you can determine the fraction of
each by heating a known mass of pennies to a specific initial temperature, say 100°C , and dumping
them into a known quantity of water, at say 20°C . The final temperature T will reveal the metal
content:
a f a
a
f
f
mpennies xcCu + 1 − x c Zn 100° C − T = mH 2O c H 2O T − 20° C .
Since all quantities are known, except x, the fraction of the penny that is copper will be found by
putting in the experimental numbers m pennies , m H 2O , T final , c Zn , and c Cu .
a
f
Q20.14
The materials used to make the support structure of the roof have a higher thermal conductivity
than the insulated spaces in between. The heat from the barn conducts through the rafters and melts
the snow.
Q20.15
The tile is a better thermal conductor than carpet. Thus, energy is conducted away from your feet
more rapidly by the tile than by the carpeted floor.
Q20.16
The question refers to baking in a conventional oven, not to microwaving. The metal has much
higher thermal conductivity than the potato. The metal quickly conducts energy from the hot oven
into the center of potato.
Q20.17
Copper has a higher thermal conductivity than the wood. Heat from the flame is conducted through
the copper away from the paper, so that the paper need not reach its kindling temperature. The
wood does not conduct the heat away from the paper as readily as the copper, so the energy in the
paper can increase enough to make it ignite.
Q20.18
In winter the interior of the house is warmer than the air outside. On a summer day we want the
interior to stay cooler than the exterior. Heavy draperies over the windows can slow down energy
transfer by conduction, by convection, and by radiation, to make it easier to maintain the desired
difference in temperature.
Q20.19
You must allow time for the flow of energy into the center of the piece of meat. To avoid burning the
outside, the meat should be relatively far from the flame. If the outer layer does char, the carbon will
slow subsequent energy flow to the interior.
Chapter 20
577
Q20.20
At night, the Styrofoam beads would decrease the overall thermal conductivity of the windows, and
thus decrease the amount of heat conducted from inside to outside. The air pockets in the Styrofoam
are an efficient insulator. During the winter day, the influx of sunlight coming through the window
warms the living space.
An interesting aside—the majority of the energy that goes into warming a home from
sunlight through a window is not the infrared light given off by the sun. Glass is a relatively good
insulator of infrared. If not, the window on your cooking oven might as well be just an open hole!
Glass is opaque to a large portion of the ultraviolet range. The glass molecules absorb ultraviolet
light from the sun and re-emit the energy in the infrared region. It is this re-emitted infrared
radiation that contributes to warming your home, along with visible light.
Q20.21
In winter the produce is protected from freezing. The heat capacity of the earth is so high that soil
freezes only to a depth of a few decimeters in temperate regions. Throughout the year the
temperature will stay nearly constant all day and night. Factors to be considered are the insulating
properties of soil, the absence of a path for energy to be radiated away from or to the vegetables, and
the hindrance to the formation of convection currents in the small, enclosed space.
Q20.22
The high mass and specific heat of the barrel of water and its high heat of fusion mean that a large
amount of energy would have to leak out of the cellar before the water and the produce froze solid.
Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.
Q20.23
The sunlight hitting the peaks warms the air immediately around them. This air, which is slightly
warmer and less dense than the surrounding air, rises, as it is buoyed up by cooler air from the
valley below. The air from the valley flows up toward the sunny peaks, creating the morning breeze.
Q20.24
Sunlight hits the earth and warms the air immediately above it. This warm, less-dense air rises,
creating an up-draft. Many raptors, like eagles, hawks and falcons use updrafts to aid in hunting.
These birds can often be seen flying without flapping their wings—just sitting in an updraft with
wings extended.
Q20.25
The bit of water immediately over the flame warms up and expands. It is buoyed up and rises
through the rest of the water. Colder, more dense water flows in to take its place. Convection
currents are set up. This effectively warms the bulk of the water all at once, much more rapidly than
it would be by heat being conducted through the water from the flame.
Q20.26
The porcelain of the teacup is a thermal insulator. That is, it is a thermal conductor of relatively low
conductivity. When you wrap your hands around a cup of hot tea, you make A large and L small in
T − Tc
for the rate of energy transfer by heat from tea into you. When you hold
the equation P = kA h
L
the cup by the handle, you make the rate of energy transfer much smaller by reducing A and
increasing L. The air around the cup handle will also reduce the temperature where you are
touching it. A paper cup can be fitted into a tubular jacket of corrugated cardboard, with the
channels running vertically, for remarkably effective insulation, according to the same principles.
Q20.27
As described in the answer to question 20.25, convection currents in the water serve to bring more of
the heat into the water from the paper cup than the specific heats and thermal conductivities of
paper and water would suggest. Since the boiling point of water is far lower than the kindling
temperature of the cup, the extra energy goes into boiling the water.
Q20.28
Keep them dry. The air pockets in the pad conduct energy by heat, but only slowly. Wet pads would
absorb some energy in warming up themselves, but the pot would still be hot and the water would
quickly conduct and convect a lot of energy right into you.
578
Heat and the First Law of Thermodynamics
Q20.29
The person should add the cream immediately when the coffee is poured. Then the smaller
temperature difference between coffee and environment will reduce the rate of energy loss during
the several minutes.
Q20.30
The cup without the spoon will be warmer. Heat is conducted from the coffee up through the metal.
The energy then radiates and convects into the atmosphere.
Q20.31
Convection. The bridge deck loses energy rapidly to the air both above it and below it.
Q20.32
The marshmallow has very small mass compared to the saliva in the teacher’s mouth and the
surrounding tissues. Mostly air and sugar, the marshmallow also has a low specific heat compared to
living matter. Then the marshmallow can zoom up through a large temperature change while
causing only a small temperature drop of the teacher’s mouth. The marshmallow is a foam with
closed cells and it carries very little liquid nitrogen into the mouth. The liquid nitrogen still on the
marshmallow comes in contact with the much hotter saliva and immediately boils into cold gaseous
nitrogen. This nitrogen gas has very low thermal conductivity. It creates an insulating thermal
barrier between the marshmallow and the teacher’s mouth (the Leydenfrost effect). A similar effect
can be seen when water droplets are put on a hot skillet. Each one dances around as it slowly
shrinks, because it is levitated on a thin film of steam. The most extreme demonstration of this effect
is pouring liquid nitrogen into one’s mouth and blowing out a plume of nitrogen gas. We strongly
recommended that you read of Jearl Walker’s adventures with this demonstration rather than trying
it.
Q20.33
(a)
Warm a pot of coffee on a hot stove.
(b)
Place an ice cube at 0°C in warm water—the ice will absorb energy while melting, but not
increase in temperature.
(c)
Let a high-pressure gas at room temperature slowly expand by pushing on a piston. Work
comes out of the gas in a constant-temperature expansion as the same quantity of heat flows
in from the surroundings.
(d)
Warm your hands by rubbing them together. Heat your tepid coffee in a microwave oven.
Energy input by work, by electromagnetic radiation, or by other means, can all alike
produce a temperature increase.
(e)
Davy’s experiment is an example of this process.
(f)
This is not necessarily true. Consider some supercooled liquid water, unstable but with
temperature below 0°C . Drop in a snowflake or a grain of dust to trigger its freezing into
ice, and the loss of internal energy measured by its latent heat of fusion can actually push its
temperature up.
Q20.34
Heat is conducted from the warm oil to the pipe that carries it. That heat is then conducted to the
cooling fins and up through the solid material of the fins. The energy then radiates off in all
directions and is efficiently carried away by convection into the air. The ground below is left frozen.
Chapter 20
579
SOLUTIONS TO PROBLEMS
Section 20.1
P20.1
Heat and Internal Energy
Taking m = 1.00 kg , we have
b
ja
ge
f
∆U g = mgh = 1.00 kg 9.80 m s 2 50.0 m = 490 J .
But
T f = Ti + ∆T =
P20.2
b
gb
a10.0 + 0.117f° C
g
∆U g = Q = mc∆T = 1.00 kg 4 186 J kg⋅° C ∆T = 490 J so ∆T = 0.117° C
The container is thermally insulated, so no energy flows by heat:
Q=0
∆Eint = Q + Winput = 0 + Winput = 2mgh
and
The work on the falling weights is equal to the work done on the
water in the container by the rotating blades. This work results in
an increase in internal energy of the water:
2mgh = ∆Eint = m water c∆T
∆T =
e
ja
f
2 × 1.50 kg 9.80 m s 2 3.00 m
2mgh
88.2 J
=
=
837 J ° C
m water c
0.200 kg 4 186 J kg ⋅° C
b
g
= 0.105° C
Section 20.2
P20.3
FIG. P20.2
Specific Heat and Calorimetry
∆Q = mc silver ∆T
b
a
g
1.23 kJ = 0.525 kg c silver 10.0° C
f
c silver = 0.234 kJ kg⋅° C
P20.4
From
Q = mc∆T
we find
∆T =
1 200 J
Q
=
= 62.0° C
mc 0.050 0 kg 387 J kg ⋅° C
b
g
Thus, the final temperature is 87.0°C .
*P20.5
We imagine the stone energy reservoir has a large area in contact with air and is always at nearly the
same temperature as the air. Its overnight loss of energy is described by
Q mc∆T
=
∆t
∆t
−6 000 J s 14 h 3 600 s h
3.02 × 10 8 J ⋅ kg⋅° C
P∆t
m=
=
=
= 1.78 × 10 4 kg
c∆ T
850 J 20° C
850 J kg ⋅° C 18° C − 38° C
P=
b
b
ga fb
ga
g
f
a
f
580
*P20.6
Heat and the First Law of Thermodynamics
The laser energy output:
e
j
P∆t = 1.60 × 10 13 J s 2.50 × 10 −9 s = 4.00 × 10 4 J .
The teakettle input:
b
g
Q = mc∆T = 0.800 kg 4 186 J kg ⋅° C 80° C = 2.68 × 10 5 J .
This is larger by 6.70 times.
P20.7
Qcold = −Q hot
amc∆T f
f
20.0 kg b 4 186 J kg ⋅° C gdT
water
a
= − mc∆T
iron
f
i b
gb
gd
− 25.0° C = − 1.50 kg 448 J kg⋅° C T f − 600° C
i
T f = 29.6° C
P20.8
Let us find the energy transferred in one minute.
Q = m cup c cup + m water c water ∆T
b
gb
g b
gb
Q = 0. 200 kg 900 J kg⋅° C + 0.800 kg 4 186 J kg⋅° C
g a−1.50° Cf = −5 290 J
If this much energy is removed from the system each minute, the rate of removal is
P=
P20.9
(a)
Q
∆t
=
5 290 J
= 88.2 J s = 88.2 W .
60.0 s
Qcold = −Q hot
bm
wcw
gd
i
d
i
d
+ m c c c T f − Tc = − mCu cCu T f − TCu − munk c unk T f − Tunk
i
where w is for water, c the calorimeter, Cu the copper sample, and unk the unknown.
g a20.0 − 10.0f° C
= −b50.0 g gb0.092 4 cal g⋅° C ga 20.0 − 80.0f° C − b70.0 g gc a 20.0 − 100 f° C
2.44 × 10 cal = e5.60 × 10 g⋅° C jc
b
g
b
250 g 1.00 cal g ⋅° C + 100 g 0.215 cal g⋅° C
unk
3
3
unk
or c unk = 0.435 cal g⋅° C .
(b)
The material of the sample is beryllium .
Chapter 20
P20.10
b f gbmghg = mc∆T
a0.600fe3.00 × 10
(a)
−3
ja
je
581
f = b3.00 g gb0.092 4 cal g⋅° Cga∆T f
kg 9.80 m s 2 50.0 m
4.186 J cal
∆T = 0.760° C ; T = 25.8° C
No . Both the change in potential energy and the heat absorbed are proportional to the
(b)
mass; hence, the mass cancels in the energy relation.
*P20.11
We do not know whether the aluminum will rise or drop in temperature. The energy the water can
J
6° C = 6 279 J . The energy the copper can put
absorb in rising to 26°C is mc∆T = 0.25 kg 4 186
kg ° C
J
74° C = 2 864 J . Since 6 279 J > 2 864 J , the final
out in dropping to 26°C is mc∆T = 0.1 kg 387
kg ° C
temperature is less than 26°C . We can write Q h = −Q c as
Q water + Q Al + QCu = 0
0.25 kg 4 186
J
J
T f − 20° C + 0.4 kg 900
T f − 26° C
kg ° C
kg ° C
d
+0.1 kg 387
i
d
i
J
T f − 100° C = 0
kg ° C
d
i
1 046.5T f − 20 930° C + 360T f − 9 360° C + 38.7T f − 3 870° C = 0
1 445.2T f = 34 160° C
T f = 23.6° C
P20.12
Qcold = −Q hot
d
i
d i
d i
bm c + m c gT − bm c + m c gT = −m c T + m c
bm c + m c + m c gT = bm c + m c gT + m c T
bm c + m c gT + m c T
T =
m Al c Al T f − Tc + m c c w T f − Tc = − m h c w T f − Th
Al Al
c w
Al Al
c w
f
P20.13
Al Al
f
Al Al
c w
f
Al Al
h w
c w
c
c
h w Th
h w f
c w
c
h w h
h w h
m Al c Al + m c c w + m h c w
e
j
The rate of collection of energy is P = 550 W m 2 6.00 m 2 = 3 300 W . The amount of energy
required to raise the temperature of 1 000 kg of water by 40.0°C is:
b
ga
f
Q = mc∆T = 1 000 kg 4 186 J kg⋅° C 40.0° C = 1.67 × 10 8 J
Thus,
P∆t = 1.67 × 10 8 J
or
∆t =
1.67 × 10 8 J
= 50.7 ks = 14.1 h .
3 300 W
582
*P20.14
Heat and the First Law of Thermodynamics
Vessel one contains oxygen according to PV = nRT :
nc =
e
j
−3
5
3
PV 1.75 1.013 × 10 Pa 16.8 × 10 m
=
= 1.194 mol .
8.314 Nm mol ⋅ K 300 K
RT
Vessel two contains this much oxygen:
nh =
(a)
e
j
2. 25 1.013 × 10 5 22.4 × 10 −3
a f
8.314 450
mol = 1.365 mol .
The gas comes to an equilibrium temperature according to
amc∆T f
n McdT
cold
c
f
a
= − mc∆T
i
f
hot
d
i
− 300 K + n h Mc T f − 450 K = 0
The molar mass M and specific heat divide out:
1.194T f − 358.2 K + 1.365T f − 614.1 K = 0
Tf =
(b)
The pressure of the whole sample in its final state is
P=
Section 20.3
P20.15
972.3 K
= 380 K
2.559
2.559 mol 8.314 J 380 K
nRT
=
= 2.06 × 10 5 Pa = 2.04 atm .
V
mol K 22.4 + 16.8 × 10 −3 m3
a
f
Latent Heat
The heat needed is the sum of the following terms:
f
b
g a
+b heat to reach melting point g + b heat to vaporizeg + a heat to reach 110° C f
Q needed = heat to reach melting point + heat to melt
Thus, we have
ga f e
j
+b 4 186 J kg⋅° C ga100° C f + e 2.26 × 10 J kg j + b 2 010 J kg⋅° C ga10.0° C f
b
Q needed = 0.040 0 kg 2 090 J kg⋅° C 10.0° C + 3.33 × 10 5 J kg
6
Q needed = 1.22 × 10 5 J
P20.16
Qcold = −Q hot
bm
wcw
gd
i
d
+ m c c c T f − Ti = − m s − L v + c w T f − 100
b
g
i
f
ga
J kg + b 4 186 J kg⋅° C ga50.0° C − 100° C f
b
0.250 kg 4 186 J kg ⋅° C + 0.050 0 kg 387 J kg⋅° C 50.0° C − 20.0° C
= − m s −2.26 × 10 6
ms =
3.20 × 10 4 J
= 0.012 9 kg = 12.9 g steam
2.47 × 10 6 J kg
Chapter 20
P20.17
583
The bullet will not melt all the ice, so its final temperature is 0°C.
Then
FG 1 mv
H2
2
+ mc ∆T
IJ
K
bullet
= mwL f
where m w is the melt water mass
mw =
mw =
P20.18
(a)
jb
e
0.500 3.00 × 10 −3 kg 240 m s
g
2
b
ga
+ 3.00 × 10 −3 kg 128 J kg⋅° C 30.0° C
3.33 × 10
5
f
J kg
86.4 J + 11.5 J
= 0.294 g
333 000 J kg
e
je
j
Q1 = heat to melt all the ice = 50.0 × 10 −3 kg 3.33 × 10 5 J kg = 1.67 × 10 4 J
b
= e50.0 × 10
Q 2 = heat to raise temp of ice to 100° C
−3
jb
ga
g
f
kg 4 186 J kg⋅° C 100° C = 2.09 × 10 4 J
Thus, the total heat to melt ice and raise temp to 100°C = 3.76 × 10 4 J
Q3 =
heat available
= 10.0 × 10 −3 kg 2.26 × 10 6 J kg = 2.26 × 10 4 J
as steam condenses
e
je
j
Thus, we see that Q3 > Q1 , but Q3 < Q1 + Q 2 .
Therefore, all the ice melts but T f < 100° C . Let us now find T f
Qcold = −Q hot
e50.0 × 10 kg je3.33 × 10 J kg j + e50.0 × 10 kg jb4 186 J kg⋅° CgdT − 0° Ci
= −e10.0 × 10 kg je −2.26 × 10 J kg j − e10.0 × 10 kg jb 4 186 J kg⋅° C gdT − 100° C i
−3
−3
5
−3
f
−3
6
f
From which, T f = 40.4° C .
(b)
Q1 = heat to melt all ice = 1.67 × 10 4 J [See part (a)]
heat given up
= 10 −3 kg 2.26 × 10 6 J kg = 2.26 × 10 3 J
Q2 =
as steam condenses
e
Q3
je
heat given up as condensed
=
= e10
steam cools to 0° C
j
−3
jb
ga
f
kg 4 186 J kg⋅° C 100° C = 419 J
Note that Q 2 + Q3 < Q1 . Therefore, the final temperature will be 0°C with some ice
remaining. Let us find the mass of ice which must melt to condense the steam and cool the
condensate to 0°C.
mL f = Q 2 + Q3 = 2.68 × 10 3 J
Thus, m =
2.68 × 10 3 J
= 8.04 × 10 −3 kg = 8.04 g .
5
3.33 × 10 J kg
Therefore, there is 42.0 g of ice left over .
584
P20.19
Heat and the First Law of Thermodynamics
e j
1.00 kg b0.092 0 cal g ⋅° C ga 293 − 77.3f° C = mb 48.0 cal g g
Q = mCu cCu ∆T = m N 2 L vap
N2
m = 0.414 kg
*P20.20
The original gravitational energy of the hailstone-Earth system changes entirely into additional
internal energy in the hailstone, to produce its phase change. No temperature change occurs, either
in the hailstone, in the air, or in sidewalk. Then
mgy = mL
y=
P20.21
F
GH
I
JK
L 3.33 × 10 5 J kg 1 kg ⋅ m 2 s 2
=
= 3.40 × 10 4 m
1J
g
9.8 m s 2
(a)
Since the heat required to melt 250 g of ice at 0°C exceeds the heat required to cool 600 g of
water from 18°C to 0°C, the final temperature of the system (water + ice) must be 0°C .
(b)
Let m represent the mass of ice that melts before the system reaches equilibrium at 0°C.
Qcold = −Q hot
b
g
J kg j = −b0.600 kg gb 4 186 J kg⋅° C ga0° C − 18.0° C f
mL f = − m w c w 0° C − Ti
e
m 3.33 × 10
5
m = 136 g, so the ice remaining = 250 g − 136 g = 114 g
P20.22
The original kinetic energy all becomes thermal energy:
FG IJ e
H K
jb
1
1
1
5.00 × 10 −3 kg 500 m s
mv 2 + mv 2 = 2
2
2
2
g
2
= 1.25 kJ .
Raising the temperature to the melting point requires
b
f
ga
Q = mc∆T = 10.0 × 10 −3 kg 128 J kg⋅° C 327° C − 20.0° C = 393 J .
Since 1 250 J > 393 J , the lead starts to melt. Melting it all requires
e
je
j
Q = mL = 10.0 × 10 −3 kg 2.45 × 10 4 J kg = 245 J .
Since 1 250 J > 393 + 245 J , it all melts. If we assume liquid lead has the same specific heat as solid
lead, the final temperature is given by
b
gd
1.25 × 10 3 J = 393 J + 245 J + 10.0 × 10 −3 kg 128 J kg⋅° C T f − 327° C
T f = 805° C
i
Chapter 20
Section 20.4
Work and Heat in Thermodynamic Processes
z
f
P20.23
P
Wif = − PdV
i
f
The work done on the gas is the negative of the area under the
curve P = αV 2 between Vi and V f .
z
f
1
Wif = − αV 2 dV = − α V f3 − Vi3
3
i
e
e
P = αV 2
i
j
O
j
V f = 2Vi = 2 1.00 m 3 = 2.00 m 3
j LNMe2.00 m j + e1.00 m j OQP =
1
5.00 atm m6 1.013 × 10 5 Pa atm
Wif = −
3
e
P20.24
585
je
z
3 3
3 3
V
3
1.00 m
3
2.00 m
FIG. P20.23
−1.18 MJ
W = − PdV
(a)
f
e
ja
− e 4.00 × 10 Pa ja3.00 − 2.00f m
− e 2.00 × 10 Paja 4.00 − 3.00f m
W = − 6.00 × 10 6 Pa 2.00 − 1.00 m3 +
6
3
6
3
+
Wi→ f = −12.0 MJ
W f →i = +12.0 MJ
(b)
FIG. P20.24
FG nR IJ dT
H PK
a
i
fa
P20.25
W = − P∆ V = − P
P20.26
W = − PdV = − P dV = − P∆V = −nR∆T = −nR T2 − T1
z
z
f
f
i
P20.27
fa f
− Ti = −nR∆T = − 0.200 8.314 280 = −466 J
f
b
i
During the heating process P =
g
FG P IJ V .
HV K
i
i
z
f
(a)
W = − PdV = −
i
W =−
Vi
2 3Vi
FG P IJ V
HV K 2
i
i
(b)
z FGH
3Vi
Vi
IJ
K
Pi
VdV
Vi
=−
Pi
9Vi2 − Vi2 = −4PV
i i
2Vi
e
j
PV = nRT
LMF P I V OPV = nRT
MNGH V JK PQ
F P IJ V
T =G
H nRV K
i
i
2
i
i
Temperature must be proportional to the square of volume, rising to nine times its original
value.
586
Heat and the First Law of Thermodynamics
Section 20.5
The First Law of Thermodynamics
P20.28
(a)
W = − P∆V = − 0.800 atm −7.00 L 1.013 × 10 5 Pa atm 10 −3 m 3 L = +567 J
(b)
∆Eint = Q + W = −400 J + 567 J = 167 J
P20.29
a
fa
fe
je
j
∆Eint = Q + W
Q = ∆Eint − W = −500 J − 220 J = −720 J
The negative sign indicates that positive energy is transferred from the system by heat.
P20.30
(a)
Q = −W = Area of triangle
Q=
(b)
ja
f
1
4.00 m 3 6.00 kPa = 12.0 kJ
2
e
Q = −W = −12.0 kJ
FIG. P20.30
P20.31
BC
CA
AB
P20.32
WBC
Q
–
W
0
∆Eint
–
–
+
–
bQ = ∆E since W = 0g
b∆E < 0 and W > 0 , so Q < 0g
+
–
+
bW < 0 , ∆E > 0 since ∆E < 0 for B → C → A ; so Q > 0g
= − P bV − V g = −3.00 atmb0.400 − 0.090 0 g m
int
BC
int
int
B
C
3
B
P(atm)
= −94.2 kJ
∆Eint = Q + W
int
3.0
a
f
Eint, C − Eint, B = 100 − 94.2 kJ
Eint, C − Eint, B = 5.79 kJ
1.0
Since T is constant,
B
C
A
0.090 0.20
D
0.40
Eint, D − Eint, C = 0
b
a
g
FIG. P20.32
f
WDA = − PD VA − VD = −1.00 atm 0.200 − 1.20 m3
= +101 kJ
a
f
Eint, A − Eint, D = −150 kJ + +101 kJ = −48.7 kJ
d
i d
i d
Now, Eint, B − Eint, A = − Eint, C − Eint, B + Eint, D − Eint, C + Eint, A − Eint, D
Eint, B − Eint, A = − 5.79 kJ + 0 − 48.7 kJ = 42.9 kJ
i
1.2
V(m 3)
587
Chapter 20
*P20.33
1 2
πr . The arrow in Figure P20.33
2
looks like a semicircle when the scale makes 1.2 L fill the same
space as 100 kPa. Its area is
The area of a true semicircle is
a
fa
f
je
500
300
1
1
π 2.4 L 200 kPa = π 2.4 × 10 −3 m3 2 × 10 5 N m 2 .
2
2
e
P(kPa)
j
The work on the gas is
0
z
A
1.2
B
3.6
B
W = − PdV = − area under the arch shown in the graph
A
FG 1 π 2.4a200f J + 3 × 10 N m
H2
= −b754 J + 1 440 Jg = −2 190 J
5
=−
2
4.8 × 10 −3 m3
FIG. P20.33
IJ
K
∆Eint = Q + W = 5 790 J − 2 190 J = 3.60 kJ
Section 20.6
P20.34
(a)
Some Applications of the First Law of Thermodynamics
FG V IJ = − P V lnFG V IJ
HV K
HV K
F W I = b0.025 0g expLM −3 000 OP =
so V = V expG +
MN 0.025 0e1.013 × 10 j PQ
H P V JK
f
W = −nRT ln
f
i
i
P20.35
f
Pf V f
f
5
f
e
j=
g
1.013 × 10 5 Pa 0.025 0 m 3
Tf =
(a)
∆Eint = Q − P∆V = 12.5 kJ − 2.50 kPa 3.00 − 1.00 m 3 = 7.50 kJ
(b)
b
1.00 mol 8.314 J K ⋅ mol
(a)
0.007 65 m 3
305 K
a
f
V1 V2
=
T1 T2
V
3.00
300 K = 900 K
T2 = 2 T1 =
1.00
V1
a
P20.36
i
(b)
nR
=
f
f
W = − P∆V = − P 3αV∆T
e
f
L
jMM e
N
= − 1.013 × 10 5 N m 2 3 24.0 × 10 −6 ° C −1
O
F
I
jGH 2.70 ×110.00 kgkg m JK a18.0° CfPP
Q
W = −48.6 mJ
b
gb
ga
f
(b)
Q = cm∆T = 900 J kg⋅° C 1.00 kg 18.0° C = 16.2 kJ
(c)
∆Eint = Q + W = 16. 2 kJ − 48.6 mJ = 16.2 kJ
3
3
6.0
V (L)
588
P20.37
Heat and the First Law of Thermodynamics
OP
a f LM
MN e
je
j PQ
F 18.0 g I =
W = −a1.00 molfb8.314 J K ⋅ molga373 K f + e1.013 × 10 N m jG
H 10 g m JK
Q = mL = 0.018 0 kg e 2.26 × 10 J kg j = 40.7 kJ
b
g
W = − P∆V = − P Vs − Vw = −
P nRT
18.0 g
+P
P
1.00 g cm 3 10 6 cm 3 m3
5
2
6
3
−3.10 kJ
6
v
∆Eint = Q + W = 37.6 kJ
P20.38
(a)
The work done during each step of the cycle equals the
negative of the area under that segment of the PV curve.
W = WDA + W AB + WBC + WCD
b
g
b
g
W = − Pi Vi − 3Vi + 0 − 3 Pi 3Vi − Vi + 0 = −4PV
i i
P20.39
(b)
The initial and final values of T for the system are equal.
Therefore, ∆Eint = 0 and Q = −W = 4PV
i i .
(c)
W = −4PV
i i = −4nRTi = −4 1.00 8.314 273 = −9.08 kJ
(a)
3
PV
i i = Pf V f = nRT = 2.00 mol 8.314 J K ⋅ mol 300 K = 4.99 × 10 J
(b)
(c)
a fa
fa f
b
ga
FIG. P20.38
f
3
Vi =
nRT 4.99 × 10 J
=
0.400 atm
Pi
Vf =
nRT 4.99 × 10 3 J 1
=
= Vi = 0.041 0 m3
1.20 atm
Pf
3
FG V IJ = −e4.99 × 10 j lnFG 1 IJ =
H 3K
HV K
z
W = − PdV = −nRT ln
f
3
+5.48 kJ
i
∆Eint = 0 = Q + W
Q = −5.48 kJ
P20.40
∆Eint, ABC = ∆Eint, AC
(a)
(conservation of energy)
∆Eint, ABC = Q ABC + WABC
(First Law)
Q ABC = 800 J + 500 J = 1 300 J
(b)
WCD = − PC ∆VCD , ∆VAB = − ∆VCD , and PA = 5 PC
1
1
Then, WCD = PA ∆VAB = − W AB = 100 J
5
5
(+ means that work is done on the system)
(c)
WCDA = WCD so that QCA = ∆Eint, CA − WCDA = −800 J − 100 J = −900 J
(– means that energy must be removed from the system by heat)
(d)
∆Eint, CD = ∆Eint, CDA − ∆Eint, DA = −800 J − 500 J = −1 300 J
and QCD = ∆Eint, CD − WCD = −1 300 J − 100 J = −1 400 J
FIG. P20.40
Chapter 20
Section 20.7
P20.41
Energy Transfer Mechanisms
∆T
L
P L 10.0 W 0.040 0 m
k=
=
= 2.22 × 10 −2 W m⋅° C
A∆T
1.20 m 2 15.0° C
P = kA
b
g
a
P20.42
f
kA∆T b0.800 W m⋅° C ge3.00 m ja 25.0° C f
P=
=
= 1.00 × 10
P20.43
In the steady state condition,
PAu = PAg
so that
k Au A Au
In this case
A Au = A Ag
2
6.00 × 10 −3 m
L
FG ∆T IJ
H ∆x K
Au
4
= k Ag A Ag
W = 10.0 kW
FG ∆T IJ
H ∆x K
Ag
∆x Au = ∆x Ag
a
f
= aT − 30.0f
∆TAu = 80.0 − T
∆TAg
and
FIG. P20.43
where T is the temperature of the junction.
Therefore,
k Au 80.0 − T = k Ag T − 30.0
a
P=
A∆T
∑k
Li
i
*P20.45
f
e6.00 m ja50.0° Cf
2
=
i
e
j
2 4.00 × 10 −3 m
0.800 W m⋅° C + 5.00 × 10 −3 m
0.023 4 W m⋅° C
= 1.34 kW
We suppose that the area of the transistor is so small that energy flow by heat from the transistor
directly to the air is negligible compared to energy conduction through the mica.
P = kA
bT − T g
h
c
L
Th = Tc +
P20.46
a
T = 51.2° C
And
P20.44
f
e
j
1.50 W 0.085 2 × 10 −3 m
PL
= 35.0° C +
= 67.9° C
kA
0.075 3 W m⋅° C 8.25 × 6.25 10 −6 m 2
b
ga
f
From Table 20.4,
(a)
R = 0.890 ft 2 ⋅° F ⋅ h Btu
(b)
The insulating glass in the table must have sheets of glass less than
estimate the R-value of a 0.250-inch air space as
Then for the double glazing
LM
N
Rb = 0.890 +
(c)
b
g
1
inch thick. So we
8
0.250
times that of the thicker air space.
3.50
FG 0.250 IJ 1.01 + 0.890OP ft ⋅° F ⋅ h =
H 3.50 K
Q Btu
2
1.85
ft 2 ⋅° F ⋅ h
.
Btu
Since A and T2 − T1 are constants, heat flow is reduced by a factor of
1.85
= 2.08 .
0.890
589
590
P20.47
Heat and the First Law of Thermodynamics
jLNM e
e
j OQPa0.965fb5 800 K g
2
P = σAeT 4 = 5.669 6 × 10 −8 W m 2 ⋅ K 4 4π 6.96 × 10 8 m
4
P = 3.77 × 10 26 W
P20.48
Suppose the pizza is 70 cm in diameter and A = 2.0 cm thick, sizzling at 100°C. It cannot lose heat by
conduction or convection. It radiates according to P = σAeT 4 . Here, A is its surface area,
a
f
A = 2π r 2 + 2π rA = 2π 0.35 m
2
a
fa
f
+ 2π 0.35 m 0.02 m = 0.81 m 2 .
Suppose it is dark in the infrared, with emissivity about 0.8. Then
e
ja fa
je
P = 5.67 × 10 −8 W m 2 ⋅ K 4 0.81 m 2 0.80 373 K
f
4
= 710 W ~ 10 3 W .
If the density of the pizza is half that of water, its mass is
ja
e
f a0.02 mf = 4 kg .
m = ρV = ρπ r 2 A = 500 kg m3 π 0.35 m
2
Suppose its specific heat is c = 0.6 cal g⋅° C . The drop in temperature of the pizza is described by:
d
Q = mc T f − Ti
P=
dT f
dt
P20.49
dT f
dQ
= mc
−0
dt
dt
=
710 J s
P
=
= 0.07 ° C s ~ 10 −1 K s
mc
4 kg 0.6 ⋅ 4 186 J kg ⋅° C
b gb
g
P = σAeT 4
2.00 W = 5.67 × 10 −8 W m 2 ⋅ K 4 0.250 × 10 −6 m 2 0.950 T 4
e
ja
je
e
T = 1.49 × 10 14 K 4
P20.50
i
j
14
f
= 3.49 × 10 3 K
We suppose the earth below is an insulator. The square meter must radiate in the infrared as much
energy as it absorbs, P = σAeT 4 . Assuming that e = 1.00 for blackbody blacktop:
e
ja f
je
1 000 W = 5.67 × 10 −8 W m 2 ⋅ K 4 1.00 m 2 1.00 T 4
e
T = 1.76 × 10 10 K
P20.51
j
4 14
= 364 K (You can cook an egg on it.)
The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle
perpendicular to the light: π R 2 . It radiates in all directions, over area 4π R 2 . Then, in steady state,
Pin = Pout
e
j
e
j
e 1 340 W m 2 π R 2 = eσ 4π R 2 T 4
The emissivity e, the radius R, and π all cancel.
L 1 340 W m
Therefore, T = M
MN 4e5.67 × 10 W m
2
−8
2
OP
⋅K jP
Q
4
14
= 277 K = 4° C .
Chapter 20
591
Additional Problems
P20.52
77.3 K = –195.8°C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heat
to vaporize:
b
ge
j
Q = mL v = 0.100 kg 2.01 × 10 5 J kg = 2.01 × 10 4 J .
The water first loses heat by cooling. Before it starts to freeze, it can lose
b
gb
ga
f
Q = mc∆T = 0.200 kg 4 186 J kg⋅° C 5.00° C = 4.19 × 10 3 J .
e
j
The remaining 2.01 × 10 4 − 4.19 × 10 3 J = 1.59 × 10 4 J that is removed from the water can freeze a
mass x of water:
Q = mL f
e
1.59 × 10 4 J = x 3.33 × 10 5 J kg
j
x = 0.047 7 kg = 47.7 g of water can be frozen
P20.53
The increase in internal energy required to melt 1.00 kg of snow is
b
ge
j
∆Eint = 1.00 kg 3.33 × 10 5 J kg = 3.33 × 10 5 J
The force of friction is
b
ge
j
f = µn = µmg = 0. 200 75.0 kg 9.80 m s 2 = 147 N
According to the problem statement, the loss of mechanical energy of the skier is assumed to be
equal to the increase in internal energy of the snow. This increase in internal energy is
a
f
∆Eint = f∆r = 147 N ∆r = 3.33 × 10 5 J
∆r = 2.27 × 10 3 m .
and
P20.54
(a)
The energy thus far gained by the copper equals the energy loss by the silver. Your down
parka is an excellent insulator.
Qcold = −Q hot
or
d i = −m c dT − T i
b9.00 g gb387 J kg⋅° Cga16.0° Cf = −b14.0 g gb234 J kg⋅° CgdT
dT − 30.0° Ci = −17.0° C
mCu c Cu T f − Ti
f
so
(b)
Ag Ag
Cu
f
i Ag
f
− 30.0° C
i
FG dT IJ
H dt K
= − mCu c Cu
Ag
T f , Ag = 13.0° C .
Differentiating the energy gain-and-loss equation gives: m Ag c Ag
FG dT IJ
H dt K
FG dT IJ
H dt K
Ag
FG IJ = − 9.00 gb387 J kg⋅° Cg b+0.500 ° C sg
H K 14.0 gb234 J kg⋅° Cg
−0.532 ° C s b negative sign ⇒ decreasing temperatureg
=−
Ag
=
Ag
mCu c Cu dT
m Ag c Ag dt
Cu
Ag
FG dT IJ
H dt K
Cu
592
P20.55
Heat and the First Law of Thermodynamics
(a)
Before conduction has time to become important, the energy lost by the rod equals the
energy gained by the helium. Therefore,
mL v He = mc ∆T Al
or
so
b g c h
bρVL g = cρVc ∆T h
cρVc ∆T h
V =
bρL g
e2.70 g cm je62.5 cm jb0.210 cal g⋅° Cga295.8° Cf
V =
e0.125 g cm je2.09 × 10 J kg jb1.00 cal 4.186 Jgb1.00 kg 1 000 g g
v He
Al
Al
He
v He
3
He
3
3
4
VHe = 1.68 × 10 4 cm3 = 16.8 liters
(b)
The rate at which energy is supplied to the rod in order to maintain constant temperatures
is given by
dT
295.8 K
P = kA
= 31.0 J s ⋅ cm ⋅ K 2.50 cm 2
= 917 W
dx
25.0 cm
This power supplied to the helium will produce a “boil-off” rate of
FG IJ b
H K
jFGH
ge
a
fe
je
j
IJ
K
917 W 10 3 g kg
P
=
= 351 cm 3 s = 0.351 L s
ρL v
0.125 g cm3 2.09 × 10 4 J kg
e
*P20.56
j
At the equilibrium temperature Teq the diameters of the sphere and ring are equal:
e
j
5.01 cm + 5.01 cme 2.40 × 10
e
j
1 ° C jeT − T j = 5.00 cm + 5.00 cme1.70 × 10
d s + d sα Al Teq − Ti = d r + d r α Cu Teq − 15° C
−5
eq
i
−5
je
1 ° C Teq − 15° C
0.01° C + 1.202 4 × 10 −4 Teq − 1.202 4 × 10 −4 Ti = 8.5 × 10 −5 Teq − 1.275 × 10 −3 ° C
1.127 5 × 10 −2 ° C + 3.524 × 10 −5 Teq = 1.202 4 × 10 −4 Ti
319.95° C + Teq = 3.412 0Ti
At the equilibrium temperature, the energy lost is equal to the energy gained:
e
j
e
m s c Al Teq − Ti = − m r cCu Teq − 15° C
e
j
j
e
10.9 g 0.215 cal g ⋅° C Teq − Ti = −25 g 0.092 4 cal g⋅° C Teq − 15° C
2.343 5Teq − 2.343 5Ti = 34.65° C − 2.31Teq
4.653 5Teq = 34.65° C + 2.343 5Ti
Solving by substitution,
b
g
4.653 5 3.412 0Ti − 319.95° C = 34.65° C + 2.343 5Ti
15.877 7Ti − 1 488.89° C = 34.65° C + 2.343 5Ti
1 523.54° C
= 113° C
13.534
(b)
Ti =
(a)
Teq = −319.95 + 3.412 0 112.57 = 64.1° C
a
f
j
j
Chapter 20
P20.57
b g
Q = mc∆T = ρV c∆T so that when a constant temperature difference ∆T is maintained,
the rate of adding energy to the liquid is P =
P
.
ρR∆T
and the specific heat of the liquid is c =
P20.58
(a)
Work done by the gas is the negative of the area under the
PV curve
W = − Pi
(b)
FG IJ
H K
dQ
dV
=ρ
c∆T = ρRc∆T
dt
dt
FG V − V IJ =
H2 K
i
i
+
PV
i i
.
2
z
In this case the area under the curve is W = − PdV . Since the
process is isothermal,
FG V IJ = nRT
H 4K
F dV IJ bPV g = − PV lnFG V 4 IJ = PV ln 4
W =− z G
HVK
HV K
PV = PV
i i = 4 Pi
i
i
Vi 4
and
i i
i i
i
Vi
i
FIG. P20.58
i i
= +1.39 PV
i i
(c)
P20.59
The area under the curve is 0 and W = 0 .
Call the initial pressure P1 . In the constant volume process 1 → 2 the work is zero.
P1V1 = nRT1
P2 V2 = nRT2
so
FG
H
IJ a f
K
P2 V2 T2
1
=
; T2 = 300 K
1 = 75.0 K
4
P1V1 T1
Now in 2 → 3
z
3
b
g
W = − PdV = − P2 V3 − V2 = − P3V3 + P2V2
2
a
fb
ga
W = −nRT3 + nRT2 = − 1.00 mol 8.314 J mol ⋅ K 300 K − 75.0 K
W = −1.87 kJ
f
593
594
*P20.60
Heat and the First Law of Thermodynamics
The initial moment of inertia of the disk is
ja
f
1
1
1
1
4
MR 2 = ρVR 2 = ρπR 2 tR 2 = 8 920 kg m3 π 28 m 1.2 m = 1.033 × 10 10 kg ⋅ m 2
2
2
2
2
e
The rotation speeds up as the disk cools off, according to
I iω i = I f ω f
2
1
1
1
MRi2ω i = MR 2f ω f = MRi2 1 − α ∆T ω f
2
2
2
1
1
ω f = ωi
= 25 rad s
2
1 − α ∆T
1 − 17 × 10 −6 1 ° C 830° C
c
c
(a)
h
h
e
j
2
= 25.720 7 rad s
The kinetic energy increases by
1
1
1
1
1
I f ω 2f − I iω i2 = I iω iω f − I iω i2 = I iω i ω f − ω i
2
2
2
2
2
1
10
2
= 1.033 × 10 kg ⋅ m 25 rad s 0.720 7 rad s = 9.31 × 10 10 J
2
d
i
b
b
g
f
ga
(b)
∆Eint = mc∆T = 2.64 × 10 7 kg 387 J kg⋅° C 20° C − 850° C = −8.47 × 10 12 J
(c)
As 8.47 × 10 12 J leaves the fund of internal energy, 9.31 × 10 10 J changes into extra kinetic
energy, and the rest, 8.38 × 10 12 J is radiated.
*P20.61
The loss of mechanical energy is
GM E m 1
1
= 670 kg 1.4 × 10 4 m s
mvi2 +
RE
2
2
e
j
2
+
6.67 × 10 −11 Nm 2 5.98 × 10 24 kg 670 kg
kg 2 6.37 × 10 6 m
= 6.57 × 10 10 J + 4.20 × 10 10 J = 1.08 × 10 11 J
One half becomes extra internal energy in the aluminum: ∆Eint = 5.38 × 10 10 J. To raise its
temperature to the melting point requires energy
mc∆T = 670 kg 900
a
fh
J
660 − −15° C = 4.07 × 10 8 J .
kg ° C
c
To melt it, mL = 670 kg 3.97 × 10 5 J kg = 2.66 × 10 8 J . To raise it to the boiling point,
b
gb
g
mc∆T = 670 1 170 2 450 − 600 J = 1.40 × 10 9 J . To boil it, mL = 670 kg 1.14 × 10 7 J kg = 7.64 × 10 9 J .
Then
b
gd
i
5.38 × 10 10 J = 9.71 × 10 9 J + 670 1 170 T f − 2 450° C J ° C
T f = 5.87 × 10 4 ° C
Chapter 20
P20.62
a
fb
g
(a)
Fv = 50.0 N 40.0 m s = 2 000 W
(b)
Energy received by each object is 1 000 W 10 s = 10 4 J = 2 389 cal . The specific heat of iron
ga f
b
is 0.107 cal g ⋅° C , so the heat capacity of each object is 5.00 × 10 3 × 0.107 = 535.0 cal ° C.
∆T =
P20.63
2 389 cal
= 4.47° C
535.0 cal ° C
The power incident on the solar collector is
j a
e
f
Pi = IA = 600 W m 2 π 0.300 m
2
= 170 W .
For a 40.0% reflector, the collected power is Pc = 67.9 W. The total
energy required to increase the temperature of the water to the
boiling point and to evaporate it is Q = cm∆T + mLV :
b
ga
f
Q = 0.500 kg 4 186 J kg ⋅° C 80.0° C + 2.26 × 10 6 J kg = 1.30 × 10 6 J .
The time interval required is ∆t =
P20.64
595
Q
Pc
=
1.30 × 10 6 J
= 5.31 h .
67.9 W
FIG. P20.63
From Q = mLV the rate of boiling is described by
P=
Q LV m
=
∆t
∆t
∴
P
m
=
∆t LV
Model the water vapor as an ideal gas
FG m IJ RT
H MK
P V m F RT I
=
G J
∆t
∆t H M K
P F RT I
P Av =
G J
L HMK
P0 V0 = nRT =
0
0
v=
V
b
ga
f
1 000 W 8.314 J mol ⋅ K 373 K
P RT
=
MLV P0 A
0.018 0 kg mol 2.26 × 10 6 J kg 1.013 × 10 5 N m 2 2.00 × 10 −4 m 2
v = 3.76 m s
b
ge
je
je
j
596
P20.65
Heat and the First Law of Thermodynamics
Energy goes in at a constant rate P . For the period from
50.0 min to 60.0 min, Q = mc∆T
a
a
f b
f
gb
T°( C)
f
ga
P 10.0 min = 10 kg + mi 4 186 J kg⋅° C 2.00° C − 0° C
(1)
P 10.0 min = 83.7 kJ + 8.37 kJ kg mi
b
g
a
f
e
Substitute P =
e
mi 3.33 × 10 5 J kg
50.0 min
e
mi 3.33 × 10 5 J kg
j
3.00
1.00
For the period from 0 to 50.0 min, Q = mi L f
P 50.0 min = mi 3.33 × 10 5 J kg
2.00
0.00
j
20.0
j into Equation (1) to find
b
40.0
60.0 t (min)
FIG. P20.65
g
= 83.7 kJ + 8.37 kJ kg mi
5.00
83.7 kJ
= 1.44 kg
mi =
66.6 − 8.37 kJ kg
a
P20.66
(a)
f
The block starts with K i =
b
gb
1
1
mvi2 = 1.60 kg 2.50 m s
2
2
g
2
= 5.00 J
All this becomes extra internal energy in ice, melting some according to “Q ” = m ice L f . Thus,
the mass of ice that melts is
m ice =
For the block:
“Q ” K i
5.00 J
=
=
= 1.50 × 10 −5 kg = 15.0 mg .
Lf
L f 3.33 × 10 5 J kg
Q = 0 (no energy flows by heat since there is no temperature difference)
W = −5.00 J
∆Eint = 0 (no temperature change)
and
∆K = −5.00 J
For the ice,
Q=0
W = +5.00 J
∆Eint = +5.00 J
and
(b)
∆K = 0
Again, K i = 5.00 J and m ice = 15.0 mg
For the block of ice: Q = 0; ∆Eint = +5.00 J ; ∆K = −5.00 J
so W = 0 .
For the copper, nothing happens: Q = ∆Eint = ∆K = W = 0 .
continued on next page
Chapter 20
Again, K i = 5.00 J. Both blocks must rise equally in temperature.
(c)
∆T =
“Q ” = mc∆T :
“Q ”
5.00 J
=
= 4.04 × 10 −3 ° C
mc 2 1.60 kg 387 J kg ⋅° C
b
gb
g
At any instant, the two blocks are at the same temperature, so for both Q = 0.
For the moving block:
∆K = −5.00 J
and
∆Eint = +2.50 J
so
W = −2.50 J
For the stationary block:
∆K = 0
and
∆Eint = +2.50 J
so
W = +2.50 J
For each object in each situation, the general continuity equation for energy, in the form
∆K + ∆Eint = W + Q , correctly describes the relationship between energy transfers and
changes in the object’s energy content.
P20.67
A = Aend walls + A ends of attic + A side walls + Aroof
a
f LMN 12 × 4.00 m × a4.00 mf tan 37.0°OPQ
F 4.00 m IJ
+2a10.0 m × 5.00 mf + 2a10.0 mfG
H cos37.0° K
A = 2 8.00 m × 5.00 m + 2 2 ×
A = 304 m 2
P=
e
ja
je
f
4.80 × 10 −4 kW m⋅° C 304 m 2 25.0° C
kA∆T
=
= 17.4 kW = 4.15 kcal s
L
0.210 m
b
gb
g
Thus, the energy lost per day by heat is 4.15 kcal s 86 400 s = 3.59 × 10 5 kcal day .
The gas needed to replace this loss is
P20.68
3.59 × 10 5 kcal day
9 300 kcal m3
= 38.6 m3 day .
FG IJ
H K
∆T
LρAdx
= kA
dt
x
Lρ
Lρ
e
z
z
8.00
∆t
4.00
0
xdx = k∆T dt
2 8.00
x
2
= k∆T∆t
4.00
je
3.33 × 10 5 J kg 917 kg m3
∆t = 3.66 × 10 4 s = 10.2 h
F b0.080 0 mg − b0.040 0 mg
jGG
2
H
2
2
I
JJ = b2.00 W m⋅° Cga10.0° Cf∆t
K
597
598
P20.69
Heat and the First Law of Thermodynamics
W = W AB + WBC + WCD + WDA
z
z
z
P
z
B
C
D
A
A
B
C
D
P2
W = − PdV − PdV − PdV − PdV
W = −nRT1
z
z
B
z
C
z
D
FG V IJ − P bV
HV K
B
2
1
C
FG V IJ − P bV
HV K
g
− VB − nRT2 ln
2
1
C
P1
A
− VD
g
VB P1
V
P
=
and 2 = 2
V1 P2
VC P1
FP I
FP I
FP I
D
A
V1
Now P1VA = P2 VB and P2 VC = P1VD , so only the logarithmic
terms do not cancel out.
Also,
C
A
dV
dV
− P2 dV − nRT2
− P1 dV
V
V
A
B
C
D
W = −nRT1 ln
B
V2
FIG. P20.69
FP I
FP I
∑ W = −nRT1 lnGH P1 JK − nRT2 lnGH P2 JK = +nRT1 lnGH P2 JK − nRT2 lnGH P2 JK = −nRbT2 − T1 g lnGH P2 JK
1
1
1
2
1
Moreover P1V2 = nRT2 and P1V1 = nRT1
∑W =
P20.70
b
g FGH PP IJK
− P1 V2 − V1 ln
2
1
For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction,
dQ
dT
= − kA
dt
dx
becomes
Under equilibrium conditions,
dT = −
F
GH
1
dQ
dt 2π kL
I FG dr IJ
JK H r K
g
dQ
is constant; therefore,
dt
Tb − Ta = −
and
F
GH
I FG IJ
JK H K
1
dQ
b
ln
dt 2π kL
a
b g
b g
dQ 2π kL Ta − Tb
=
dt
ln b a
But Ta > Tb , so
P20.71
b
dQ
dT
= − k 2π rL
dt
dr
From problem 70, the rate of energy flow through the wall is
b g
b g
dQ 2π kL Ta − Tb
=
dt
ln b a
e
jb
ga
−5
dQ 2π 4.00 × 10 cal s ⋅ cm⋅° C 3 500 cm 60.0° C
=
dt
ln 256 cm 250 cm
b
g
f
dQ
= 2. 23 × 10 3 cal s = 9.32 kW
dt
This is the rate of energy loss from the plane by heat, and consequently is
the rate at which energy must be supplied in order to maintain a constant
temperature.
FIG. P20.71
V
Chapter 20
P20.72
599
Qcold = −Q hot
b
dT
Q Al = − Q water + Qcalo
or
m Al c Al
f
− Ti
i
Al
g
b
gd
= − m w c w + m c c c T f − Ti
i
w
b0.200 kg gc a+39.3° Cf = − 0.400 kgb4 186 J kg⋅° Cg + 0.040 0 kgb630 J kg⋅° Cg a−3.70° Cf
Al
c Al =
*P20.73
6.29 × 10 3 J
= 800 J kg⋅° C
7.86 kg⋅° C
e
a
j
fb
(a)
P = σAeT 4 = 5.67 × 10 −8 W m 2 K 4 5.1 × 10 14 m 2 0.965 5 800 K
(b)
Tavg = 0.1 4 800 K + 0.9 5 890 K = 5.78 × 10 3 K
b
g
b
e
= 3.16 × 10 22 W
5 800 − 5 781
= 0.327% .
5 800
g
j e
j b
W 0.9e5.1 × 10 j0.965b5 890g = 3.17 × 10 W
P = 5.67 × 10 −8 W m 2 K 4 0.1 5.1 × 10 14 m 2 0.965 4 800 K
+5.67 × 10 −8
4
g
This is cooler than 5 800 K by
(c)
g
14
This is larger than 3.158 × 10 22 W by
4
4
22
1.29 × 10 20 W
= 0.408%.
3.16 × 10 22 W
ANSWERS TO EVEN PROBLEMS
P20.2
0.105°C
P20.22
liquid lead at 805°C
P20.4
87.0°C
P20.24
(a) −12.0 MJ ; (b) +12.0 MJ
P20.6
The energy input to the water is 6.70 times
larger than the laser output of 40.0 kJ.
P20.26
−nR T2 − T1
P20.8
88.2 W
P20.28
(a) 567 J ; (b) 167 J
P20.10
(a) 25.8°C ; (b) no
P20.30
(a) 12.0 kJ; (b) −12.0 kJ
P20.32
42.9 kJ
P20.34
(a) 7.65 L; (b) 305 K
bm
=
Al c Al
g
+ m c c w Tc + m h c w Th
b
g
P20.12
Tf
P20.14
(a) 380 K ; (b) 206 kPa
P20.36
(a) −48.6 mJ ; (b) 16.2 kJ; (c) 16.2 kJ
P20.16
12.9 g
P20.38
(a) −4PV
i i ; (b) +4PV
i i ; (c) −9.08 kJ
P20.18
(a) all the ice melts; 40.4°C ;
(b) 8.04 g melts; 0°C
P20.40
(a) 1 300 J ; (b) 100 J ; (c) −900 J ; (d) −1 400 J
P20.20
34.0 km
P20.42
10.0 kW
m Al c Al + m c c w + m h c w
600
Heat and the First Law of Thermodynamics
P20.44
1.34 kW
P20.46
(a) 0.890 ft 2 ⋅° F ⋅ h Btu ; (b) 1.85
(c) 2.08
P20.48
(a) ~ 10 3 W ; (b) ~ −10 −1 K s
P20.50
364 K
P20.52
47.7 g
P20.54
(a) 13.0°C ; (b) −0.532 ° C s
P20.56
(a) 64.1°C ; (b) 113°C
P20.58
see the solution (a)
P20.60
10
ft 2 ⋅° F ⋅ h
;
Btu
1
PV
i i ; (c) 0
i i ; (b) 1.39 PV
2
(a) 9.31 × 10 J ; (b) −8.47 × 10
(c) 8.38 × 10 12 J
12
J;
P20.62
(a) 2 000 W ; (b) 4.47°C
P20.64
3.76 m s
P20.66
(a) 15.0 mg ; block: Q = 0; W = −5.00 J ;
∆Eint = 0 ; ∆K = −5.00 J ;
ice: Q = 0; W = 5.00 J ; ∆Eint = 5.00 J ; ∆K = 0
(b) 15.0 mg ; block: Q = 0; W = 0 ;
∆Eint = 5.00 J ; ∆K = −5.00 J ;
metal: Q = 0; W = 0 ; ∆Eint = 0 ; ∆K = 0
(c) 0.004 04°C ; moving block: Q = 0;
W = −2.50 J ; ∆Eint = 2.50 J ; ∆K = −5.00 J ;
stationary block: Q = 0; W = 2.50 J ;
∆Eint = 2.50 J ; ∆K = 0
P20.68
10.2 h
P20.70
see the solution
P20.72
800 J kg ⋅° C
21
The Kinetic Theory of Gases
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
21.1
21.2
21.3
21.4
21.5
21.6
21.7
Molecular Model of an Ideal
Gas
Molar Specific Heat of an
Ideal Gas
Adiabatic Processes for an
Ideal Gas
The Equipartition of Energy
The Boltzmann Distribution
Law
Distribution of Molecular
Speeds
Mean Free Path
Q21.1
The molecules of all different kinds collide with the walls of the
container, so molecules of all different kinds exert partial
pressures that contribute to the total pressure. The molecules
can be so small that they collide with one another relatively
rarely and each kind exerts partial pressure as if the other kinds
of molecules were absent. If the molecules collide with one
another often, the collisions exactly conserve momentum and
so do not affect the net force on the walls.
Q21.2
The helium must have the higher rms speed. According to
Equation 21.4, the gas with the smaller mass per atom must
have the higher average speed-squared and thus the higher
rms speed.
Q21.3
Yes. As soon as the gases are mixed, they come to thermal
equilibrium. Equation 21.4 predicts that the lighter helium
atoms will on average have a greater speed than the heavier
nitrogen molecules. Collisions between the different kinds of
molecules gives each kind the same average kinetic energy of
translation.
Q21.4
If the average velocity were non-zero, then the bulk sample of gas would be moving in the direction
of the average velocity. In a closed tank, this motion would result in a pressure difference within the
tank that could not be sustained.
Q21.5
The alcohol evaporates, absorbing energy from the skin to lower the skin temperature.
Q21.6
Partially evacuating the container is equivalent to letting the remaining gas expand. This means that
the gas does work, making its internal energy and hence its temperature decrease. The liquid in the
container will eventually reach thermal equilibrium with the low pressure gas. This effect of an
expanding gas decreasing in temperature is a key process in your refrigerator or air conditioner.
Q21.7
Since the volume is fixed, the density of the cooled gas cannot change, so the mean free path does
not change. The collision frequency decreases since each molecule of the gas has a lower average
speed.
Q21.8
The mean free path decreases as the density of the gas increases.
Q21.9
The volume of the balloon will decrease. The pressure inside the balloon is nearly equal to the
constant exterior atmospheric pressure. Then from PV = nRT , volume must decrease in proportion
to the absolute temperature. Call the process isobaric contraction.
601
602
The Kinetic Theory of Gases
Q21.10
The dry air is more dense. Since the air and the water vapor are at the same temperature, they have
the same kinetic energy per molecule. For a controlled experiment, the humid and dry air are at the
same pressure, so the number of molecules per unit volume must be the same for both. The water
molecule has a smaller molecular mass (18.0 u) than any of the gases that make up the air, so the
humid air must have the smaller mass per unit volume.
Q21.11
Suppose the balloon rises into air uniform in temperature. The air cannot be uniform in pressure
because the lower layers support the weight of all the air above them. The rubber in a typical balloon
is easy to stretch and stretches or contracts until interior and exterior pressures are nearly equal. So
as the balloon rises it expands. This is an isothermal expansion, with P decreasing as V increases by
the same factor in PV = nRT . If the rubber wall is very strong it will eventually contain the helium at
higher pressure than the air outside but at the same density, so that the balloon will stop rising.
More likely, the rubber will stretch and break, releasing the helium to keep rising and “boil out” of
the Earth’s atmosphere.
Q21.12
A diatomic gas has more degrees of freedom—those of vibration and rotation—than a monatomic
gas. The energy content per mole is proportional to the number of degrees of freedom.
Q21.13
(a)
Average molecular kinetic energy increases by a factor of 3.
(b)
The rms speed increases by a factor of
(c)
Average momentum change increases by
(d)
Rate of collisions increases by a factor of
(e)
Pressure increases by a factor of 3.
3.
3.
3 since the mean free path remains unchanged.
Q21.14
They can, as this possibility is not contradicted by any of our descriptions of the motion of gases. If
the vessel contains more than a few molecules, it is highly improbable that all will have the same
speed. Collisions will make their speeds scatter according to the Boltzmann distribution law.
Q21.15
Collisions between molecules are mediated by electrical interactions among their electrons. On an
atomic level, collisions of billiard balls work the same way. Collisions between gas molecules are
perfectly elastic. Collisions between macroscopic spheres can be very nearly elastic. So the hardsphere model is very good. On the other hand, an atom is not ‘solid,’ but has small-mass electrons
moving through empty space as they orbit the nucleus.
Q21.16
As a parcel of air is pushed upward, it moves into a region of lower pressure, so it expands and does
work on its surroundings. Its fund of internal energy drops, and so does its temperature. As
mentioned in the question, the low thermal conductivity of air means that very little heat will be
conducted into the now-cool parcel from the denser but warmer air below it.
Q21.17
A more massive diatomic or polyatomic molecule will generally have a lower frequency of vibration.
At room temperature, vibration has a higher probability of being excited than in a less massive
molecule. The absorption of energy into vibration shows up in higher specific heats.
SOLUTIONS TO PROBLEMS
Section 21.1
P21.1
Molecular Model of an Ideal Gas
F = Nm
P=
a
8.00 sin 45.0°− −8.00 sin 45.0°
∆v
= 500 5.00 × 10 −3 kg
∆t
30.0 s
e
F
= 1.57 N m 2 = 1.57 Pa
A
j
f
ms
= 0.943 N
Chapter 21
e5.00 × 10 j 2e4.68 × 10
−26
23
P21.2
P21.3
F=
jb
kg 300 m s
1.00 s
14.0 N
F
and P = =
= 17.6 kPa .
A 8.00 × 10 −4 m 2
603
g = 14.0 N
We first find the pressure exerted by the gas on the wall of the container.
NkT 3 N A k BT 3 RT 3 8.314 N ⋅ m mol ⋅ K 293 K
=
=
=
= 9.13 × 10 5 Pa
P=
3
−3
V
V
V
8.00 × 10 m
Thus, the force on one of the walls of the cubical container is
b
e
f
ga
je
j
F = PA = 9.13 × 10 5 Pa 4.00 × 10 −2 m 2 = 3.65 × 10 4 N .
P21.4
F
GH
I
JK
2 N mv 2
, so that
3V
2
Use Equation 21.2, P =
K av =
mv 2 3 PV
=
where N = nN A = 2 N A
2
2N
K av =
3 8.00 atm 1.013 × 10 5 Pa atm 5.00 × 10 −3 m 3
3 PV
=
2 2N A
2 2 mol 6.02 × 10 23 molecules mol
b
a
g
a
fe
je
fe
j
j
K av = 5.05 × 10 −21 J molecule
P21.5
P=
2N
KE
3V
d i
Equation 21.2
e
je
j
−3
5
3 PV
3 1.20 × 10 4.00 × 10
=
= 2.00 × 10 24 molecules
N=
2 KE
2
3.60 × 10 −22
d i
n=
P21.6
e
j
24
2.00 × 10 molecules
N
=
= 3.32 mol
N A 6.02 × 10 23 molecules mol
One mole of helium contains Avogadro’s number of molecules and has a mass of 4.00 g. Let us call m
the mass of one atom, and we have
N A m = 4.00 g mol
4.00 g mol
m=
= 6.64 × 10 −24 g molecule
or
6.02 × 10 23 molecules mol
m = 6.64 × 10 −27 kg
P21.7
a
f
f
5
4
PV 1.013 × 10 Pa 3 π 0.150 m
=
N=
k BT
1.38 × 10 −23 J K 293 K
(a)
PV = Nk BT :
(b)
K=
(c)
For helium, the atomic mass is
e
ja
3
= 3.54 × 10 23 atoms
ja f
3
3
k BT = 1.38 × 10 −23 293 J = 6.07 × 10 −21 J
2
2
e
m=
4.00 g mol
molecules mol
−27
kg molecule
6.02 × 10
m = 6.64 × 10
1
3
mv 2 = k BT :
2
2
23
∴ v rms =
3 k BT
= 1.35 km s
m
= 6.64 × 10 −24 g molecule
604
P21.8
The Kinetic Theory of Gases
v=
vO
v He
3 k BT
m
M He
4.00
1
=
=
=
32.0
8.00
MO
vO =
P21.9
1 350 m s
8.00
= 477 m s
ja
f
(a)
K=
3
3
k BT = 1.38 × 10 −23 J K 423 K = 8.76 × 10 −21 J
2
2
(b)
K=
1
2
= 8.76 × 10 −21 J
mv rms
2
e
so
v rms =
For helium,
m=
1.75 × 10 −20 J
m
(1)
4.00 g mol
6.02 × 10 23 molecules mol
= 6.64 × 10 −24 g molecule
m = 6.64 × 10 −27 kg molecule
39.9 g mol
m=
Similarly for argon,
6.02 × 10 23 molecules mol
= 6.63 × 10 −23 g molecule
m = 6.63 × 10 −26 kg molecule
Substituting in (1) above,
P21.10
(a)
we find for helium,
v rms = 1.62 km s
and for argon,
v rms = 514 m s
PV = nRT =
Nmv 2
3
The total translational kinetic energy is
3
3
PV = 3.00 × 1.013 × 10 5 5.00 × 10 −3 = 2.28 kJ
2
2
je
j
3 k T 3 RT 3a8.314fa300f
mv
=
=
=
= 6. 21 × 10
2
2
2N
2e6.02 × 10 j
F 1 N m I FG 1 J IJ = 1 J m
1 Pa = a1 PafG
H 1 Pa JK H 1 N ⋅ m K
Etrans =
(b)
2
e
B
23
A
P21.11
(a)
(b)
Nmv 2
= Etrans :
2
2
−21
J
3
For a monatomic ideal gas, Eint =
3
nRT
2
For any ideal gas, the energy of molecular translation is the same,
Etrans =
Thus, the energy per volume is
3
3
nRT = PV .
2
2
Etrans
3
=
P .
2
V
Chapter 21
Section 21.2
P21.12
Eint =
∆Eint
P21.13
P21.14
Molar Specific Heat of an Ideal Gas
3
nRT
2
3
3
= nR∆T = 3.00 mol 8.314 J mol ⋅ K 2.00 K = 74.8 J
2
2
a
fb
f
ga
We us the tabulated values for C P and C V
b
f
ga
(a)
Q = nC P ∆T = 1.00 mol 28.8 J mol ⋅ K 420 − 300 K = 3.46 kJ
(b)
∆Eint = nCV ∆T = 1.00 mol 20.4 J mol ⋅ K 120 K = 2.45 kJ
(c)
W = −Q + ∆Eint = −3.46 kJ + 2.45 kJ = −1.01 kJ
b
f
ga
The piston moves to keep pressure constant. Since V =
nRT
, then
P
nR∆T
for a constant pressure process.
P
2Q
Q
Q
=
=
Q = nC P ∆T = n CV + R ∆T so ∆T =
7nR
n CV + R n 5 R 2 + R
∆V =
b g
b g b
nR F 2Q I 2Q 2 QV
∆V =
G J= =
P H 7nR K 7 P 7 nRT
e4.40 × 10 Jja5.00 Lf
2
∆V =
= 2.52 L
7 a1.00 molfb8.314 J mol ⋅ K ga300 K f
and
g
3
Thus,
P21.15
V f = Vi + ∆V = 5.00 L + 2.52 L = 7.52 L
n = 1.00 mol, Ti = 300 K
(b)
Since V = constant, W = 0
(a)
∆Eint = Q + W = 209 J + 0 = 209 J
(c)
∆Eint = nCV ∆T = n
so
FG 3 RIJ ∆T
H2 K
∆T =
a
fb
f
2 209 J
2 ∆Eint
=
= 16.8 K
3nR
3 1.00 mol 8.314 J mol ⋅ K
a
T = Ti + ∆T = 300 K + 16.8 K = 317 K
g
605
606
P21.16
The Kinetic Theory of Gases
(a)
Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so C P =
Q = nC P ∆T =
e
FG IJ
H K
N m je100 m j
a1.00 K f =
7
7 PV
∆T
nR∆T =
2
2 T
5
2
7 1.013 × 10
Q=
2
300 K
(b)
(a)
Ug
gy
=
e
1.18 × 10 5 J
j
9.80 m s 2 2.00 m
We assume that the bulb does not expand. Then this is a constant-volume heating process.
PV
The quantity of the gas is n = i . The energy input is Q = P ∆t = nCV ∆T so
RTi
P∆t P∆tRTi
=
.
nCV
PVC
i
V
FG
H
The final temperature is T f = Ti + ∆T = Ti 1 +
The final pressure is Pf = Pi
P21.18
(a)
(b)
F
GH
FG
H
Tf
= Pi 1 +
Ti
IJ
K
P∆tR
.
PVC
i
V
IJ
K
P∆tR
.
PVC
i
V
I = 1.18 atm
J
s ⋅ mol ⋅ K 1.013 × 10 N 4π a0.05 mf 12.5 J K
F 1.00 mol I = 719 J kg ⋅ K = 0.719 kJ kg ⋅ K
5
5
C = R = b8.314 J mol ⋅ K gG
2
2
H 0.028 9 kg JK
F PV IJ
m = Mn = M G
H RT K
F 200 × 10 Pae0.350 m j I
m = b0.028 9 kg molgG
GH b8.314 J mol ⋅ K ga300 K f JJK = 0.811 kg
Pf = 1 atm 1 +
3.60 J 4 s 8.314 J ⋅ m 2 3 mol ⋅ K
3
5
V
3
(c)
118 kJ
= 6.03 × 10 3 kg
∆T =
(b)
3
U g = mgy
m=
*P21.17
3
We consider a constant volume process where no work is done.
b
f
ga
Q = mCV ∆T = 0.811 kg 0.719 kJ kg ⋅ K 700 K − 300 K = 233 kJ
(d)
7R
2
We now consider a constant pressure process where the internal energy of the gas is
increased and work is done.
FG 7R IJ ∆T = mFG 7C IJ ∆T
b g
H2K
H 5K
L7
O
Q = 0.811 kg M b0.719 kJ kg ⋅ K gPa 400 K f = 327 kJ
N5
Q
Q = mC P ∆T = m CV + R ∆T = m
V
Chapter 21
P21.19
Consider 800 cm3 of (flavored) water at 90.0 °C mixing with 200 cm3 of diatomic ideal gas at 20.0°C:
Qcold = −Q hot
d
i
dT
a f
i = −bρV g
m air c P , air T f − Ti , air = − m w c w ∆T
or
a∆T f
w
− m air c P , air
=
− Ti , air
f
mwcw
w
a90.0° C − 20.0° Cf
b ρ V gc
c
air P , air
w w
w
where we have anticipated that the final temperature of the mixture will be close to 90.0°C.
7
The molar specific heat of air is
C P, air = R
2
7 R
7
1.00 mol
So the specific heat per gram is
= 8.314 J mol ⋅ K
= 1.01 J g⋅° C
c P, air =
2 M
2
28.9 g
F
I
FG IJ b
g
G
JK
H K
H
1.20 × 10 g cm je 200 cm j b1.01 J g⋅° C ga70.0° C f
a∆T f = − e
e1.00 g cm je800 cm j b4.186 J kg⋅° Cg
a∆T f ≈ −5.05 × 10 ° C
−3
3
w
or
3
3
3
−3
w
The change of temperature for the water is between 10 −3 ° C and 10 −2 ° C .
P21.20
b
Q = nC P ∆T
g
isobaric
b
+ nCV ∆T
g
isovolumetric
In the isobaric process, V doubles so T must double, to 2Ti .
In the isovolumetric process, P triples so T changes from 2Ti to 6Ti .
Q =n
P21.21
607
FG 7 RIJ b2T − T g + nFG 5 RIJ b6T − 2T g = 13.5nRT =
H2 K
H2 K
i
i
i
i
i
13.5 PV
In the isovolumetric process A → B , W = 0 and Q = nCV ∆T = 500 J
500 J = n
FG 3R IJ bT
H2K
B
TB = 300 K +
g
− TA or TB = TA +
a
2 500 J
3nR
f
a f
= 340 K
3a1.00 molfb8.314 J mol ⋅ K g
2 500 J
In the isobaric process B → C ,
Q = nC P ∆T =
Thus,
b
g
5nR
TC − TB = −500 J .
2
a
f
2 500 J
1 000 J
= 340 K −
= 316 K
5nR
5 1.00 mol 8.314 J mol ⋅ K
(a)
TC = TB −
(b)
The work done on the gas during the isobaric process is
a
b
fb
g a
g
fb
ga
WBC = − PB ∆V = −nR TC − TB = − 1.00 mol 8.314 J mol ⋅ K 316 K − 340 J
or
WBC = +200 J
The work done on the gas in the isovolumetric process is zero, so in total
Won gas = +200 J .
f
608
*P21.22
The Kinetic Theory of Gases
(a)
At any point in the heating process, Pi = kVi and P = kV =
nRTi
Pf =
Vi
2
2Vi = 2 Pi and T f =
Pf V f
nR
z
f
(b)
The work input is W = − PdV = −
i
=
2 Pi 2Vi
= 4Ti .
nR
z
nRTi
2Vi
Vi
Vi
2
nRT V 2
VdV = − 2 i
2
Vi
b
P21.23
(a)
a
2Vi
=−
Vi
nRTi
2Vi
2
e4V
i
2
3
− Vi2 = − nRTi .
2
j
g
5
15
R 4Ti − Ti = + nRTi . The heat input
2
2
The change in internal energy, is ∆Eint = nCV ∆T = n
is Q = ∆Eint − W =
Pi
nRTi
V . At the end,
V=
Vi
Vi2
f
18
nRTi = 9 1 mol RTi .
2
The heat required to produce a temperature change is
Q = n1C 1 ∆T + n 2 C 2 ∆T
The number of molecules is N 1 + N 2 , so the number of “moles of the mixture” is n1 + n 2 and
Q = n1 + n 2 C∆T ,
b
g
n C + n 2C 2
C= 1 1
.
n1 + n 2
so
(b)
m
Q = ∑ n i C i ∆T =
i =1
F ∑ n I C∆T
GH JK
m
i =1
i
m
∑ n i Ci
i =1
m
C=
∑ ni
i =1
Section 21.3
P21.24
(a)
(b)
(c)
Adiabatic Processes for an Ideal Gas
γ
PV
i i
=
Tf
Pf V f
Ti
=
Pf V fγ
PV
i i
=
so
f
f
Tf
i
i
Ti
Since the process is adiabatic,
C P R + CV
=
,
CV
CV
b
∆Eint = nCV ∆T = 0.016 0 mol
and
i
Vi
FG P IJ FG V IJ = a20.0fa0.118f
H P KH V K
Since γ = 1.40 =
FPI
=G J
HP K
Vf
f
1γ
=
FG 1.00 IJ
H 20.0 K
57
= 0.118
= 2.35
Q=0
CV =
5
R and ∆T = 2.35Ti − Ti = 1.35Ti
2
gFGH 52 IJK b8.314 J mol ⋅ K g 1.35a300 K f =
W = −Q + ∆Eint = 0 + 135 J = +135 J .
135 J
Chapter 21
P21.25
(a)
γ
γ
PV
i i = Pf V f
Pf
(b)
FV I
=PG J
HV K
i
Ti =
FG 12.0 IJ
H 30.0 K
= 5.00 atm
f
e
1.40
= 1.39 atm
je
j
5.00 1.013 × 10 5 Pa 12.0 × 10 −3 m 3
PV
i i
=
= 365 K
nR
2.00 mol 8.314 J mol ⋅ K
Tf =
(c)
γ
i
Pf V f
=
nR
b
e
g
je
1.39 1.013 × 10 Pa 30.0 × 10 −3 m 3
5
b
2.00 mol 8.314 J mol ⋅ K
g
j=
253 K
The process is adiabatic: Q = 0
C P R + CV
5
=
, CV = R
2
CV
CV
5
= nCV ∆T = 2.00 mol
8.314 J mol ⋅ K
2
γ = 1.40 =
∆Eint
FG b
H
gIJK a253 K − 365 K f =
−4.66 kJ
W = ∆Eint − Q = −4.66 kJ − 0 = −4.66 kJ
P21.26
Vi = π
F 2.50 × 10
GH 2
−2
m
I
JK
2
0.500 m = 2.45 × 10 −4 m 3
The quantity of air we find from PV
i i = nRTi
n=
e
je
1.013 × 10 5 Pa 2.45 × 10 −4 m 3
PV
i i
=
RTi
8.314 J mol ⋅ K 300 K
b
ga
f
j
n = 9.97 × 10 −3 mol
Adiabatic compression: Pf = 101.3 kPa + 800 kPa = 901.3 kPa
(a)
γ
γ
PV
i i = Pf V f
V f = Vi
FPI
GH P JK
1γ
i
= 2. 45 × 10 −4 m 3
f
FG 101.3 IJ
H 901.3 K
57
V f = 5.15 × 10 −5 m 3
(b)
Pf V f = nRT f
Tf
FPI
=T
=T
G J
PV
P HP K
F 101.3 IJ b g =
= 300 K G
H 901.3 K
i
Pf V f
i i
i
1γ
Pf
i
i
f
F P Ib
=T G J
HP K
i
1 γ −1
g
i
f
5 7 −1
Tf
(c)
560 K
The work put into the gas in compressing it is ∆Eint = nCV ∆T
e
W = 9.97 × 10 −3 mol
W = 53.9 J
continued on next page
j 52 b8.314 J mol ⋅ K ga560 − 300f K
609
610
The Kinetic Theory of Gases
Now imagine this energy being shared with the inner wall as the gas is held at constant
volume. The pump wall has outer diameter 25.0 mm + 2.00 mm + 2.00 mm = 29.0 mm , and
volume
LMπ e14.5 × 10 mj − π e12.5 × 10 mj OP4.00 × 10
N
Q
and mass ρV = e7.86 × 10 kg m je6.79 × 10 m j = 53.3 g
2
−3
2
−3
3
−6
3
−2
m = 6.79 × 10 −6 m 3
3
The overall warming process is described by
53.9 J = nC V ∆T + mc∆T
e
j 52 b8.314 J mol ⋅ K gdT − 300 K i
+e53.3 × 10 kg jb 448 J kg ⋅ K gdT − 300 K i
53.9 J = b0.207 J K + 23.9 J K gdT − 300 K i
53.9 J = 9.97 × 10 −3 mol
ff
−3
ff
ff
T ff − 300 K = 2.24 K
P21.27
Tf
Ti
FV I
=G J
HV K
γ −1
i
=
f
FG 1 IJ
H 2K
0. 400
If Ti = 300 K , then T f = 227 K .
*P21.28
(a)
In
γ
PV
i i
=
Pf V fγ
we have Pf
FV I
=PG J
HV K
i
f
Pf
Then
Pf V f
PV
i i
=
Ti
Tf
γ
i
F 0.720 m I
=PG
H 0.240 m JK
3
i
T f = Ti
1.40
3
Pf V f
PV
i i
= 4.66 Pi
a f 13 = 1.55
= Ti 4.66
The factor of increase in temperature is the same as the factor of increase in internal energy,
Eint, f
= 1.55 .
according to Eint = nCV T . Then
Eint, i
(b)
FV I
=
=G J
In
T
PV
HV K
F 0.720 m I
2=G
H V JK
γ
Tf
Pf V f
i
Vf
i
i i
f
Vi
3
FV I
=G J
HV K
0. 40
f
3
0.720 m
= 2 1 0 . 4 = 2 2.5 = 5.66
Vf
Vf =
0.720 m 3
= 0.127 m3
5.66
i
f
γ −1
we have
611
Chapter 21
P21.29
(a)
See the diagram at the right.
(b)
PBVBγ = PC VCγ
P
B
3 Pi
γ
γ
3 PV
i i = PV
i C
Adiabatic
e j e j
= 2.19a 4.00 L f = 8.77 L
VC = 3 1 γ Vi = 3 5 7 Vi = 2.19Vi
VC
(c)
Pi
PBVB = nRTB = 3 PV
i i = 3nRTi
a
f
C
A
VC
Vi = 4 L
TB = 3Ti = 3 300 K = 900 K
(d)
After one whole cycle, TA = Ti = 300 K .
(e)
In AB, Q AB = nCV ∆V = n
V(L)
FIG. P21.29
FG 5 RIJ b3T − T g = a5.00fnRT
H2 K
i
i
i
QBC = 0 as this process is adiabatic
g a f
b
PC VC = nRTC = Pi 2.19Vi = 2.19 nRTi
so
TC = 2.19Ti
QCA = nC P ∆T = n
FG 7 RIJ bT − 2.19T g = a−4.17fnRT
H2 K
i
i
For the whole cycle,
i
a
f
a
f
Q ABCA = Q AB + QBC + QCA = 5.00 − 4.17 nRTi = 0.829 nRTi
b ∆E g
int ABCA
= 0 = Q ABCA + W ABCA
a
f
a
f
W ABCA = −Q ABCA = − 0.829 nRTi = − 0.829 PV
i i
a
fe
je
j
W ABCA = − 0.829 1.013 × 10 5 Pa 4.00 × 10 −3 m 3 = −336 J
P21.30
(a)
See the diagram at the right.
(b)
PBVBγ = PC VCγ
P
B
3Pi
Adiabatic
γ
γ
3 PV
i i = PV
i C
VC = 3 1 γ Vi = 3 5 7 Vi = 2.19Vi
(c)
PBVB = nRTB = 3 PV
i i = 3nRTi
TB = 3Ti
(d)
After one whole cycle, TA = Ti
Pi
A
C
VC
Vi
FIG. P21.30
continued on next page
af
V L
612
The Kinetic Theory of Gases
(e)
In AB, Q AB = nCV ∆T = n
FG 5 RIJ b3T − T g = a5.00fnRT
H2 K
i
i
i
QBC = 0 as this process is abiabatic
b g
F7 I
= nC ∆T = nG RJ bT − 2.19T g = −4.17nRT
H2 K
PC VC = nRTC = Pi 2.19Vi = 2.19nRTi so TC = 2.19Ti
QCA
P
i
i
i
For the whole cycle,
a
f
Q ABCA = Q AB + QBC + QCA = 5.00 − 4.17 nRTi = 0.830nRTi
b ∆E g
int ABCA
= 0 = Q ABCA + W ABCA
W ABCA = −Q ABCA = −0.830nRTi = −0.830 PV
i i
P21.31
(a)
The work done on the gas is
z
Vb
Wab = − PdV .
Va
For the isothermal process,
Wab ′ = −nRTa
z FGH
Vb′
Va
IJ
K
1
dV
V
FG V IJ = nRT lnFG V IJ .
HV K
HV K
= 5.00 molb8.314 J mol ⋅ K ga 293 K f lna10.0 f
Wab ′ = −nRTa ln
Thus, Wab′
b′
a
a
b′
FIG. P21.31
Wab′ = 28.0 kJ .
(b)
For the adiabatic process, we must first find the final temperature, Tb . Since air consists
primarily of diatomic molecules, we shall use
γ air = 1.40 and C V , air =
a
f
5 R 5 8.314
=
= 20.8 J mol ⋅ K .
2
2
Then, for the adiabatic preocess
Tb
FV I
=T G J
HV K
γ −1
a
a
a f
= 293 K 10.0
b
0. 400
= 736 K .
Thus, the work done on the gas during the adiabatic process is
b
g = b−0 + nC ∆T g = nC bT − T g
= 5.00 molb 20.8 J mol ⋅ K ga736 − 293f K = 46.0 kJ
Wab −Q + ∆Eint
or
Wab
continued on next page
ab
V
ab
V
b
a
.
613
Chapter 21
(c)
For the isothermal process, we have
Pb ′ Vb ′ = PaVa .
Thus, Pb ′ = Pa
FG V IJ = 1.00 atma10.0f =
HV K
a
10.0 atm .
b′
For the adiabatic process, we have Pb Vbγ = Pa Vaγ .
Thus, Pb = Pa
P21.32
FG V IJ
HV K
γ
a
a f
= 1.00 atm 10.0
b
1.40
= 25.1 atm .
We suppose the air plus burnt gasoline behaves like a diatomic
ideal gas. We find its final absolute pressure:
e
21.0 atm 50.0 cm3
FG 1 IJ
H 8K
Pf = 21.0 atm
j
75
75
e
= Pf 400 cm3
j
75
= 1.14 atm
Now Q = 0
d
and W = ∆Eint = nCV T f − Ti
∴W =
i
5
5
5
nRT f − nRTi = Pf V f − PV
i i
2
2
2
d
i
FIG. P21.32
F 1.013 × 10 N m I 10
5
W = 1.14 atme 400 cm j − 21.0 atme50.0 cm j G
2
H 1 atm JK e
3
3
W = −150 J
The output work is −W = +150 J
The time for this stroke is
P=
F
GH
1 1 min
4 2 500
−W
150 J
=
= 25.0 kW
∆t
6.00 × 10 −3 s
I FG 60 s IJ = 6.00 × 10
JK H 1 min K
−3
s
5
2
−6
m3 cm 3
j
614
The Kinetic Theory of Gases
Section 21.4
P21.33
The Equipartition of Energy
The heat capacity at constant volume is nC V . An ideal gas of diatomic molecules has three degrees of
freedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule,
then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisions
will set the molecule spinning only about the x and z axes.
(a)
If the molecules do not vibrate, they have five degrees of freedom. Random collisions put
1
equal amounts of energy k BT into all five kinds of motion. The average energy of one
2
5
molecule is k BT . The internal energy of the two-mole sample is
2
N
FG 5 k TIJ = nN FG 5 k TIJ = nFG 5 RIJ T = nC T .
H2 K
H2 K H2 K
B
The molar heat capacity is C V =
A
B
V
5
R and the sample’s heat capacity is
2
nCV = n
FG 5 RIJ = 2 molFG 5 b8.314 J mol ⋅ K gIJ
H2 K
H2
K
nCV = 41.6 J K
For the heat capacity at constant pressure we have
g FGH 52 R + RIJK = 72 nR = 2 molFGH 72 b8.314 J mol ⋅ K gIJK
b
nC P = n C V + R = n
nC P = 58.2 J K
(b)
In vibration with the center of mass fixed, both atoms are always moving in opposite
directions with equal speeds. Vibration adds two more degrees of freedom for two more
terms in the molecular energy, for kinetic and for elastic potential energy. We have
and
P21.34
FG 7 RIJ =
H2 K
F9 I
= nG RJ =
H2 K
nCV = n
58.2 J K
nC P
74.8 J K
FG k T IJ = f FG nRT IJ
H2K H 2K
1 F dE I 1
= G
J = fR
n H dT K 2
(1)
Eint = Nf
B
(2)
CV
int
(3)
C P = CV + R =
(4)
γ =
CP f + 2
=
CV
f
b
g
1
f +2 R
2
Chapter 21
P21.35
Rotational Kinetic Energy =
1 2
Iω
2
615
Cl
I = 2mr 2 , m = 35.0 × 1.67 × 10 −27 kg , r = 10 −10 m
I = 1.17 × 10 −45 kg ⋅ m 2
∴ K rot =
FIG. P21.35
1 2
Iω = 2.33 × 10 −21 J
2
Section 21.5
The Boltzmann Distribution Law
Section 21.6
Distribution of Molecular Speeds
P21.36
(a)
Cl
ω = 2.00 × 10 12 s −1
The ratio of the number at higher energy to the number at lower energy is e −∆E kBT where
∆E is the energy difference. Here,
a
∆E = 10.2 eV
10
JI
fFGH 1.601×eV
JK = 1.63 × 10
−19
−18
J
and at 0°C,
ja
e
f
k BT = 1.38 × 10 −23 J K 273 K = 3.77 × 10 −21 J .
Since this is much less than the excitation energy, nearly all the atoms will be in the ground
state and the number excited is
e
JI
1.63 × 10
j FGH −3.77
J = e2.70 × 10 je
JK
× 10
−18
2.70 × 10 25 exp
25
−21
−433
.
This number is much less than one, so almost all of the time no atom is excited .
(b)
At 10 000°C,
e
j
k BT = 1.38 × 10 −23 J K 10 273 K = 1.42 × 10 −19 J .
The number excited is
JI
1.63 × 10
e2.70 × 10 j expFGH −1.42
J = e2.70 × 10 je
JK
× 10
25
−18
−19
25
−11.5
= 2.70 × 10 20 .
616
P21.37
The Kinetic Theory of Gases
(a)
v av =
(b)
ev j
∑ ni v i
N
2
av
=
∑ ni vi2
P21.38
(a)
(b)
P21.39
= 54.9 m 2 s 2
N
ev j
2
so v rms =
(c)
af af af af af a f
1
1 2 + 2 3 + 3 5 + 4 7 + 3 9 + 2 12 = 6.80 m s
15
=
av
= 54.9 = 7.41 m s
v mp = 7.00 m s
Vrms, 35
Vrms, 37
3 RT
M 35
=
3 RT
M 37
=
F 37.0 g mol I
GH 35.0 g mol JK
35
The lighter atom,
12
= 1.03
Cl , moves faster.
dN v
= 0 to find
dv
In the Maxwell Boltzmann speed distribution function take
F m I
4π N G
H 2π k T JK
F
GH
32
exp −
B
mv 2
2 k BT
I F 2v − 2mv I = 0
JK GH 2k T JK
3
B
and solve for v to find the most probable speed.
Reject as solutions
v = 0 and v = ∞
Retain only
2−
Then
v mp =
mv 2
=0
k BT
2 k BT
m
P21.40
The most probable speed is v mp =
P21.41
(a)
From v av =
P21.42
6.64 × 10
−27
kg
f=
132 m s .
8 k BT
πm
we find the temperature as T =
e
ja
e
2 1.38 × 10 −23 J K 4.20 K
2 k BT
=
m
e
e
8 1.38 × 10
je
π 6.64 × 10 −27 kg 2.37 × 10 3 m s
(b)
T=
At 0°C,
1
3
2
= k BT0
mv rms0
2
2
e
8 1.38 × 10 −23 J mol ⋅ K
At the higher temperature,
b
1
m 2 v rms0
2
g
je
π 6.64 × 10 −27 kg 1.12 × 10 4 m s
2
j
=
a
j
−23
J mol ⋅ K
j
2
= 1.06 × 10 3 K
3
k BT
2
f
T = 4T0 = 4 273 K = 1 092 K = 819° C .
j
2
= 2.37 × 10 4 K
Chapter 21
*P21.43
(a)
(b)
From the Boltzmann distribution law, the number density of molecules with gravitational
energy mgy is n 0 e − mgy k BT . These are the molecules with height y, so this is the number per
volume at height y as a function of y.
b g=e
n y
n0
− mgy k BT
=e
= e − Mgy
e
N A k BT
= e − Mgy
je
RT
je
− 28 .9 × 10 −3 kg mol 9.8 m s 2 11 × 10 3 m
j b8.314 J mol⋅K ga 293 K f
= e −1.279 = 0.278
*P21.44
(a)
We calculate
z
∞
e − mgy
k BT
z
∞
dy =
e − mgy
k BT
=−
FG − mgdy IJ F − k T I
H k T K GH mg JK
k T
k T
=−
0 − 1f =
a
mg
mg
B
B
y=0
0
k BT − mgy
e
mg
∞
k BT
B
B
0
Using Table B.6 in the appendix
z
∞
ye − mgy
k BT
dy =
0
z
z
∞
Then y =
ye − mgy
0
∞
k BT
dy
=
e − mgy
k BT
dy
F k TI
bmg k T g GH mg JK
1!
2
=
2
B
.
B
bk T mg g
B
k BT mg
2
=
k BT
.
mg
0
(b)
Section 21.7
P21.45
(a)
y=
b
k BT
8.314 J 283 K s 2
RT
=
=
= 8.31 × 10 3 m
M N A g Mg mol ⋅ K 28.9 × 10 −3 kg 9.8 m
g
Mean Free Path
FG N IJ RT and N = PVN so that
RT
HN K
e1.00 × 10 ja133fa1.00fe6.02 × 10 j =
N=
a8.314fa300f
A
PV =
A
−10
(b)
A=
f=
23
3.21 × 10 12 molecules
1
1.00 m3
V
=
=
2 12
2 12
nV π d 2
Nπ d 2
3.21 × 10 12 molecules π 3.00 × 10 −10 m
e
A = 779 km
(c)
617
v
= 6.42 × 10 −4 s −1
A
je
j a 2f
2
12
618
P21.46
The Kinetic Theory of Gases
The average molecular speed is
v=
8 k BT
8 k B N AT
=
πm
π N Am
v=
8 RT
πM
b
π e 2.016 × 10
g
8 8.314 J mol ⋅ K 3.00 K
v=
−3
kg mol
j
v = 178 m s
(a)
The mean free path is
1
A=
2
2π d n V
=
1
e
j
2π 0. 200 × 10 −9 m
2
1 m3
A = 5.63 × 10 18 m
The mean free time is
A 5.63 × 10 18 m
=
= 3.17 × 10 16 s = 1.00 × 10 9 yr .
178 m s
v
(b)
Now nV is 10 6 times larger, to make A smaller by 10 6 times:
A = 5.63 × 10 12 m .
Thus,
P21.47
A
= 3.17 × 10 10 s = 1.00 × 10 3 yr .
v
From Equation 21.30, A =
For an ideal gas, nV =
k BT
Therefore, A =
P21.48
A=
2π d 2 nV
d = 3.60 × 10 −10 m
2π d 2 nV
N
P
=
V k BT
2π d 2 P
−1
1
, as required.
nV =
nV =
P
k BT
1.013 × 10 5
e1.38 × 10 ja293f
−23
= 2.51 × 10 25 m3
∴ A = 6.93 × 10 −8 m, or about 193 molecular diameters .
Chapter 21
P21.49
k BT
Using P = nV k BT , Equation 21.30 becomes A =
e1.38 × 10
−23
ja
J K 293 K
(1)
2π Pd 2
f
= 9.36 × 10 −8 m
(a)
A=
(b)
Equation (1) shows that P1 A1 = P2 A 2 . Taking P1 A1 from (a) and with A 2 = 1.00 m, we find
e
je
5
2π 1.013 × 10 Pa 3.10 × 10
P2 =
(c)
−10
j
m
2
a1.00 atmfe9.36 × 10
−8
j=
m
1.00 m
9.36 × 10 −8 atm .
For A 3 = 3.10 × 10 −10 m , we have
a1.00 atmfe9.36 × 10
P =
3
3.10 × 10
−10
−8
j=
m
m
302 atm .
Additional Problems
P21.50
(a)
n=
PV (1.013 × 10 5 Pa)( 4.20 m × 3.00 m × 2.50 m)
=
= 1.31 × 10 3 mol
( 8.314 J mol ⋅ K )( 293 K )
RT
e
je
N = nN A = 1.31 × 10 3 mol 6.02 × 10 23 molecules mol
j
N = 7.89 × 10 26 molecules
jb
e
g
(b)
m = nM = 1.31 × 10 3 mol 0.028 9 kg mol = 37.9 kg
(c)
1
3
3
m 0 v 2 = k BT = 1.38 × 10 −23 J k 293 K = 6.07 × 10 −21 J molecule
2
2
2
(d)
For one molecule,
m0 =
f
0.028 9 kg mol
M
=
= 4.80 × 10 −26 kg molecule
N A 6.02 × 10 23 molecules mol
v rms =
(e),(f)
ja
e
e
j=
2 6.07 × 10 −21 J molecule
4.80 × 10
−26
kg molecule
FG 5 RIJ T = 5 PV
H2 K 2
5
= e1.013 × 10 Paje31.5 m j =
2
503 m s
Eint = nCV T = n
Eint
5
3
7.98 MJ
619
620
P21.51
The Kinetic Theory of Gases
(a)
Pf = 100 kPa
Vf =
nRT f
=
Pf
T f = 400 K
b
ga
2.00 mol 8.314 J mol ⋅ K 400 K
3
100 × 10 Pa
a f
f = 0.066 5 m
3
= 66.5 L
a
fb
ga f
W = − P∆V = −nR∆T = −a 2.00 molfb8.314 J mol ⋅ K ga100 K f =
∆Eint = 3.50 nR∆T = 3.50 2.00 mol 8.314 J mol ⋅ K 100 K = 5.82 kJ
−1.66 kJ
Q = ∆Eint − W = 5.82 kJ + 1.66 kJ = 7.48 kJ
(b)
T f = 400 K
Pf = Pi
(c)
V f = Vi =
b
f
ga
nRTi 2.00 mol 8.314 J mol ⋅ K 300 K
=
= 0.049 9 m 3 = 49.9 L
Pi
100 × 10 3 Pa
F T I = 100 kPaFG 400 K IJ =
GH T JK
H 300 K K
f
z
W = − PdV = 0 since V = constant
133 kPa
i
∆Eint = 5.82 kJ as in part (a)
Q = ∆Eint − W = 5.82 kJ − 0 = 5.82 kJ
Pf = 120 kPa
T f = 300 K
V f = Vi
F P I = 49.9 LFG 100 kPa IJ =
GH P JK
H 120 kPa K
i
a f
∆Eint = 3.50 nR∆T = 0 since T = constant
41.6 L
f
F I = −nRT lnFG P IJ
GH JK
HP K
F 100 kPa IJ = +909 J
W = −a 2.00 molfb8.314 J mol ⋅ K ga300 K f lnG
H 120 kPa K
z
z
Vf
W = − PdV = −nRTi
Vi
Vf
dV
= −nRTi ln
V
Vi
i
i
f
Q = ∆Eint − W = 0 − 910 J = −909 J
(d)
Pf = 120 kPa
γ =
C P CV + R 3.50 R + R 4.50 9
=
=
=
=
3.50 R
3.50 7
CV
CV
FPI
F 100 kPa IJ =
=
: so
V = V G J = 49.9 L G
H 120 kPa K
HP K
F P V IJ = 300 K FG 120 kPa IJ FG 43.3 L IJ = 312 K
T =T G
H 100 kPa K H 49.9 L K
H PV K
∆E = a3.50fnR∆T = 3.50a 2.00 molfb8.314 J mol ⋅ K ga12.4 K f = 722 J
Q = 0 badiabatic process g
1γ
Pf V fγ
f
γ
PV
i i
i
f
f
f
i i
int
W = −Q + ∆Eint = 0 + 722 J = +722 J
i
i
f
79
43.3 L
Chapter 21
P21.52
(a)
The average speed v av is just the weighted average of all the speeds.
af a f a f a f a f a f a f=
a 2 + 3 + 5 + 4 + 3 + 2 + 1f
2 v + 3 2 v + 5 3 v + 4 4v + 3 5 v + 2 6 v + 1 7 v
v av =
(b)
3.65 v
First find the average of the square of the speeds,
2
v av
=
a f + 3a2vf + 5a3vf + 4a4vf + 3a5vf + 2a6 vf + 1a7 vf
2v
2
2
2
2
2
2
2+3+5+4+3+2+1
2
= 15.95 v 2 .
2
= 3.99 v .
The root-mean square speed is then v rms = v av
(c)
The most probable speed is the one that most of the particles have;
i.e., five particles have speed 3.00 v .
(d)
PV =
1
2
Nmv av
3
a
f
F
GH
2
20 m 15.95 v
mv 2
= 106
Therefore, P =
3
V
V
(e)
1
1
2
= m 15.95 v 2 = 7.98mv 2 .
mv av
2
2
e
j
z
f
(a)
PV γ = k . So, W = − PdV = − k
i
(b)
.
The average kinetic energy for each particle is
K=
P21.53
I
JK
z
f
i
dV Pf V f − PV
i i
=
γ
γ −1
V
dEint = dQ + dW and dQ = 0 for an adiabatic process.
d
i
Therefore, W = + ∆Eint = nCV T f − Ti .
To show consistency between these 2 equations, consider that γ =
Therefore,
C
1
= V.
γ −1 R
Using this, the result found in part (a) becomes
d
W = Pf V f − PV
i i
Also, for an ideal gas
i CR .
V
PV
= nT so that W = nCV T f − Ti .
R
d
i
CP
and C P − CV = R .
CV
621
622
*P21.54
The Kinetic Theory of Gases
d
W = nCV T f − Ti
(a)
i
3
−2 500 J = 1 mol 8.314 J mol ⋅ K T f − 500 K
2
d
i
T f = 300 K
γ
γ
PV
i i = Pf V f
(b)
F nRT IJ
PG
H P K
γ
i
i
b g
Pi
Pf = Pi
*P21.55
b g
γ γ −1
γ γ −1
=
Tf
I b5 3gb3 2
FT
GH T JK
f
Pf
i
γ
f
f
i
Ti
F nRT I
=P G
H P JK
Tiγ Pi1 −γ = T fγ Pf1−γ
f
T I b g
F
P =PG J
HT K
g
F 300 IJ = 1.00 atm
= 3.60 atmG
H 500 K
f
f
i
γ γ −1
i
5 2
Let the subscripts ‘1’ and ‘2’ refer to the hot and cold compartments, respectively. The pressure is
higher in the hot compartment, therefore the hot compartment expands and the cold compartment
contracts. The work done by the adiabatically expanding gas is equal and opposite to the work done
by the adiabatically compressed gas.
nR
nR
T1i − T1 f = −
T2i − T2 f
γ −1
γ −1
d
i
d
∴ T1 f + T2 f = T1i + T2 i = 800 K
Consider the adiabatic changes of the gases.
P1i V1γi = P1 f V1γf and P2i V2γi = P2 f V2γ f
∴
∴
P1i V1γi
P2 iV2γi
=
P1 f V1γf
P2 f V2γ f
F I
GH JK
V1 f
P1i
=
P2 i
V2 f
γ
, since V1i = V2i and P1 f = P2 f
F
GH
nRT1 f P1 f
nRT1i V1i
∴
=
nRT2 i V2 i
nRT2 f P2 f
∴
∴
F
GH
FT
=G
HT
T1 f
T1i
=
T2i
T2 f
T1 f
T2 f
1i
2i
I
JK
IJ
K
I
JK
γ
, using the ideal gas law
γ
, since V1i = V2i and P1 f = P2 f
1γ
=
FG 550 K IJ
H 250 K K
1 1.4
= 1.756 (2)
Solving equations (1) and (2) simultaneously gives
T1 f = 510 K, T2 f = 290 K .
i
(1)
Chapter 21
*P21.56
623
The work done by the gas on the bullet becomes its kinetic energy:
b
1
1
mv 2 = 1.1 × 10 −3 kg 120 m s
2
2
The work on the gas is
2
= 7.92 J .
1
Pf V f − PV
i i = −7.92 J.
γ −1
d
FV I .
GH V JK
L FV I O
1
So −7.92 J =
P MV G J − V P .
PQ
0. 40 M H V K
N
g
i
γ
γ
Also Pf V fγ = PV
i i
i
Pf = Pi
f
γ
i
i
f
i
f
And V f = 12 cm3 + 50 cm 0.03 cm 2 = 13.5 cm3 .
Then Pi =
P21.57
a f
LM13.5 cm c h
N
−7.92 J 0.40 10 6 cm 3 m 3
3
12 1.40
13 .5
=
− 12 cm O
QP
3
5.74 × 10 6 Pa = 56.6 atm .
The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of the
water at this depth of 50.0 meters. This is:
je
ja
F 1.00 atm IJ = 5.98 atm
PaG
H 1.013 × 10 Pa K
e
f
P = P0 + ρgh = 1.00 atm + 1.03 × 10 3 kg m 3 9.80 m s 2 50.0 m
or
P = 1.00 atm + 5.05 × 10 5
5
If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction
1
1
of the total pressure) oxygen molecules should make up only
of the total number of
5.98
5.98
molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio by
weight is then
a4.98 mol Hefb4.003 g mol Hegg =
b1.00 mol O gb2 × 15.999 g mol O gg
2
P21.58
(a)
0.623 .
2
Maxwell’s speed distribution function is
Nv
F m I
= 4π N G
H 2π k T JK
3 2
v 2 e − mv
2
2 k BT
B
With
N = 1.00 × 10 4 ,
0.032 kg
M
=
= 5.32 × 10 −26 kg
m=
N A 6.02 × 10 23
T = 500 K
and
k B = 1.38 × 10 −23 J molecule ⋅ K
e
j
this becomes N v = 1.71 × 10 −4 v 2 e
e
j
− 3 .85 × 10 −6 v 2
To the right is a plot of this function for the range
0 ≤ v ≤ 1 500 m s.
FIG. P21.58(a)
continued on next page
624
The Kinetic Theory of Gases
(b)
The most probable speed occurs where N v is a maximum.
From the graph, v mp ≈ 510 m s
(c)
v av =
8 k BT
=
πm
e
ja f =
π e5.32 × 10 j
8 1.38 × 10 −23 500
−26
575 m s
Also,
v rms =
(d)
3 k BT
=
m
e
ja f =
3 1.38 × 10 −23 500
5.32 × 10
−26
624 m s
300 m s ≤ v ≤ 600 m s
The fraction of particles in the range
z
600
N v dv
is
300
where
N = 10 4
N
and the integral of N v is read from the graph as the area under the curve.
This is approximately 4 400 and the fraction is 0.44 or 44% .
P21.59
(a)
Since pressure increases as volume decreases (and vice versa),
LM OP
N Q
1 dV
dV
< 0 and −
> 0.
V dP
dP
(b)
For an ideal gas, V =
FG
H
IJ
K
1 d nRT
nRT
and κ 1 = −
.
P
V dP P
If the compression is isothermal, T is constant and
κ1 = −
(c)
IJ
K
For an adiabatic compression, PV γ = C (where C is a constant) and
κ2 =−
(d)
FG
H
1
1
nRT
− 2 = .
V
P
P
κ1 =
γ =
FG IJ
H K
1 d C
V dP P
1γ
=
FG IJ
H K
1 1 C1 γ
1
P1 γ
= 1 γ +1 =
.
γP
V γ P b1 γ g+1 γP
1
1
=
= 0.500 atm −1
2.00 atm
P
a
f
CP
5
and for a monatomic ideal gas, γ = , so that
3
CV
κ2 =
1
=
γP
5
3
a
1
= 0.300 atm −1
2.00 atm
f
Chapter 21
P21.60
(a)
B
The speed of sound is v =
ρ
where B = −V
dP
.
dV
According to Problem 59, in an adiabatic process, this is B =
a f
a f
1
κ2
= γP .
nRT M PM
m s nM
=
=
=
where m s is the sample mass. Then, the speed of sound
V
V
V RT
RT
Also, ρ =
B
in the ideal gas is v =
ρ
b
= γP
ga
FG RT IJ =
H PM K
1.40 8.314 J mol ⋅ K 293 K
v=
γRT
.
M
f=
(b)
344 m s
0.028 9 kg mol
This nearly agrees with the 343 m/s listed in Table 17.1.
(c)
We use k B =
γ kB N AT
γ k BT
γRT
R
and M = mN A : v =
=
=
.
M
mN A
m
NA
The most probable molecular speed is
2k BT
,
m
8k BT
, and the rms speed is
πm
the average speed is
3k BT
.
m
All are somewhat larger than the speed of sound.
P21.61
n=
(a)
(b)
(c)
(d)
1.20 kg
m
=
= 41.5 mol
M 0.028 9 kg mol
Vi =
Pf
Pi
a
fb
Vf
Vi
so V f
FP I
F 400 IJ
= V G J = e0.514 m jG
H 200 K
HPK
f
i
2
3
2
= 2.06 m 3
i
e400 × 10 Paje2.06 m j = 2.38 × 10 K
nR
a41.5 molfb8.314 J mol ⋅ K g
F P I 2V = − 2 F P I V − V
W = − z PdV = −C z V dV = − G
j
G Je
3 HV K
H V JK 3
2 F 200 × 10 Pa I L
W =− G
e2.06 m j − a0.514 mf OQP = −4.80 × 10 J
3 H 0.514 m JK NM
Tf =
Pf V f
3
3
3
=
Vf
Vf
Vi
Vi
12
3
(e)
f
ga
41.5 mol 8.314 J mol ⋅ K 298 K
nRTi
=
= 0.514 m 3
Pi
200 × 10 3 Pa
=
625
a
∆Eint = nCV ∆T = 41.5 mol
i
12
i
3 3 2
3 2 Vf
i
12
i
Vi
32
f
3 2
i
3 2
fLMN 52 b8.314 J mol ⋅ K gOPQe2.38 × 10
5
3
j
− 298 K
∆Eint = 1.80 × 10 6 J
Q = ∆Eint − W = 1.80 × 10 6 J + 4.80 × 10 5 J = 2.28 × 10 6 J = 2.28 MJ
626
P21.62
The Kinetic Theory of Gases
b
1
1
1
mvi2 − mv 2f = 0.142 kg
2
2
2
The ball loses energy
g a47.2f − a42.5f
2
2
m 2 s 2 = 29.9 J
g a19.4 mf = 0.083 4 m
PV 1.013 × 10 Pae0.083 4 m j
=
n=
RT
b8.314 J mol ⋅ K ga293 K f = 3.47 mol
b
V = π 0.037 0 m
The air volume is
2
3
5
and its quantity is
3
The air absorbs energy according to
Q = nC P ∆T
∆T =
So
P21.63
af
N v v = 4π N
Note that
Thus,
Q
=
nC P 3. 47 mol
F −mv I
GH 2k T JK
F 2k T IJ
v =G
H m K
F m I
N a vf = 4π N G
H 2π k T JK
F v I ee
N avf
=G
J
N ev j H v K
F m I
GH 2π k T JK
32
B
For
B
12
v
v2e
e− v
2
2
v mp
j
2
1 − v 2 v mp
j
mp
mp
v mp
50
a f = FG 1 IJ
N e v j H 50 K
Nv v
v
3 2
B
v
v=
0. 296° C
B
mp
v
c hb8.314 J mol ⋅ K g =
2
v 2 exp
2
And
29.9 J
7
2
2
e
b g
1 − 1 50
2
= 1.09 × 10 −3
mp
The other values are computed similarly, with the
following results:
v
v mp
1
50
1
10
1
2
1
2
10
50
af
Nv v
e j
N v v mp
1.09 × 10 −3
2.69 × 10 −2
0.529
1.00
0.199
1.01 × 10 −41
1.25 × 10 −1 082
To find the last value, note:
a50f e
2 1 − 2 500
10 log 2 500 e
= 2 500 e −2 499
aln10 fb−2 499 ln10 g = 10 log 2 50010 −2 499 ln10 = 10 log 2 500 − 2 499 ln 10 = 10 −1 081.904
Chapter 21
P21.64
(a)
627
The effect of high angular speed is like the effect of a very high gravitational field on an
atmosphere. The result is:
The larger-mass molecules settle to the outside while the region at smaller r has a higher
concentration of low-mass molecules.
(b)
Consider a single kind of molecules, all of mass m. To cause the centripetal acceleration of
the molecules between r and r + dr , the pressure must increase outward according to
∑ Fr = mar . Thus,
a
f
b
ge j
PA − P + dP A = − nmA dr rω 2
where n is the number of molecules per unit volume and A is the area of any cylindrical
surface. This reduces to dP = nmω 2 rdr .
But also P = nk BT , so dP = k BTdn . Therefore, the equation becomes
z
z
n
r
dn mω 2
dn mω 2
=
=
rdr giving
rdr or
n
k BT
n
k BT 0
n
0
af
ln n
n
n0
F I
GH JK
mω 2 r 2
=
k BT 2
FG n IJ = mω r
H n K 2k T
2
ln
0
P21.65
Then,
=
0
and solving for n: n = n 0 e mr
B
2
2
=
as v av
First find v av
2
v av
2
r
2
ω 2 2 k BT
.
z
∞
1
m
.
v 2 N v dv . Let a =
N0
2 k BT
4 Nπ − 1 2 a 3 2
∞
N
0
z
v 4 e − av
2
dv
= 4 a 3 2 π −1 2
π 3 k BT
=
a
m
3
8a2
2
=
The root-mean square speed is then v rms = v av
3 k BT
.
m
To find the average speed, we have
v av
*P21.66
e
∞
4Na 3 2 π −1 2
1
=
vN v dv =
N0
N
z
j
z
∞
2
v 3 e − av dv =
0
4 a 3 2 π −1 2
=
2a 2
8 k BT
.
πm
dP
for the function implied by PV = nRT = constant , and also for the different
dV
function implied by PV γ = constant . We can use implicit differentiation:
We want to evaluate
From PV = constant
From PV γ = constant
Therefore,
The theorem is proved.
P
FG dP IJ
H dV K
FG dP IJ
H dV K
dV
dP
+V
=0
dV
dV
PγV γ −1 + V γ
FG dP IJ
H dV K
dP
=0
dV
=γ
adiabat
FG dP IJ
H dV K
isotherm
=−
isotherm
=−
adiabat
P
V
γP
V
628
P21.67
The Kinetic Theory of Gases
e
je
j
1.013 × 10 5 Pa 5.00 × 10 −3 m 3
PV
=
= 0.203 mol
RT
8.314 J mol ⋅ K 300 K
(a)
n=
(b)
TB = TA
b
f
ga
FG P IJ = 300 K FG 3.00 IJ =
H 1.00 K
HP K
B
900 K
A
TC = TB = 900 K
VC = VA
(c)
C
15.0 L
A
a
FIG. P21.67
fb
f
ga
ga
3
3
nRTA = 0.203 mol 8.314 J mol ⋅ K 300 K = 760 J
2
2
3
3
= Eint, C = nRTB = 0.203 mol 8.314 J mol ⋅ K 900 K = 2.28 kJ
2
2
Eint, A =
Eint, B
FG T IJ = 5.00 LFG 900 IJ =
H 300 K
HT K
a
(d)
A
B
C
P (atm)
1.00
3.00
1.00
fb
V(L)
5.00
5.00
15.00
f
Eint (kJ)
0.760
2.28
2.28
T(K)
300
900
900
(e)
For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For
BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the
piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas
cool without touching the piston.
(f)
For AB:
a
f
∆Eint = Eint, B − Eint, A = 2.28 − 0.760 kJ = 1.52 kJ
W= 0
Q = ∆Eint − W = 1.52 kJ
For BC:
FG V IJ
HV K
∆Eint = 0 , W = −nRTB ln
a
fb
C
B
ga
f a f
W = − 0. 203 mol 8.314 J mol ⋅ K 900 K ln 3.00 = −1.67 kJ
Q = ∆Eint − W = 1.67 kJ
For CA:
a
f
W = − P∆V = −nR∆T = −a0.203 molfb8.314 J mol ⋅ K ga −600 K f =
∆Eint = Eint, A − Eint, C = 0.760 − 2.28 kJ = −1.52 kJ
1.01 kJ
Q = ∆Eint − W = −1.52 kJ − 1.01 kJ = −2.53 kJ
(g)
We add the amounts of energy for each process to find them for the whole cycle.
Q ABCA = +1.52 kJ + 1.67 kJ − 2.53 kJ = 0.656 kJ
W ABCA = 0 − 1.67 kJ + 1.01 kJ = −0.656 kJ
b ∆E g
int ABCA
= +1.52 kJ + 0 − 1.52 kJ = 0
P21.68
(a)
00 mol I F 6.02 × 10 molecules I
b10 000 g gFGH 1.18.0
JK =
g JK GH
1.00 mol
23
Chapter 21
629
3.34 × 10 26 molecules
(b)
After one day, 10 −1 of the original molecules would remain. After two days, the fraction
would be 10 −2 , and so on. After 26 days, only 3 of the original molecules would likely
remain, and after 27 days , likely none.
(c)
The soup is this fraction of the hydrosphere:
F 10.0 kg I .
GH 1.32 × 10 kg JK
21
Therefore, today’s soup likely contains this fraction of the original molecules. The number of
original molecules likely in the pot again today is:
F 10.0 kg I 3.34 × 10
GH 1.32 × 10 kg JK e
21
P21.69
26
j
molecules = 2.53 × 10 6 molecules .
1
GmM
GM
. Since the free-fall acceleration at the surface is g = 2 , this can
mv 2 =
2
RE
RE
1
GmM
2
also be written as: mv =
= mgRE .
2
RE
(a)
For escape,
(b)
For O 2 , the mass of one molecule is
m=
0.032 0 kg mol
6.02 × 10 23 molecules mol
= 5.32 × 10 −26 kg molecule .
FG 3 k T IJ , the temperature is
H 2 K
e5.32 × 10 kgje9.80 m s je6.37 × 10 mj =
mgR
=
T=
15 k
15e1.38 × 10
J mol ⋅ K j
B
Then, if mgRE = 10
−26
6
2
E
−23
B
P21.70
(a)
1.60 × 10 4 K .
For sodium atoms (with a molar mass M = 32.0 g mol )
1
3
mv 2 = k BT
2
2
FG IJ
H K
1 M
3
v 2 = k BT
2 NA
2
v rms =
(b)
t=
d
v rms
=
3 RT
=
M
0.010 m
= 20 ms
0.510 m s
b
ge
3 8.314 J mol ⋅ K 2.40 × 10 −4 K
23.0 × 10
−3
kg
j=
0.510 m s
630
The Kinetic Theory of Gases
ANSWERS TO EVEN PROBLEMS
P21.2
17.6 kPa
P21.42
819°C
P21.4
5.05 × 10 −21 J molecule
P21.44
(a) see the solution; (b) 8.31 km
P21.6
6.64 × 10 −27 kg
P21.46
(a) 5.63 × 10 18 m; 1.00 × 10 9 yr ;
(b) 5.63 × 10 12 m; 1.00 × 10 3 yr
P21.8
477 m s
P21.48
193 molecular diameters
P21.10
(a) 2.28 kJ; (b) 6.21 × 10 −21 J
P21.50
(a) 7.89 × 10 26 molecules; (b) 37.9 kg ;
P21.12
74.8 J
P21.14
7.52 L
P21.16
(a) 118 kJ ; (b) 6.03 × 10 3 kg
P21.18
(a) 719 J kg ⋅ K ; (b) 0.811 kg ; (c) 233 kJ;
(d) 327 kJ
(c) 6.07 × 10 −21 J molecule ; (d) 503 m s;
(e) 7.98 MJ ; (f) 7.98 MJ
P21.20
13.5 PV
P21.22
(a) 4Ti ; (b) 9 1 mol RTi
P21.24
(a) 0.118 ; (b) 2.35 ; (c) 0; 135 J ; 135 J
P21.26
(a) 5.15 × 10 −5 m 3 ; (b) 560 K ; (c) 2.24 K
P21.28
(a) 1.55 ; (b) 0.127 m3
P21.30
(a) see the solution; (b) 2.19Vi ; (c) 3Ti ;
(d) Ti ; (e) −0.830 PV
i i
P21.32
25.0 kW
P21.34
see the solution
P21.36
(a) No atom, almost all the time;
(b) 2.70 × 10 20
P21.38
(a) 1.03 ; (b)
P21.40
132 m s
a
35
f
Cl
P21.52
(a) 3.65 v; (b) 3.99 v; (c) 3.00 v;
mv 2
; (e) 7.98mv 2
(d) 106
V
P21.54
(a) 300 K ; (b) 1.00 atm
P21.56
5.74 × 10 6 Pa
P21.58
(a) see the solution; (b) 5.1 × 10 2 m s;
(c) v av = 575 m s ; v rms = 624 m s ; (d) 44%
P21.60
(a) see the solution; (b) 344 m s nearly
agreeing with the tabulated value;
(c) see the solution; somewhat smaller
than each
P21.62
0.296°C
P21.64
see the solution
P21.66
see the solution
P21.68
(a) 3.34 × 10 26 molecules ; (b) during the
27th day; (c) 2.53 × 10 6 molecules
P21.70
(a) 0.510 m s ; (b) 20 ms
F
GH
I
JK
22
Heat Engines, Entropy, and the
Second Law of Thermodynamics
CHAPTER OUTLINE
22.1
22.2
22.3
22.4
22.5
22.6
22.7
22.8
Heat Engines and the
Second Law of
Thermodynamics
Heat Pumps and
Refrigerators
Reversible and Irreversible
Processes
The Carnot Engine
Gasoline and Diesel Engines
Entropy
Entropy Changes in
Irreversible Processes
Entropy on a Microscopic
Scale
ANSWERS TO QUESTIONS
Q22.1
First, the efficiency of the automobile engine cannot exceed the
Carnot efficiency: it is limited by the temperature of burning
fuel and the temperature of the environment into which the
exhaust is dumped. Second, the engine block cannot be
allowed to go over a certain temperature. Third, any practical
engine has friction, incomplete burning of fuel, and limits set
by timing and energy transfer by heat.
Q22.2
It is easier to control the temperature of a hot reservoir. If it
cools down, then heat can be added through some external
means, like an exothermic reaction. If it gets too hot, then heat
can be allowed to “escape” into the atmosphere. To maintain
the temperature of a cold reservoir, one must remove heat if
the reservoir gets too hot. Doing this requires either an “even
colder” reservoir, which you also must maintain, or an
endothermic process.
Q22.3
A higher steam temperature means that more energy can be extracted from the steam. For a
constant temperature heat sink at Tc , and steam at Th , the efficiency of the power plant goes as
Th − Tc
T
= 1 − c and is maximized for a high Th .
Th
Th
Q22.4
No. Any heat engine takes in energy by heat and must also put out energy by heat. The energy that
is dumped as exhaust into the low-temperature sink will always be thermal pollution in the outside
environment. So-called ‘steady growth’ in human energy use cannot continue.
Q22.5
No. The first law of thermodynamics is a statement about energy conservation, while the second is a
statement about stable thermal equilibrium. They are by no means mutually exclusive. For the
particular case of a cycling heat engine, the first law implies Q h = Weng + Q c , and the second law
implies Q c > 0.
Q22.6
Take an automobile as an example. According to the first law or the idea of energy conservation, it
must take in all the energy it puts out. Its energy source is chemical energy in gasoline. During the
combustion process, some of that energy goes into moving the pistons and eventually into the
mechanical motion of the car. Clearly much of the energy goes into heat, which, through the cooling
system, is dissipated into the atmosphere. Moreover, there are numerous places where friction, both
mechanical and fluid, turns mechanical energy into heat. In even the most efficient internal
combustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car.
The rest ends up as useless heat in the atmosphere.
631
632
Heat Engines, Entropy, and the Second Law of Thermodynamics
Q22.7
Suppose the ambient temperature is 20°C. A gas can be heated to the temperature of the bottom of
the pond, and allowed to cool as it blows through a turbine. The Carnot efficiency of such an engine
∆T 80
=
= 22%.
is about e c =
Th 373
Q22.8
No, because the work done to run the heat pump represents energy transferred into the house by
heat.
Q22.9
A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and
shatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in a
videotape running backwards. The free flight of a projectile is nearly reversible.
Q22.10
Below the frost line, the winter temperature is much higher than the air or surface temperature. The
earth is a huge reservoir of internal energy, but digging a lot of deep trenches is much more
expensive than setting a heat-exchanger out on a concrete pad. A heat pump can have a much
higher coefficient of performance when it is transferring energy by heat between reservoirs at close
to the same temperature.
Q22.11
(a)
When the two sides of the semiconductor are at different temperatures, an electric potential
(voltage) is generated across the material, which can drive electric current through an
external circuit. The two cups at 50°C contain the same amount of internal energy as the pair
of hot and cold cups. But no energy flows by heat through the converter bridging between
them and no voltage is generated across the semiconductors.
(b)
A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb
output or wasted energy by heat, which has nowhere to go between two cups of equally
warm water.
Q22.12
Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hot
stuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than the
entropy increase of the cold stuff. As you inflate a soft car tire at a service station, air from a tank at
high pressure expands to fill a larger volume. That air increases in entropy and the surrounding
atmosphere undergoes no significant entropy change. The brakes of your car get warm as you come
to a stop. The shoes and drums increase in entropy and nothing loses energy by heat, so nothing
decreases in entropy.
Q22.13
(a)
For an expanding ideal gas at constant temperature, ∆S =
(b)
For a reversible adiabatic expansion ∆Q = 0 , and ∆S = 0 . An ideal gas undergoing an
irreversible adiabatic expansion can have any positive value for ∆S up to the value given in
part (a).
FG IJ
H K
V
∆Q
= nR ln 2 .
V1
T
Q22.14
The rest of the Universe must have an entropy change of +8.0 J/K, or more.
Q22.15
Even at essentially constant temperature, energy must flow by heat out of the solidifying sugar into
the surroundings, to raise the entropy of the environment. The water molecules become less
ordered as they leave the liquid in the container to mix into the whole atmosphere and
hydrosphere. Thus the entropy of the surroundings increases, and the second law describes the
situation correctly.
Chapter 22
633
Q22.16
To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature.
“Remove energy from it by heat” is not such a good answer, for if you hammer on it or rub it with a
blunt file and at the same time remove energy from it by heat into a constant temperature bath, its
entropy can stay constant.
Q22.17
An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into
internal energy. It continuously creates entropy as the organized motion of the falling water turns
into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of
the energy stream to our use. Water flows spontaneously from high to low elevation and energy
spontaneously flows by heat from high to low temperature. Into the great flow of solar radiation
from Sun to Earth, living things put themselves. They live on energy flow, more than just on energy.
A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily,
before the energy inevitably is radiated from the body of the snake to a low-temperature sink (outer
space). A tree builds organized cellulose molecules and we build libraries and babies who look like
their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down
to disorder. We do not violate the second law, for we build local reductions in the entropy of one
thing within the inexorable increase in the total entropy of the Universe. Your roommate’s exercise
puts energy into the room by heat.
Q22.18
(a)
Entropy increases as the yeast dies and as energy is transferred from the hot oven into the
originally cooler dough and then from the hot bread into the surrounding air.
(b)
Entropy increases some more as you metabolize the starches, converting chemical energy
into internal energy.
Q22.19
Either statement can be considered an instructive analogy. We choose to take the first view. All
processes require energy, either as energy content or as energy input. The kinetic energy which it
possessed at its formation continues to make the Earth go around. Energy released by nuclear
reactions in the core of the Sun drives weather on the Earth and essentially all processes in the
biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how
warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous
energy input is required for life because energy tends to be continuously degraded, as heat flows
into lower-temperature sinks. The continuously increasing entropy of the Universe is the index to
energy-transfers completed.
Q22.20
The statement is not true. Although the probability is not exactly zero that this will happen, the
probability of the concentration of air in one corner of the room is very nearly zero. If some billions
of molecules are heading toward that corner just now, other billions are heading away from the
corner in their random motion. Spontaneous compression of the air would violate the second law of
thermodynamics. It would be a spontaneous departure from thermal and mechanical equilibrium.
Q22.21
Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spaces
below them. The accumulation of larger candies on top and smaller ones on the bottom implies a
small increase in order, a small decrease in one contribution to the total entropy, but the second law
is not violated. The total entropy increases as the system warms up, its increase in internal energy
coming from the work put into shaking the box and also from a bit of gravitational energy loss as the
beans settle compactly together.
634
Heat Engines, Entropy, and the Second Law of Thermodynamics
SOLUTIONS TO PROBLEMS
Section 22.1
P22.1
P22.2
Heat Engines and the Second Law of Thermodynamics
Weng
e=
(b)
Q c = Q h − Weng = 360 J − 25.0 J = 335 J
Qh
=
25.0 J
= 0.069 4 or 6.94%
360 J
(a)
Weng = Q h − Q c = 200 J
e=
Weng
Qc
=1−
Qh
Qh
(1)
= 0.300
(2)
From (2), Q c = 0.700 Q h
(3)
Solving (3) and (1) simultaneously,
we have
P22.3
(a)
Q h = 667 J and
(b)
Q c = 467 J .
(a)
We have e =
Weng
Qh
=
Qh − Qc
Qh
=1−
Qc
Qh
= 0.250
with Q c = 8 000 J, we have Q h = 10.7 kJ
(b)
Weng = Q h − Q c = 2 667 J
and from P =
*P22.4
Weng
∆t
, we have ∆t =
Weng
P
=
2 667 J
= 0.533 s .
5 000 J s
We have Q hx = 4Q hy , Weng x = 2Weng y and Q cx = 7Q cy . As well as Q hx = Weng x + Q cx and
Q hy = Weng y + Q cy . Substituting, 4Q hy = 2Weng y + 7Q cy
4Q hy = 2Weng y + 7Q hy − 7Weng y
5Weng y = 3Q hy
(b)
ey =
(a)
ex =
Weng y
Q hy
Weng x
Q hx
=
=
3
= 60.0%
5
2Weng y
4Q hy
=
a
f
2
0.600 = 0.300 = 30.0%
4
Chapter 22
*P22.5
(a)
The input energy each hour is
e7.89 × 10
3
jb
= 1.18 × 10 J h
g 601min
h
F 1 L IJ = 29.4 L h
J hjG
H 4.03 × 10 J K
9
J revolution 2 500 rev min
e
implying fuel input 1.18 × 10 9
7
Q h = Weng + Q c . For a continuous-transfer process we may divide by time to have
(b)
Q h Weng Q c
=
+
∆t
∆t
∆t
Weng Q h Q c
Useful power output =
=
−
∆t
∆t
∆t
F 7.89 × 10 J − 4.58 × 10 J I 2 500 rev 1 min = 1.38 × 10
GH revolution revolution JK 1 min 60 s
F 1 hp IJ = 185 hp
P = 1.38 × 10 WG
H 746 W K
P
1.38 × 10 J s F 1 rev I
P = τω ⇒ τ =
=
ω
b2 500 rev 60 sg GH 2π rad JK = 527 N ⋅ m
Q
4.58 × 10 J F 2 500 rev I
=
G
J = 1.91 × 10 W
∆t
revolution H 60 s K
3
=
5
eng
3
c
5
e
je
j
Q c = mL f = 15 × 10 −3 kg 1.18 × 10 4 J kg = 177 J
The heat to melt 15.0 g of Hg is
The energy absorbed to freeze 1.00 g of aluminum is
e
je
j
Q h = mL f = 10 −3 kg 3.97 × 10 5 J / kg = 397 J
Weng = Q h − Q c = 220 J
and the work output is
e=
Section 22.2
Weng
Qh
=
220 J
= 0.554 , or 55.4%
397 J
Th − Tc 933 K − 243.1 K
=
= 0.749 = 74.9%
933 K
Th
The theoretical (Carnot) efficiency is
P22.7
W
eng
(d)
P22.6
5
5
eng
(c)
3
Heat Pumps and Refrigerators
b
g
COP refrigerator =
Qc
W
(a)
If Q c = 120 J and COP = 5.00 , then W = 24.0 J
(b)
Heat expelled = Heat removed + Work done.
Q h = Q c + W = 120 J + 24 J = 144 J
635
636
P22.8
Heat Engines, Entropy, and the Second Law of Thermodynamics
Qc
Q
. Therefore, W = c .
W
3.00
The heat removed each minute is
COP = 3.00 =
b
b
gb
gb
f b
f
ga
ga
ge
QC
= 0.030 0 kg 4 186 J kg ° C 22.0° C + 0.030 0 kg 3.33 × 10 5 J kg
t
+ 0.030 0 kg 2 090 J kg ° C 20.0° C = 1.40 × 10 4 J min
or,
Qc
= 233 J s.
t
Thus, the work done per sec = P =
P22.9
j
233 J s
= 77.8 W .
3.00
(a)
FG 10.0 Btu IJ FG 1055 J IJ FG 1 h IJ FG 1 W IJ =
H h ⋅ W K H 1 Btu K H 3 600 s K H 1 J s K
(b)
Coefficient of performance for a refrigerator:
(c)
With EER 5, 5
2.93
Btu 10 000 Btu h
=
:
P
h⋅ W
aCOPf
P=
refrigerator
10 000 Btu h
5 Btu
h⋅W
= 2 000 W = 2.00 kW
a
fb
g
Cost = e3.00 × 10 kWhjb0.100 $ kWhg = $300
P ∆t = 2.00 kW 1 500 h = 3.00 × 10 3 kWh
Energy purchased is
3
With EER 10, 10
Btu 10 000 Btu h
=
:
P
h⋅ W
P=
10 000 Btu h
10 Btu
h⋅W
= 1 000 W = 1.00 kW
a
fb g
Cost = e1.50 × 10 kWhjb0.100 $ kWhg = $150
P ∆t = 1.00 kW 1 500 h = 1.50 × 10 3 kWh
Energy purchased is
3
Thus, the cost for air conditioning is
Section 22.3
half as much with EER 10
Reversible and Irreversible Processes
No problems in this section
Section 22.4
P22.10
The Carnot Engine
Weng
T
and
When e = e c , 1 − c =
Th
Qh
(a)
Qh
Weng
∆t
Qh
∆t
=1−
Tc
Th
FH IK ∆t e1.50 × 10 Wjb3 600 sg
=
=
Weng
∆t
5
T
1−
1 − Tc
h
293
773
Q h = 8.69 × 10 8 J = 869 MJ
(b)
Qc = Qh −
F W I ∆t = 8.69 × 10 − e1.50 × 10 jb3 600g = 3.30 × 10
GH ∆t JK
eng
8
5
8
J = 330 MJ
Chapter 22
P22.11
Tc = 703 K
P22.13
∆T 1 440
=
= 67.2%
Th 2 143
(a)
ec =
(b)
Q h = 1.40 × 10 5 J , Weng = 0.420 Q h
P=
P22.12
Th = 2 143 K
Weng
∆t
=
5.88 × 10 4 J
= 58.8 kW
1s
∆T 120 K
=
= 0.253
Th 473 K
The Carnot efficiency of the engine is
ec =
At 20.0% of this maximum efficiency,
e = 0.200 0.253 = 0.050 6
From the definition of efficiency
Weng = Q h e
and
Qh =
a
Isothermal expansion at
Th = 523 K
Isothermal compression at
Tc = 323 K
f
Weng
e
=
10.0 kJ
= 197 kJ
0.050 6
Gas absorbs 1 200 J during expansion.
(a)
(b)
P22.14
*P22.15
FG T IJ = 1 200 JFG 323 IJ = 741 J
H 523 K
HT K
= Q − Q = b1 200 − 741g J = 459 J
c
Qc = Qh
Weng
h
h
c
Tc
Th
We use
ec = 1 −
as
0.300 = 1 −
From which,
Th = 819 K = 546° C
573 K
Th
The efficiency is
ec = 1 −
Then
Tc
=
Th
Qh
∆t
(a)
Qc
∆t
Qh
∆t
=
a
a
f
f
Q c Th
273 + 100 K
= 15. 4 W
= 19.6 W
∆t Tc
273 + 20 K
Q h = Weng + Q c
The useful power output is
(b)
Qc
Tc
=1−
Th
Qh
Qh =
F Q I ∆t = mL
GH ∆t JK
h
V
Weng
∆t
m=
=
Qh
∆t
−
Qc
∆t
= 19.6 W − 15.4 W = 4.20 W
gFGH
I
JK
Q h ∆t
3 600 s
= 19.6 J s
= 3.12 × 10 −2 kg
∆t LV
2.26 × 10 6 J kg
b
637
638
P22.16
Heat Engines, Entropy, and the Second Law of Thermodynamics
a
a
f
f
The Carnot summer efficiency is
e c ,s = 1 −
273 + 20 K
Tc
=1−
= 0.530
Th
273 + 350 K
And in winter,
ec ,w = 1 −
283
= 0.546
623
FG 0.546 IJ = 0.330 or 33.0%
H 0.530 K
F P V I = F PV I .
In an adiabatic process, P V = PV . Also, G
H T JK GH T JK
F P Ib g
Dividing the second equation by the first yields T = T G J
HPK
Then the actual winter efficiency is
0.320
γ
P22.17
(a)
f
γ
f
f
γ
i i
γ
f
i i
f
i
f
Since γ =
γ −1 γ
.
i
γ −1 2
5
= = 0.400 and we have
for Argon,
γ
5
3
b
T f = 1 073 K
(b)
i
f
× 10 Pa I
gFGH 1300
J
.50 × 10 Pa K
3
0. 400
6
= 564 K .
∆Eint = nCV ∆T = Q − Weng = 0 − Weng , so Weng = −nCV ∆T ,
and the power output is
P=
=
Weng
t
=
−nCV ∆T
or
t
b−80.0 kg ge
1.00 mol
0 .039 9 kg
jc hb8.314 J mol ⋅ K gb564 − 1 073gK
3
2
60.0 s
5
P = 2.12 × 10 W = 212 kW
P22.18
Tc
564 K
=1−
= 0.475 or 47.5%
1 073 K
Th
(c)
eC = 1 −
(a)
emax = 1 −
(b)
P=
Weng
∆t
Tc
278
=1−
= 5.12 × 10 −2 = 5.12%
Th
293
= 75.0 × 10 6 J s
From e =
(c)
jb
e
g
Weng = 75.0 × 10 6 J s 3 600 s h = 2.70 × 10 11 J h
Therefore,
Weng
Qh
we find
Qh =
Weng
e
=
2.70 × 10 11 J h
5.12 × 10 −2
= 5.27 × 10 12 J h = 5.27 TJ h
As fossil-fuel prices rise, this way to use solar energy will become a good buy.
Chapter 22
*P22.19
(a)
e=
Weng1 + Weng2
e1 Q1 h + e 2 Q 2 h
Q h1
=
Q h1
Now Q 2 h = Q1 c = Q1 h − Weng 1 = Q h1 − e1Q1 h .
So e =
(b)
b
e 1 Q 1 h + e 2 Q 1 h − e1 Q 1 h
Q1 h
e = e 1 + e 2 − e1 e 2 = 1 −
g=
e1 + e 2 − e 1 e 2 .
FG
H
Ti
T
T
+1− c − 1− i
Th
Ti
Th
IJ FG 1 − T IJ = 2 − T
KH T K T
c
i
i
h
−
Tc
T T T
T
−1+ i + c − c = 1− c
Ti
Th Ti Th
Th
The combination of reversible engines is itself a reversible engine so it has the Carnot
efficiency.
(c)
With Weng2 = Weng1 , e =
1−
FG
H
Tc
T
= 2 1− i
Th
Th
Weng1 + Weng2
Q1 h
IJ
K
2T
Tc
=1− i
Th
Th
2Ti = Th + Tc
0−
Ti =
(d)
b
1
Th + Tc
2
e1 = e 2 = 1 −
g
Ti
T
=1− c
Th
Ti
Ti2 = Tc Th
b g
Ti = ThTc
P22.20
12
The work output is Weng =
We are told e =
0.200 =
and
Weng
Qh
b
b
5.00 m s
1
mt
Qh
2
g
b
b
5.00 m s
1
Substitute Q h = m t
2
0.200
Then,
2
6.50 m s
300 K 1
= mt
eC = 1 −
2
Th
Qh
300 K
= 0.200
1−
Th
F
GG
H
1
2
1
2
g
g
2
.
b
g
m b5.00 m sg
t
.
2
m t 6.50 m s
300 K
= 0.338
Th
300 K
Th =
= 453 K
0.662
1−
g
1
2
m train 5.00 m s .
2
2
2
I
JJ
K
=
2Weng1
Q1 h
= 2 e1
639
640
P22.21
Heat Engines, Entropy, and the Second Law of Thermodynamics
For the Carnot engine, e c = 1 −
Tc
300 K
=1−
= 0.600 .
750 K
Th
Weng
Also,
ec =
so
Qh =
and
Q c = Q h − Weng = 250 J − 150 J = 100 J .
Qh
.
Weng
ec
=
150 J
= 250 J .
0.600
FIG. P22.21
(a)
Qh =
Weng
eS
=
150 J
= 214 J
0.700
Q c = Q h − Weng = 214 J − 150 J = 64.3 J
(b)
Q h, net = 214 J − 250 J = −35.7 J
Q c, net = 64.3 J − 100 J = −35.7 J
The net flow of energy by heat from the cold to the hot
reservoir without work input, is impossible.
(c)
(d)
Weng
For engine S:
Q c = Q h − Weng =
so
Weng =
and
Q h = Q c + Weng = 233 J + 100 J = 333 J .
Qc
1
eS
−1
=
eS
FIG. P22.21(b)
− Weng .
100 J
= 233 J .
1
0 .700 − 1
Q h, net = 333 J − 250 J = 83.3 J
Wnet = 233 J − 150 J = 83.3 J
Q c,net = 0
The output of 83.3 J of energy from the heat engine
by work in a cyclic process without any exhaust by
heat is impossible.
FIG. P22.21(d)
(e)
Both engines operate in cycles, so
∆SS = ∆SCarnot = 0 .
For the reservoirs,
∆S h = −
Qh
Th
Thus, ∆Stotal = ∆SS + ∆SCarnot + ∆S h + ∆S c = 0 + 0 −
A decrease in total entropy is impossible.
and ∆S c = +
Qc
Tc
.
83.3 J
0
+
= −0.111 J K .
750 K 300 K
Chapter 22
P22.22
(a)
641
First, consider the adiabatic process D → A :
FG V IJ = 1 400 kPaFG 10.0 L IJ
H 15.0 L K
HV K
FG nRT IJ V = FG nRT IJ V
HV K HV K
F V IJ = 720 K FG 10.0 IJ =
T =T G
H 15.0 K
HV K
γ
PD VDγ = PA VAγ so
PD = PA
D
Also
D
53
A
γ
γ
A
D
A
A
γ −1
or
D
A
= 712 kPa .
D
23
A
549 K .
D
Now, consider the isothermal process C → D : TC = TD = 549 K .
FG V IJ = LMP FG V IJ OPFG V IJ = P V
H V K MN H V K PQH V K V V
1 400 kPaa10.0 L f
=
= 445 kPa
24.0 La15.0 L f
γ
PC = PD
D
C
A
A
D
D
C
A
C
γ
A
γ −1
D
53
PC
23
Next, consider the adiabatic process B → C : PBVBγ = PC VCγ .
But, PC =
PA VAγ
VC VDγ −1
FG V IJ V = F P V I V which reduces to V = V V
V
H V K GH V V JK
F V IJ = 1 400 kPaFG 10.0 L IJ = 875 kPa .
Finally, P = P G
H 16.0 L K
HV K
A
Hence, PA
B
B
A
γ
B
A
C
γ
A
γ −1
D
γ
B
C
A C
15.0 L
D
A
B
State
A
B
C
D
(b)
FG V IJ .
HV K
10.0 La 24.0 L f
=
= 16.0 L
from above. Also considering the isothermal process, PB = PA
P(kPa)
1 400
875
445
712
For the isothermal process A → B :
so Q = −W = nRT ln
V(L)
10.0
16.0
24.0
15.0
T(K)
720
720
549
549
∆Eint = nCV ∆T = 0
FG V IJ = 2.34 molb8.314 J mol ⋅ K ga720 K f lnFG 16.0 IJ =
H 10.0 K
HV K
B
+6.58 kJ .
A
For the adiabatic process B → C :
b
LM 3 b8.314 J mol ⋅ K gOPa549 − 720f K =
N2
Q
= 0 + a −4.98 kJf = −4.98 kJ .
g
∆Eint = nCV TC − TB = 2.34 mol
and W = −Q + ∆Eint
continued on next page
Q= 0
−4.98 kJ
A
B
.
642
Heat Engines, Entropy, and the Second Law of Thermodynamics
∆Eint = nCV ∆T = 0
For the isothermal process C → D :
and Q = −W = nRT ln
FG V IJ = 2.34 molb8.314 J mol ⋅ K ga549 K f lnFG 15.0 IJ =
H 24.0 K
HV K
D
−5.02 kJ .
C
Finally, for the adiabatic process D → A :
b
g
∆Eint = nCV TA − TD = 2.34 mol
Q= 0
LM 3 b8.314 J mol ⋅ K gOPa720 − 549f K =
N2
Q
+4.98 kJ
and W = −Q + ∆Eint = 0 + 4.98 kJ = +4.98 kJ .
Process
A→B
B→C
C→D
D→ A
ABCDA
Q(kJ)
+6.58
0
–5.02
0
+1.56
W(kJ)
–6.58
–4.98
+5.02
+4.98
–1.56
∆Eint (kJ)
0
–4.98
0
+4.98
0
The work done by the engine is the negative of the work input. The output work Weng is
given by the work column in the table with all signs reversed.
(c)
e=
Weng
Qh
ec = 1 −
P22.23
aCOPf
refrig
P22.24
aCOPf
heat pump
P22.25
(a)
=
=
−W ABCD 1.56 kJ
=
= 0.237 or 23.7%
6.58 kJ
Q A→B
Tc
549
=1−
= 0.237 or 23.7%
720
Th
Tc
270
=
= 9.00
∆T 30.0
=
Qc + W
W
=
Th 295
=
= 11.8
∆T
25
For a complete cycle, ∆Eint = 0 and
W = Qh − Q c = Q c
LM bQ g OP
MN Q − 1PQ .
h
c
We have already shown that for a Carnot cycle (and only for a Carnot cycle)
Therefore,
(b)
W = Qc
LM T − T OP
N T Q
h
c
COP =
Qc
=
Th
.
Tc
.
c
We have the definition of the coefficient of performance for a refrigerator,
Using the result from part (a), this becomes
Qh
Tc
.
Th − Tc
COP =
Qc
W
.
Chapter 22
P22.26
643
COP = 0.100COPCarnot cycle
Qh
or
W
Qh
W
FQ I
F
I
1
= 0.100G
GH W JK
H Carnot efficiency JK
F T IJ = 0.100FG 293 K IJ = 1.17
= 0.100G
H 293 K − 268 K K
HT −T K
h
= 0.100
Carnot cycle
h
h
c
FIG. P22.26
Thus, 1.17 joules of energy enter the room by heat for each joule of work done.
P22.27
P22.28
Qc
Tc
4.00
=
= 0.013 8 =
∆T 289
W
∴ W = 72.2 J per 1 J energy removed by heat.
aCOPf
Carnot refrig
=
A Carnot refrigerator runs on minimum power.
Q t Q t
Q
Q
For it: h = c so h = c .
Th
Tc
Th
Tc
Solving part (b) first:
P22.29
FG IJ b
H K
IJ e
K
gFGH
jFGH
IJ
K
(b)
Q h Q c Th
298 K
1h
=
= 8.00 MJ h
= 8.73 × 10 6 J h
= 2. 43 kW
273 K
3 600 s
t
t Tc
(a)
8.00 × 10 6 J h
W Qh Qc
=
−
= 2.43 kW −
= 204 W
t
t
t
3 600 s h
e=
W
= 0.350
Qh
W = 0.350Q h
Qh = W + Qc
Q c = 0.650Q h
Q c 0.650Q h
=
= 1.86
COP refrigerator =
W 0.350Q h
b
*P22.30
g
To have the same efficiencies as engines, 1 −
between reservoirs with the same ratio
becomes
Thp
Thp − Tcp
=
Tcp
Thp
Tcp
Thp
=
=1−
Tcr
the pump and refrigerator must operate
Thr
Tcr
, which we define as r. Now COPp = 1.50COPr
Thr
Thp
3 Tcr
2
3r
2
3 rThr
or
=
,r= .
,
=
2 Thr − Tcr
3
Thp − rThp 2 Thr − rThr 1 − r 1 − r
(a)
COPr =
2
r
= 3
1−r 1−
(b)
COPp =
1
1
=
= 3.00
1 − r 1 − 32
(c)
e=1−r =1−
2
3
= 2.00
2
= 33.3%
3
644
Heat Engines, Entropy, and the Second Law of Thermodynamics
Section 22.5
P22.31
(a)
Gasoline and Diesel Engines
γ
γ
PV
i i = Pf V f
FV I
F 50.0 cm I
= P G J = e3.00 × 10 PajG
H 300 cm JK
HV K
γ
Pf
(b)
i
3
6
i
z
Vi
= 244 kPa
3
f
W = PdV
1.40
P = Pi
Vi
FG V IJ
HVK
i
γ
Integrating,
F 1 IJ PV LM1 − FG V IJ
W =G
H γ − 1 K MN H V K
γ −1
i
i i
f
OP
PQ = a2.50fe3.00 × 10 Paje5.00 × 10
6
−5
= 192 J
P22.32
Compression ratio = 6.00 , γ = 1.40
FG V IJ
HV K
F 1 IJ
e=1−G
H 6.00 K
γ −1
(a)
2
Efficiency of an Otto-engine e = 1 −
1
(b)
P22.33
0 . 400
= 51. 2% .
If actual efficiency e ′ = 15.0% losses in system are e − e ′ = 36.2% .
eOtto = 1 −
1
bV V g
1
γ −1
=1−
2
1
a6.20fb7 5−1g
=1−
1
a6.20f
0. 400
eOtto = 0.518
We have assumed the fuel-air mixture to behave like a diatomic gas.
Now e =
Weng
Qh
=
Weng t
Qh t
746 W 1 hp
Q h Weng t
=
= 102 hp
t
e
0.518
Qh
= 146 kW
t
Q h = Weng + Q c
Qc
t
Qc
t
=
Q h Weng
−
t
t
F 746 W I =
GH 1 hp JK
= 146 × 10 3 W − 102 hp
70.8 kW
L F 50.0 cm I OP
m jM1 − G
MN H 300 cm JK PQ
3
3
3
0. 400
Chapter 22
P22.34
(a), (b) The quantity of gas is
n=
e
je
j
100 × 10 3 Pa 500 × 10 −6 m 3
PA VA
= 0.020 5 mol
=
RTA
8.314 J mol ⋅ K 293 K
Eint, A =
b
f
ga
5
5
5
nRTA = PA VA = 100 × 10 3 Pa 500 × 10 −6 m 3 = 125 J
2
2
2
e
je
FG V IJ = e100 × 10 Paja8.00f
HV K
γ
In process AB, PB = PA
e
3
A
j
1.40
= 1.84 × 10 6 Pa
B
je
j
1.84 × 10 6 Pa 500 × 10 −6 m3 8.00
PBVB
= 673 K
=
TB =
nR
0.020 5 mol 8.314 J mol ⋅ K
Eint, B =
so
b
b
gb
g
gb
f
ga
5
5
nRTB = 0.020 5 mol 8.314 J mol ⋅ K 673 K = 287 J
2
2
∆Eint, AB = 287 J − 125 J = 162 J = Q − Wout = 0 − Wout
W AB = −162 J
Process BC takes us to:
b
gb
gb
g
0.020 5 mol 8.314 J mol ⋅ K 1 023 K
nRTC
=
= 2.79 × 10 6 Pa
VC
62.5 × 10 −6 m 3
5
5
Eint, C = nRTC = 0.020 5 mol 8.314 J mol ⋅ K 1 023 K = 436 J
2
2
Eint, BC = 436 J − 287 J = 149 J = Q − Wout = Q − 0
PC =
b
gb
gb
g
QBC = 149 J
In process CD:
FG V IJ = e2.79 × 10 PajFG 1 IJ = 1.52 × 10 Pa
H 8.00 K
HV K
e1.52 × 10 Paje500 × 10 m j = 445 K
P V
=
=
nR
b0.020 5 molgb8.314 J mol ⋅ K g
5
5
= nRT = b0.020 5 molgb8.314 J mol ⋅ K ga 445 K f =
2
2
γ
PD = PC
6
C
5
D
5
TD
1.40
−6
3
D D
Eint, D
D
190 J
∆Eint, CD = 190 J − 436 J = −246 J = Q − Wout = 0 − Wout
WCD = 246 J
and
∆Eint, DA = Eint, A − Eint, D = 125 J − 190 J = −65.0 J = Q − Wout = Q − 0
QDA = −65.0 J
continued on next page
645
646
Heat Engines, Entropy, and the Second Law of Thermodynamics
For the entire cycle, ∆Eint, net = 162 J + 149 − 246 − 65.0 = 0 . The net work is
Weng = −162 J + 0 + 246 J + 0 = 84.3 J
Q net = 0 + 149 J + 0 − 65.0 J = 84.3 J
The tables look like:
State
A
B
C
D
A
T(K)
293
673
1 023
445
293
P(kPa)
100
1 840
2 790
152
100
V(cm3 )
500
62.5
62.5
500
500
Process
AB
BC
CD
DA
ABCDA
Q(J)
0
149
0
–65.0
84.3
output W(J)
–162
0
246
0
84.3
∆Eint (J)
162
149
–246
–65.0
0
(c)
The input energy is Q h = 149 J , the waste is Q c = 65.0 J , and Weng = 84.3 J .
(d)
The efficiency is: e =
(e)
Let f represent the angular speed of the crankshaft. Then
Weng
Qh
=
84.3 J
= 0.565 .
149 J
obtain work in the amount of 84.3 J/cycle:
1 000 J s =
f=
Section 22.6
P22.35
Eint (J)
125
287
436
190
125
FG f IJ b84.3 J cycleg
H 2K
2 000 J s
= 23.7 rev s = 1.42 × 10 3 rev min
84.3 J cycle
Entropy
For a freezing process,
∆S =
b
ge
f
is the frequency at which we
2
j
5
∆Q − 0.500 kg 3.33 × 10 J kg
=
= −610 J K .
273 K
T
Chapter 22
P22.36
647
At a constant temperature of 4.20 K,
20.5 kJ kg
Lv
∆Q
=
=
T
4.20 K
4.20 K
∆S = 4.88 kJ kg ⋅ K
∆S =
FG IJ
H K
F 353 IJ = 46.6 cal K =
∆S = 250 g b1.00 cal g⋅° C g lnG
H 293 K
z z
f
P22.37
∆S =
i
*P22.38
(a)
T
f
Tf
dQ
mcdT
=
= mc ln
T
T
Ti
T
i
195 J K
The process is isobaric because it takes place under constant atmospheric pressure. As
described by Newton’s third law, the stewing syrup must exert the same force on the air as
the air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal
(T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), or
isentropic (entropy increases). It could be made as nearly reversible as you wish, by not
using a kitchen stove but a heater kept always just incrementally higher in temperature
than the syrup. The process would then also be eternal, and impractical for food production.
(b)
The final temperature is
220° F = 212° F + 8° F = 100° C + 8° F
FG 100 − 0° C IJ = 104° C .
H 212 − 32° F K
For the mixture,
b
ga
Q = m1 c 1 ∆T + m 2 c 2 ∆T = 900 g 1 cal g⋅° C + 930 g 0.299 cal g⋅° C 104.4° C − 23° C
= 9.59 × 10 4 cal = 4.02 × 10 5 J
(c)
Consider the reversible heating process described in part (a):
z zb
f
f
g
T
b
g
T
T
F 4.186 J IJ FG 1° C IJ lnFG 273 + 104 IJ
= 900a1f + 930a0.299f bcal ° C gG
H 1 cal K H 1 K K H 273 + 23 K
= b 4 930 J K g0.243 = 1.20 × 10 J K
∆S =
i
dQ
=
T
m1 c 1 + m 2 c 2 dT
= m1 c 1 + m 2 c 2 ln
i
3
f
i
f
648
*P22.39
Heat Engines, Entropy, and the Second Law of Thermodynamics
We take data from the description of Figure 20.2 in section 20.3, and we assume a constant specific
heat for each phase. As the ice is warmed from –12°C to 0°C, its entropy increases by
z
f
∆S =
i
z
z
273 K
273 K
mc ice dT
dQ
273 K
=
= mc ice T −1 dT = mc ice ln T 261 K
T
T
261 K
261 K
b
f
ga
b
IJ IJ
gFGH FGH 273
261 K K
∆S = 0.027 0 kg 2 090 J kg ⋅° C ln 273 K − ln 261 K = 0.027 0 kg 2 090 J kg⋅° C ln
∆S = 2.54 J K
As the ice melts its entropy change is
e
j
5
Q mL f 0.027 0 kg 3.33 × 10 J kg
∆S = =
=
= 32.9 J K
T
T
273 K
As liquid water warms from 273 K to 373 K,
z
f
∆S =
i
FG T IJ = 0.027 0 kgb4 186 J kg⋅° Cg lnFG 373 IJ = 35.3 J K
H 273 K
HT K
mc liquid dT
f
= mc liquid ln
T
i
As the water boils and the steam warms,
∆S =
∆S =
FG IJ
H K
Tf
mL v
+ mc steam ln
T
Ti
e
0.027 0 kg 2.26 × 10 6 J kg
373 K
j + 0.027 0 kgb2 010 J kg⋅° Cg lnFG 388 IJ = 164 J K + 2.14 J K
H 373 K
The total entropy change is
a2.54 + 32.9 + 35.3 + 164 + 2.14f J K =
236 J K .
We could equally well have taken the values for specific heats and latent heats from Tables 20.1 and
20.2. For steam at constant pressure, the molar specific heat in Table 21.2 implies a specific heat of
1 mol
= 1 970 J kg ⋅ K , nearly agreeing with 2 010 J kg ⋅ K .
35.4 J mol ⋅ K
0.018 kg
b
Section 22.7
gFGH
I
JK
Entropy Changes in Irreversible Processes
F
GH
1 000 1 000
Q 2 Q1
−
=
−
T2 T1
290 5 700
I
JK
P22.40
∆S =
J K = 3.27 J K
P22.41
The car ends up in the same thermodynamic state as it started, so it undergoes zero changes in
entropy. The original kinetic energy of the car is transferred by heat to the surrounding air, adding
to the internal energy of the air. Its change in entropy is
∆S =
1
2
mv 2
T
=
a f
750 20.0
293
2
J K = 1.02 kJ K .
Chapter 22
P22.42
649
c iron = 448 J kg⋅° C ; c water = 4 186 J kg⋅° C
b
gd
i b
gb
gd
Qcold = −Q hot :
4.00 kg 4 186 J kg⋅° C T f − 10.0° C = − 1.00 kg 448 J kg⋅° C T f − 900° C
which yields
T f = 33.2° C = 306.2 K
∆S =
z
306.2 K
283 K
i
z
c water m water dT 306.2 K c iron m iron dT
+
T
T
1 173 K
FG 306.2 IJ + c m lnFG 306.2 IJ
H 283 K
H 1 173 K
∆S = b 4 186 J kg ⋅ K gb 4.00 kg gb0.078 8g + b 448 J kg ⋅ K gb1.00 kg ga −1.34f
∆S = c water m water ln
iron
iron
∆S = 718 J K
P22.43
Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy
that leaves my body by heat into the room-temperature surroundings. My rate of energy output is
equal to my metabolic rate,
2 500 kcal d =
FG
H
IJ
K
2 500 × 10 3 cal 4.186 J
= 120 W .
86 400 s
1 cal
My body is in steady state, changing little in entropy, as the environment increases in entropy at the
rate
∆S Q T Q ∆t 120 W
=
=
=
= 0. 4 W K ~ 1 W K .
293 K
∆t
∆t
T
When using powerful appliances or an automobile, my personal contribution to entropy production
is much greater than the above estimate, based only on metabolism.
P22.44
b
V=
(b)
∆Eint = nCV ∆T =
(c)
W =0
(d)
∆Sargon =
b
so
ga
f
ge
j
F 40.0 gm I L 3 b8.314 J mol ⋅ K gOa−200° Cf =
GH 39.9 g mol JK MN 2
PQ
i
∆S bath =
−2.50 kJ
Q = ∆Eint = −2.50 kJ
FG IJ
H K
F 40.0 g I L 3 b8.314 J mol ⋅ K gO lnFG 273 IJ =
=G
PQ H 473 K
H 39.9 g mol JK MN 2
z
f
(e)
gb
40.0 g 8.314 J mol ⋅ K 473 K
nRTi
= 39.4 × 10 −3 m3 = 39.4 L
=
3
Pi
39.9 g mol 100 × 10 Pa
(a)
Tf
dQ
= nCV ln
T
Ti
−6.87 J K
2.50 kJ
= +9.16 J K
273 K
The total change in entropy is
∆Stotal = ∆Sargon + ∆S bath = −6.87 J K + 9.16 J K = +2.29 J K
∆Stotal > 0 for this irreversible process.
650
P22.45
Heat Engines, Entropy, and the Second Law of Thermodynamics
∆S = nR ln
F V I = R ln 2 =
GH V JK
f
5.76 J K
i
There is no change in temperature .
FIG. P22.45
P22.46
F V I = b0.044 0ga2fR ln 2
GH V JK
∆S = 0.088 0a8.314f ln 2 = 0.507 J K
∆S = nR ln
f
i
FIG. P22.46
P22.47
For any infinitesimal step in a process on an ideal gas,
dEint = dQ + dW :
dQ = dEint − dW = nCV dT + PdV = nCV dT +
and
dQ
dT
dV
= nCV
+ nR
T
T
V
If the whole process is reversible,
∆S =
z
f
i
Also, from the ideal gas law,
a
∆S = 1.00 mol
Tf
Ti
=
dQr
=
T
z FGH
f
i
nCV
IJ
K
nRTdV
V
FG IJ
H K
Pf V f
PV
i i
040 0g I
F 0.040 0 I
+ a1.00 molfb8.314 J mol ⋅ K g lnG
fLMN 32 b8.314 J mol ⋅ K gOPQ lnFGH aa12..0000fbfb00..025
J
0g K
H 0.025 0 JK
= 18.4 J K
P22.48
F T I + nR lnF V I
GH T JK
GH V JK
L5
O F 2P ⋅ 2V IJ + a1.00 molfb8.314 J mol ⋅ K g lnFG 2V IJ
= a1.00 molfM b8.314 J mol ⋅ K gP lnG
HVK
N2
Q H PV K
∆S = nC V ln
FG IJ
H K
Tf
Vf
dT
dV
+ nR
= nCV ln
+ nR ln
T
V
Ti
Vi
f
f
i
i
∆S = 34.6 J K
Chapter 22
Section 22.8
P22.49
P22.50
P22.51
651
Entropy on a Microscopic Scale
(a)
A 12 can only be obtained one way 6 + 6
(b)
A 7 can be obtained six ways: 6 + 1 , 5 + 2 , 4 + 3 , 3 + 4 , 2 + 5 , 1 + 6
(a)
The table is shown below. On the basis of the table, the most probable result of a toss is
2 heads and 2 tails .
(b)
The most ordered state is the least likely state. Thus, on the basis of the table this is
either all heads or all tails .
(c)
The most disordered is the most likely state. Thus, this is 2 heads and 2 tails .
Result
All heads
3H, 1T
2H, 2T
1H, 3T
All tails
Possible Combinations
HHHH
THHH, HTHH, HHTH, HHHT
TTHH, THTH, THHT, HTTH, HTHT, HHTT
HTTT, THTT, TTHT, TTTH
TTTT
Total
1
4
6
4
1
(a)
Result
All red
2R, 1G
1R, 2G
All green
Possible Combinations
RRR
RRG, RGR, GRR
RGG, GRG, GGR
GGG
Total
1
3
3
1
(b)
Result
All red
4R, 1G
3R, 2G
Possible Combinations
RRRRR
RRRRG, RRRGR, RRGRR, RGRRR, GRRRR
RRRGG, RRGRG, RGRRG, GRRRG, RRGGR,
RGRGR, GRRGR, RGGRR, GRGRR, GGRRR
GGGRR, GGRGR, GRGGR, RGGGR, GGRRG,
GRGRG, RGGRG, GRRGG, RGRGG, RRGGG
RGGGG, GRGGG, GGRGG, GGGRG, GGGGR
GGGGG
Total
1
5
2R, 3G
1R, 4G
All green
10
10
5
1
Additional Problems
P22.52
The conversion of gravitational potential energy into kinetic energy as the water falls is reversible.
But the subsequent conversion into internal energy is not. We imagine arriving at the same final
state by adding energy by heat, in amount mgy, to the water from a stove at a temperature
infinitesimally above 20.0°C. Then,
∆S =
z
e
je
ja
f
3
3
2
dQ Q mgy 5 000 m 1 000 kg m 9.80 m s 50.0 m
= =
=
= 8.36 × 10 6 J K .
T
T
T
293 K
652
P22.53
Heat Engines, Entropy, and the Second Law of Thermodynamics
H ET
so if all the electric energy is converted into internal energy, the steady-state
∆t
condition of the house is described by H ET = Q .
Pelectric =
(a)
(b)
Therefore,
Pelectric =
For a heat pump,
aCOPf
Q
= 5 000 W
∆t
=
Carnot
Th 295 K
=
= 10.92
∆T
27 K
a
f
Actual COP = 0.6 10.92 = 6.55 =
Qh
W
=
Q h ∆t
W ∆t
Therefore, to bring 5 000 W of energy into the house only requires input power
P22.54
Pheat pump =
W Q h ∆t 5 000 W
=
=
= 763 W
∆t
COP
6.56
f
e
Q c = mc∆T + mL + mc∆T =
b
ga
j
b
ga
Q c = 0.500 kg 4 186 J kg ⋅° C 10° C + 0.500 kg 3.33 × 10 5 J kg + 0.500 kg 2 090 J kg⋅° C 20° C
Q c = 2.08 × 10 5 J
Qc
W
W=
P22.55
b
g
= COPc refrigerator =
b
Q c Th − Tc
Tc
Tc
Th − Tc
g = e2.08 × 10 Jj 20.0° C − a−20.0° Cf =
a273 − 20.0f K
5
∆S hot =
−1 000 J
600 K
∆Scold =
+750 J
350 K
(a)
∆SU = ∆S hot + ∆Scold = 0.476 J K
(b)
ec = 1 −
T1
= 0.417
T2
b
g
Weng = e c Q h = 0.417 1 000 J = 417 J
(c)
Wnet = 417 J − 250 J = 167 J
b
g
T1 ∆SU = 350 K 0.476 J K = 167 J
32.9 kJ
f
Chapter 22
*P22.56
(a)
653
The energy put into the engine by the hot reservoir is dQ h = mcdTh . The energy put into the
LM F
MN GH
a f
cold reservoir by the engine is dQ c = − mcdTc = 1 − e dQ h = 1 − 1 −
−
dTc dTh
=
Tc
Th
z
dT
dT
−
=
T T T
Tf
Tc
IJ OPmcdT . Then
K PQ
h
z
Tf
h
− ln T
ln
Tc
Th
Tf
Tc
T
= ln T T f
h
Tf
Tc
= ln
Tf
Th
T f2 = Tc Th
b g
T f = ThTc
(b)
d
12
i
d
Then Q h = Weng + Q c .
d i d i
= mceT − T T − T T + T j
= mceT − 2 T T + T j = mce T − T j
FV I
For an isothermal process,
Q = nRT lnG J
HV K
Therefore,
Q = nRb3T g ln 2
F 1I
Q = nRbT g lnG J
and
H 2K
3
= nRbT − 3T g
For the constant volume processes, Q = ∆E
2
3
= nRb3T − T g
and
Q = ∆E
2
Weng = mc Th − T f − mc T f − Tc
P22.57
(a)
h
h c
h
h
h c
c
c
c
h
2
c
2
1
1
i
3
i
2
int, 2
4
int, 4
i
i
i
i
The net energy by heat transferred is then
Q = Q1 + Q 2 + Q 3 + Q 4
or
(b)
i
The hot reservoir loses energy Q h = mc Th − T f . The cold reservoir gains Q c = mc T f − Tc .
FIG. P22.57
Q = 2nRTi ln 2 .
A positive value for heat represents energy transferred into the system.
Therefore,
a
Q h = Q1 + Q 4 = 3nRTi 1 + ln 2
f
Since the change in temperature for the complete cycle is zero,
∆Eint = 0 and Weng = Q
Therefore, the efficiency is
ec =
Weng
Qh
=
2 ln 2
Q
=
= 0.273
Q h 3 1 + ln 2
a
f
654
P22.58
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
Weng
t
8
= 1.50 × 10 Waelectrical
LM OP
=
=
,
Q
mL
f
MN 0.150 PQ∆t ,
Weng
t
and L = 33.0 kJ g = 33.0 × 10 6 J kg
LM W t OP ∆t
N 0.150 Q L
e1.50 × 10 Wjb86 400 s dayg
m=
= 2 620 metric tons day
0.150e33.0 × 10 J kg je10 kg metric tonj
Cost = b$8.00 metric tongb 2 618 metric tons day gb365 days yr g
m=
eng
8
6
(b)
3
Cost = $7.65 million year
(c)
First find the rate at which heat energy is discharged into the water. If the plant is 15.0%
efficient in producing electrical energy then the rate of heat production is
Qc
t
Then,
Qc
t
=
=
FW
GH t
eng
I FG 1 − 1IJ = e1.50 × 10 WjFG 1 − 1IJ = 8.50 × 10
JK H e K
H 0.150 K
8
8
W.
mc∆T
and
t
Q
c
8.50 × 10 8 J s
m
= t =
= 4.06 × 10 4 kg s .
t c∆T
4 186 J kg ⋅° C 5.00° C
b
P22.59
Weng
T
=
ec = 1 − c =
Th
Qh
Q h = Weng + Q c :
Weng
∆t
Qh
∆t
:
Qh
∆t
Qc
∆t
Qc
∆t
Q c = mc∆T :
Qc
∆t
=
=
f
ga
P Th
P
=
Th − Tc
1 − Tc Th
b
Qh
∆t
g
−
Weng
∆t
=
P Th
P Tc
−P =
Th − Tc
Th − Tc
=
FG ∆m IJ c∆T = P T
H ∆t K T − T
c
h
c
P Tc
∆m
=
Th − Tc c∆T
∆t
b
g
e
ja
f
1.00 × 10 9 W 300 K
∆m
=
= 5.97 × 10 4 kg s
∆t 200 K 4 186 J kg ⋅° C 6.00° C
b
ga
f
Chapter 22
P22.60
Weng
T
=
ec = 1 − c =
Th
Qh
Weng
∆t
Qh
∆t
Qh
P
=
=
P Th
Th − Tc
e1 − j
F Q I −P = P T
Q
=G
∆t H ∆t JK
T −T
∆t
Tc
Th
c
h
c
h
c
Q c = mc∆T , where c is the specific heat of water.
Qc
Therefore,
∆t
∆m
=
∆t
and
P22.61
(a)
=
FG ∆m IJ c∆T = P T
H ∆t K T − T
c
h
c
P Tc
Th − Tc c∆T
b
g
a
f a
f
5
98.6° F = a98.6 − 32.0f° C = a37.0 + 273.15f K = 310.15 K
9
dQ
dT
F 310.15 IJ = 54.86 cal K
∆S
=z
= b 453.6 g gb1.00 cal g ⋅ K g × z
= 453.6 lnG
H 274.82 K
T
T
Q
a310.15 − 274.82f = −51.67 cal K
∆S
=−
= −a 453.6 fa1.00f
35.0° F =
5
35.0 − 32.0 ° C = 1.67 + 273.15 K = 274.82 K
9
310.15
ice water
274.82
body
Tbody
310.15
∆Ssystem = 54.86 − 51.67 = 3.19 cal K
(b)
a453.6fa1fbT
F
g e
ja fb
− 274.82 = 70.0 × 10 3 1 310.15 − TF
g
Thus,
b70.0 + 0.453 6g × 10 T = a70.0fa310.15f + b0.453 6ga274.82f × 10
3
F
3
and TF = 309.92 K = 36.77° C = 98.19° F
FG 309.92 IJ = 54.52 cal K
H 274.82 K
F 310.15 IJ = −51.93 cal K
= −e70.0 × 10 j lnG
H 309.92 K
∆Sice
′ water = 453.6 ln
∆S body
′
3
∆Ssys
′ = 54.52 − 51.93 = 2.59 cal K which is less than the estimate in part (a).
655
656
P22.62
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
For the isothermal process AB, the work on the gas is
W AB = − PA VA ln
FG V IJ
HV K
B
A
e
j FGH 1050..00 IJK
je
W AB = −5 1.013 × 10 5 Pa 10.0 × 10 −3 m3 ln
W AB = −8.15 × 10 3 J
where we have used 1.00 atm = 1.013 × 10 5 Pa
1.00 L = 1.00 × 10 −3 m 3
and
ja
e
FIG. P22.62
f
WBC = − PB ∆V = − 1.013 × 10 5 Pa 10.0 − 50.0 × 10 −3 m3 = +4.05 × 10 3 J
WCA = 0 and Weng = −W AB − WBC = 4.11 × 10 3 J = 4.11 kJ
(b)
Since AB is an isothermal process, ∆Eint, AB = 0
and
Q AB = −W AB = 8.15 × 10 3 J
For an ideal monatomic gas,
CV =
3R
5R
and C P =
2
2
TB = TA =
e
je
j
1.013 × 10 5 50.0 × 10 −3
PBVB
5.05 × 10 3
=
=
nR
R
R
e
je
j
1.013 × 10 5 10.0 × 10 −3
PC VC
1.01 × 10 3
=
=
TC =
nR
R
R
Also,
QCA = nC V ∆T = 1.00
FG 3 RIJ FG 5.05 × 10
H 2 KH
3
so the total energy absorbed by heat is Q AB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ .
(c)
QBC = nC P ∆T =
QBC =
(d)
e=
a
5
1.013 × 10 5
2
Weng
Qh
e
=
f
5
5
nR∆T = PB ∆VBC
2
2
j a10.0 − 50.0f × 10
Weng
Q AB + QCA
=
−3
= −1.01 × 10 4 J = −10.1 kJ
4.11 × 10 3 J
= 0.289 or 28.9%
1.42 × 10 4 J
I
JK
− 1.01 × 10 3
= 6.08 kJ
R
657
Chapter 22
*P22.63
Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold
reservoir.
Tc
280 K
=
= 14.0
20 K
Th − Tc
Its ideal COP is
COPCarnot =
(a)
0.400 14.0 = 5.60 =
a f
Its actual COP is
5.60
Qc
Qh − Qc
=
Q c ∆t
Q h ∆t − Q c ∆t
Qh
Q
Q
− 5.60 c = c
∆t
∆t
∆t
a
f
5.60 10.0 kW = 6.60
Weng
Q
Qc
and c = 8.48 kW
∆t
∆t
Qh Qc
−
= 10.0 kW − 8.48 kW = 1.52 kW
∆t
∆t
(b)
Q h = Weng + Q c :
(c)
The air conditioner operates in a cycle, so the entropy of the working fluid does not change.
The hot reservoir increases in entropy by
=
∆t
Qh
Th
=
e10.0 × 10
3
jb
J s 3 600 s
300 K
g = 1.20 × 10
5
J K
The cold room decreases in entropy by
∆S = −
Qc
Tc
=−
e8.48 × 10
3
jb
J s 3 600 s
280 K
g = −1.09 × 10
The net entropy change is positive, as it must be:
+1.20 × 10 5 J K − 1.09 × 10 5 J K = 1.09 × 10 4 J K
(d)
COPCarnot =
We suppose the actual COP is
0.400 11.2 = 4.48
a f
As a fraction of the original 5.60, this is
drop by 20.0% .
z
Vf
P22.64
(a)
W=
Vi
(b)
Tc
280 K
=
= 11.2
25 K
Th − Tc
The new ideal COP is
PdV = nRT
z
2 Vi
Vi
a f
4. 48
= 0.800 , so the fractional change is to
5.60
FG IJ
H K
2Vi
dV
= 1.00 RT ln
= RT ln 2
V
Vi
The second law refers to cycles.
5
J K
658
P22.65
Heat Engines, Entropy, and the Second Law of Thermodynamics
At point A, PV
i i = nRTi
and
n = 1.00 mol
At point B, 3PV
i i = nRTB
so
TB = 3Ti
and
TC = 6Ti
so
TD = 2Ti
b3 P gb2V g = nRT
At point D, P b 2V g = nRT
At point C,
i
i
i
C
i
D
The heat for each step in the cycle is found using C V =
CP =
5R
:
2
3R
and
2
b
g
= nC b6T − 3T g = 7.50nRT
= nC b 2T − 6T g = −6nRT
= nC bT − 2T g = −2.50nRT
Q AB = nCV 3Ti − Ti = 3nRTi
QBC
QCD
QDA
i
V
i
i
P
i
Qleaving = Q c = QCD + QDA = 8.50nRTi
(c)
Actual efficiency,
e=
(d)
Carnot efficiency,
ec = 1 −
i
z z
Qh − Qc
Qh
= 0.190
Tc
T
= 1 − i = 0.833
6Ti
Th
z
FG IJ
H K
f
f
Tf
nC P dT
dQ
T
=
= nC P T −1 dT = nC P ln T T f = nC P ln T f − ln Ti = nC P ln
i
T
T
Ti
i
i
∆S = nC P ln
(a)
i
Qentering = Q h = Q AB + QBC = 10.5nRTi
(b)
∆S =
FIG. P22.65
i
i
Therefore,
i
*P22.67
i
(a)
f
*P22.66
P
F PV
GH nR
f
d
i
I
JK
nR
= nC P ln 3
PVi
The ideal gas at constant temperature keeps constant internal energy. As it puts out energy
by work in expanding it must take in an equal amount of energy by heat. Thus its entropy
increases. Let Pi , Vi , Ti represent the state of the gas before the isothermal expansion. Let
PC , VC , Ti represent the state after this process, so that PV
i i = PC VC . Let Pi , 3Vi , T f represent
the state after the adiabatic compression.
b g
γ
Then
PC VCγ = Pi 3Vi
Substituting
PC =
gives
γ −1
= Pi 3 γ Viγ
PV
i i VC
Then
VCγ −1 = 3 γ Viγ −1 and
continued on next page
PV
i i
VC
e
j
VC
= 3γ
Vi
bγ −1g
Chapter 22
The work output in the isothermal expansion is
z
z
C
C
W = PdV = nRTi V −1 dV = nRTi ln
i
i
FG V IJ = nRT lne3
HV K
C
i
i
659
γ I
H γ − 1 JK ln 3
b g j = nRTi FG
γ γ −1
This is also the input heat, so the entropy change is
∆S =
P22.68
IJ
K
Since
C P = γ C V = CV + R
we have
bγ − 1gC
and
CP =
V
= R , CV =
R
γ −1
γR
γ −1
∆S = nC P ln 3
Then the result is
(b)
FG
H
γ
Q
= nR
ln 3
γ −1
T
The pair of processes considered here carry the gas from the initial state in Problem 66 to the
final state there. Entropy is a function of state. Entropy change does not depend on path.
Therefore the entropy change in Problem 66 equals ∆Sisothermal + ∆S adiabatic in this problem.
Since ∆Sadiabatic = 0, the answers to Problems 66 and 67 (a) must be the same.
Simply evaluate the maximum (Carnot) efficiency.
eC =
∆T 4.00 K
=
= 0.014 4
277 K
Th
The proposal does not merit serious consideration.
P22.69
The heat transfer over the paths CD and BA is zero
since they are adiabatic.
b
g
Over path BC: QBC = nC P TC − TB > 0
b
g
Over path DA: QDA = nCV TA − TD < 0
P
Adiabatic
Processes
B
C
Therefore, Q c = QDA and Q h = Q BC
D
The efficiency is then
bT
Q
bT
1 LT − T O
e=1− M
P
γ NT −T Q
e=1−
Qc
=1−
h
D
A
C
B
g
− T gC
A
D
− TA C V
C
B
P
Vi
3Vi
FIG. P22.69
V
660
P22.70
Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)
Use the equation of state for an ideal gas
nRT
P
1.00 8.314 600
VA =
= 1.97 × 10 −3 m 3
5
25.0 1.013 × 10
V=
VC
a fa f
e
j
1.00a8.314fa 400f
=
=
1.013 × 10 5
32.8 × 10 −3 m3
FIG. P22.70
Since AB is isothermal,
PA VA = PBVB
and since BC is adiabatic,
PBVBγ = PC VCγ
Combining these expressions, VB
LF P I V OP
= MG J
MNH P K V PQ
C
γ
C
A
A
b g
1 γ −1
LF 1.00 I e32.8 × 10 m j
= MG
MMH 25.0 JK 1.97 × 10 m
N
OPb
PP
Q
LF 25.0 I e1.97 × 10 m j
= MG
MMH 1.00 JK 32.8 × 10 m
N
OPb
PP
Q
3 1.40
−3
−3
3
1 0. 400
g
VB = 11.9 × 10 −3 m3
Similarly,
VD
LF P I V OP
= MG J
MNH P K V PQ
b g
A
γ 1 γ −1
A
C
C
or
VD = 5.44 × 10 −3 m3
Since AB is isothermal,
PA VA = PBVB
and
PB = PA
Also, CD is an isothermal and PD
−3
FG V IJ = 25.0 atmF 1.97 × 10
GH 11.9 × 10
HV K
F V IJ = 1.00 atmF 32.8 × 10
=P G
GH 5.44 × 10
HV K
A
B
C
C
3 1.40
−3
D
−3
−3
−3
−3
I
JK
m I
J=
m K
3
1 0 . 400
g
m3
= 4.14 atm
m3
3
3
6.03 atm
Solving part (c) before part (b):
ec = 1 −
Tc
400 K
=1−
= 0.333
600 K
Th
(c)
For this Carnot cycle,
(b)
Energy is added by heat to the gas during the process AB. For the isothermal process,
∆Eint = 0 .
and the first law gives
Q AB = −W AB = nRTh ln
FG V IJ
HV K
B
A
b
ga
f FGH 111.97.9 IJK = 8.97 kJ
or
Q h = Q AB = 1.00 mol 8.314 J mol ⋅ K 600 K ln
Then, from
e=
Weng
Qh
a
f
the net work done per cycle is Weng = e c Q h = 0.333 8.97 kJ = 2.99 kJ .
Chapter 22
P22.71
(a)
20.0°C
(b)
∆S = mc ln
(c)
(d)
Tf
T1
+ mc ln
Tf
T2
gLM
N
b
= 1.00 kg 4.19 kJ kg ⋅ K ln
Tf
T1
+ ln
661
OP = b4.19 kJ K g lnFG 293 ⋅ 293 IJ
H 283 303 K
T Q
Tf
2
∆S = +4.88 J K
Yes . Entropy has increased.
ANSWERS TO EVEN PROBLEMS
P22.2
(a) 667 J ; (b) 467 J
P22.4
(a) 30.0% ; (b) 60.0%
P22.6
55.4%
P22.8
77.8 W
P22.10
(a) 869 MJ ; (b) 330 MJ
P22.12
197 kJ
P22.14
546°C
P22.16
33.0%
P22.18
(a) 5.12%; (b) 5.27 TJ h;
(c) see the solution
P22.34
(a), (b) see the solution;
(c) Q h = 149 J ; Q c = 65.0 J ; Weng = 84.3 J ;
(d) 56.5%; (e) 1.42 × 10 3 rev min
P22.36
4.88 kJ kg ⋅ K
P22.38
(a) isobaric; (b) 402 kJ; (c) 1.20 kJ K
P22.40
3.27 J K
P22.42
718 J K
P22.44
(a) 39.4 L ; (b) −2.50 kJ; (c) −2.50 kJ;
(d) −6.87 J K ; (e) +9.16 J K
P22.46
0.507 J K
P22.48
34.6 J K
P22.50
(a) 2 heads and 2 tails ;
(b) All heads or all tails;
(c) 2 heads and 2 tails
P22.20
453 K
P22.22
(a), (b) see the solution;
(c) 23.7%; see the solution
P22.24
11.8
P22.52
8.36 MJ K
P22.26
1.17 J
P22.54
32.9 kJ
P22.28
(a) 204 W ; (b) 2.43 kW
P22.56
see the solution
P22.30
(a) 2.00 ; (b) 3.00 ; (c) 33.3%
P22.58
(a) 2.62 × 10 3 tons d ; (b) $7.65 million yr ;
P22.32
(a) 51.2% ; (b) 36.2%
(c) 4.06 × 10 4 kg s
662
P22.60
Heat Engines, Entropy, and the Second Law of Thermodynamics
P Tc
Th − Tc c∆T
b
g
P22.62
(a) 4.11 kJ ; (b) 14.2 kJ; (c) 10.1 kJ; (d) 28.9%
P22.64
see the solution
P22.66
nC P ln 3
P22.68
no; see the solution
P22.70
(a)
A
B
C
D
P, atm
25.0
4.14
1.00
6.03
(b) 2.99 kJ ; (c) 33.3%
V, L
1.97
11.9
32.8
5.44
23
Electric Fields
CHAPTER OUTLINE
23.1
23.2
23.3
23.4
23.5
23.6
23.7
Properties of Electric
Charges
Charging Objects by
Induction
Coulomb’s Law
The Electric Field
Electric Field of a
Continuous Charge
Distribution
Electric Field Lines
Motion of Charged Particles
in a Uniform Electric Field
ANSWERS TO QUESTIONS
Q23.1
A neutral atom is one that has no net charge. This means that it
has the same number of electrons orbiting the nucleus as it has
protons in the nucleus. A negatively charged atom has one or
more excess electrons.
Q23.2
When the comb is nearby, molecules in the paper are polarized,
similar to the molecules in the wall in Figure 23.5a, and the
paper is attracted. During contact, charge from the comb is
transferred to the paper by conduction. Then the paper has the
same charge as the comb, and is repelled.
Q23.3
The clothes dryer rubs dissimilar materials together as it
tumbles the clothes. Electrons are transferred from one kind of
molecule to another. The charges on pieces of cloth, or on
nearby objects charged by induction, can produce strong
electric fields that promote the ionization process in the
surrounding air that is necessary for a spark to occur. Then you
hear or see the sparks.
Q23.4
To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could
discharge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere.
Q23.5
Electrons are less massive and more mobile than protons. Also, they are more easily detached from
atoms than protons.
Q23.6
The electric field due to the charged rod induces charges on near and far sides of the sphere. The
attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger
than the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. The
result is a net attraction of the sphere to the rod. When the sphere touches the rod, charge is
conducted between the rod and the sphere, leaving both the rod and the sphere like-charged. This
results in a repulsive Coulomb force.
Q23.7
All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite
readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics
demonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts of
excess charge on an object in a humid climate. During a North American winter, the cold, dry air
allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking
(pun also intended) introduction to static electricity sparks.
1
2
Electric Fields
Q23.8
Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of
two particles, and inversely proportional to the square of the separation distance. An electrical force
exhibits the same proportionalities, with charge as the intrinsic property.
Differences: The electrical force can either attract or repel, while the gravitational force as
described by Newton’s law can only attract. The electrical force between elementary particles is
vastly stronger than the gravitational force.
Q23.9
No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge
exists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of the
charges are reversed. The attraction between these charges and the negative charges on the balloon
is stronger than the repulsion between the negative charges on the balloon and the negative charges
in the polarized molecules (because they are farther from the balloon), so that there is a net attractive
force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which
excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually
causing the attractive force to be insufficient to support the weight of the balloon.
Q23.10
The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is brought
towards the aluminum from above, the top of the aluminum will have a negative charge induced on
it, while the parts draping over the pencil can have a positive charge induced on them. These
positive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is a
good insulator, the net charge on the aluminum can be zero.
Q23.11
So the electric field created by the test charge does not distort the electric field you are trying to
measure, by moving the charges that create it.
Q23.12
With a very high budget, you could send first a proton and then an electron into an evacuated
region in which the field exists. If the field is gravitational, both particles will experience a force in
the same direction, while they will experience forces in opposite directions if the field is electric.
On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pith
ball + and the other –, creating a large-scale dipole. Carefully suspend this dipole about its center of
mass so that it can rotate freely. When suspended in the field in question, the dipole will rotate to
align itself with an electric field, while it will not for a gravitational field. If the test device does not
rotate, be sure to insert it into the field in more than one orientation in case it was aligned with the
electric field when you inserted it on the first trial.
Q23.13
The student standing on the insulating platform is held at the same electrical potential as the
generator sphere. Charge will only flow when there is a difference in potential. The student who
unwisely touches the charged sphere is near zero electrical potential when compared to the charged
sphere. When the student comes in contact with the sphere, charge will flow from the sphere to him
or her until they are at the same electrical potential.
Q23.14
An electric field once established by a positive or negative charge extends in all directions from the
charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at
point A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if a
charge were present at point A, however. A field does exist at point A.
Q23.15
If a charge distribution is small compared to the distance of a field point from it, the charge
distribution can be modeled as a single particle with charge equal to the net charge of the
distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior
points just as if all of its charge were a point charge at its center.
Chapter 23
3
Q23.16
The direction of the electric field is the direction in which a positive test charge would feel a force
when placed in the field. A charge will not experience two electrical forces at the same time, but the
vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which
they cross would feel a force in two directions. Furthermore, the path that the test charge would
follow if released at the point where the field lines cross would be indeterminate.
Q23.17
Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the point
charges q1 and q 2 in Figure 23.14 or q and –q in Figure 23.15, respectively. The net electric field is the
vector sum of E1 and E 2 , shown as E. Figure 23.21 shows only one electric field line at each point
away from the charge. At the point location of an object modeled as a point charge, the direction of
the field is undefined, and so is its magnitude.
Q23.18
The electric forces on the particles have the same magnitude, but are in opposite directions. The
electron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to its
much smaller mass.
Q23.19
The electric field around a point charge approaches infinity as r approaches zero.
Q23.20
Vertically downward.
Q23.21
Four times as many electric field lines start at the surface of the larger charge as end at the smaller
charge. The extra lines extend away from the pair of charges. They may never end, or they may
terminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and –q.
Q23.22
At a point exactly midway between the two changes.
Q23.23
Linear charge density, λ, is charge per unit length. It is used when trying to determine the electric
field created by a charged rod.
Surface charge density, σ, is charge per unit area. It is used when determining the electric field
above a charged sheet or disk.
Volume charge density, ρ, is charge per unit volume. It is used when determining the electric
field due to a uniformly charged sphere made of insulating material.
Q23.24
Yes, the path would still be parabolic. The electrical force on the electron is in the downward
direction. This is similar to throwing a ball from the roof of a building horizontally or at some angle
with the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolic
trajectory.
Q23.25
No. Life would be no different if electrons were + charged and protons were – charged. Opposite
charges would still attract, and like charges would repel. The naming of + and – charge is merely a
convention.
Q23.26
If the antenna were not grounded, electric charges in the atmosphere during a storm could place the
antenna at a high positive or negative potential. The antenna would then place the television set
inside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps the
antenna, the television set, and even the air around the antenna at close to zero potential.
Q23.27
People are all attracted to the Earth. If the force were electrostatic, people would all carry charge
with the same sign and would repel each other. This repulsion is not observed. When we changed
the charge on a person, as in the chapter-opener photograph, the person’s weight would change
greatly in magnitude or direction. We could levitate an airplane simply by draining away its electric
charge. The failure of such experiments gives evidence that the attraction to the Earth is not due to
electrical forces.
4
Electric Fields
Q23.28
In special orientations the force between two dipoles can be zero or a force of repulsion. In general
each dipole will exert a torque on the other, tending to align its axis with the field created by the first
dipole. After this alignment, each dipole exerts a force of attraction on the other.
SOLUTIONS TO PROBLEMS
Section 23.1
*P23.1
(a)
Properties of Electric Charges
The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its
mass by a negligible amount, to
e
j
1.007 9 1.660 × 10 −27 kg − 9.11 × 10 −31 kg = 1.67 × 10 −27 kg .
Its charge, due to loss of one electron, is
e
j
0 − 1 −1.60 × 10 −19 C = +1.60 × 10 −19 C .
(b)
By similar logic, charge = +1.60 × 10 −19 C
e
j
mass = 22.99 1.66 × 10 −27 kg − 9.11 × 10 −31 kg = 3.82 × 10 −26 kg
(c)
charge of Cl − = −1.60 × 10 −19 C
e
j
mass = 35.453 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 5.89 × 10 −26 kg
(d)
e
j
charge of Ca ++ = −2 −1.60 × 10 −19 C = +3.20 × 10 −19 C
e
j e
j
mass = 40.078 1.66 × 10 −27 kg − 2 9.11 × 10 −31 kg = 6.65 × 10 −26 kg
(e)
e
e
mass = 14.007 1.66 × 10 −27
(f)
j
kg j + 3e9.11 × 10
charge of N 3 − = 3 −1.60 × 10 −19 C = −4.80 × 10 −19 C
e
−31
j
j
kg = 2.33 × 10 −26 kg
charge of N 4 + = 4 1.60 × 10 −19 C = +6.40 × 10 −19 C
e
j e
j
mass = 14.007 1.66 × 10 −27 kg − 4 9.11 × 10 −31 kg = 2.32 × 10 −26 kg
(g)
We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.
e
j
charge = 7 1.60 × 10 −19 C = 1.12 × 10 −18 C
e
mass = 14.007 1.66 × 10
(h)
−27
j e
j
kg − 7 9.11 × 10 −31 kg = 2.32 × 10 −26 kg
charge = −1.60 × 10 −19 C
b
g
mass = 2 1.007 9 + 15.999 1.66 × 10 −27 kg + 9.11 × 10 −31 kg = 2.99 × 10 −26 kg
Chapter 23
P23.2
F 10.0 grams I FG 6.02 × 10
GH 107.87 grams mol JK H
(a)
N=
(b)
# electrons added =
IJ FG 47 electrons IJ =
K H atom K
2.62 × 10 24
Q
1.00 × 10 −3 C
=
= 6.25 × 10 15
e 1.60 × 10 −19 C electron
2.38 electrons for every 10 9 already present .
or
Section 23.2
Charging Objects by Induction
Section 23.3
Coulomb’s Law
P23.3
atoms
mol
23
If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains
N≅
F 70 000 grams I FG 6.02 × 10
GH 18 grams mol JK H
molecules
mol
23
IJ FG 10 protons IJ ≅ 2.3 × 10
K H molecule K
28
protons .
With an excess of 1% electrons over protons, each person has a charge
e
je
j
e3.7 × 10 j
= e9 × 10 j
q = 0.01 1.6 × 10 −19 C 2.3 × 10 28 = 3.7 × 10 7 C .
So
F = ke
q1 q 2
r2
7 2
9
0.6 2
N = 4 × 10 25 N ~ 10 26 N .
This force is almost enough to lift a weight equal to that of the Earth:
e
j
Mg = 6 × 10 24 kg 9.8 m s 2 = 6 × 10 25 N ~ 10 26 N .
*P23.4
The force on one proton is F =
e8.99 × 10
P23.5
(a)
(b)
9
F 1.6 × 10
N ⋅ m C jG
H 2 × 10
Fe =
Fg =
2
k e q1 q 2
r2
r2
r2
−19
−15
e8.99 × 10
=
Gm1 m 2
k e q1 q 2
C
m
9
e6.67 × 10
=
I
JK
away from the other proton. Its magnitude is
2
= 57.5 N .
je
N ⋅ m 2 C 2 1.60 × 10 −19 C
e
3.80 × 10
−11
−10
j
m
2
je
j
2
= 1.59 × 10 −9 N
N ⋅ m 2 C 2 1.67 × 10 −27 kg
e3.80 × 10
−10
j
m
2
j
2
= 1.29 × 10 −45 N
The electric force is larger by 1.24 × 10 36 times .
(c)
If k e
q
=
m
q1 q 2
r
2
=G
m1 m 2
r2
brepulsiong
with q1 = q 2 = q and m1 = m 2 = m , then
6.67 × 10 −11 N ⋅ m 2 kg 2
G
=
= 8.61 × 10 −11 C kg .
ke
8.99 × 10 9 N ⋅ m 2 C 2
5
6
Electric Fields
P23.6
We find the equal-magnitude charges on both spheres:
q1 q 2
F = ke
= ke
r2
q2
r2
a
1.00 × 10 4 N
= 1.05 × 10 −3 C .
8.99 × 10 9 N ⋅ m 2 C 2
f
F
= 1.00 m
ke
q=r
so
The number of electron transferred is then
N xfer =
1.05 × 10 −3 C
= 6.59 × 10 15 electrons .
1.60 × 10 −19 C e −
The whole number of electrons in each sphere is
F 10.0 g I e6.02 × 10
GH 107.87 g mol JK
N tot =
23
je
j
atoms mol 47 e − atom = 2.62 × 10 24 e − .
The fraction transferred is then
f=
P23.7
F
GH
I
JK
N xfer
6.59 × 10 15
=
= 2.51 × 10 −9 = 2.51 charges in every billion.
24
N tot
2.62 × 10
F1 = k e
F2 = k e
e8.99 × 10
=
q1 q 2
r
2
q1 q 2
r2
=
e8.99 × 10
je
je
N ⋅ m 2 C 2 7.00 × 10 −6 C 2.00 × 10 −6 C
9
9
N ⋅ m2
j = 0.503 N
a0.500 mf
C je7.00 × 10 C je 4.00 × 10 C j
= 1.01 N
a0.500 mf
2
−6
2
−6
2
Fx = 0.503 cos 60.0°+1.01 cos 60.0° = 0.755 N
Fy = 0.503 sin 60.0°−1.01 sin 60.0° = −0.436 N
a
f a
f
F = 0.755 N i − 0.436 N j = 0.872 N at an angle of 330°
FIG. P23.7
P23.8
F = ke
P23.9
(a)
q1 q 2
r2
e8.99 × 10
=
9
je
N ⋅ m 2 C 2 1.60 × 10 −19 C
e
j
2 6.37 × 10 6 m
j e6.02 × 10 j
2
23 2
2
= 514 kN
The force is one of attraction . The distance r in Coulomb’s law is the distance between
centers. The magnitude of the force is
F=
(b)
k e q1 q 2
r2
e
= 8.99 × 10 9 N ⋅ m 2 C 2
12.0 × 10 C je18.0 × 10 C j
=
je
a0.300 mf
−9
−9
2
2.16 × 10 −5 N .
The net charge of −6.00 × 10 −9 C will be equally split between the two spheres, or
−3.00 × 10 −9 C on each. The force is one of repulsion , and its magnitude is
F=
k e q1 q 2
r
2
e
= 8.99 × 10 9 N ⋅ m 2 C 2
3.00 × 10 C je3.00 × 10 C j
=
je
a0.300 mf
−9
−9
2
8.99 × 10 −7 N .
Chapter 23
P23.10
x from the left end of the rod. This bead
Let the third bead have charge Q and be located distance
will experience a net force given by
F=
b g
k e 3q Q
x
2
i+
7
b g e− i j .
ad − x f
ke q Q
2
The net force will be zero if
3
1
=
2
x
d−x
a
f
2
, or d − x =
x
3
.
This gives an equilibrium position of the third bead of x = 0.634d .
The equilibrium is stable if the third bead has positive charge .
P23.11
kee2
e
(a)
F=
(b)
We have F =
r2
= 8.99 × 10 N ⋅ m
2
2
−10
C
j
m
2
j
2
= 8.22 × 10 −8 N
mv 2
from which
r
v=
P23.12
e1.60 × 10
C j
e0.529 × 10
−19
9
e
j=
8.22 × 10 −8 N 0.529 × 10 −10 m
Fr
=
m
9.11 × 10
−31
The top charge exerts a force on the negative charge
kg
k e qQ
d 2 + 2
x
2
ch
2.19 × 10 6 m s .
which is directed upward and to the
FG d IJ to the x-axis. The two positive charges together exert force
H 2x K
F 2 k qQ I FG a− xfi IJ
−2 k qQ
d
GG
JJ G
= ma or for x << , a ≈
x.
2
md 8
eH + x j K GH e + x j JJK
left, at an angle of tan −1
e
d2
4
(a)
d2
4
2
12
e
3
The acceleration is equal to a negative constant times the excursion from equilibrium, as in
16 k e qQ
a = −ω 2 x , so we have Simple Harmonic Motion with ω 2 =
.
md 3
T=
(b)
2
2π
ω
=
π
2
md 3
, where m is the mass of the object with charge −Q .
k e qQ
v max = ωA = 4a
k e qQ
md 3
8
Electric Fields
Section 23.4
P23.13
P23.14
The Electric Field
For equilibrium,
Fe = −Fg
or
qE = − mg − j .
Thus,
E=
e j
je
j
9.11 × 10 −31 kg 9.80 m s 2
mg
j=
j = − 5.58 × 10 −11 N C j
−19
q
−1.60 × 10
C
e
(a)
E=
(b)
1.67 × 10 −27 kg 9.80 m s 2
mg
E=
j=
j=
q
1.60 × 10 −19 C
e
e
e
jb
e
j
je
∑ Fy = 0 : QEj + mge− jj = 0
∴m =
P23.15
e
mg
j.
q
j
j
e1.02 × 10
j
−7
j
NC j
g
24.0 × 10 −6 C 610 N C
QE
=
= 1.49 grams
g
9.80 m s 2
The point is designated in the sketch. The magnitudes of the electric fields,
E1 , (due to the −2.50 × 10 −6 C charge) and E 2 (due to the 6.00 × 10 −6 C
charge) are
E1 =
E2 =
ke q
r
2
keq
r
2
e8.99 × 10
=
=
e8.99 × 10
9
je
N ⋅ m 2 C 2 2.50 × 10 −6 C
d
9
j
2
je
ad + 1.00 mf
N ⋅ m 2 C 2 6.00 × 10 −6 C
j
2
(1)
FIG. P23.15
(2)
Equate the right sides of (1) and (2)
to get
ad + 1.00 mf
or
d + 1.00 m = ±1.55d
which yields
d = 1.82 m
or
d = −0.392 m .
2
= 2.40d 2
The negative value for d is unsatisfactory because that locates a point between the charges where
both fields are in the same direction.
Thus,
d = 1.82 m to the left of the − 2.50 µC charge .
9
Chapter 23
P23.16
If we treat the concentrations as point charges,
E+ = ke
E− = ke
q
e
j b1a40000.0 mCgf e− jj = 3.60 × 10
a40.0 Cf e− jj = 3.60 × 10
C j
b1 000 mg
= 8.99 × 10 9 N ⋅ m 2 C 2
r2
q
e
= 8.99 × 10 9 N ⋅ m 2
r2
2
2
2
e ja
f
e ja
f
5
N C − j downward
5
N C − j downward
E = E + + E − = 7.20 × 10 5 N C downward
*P23.17
The first charge creates at the origin field
k eQ
+Q x = 0
to the right.
a2
Suppose the total field at the origin is to the right. Then q must
be negative:
k eQ
a
2
i+
keq
2k Q
− ij =
i
e
a
a3 a f
e
2
2
q
x
FIG. P23.17
q = −9Q .
In the alternative, the total field at the origin is to the left:
k eQ
a
P23.18
2
(a)
i+
ke q
9a
2
e− ij = 2 ka Q e− ij
e
2
q = +27Q .
e8.99 × 10 je7.00 × 10 j = 2.52 × 10
r
a0.500f
k q e8.99 × 10 je 4.00 × 10 j
=
=
= 1.44 × 10
r
a0.500f
E1 =
−6
9
ke q
=
2
2
NC
5
NC
−6
9
E2
5
e
2
2
Ex = E2 − E1 cos 60° = 1. 44 × 10 5 − 2.52 × 10 5 cos 60.0° = 18.0 × 10 3 N C
5
FIG. P23.18
3
Ey = − E1 sin 60.0° = −2.52 × 10 sin 60.0° = −218 × 10 N C
E = 18.0 i − 218 j × 10 3 N C = 18.0 i − 218 j kN C
(b)
P23.19
(a)
e
j
e
j
e8.99 × 10 je3.00 × 10 j e− jj = −e2.70 × 10
− jj =
E =
e
r
a0.100f
k q
e8.99 × 10 je6.00 × 10 j e− ij = −e5.99 × 10
−ij =
E =
e
r
a0.300f
E = E + E = −e5.99 × 10 N C ji − e 2.70 × 10 N C j j
1
2
1
e
−9
9
k e q1
2
2
2
2
2
2
1
e
3
je
j
e−3.00i − 13.5 jj µN
F = qE = 5.00 × 10 −9 C −599 i − 2 700 j N C
e
j
F = −3.00 × 10 −6 i − 13.5 × 10 −6 j N =
e36.0i − 436 jj mN
j
3
NC j
2
NCi
−9
9
2
2
(b)
je
F = qE = 2.00 × 10 −6 C 18.0 i − 218 j × 10 3 N C = 36.0 i − 436 j × 10 −3 N =
j
FIG. P23.19
10
Electric Fields
P23.20
E=
(a)
keq
r2
e8.99 × 10 je2.00 × 10 j = 14 400 N C
a1.12f
and
E = 2b14 400 g sin 26.6° = 1.29 × 10
−6
9
=
2
Ex = 0
y
e
je
j
F = qE = −3.00 × 10 −6 1. 29 × 10 4 j = −3.86 × 10 −2 j N
(a)
E=
r1 +
r12
E = 3.06
ke q
a
2
ke q2
r22
r2 +
i + 5.06
F = qE = 5.91
(b)
P23.22
ke q 2
a2
k e q3
keq
a
2
r32
r3 =
b g i + k b3 qg ei cos 45.0°+ j sin 45.0°j + k b4qg j
ke 2q
e
a2
j = 5.91
keq
a2
e
2a2
a2
at 58.8°
at 58.8°
The electric field at any point x is
E=
a f
ax − af cx − a− afh ex − a j
ke q
2
ke q
−
2
=
k e q 4ax
2
2 2
.
b g
4a k e q
When x is much, much greater than a, we find E ≅
P23.23
FIG. P23.20
(b)
k e q1
NC
E = 1.29 × 10 4 j N C .
so
P23.21
4
(a)
One of the charges creates at P a field E =
the x-axis as shown.
x3
k eQ n
R2 + x2
.
at an angle θ to
When all the charges produce field, for n > 1 , the components
perpendicular to the x-axis add to zero.
The total field is
(b)
P23.24
E=∑
b g cosθ =
nk e Q n i
2
R +x
2
e
k eQx i
2
R + x2
j
3 2
.
FIG. P23.23
A circle of charge corresponds to letting n grow beyond all bounds, but the result does not
depend on n. Smearing the charge around the circle does not change its amount or its
distance from the field point, so it does not change the field .
keq
r2
r=
keq
a2
e− ij + a2kaqf e− ij + a3kaqf e− ij + … = − ka qi FGH1 + 21 + 31 + …IJK =
e
2
e
2
e
2
2
2
−
π 2ke q
i
6a2
Chapter 23
Section 23.5
P23.25
Electric Field of a Continuous Charge Distribution
e
b g
a f a f a f a
je
j
8.99 × 10 9 22.0 × 10 −6
ke Q
keλ
k eQ
=
=
=
E=
0.290 0.140 + 0.290
d +d
d +d
d +d
fa
f
E = 1.59 × 10 6 N C , directed toward the rod.
P23.26
E=
z
k e dq
x2
E = keλ 0
z
∞
x0
P23.27
P23.28
P23.29
E=
ex
, where dq = λ 0 dx
FG 1 IJ
H xK
dx
x
=
x0
keλ 0
x0
The direction is − i or left for λ 0 > 0
e8.99 × 10 je75.0 × 10 jx = 6.74 × 10 x
ex + 0.100 j
ex + 0.010 0j
−6
9
j
+ a2
∞
= keλ 0 −
2
k e xQ
2
FIG. P23.25
3 2
=
5
2 32
2
32
2
(a)
At x = 0.010 0 m ,
E = 6.64 × 10 6 i N C = 6.64i MN C
(b)
At x = 0.050 0 m ,
E = 2.41 × 10 7 i N C = 24.1i MN C
(c)
At x = 0.300 m ,
E = 6.40 × 10 6 i N C = 6. 40 i MN C
(d)
At x = 1.00 m ,
E = 6.64 × 10 5 i N C = 0.664i MN C
z
E = dE =
E=
LM k λ x dxe− ij OP
z MMN x PPQ = −k λ x i z x
∞
e
ex
k e Qx
+ a2
e
3
x0
2
∞
0 0
j
0 0
−3
F
GH
dx = − k e λ 0 x 0 i −
x0
3 2
For a maximum,
dE
= Qk e
dx
LM
MM ex
N
a
x 2 + a 2 − 3 x 2 = 0 or x =
2
1
2
+ a2
j
3 2
−
OP
=0
P
+
x
a
e
j PQ
3x 2
2 52
2
.
Substituting into the expression for E gives
E=
k eQa
2
e aj
3
2
2 32
=
k eQ
3
3
2
a
2
=
2 k eQ
3 3a
2
=
Q
6 3π ∈0 a 2
.
∞
1
2x
2
x0
I=
JK
keλ 0
−i
2 x0
e j
11
12
Electric Fields
P23.30
F
GH
e
E = 2π 8.99 × 10
P23.31
x
E = 2π k eσ 1 −
2
x +R
9
2
I
JK
je7.90 × 10
−3
F
jGG 1 −
H
I
JJ = 4.46 × 10 FG 1 −
H
+ a0.350f K
x
x2
2
(a)
At x = 0.050 0 m ,
E = 3.83 × 10 8 N C = 383 MN C
(b)
At x = 0.100 m ,
E = 3.24 × 10 8 N C = 324 MN C
(c)
At x = 0.500 m ,
E = 8.07 × 10 7 N C = 80.7 MN C
(d)
At x = 2.00 m ,
E = 6.68 × 10 8 N C = 6.68 MN C
(a)
From Example 23.9: E = 2π k eσ 1 −
F
GH
σ=
I
J
+ 0.123 K
x
8
x2
I
JK
x
x2 + R2
Q
= 1.84 × 10 −3 C m 2
πR 2
ja
e
f
E = 1.04 × 10 8 N C 0.900 = 9.36 × 10 7 N C = 93.6 MN C
b
appx: E = 2π k eσ = 104 MN C about 11% high
(b)
F
jGH
e
E = 1.04 × 10 8 N C 1 −
I = 1.04 × 10
J e
cm K
30.0 cm
jb
g
8
N C 0.004 96 = 0.516 MN C
30.0 + 3.00
Q
5.20 × 10 −6
appx: E = k e 2 = 8.99 × 10 9
= 0.519 MN C about 0.6% high
2
r
0.30
e
P23.32
g
2
2
b
j a f
LM
MN
L
= 2π k σ M1 −
MN
The electric field at a distance x is
Ex = 2π k eσ 1 −
This is equivalent to
Ex
For large x,
R2
x2
x2 + R2
OP
PQ
1
e
1+
<< 1 and
x
g
1 + R2 x2
OP
PQ
R2
R2
≈
+
1
x2
2x 2
I
F
e1 + R e2 x j − 1j
1
J
= 2π k σ G 1 −
= 2π k σ
GH 1 + R e2 x j JK
1 + R e2 x j
k Qe1 x j
F RI
=
= k QG x +
H 2 JK
1 + R e2x j
2
Ex
so
Q
Substitute σ =
,
π R2
But for x >> R ,
1
x2 + R2 2
Ex
≈
1
x2
, so
Ex ≈
e
2
2
e
2
k eQ
x2
2
e
2
e
2
2
for a disk at large distances
2
2
2
Chapter 23
P23.33
z
z
Due to symmetry
Ey = dEy = 0 , and Ex = dE sin θ = k e
where
dq = λds = λrdθ ,
so that,
Ex =
z
keλ π
k λ
sin θdθ = e − cos θ
r 0
r
a
f
π
=
0
z
dq sin θ
r2
2k e λ
r
q
L
and r = .
π
L
2
8.99 × 10 9 N ⋅ m 2 C 2 7.50 × 10 −6 C π
2 k e qπ
=
Ex =
.
2
L2
0.140 m
FIG. P23.33
λ=
where
e
Thus,
je
f
a
13
j
7
Ex = 2.16 × 10 N C .
Solving,
e
j
Since the rod has a negative charge, E = −2.16 × 10 7 i N C = −21.6 i MN C .
P23.34
(a)
We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has
Qdx
and produces, at the chosen point, a field
charge
h
kex
Qdx
dE =
i.
3
2
h
x2 + R2
e
j
The total field is
z z
E=
dE =
d+h
e
k Qi ex + R j
E=
2h
b− 1 2g
d
all charge
k eQxdx
h x2 + R
j
2 3 2
d+h
2 −1 2
2
e
(b)
i=
ze
k eQi d + h 2
x + R2
2h x=d
k Qi
= e
h
x=d
LM
MM d
Ne
j
−3 2
2 xdx
OP
−
P
+R j
ead + hf + R j PQ
1
2
1
2 12
2
2
12
Think of the cylinder as a stack of disks, each with thickness dx, charge
per-area σ =
dE =
So,
Qdx
. One disk produces a field
π R2h
2π k eQdx
π R2h
F
GG 1 −
H ex
I
Ji .
+ R j JK
F
2 k Qdx G
1−
R h G
H ex
x
2
2 12
I
JJ i
E=
z dE = z
+R j K
L
O
2 k Qi M
2 k Qi L
1
E=
M z dx − 2 z ex + R j 2xdxPP = R h Mx
R h MN
MN
Q
2 k Qi L
O
d + h − d − ead + hf + R j + ed + R j P
E=
M
R h N
Q
2 k Qi L
O
h + ed + R j − ead + hf + R j P
E=
R h MN
Q
d+h
all charge
e
2
x =d
e
2
d+h
d+h
d
x=d
2
2
2 12
x
2 12
2
2 −1 2
2
e
2
e
2
Qdx
, and chargeh
2
e
2
12
2
2 12
2
2
12
e
2
2
1 x +R
d+h
−
d
2
12
j
1 2 d+h
d
OP
PP
Q
14
Electric Fields
P23.35
(a)
The electric field at point P due to each element of length dx, is
k dq
dE = 2 e 2 and is directed along the line joining the element to
x +y
point P. By symmetry,
Ex = dEx = 0
and since
dq = λdx ,
E = Ey = dEy = dE cos θ
where
cos θ =
z
z
z
ze
2
Therefore,
E = 2 k e λy
0
P23.36
dx
2
x +y
j
2 32
y
2
x + y2
.
FIG. P23.35
2 k e λ sin θ 0
.
y
=
θ 0 = 90°
Ey =
2k e λ
.
y
(b)
For a bar of infinite length,
(a)
The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π r r + L .
j b
e
and
a f
g
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 × 10 −10 C
(b)
For the curved lateral surface only, A = 2π rL .
j b
e
f
ga
Q = σA = 15.0 × 10 −9 C m 2 2π 0.025 0 m 0.060 0 m = 1.41 × 10 −10 C
P23.37
j b
e
g b0.060 0 mg =
2
(c)
Q = ρV = ρπ r 2 L = 500 × 10 −9 C m 3 π 0.025 0 m
(a)
Every object has the same volume, V = 8 0.030 0 m
e
je
a
f
3
5.89 × 10 −11 C
= 2.16 × 10 −4 m3 .
j
For each, Q = ρV = 400 × 10 −9 C m 3 2.16 × 10 −4 m3 = 8.64 × 10 −11 C
(b)
We must count the 9.00 cm 2 squares painted with charge:
(i)
6 × 4 = 24 squares
e
j e
j
j e
j
j e
j
j e
j
Q = σA = 15.0 × 10 −9 C m 2 24.0 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C
(ii)
34 squares exposed
e
Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C
(iii)
34 squares
e
Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C
(iv)
32 squares
e
Q = σA = 15.0 × 10 −9 C m 2 32.0 9.00 × 10 −4 m 2 = 4.32 × 10 −10 C
(c)
(i)
total edge length:
e
= e80.0 × 10
Q = λ = 80.0 × 10 −12
(ii)
Q=λ
continued on next page
b
g
C mj24 × b0.030 0 mg =
C mj44 × b0.030 0 mg =
= 24 × 0.030 0 m
−12
5.76 × 10 −11 C
1.06 × 10 −10 C
Chapter 23
Section 23.6
e
j b
g
e
j b
g
(iii)
Q = λ = 80.0 × 10 −12 C m 64 × 0.030 0 m = 1.54 × 10 −10 C
(iv)
Q = λ = 80.0 × 10 −12 C m 40 × 0.030 0 m = 0.960 × 10 −10 C
Electric Field Lines
P23.39
P23.38
FIG. P23.38
P23.40
(a)
(b)
P23.41
(a)
FIG. P23.39
q1 −6
1
=
= −
3
q 2 18
q1 is negative, q 2 is positive
The electric field has the general appearance shown. It is zero
at the center , where (by symmetry) one can see that the three
charges individually produce fields that cancel out.
In addition to the center of the triangle, the electric field lines in the
second figure to the right indicate three other points near the
middle of each leg of the triangle where E = 0 , but they are more
difficult to find mathematically.
(b)
You may need to review vector addition in Chapter Three. The
electric field at point P can be found by adding the electric field
vectors due to each of the two lower point charges: E = E 1 + E 2 .
The electric field from a point charge is E = k e
q
r2
r.
As shown in the solution figure at right,
E1 = k e
E2 = ke
q
a2
q
a2
to the right and upward at 60°
to the left and upward at 60°
E = E1 + E 2 = k e
= 1.73 k e
q
a2
j
q
a
2
ecos 60° i + sin 60° jj + e− cos 60° i + sin 60° jj = k
FIG. P23.41
e
q
a2
e
j
2 sin 60° j
15
16
Electric Fields
Section 23.7
P23.42
Motion of Charged Particles in a Uniform Electric Field
qE
m
F = qE = ma
a=
v f = vi + at
vf =
qEt
m
e1.602 × 10 ja520fe48.0 × 10 j =
=
4.39 × 10 6 m s
e1.602 × 10 ja520fe48.0 × 10 j =
2.39 × 10 3 m s
−19
ve
electron:
−9
−31
9.11 × 10
in a direction opposite to the field
−19
vp =
proton:
P23.43
P23.44
−9
1.67 × 10 −27
in the same direction as the field
a f
qE 1.602 × 10 −19 640
=
= 6.14 × 10 10 m s 2
m
1.67 × 10 −27
(a)
a=
(b)
v f = vi + at
(c)
x f − xi =
(d)
K=
(a)
−19
6.00 × 10 5
qE 1.602 × 10
a=
=
= 5.76 × 10 13 m s so a = −5.76 × 10 13 i m s 2
m
1.67 × 10 −27
(b)
v f = vi + 2 a x f − xi
e
1
vi + v f t
2
d
i
xf =
e
e
je
je
e
d
e
je
j
2
= 1.20 × 10 −15 J
j
j
jb
g
v i = 2.84 × 10 6 i m s
v f = vi + at
e
j
0 = 2.84 × 10 6 + −5.76 × 10 13 t
t = 4.93 × 10 −8 s
The required electric field will be in the direction of motion .
Work done = ∆K
so,
− Fd = −
which becomes
eEd = K
and
E=
j
i
e
P23.45
t = 1.95 × 10 −5 s
1
1.20 × 10 6 1.95 × 10 −5 = 11.7 m
2
1
1
mv 2 = 1.67 × 10 −27 kg 1.20 × 10 6 m s
2
2
0 = vi2 + 2 −5.76 × 10 13 0.070 0
(c)
j
1.20 × 10 6 = 6.14 × 10 10 t
1
mvi2 (since the final velocity = 0 )
2
K
.
ed
Chapter 23
P23.46
17
The acceleration is given by
d
a f
i
v 2f = vi2 + 2 a x f − x i or
v 2f = 0 + 2 a − h .
Solving
a=−
Now
∑ F = ma :
Therefore
qE = −
(a)
v 2f
2h
F
GH
.
mv 2f
2h
− mg j + qE = −
mv 2f j
I
JK
2h
.
+ mg j .
Gravity alone would give the bead downward impact velocity
ja
e
f
2 9.80 m s 2 5.00 m = 9.90 m s .
To change this to 21.0 m/s down, a downward electric field must exert a downward
electric force.
P23.47
F
GH
I
JK
(b)
(a)
t=
(b)
ay =
(c)
I LM b21.0 m sg
JK M 2a5.00 mf
N
OP
PQ
2
− 9.80 m s 2 = 3.43 µC
0.050 0
x
=
= 1.11 × 10 −7 s = 111 ns
v x 4.50 × 10 5
e
je
j
−19
9.60 × 10 3
qE 1.602 × 10
=
= 9.21 × 10 11 m s 2
−27
m
1.67 × 10
e
y f − yi = v yi t +
*P23.48
F
GH
2
1.00 × 10 −3 kg N ⋅ s 2
m vf
−g =
q=
E 2h
1.00 × 10 4 N C kg ⋅ m
j
1
ayt 2 :
2
yf =
1
9.21 × 10 11 1.11 × 10 −7
2
e
je
e
j
2
= 5.68 × 10 −3 m = 5.68 mm
je
j
v yf = v yi + a y t = 9.21 × 10 11 1.11 × 10 −7 = 1.02 × 10 5 m s
v x = 4.50 × 10 5 m s
jb
ge j
e j
∑ F = e 2 × 10 kg ⋅ m s je− jj = 1 × 10 m s − j .
a=
and moves with acceleration:
e
je j
m
2 × 10
kg
Its x-component of velocity is constant at e1.00 × 10 m sj cos 37° = 7.99 × 10 m s . Thus it moves in a
e
The particle feels a constant force: F = qE = 1 × 10 −6 C 2 000 N C − j = 2 × 10 −3 N − j
−3
2
13
−16
5
2
4
parabola opening downward. The maximum height it attains above the bottom plate is described by
d
i
2
2
= v yi
+ 2 a y y f − yi :
v yf
e
0 = 6.02 × 10 4 m s
y f = 1.81 × 10 −4 m .
continued on next page
j − e2 × 10
2
13
jd
m s2 y f − 0
i
18
Electric Fields
Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric
parabola and strikes the bottom plate after a time given by
y f = yi + v yi t +
1
ayt 2
2
e
j
0 = 0 + 6.02 × 10 4 m s t +
1
−1 × 10 13 m s 2 t 2
2
e
j
since t > 0 ,
t = 1.20 × 10 −8 s .
The particle’s range is
x f = xi + v x t = 0 + 7.99 × 10 4 m s 1.20 × 10 −8 s = 9.61 × 10 −4 m .
e
je
j
In sum,
The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm
and a width of 0.961 mm.
P23.49
vi = 9.55 × 10 3 m s
ay =
(a)
e
^
ja f
j
−19
720
eE 1.60 × 10
=
= 6.90 × 10 10 m s 2
−
27
m
1.67 × 10
e
vi2 sin 2θ
= 1. 27 × 10 −3 m so that
ay
R=
e9.55 × 10 j
3 2
6.90 × 10
sin 2θ
10
sin 2θ = 0.961
t=
(b)
FIG. P23.49
= 1.27 × 10 −3
θ = 36.9°
R
R
=
vix vi cos θ
90.0°−θ = 53.1°
If θ = 36.9° , t = 167 ns .
If θ = 53.1° , t = 221 ns .
Additional Problems
*P23.50
The two given charges exert equal-size forces of attraction on each
other. If a third charge, positive or negative, were placed between
them they could not be in equilibrium. If the third charge were at a
point x > 15 cm , it would exert a stronger force on the 45 µ C than
on the −12 µ C , and could not produce equilibrium for both. Thus
the third charge must be at x = − d < 0 . Its equilibrium requires
b
k e q 12 µ C
d
2
g = k qb45 µ Cg
a15 cm + df
e
2
15 cm + d = 1.94d
FG 15 cm + d IJ
H d K
2
=
d
q
x=0
–
–12 µC
15 cm
x
+
45 µ C
FIG. P23.50
45
= 3.75
12
d = 16.0 cm .
The third charge is at x = −16.0 cm . The equilibrium of the −12 µ C requires
b
g = k b45 µ Cg12 µ C
a16.0 cmf
a15 cmf
k e q 12 µ C
2
e
2
q = 51.3 µ C .
All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is
the only solution.
Chapter 23
P23.51
jb
e
while the e − has acceleration
ae =
(a)
e1.60 × 10
jb
C 640 N C
kg
FG
H
IJ
K
1
1
1
a p t 2 + a e t 2 = 1 837 a p t 2 .
2
2
2
1
4.00 cm
2
= 21.8 µm .
d = apt =
2
1 837
g = 1.12 × 10
14
m s 2 = 1 836 a p .
1
a p t 2 ), knowing:
2
The distance from the positive plate to where the meeting occurs equals the distance the
1
sodium ion travels (i.e., d Na = a Na t 2 ). This is found from:
2
eE
eE
1
1
1
1
2
2
t2 +
t2 .
4.00 cm = a Na t + aCl t :
4.00 cm =
2
2
2 22.99 u
2 35. 45 u
1
1
1
This may be written as
4.00 cm = a Na t 2 + 0.649 a Na t 2 = 1.65 a Na t 2
2
2
2
1
4.00 cm
2
= 2.43 cm .
so
d Na = a Na t =
2
1.65
FG
H
(a)
−31
We want to find the distance traveled by the proton (i.e., d =
Thus,
(b)
−19
9.110 × 10
4.00 cm =
P23.52
g
The proton moves with acceleration
−19
C 640 N C
qE 1.60 × 10
=
= 6.13 × 10 10 m s 2
ap =
m
1.673 × 10 −27 kg
IJ
K
b
FG
H
IJ
K
g
FG
H
IJ
K
The field, E1 , due to the 4.00 × 10 −9 C charge is in the –x
direction.
8.99 × 10 9 N ⋅ m 2 C 2 −4.00 × 10 −9 C
keq
E1 = 2 r =
i
2
r
2.50 m
e
a
je
f
j
FIG. P23.52(a)
= −5.75 i N C
Likewise, E 2 and E3 , due to the 5.00 × 10 −9 C charge and the 3.00 × 10 −9 C charge are
E2 =
E3
ke q
r
2
e8.99 × 10
r=
e8.99 × 10
=
9
je
j i = 11.2 N C i
a2.00 mf
N ⋅ m C je3.00 × 10 C j
i = 18.7 N C i
a1.20 mf
9
N ⋅ m 2 C 2 5.00 × 10 −9 C
2
2
−9
2
2
E R = E1 + E 2 + E 3 = 24.2 N C in +x direction.
(b)
E2
E3
ke q
b
ge
j
=
r = b11. 2 N C ge + jj
r
k q
=
r = b5.81 N C ge −0.371i +0.928 jj
r
E1 =
r2
ke q
r = −8.46 N C 0.243 i + 0.970 j
2
e
2
Ex = E1 x + E3 x = −4.21i N C
ER = 9.42 N C
Ey = E1 y + E2 y + E3 y = 8.43 j N C
θ = 63.4° above − x axis
FIG. P23.52(b)
19
20
Electric Fields
*P23.53
(a)
Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.
∑ F = ma
(b)
qE =
mv 2
R
E=
d
For the x-motion,
2
v xf2 = v xi
+ 2 a x x f − xi
0 = v 2 + 2ax R
ax = −
Ex = −
mv 2
qR
i
v 2 Fx qE x
=
=
m
2R m
mv 2
. Similarly for the y-motion,
2 qR
v 2 = 0 + 2ay R
ay = +
v 2 qE y
=
m
2R
Ey =
mv 2
2 qR
The magnitude of the field is
E x2 + E y2 =
P23.54
at 135° counterclockwise from the x -axis .
2 qR
From the free-body diagram shown,
∑ Fy = 0 :
T cos 15.0° = 1.96 × 10 −2 N .
So
T = 2.03 × 10 −2 N .
From
or
P23.55
mv 2
(a)
∑ Fx = 0 , we have
q=
qE = T sin 15.0°
e
j
2.03 × 10 −2 N sin 15.0°
T sin 15.0°
=
= 5.25 × 10 −6 C = 5.25 µC .
E
1.00 × 10 3 N C
FIG. P23.54
Let us sum force components to find
∑ Fx = qEx − T sin θ = 0 , and ∑ Fy = qEy + T cos θ − mg = 0 .
Combining these two equations, we get
q=
e1.00 × 10 ja9.80f
eE cot θ + E j a3.00 cot 37.0°+5.00f × 10
−3
mg
x
=
y
5
= 1.09 × 10 −8 C
= 10.9 nC
(b)
From the two equations for
T=
Free Body Diagram
FIG. P23.55
∑ Fx
and
∑ Fy
qEx
= 5.44 × 10 −3 N = 5.44 mN .
sin 37.0°
we also find
Chapter 23
P23.56
21
This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ. The
unknowns are q and T.
The approach to this problem should be the same as for the last problem, but without
numbers to substitute for the variables. Likewise, we can use the free body diagram given in the
solution to problem 55.
Again, Newton’s second law:
and
qA
, into Eq. (2),
sin θ
Substituting T =
(a)
(b)
∑ Fx = −T sin θ + qA = 0
(1)
∑ Fy = +T cos θ + qB − mg = 0
(2)
qA cos θ
+ qB = mg .
sin θ
mg
A cot θ + B
Isolating q on the left,
q=
a
Substituting this value into Eq. (1),
T=
a A cosθ + B sinθ f
mgA
f
.
.
If we had solved this general problem first, we would only need to substitute the
appropriate values in the equations for q and T to find the numerical results needed for
problem 55. If you find this problem more difficult than problem 55, the little list at the first
step is useful. It shows what symbols to think of as known data, and what to consider
unknown. The list is a guide for deciding what to solve for in the analysis step, and for
recognizing when we have an answer.
P23.57
F=
k e q1 q 2
r
2
tan θ =
:
15.0
60.0
θ = 14.0°
e8.99 × 10 je10.0 × 10 j
F =
a0.150f
e8.99 × 10 je10.0 × 10 j
F =
a0.600f
e8.99 × 10 je10.0 × 10 j
F =
a0.619f
−6 2
9
1
2
−6 2
9
3
2
= 2.50 N
−6 2
9
2
= 40.0 N
2
= 2.35 N
Fx = − F3 − F2 cos 14.0° = −2.50 − 2.35 cos 14.0° = −4.78 N
Fy = − F1 − F2 sin 14.0° = −40.0 − 2.35 sin 14.0° = −40.6 N
Fnet = Fx2 + Fy2 =
tan φ =
Fy
Fx
φ = 263°
=
−40.6
−4.78
a4.78f + a40.6f
2
2
= 40.9 N
FIG. P23.57
22
Electric Fields
P23.58
d cos 30.0° = 15.0 cm,
15.0 cm
d=
cos 30.0°
From Figure A:
or
θ = sin −1
From Figure B:
θ = sin −1
FG d IJ
H 50.0 cm K
F 15.0 cm I = 20.3°
GH 50.0 cmacos 30.0°f JK
Figure A
Fq
or
= tan θ
mg
Fq = mg tan 20.3°
From Figure C:
Fq = 2 F cos 30.0°
(1)
LM k q OP cos 30.0°
MN a0.300 mf PQ
Combining equations (1) and (2),
L k q OP cos 30.0° = mg tan 20.3°
2M
MN a0.300 mf PQ
mg a0.300 mf tan 20.3°
q =
2
e
Fq = 2
e
2
(2)
Figure B
2
2
2
2
2 k e cos 30.0°
e2.00 × 10 kg je9.80 m s ja0.300 mf tan 20.3°
=
2e8.99 × 10 N ⋅ m C j cos 30.0°
−3
q
2
2
2
9
2
2
Figure C
q = 4.20 × 10 −14 C 2 = 2.05 × 10 −7 C = 0.205 µC
P23.59
Charge
FIG. P23.58
Q
resides on each block, which repel as point charges:
2
F=
Q=
Solving for Q,
*P23.60
P23.61
b gb g = kbL − L g .
L
k bL − L g
.
2L
ke Q 2 Q 2
i
2
i
ke
If we place one more charge q at the 29th vertex, the total force on the central charge will add up to
k qQ
k e qQ
toward vertex 29 .
F28 charges =
zero: F28 charges + e 2 away from vertex 29 = 0
a
a2
According to the result of Example 23.7, the left-hand rod creates
this field at a distance d from its right-hand end:
k eQ
E=
d 2a + d
a
dF =
F=
F=
f
k eQQ
dx
2a d d + 2a
a
k eQ
2a
2
+ k eQ
4a
2
FIG. P23.61
f
IJ
z a f FGH
K
F k Q I lnF
FG − ln 2a + b + ln b IJ = k Q ln b
= G
H b
K
b − 2a
ab − 2 afab + 2af H 4a JK GH b
4a
b
k Q2
dx
1
2a + x
= e
− ln
x x + 2a
x
2a
2a
x=b− 2a
2
e
2
2
b
b−2a
2
e
2
2
2
b2
− 4a 2
I
JK
23
Chapter 23
P23.62
b
g
At equilibrium, the distance between the charges is r = 2 0.100 m sin 10.0° = 3.47 × 10 −2 m
Now consider the forces on the sphere with charge +q , and use
∑ Fy = 0 :
∑ Fx = 0 :
∑ Fy = 0 :
mg
cos 10.0°
= F2 − F1 = T sin 10.0°
T cos 10.0° = mg , or T =
(1)
Fnet
(2)
θθ L
–q
+q
r
Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1).
mg sin 10.0°
Fnet =
= mg tan 10.0° = 2.00 × 10 −3 kg 9.80 m s 2 tan 10.0° = 3.46 × 10 −3 N
cos 10.0°
e
je
j
Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted
by −q , and F2 is the force exerted on +q by the external electric field.
FIG. P23.62
Fnet = F2 − F1 or F2 = Fnet + F1
e5.00 × 10 Cje5.00 × 10 Cj = 1.87 × 10
C j
e3.47 × 10 mj
−8
e
9
F1 = 8.99 × 10 N ⋅ m
2
−8
2
−3
2
−2
N
Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10 −3 N + 1.87 × 10 −2 N = 2.21 × 10 −2 N
and F2 = qE , or E =
P23.63
z
Q = λd =
z
F2 2. 21 × 10 −2 N
=
= 4. 43 × 10 5 N C = 443 kN C .
q
5.00 × 10 −8 C
90 .0 °
λ 0 cos θRdθ = λ 0 R sin θ
−90 .0 °
90 .0 °
−90.0 °
a f
= λ 0 R 1 − −1 = 2 λ 0 R
λ = 10.0 µC m
so
b ga f
F b3.00 µCgeλ cos θRdθ j I
1
1 F b3.00 µC gbλd g I
cos θ =
dF =
G
JJ
G
J
4π ∈ G
4π ∈ H
R
R
K
H
K
e3.00 × 10 Cje10.0 × 10 C mj cos θdθ
F = z e8.99 × 10 N ⋅ m C j
a0.600 mf
8.99a30.0f
F =
e10 Nj z FGH 12 + 12 cos 2θ IJK dθ
0.600
F1 1
I = 0.707 N Downward.
F = a0.450 N fG θ + sin 2θ
JK
H2 4
Q = 12.0 µC = 2 λ 0 0.600 m = 12.0 µC
0
2
0
y
2
0
90.0 °
y
2
0
−6
−6
9
2
2
2
−90 .0 °
y
−3
−π 2
−π 2
cosθ
0
–1
π 2
π 2
y
1
1
0
0°
360°
cos2θ
0°
360°
FIG. P23.63
Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 .
P23.64
At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be
located by determining the angle θ corresponding to equilibrium.
In terms of lengths s,
attractive force
1
a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an
2
es +
continued on next page
k eQq
1
2
a 3
j
2
24
Electric Fields
The other two charges exert equal repulsive forces of magnitude
of the two repulsive forces add, balancing the attractive force,
Fnet = k eQq
1
2
k eQq
r2
. The horizontal components
LM 2 cosθ
MM r − es +
N
2
a
1
2
r=
The equilibrium condition, in terms of θ, is
Fnet =
Thus the equilibrium value of θ satisfies
2 cos θ sin 2 θ
sin θ
2
a
s=
From Figure P23.64
OP
P=0
3j P
Q
1
1
a cot θ
2
FG 4 IJ k QqFG 2 cos θ sin θ −
H a K GH
e
2
2
e
e
3 + cot θ
j
2
I
JJ = 0 .
3 + cot θ j K
1
2
=1.
One method for solving for θ is to tabulate the left side. To three significant figures a value of θ
corresponding to equilibrium is 81.7°.
The distance from the vertical side of the triangle to the equilibrium position is
1
s = a cot 81.7° = 0.072 9 a .
2
θ
2 cos θ sin 2 θ
60°
FIG. P23.64
e
3 + cot θ
j
2
4
70°
2.654
80°
1.226
90°
0
81°
1.091
81.5°
1.024
81.7°
0.997
A second zero-field point is on the negative side of the x-axis, where θ = −9.16° and s = −3.10 a .
P23.65
(a)
(b)
From the 2Q charge we have
Fe − T2 sin θ 2 = 0 and mg − T2 cos θ 2 = 0 .
Combining these we find
Fe
T sin θ 2
= 2
= tan θ 2 .
mg T2 cos θ 2
From the Q charge we have
Fe = T1 sin θ 1 = 0 and mg − T1 cos θ 1 = 0 .
Combining these we find
Fe
T sin θ 1
= 1
= tan θ 1 or θ 2 = θ 1 .
mg T1 cos θ 1
Fe =
k e 2QQ
r
2
=
2 k eQ
FIG. P23.65
2
r2
If we assume θ is small then
tan θ ≈
r 2
.
Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r.
F I
GH JK
F
GH
Fe
2k Q 2 1
4k eQ 2
r
= tan θ then e 2
≈
and solving for r we find r ≈
mg
2
mg
mg
r
I
JK
13
.
25
Chapter 23
P23.66
(a)
The distance from each corner to the center of the square is
FG L IJ + FG L IJ
H 2K H 2K
2
2
=
L
2
z
+q
.
+q
L/2
L/2
+q
The distance from each positive charge to −Q is then
z2 +
x
–Q
+q
L2
. Each positive charge exerts a force directed
2
FIG. P23.66
k eQq
along the line joining q and −Q , of magnitude
z 2 + L2 2
.
z
The line of force makes an angle with the z-axis whose cosine is
2
z + L2 2
The four charges together exert forces whose x and y components add to zero, while the
4k e Qqz
F= −
k
z-components add to
3 2
2
z + L2 2
e
(b)
For z >> L , the magnitude of this force is Fz = −
4k eQqz
eL 2 j
2
32
F 4a2f k Qq I z = ma
GH L JK
3 2
=−
3
e
j
z
Therefore, the object’s vertical acceleration is of the form a z = −ω 2 z
with ω 2 =
af
42
32
k eQq
3
mL
=
k eQq 128
mL3
.
Since the acceleration of the object is always oppositely directed to its excursion from
equilibrium and in magnitude proportional to it, the object will execute simple harmonic
motion with a period given by
T=
P23.67
(a)
2π
ω
=
2π
a128f
14
mL3
=
k e Qq
mL3
.
k eQq
π
a8 f
14
FG
H
IJ
K
qE
,
m
which is constant and directed downward. Therefore, it behaves like a simple pendulum in
the presence of a modified uniform gravitational field with a period given by:
The total non-contact force on the cork ball is: F = qE + mg = m g +
T = 2π
L
0.500 m
= 2π
g + qE m
9.80 m s 2 + 2.00 × 10 −6 C 1.00 × 10 5 N C 1.00 × 10 −3 kg
e
je
j
= 0.307 s
(b)
Yes . Without gravity in part (a), we get T = 2π
T = 2π
0.500 m
e2.00 × 10 Cje1.00 × 10
−6
5
j
L
qE m
N C 1.00 × 10 −3 kg
= 0.314 s (a 2.28% difference).
26
Electric Fields
P23.68
The bowl exerts a normal force on each bead, directed along the
radius line or at 60.0° above the horizontal. Consider the free-body
diagram of the bead on the left:
∑ Fy = n sin 60.0°− mg = 0 ,
mg
.
sin 60.0°
or
n=
Also,
∑ Fx = − Fe + n cos 60.0° = 0 ,
keq 2
or
R
2
F mg I
RG
H k 3 JK
q=
Thus,
= n cos 60.0° =
n
mg
mg
=
.
tan 60.0°
3
Fe
mg
12
.
FIG. P23.68
e
P23.69
(a)
60.0°
There are 7 terms which contribute:
3 are s away (along sides)
3 are
1 is
1
2s away (face diagonals) and sin θ =
3s away (body diagonal) and sin φ =
= cos θ
2
1
3
.
FIG. P23.69
The component in each direction is the same by symmetry.
F=
P23.70
ke q2
s2
LM1 + 2 + 1 OPei + j + kj =
N 2 2 3 3Q
ke q2
keq 2
s2
a1.90fei + j + kj
(b)
F = Fx2 + Fy2 + Fz2 = 3. 29
(a)
Zero contribution from the same face due to symmetry, opposite
face contributes
FG k q sin φIJ where
Hr K
4
e
2
sin φ =
s
r
r=
s2
away from the origin
FG s IJ + FG s IJ
H 2K H 2K
E=4
2
k e qs
r3
=
2
+ s 2 = 1.5 s = 1.22s
4
a1.22f
3
keq
s2
= 2.18
keq
s2
FIG. P23.70
(b)
The direction is the k direction.
27
Chapter 23
P23.71
k e xQ
E = Ex =
The field on the axis of the ring is calculated in Example 23.8,
The force experienced by a charge −q placed along the axis of the ring is
ex + a j
LM x
F = − k Qq
MM ex + a j
N
F k Qq IJ x
F = −G
Ha K
e
2 3 2
2
e
and when x << a , this becomes
2 3 2
2
OP
PP
Q
3
This expression for the force is in the form of Hooke’s law, with an
k=
effective spring constant of
k
, we have
m
Since ω = 2π f =
P23.72
dE =
F
G
+ a0.150 mf GH
k e dq
x
2
z
E=
dE = k e λ
2
z
0. 400 m
x=0
all charge
f=
I k λe− xi + 0.150 mjjdx
J=
x + a0.150 mf JK
x + a0.150 mf
e− xi + 0.150 mjjdx
x + a0.150 mf
− x i + 0.150 m j
2
2
2
2 3 2
2
2
.
FIG. P23.72
2
2
0
2
ma 3
0. 400 m
2
9
k eQq
1
2π
2 32
2
0. 400 m
2
a3
e
LM
OP
0.150 mf j x
a
+i
E = k λM
+
PP
0
150
m
.
x
+
a
f
a0.150 mf x + a0.150 mf
MN
Q
E = e8.99 × 10 N ⋅ m C je35.0 × 10 C mj ia 2.34 − 6.67f m + ja6.24 − 0 f m
E = e −1.36 i + 1.96 jj × 10 N C = e −1.36 i + 1.96 jj kN C
e
k eQq
0
−9
−1
−1
3
P23.73
The electrostatic forces exerted on the two charges result in a net torque
τ = −2 Fa sin θ = −2Eqa sin θ .
For small θ, sin θ ≈ θ and using p = 2 qa , we have
τ = −Epθ .
The torque produces an angular acceleration given by τ = Iα = I
Combining these two expressions for torque, we have
This equation can be written in the form
d 2θ
dt 2
.
FG IJ
H K
Ep
d 2θ
θ =0.
+
2
I
dt
d 2θ
dt
2
FIG. P23.73
= −ω 2θ where ω 2 =
Ep
.
I
This is the same form as Equation 15.5 and the frequency of oscillation is found by comparison with
Equation 15.11, or
f=
1
2π
pE
1
=
I
2π
2 qaE
.
I
28
Electric Fields
ANSWERS TO EVEN PROBLEMS
P23.2
P23.4
(a) 2.62 × 10 24 ; (b) 2.38 electrons for every
10 9 present
P23.36
(a) 200 pC; (b) 141 pC; (c) 58.9 pC
P23.38
see the solution
P23.40
(a) −
57.5 N
−9
1
; (b) q1 is negative and q 2 is positive
3
P23.6
2.51 × 10
P23.8
514 kN
P23.42
electron: 4.39 Mm s ; proton: 2.39 km s
P23.10
x = 0.634d . The equilibrium is stable if the
third bead has positive charge.
P23.44
(a) −57.6 i Tm s 2 ; (b) 2.84i Mm s; (c) 49.3 ns
P23.46
(a) down; (b) 3.43 µC
P23.48
The particle strikes the negative plate after
moving in a parabola 0.181 mm high and
0.961 mm.
P23.50
Possible only with +51.3 µ C at
x = −16.0 cm
P23.12
md 3
where m is the mass
k e qQ
π
(a) period =
2
k e qQ
of the object with charge −Q ; (b) 4a
md 3
P23.14
1.49 g
P23.16
720 kN C down
P23.52
(a) 24.2 N C at 0°; (b) 9. 42 N C at 117°
P23.18
(a) 18.0 i − 218 j kN C ;
P23.54
5.25 µC
P23.56
(a)
P23.58
0.205 µC
e
j
(b) 36.0 i − 436 j mN
mg
mgA
; (b)
A cot θ + B
A cos θ + B sin θ
P23.20
(a) 12.9 j kN C ; (b) −38.6 j mN
P23.22
see the solution
P23.24
π 2 ke q
−
i
6a2
P23.62
443 i kN C
P23.26
keλ 0
−i
x0
P23.64
0.072 9 a
P23.28
keλ 0
−i
2x0
P23.66
see the solution; the period is
P23.68
R
P23.30
P23.32
P23.34
P23.60
e j
e j
(a) 383 MN C away; (b) 324 MN C away;
(c) 80.7 MN C away; (d) 6.68 MN C away
LMe
j − ead + hf + R j
N
2 k Qi L
(b)
h + ed + R j − ead + hf + R j
R h MN
k eQ i
d 2 + R2
h
e
2
2
a2
toward the 29th vertex
F mg I
GH k 3 JK
12
e
see the solution
(a)
k e qQ
−1 2
2 12
2
2
2
−1 2
2
OP;
Q
OP
Q
12
P23.70
(a) see the solution; (b) k
P23.72
e−1.36i + 1.96 jj kN C
π
81 4
mL3
k eQq
24
Gauss’s Law
CHAPTER OUTLINE
24.1
24.2
24.3
24.4
24.5
Electric Flux
Gauss’s Law
Application of Gauss’s Law
to Various Charge
Distributions
Conductors in Electrostatic
Equilibrium
Formal Derivation of
Gauss‘s Law
ANSWERS TO QUESTIONS
Q24.1
The luminous flux on a given area is less when the sun is low in
the sky, because the angle between the rays of the sun and the
local area vector, dA, is greater than zero. The cosine of this
angle is reduced. The decreased flux results, on the average, in
colder weather.
Q24.2
If the region is just a point, line, or plane, no. Consider two
protons in otherwise empty space. The electric field is zero at
the midpoint of the line joining the protons. If the field-free
region is three-dimensional, then it can contain no charges, but
it might be surrounded by electric charge. Consider the interior
of a metal sphere carrying static charge.
Q24.3
The surface must enclose a positive total charge.
Q24.4
The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains
zero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field lines entering
one side of the closed surface come out the other side and the net flux is zero.
Q24.5
Gauss’s law cannot tell the different values of the electric field at different points on the surface.
When E is an unknown number, then we can say E cos θdA = E cos θdA . When E x , y , z is an
z
unknown function, then there is no such simplification.
z
b
g
Q24.6
The electric flux through a sphere around a point charge is independent of the size of the sphere. A
sphere of larger radius has a larger area, but a smaller field at its surface, so that the product of field
strength and area is independent of radius. If the surface is not spherical, some parts are closer to the
charge than others. In this case as well, smaller projected areas go with stronger fields, so that the
net flux is unaffected.
Q24.7
Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a
1
field lines pointing out from it horizontally to
vertical sheet carrying charge +1 coulomb. It has
∈0
the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet
and far away, showing that the field has the same value at all distances.
29
30
Gauss’s Law
Q24.8
Consider any point, zone, or object where electric field lines begin. Surround it with a close-fitting
gaussian surface. The lines will go outward through the surface to constitute positive net flux. Then
Gauss’s law asserts that positive net charge must be inside the surface: it is where the lines begin.
Similarly, any place where electric field lines end must be just inside a gaussian surface passing net
negative flux, and must be a negative charge.
Q24.9
Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels
every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface
of the conductor.
Q24.10
If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell
carries no charge. The person is not harmed by touching this wall. If the person carries a (small)
charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner wall
of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his
body jumps to the metal.
Q24.11
The electric fields outside are identical. The electric fields inside are very different. We have E = 0
everywhere inside the conducting sphere while E decreases gradually as you go below the surface of
the sphere with uniform volume charge density.
Q24.12
There is zero force. The huge charged sheet creates a uniform field. The field can polarize the
neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal
amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of
charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces add
to zero.
Q24.13
Gauss’s law predicts, as described in section 24.4, that excess charge on a conductor will reside on
the surface of the conductor. If a car is left charged by a lightning strike, then that charge will remain
on the outside of the car, not harming the occupants. It turns out that during the lightning strike, the
current also remains on the outside of the conductor. Note that it is not necessarily safe to be in a
fiberglass car or a convertible during a thunderstorm.
SOLUTIONS TO PROBLEMS
Section 24.1
P24.1
Electric Flux
ja
e
f
(a)
Φ E = EA cos θ = 3.50 × 10 3 0.350 × 0.700 cos 0° = 858 N ⋅ m 2 C
(b)
θ = 90.0°
(c)
Φ E = 3.50 × 10 3 0.350 × 0.700 cos 40.0° = 657 N ⋅ m 2 C
ΦE = 0
ja
e
e
f
je
j
P24.2
Φ E = EA cos θ = 2.00 × 10 4 N C 18.0 m 2 cos 10.0° = 355 kN ⋅ m 2 C
P24.3
Φ E = EA cos θ
a
a
A = π r 2 = π 0.200
f
5.20 × 10 5 = E 0.126 cos 0°
f
2
= 0.126 m 2
E = 4.14 × 10 6 N C = 4.14 MN C
Chapter 24
P24.4
(a)
a
fa
f
A ′ = 10.0 cm 30.0 cm
30.0 cm
2
A ′ = 300 cm = 0.030 0 m 2
Φ E , A ′ = EA ′ cos θ
jb
e
31
g
Φ E , A ′ = 7.80 × 10 4 0.030 0 cos 180°
10.0 cm
60.0Þ
Φ E , A ′ = −2.34 kN ⋅ m 2 C
(b)
e
ja f
F 10.0 cm IJ = 600 cm = 0.060 0 m
A = a30.0 cmfa wf = a30.0 cmfG
H cos 60.0° K
Φ
= e7.80 × 10 jb0.060 0g cos 60.0° = +2.34 kN ⋅ m C
2
2
2
4
E, A
(c)
FIG. P24.4
Φ E , A = EA cos θ = 7.80 × 10 4 A cos 60.0°
The bottom and the two triangular sides all lie parallel to E, so Φ E = 0 for each of these. Thus,
Φ E, total = −2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0 .
P24.5
P24.6
j
Φ E = E ⋅ A = a i + bj ⋅ A i = aA
(b)
Φ E = a i + bj ⋅ Aj = bA
(c)
Φ E = a i + bj ⋅ Ak = 0
e
j
e
j
Only the charge inside radius R contributes to the total flux.
ΦE =
P24.7
e
(a)
q
∈0
Φ E = EA cos θ through the base
a fa f
Φ E = 52.0 36.0 cos 180° = −1.87 kN ⋅ m 2 C .
Note the same number of electric field lines go through the base as go through the
pyramid’s surface (not counting the base).
FIG. P24.7
For the slanting surfaces, Φ E = +1.87 kN ⋅ m 2 C .
P24.8
The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side
of the cone is Φ E = E ⋅ dA = ERh . This is the same as the flux that exits the right side of the cone.
z
Note that for a uniform field only the cross sectional area matters, not shape.
32
Gauss’s Law
Section 24.2
P24.9
(a)
Gauss’s Law
ΦE =
b
g
+5.00 µC − 9.00 µC + 27.0 µC − 84.0 µC
qin
=
= −6.89 × 10 6 N ⋅ m 2 C 2
∈0
8.85 × 10 −12 C 2 N ⋅ m 2
Φ E = −6.89 MN ⋅ m 2 C
(b)
P24.10
(a)
Since the net electric flux is negative, more lines enter than leave the surface.
E=
k eQ
r2
8.90 × 10
:
But Q is negative since E points inward.
(b)
P24.11
P24.12
ΦE =
e8.99 × 10 jQ
=
a0.750f
9
2
2
Q = −5.56 × 10 −8 C = −55.6 nC
The negative charge has a spherically symmetric charge distribution.
qin
∈0
Through S1
ΦE =
−2Q + Q
Q
= −
∈0
∈0
Through S 2
ΦE =
+Q − Q
= 0
∈0
Through S3
ΦE =
−2Q + Q − Q
2Q
= −
∈0
∈0
Through S 4
ΦE = 0
(a)
One-half of the total flux created by the charge q goes through the plane. Thus,
q
1
1 q
Φ E , plane = Φ E , total =
=
.
2
2 ∈0
2 ∈0
(b)
The square looks like an infinite plane to a charge very close to the surface. Hence,
q
Φ E , square ≈ Φ E , plane =
.
2 ∈0
(c)
FG IJ
H K
The plane and the square look the same to the charge.
P24.13
The flux through the curved surface is equal to the flux through the flat circle, E0 π r 2 .
P24.14
(a)
Φ E , shell =
(b)
Φ E, half shell =
(c)
qin 12.0 × 10 −6
=
= 1.36 × 10 6 N ⋅ m 2 C = 1.36 MN ⋅ m 2 C
∈0 8.85 × 10 −12
1
1.36 × 10 6 N ⋅ m 2 C = 6.78 × 10 5 N ⋅ m 2 C = 678 kN ⋅ m 2 C
2
e
j
No, the same number of field lines will pass through each surface, no matter how the
radius changes.
Chapter 24
P24.15
With δ very small, all points on the hemisphere are nearly at
a distance R from the charge, so the field everywhere on the
kQ
curved surface is e 2 radially outward (normal to the
R
surface). Therefore, the flux is this field strength times the
area of half a sphere:
(a)
z
δ →0
33
Q
Φ curved = E ⋅ dA = Elocal A hemisphere
FG
H
Φ curved = k e
(b)
Q
R2
IJ FG 1 4π R IJ = 1 Qa2π f =
K H 2 K 4π ∈
2
0
FIG. P24.15
The closed surface encloses zero charge so Gauss’s law gives
Φ curved + Φ flat = 0
*P24.16
+Q
2 ∈0
Φ flat = − Φ curved =
or
−Q
.
2 ∈0
Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.
z
E ⋅ dA =
mf
b+120 N CgA + b−100 N CgA = ρAa100
∈
b20 N Cge8.85 × 10 C N ⋅ m j = 1.77 × 10
ρ=
q
:
∈0
0
−12
2
2
100 m
−12
C m3
The charge is positive , to produce the net outward flux of electric field.
P24.17
The total charge is Q − 6 q . The total outward flux from the cube is
Q−6q
∈0
, of which one-sixth goes
through each face:
bΦ g
=
bΦ g
=
E one face
E one face
P24.18
Q−6q
6 ∈0
Q−6q
6 ∈0
=
a5.00 − 6.00f × 10
−6
6 × 8.85 × 10
C ⋅ N ⋅ m2
−12
C
2
= −18.8 kN ⋅ m 2 C .
The total charge is Q − 6 q . The total outward flux from the cube is
through each face:
bΦ g
E one face
P24.19
=
Q−6q
6 ∈0
.
If R ≤ d , the sphere encloses no charge and Φ E =
qin
= 0 .
∈0
If R > d , the length of line falling within the sphere is 2 R 2 − d 2
so
ΦE =
2λ R 2 − d 2
∈0
.
Q−6q
∈0
, of which one-sixth goes
34
Gauss’s Law
P24.20
Φ E , hole = E ⋅ A hole =
FG k Q IJ eπ r
HR K
e
2
F e8.99 × 10 N ⋅ m C je10.0 × 10 Cj I
JJ π e1.00 × 10
j GG
a0.100 mf
H
K
2
9
2
=
−6
2
2
−3
j
m
2
Φ E, hole = 28.2 N ⋅ m 2 C
P24.21
ΦE =
qin
170 × 10 −6 C
=
= 1.92 × 10 7 N ⋅ m 2 C
∈0 8.85 × 10 −12 C 2 N ⋅ m 2
1.92 × 10 7 N ⋅ m 2 C
1
ΦE =
6
6
(a)
bΦ g
(b)
Φ E = 19.2 MN ⋅ m 2 C
(c)
E one face
=
bΦ g
E one face
= 3.20 MN ⋅ m 2 C
The answer to (a) would change because the flux through each face of the cube would
not be equal with an asymmetric charge distribution. The sides of the cube nearer the
charge would have more flux and the ones further away would have less. The answer
to (b) would remain the same, since the overall flux would remain the same.
P24.22
No charge is inside the cube. The net flux through the cube is zero. Positive flux
comes out through the three faces meeting at g. These three faces together fill
solid angle equal to one-eighth of a sphere as seen from q, and together pass
1 q
. Each face containing a intercepts equal flux going into the cube:
flux
8 ∈0
FG IJ
H K
0 = Φ E , net
Φ E , abcd =
Section 24.3
P24.23
FIG. P24.22
q
= 3 Φ E , abcd +
8 ∈0
−q
24 ∈0
Application of Gauss’s Law to Various Charge Distributions
The charge distributed through the nucleus creates a field at the surface equal to that of a point
k q
charge at its center: E = e2
r
e8.99 × 10 Nm C je82 × 1.60 × 10
E=
a208f 1.20 × 10 m
9
2
13
E = 2.33 × 10 21 N C
2
−15
−19
C
j
2
away from the nucleus
Chapter 24
P24.24
(a)
E=
(b)
E=
k eQr
a3
= 0
e8.99 × 10 je26.0 × 10 ja0.100f = 365 kN C
a
a0.400f
k Q e8.99 × 10 je 26.0 × 10 j
=
= 1.46 MN C
E=
r
a0.400f
k eQr
3
−6
9
=
3
−6
9
e
(c)
E=
(d)
2
2
k eQ
r2
e8.99 × 10 je26.0 × 10 j =
=
a0.600f
−6
9
649 kN C
2
The direction for each electric field is radially outward .
*P24.25
mg = qE = q
0
P24.26
−12
0.01 9.8
Q 2 ∈0 mg 2 8.85 × 10
=
=
= −2.48 µC m 2
A
q
−0.7 × 10 −6
0
jb
e
2 8.99 × 10 9 Q 2.40
2k e λ
4
E=
3.60 × 10 =
0.190
r
Q = +9.13 × 10 −7 C = +913 nC
(a)
g
E= 0
(b)
*P24.27
ja fa f
e
FG σ IJ = qFG Q A IJ
H2∈ K H 2∈ K
The volume of the spherical shell is
a
f − a0.20 mf
4
π 0.25 m
3
3
3
= 3.19 × 10 −2 m3 .
Its charge is
e
je
j
ρV = −1.33 × 10 −6 C m 3 3.19 × 10 −2 m3 = −4.25 × 10 −8 C .
The net charge inside a sphere containing the proton’s path as its equator is
−60 × 10 −9 C − 4. 25 × 10 −8 C = −1.02 × 10 −7 C .
The electric field is radially inward with magnitude
ke q
r
2
=
q
∈0 4π r
2
=
e
8.99 × 10 9 Nm 2 1.02 × 10 −7 C
2
a
f
C 0.25 m
2
j = 1.47 × 10
4
N C.
For the proton
∑ F = ma
F eEr IJ
v=G
HmK
eE =
12
mv 2
r
F 1.60 × 10
=G
GH
−19
e
j
C 1.47 × 10 4 N C 0.25 m
1.67 × 10 −27 kg
I
JJ
K
12
= 5.94 × 10 5 m s .
35
36
Gauss’s Law
P24.28
e
σ = 8.60 × 10 −6 C cm 2
E=
jFGH 100mcm IJK
2
= 8.60 × 10 −2 C m 2
σ
8.60 × 10 −2
=
= 4.86 × 10 9 N C away from the wall
2 ∈0 2 8.85 × 10 −12
e
j
The field is essentially uniform as long as the distance from the center of the wall to the field point is
much less than the dimensions of the wall.
P24.29
If ρ is positive, the field must be radially outward. Choose as the
gaussian surface a cylinder of length L and radius r, contained inside
the charged rod. Its volume is π r 2 L and it encloses charge ρπ r 2 L .
Because the charge distribution is long, no electric flux passes
through the circular end caps; E ⋅ dA = EdA cos 90.0° = 0 . The curved
surface has E ⋅ dA = EdA cos 0° , and E must be the same strength
everywhere over the curved surface.
q
ρπ r 2 L
Gauss’s law, E ⋅ dA =
,
becomes
E
dA =
.
∈0
∈0
Curved
z
FIG. P24.29
z
Surface
Now the lateral surface area of the cylinder is 2π rL :
b g
E 2π r L =
*P24.30
ρπ r 2 L
.
∈0
ρr
radially away from the cylinder axis .
2 ∈0
E=
Thus,
Let ρ represent the charge density. For the field inside the sphere at r1 = 5 cm we have
E1 4π r12 =
q inside 4π r13 ρ
=
∈0
3 ∈0
E1 =
e
je
r1 ρ
3 ∈0
j
−12
C 2 −86 × 10 3 N
3 ∈0 E1 3 8.85 × 10
ρ=
=
= −4.57 × 10 −5 C m 3 .
r1
0.05 m
Nm 2
C
Now for the field outside at r3 = 15 cm
4π r23 ρ
E3 4π r32 =
3 ∈0
a
f e−4.57 × 10 Cj = 8.99 × 10
k 4 π 0.10 m
E3 = 2e
r3 3
3
−5
m
3
9
e
Nm 2 −1.91 × 10 −7 C
a0.15 mf C
2
2
j = −7.64 × 10
4
NC
E 3 = 76. 4 kN C radially inward
P24.31
E= 0
(a)
E=
(b)
P24.32
k eQ
r2
e8.99 × 10 je32.0 × 10 j = 7.19 MN C
=
a0.200f
−6
9
E = 7.19 MN C radially outward
2
The distance between centers is 2 × 5.90 × 10 −15 m . Each produces a field as if it were a point charge
at its center, and each feels a force as if all its charge were a point at its center.
F=
k e q1 q 2
r2
e
a46f e1.60 × 10 Cj
C j
e2 × 5.90 × 10 mj
2
9
= 8.99 × 10 N ⋅ m
2
−19
2
−15
2
2
= 3.50 × 10 3 N = 3.50 kN
Chapter 24
P24.33
Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a
0.05 m separation on the ends of strings making angles of 10° with the vertical.
(a)
mg
∑ Fy = T cos 10°−mg = 0 ⇒ T = cos 10°
∑ Fx = T sin 10°− Fe = 0 ⇒ Fe = T sin 10° , so
Fe =
FG mg IJ sin 10° = mg tan 10° = b0.001 kgge9.8 m s j tan 10°
H cos 10° K
2
Fe ≈ 2 × 10 −3 N ~ 10 −3 N or 1 mN
(b)
Fe =
keq 2
r2
e8.99 × 10 N ⋅ m
N≈
a0.25 mf
9
2 × 10
−3
2
j
C 2 q2
2
q ≈ 1. 2 × 10 −7 C ~ 10 −7 C or 100 nC
*P24.34
37
keq
(c)
E=
(d)
ΦE =
r
≈
2
e8.99 × 10
9
je
N ⋅ m 2 C 2 1.2 × 10 −7 C
a0.25 mf
2
N C ~ 10 kN C
4 3
πa ρ
3
ρ=
3Q
4π a 3
The flux is that created by the enclosed charge within radius r:
ΦE =
(b)
4
q
1.2 × 10 −7 C
≈
= 1. 4 × 10 4 N ⋅ m 2 C ~ 10 kN ⋅ m 2 C
∈0 8.85 × 10 −12 C 2 N ⋅ m 2
The charge density is determined by Q =
(a)
j ≈ 1.7 × 10
ΦE =
q in 4π r 3 ρ
4π r 3 3Q
Qr 3
=
=
=
∈0
3 ∈0
3 ∈0 4π a 3
∈0 a 3
Q
. Note that the answers to parts (a) and (b) agree at r = a .
∈0
(c)
ΦE
Q
∈0
0
0
a
FIG. P24.34(c)
r
FIG. P24.33
38
Gauss’s Law
P24.35
(a)
e
je
j
9
2
2
2.00 × 10 −6 C 7.00 m
2 k e λ 2 8.99 × 10 N ⋅ m C
=
E=
0.100 m
r
E = 51.4 kN C , radially outward
(b)
b
g
Φ E = EA cos θ = E 2π rA cos 0°
j a
e
fb
ga f
Φ E = 5.14 × 10 4 N C 2π 0.100 m 0.020 0 m 1.00 = 646 N ⋅ m 2 C
P24.36
(a)
(b)
ρ=
Q
5.70 × 10 −6
=
= 2.13 × 10 −2 C m 3
3
4π 3
4π
a
0.040 0
3
3
b
g
FG 4 π r IJ = e2.13 × 10
H3 K
F4 I
= ρ G π r J = e 2.13 × 10
H3 K
qin = ρ
3
−2
qin
3
−2
jFGH 34 π IJK b0.020 0g
jFGH 34 π IJK b0.040 0g
3
= 7.13 × 10 −7 C = 713 nC
3
= 5.70 µC
9.00 × 10 −6 C m 2
σ
=
= 508 kN C , upward
2 ∈0 2 8.85 × 10 −12 C 2 N ⋅ m 2
P24.37
E=
P24.38
Note that the electric field in each case is directed radially inward, toward the filament.
j
e
je
j
e
je
j
e
je
j
(a)
E=
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 16.2 MN C
0.100 m
r
(b)
E=
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 8.09 MN C
0.200 m
r
(c)
−6
9
2
2
2 k e λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C m
=
= 1.62 MN C
E=
1.00 m
r
Section 24.4
P24.39
e
z
Conductors in Electrostatic Equilibrium
b g
EdA = E 2π rl =
qin
∈0
E=
q in l
λ
=
2π ∈0 r 2π ∈0 r
(a)
r = 3.00 cm
E= 0
(b)
r = 10.0 cm
E=
(c)
r = 100 cm
E=
e
30.0 × 10 −9
ja
2π 8.85 × 10 −12 0.100
e
30.0 × 10 −9
f=
=
ja f
2π 8.85 × 10 −12 1.00
5 400 N C , outward
540 N C , outward
Chapter 24
P24.40
σ=
P24.41
EA =
From Gauss’s Law,
ja
Q
∈0
39
f
Q
=∈0 E = 8.85 × 10 −12 −130 = −1.15 × 10 −9 C m 2 = −1.15 nC m 2
A
e
σ conductor
for the field outside the aluminum looks
∈0
The fields are equal. The Equation 24.9 E =
σ insulator
for the field around glass. But its charge will spread out to
2 ∈0
Q
cover both sides of the aluminum plate, so the density is σ conductor =
. The glass carries charge
2A
Q
Q
only on area A, with σ insulator = . The two fields are
the same in magnitude, and both are
A
2 A ∈0
perpendicular to the plates, vertically upward if Q is positive.
different from Equation 24.8 E =
*P24.42
(a)
All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside is
zero .
(b)
The charged sphere creates field at exterior points as if it were a point charge at the center:
E=
P24.43
ke q
r
2
away =
e8.99 × 10
9
je
Nm 2 40 × 10 −9 C
2
a
f
C 0.17 m
(c)
e8.99 × 10
E=
(d)
All three answers would be the same.
(a)
E=
9
je
2
Nm 2 40 × 10 −9 C
2
a
f
C 0.75 m
σ
∈0
2
e
j outward =
j outward =
je
1.24 × 10 4 N C outward
639 N C outward
j
σ = 8.00 × 10 4 8.85 × 10 −12 = 7.08 × 10 −7 C m 2
σ = 708 nC m 2 , positive on one face and negative on the other.
P24.44
e
σ=
(a)
E= 0
k eQ
(b)
E=
(c)
E= 0
(d)
ja
f
Q
2
Q = σA = 7.08 × 10 −7 0.500 C
A
Q = 1.77 × 10 −7 C = 177 nC , positive on one face and negative on the other.
(b)
E=
r2
k eQ
r2
e8.99 × 10 je8.00 × 10 j = 7.99 × 10
=
b0.030 0g
7
NC
E = 79.9 MN C radially outward
e8.99 × 10 je4.00 × 10 j = 7.34 × 10
=
b0.070 0g
6
NC
E = 7.34 MN C radially outward
−6
9
2
−6
9
2
40
Gauss’s Law
P24.45
The charge divides equally between the identical spheres, with charge
like point charges at their centers:
F=
P24.46
b gb g = k Q
aL + R + Rf 4aL + 2Rf
ke Q 2 Q 2
2
e
2
2
=
e
8.99 × 10 9 N ⋅ m 2 60.0 × 10 −6 C
a
f
4 C 2 2.01 m
j
Q
on each. Then they repel
2
2
2
= 2.00 N .
The electric field on the surface of a conductor varies inversely with the radius of curvature of the
surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. The
local charge density and the electric field intensity are related by
E=
(a)
σ
σ =∈0 E .
or
∈0
Where the radius of curvature is the greatest,
e
je
j
σ =∈0 Emin = 8.85 × 10 −12 C 2 N ⋅ m 2 2.80 × 10 4 N C = 248 nC m 2 .
(b)
Where the radius of curvature is the smallest,
e
je
j
σ =∈0 Emax = 8.85 × 10 −12 C 2 N ⋅ m 2 5.60 × 10 4 N C = 496 nC m 2 .
P24.47
(a)
(b)
P24.48
(a)
(b)
P24.49
(a)
Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting
shell is zero, the total charge inside the gaussian surface must be zero, so the inside
charge/length = − λ .
0 = λA + qin
so
qin
= −λ
A
Outside surface:
The total charge on the metal cylinder is
2λA = qin + qout
qout = 2λA + λA
so the outside charge/length is
E=
E=
e
r
k eQ
r2
r
e8.99 × 10 je6.40 × 10 j =
=
a0.150f
−6
9
2
2.56 MN C , radially inward
The charge density on each of the surfaces (upper and lower) of the plate is:
E=
FG IJ
H K
e
E=
j
−8
1 q
1 4.00 × 10 C
=
= 8.00 × 10 −8 C m 2 = 80.0 nC m 2 .
2 A
2 0.500 m 2
a
f
FG σ IJ k = F 8.00 × 10 C m I k = b9.04 kN Cgk
H ∈ K GH 8.85 × 10 C N ⋅ m JK
−8
0
(c)
3λ
radially outward
2π ∈0 r
E=0
σ=
(b)
b g = 6k λ =
2 k e 3λ
3λ .
b−9.04 kN Cgk
−12
2
2
2
Chapter 24
P24.50
(a)
The charge +q at the center induces charge −q on the inner surface of the conductor,
where its surface density is:
−q
σa =
.
4π a 2
(b)
The outer surface carries charge Q + q with density
σb =
P24.51
Q+q
41
.
4π b 2
Use Gauss’s Law to evaluate the electric field in each region, recalling that the electric field is zero
everywhere within conducting materials. The results are:
E = 0 inside the sphere and within the material of the shell
P24.52
E = ke
Q
between the sphere and shell, directed radially inward
r2
E = ke
2Q
outside the shell, directed radially outward .
r2
Charge
−Q is on the outer surface of the sphere .
Charge
+Q is on the inner surface of the shell ,
and
+2Q is on the outer surface of the shell.
An approximate sketch is given at the right. Note that the electric field lines
should be perpendicular to the conductor both inside and outside.
FIG. P24.52
Section 24.5
P24.53
(a)
Formal Derivation of Gauss‘s Law
Uniform E, pointing radially outward, so Φ E = EA . The arc length is ds = Rdθ ,
and the circumference is 2π r = 2π R sin θ
z
A = 2π rds =
zb
θ
z
θ
g
a
2π R sin θ Rdθ = 2π R 2 sin θdθ = 2π R 2 − cos θ
0
f
θ
0
0
a
f
a
1
Q
Q
ΦE =
⋅ 2π R 2 1 − cos θ =
1 − cos θ
4π ∈0 R 2
2 ∈0
a
f
[independent of R!]
f
(b)
For θ = 90.0° (hemisphere): Φ E =
Q
Q
1 − cos 90° =
.
2 ∈0
2 ∈0
(c)
For θ = 180° (entire sphere): Φ E =
Q
Q
1 − cos 180° =
2 ∈0
∈0
a
b
= 2π R 2 1 − cos θ
f
[Gauss’s Law].
g
FIG. P24.53
42
Gauss’s Law
Additional Problems
P24.54
E = ay i + bzj + cxk
E = ay i + cxk
In general,
In the xy plane, z = 0 and
z
ze
Φ E = E ⋅ dA =
w
z
(b)
x=0
y =h
x=0
j
x2
Φ E = ch xdx = ch
2
x=0
(a)
y=0
ay i + cxk ⋅ k dA
w
P24.55
z
y
x=w
chw 2
=
2
dA = hdx
x
FIG. P24.54
qin = +3Q − Q = +2Q
The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed
radially outward .
k e qin
for r ≥ c .
E=
(d)
Since all points within this region are located inside conducting material, E = 0 for
r
2
=
2 k eQ
(c)
r2
b < r < c.
z
(e)
Φ E = E ⋅ dA = 0 ⇒ qin =∈0 Φ E = 0
(f)
qin = +3Q
(g)
E=
(h)
qin = ρV =
(i)
E=
(j)
From part (d), E = 0 for b < r < c . Thus, for a
spherical gaussian surface with b < r < c ,
qin = +3Q + qinner = 0 where qinner is the
charge on the inner surface of the
conducting shell. This yields qinner = −3Q .
(k)
Since the total charge on the conducting
shell is q net = qouter + qinner = −Q , we have
k e qin
r
2
k e qin
r
2
=
3 k eQ
r2
(radially outward) for a ≤ r < b .
F +3Q I FG 4 π r IJ = +3Q r
GH π a JK H 3 K
a
k F
r I
r
=
+3Q J = 3 k Q
(radially outward) for 0 ≤ r ≤ a .
G
r H
a K
a
4
3
e
2
3
3
3
3
3
3
e
3
b g
qouter = −Q − qinner = −Q − −3Q = +2Q .
(l)
This is shown in the figure to the right.
E
a
b
c
FIG. P24.55(l)
r
Chapter 24
43
P24.56
The sphere with large charge creates a strong field to polarize the other sphere. That means it
pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.
This patch of opposite charge is smaller in amount but located in a stronger external field, so it can
feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker
field on the other side.
P24.57
(a)
(b)
z
e
j
E ⋅ dA = E 4π r 2 =
qin
∈0
FG 4 π r IJ
H3 K
3
For r < a ,
qin = ρ
so
E=
For a < r < b and c < r ,
qin = Q .
So
Q
E=
.
4π r 2 ∈0
For b ≤ r ≤ c ,
E = 0 , since E = 0 inside a conductor.
ρr
.
3 ∈0
FIG. P24.57
Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the
conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.
q1 + Q = 0
Therefore,
and
σ1 =
q1
4π b
2
=
−Q
.
4π b 2
Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollow
sphere is uncharged, we require
q
Q
σ2 = 1 2 =
.
q1 + q 2 = 0
and
4π c
4π c 2
P24.58
z
e
j
E ⋅ dA = E 4π r 2 =
(a)
e−3.60 × 10
3
qin
∈0
j a
f
N C 4π 0.100 m
2
=
Q
8.85 × 10
−12
C 2 N ⋅ m2
aa < r < bf
Q = −4.00 × 10 −9 C = −4.00 nC
(b)
We take Q ′ to be the net charge on the hollow sphere. Outside c,
Q + Q′
2
+2.00 × 10 2 N C 4π 0.500 m =
r>c
8.85 × 10 −12 C 2 N ⋅ m 2
e
j a
f
a f
Q + Q ′ = +5.56 × 10 −9 C , so Q ′ = +9.56 × 10 −9 C = +9.56 nC
(c)
For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of
the hollow sphere. Thus, Q1 = −Q = +4.00 nC .
Then, if Q 2 is the total charge on the outer surface of the hollow sphere,
Q 2 = Q ′ − Q1 = 9.56 nC − 4.0 nC = +5.56 nC .
44
Gauss’s Law
*P24.59
y
The vertical velocity component of the moving charge
increases according to
dv y
m
dv y dx
= qE y .
m
dx dt
= Fy
dt
v
0
q
Now
dx
= v x has the nearly constant value v. So
dt
dv y =
q
Ey dx
mv
z
vy
x
Q
FIG. P24.59
q ∞
Ey dx .
mv −∞
0
vx
d
z
vy
v y = dv y =
θ
The radially outward compnent of the electric field varies along the x axis, but is described by
z
∞
z
∞
Ey dA =
−∞
−∞
z
∞
So
Ey dx =
−∞
tan θ =
P24.60
vy
v
b g
Ey 2π d dx =
=
Q
.
∈0
qQ
Q
and v y =
. The angle of deflection is described by
2π d ∈0
mv 2π d ∈0
qQ
2π ∈0 dmv
θ = tan −1
2
qQ
2π ∈0 dmv 2
.
First, consider the field at distance r < R from the center of a uniform sphere of positive charge
Q = + e with radius R.
b
e
g
j
4π r 2 E =
(a)
qin ρV
=
=
∈0
∈0
F
GH
+e
4π 3
3 R
I
JK
4
3π
r3
∈0
so E =
F e I r directed outward
GH 4π ∈ R JK
0
3
The force exerted on a point charge q = − e located at distance r from the center is then
F = qE = − e
F e I r = −F e I r =
GH 4π ∈ R JK GH 4π ∈ R JK
2
0
3
3
0
− Kr .
k e2
e2
= e3
3
4π ∈0 R
R
(b)
K=
(c)
Fr = m e a r = −
F k e I r , so a
GH R JK
e
2
3
r
=−
F k e I r = −ω r
GH m R JK
e
e
2
3
2
Thus, the motion is simple harmonic with frequency
(d)
f = 2.47 × 10
15
1
Hz =
2π
e8.99 × 10
9
je
f=
ω
1
=
2π
2π
N ⋅ m 2 C 2 1.60 × 10 −19 C
e9.11 × 10
−31
j
j
2
kg R 3
which yields R 3 = 1.05 × 10 −30 m3 , or R = 1.02 × 10 −10 m = 102 pm .
kee2
me R3
.
Chapter 24
P24.61
The field direction is radially outward perpendicular to the axis. The field strength depends on r but
not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L.
If r < a ,
ΦE =
E=
qin
∈0
and
λ
2π r ∈0
or
b
g
E 2π rL =
E=
b
E=
λ
g
b
E=
r
ar < a f
.
e
j
λL + ρπ r 2 − a 2 L
∈0
e
λ + ρπ r 2 − a 2
2π r ∈0
g
E 2π rL =
If r > b ,
λL
∈0
2π r ∈0
E 2π rL =
If a < r < b ,
P24.62
45
j r
e
aa < r < bf .
j
λL + ρπ b 2 − a 2 L
∈0
e
λ + ρπ b 2 − a 2
2π r ∈0
j r
ar > b f
.
Consider the field due to a single sheet and let E+ and E−
represent the fields due to the positive and negative sheets. The
field at any distance from each sheet has a magnitude given by
Equation 24.8:
E+ = E− =
(a)
σ
.
2 ∈0
To the left of the positive sheet, E+ is directed toward the
left and E− toward the right and the net field over this
region is E = 0 .
(b)
In the region between the sheets, E+ and E− are both
directed toward the right and the net field is
E=
(c)
σ
∈0
to the right .
FIG. P24.62
To the right of the negative sheet, E+ and E− are again oppositely directed and E = 0 .
46
Gauss’s Law
P24.63
The magnitude of the field due to the each sheet given by
Equation 24.8 is
E=
(a)
σ
directed perpendicular to the sheet.
2 ∈0
In the region to the left of the pair of sheets, both fields are
directed toward the left and the net field is
σ
E=
∈0
to the left .
(b)
In the region between the sheets, the fields due to the individual sheets are oppositely
directed and the net field is
E= 0 .
(c)
In the region to the right of the pair of sheets, both are fields are directed toward the right
and the net field is
σ
to the right .
∈0
E=
P24.64
FIG. P24.63
The resultant field within the cavity is the superposition of two
fields, one E + due to a uniform sphere of positive charge of radius
2a, and the other E − due to a sphere of negative charge of radius a
centered within the cavity.
F
GH
I
JK
4 π r 3ρ
= 4π r 2 E+
3 ∈0
–−
F
GH
I
JK
4 π r13 ρ
= 4π r12 E−
3 ∈0
Since r = a + r1 ,
so
E+ =
ρr
ρr
r =
3 ∈0
3 ∈0
so
E− =
ρ r1
−ρ
− r1 =
r1 .
3 ∈0
3 ∈0
a f
−ρ r − a
E− =
3 ∈0
E = E+ + E− =
*P24.65
b g
Thus,
Ex = 0
and
Ey =
ρa
3 ∈0
FIG. P24.64
ρr
ρr
ρa
ρa
ρa −
+
=
= 0 i +
j.
3 ∈0 3 ∈0 3 ∈0 3 ∈0
3 ∈0
at all points within the cavity.
Consider the charge distribution to be an unbroken charged spherical shell with uniform charge
density σ and a circular disk with charge per area −σ . The total field is that due to the whole sphere,
4π R 2σ
σ
σ
σ
Q
=
=
outward plus the field of the disk −
=
radially inward. The total
2
2
∈
∈
∈0
2
2
4π ∈0 R
4πε 0 R
0
0
field is
σ
σ
σ
outward .
−
=
∈0 2 ∈0
2 ∈0
Chapter 24
P24.66
The electric field throughout the region is directed along x; therefore, E will be
perpendicular to dA over the four faces of the surface which are perpendicular
to the yz plane, and E will be parallel to dA over the two faces which are parallel
to the yz plane. Therefore,
e
Φ E = − Ex
x=a
j A + eE
x x=a+c
jA = −e3 + 2 a jab + e3 + 2aa + cf jab = 2abca2 a + cf .
2
2
Substituting the given values for a, b, and c, we find Φ E = 0.269 N ⋅ m 2 C .
FIG. P24.66
Q =∈0 Φ E = 2.38 × 10 −12 C = 2.38 pC
P24.67
z
e
j
E ⋅ dA = E 4π r 2 =
qin
∈0
z
R
e
0
(b)
AR 5
5
j
4π Ar 5
5
AR 5
.
5 ∈0 r 2
and
E=
For r < R ,
qin = Ar 2 4π r 2 dr =
z
r
e
0
Ar 3
.
5 ∈0
E=
and
P24.68
j
qin = Ar 2 4π r 2 dr = 4π
For r > R ,
(a)
The total flux through a surface enclosing the charge Q is
disk is
Q
. The flux through the
∈0
z
Φ disk = E ⋅ dA
where the integration covers the area of the disk. We must evaluate this integral
1Q
to find how b and R are related. In the figure, take dA to be
and set it equal to 4
∈0
the area of an annular ring of radius s and width ds. The flux through dA is
b
FIG. P24.68
g
E ⋅ dA = EdA cos θ = E 2π sds cos θ .
The magnitude of the electric field has the same value at all points within the annular ring,
E=
1 Q
1
Q
=
4π ∈0 r 2 4π ∈0 s 2 + b 2
and
cos θ =
b
b
=
2
r
s + b2
e
j
12
.
Integrate from s = 0 to s = R to get the flux through the entire disk.
Φ E , disk =
Qb
2 ∈0
ze
R
0
sds
s2 + b 2
j
=
32
The flux through the disk equals
This is satisfied if R = 3 b .
LM e
N
Qb
− s2 + b 2
2 ∈0
j OPQ
12
R
=
0
LM
MM e
N
Q
b
1−
2 ∈0
R2 + b2
Q
b
provided that
4 ∈0
R2 + b2
e
j
12
=
OP
j PPQ
12
1
.
2
47
48
Gauss’s Law
P24.69
z
E ⋅ dA =
qin
1
=
∈0 ∈0
z
z
r
0
a
4π r 2 dr
r
4π a r
4π a r 2
E 4π r 2 =
rdr =
∈0 0
∈0 2
E=
a
= constant magnitude
2 ∈0
(The direction is radially outward from center for positive a; radially inward for negative a.)
P24.70
z
z
1
ρdV . We
∈0
use a gaussian surface which is a cylinder of radius r, length A , and is coaxial with the charge
distribution.
In this case the charge density is not uniform, and Gauss’s law is written as
(a)
b
g
When r < R , this becomes E 2π rA =
ρ0
∈0
z FGH
r
a−
0
E ⋅ dA =
IJ
K
r
dV . The element of volume is a cylindrical
b
shell of radius r, length A , and thickness dr so that dV = 2π rAdr .
b
g FGH 2π r∈ Aρ
2
E 2π rA =
(b)
0
b
(a)
I FG a − r IJ so inside the cylinder, E =
JK H 2 3b K
FG
H
ρ 0r
2r
a−
2 ∈0
3b
When r > R , Gauss’s law becomes
g
E 2π rA =
P24.71
0
ρ0
∈0
z FGH a − br IJK b2π rAdrg or outside the cylinder, E =
FG
H
R
IJ
K
.
ρ0R2
2R
a−
2 ∈0 r
3b
0
IJ
K
Consider a cylindrical shaped gaussian surface perpendicular
to the yz plane with one end in the yz plane and the other end
containing the point x:
Use Gauss’s law:
z
E ⋅ dA =
.
y
qin
∈0
gaussian
surface
By symmetry, the electric field is zero in the yz plane and is
perpendicular to dA over the wall of the gaussian cylinder.
Therefore, the only contribution to the integral is over the end
cap containing the point x :
z
a f
(b)
z
ρ Ax
q
E ⋅ dA = in or EA =
∈0
∈0
so that at distance x from the mid-line of the slab, E =
a=
a f
FG
H
x
x
ρx
.
∈0
IJ
K
−e E
ρe
F
=
=−
x
me
me
m e ∈0
FIG. P24.71
The acceleration of the electron is of the form
a = −ω 2 x with ω =
Thus, the motion is simple harmonic with frequency
f=
ω
1
=
2π
2π
ρe
.
m e ∈0
ρe
.
m e ∈0
Chapter 24
P24.72
Consider the gaussian surface described in the solution to problem 71.
(a)
d
,
2
1
E ⋅ dA =
dq
∈0
z
CA
∈0
EA =
E=
(b)
3
Cd
24 ∈0
For −
E=
(a)
dq = ρ dV = ρAdx = CAx 2 dx
For x >
z
P24.73
49
z
d 2
x 2 dx =
0
FG IJ F d I
H K GH 8 JK
3
1 CA
3 ∈0
E=
or
d
d
<x<
2
2
z
Cd 3 d
i for x > ;
24 ∈0
2
E ⋅ dA =
Cx 3 i for x > 0 ;
3 ∈0
E=−
z
E=−
Cd 3 d
i for x < −
24 ∈0
2
z
1
CA x 2
CAx 3
dq =
x dx =
3 ∈0
∈0
∈0 0
Cx 3 i for x < 0
3 ∈0
A point mass m creates a gravitational acceleration
g=−
z
Gm
r2
r at a distance r.
Gm
e
j
4π r 2 = −4π Gm .
r2
Since the r has divided out, we can visualize the field as unbroken field lines. The same flux
would go through any other closed surface around the mass. If there are several or no
masses inside a closed surface, each creates field to make its own contribution to the net flux
according to
The flux of this field through a sphere is
z
(b)
g ⋅ dA = −
g ⋅ dA = −4π Gm in .
Take a spherical gaussian surface of radius r. The field is inward so
g ⋅ dA = g 4π r 2 cos 180° = − g 4π r 2
z
and
Then,
Or, since
4
−4π Gm in = −4π G π r 3 ρ .
3
4
4 3
2
− g 4π r = −4π G π r ρ and g = π rρG .
3
3
M EGr
ME
M EGr
, g=
or g =
inward .
ρ= 4
3
RE3
RE3
3 π RE
ANSWERS TO EVEN PROBLEMS
P24.2
355 kN ⋅ m 2 C
P24.10
(a) −55.6 nC ; (b) The negative charge has a
spherically symmetric distribution.
P24.4
(a) −2.34 kN ⋅ m 2 C ; (b) +2.34 kN ⋅ m 2 C ;
(c) 0
P24.12
(a)
P24.14
(a) 1.36 MN ⋅ m 2 C ; (b) 678 kN ⋅ m 2 C ;
(c) No; see the solution.
P24.6
q
∈0
P24.8
ERh
q
q
; (b)
; (c) Plane and square
2 ∈0
2 ∈0
both subtend a solid angle of a hemisphere
at the charge.
50
Gauss’s Law
P24.16
P24.18
1.77 pC m3 positive
P24.46
(a) 248 nC m 2 ; (b) 496 nC m 2
Q−6q
P24.48
(a) 2.56 MN C radially inward; (b) 0
P24.50
(a)
P24.52
see the solution
P24.54
chw 2
2
P24.56
see the solution
P24.58
(a) −4.00 nC; (b) +9.56 nC ; (c) +4.00 nC
and +5.56 nC
P24.60
(a, b) see the solution; (c)
6 ∈0
P24.20
28. 2 N ⋅ m 2 C
P24.22
−q
24 ∈0
P24.24
(a) 0; (b) 365 kN C ; (c) 1.46 MN C;
(d) 649 kN C
P24.26
(a) 913 nC ; (b) 0
P24.28
4.86 GN C away from the wall. It is
constant close to the wall
P24.30
76.4 kN C radially inward
P24.32
3.50 kN
−q
4π a
2
; (b)
Q+q
4π b 2
1
2π
ke e2
me R3
;
(d) 102 pm
P24.34
P24.36
P24.38
(a)
3
Qr
Q
; (c) see the solution
; (b)
∈0
∈0 a 3
713 nC ; (b) 5.70 µC
(a) 16.2 MN C toward the filament;
(b) 8.09 MN C toward the filament;
(c) 1.62 MN C toward the filament
2
P24.40
−1.15 nC m
P24.42
(a) 0; (b) 12.4 kN C radially outward;
(c) 639 N C radially outward; (d) Nothing
would change.
P24.44
(a) 0; (b) 79.9 MN C radially outward;
(c) 0; (d) 7.34 MN C radially outward
σ
to the right; (c) 0
∈0
P24.62
(a) 0; (b)
P24.64
see the solution
P24.66
0.269 N ⋅ m 2 C ; 2.38 pC
P24.68
see the solution
P24.70
(a)
P24.72
(a) E =
FG
H
IJ
K
FG
H
ρ 0r
ρ R2
2r
2R
; (b) 0
a−
a−
2 ∈0
3b
2 ∈0 r
3b
IJ
K
Cd 3 d
i for x > ;
24 ∈0
2
3
Cd d
E=−
i for x < − ;
24 ∈0
2
3
Cx Cx 3 i for x > 0 ; E = −
i for x < 0
(b) E =
3 ∈0
3 ∈0
25
Electric Potential
CHAPTER OUTLINE
25.1
25.2
25.3
25.4
25.5
25.6
25.7
25.8
Potential Difference and
Electric Potential
Potential Difference in a
Uniform Electric Field
Electric Potential and
Potential Energy Due to
Point Charges
Obtaining the Value of the
Electric Field from the
Electric Potential
Electric Potential Due to
Continuous Charge
Distributions
Electric Potential Due to a
Charged Conductor
The Milliken Oil Drop
Experiment
Application of Electrostatistics
ANSWERS TO QUESTIONS
Q25.1
When one object B with electric charge is immersed in the
electric field of another charge or charges A, the system
possesses electric potential energy. The energy can be measured
by seeing how much work the field does on the charge B as it
moves to a reference location. We choose not to visualize A’s
effect on B as an action-at-a-distance, but as the result of a twostep process: Charge A creates electric potential throughout the
surrounding space. Then the potential acts on B to inject the
system with energy.
Q25.2
The potential energy increases. When an outside agent makes it
move in the direction of the field, the charge moves to a region
of lower electric potential. Then the product of its negative
charge with a lower number of volts gives a higher number of
joules. Keep in mind that a negative charge feels an electric force
in the opposite direction to the field, while the potential is the
work done on the charge to move it in a field per unit charge.
Q25.3
To move like charges together from an infinite separation, at which the potential energy of the
system of two charges is zero, requires work to be done on the system by an outside agent. Hence
energy is stored, and potential energy is positive. As charges with opposite signs move together
from an infinite separation, energy is released, and the potential energy of the set of charges
becomes negative.
Q25.4
The charge can be moved along any path parallel to the y-z plane, namely perpendicular to the field.
Q25.5
The electric field always points in the direction of the greatest change in electric potential. This is
∂V
∂V
∂V
, Ey = −
and Ez = −
.
implied by the relationships Ex = −
∂x
∂y
∂z
Q25.6
(a)
The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.
(b)
The equipotential surfaces are nesting concentric spheres around a uniformly charged
sphere.
Q25.7
If there were a potential difference between two points on the conductor, the free electrons in the
conductor would move until the potential difference disappears.
51
52
Electric Potential
Q25.8
No. The uniformly charged sphere, whether hollow or solid metal, is an equipotential volume. Since
there is no electric field, this means that there is no change in electrical potential. The potential at
every point inside is the same as the value of the potential at the surface.
Q25.9
Infinitely far away from a line of charge, the line will not look like a point. In fact, without any
distinguishing features, it is not possible to tell the distance from an infinitely long line of charge.
Another way of stating the answer: The potential would diverge to infinity at any finite distance, if it
were zero infinitely far away.
Q25.10
The smaller sphere will. In the solution to the example referred to, equation 1 states that each will
q
have the same ratio of charge to radius, . In this case, the charge density is a surface charge
r
q
, so the smaller-radius sphere will have the greater charge density.
density,
4π r 2
Q25.11
The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the
air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric
potential. If other grounded objects are nearby, the maximum potential might be reduced.
Q25.12
The intense—often oscillating—electric fields around high voltage lines is large enough to ionize the
air surrounding the cables. When the molecules recapture their electrons, they release that energy in
the form of light.
Q25.13
A sharp point in a charged conductor would imply a large electric field in that region. An electric
discharge could most easily take place at that sharp point.
Q25.14
Use a conductive box to shield the equipment. Any stray electric field will cause charges on the outer
surface of the conductor to rearrange and cancel the stray field inside the volume it encloses.
Q25.15
No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric
field in the wire to push more charge to the outer sphere. All of the charge is on the outer sphere.
Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere.
Q25.16
The grounding wire can be touched equally well to any point on the sphere. Electrons will drain
away into the ground and the sphere will be left positively charged. The ground, wire, and sphere
are all conducting. They together form an equipotential volume at zero volts during the contact.
However close the grounding wire is to the negative charge, electrons have no difficulty in moving
within the metal through the grounding wire to ground. The ground can act as an infinite source or
sink of electrons. In this case, it is an electron sink.
SOLUTIONS TO PROBLEMS
Section 25.1
P25.1
Potential Difference and Electric Potential
∆V = −14.0 V
∆V =
W
,
Q
e
je
j
jb
g
and
Q = − N A e = − 6.02 × 10 23 1.60 × 10 −19 = −9.63 × 10 4 C
so
W = Q∆V = −9.63 × 10 4 C −14.0 J C = 1.35 MJ
e
Chapter 25
P25.2
53
a f
7.37 × 10 −17 = q 115
∆K = q ∆V
q = 6. 41 × 10 −19 C
P25.3
(a)
Energy of the proton-field system is conserved as the proton moves from high to low
potential, which can be defined for this problem as moving from 120 V down to 0 V.
K i + Ui + ∆Emech = K f + U f
0 + qV + 0 =
1
mv p2 + 0
2
e1.60 × 10
C 120 V
−19
ja
fFGH 1 V1 J⋅ C IJK = 12 e1.67 × 10
−27
j
kg v p2
v p = 1.52 × 10 5 m s
(b)
The electron will gain speed in moving the other way,
from Vi = 0 to V f = 120 V :
K i + Ui + ∆Emech = K f + U f
0+0+0=
0=
1
mv e2 + qV
2
jb
1
9.11 × 10 −31 kg v e2 + −1.60 × 10 −19 C 120 J C
2
e
j e
g
v e = 6.49 × 10 6 m s
P25.4
W = ∆K = − q∆V
0−
1
9.11 × 10 −31 kg 4.20 × 10 5 m s
2
e
je
j
2
e
j
= − −1.60 × 10 −19 C ∆V
From which, ∆V = −0.502 V .
Section 25.2
P25.5
(a)
Potential Difference in a Uniform Electric Field
We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm).
∆U = − (work done)
∆U = − (work from origin to (20.0 cm, 0)) – (work from (20.0 cm, 0) to (20.0 cm, 50.0 cm))
Note that the last term is equal to 0 because the force is perpendicular to the displacement.
b g
e
jb
ga
f
∆U = − qEx ∆x = − 12.0 × 10 −6 C 250 V m 0.200 m = −6.00 × 10 −4 J
(b)
P25.6
E=
∆V =
∆U
6.00 × 10 −4 J
=−
= −50.0 J C = −50.0 V
q
12.0 × 10 −6 C
∆V 25.0 × 10 3 J C
=
= 1.67 × 10 6 N C = 1.67 MN C
d
1.50 × 10 −2 m
54
Electric Potential
P25.7
∆U = −
jLNMe1.40 × 10 m sj − e3.70 × 10
= e−1.60 × 10 j∆V
1
1
m v 2f − vi2 = − 9.11 × 10 −31 kg
2
2
e
j
e
+6.23 × 10 −18
∆U = q∆V :
5
2
6
ms
j OQP = 6.23 × 10
2
−18
J
−19
∆V = −38.9 V. The origin is at highest potential.
P25.8
jb
e
g
(a)
∆V = Ed = 5.90 × 10 3 V m 0.010 0 m = 59.0 V
(b)
1
mv 2f = q∆V :
2
ja f
1
9.11 × 10 −31 v 2f = 1.60 × 10 −19 59.0
2
e
j
e
v f = 4.55 × 10 6 m s
P25.9
z
z
z
B
C
B
A
A
C
VB − VA = − E ⋅ ds = − E ⋅ ds − E ⋅ ds
a
f z dy − aE cos 90.0°f
= a325fa0.800 f = +260 V
VB − VA = − E cos 180°
VB − VA
z
0 .500
0. 400
−0.300
−0 . 200
dx
FIG. P25.9
*P25.10
Assume the opposite. Then at some point A on some equipotential surface the electric field has a
nonzero component Ep in the plane of the surface. Let a test charge start from point A and move
z
B
some distance on the surface in the direction of the field component. Then ∆V = − E ⋅ ds is nonzero.
A
The electric potential charges across the surface and it is not an equipotential surface. The
contradiction shows that our assumption is false, that Ep = 0 , and that the field is perpendicular to
the equipotential surface.
P25.11
(a)
Arbitrarily choose V = 0 at 0. Then at other points
V = − Ex
and
U e = QV = −QEx .
Between the endpoints of the motion,
bK + U
s
+ Ue
g = bK + U
i
s
+ Ue
g
f
1 2
2QE
0 + 0 + 0 = 0 + kx max
− QEx max so x max =
.
2
k
(b)
At equilibrium,
∑ Fx = − Fs + Fe = 0 or
kx = QE .
So the equilibrium position is at x =
continued on next page
QE
.
k
FIG. P25.11
Chapter 25
(c)
d2x
∑ Fx = − kx + QE = m dt 2
The block’s equation of motion is
.
QE
QE
,
, or x = x ′ +
k
k
so the equation of motion becomes:
d 2 x + QE k
QE
d 2 x′
k
=−
−k x′ +
+ QE = m
x′ .
, or
2
k
m
dt
dt 2
x′ = x −
Let
FG
H
b
IJ
K
g
FG IJ
H K
This is the equation for simple harmonic motion a x′ = −ω 2 x ′
(d)
with
ω=
The period of the motion is then
T=
bK + U
g
b
+ U e i + ∆Emech = K + U s + U e
s
0 + 0 + 0 − µ k mgx max = 0 +
b
2 QE − µ k mg
x max =
P25.12
g
g
k
.
m
2π
ω
= 2π
m
.
k
f
1 2
kx max − QEx max
2
k
1
ayt 2
2
y f − yi = v yi t +
For the entire motion,
1
ayt 2
2
2mvi
− mg − qE = −
t
m 2 vi
−g
E=
q t
0 − 0 = vi t +
∑ Fy = ma y :
FG
H
d
IJ
K
2
2
= v yi
+ 2 a y y f − yi
v yf
For the upward flight:
FG
H
0 = vi2 + 2 −
2 vi
t
IJ by
K
so
ay = −
and
E=−
and
y max =
i
max
−0
g
FG
IJ
FG
IJ FG 1 v tIJ
z
H
K
H
KH 4 K
I L1
2.00 kg F 2b 20.1 m sg
O
∆V =
− 9.80 m s J M b 20.1 m sga 4.10 sfP =
G
4.10 s
5.00 × 10 C H
Q
KN4
∆V = −
FG
H
IJ
K
m 2 vi
− g j.
q t
1
vi t
4
y
ymax
E ⋅ dy = +
0
max
m 2 vi
m 2 vi
−g y
=
−g
q t
q t
0
i
2
−6
P25.13
2 vi
t
40.2 kV
Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length,
V = − Ed and U e = − λLEd .
(a)
aK + U f = a K + U f
i
0+0=
v=
(b)
f
1
µLv 2 − λLEd
2
2 λEd
µ
The same.
=
e
jb
ga
f=
2 40.0 × 10 −6 C m 100 N C 2.00 m
b0.100 kg mg
0.400 m s
55
56
Electric Potential
P25.14
Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of
the charge is − E ⋅ s = − EL cos θ . At the final point a, V = − EL . Suppose the table is frictionless:
K +U i = K +U f
a
f a
f
0 − qEL cos θ =
v=
Section 25.3
P25.15
1
mv 2 − qEL
2
a
f
2 qEL 1 − cos θ
=
m
jb
e
fa
ga
2 2.00 × 10 −6 C 300 N C 1.50 m 1 − cos 60.0°
0.010 0 kg
f=
0.300 m s
Electric Potential and Potential Energy Due to Point Charges
e
je
j
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
q
=
= 1.44 × 10 −7 V .
r
1.00 × 10 −2 m
(a)
The potential at 1.00 cm is V1 = k e
(b)
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C
q
The potential at 2.00 cm is V2 = k e =
= 0.719 × 10 −7 V .
r
2.00 × 10 −2 m
Thus, the difference in potential between the two points is ∆V = V2 − V1 = −7.19 × 10 −8 V .
(c)
The approach is the same as above except the charge is −1.60 × 10 −19 C . This changes the
sign of each answer, with its magnitude remaining the same.
That is, the potential at 1.00 cm is −1.44 × 10 −7 V .
The potential at 2.00 cm is −0.719 × 10 −7 V , so ∆V = V2 − V1 = 7.19 × 10 −8 V .
P25.16
(a)
Since the charges are equal and placed symmetrically, F = 0 .
(b)
Since F = qE = 0 , E = 0 .
(c)
q
V = 2 k e = 2 8.99 × 10 9 N ⋅ m 2 C 2
r
e
× 10 C I
jFGH 2.000.800
m JK
−6
V = 4.50 × 10 4 V = 45.0 kV
P25.17
(a)
(b)
E=
Q
4π ∈0 r 2
V=
Q
4π ∈0 r
r=
V
3 000 V
=
= 6.00 m
E 500 V m
V = −3 000 V =
Q=
Q
4π ∈0 6.00 m
a
−3 000 V
e8.99 × 10
9
V ⋅m C
f
j
a6.00 mf =
−2.00 µC
FIG. P25.16
Chapter 25
P25.18
Ex =
(a)
k e q1
x2
+
ke q2
ax − 2.00f
2
=0
Ex = k e
becomes
a
Dividing by k e ,
2 qx 2 = q x − 2.00
Therefore E = 0
when
f
2
F + q + −2 q I = 0 .
GH x ax − 2.00f JK
2
2
x 2 + 4.00 x − 4.00 = 0 .
x=
−4.00 ± 16.0 + 16.0
= −4.83 m .
2
(Note that the positive root does not correspond to a physically valid situation.)
V=
(b)
k e q1
k q
+ e 2 =0
x
2.00 − x
FG + q − 2 q IJ = 0 .
H x 2.00 − x K
2 qx = qa 2.00 − xf .
or
V = ke
when
x = 0.667 m
For x < 0
x = −2.00 m .
Again solving for x,
For 0 ≤ x ≤ 2.00 V = 0
and
P25.19
V = ∑k
i
−2 q
q
=
.
x
2−x
qi
ri
e
je
V = 8.99 × 10 9 7.00 × 10 −6
1
1 O
−1
−
+
jLMN 0.010
P
0 0.010 0 0.038 7 Q
V = −1.10 × 10 7 V = −11.0 MV
FIG. P25.19
P25.20
(a)
e
je
je
f
j
5.00 × 10 −9 C −3.00 × 10 −9 C 8.99 × 10 9 V ⋅ m C
qQ
U=
=
= −3.86 × 10 −7 J
4π ∈0 r
0.350 m
a
The minus sign means it takes 3.86 × 10 −7 J to pull the two charges apart from 35 cm to a
much larger separation.
(b)
V=
Q1
Q2
+
4π ∈0 r1 4π ∈0 r2
e5.00 × 10 Cje8.99 × 10
=
−9
0.175 m
V = 103 V
9
V ⋅m C
57
j + e−3.00 × 10 Cje8.99 × 10
−9
0.175 m
9
V ⋅m C
j
58
Electric Potential
P25.21
U e = q 4V1 + q 4V2 + q 4V3 = q 4
e
U e = 10.0 × 10 −6 C
je
2
FG 1 IJ FG q
H 4π ∈ K H r
0
1
+
1
8.99 × 10 9 N ⋅ m 2 C 2
q 2 q3
+
r2 r3
IJ
K
F 1
jGG 0.600 m + 0.1501 m + 0.600 m 1+ 0.150 m
a
f a
f
H
2
2
I
JJ
K
U e = 8.95 J
P25.22
V=
(a)
FG IJ
H K
k e q1 k e q 2
k q
+
=2 e
r1
r2
r
F e8.99 × 10 N ⋅ m C je2.00 × 10 Cj I
GG
JJ
1
00
+
0
500
.
m
.
m
a
a
f
f
H
K
9
V=2
2
−6
2
2
2
V = 3.22 × 10 4 V = 32.2 kV
e
je
j
U = qV = −3.00 × 10 −6 C 3.22 × 10 4 J C = −9.65 × 10 −2 J
(b)
P25.23
FIG. P25.22
U = U1 + U 2 + U 3 + U 4
b
g b
U = 0 + U12 + U 13 + U 23 + U 14 + U 24 + U 34
U =0+
U=
k eQ 2 k e Q 2
+
s
s
k eQ
s
2
FG 4 + 2 IJ =
H 2K
FG 1 + 1IJ + k Q FG 1 +
H 2 K s H
e
5.41
k eQ
s
2
g
1
2
IJ
K
+1
2
FIG. P25.23
FG
H
An alternate way to get the term 4 +
IJ is to recognize that there are 4 side pairs and 2 face
2K
2
diagonal pairs.
P25.24
Each charge creates equal potential at the center. The total potential is:
V =5
P25.25
(a)
LM k b− qg OP =
MN R PQ
e
−
5keq
.
R
Each charge separately creates positive potential everywhere. The total potential produced
by the three charges together is then the sum of three positive terms. There is no point
located at a finite distance from the charges, at which this total potential is zero.
(b)
V=
2k e q
ke q keq
+
=
a
a
a
Chapter 25
P25.26
59
Consider the two spheres as a system.
(a)
e j
m1 v1
m2
Conservation of momentum:
0 = m1 v1 i + m 2 v 2 − i or v 2 =
By conservation of energy,
0=
and
k e q1 q 2 k e q1 q 2 1
1 m12 v12
−
= m1 v12 +
2
2 m2
r1 + r2
d
v1 =
d
=
b g
k e − q1 q 2
1
1
m1 v12 + m 2 v 22 +
r1 + r2
2
2
FG 1 − 1 IJ
b
g H r + r dK
2b0.700 kg ge8.99 × 10 N ⋅ m C je 2 × 10 C je3 × 10 C j F
1
G
H
8 × 10
b0.100 kg gb0.800 kg g
2 m 2 k e q1 q 2
m1 m1 + m 2
1
2
9
v1 =
b g
k e − q1 q 2
2
−6
2
−6
−3
m
−
1
1.00 m
IJ
K
= 10.8 m s
v2 =
(b)
b
g
m1 v1 0.100 kg 10.8 m s
=
= 1.55 m s
0.700 kg
m2
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really be
moving faster than calculated in (a) .
P25.27
Consider the two spheres as a system.
(a)
e j
Conservation of momentum:
0 = m 1 v1 i + m 2 v 2 − i
or
v2 =
By conservation of energy,
0=
and
k e q1 q 2 k e q1 q 2 1
1 m12 v12
−
= m1 v12 +
.
2
2 m2
r1 + r2
d
b g
k e − q1 q 2
v1 =
v2 =
(b)
m1 v1
.
m2
d
=
b g
k e − q1 q 2
1
1
m1 v12 + m 2 v 22 +
r1 + r2
2
2
IJ
K
2m k q q F 1
G
m bm + m g H r + r
2 m 2 k e q1 q 2
m 1 m1 + m 2
b
FG m IJ v
Hm K
1
2
1
=
FG 1
gHr +r
1
−
2
1
d
1 e 1 2
2
1
2
1
2
−
1
d
IJ
K
If the spheres are metal, electrons will move around on them with negligible energy loss to
place the centers of excess charge on the insides of the spheres. Then just before they touch,
the effective distance between charges will be less than r1 + r2 and the spheres will really be
moving faster than calculated in (a) .
60
Electric Potential
*P25.28
(a)
In an empty universe, the 20-nC charge can be placed at its location with no energy
investment. At a distance of 4 cm, it creates a potential
V1 =
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 20 × 10 −9 C
k e q1
=
= 4.50 kV .
0.04 m
r
To place the 10-nC charge there we must put in energy
e
je
j
U12 = q 2 V1 = 10 × 10 −9 C 4.5 × 10 3 V = 4.50 × 10 −5 J .
Next, to bring up the –20-nC charge requires energy
b
U 23 + U 13 = q3 V2 + q3 V1 = q 3 V2 + V1
g
e
= −20 × 10 −9 C 8.99 × 10 9 N ⋅ m 2 C 2
× 10 C 20 × 10 C I
+
jFGH 100.04
m
0.08 m JK
−9
−9
= −4.50 × 10 −5 J − 4.50 × 10 −5 J
The total energy of the three charges is
U12 + U 23 + U13 = −4.50 × 10 −5 J .
(b)
The three fixed charges create this potential at the location where the fourth is released:
e
V = V1 + V2 + V3 = 8.99 × 10 9 N ⋅ m 2 C 2
F
jGH
20 × 10 −9
0.04 2 + 0.03 2
+
10 × 10 −9 20 × 10 −9
−
0.03
0.05
I Cm
JK
V = 3.00 × 10 3 V
Energy of the system of four charged objects is conserved as the fourth charge flies away:
FG 1 mv + qV IJ = FG 1 mv + qV IJ
H2
K H2
K
1
0 + e 40 × 10 C je3.00 × 10 V j = e 2.00 × 10
2
2e1.20 × 10 Jj
= 3.46 × 10 m s
v=
2
2
i
f
−9
3
−13
j
kg v 2 + 0
−4
4
2 × 10 −13 kg
*P25.29
The original electrical potential energy is
U e = qV = q
ke q
.
d
In the final configuration we have mechanical equilibrium. The spring and electrostatic forces on
k q2
k q
each charge are − k 2d + q e 2 = 0 . Then k = e 3 . In the final configuration the total potential
18d
3d
a f a f
1 k q
+ qV =
a 2d f
2 18d
keq 4 keq2
=
. The missing energy must have become internal
3
3d 9 d
k q 2 4k q 2
energy, as the system is isolated: e = e + ∆Eint
d
9d
energy is
∆Eint =
1 2
kx
2
5 keq2
.
9 d
e
2
2
+q
Chapter 25
P25.30
af
V x =
(a)
b g
k +Q
k e Q1 k e Q 2
+
= e
+
r1
r2
x2 + a2
af
2 k eQ
kQ
= e
2
2
a
x +a
V x =
af
2
b k Q ag = b x ag
F 2
GG
H b x ag
2
b g
+ a − af
k e +Q
x2
I
JJ
+1K
2
V x
e
2
+1
FIG. P25.30(a)
bg
V y =
(b)
b g
b g
k e Q 1 k e Q 2 k e +Q k e −Q
+
=
+
r1
r2
y−a
y+a
b g k aQ FGH y a1− 1 − y a1+ 1 IJK
F 1 − 1 I
V b yg
= G
bk Q ag H y a − 1 y a + 1 JK
e
V y =
e
FIG. P25.30(b)
P25.31
V=
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 8.00 × 10 −9 C
kQ
k eQ
72.0 V ⋅ m
=
so r = e =
.
V
V
V
r
For V = 100 V , 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m .
The radii are inversely proportional to the potential.
P25.32
Using conservation of energy for the alpha particle-nucleus system,
we have
K f + U f = K i + Ui .
But
Ui =
and
ri ≈ ∞.
Thus,
Ui = 0 .
Also
K f = 0 ( v f = 0 at turning point),
so
U f = Ki
ri
k e qα qgold
or
rmin =
k e qα qgold
rmin
2 k e qα qgold
mα vα2
=
=
1
mα vα2
2
e
ja fa fe
kg je 2.00 × 10
2 8.99 × 10 9 N ⋅ m 2 C 2 2 79 1.60 × 10 −19 C
e6.64 × 10
−27
7
ms
j
2
j
2
= 2.74 × 10 −14 m = 27.4 fm .
61
62
Electric Potential
P25.33
P25.34
Using conservation of energy
k e eQ k e qQ 1
=
+ mv 2
we have:
2
r1
r2
2 k e eQ 1 1
−
m
r1 r2
FG
H
IJ
K
a2fe8.99 × 10
N ⋅ m 2 C 2 −1.60 × 10 −19 C 10 −9 C
which gives:
v=
or
v=
Thus,
v = 7.26 × 10 6 m s .
k e qi q j
9
je
9.11 × 10
e
, summed over all pairs of i , j where i ≠ j .
rij
2
2
2
FIG. P25.34
−6 2
Each charge moves off on its diagonal line. All charges have equal speeds.
∑ K +U i = ∑ K +U f
a
0+
f
a
2
2
4k e q
2k q
+ e
L
2L
f
F 1 I 4k q
= 4G mv J +
H 2 K 2L
2
FG 2 + 1 IJ k q = 2mv
H 2K L
F 1 IJ k q
v = G1 +
H 8 K mL
e
2
e
e
2
+
2k e q 2
2 2L
2
2
A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 face
diagonal pairs separated by 2s and 4 interior diagonal pairs separated 3s .
U=
Section 25.4
P25.37
2
2
9
P25.36
kg
jF 1 − 1 I .
GH 0.030 0 m 0.020 0 m JK
b g
L qb−2 qg + b−2 qgb3qg + b2 qgb3qg + qb2 qg + qb3qg + 2 qb−2 qg OP
U=k M
a
b
a
MN b
a +b
a + b PQ
L −2 − 6 + 6 + 2 + 3 − 4 OP
U=k q M
N 0.400 0.200 0.400 0.200 0.447 0.447 Q
L 4 − 4 − 1 OP = −3.96 J
U = e8.99 × 10 je6.00 × 10 j M
N 0.400 0.200 0.447 Q
U=∑
e
P25.35
je
−31
LM
N
OP
Q
keq2
k q2
12
4
+
= 22.8 e
12 +
s
s
2
3
Obtaining the Value of the Electric Field from the Electric Potential
b
g
V = a + bx = 10.0 V + −7.00 V m x
(a)
(b)
At x = 0 ,
V = 10.0 V
At x = 3.00 m ,
V = −11.0 V
At x = 6.00 m ,
V = −32.0 V
E=−
b
g
dV
= − b = − −7.00 V m = 7.00 N C in the + x direction
dx
Chapter 25
P25.38
k eQ
R
dV
= 0
Er = −
dr
(a)
For r < R
V=
(b)
For r ≥ R
V=
k eQ
r
kQ
kQ
dV
= − − e2 = e2
Er = −
dr
r
r
FG
H
P25.39
IJ
K
V = 5 x − 3 x 2 y + 2 yz 2
b
Evaluate E at 1, 0 , − 2
g
a fa f
af a f
−4yz = −4a0fa −2f = 0
+ E = a −5 f + a −5 f + 0 =
∂V
= −5 + 6 xy = −5 + 6 1 0 = −5
∂x
∂V
2
2
= +3 x 2 − 2 z 2 = 3 1 − 2 −2 = −5
Ey = −
∂y
Ex = −
Ez = −
∂V
=
∂z
E = E x2 + E y2
P25.40
2
2
z
2
7.07 N C
∆V
∆s
E A > EB since E =
(a)
2
a f
6−2 V
∆V
=−
= 200 N C down
2 cm
∆s
(b)
EB = −
(c)
The figure is shown to the right, with sample field lines
sketched in.
FIG. P25.40
P25.41
Ey = −
Ey =
Section 25.5
P25.42
LM
MMN
∂V
∂ k eQ
=−
ln
∂y
∂y
LM
MN
k eQ
1−
y
F
GG
H
2
+
y
y2
2
+ y2 +
+ y2
2
+ y2
OP
PQ =
I OP
JJ P
K PQ
k eQ
y
2
+ y2
Electric Potential Due to Continuous Charge Distributions
∆V = V2 R − V0 =
k eQ
a f
R2 + 2R
2
−
k eQ k eQ
=
R
R
FG 1 − 1IJ =
H 5 K
−0.553
k eQ
R
63
64
Electric Potential
P25.43
LM λ OP = C ⋅ FG 1 IJ =
N x Q m H mK
(a)
α =
(b)
V = ke
z
C
m2
z
LM
N
z
FG
H
L
dq
λdx
xdx
L
= ke
= k eα
= k eα L − d ln 1 +
r
r
d+x
d
0
IJ OP
KQ
FIG. P25.43
P25.44
V=
z
k e dq
= ke
r
z
αxdx
b
b2 + L 2 − x
g
2
L
−x.
2
Let z =
Then x =
V = k eα
L
− z , and dx = − dz
2
zb
ga f = − k αL
L 2 − z − dz
2
b +z
e
2
2
z
dz
2
b +z
2
+ k eα
z
zdz
2
b +z
2
=−
FH
LMF L I F L I OP
GMH 2 − xJK + GH 2 − xJK + b P + k α FGH L2 − xIJK + b
N
Q
L
O L FL I
k αL M L 2 − L + b L 2 g + b P
F LI
V=−
ln M
+ k α M G − LJ + b − G J
P
H
K
H 2K
2
2
M
MN L 2 + bL 2g + b PQ N
L
O
k αL M b + eL 4j − L 2 P
V= −
ln M
2
MN b + eL 4j + L 2 PPQ
L
2
k αL
V = − e ln
2
2
L
2
2
e
0
0
2
2
e
2
2
2
2
2
2
2
2
e
IK
k eαL
ln z + z 2 + b 2 + k eα z 2 + b 2
2
2
+ b2
OP
PQ
e
P25.45
z
1
4π ∈0
V = dV =
z
dq
r
All bits of charge are at the same distance from O.
So V =
P25.46
dV =
FG IJ e
H K
1
Q
= 8.99 × 10 9 N ⋅ m 2 C 2
4π ∈0 R
k e dq
z
b
a
−6
−1.51 MV .
where dq = σdA = σ 2π rdr
r 2 + x2
V = 2πσk e
.50 × 10 C I
=
jFGH −70.140
m π JK
rdr
2
r + x2
= 2π k eσ
LM
N
x2 + b2 − x2 + a2
OP
Q
FIG. P25.46
Chapter 25
P25.47
V = ke
z
z
a f
V = − k e λ ln − x
V = k e ln
Section 25.6
P25.48
z
z
3R
−R
dq
λdx
λds
λdx
= ke
+ ke
+ ke
−x
r
R
x
R
all charge
semicircle
−3 R
−R
−3 R
+
keλ
3R
π R + k e λ ln x R
R
a
3R
+ k e λ π + k e ln 3 = k e λ π + 2 ln 3
R
f
Electric Potential Due to a Charged Conductor
Substituting given values into V =
ke q
r
7.50 × 10 3 V =
e8.99 × 10
9
j
N ⋅ m2 C 2 q
0.300 m
.
Substituting q = 2.50 × 10 −7 C ,
N=
P25.49
(a)
2.50 × 10 −7 C
= 1.56 × 10 12 electrons .
1.60 × 10 −19 C e −
E= 0 ;
e
je
j
8.99 × 10 9 26.0 × 10 −6
ke q
=
= 1.67 MV
V=
0.140
R
(b)
e8.99 × 10 je26.0 × 10 j = 5.84 MN C
r
a0.200f
k q e8.99 × 10 je 26.0 × 10 j
=
= 1.17 MV
V=
E=
keq
2
−6
9
=
2
away
−6
9
e
(c)
E=
V=
R
0.200
keq
e8.99 × 10 je26.0 × 10 j =
=
a0.140f
R2
−6
9
ke q
= 1.67 MV
R
2
11.9 MN C away
65
66
Electric Potential
*P25.50
(a)
Both spheres must be at the same potential according to
where also
q1 + q 2 = 1.20 × 10 −6 C .
Then
q1 =
k e q1 k e q 2
=
r1
r2
q 2 r1
r2
q 2 r1
+ q 2 = 1.20 × 10 −6 C
r2
1.20 × 10 −6 C
= 0.300 × 10 −6 C on the smaller sphere
1 + 6 cm 2 cm
q2 =
q1 = 1.20 × 10 −6 C − 0.300 × 10 −6 C = 0.900 × 10 −6 C
V=
(b)
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 0.900 × 10 −6 C
k e q1
=
= 1.35 × 10 5 V
r1
6 × 10 −2 m
Outside the larger sphere,
E1 =
k e q1
r12
r=
V1
1.35 × 10 5 V
r=
r = 2. 25 × 10 6 V m away .
0.06 m
r1
Outside the smaller sphere,
1.35 × 10 5 V
r = 6.74 × 10 6 V m away .
0.02 m
E2 =
The smaller sphere carries less charge but creates a much stronger electric field than the
larger sphere.
Section 25.7
The Milliken Oil Drop Experiment
Section 25.8
Application of Electrostatistics
P25.51
(a)
Emax = 3.00 × 10 6 V m =
6
k eQ
a
r
2
=
f
FG IJ
HK
FG IJ
HK
k eQ 1
1
= Vmax
r r
r
Vmax = Emax r = 3.00 × 10 0.150 = 450 kV
(b)
P25.52
V=
k eQmax
r2
= Emax
RSor k Q
T r
e
max
= Vmax
UV
W
ke q
k q
V
and E = e2 . Since E = ,
r
r
r
(b)
r=
6.00 × 10 5 V
V
=
= 0.200 m and
E 3.00 × 10 6 V m
(a)
q=
Vr
= 13.3 µC
ke
Q max =
a
6
Emax r 2 3.00 × 10 0.150
=
ke
8.99 × 10 9
f
2
= 7.51 µC
Chapter 25
67
Additional Problems
P25.53
q q
U = qV = k e 1 2 = 8.99 × 10 9
r12
P25.54
(a)
e
a38fa54f 1.60 × 10
j a5.50 +e6.20f × 10 j
−19 2
−15
= 4.04 × 10 −11 J = 253 MeV
To make a spark 5 mm long in dry air between flat metal plates requires potential difference
e
je
j
∆V = Ed = 3 × 10 6 V m 5 × 10 −3 m = 1.5 × 10 4 V ~ 10 4 V