A N ote To The
Instructor
.
The solutions here are somewhat brief, as they are designedfor the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server accessmust be restricted to your students.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it wi11sometimes obfuscate the approach
if viewed by a novice.
There are some tmditional formula, such as
2-
Vx -VOx
2
+
2 axx,
which are not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercisesand problems in the text
which build upon previous exercisesand problems. Instead of rederiving expressions,I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in particular , I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercisesand Problems which are enclosedin a box also appear in the Student's Solution Manual
with considerably more detail and, when appropriate, include discussionon any physical implications
of the answer. These student solutions carefully discussthe steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented. You are
encouraged to refer students to the Student 's Solution Manual for these exercises and problems.
However, the material from the Student's Solution Manual must not be copied.
Paul
Stanley
Beloit
College
stanley@clunet.
1
edu
El-l
(a) Megaphones;
(e) Terabulls (Terribles);
Boo); (j) Attoboys ('atta
(To Kill A Mockingbird,
(b) Microphones; (c) Decacards (Deck of Cards); (d) Gigalows (Gigolos);
(f) Decimates; (g) Centipedes; (h) Nanonanettes (?); (i) Picoboos (Peek-aboy); (k) Two Hectowithits (To Heck With It); (1) Two Kilomockingbirds
or Tequila Mockingbird).
El-2
(a) $36,000152 week = $692¡week. (b) $10,000,0001(20 x 12 month) = $41, 700¡month.
30 x 109/8 = 3.75 x 109.
~
Multiply
out the factors
1 century
which
= 100 years
(c)
make up a century.
365 days
1 year
60 minutes .
24 hours
I day
1 hour
This gives 5.256 x 107 minutes in a century, so a microcentury is 52.56 minutes.
The percentage difference from Fermi's approximation is (2.56 min)/(50 min) x 100% or 5.12%.
There are 24 time-zones, so the circumference
El-4
(3000 mi)/(3 hr) = 1000 mi/timezone-hour.
is approximately 24 x 1000 mi = 24,000 miles.
El-5
Actual number oí seconds in a year is
{365.25 days)
24 hr
1 day
60 min
lhr
608'
=
1
3.1558
x
107 s.
min
The percentage error of the approximation is then
El-6
(a) 10-8 secondsper shake means 108 shakesper second. There are
365 days
24 hr
1 year
1 day
60 min
I hr
60 s
1 min
= 3.1536
x 107 s/year.
This means there are more shakesin a second.
(b) Humans have existed for a fraction of
106 years/101O years = 10-4.
That fraction of a day is
10-4 (24 hr)
60 min
lhr
60 s
=
8.64
s.
1 min
~
We'll 8Ssume,for convenienceonly, that the runner with the longer time ran exactly one
mile. Let the speedof the runner with the shorter time be given by Vl , and call the distance actually
ran by this runner d1. Then Vl = dl/tl. Similarly, V2= d2/t2 for the other runner, and d2 = 1 mile.
We want to know when Vl > V2. Substitute our expressionsfor speed, and get dl/tl > d2/t2.
Rearrange, and d1/d2 > tl/t2 or d1/d2 > 0.99937. Then d1 > 0.99937 mile x (5280 feet/l mile) or
d1 > 5276.7 feet is the condition that the first runner W8Sindeed f8Ster. The first track can be no
more than 3.3 feet too short to guarantee that the first runner W8Sf8Ster.
2
~
The volume of Antarctica is approximated
area of the base is the area of a semicircle. Then
by the area of the base time the height; the
'1
-7rr
2
V=Ah=
2
h.
The volume is
v
El-16
-
The volume is (77x104m2)(26m)
= 2.00x107m3.
This is equivalent to
(2.00 x 107 m3)(10-3 km/m)3 = 0.02 km3
El-17
(a) C = 27rr = 27r(6.37 x 103 km) = 4.00 x 104 km. (b) A = 47rr2 = 47r(6.37 x 103 km)2 =
5.10 x 108 km. (c) V = ~7r(6.37 x 103 km)3 = 1.08 x 1012 km3.
El-18
The conversions: squirrel, 19 km/hr(1000 m/km)/(3600 s/hr) = 5.3m/s;
rabbit, 30 knots(1.688ft/s/knot)(0.3048 m/ft)= 15m/s;
snail, 0.030 mi/hr(1609 m/mi)/(3600 s/hr) = O.013m/s;
spider, 1.8 ft/s(0.3048 m/ft) = 0.55m/s;
cheetah, 1.9 km/min(1000 m/km)/(60 s/min) = 32m/s;
human, 1000 cm/s/(100 cm/m) = 10m/s;
fox, 1100 m/min/(60 s/min) =18m/s;
lion, 1900 km/day(1000 m/km)/(86,400 s/day} = 22m/s.
The order is snail, spider, squirrel, human, rabbit, fox, lion, cheetah.
~
Then
One light-year is the distance traveled by light in one year, or (3 x 108 mis) x (1 year).
1609 m
1 mi
El-20
Start
with
the British
gal
30.0 mi
El-21
~
gal
units
)(
parsec
inverted,
1.639
x
10-2
L
mi
1.609
in3
= 7.84 x 10-2 L/km.
km
(b) A light-year is
3600 s'
(3.00 x 105 km/s)
A
100 year
1 century ,
1 hr
24
hr
(365 days) = 9.46 x 1012 km.
1 day,
is
1.50 x 108 km
0° 0' 1"
(a) (1.5 x 108 km)/(3.09
1.59 x 10-5 ly.
360°
=
27r
3.09
x
1013
km.
rad
x 1013 km/pc = 4.85 x 10-6 pc. (1.5 x 108 km)/(9.46
4
x 1012 km/ly)
=
~
El-22
First
find the "logarithmic
average"
by
1
2 (log(2 x 1026)+ log(l x 10-15)) ,
log dav
1
21og (2 x 1026x 1 x 10-15) ,
~ log2 x 1011= log ( /2XWU)
.
Salve, and dav = 450 km.
El-24
(a) (2 x 1.0 + 16)u(1:661 x 10-27kg) = 3.0 x 10-26kg.
(b) (1.4 x 1021kg)/(3.0 X 10-26kg) = 4.7 x 1046 molecules.
El-25
It is cheaper to buy coffee in New York (at least according to the physics textbook, that is.)
El-26
The room volume is (21 x 13 x 12)ft3(0.3048 m/ft)3 = 92.8m3. The mass contained in the
room is
(92.8m3)(1.2Lkg/m3)
= 112 kg.
~
One mole ofsugar cubes would have avolume ofNA x 1.0 cm3, where NA is the Avogadro
constant. Since the volume of a cube is equal to the length cubed, V = l3, then l = ~
cm
= 8.4 x 107 cm.
El-28
The number of 8econd8 in a week i8 60 x 60 x 24 x 7 = 6.05 x 105, The "weight" 1088per
8econd i8 then
(0.23kg)/(6.05
x 105 s) = 3.80 x 10-1 mg/s.
~
The definition of the meter WBSwavelengths per meter; the question BSksfor meters per
wavelength, so we want to take the reciprocal. The definition is accurate to 9 figures, so the reciprocal
should be written BS1/1,650,763.73 = 6.05780211x 10-7 m = 605.780211nm.
El-30
(a) 37.76 + 0.132 = 37.89. (b) 16.264-
16.26325 = 0.001,
~
The easiest appraach is ta first salve Darcy's Law far K, and then substitute the knawn
SI units far the ather quantities. Then
K=
~
AHt
has
units
which can be simplified to m/s.
fi
of
(m3) (m)(~) (S)
(m2)
El-32
The Planck
length,
lp, is found
=
[lp]
L
=
from
lci][Gj][hk],
(LT-l)i(L3T-2M-l)j(ML2T-1)k,
Li+3j+2kT-i-2j-kM-j+k.
Equate powers on each sirle,
L: 1 =
T: O =
M: O =
i + 3j + 2k,
-i -2j -k,
-j + k.
Then j = k"and i = -3k, and 1 = 2k; so k = 1/2, j = 1/2, and i = -3/2. Then
[lp}
=
[C-3/2][G1/2][h1/2],
{3.00 x 108 m/8)-3/2{6.67
x 10-11 m3/82 .kg)1/2{6.63
x 10-34 kg .m2/8)1/2,
4.05 x 10-35 m.
El-33
The Planck mass, mp, is found fron
[mp]
M
=
=
[Ci][Gj][hk],
(LT-l)i(L3T-2M-l)j(ML2T-l)k,
Li+3j+2kT-i-2j-kM-j+k.
Equate powers on each side,
L: O
-
i + 3j + 2k,
T: O
-
-i
M: I
Then k = j + 1, and i = -3j
[mp]
~
-1,
-2j
-k,
-j+k.
and O = -1 + 2kj so k = 1/2, and j = -1/2,
=
[C1/2][G-1/2][h1/2],
=
{3.00 x 108 m/s)1/2{6.67 X 10-11 m3 /S2. kg)-1/2{6.63
=
5.46 x 10-8 kg.
There are 24 x 60 = 1440 traditional
straightforward
and i = 1/2. Then
X 10-34 kg .m2/s)1/2,
minutes in a day. The conversion plan is then fairly
" 1440trad.
min
822.8 dec. min (
=
1184.8
trad.
min.
, lUUUdec. min )
This is traditional minutes since midnight, the time in traditional hours can be found by dividing
by 60 min/hr , the integer part of the quotient is the hours, while the remainder is the minutes. So
the time is 19 hours, 45 minutes, which would be 7:45 pm.
Pl-2
(a) By similar triangles, the ratio ofthe distances is the same as the ratio ofthe diameters390:1.
(b) Volume is proportionalto the radius (diameter) cubed, or 3903 = 5.93 x 107.
(c) 0.52°(27r/360°) = 9.1 x 10-3 rad. The diameter is then (9.1 x 10-3 rad)(3.82 x 105 km) =
3500 km.
6
Pl-3
(a) The circumference of the Earth is approximately 40,000 km; 0.5 secondsof an arc is
0.5/(60 x 60 x 360) = 3.9 x 10-7 of a circumference, so the north-south error is ::1:(3.9x 10-7)(4 x
107m) = ::1:15.6
m. This is a range of 31 m.
(b) The east-westrange is smaller, becausethe distance measuredalong a latitude is smaller than
the circumference by a factor of the cosine of the latitude. Then the range is 31 cos43.6° = 22 m.
(c) The tanker is in Lake Ontario, s(jme 20 km off the coast of Hamlin?
Pl-4
Your position is determined by the time it takes for your longitude to rotate "underneath"
the sun (in fact, that's the way longitude was measuredoriginally as in 5 hours west ofthe Azores...)
the rate the sun sweepover at equator is 25,000 miles/86,400 s = 0.29 miles/second. The correction
factor becauseof latitude is the cosine of the latitude, so the sun sweepsoverhead near England at
approximately 0.19 mi/s. Consequently a 30 mile accuracy requires an error in time of no more than
(30 mi)/(0.19 lI!i/s) = 158 seconds.
Trip takes about 6 m~nths, so clock accuracy needs to be within (158s)/(180 day) = 1.2 seconds/day.
(b) Same, except 0.5 miles accuracy requires 2.6 s accuracy, so clock needs to be within 0.007
s/day!
Pl-5
Let B be breaths/minute while sleeping. Each breath takes in (1.43 g/L)(0.3 L) = 0.429 gj
and lets out (1.96 g/L)(0.3 L) = 0.288 g. The net loss is 0.141 g. Multiply by the number ofbreaths,
(8 hr)(60 min./hr)B(0.141 g) = B(67.68 g). I'll take a short nap, and count my breaths, then finish
the problem.
I'm back now, and I foundmy breaths to be 8/minute. So I lose 541 g/night, or about 1 pound.
~
Let the radius of the grain be given by r 9 .Then the surface area of the grain is Ag = 47rr~,
and the volume is given by Vg = (4/3)7rr:.
If N grains of sand have a total surface area equal to that of a cube 1 m on a edge, then
N Ag = 6 m2. The total volume Vt of this number of grains of sand is NVg. Eliminate N from these
two expressions and get
Yt = NVg = ~Vg
Ag
Then
Yt = (2 m2)(50
=
~.
3
X 10-6 m) = 1 x 10-4 m3.
The mass oí a volume
Yt is given by
=
0.26
kg.
Pl-8
For a cylinder V = 7rr2h, and A = 27rr2 + 27rrh. We want to minimize A with respect to
changes in r, so
dA
dr
=
1; ( 27rr2
+ 27rr~ )
47rr -22.
v
r
Set this equal to zero; then V = 2?rr3. Notice that h = 2r in this expression.
7
Pl-9
(a) The volume
per particle
is
{9.27 x 10-26kg)/{7870kg/m3)
= 1.178 x 10-28m3.
The radius of the corresponding sphere is
r = V.
3/3(1.178
x47r
lO-28m3) .
=
1.41
x
10-1Om.
Double this, and the spacing is 282 pm.
(b) The volume per particle is
{3.82 x lO-26kg)/{lOl3kg/m3)
= 3.77 x lO-29m3.
The r~ius of the corresponding sphere is
r = V.
3/3(3.77
Double
PI-I0
this, and the spacing
x 411"
10-29m3) ,
=
2.08
x
10-1om.
is 416 pm.
(a) The area of the plate is (8.43 cm)(5.12 cm) = 43.2 cm2. (b) (3.14)(3.7 cm)2 = 43 cm2.
8
~
Add the vectors as is shown in Fig. 2-4. If a has length a = 4 m and b has length b = 3 m
then the sum is given by s. The cosine law can be used to find the magnitude s of s,
82 =
a2 +
b2 -2abcos(},
where () is the angle between sides a and b in the figure.
(a) (7 m)2 = (4 m)2 + (3 m)2 -2(4 m)(3 m) cos(), so cos() = -1.0, and () = 180°. This means
that a and b are pointing in the same direction.
(b) (1 m)2 = (4 m)2 + (3 m)2 -2(4 m)(3 m)cos(), so cos() = 1.0, and () = 0°. This means that
a and b are pointing in the opposite direction.
(c) (5 m)2 = (4 m)2 + (3 m)2 -2(4 m)(3 m) cos(), so cos(J = 0, and () = 90°. This means that a
and b are pointing at right angles to each other .
( & ) Consider
the
figures
below.
= 2.1)
{b) Net displacement is 2.4 km west, {5.2 -3.
.¡2.42
+
2.12
km
=
km
3.2
south.
A bird
would
By
km.
Consider the figure below.
E2-4 (a) The components are (7.34) cos(252°) = -2.27i and (7.34) sin(252°)= -6.98j.
(b) The magnitude is V( -25)2 + (43)2 = 50; the direction is (} = tan-l(43/25) = 120°. We
did need to choosethe correct quadrant.
~
The components
are given by the trigonometry
o =Hsin{}
relations
= {3.42 km) sin 35.0° = 1.96 km
A = H cos (} = (3.42
km)
9
cos 35.0°
= 2.80
km.
The stated angle is measuredfrom the east-west axis, counter clockwise from east. So O is measured
against the north-south axis, with north being positive; A is measured against east-west with east
being positive.
Since her individual steps are displacement vectors which are only north-south or east-west, she
must eventually take enough north-south steps to equal 1.96 km, and enough east-west steps to
equal 2.80 km. Any individual step can only be along one or the other direction, so the minimum
total will be 4.76 km.
Let rr = 124¡ km and ri = (72.6¡ + 31.4]) km. Then the ship needs to travel
E2-6
~r = rf -ri
= (51.4i + 31.4]) km.
Ship needs to travel v51.42 + 31.42 km = 60.2 km in a direction (} = tan-l(31.4/51.4)
= 31.4° west
of north.
~
(a) In unit vector notation we need only add the components; a+b = (5¡+3j)+(
(5-
-3¡+2j)
=
3)¡ + (3 + 2)j = 2¡ + 5j.
(b) If we define c = a + b and write the magnitude of c as c, then c = ~
= V22 + 52 =
5.39. The direction is given by tan() = Cy/cx which gives an angle of 68.2°, measured counterclockwise from the positive x-axis.
...A
A
A
A
A
A
(a) a + b = (4- l)i + ( -3 + l)j + (1 + 4)k = 3i -2j + 5k.
...A
A
A
A
A
A
(b) a- b = (4- -1)i + ( -3l)j + (1- 4)k = 5i -4j
-3k.
E2-8
A
(c)
E2-9
Rearrange,
and
c=
b -a,
or
b -a=
(-1-4)i+
(a) The magnitude of a is V4.02 + (-3.0)2
A
(1-
-3)j
A
+
(4 -1)k
=
A
-5i+4j
A
A
+3k.
= 5.0j the direction is 9 = tan-l(-3.0/4.0)
=
323°.
(b) The magnitude of b is V6.02 + 8.03 = 10.0j the direction is 9 = tan-l(6.0/8.0)
= 36.9°.
-A
A
(c) The resultant vector is a + b = (4.0 + 6.0)i + ( -3.0 + 8.0)j. The magnitude of a + b is
V(10.0)2 + (5.0)2 = 11.2j the direction is 9 = tan-l(5.0/10.0)
= 26.6°.
(d)
The
resultant
vector
is
a-
-A
b =
A
(4.0-
6.0)i
+
(-3.0-
8.0)j.
The
magnitude
of
a-
V( -2.0)2 + ( -11.0)2 = 11.2j the direction is 9 = tan-l( -11.0/ -2.0) = 260°.
-A
A
(e) The resultant vector is b- a = (6.0- 4.0)i + (8.0- -3.0)j.
The magnitude of b -a
V(2-:0)2 + (11.0)2 = 11.2j the direction is 9 = tan-l(11.0/2.0)
= 79.7°.
E2-10
(a) Find components of a; ax = (12.7) cos(28.2°) = 11.2, ay = (12.7) sin(28.2°) =
Find components of b; bx = (12.7) cos(133°) = -8.66, by = (12.7) sin(133°) = 9.29. Then
r =
a +
...A
b =
A
{11.2
-8.66)i
+
{6.00
(b) The magnitude oí r is v2.542 + 15.292 = 15.5.
(c) The angle is (} = tan-l(15.29/2.54)
= 80.6°.
E2-11
Consider the figure below.
10
+
9.29)j
A
=
2.54i
A
+
15.29j.
b
is
is
E2-12
~
Consider
the figure below.
Our axes will be chosenso that i points toward 3 O'clock and j points toward 12 O'clock.
(a)
The two relevant positions are ri = (11.3 cm)i and rf = (11.3 cm)j. Then
~i!
=
i!f -i!i
=
{11.3
cm)j
-{11.3
cm)i.
{b)
The two relevant positions are now ri = {11.3 cm)] and rr = { -11.3 cm)]. Then
j;}.r
=
rf -ri
=
{11.3 cm)j -{ -11.3 cm)j
-{22.6
{c)
The two relevant
,
positions
are now rj = (-11.3
~i
E2-14
cm)j.
=
if
=
{-11.3
=
{0 cm)].
cm)j
and rf = {-11.3
-ii
cm)]
-{-11.3
cm)]
(a) The components of rl are
Tlx = (4.13 m) cos(225°)
= -2.92
m
rly
= -2.92
m.
and
= (4.13 m) sin(225°)
11
,
cm)j.
Then
The components oí r2 are
rlx
=
{5.26
m)
cos{OO)
=
5.26
m
and
rly
=
(5.26m)sin(OO)
= Om.
The components of r3 are
rlx
= (5.94m)cos(64.00)
= 2.60m
rly
= (5.94 m) sin(64.00)
and
= 5.34 m.
(b) The resulting displacement is
[ (-2.92 + 5.26 + 2.60)i + ( -2.92 + O+ 5.34)1] m = (4.94i + 2.421)m.
(c) The magnitude of the re8ulting di8placement i8 v 4.942+ 2.422m = 5.5 m. The direction of
the re8ulting di8placement i8 (J = tan-l(2.42/4.94) = 26.1°. (d) To bring the particle back to the
8tarting point we need only rever8ethe an8Werto (c); the magnitude will be the 8ame,but the angle
will be 206°.
E2-15
The components oí the initial position are
rlx
=
(12,000
ft) cos(40°)
= 9200
ft
rly = (12,000 ft) sin(40°) = 7700 ft.
The co~ponents of the final position are
r2x
= (25,8000
r2y =
(25,800
ft) cos(163°)
ft) sin(163°)
= -24,700
= 7540
ft
ft.
The displacement is
r = r2 -r1
= [ ( -24,700
-9,
200)i + (7,540
-9,200)])
] = ( -33900i
-1660])
ft.
E2-16 (a) The displacement vector is r = (410i -820j) mi, where positive x is ea.stand positive
y is north. The magnitude ofthe displacement is V(410)2 + (-820)2 mi = 920 mi. The direction is
(} = tan-l(-820/410) = 300°.
(b) The averagevelocity is the displacement divided by the total time, 2.25 hours. Then
"av = (1801- 360j) mi/hr.
(c) The averagespeed is total distance over total time, or (410+820)/(2.25) mi/hr = 550 mi/hr.
12
~
~
(a) Evaluate
r
r when t = 2 s.
-
[(2 m/s3)t~ -(5
m/s)t]i+
[(2 m/s3)(2 8)3[(16 m) -(10
[(6 m) -(7
m/s4)t4]]
(5 m/s)(2 s»)i+ 1(6 m)-
m)]i + {(6 m) -(112
[(6 m)]i + [-(106
(7 m/s4)(2 8)4]]
m)]]
m)]].
(b) Evaluate:
dr
dt
v-
[(2 m/s3)3t2 -(5
m/s)]i + [-(7 m/s4)4t3]j
[(6 m/s3)t2 -(5
m/s)]i + [--(28 m/s4)t3Jj.
Into this last expression we now evaluate v(t = 2 s) and get
v
=
[(6 m/s3)(2
[(24 m/s)
[(19 m/s)]i
sf
-(5
-(5
m/s)]i
+ [-(224
m/s)]i
+ [-(28
+ [-(224
m/s4)(2
s)3]j
m/s)]j
m/s)ti,
for the velocity v when t = 2 s.
(c) Evaluate
dv
--a-
[(6 m/s3)2t]i + l-(28 m/s4)3t2]j
dt
[(12 m/s3)t]i +{-(84
m/s4)t2]j,
Into this last expression we now evaluate a( t = 2 s) and get
a =
[(12 mis3)(2 s)]i + [-(84 mis4)(2 2)2]j
2 A
2 A
[(24 mis )]i + [-(336 mis )]j.
E2-18 (a) Let ui point north, j point e8St, and k point up. The displacement is (8.7i + 9.7j +
2.9k) km. The averagevelocity is found by dividing each term by 3.4 hr; then
Vav = (2.6i + 2.9j + 0.85) km/hr .
The magnitude of the average velocity is v2.62 + 2.92 + 0.852 km/hr = 4.0 km/hr.
(b) The horizontal velocity has a magnitude of v2.62 + 2.92 km/hr = 3.9 km/hr.
the horizontal is given by (} = tan-l(0.85/3.9)
= 13°.
E2-19
(a) The derivative
of the velocity
The angle with
is
a = [(6,Om/s2)-(8.Om/s3)t]i
so the acceleration at t = 3 s is a = ( -18.0 m/s2)i. (b) The acceleration is zero when (6.0 m/s2) (8.0m/s3)t = 0, or t = 0.75s. (c) The velocity is never zero; there is no way to "cancel" out the
y component. (d) The speed equals 10m/s when 10 = ~,
or Vx = :!::6.0m/s. This happens
when (6.0m/s2) -(8.0m/s3)t = :!::6.0m/s, or when t = Os.
13
E2-20
If v is constant then so is V2 = V~ + V~. Take the derivative;
d
d
2vxdtvx + 2v1ldtv1l = 2(vxax + v1la1l).
But if the value is constant the derivative is zero.
~
Let the actual flight time, as measured by the passengers, be T. There is some time
difference between the two cities, call it ~T = Namulevu time -Los Angeles time. The ~T will be
positive ií Namulevu is east oí Los Angeles. The time in Los Angeles can then be íound from the
time in Namulevu by subtracting ~T.
The actual time oí flight from Los Angeles to Namulev\¡ is then the difference between when the
plane lands (LA times) and when the plane takes off (LA time) :
T
=
(18:50 -~T)
=
6:00-~T,
-(12:50)
where we have written times in 24 hour format to avoid the AM/PM issue. The return flight time
can be found from
T
=
(18:50) -(1:50-
=
17:00 + 6.T,
6.T)
where we have again changed to LA time for the purpose of the calculation.
(b) Now we just need to solve the two equations and two unknowns.
17:00 + 6.T
=
6:00 -6.T
26.T
=
6:00 -17:00
6.T
=
-5:30.
Since this is a negative number, Namulevu is located west of Los Angeles.
(a) T = 6:00- 6.T = 11: 30, or eleven and a half hours.
(c) The distance traveled by the plane is given by d = vt = (520 mi/hr)(11.5 hr) = 5980 mi.
We'll draw a circle around Los Angeles with a radius of 5980 mi, and then we look for where it
intersects with longitudes that would belong to a time zone 6.T away from Los Angeles. Since the
Earth rotates once every 24 hours and there are 360 longitude degrees,then each hour corresponds
to 15 longitude degrees,and then Namulevu must be located approximately 15° x 5.5 = 83° west of
Los Angeles, or at about longitude 160 east. The location on the globe is then latitude 5°, in the
vicinity of Vanuatu.
When this exercisewas originally typeset the times for the outbound and the inbound fiights were
inadvertentlyswitched. I supposethat we could blame this on the airlines; nonetheless,when the answers
werepreparedfor the backof the book the reversednumbersput Namulevueastof Los Angeles.That would
put it in either the North Atlantic or Brazil.
E2-22 There is a three hour time zone difference. So the flight is seven hours long, but it takes
3 hr 51 min for the sun to travel same distance. Look forwhen the sunset distance has caught up
with plane:
dsunset
Vsunset(t(t-
so t =
3:31
into
1:35)
1:35)/3:51
flight.
14
=
=
=
dplane'
Vplanet,
t/7:00,
E2-23
The distance
is
d = vt =
E2-24
(112
km/hi)(1
8)/(3600
8/hr)
= 31 m.
The time taken for the hall to reach the plate is
~
Speed is distance traveled divided by time taken; this is equivalent to the inverse of the
slope of the line in Fig. 2-32. The line appears to pass through the origin and through the point
(1600 km, 80 x 106 y), so the speed is v = 1600 km/80 x 106 y= 2 x 10-5 km/y. Converting,
v = 2 x lO-5km/y
E2-26
1000 m
1km
100 cm
1m
{a) For Maurice Greene vav = {100m)/{9.81
vav = {26.219
{2.0950 mi)
hr)
= 2 cm/y
m) = 10.2m/.s. For Khalid Khannouchi,
1609m
1 mi
) = 5.594 m/s.
lhr
3600"8
(b) Ir Maurice Greene ran tbe maratbon witb an averagespeed equal to bis averagesprint speed
tben it would take bim
t=
(26.219 mi)
1609m
I mi
lhr
=
1.149
hr,
36008
or 1 hour, 9 minutes.
E2-27
The time saved is the difference,
{700km/hr)
km)
A.t = {88.5
{700 km/hr)
km)
-{104.6
= 1.22 hr,
which is about 1 hour 13 minutes.
E2-28
The ground elevation will increase by 35 m in a horizontal distance of
x = {35.0m)/tan{4.3°)
= 465 m.
The plane will cover that distance in
t=
(0.465 km)
(1300 km/hr)
~
)
= 1.38.
lhr
~
Let Vl = 40 km/hr be the speed up the hill, tl be the time taken, and d1 be the distance
traveled in that time. We similarly define V2 = 60 km/hr for the down hill trip, as well as t2 and d2.
Note that d2 = d1.
15
Vl = d1/tl and V2 = d2/t2. vav = d/t, where d total distance and t is the total time. The total
distance is d1 + d2 = 2d1. The total time t is just the sum of tl and t2, so
=
vav
d
t
2dl
tl + t2
2dl
dl/Vl
+ d2/V2
2
l/v1'+
1/v2 ,
Take the reciprocal of both sides to get a simpler looking expression
2
-=
vav
1
-+
VI
1
-,
V2
Then the average speed is 48 km/hr .
E2-30
(a) Average speed is total distance divided by total time. Then
(b) Same approach, but different information
given, so
E2-31 The distance traveled is the total area under the curve. The "curve" has four regions: (I)
a triangle from O to 2 s; (II) a rectangle from 2 to 10 s; (III) a trapezoid from 10 to 12 s; and (IV)
a rectangle from 12 to 16 s.
The area underneath the curve is the sum of the areas of the four regions.
E2-32
The acceleration is the slope of a velocity-time
a= (8m/s) -(4m/s)
curve,
=
-2
m/s2,
(IOs) -(12s)
~
The initial velocity is Vi = (18 m/s)i, the final velocity is Vf = ( -30 m/s)i.
acceleration is then
6.v
Rav-
which gives aav =
-20.0
At
--
=
Vf
-Vi
.( -30
~t
m!s)i
-(18
2.4
m/s2)i.
16
s
m!s)i
The average
E2-34
Consider the figure below.
E2-35 (a) Up to A Vx > O and is constant. From A to B Vx is decreasing,but still positive. From
B to C Vx = O. From C to D Vx < 0, but Ivxl is decreasing.
(b) No. Constant acceleration would appear as (part of) a parabola; but it would be challenging
to distinguish between a parabola and an almost parabola.
E2-36 (a) Up to A Vx > O and is decreasing. From A to B Vx = O. From B to C Vx > 0 and is
increasing. From C to D Vx > 0 and is constant.
(b) No. Constant acceleration would appear as (part of) a parabola; but it would be challenging
to distinguish between a parabola and an almost parabola.
~
E2-38
Consider
the
figure
below.
Consider the figure below.
17
The acceleration
is a constant
2 cm/s/s during
the entire time
interval.
E2-39
(a) A must have units of m/s2, B must have units of m/s3.
(b) The maximum positive x position occurs when v"' = O, so
v:&
=
dx
=
2At
-3Bt2
dt
implies Vx = O when either t = O or t = 2A/3B = 2(3.0m/s2)/3(1.0m/s3)
= 2.0s.
( c ) Particle starts from rest, then travels in positive direction until t = 2 s, a distance of
x =
(3.0m/s2)(2.0S)2
-(1.0m/s3)(2.0s)3
= 4.0m.
Then the particle moves back to a final position of
x = (3.0 m/s2)(4.0S)2
-(1.0m/s3)(4.0s)3
= -16.0m.
The total path followed was 4.0 m + 4.0 m + 16.0 m = 24.0 m.
(d) The displacement is -16.0m as was found in part (c).
(e) The velocity is Vx = (6.0m/s2)t -(3.0m/s3)t2.
When t = 0, Vx = O.Om/s. When t = 1.0s,
Vx = 3.0m/s. When t = 2.0s, Vx = O.Om/s. When t = 3.0s, Vx = -9.0m/s.
When t = 4.0s,
Vx = -24.0m/s.
(f) The acceleration is the time derivative of the velocity,
ax = dvx
= (6.0m/s2)-(6.0m/s3)t.
dt
When t = Os, ax = 6.Om/s2. When t = 1.Os, ax = O.Om/s2. When t = 2.Os, ax = -6.Om/s2.
When t = 3.Os, ax = -12.Om/s2. When t = 4.Os, ax = -18.Om/s2.
(g) The distance traveled was found in part (a) to be -2Om The averagespeed during the time
interval is then Vx,av= (-2Om)/(2.Os) = -lOm/s.
E2-40
will be
VOz= O, Vz = 360 km/hr
= 100 m/s. Assuming constant acceleration the average velocity
Vx,av= i(lOOm/s + O) = 50m/s.
The time to travel the distance oí the runway at this averagevelocity is
t = (1800m)/(50m/8)
= 368.
The acceleration is
ax = 2x/t2
= 2(1800m)/(36.0s)2
18
= 2.78m/s2.
~
(a) Apply
Eq.
2-26,
Vz
{3.0 x 107 mis)
3.1 x 106 s
=
VOz+ azt,
=
{0) + {9.8 mis2)t,
=
t.
(b) Apply Eq. 2-28 using an initial position of Xo = O,
E2-42
X
=
1
Xo + VOx + 2axt
x
=
(0) + (0) + ~(9.8 m/s~)(3.1 x 106 S)2,
X
=
4.7 x 1013 m.
that
E2-43
,
VOx= o and Vx = 27.8m/s. Then
t = (vx -vox)/a
I want
2
= ((27.8m/s)
-(O))
/(50m/s2)
= 0~56s.
car .
The muon will travel for t seconds before it comes to a rest, where t is given by
t = (vx -vox)/a
= ((O) -(5.20
x 106m/s)) /(-1.30
x 1014m/s2) = 4 x 10-8s.
The distance traveled will be
1
1 -1.30 x 1014m/s2)(4 X 10-8s)2 + (5.20 x 106m/s)(4 x 10-8s) = 0.104m.
x = 2axt2
+ vOxt = 2(
E2-44
The averagevelocity of the electron was
1
V:1:,av
= 2(1.5
x 105m/s + 5.8 x 106m/s) = 3.0 x 106m/s.
The time to travel the distance of the runway at this averagevelocity is
t = (0.012m)/(3.0 x 106m/s) = 4.0 x 10-9s.
The acceleration is
ax = (v x -vOx)/t
~
= ((5.8 x 106 m/s)
-(1.5
x 105 m/s))/(4.0
x 10-9 s) = 1.4 x 1015m/s2,
It will be easier to solve the problem ir we change the units ror the initial velocity,
1000 m
km
and then applying Eq. 2-26,
v x
(O)
1
-202
m
:=
VOx +
:=
(283
=
ax.
2
s
:
axt,
mis)
+
ax(1.4
The problem asks íor this in terms oí g, so
-202
9
m/s2
9:8-;;;¡;;-
19
= 21g.
s),
E2-50
(a) The deceleration is found from
2
ax = "i2(x
-vot)
= ~«34m)
--- 2
-(16
m(s)(4.0
s)) = -3.75m/s2.
(b) The impact speed is
v x = VOx+ axt = (16m/s) + (-3.75m/s2)(4.0s)
-
19
~S2(-1700m)
J
-
---s.
t
-
2y
Assuming the drops fall from rest, the time to fall is
{if;
E2-51
= 1.0m/s.
ay
(-9.8 m/s2)
The velocity of the falling drops would be
Vy = ayt = (-9.8m/s2)(19s) = 190m/s,
or about 2/3 the speed of sound.
E2-52
Solve the problem out of order .
(b) The time to fall is
t- -fiYV ~
-V 12(-120m)(-9.8m/s2)
-4.9s.
(a) The speed at which the elevator hits the ground is
Vy = ayt = (-9.8m/s2)(4.9s)
= 48m/s.
( d) The time to fall half-way is
t (C) The speed at the half-way
{jj.
2y
-
ay
point
J
~s2(-60m)
- 3 5 s.
( -9.8
m/s2)
is
Vy = ayt = (-9.8m/s2)(3.5s)
= 34m/s.
~
The initial velocity of the "dropped" wrench would be zero. I choosevertical to be along
the y axis with up as positive, which is the convention of Eq. 2-29 and Eq. 2-30. It turns out that
it is much easier to solve part (b) before solving part (a).
(b) We solve Eq. 2-29 for the time of the fall.
Vy
( -24.0 m/s)
2.45 s
=
VOy-gt,
=
(0) -(9.8
=
t.
m/s2)t,
(a) Now we can use Eq. 2-30 to find the height from which the wrench fell.
y
=
(0)
=
0
=
1
Yo + VOyt --gt
2
2
,
1
Yo + (0)(2.45 s) --(9.8
2
Yo -29.4 m
2
2
m/s )(2.45 s) ,
We have set y = 0 to correspond to the final position ofthe wrench: on the ground. This results in
an initial position of Yo = 29.4 m; it is positive because the wrench was dropped from a point above
where it landed.
21
E2-54 (a) It is easier to solve the problem from the point of view of an object which falls from the
highest point. The time to fall from the highest point is
t-
fj;
V~~2(-53.7m)
2y -
ay
( -9.81 m/s2)
- 331 s.
The speed at which the object hits the ground is
Vy = ayt = ( -9.81 m/s2)(3.31 s) = -32.5 m/s.
But the motion is symmetric, so the object must have been launched up with a velocity of Vy =
32.5m/s.
(b) Double the previous answer; the time of flight is 6.62 s.
E2-55
(a) The time to fall the first 50 meters is
E2-56
then
The rock returns to the ground with an equal, but opposite, velocity. The acceleration is
ay = (( -14.6m/s)
-(14.6
m/s))/(7.72s)
= 3.78m/s2,
That would put them on Mercury.
~
(a) Solve Eq. 2-30 for the initial velocity. Let the distances be measured from the ground
so that Yo = 0.
y
(36.8 m)
36.8 m
27.4 mis
=
1
Yo + VOyt -2gt
==
(O) + voy(2.25
=
voy(2.25
=
VOy.
(b) Salve Eq. 2-29 far the velacity,
2
,
s) -~(9.8
s) -24.8
using the result
mis2)(2.25
8)2,
m,
fram
part
(a).
Vy =
VOy-gt,
Vy =
Vy =
(27.4 m/s) -(9.8 m/s2)(2.25 s),
5,4 m/s.
(c) We need to solve Eq. 2-30 to find the height to which the ball rises, but we don't know how
long it takes to get there. So we first solve Eq. 2-29, becausewe do know the velocity at the highest
point (Vy = 0).
vy
=
VOy -gt,
(O)
=
2.8 s
=
(27.4 mis)
f
22
-(9.8
mis2)t,
And then we find the height to which the object rises,
y
=
1
Yo + vOyt -2gt
y
=
{0) + {27.4
y
=
38.3m.
2
,
I
s) -2{9.8
m/s){2.8
m/s2){2.8
8)2,
This is the height as measuredfrom the ground; so the hall rises 38.3- 36.8 = 1.5 m ahove the point
specified in the prohlem.
E2-58
The time it takes for the hall to fall 2.2 m is
The hall hits the ground with a velocity of
Vy = ayt = (-9.8m/s2)(0.67s) = -6.6m/s.
The ball then bounces up to a height of 1.9 m. It is easier to solve the falling part of the motion,
and then apply symmetry. The time is would take to fall is
/2Y
I
= v , ( 2(-1.9m)
-9.8 m/s2)
=
0.628.
t"=v~
The hall hits the ground with a velocity oí
V1l= a1lt = (-9.8m/s2)(0.62s)
= -6.1m/s.
But we are interested in when the ball moves up, so v1l = 6.1 m/s.
The"acceleration while in contact with the ground is
ay = ((6.1 m/s) -( -6.6m/s))/(O.O96s)
E2-59
= 130m/s2.
The position as a function of time for the first object is
1 2
yl = -2gt ,
The position as a function of time for the second object is
y2 = -.!g(t
2
-1 s)2
The difference,
~y=
Y2 -yl
= frac12g
((2s)t
-1)
,
is the set equal to 10 m, so t = 1.52 s.
E2-60
Answer part (b) first.
(b) Use the quadratic equation to solve
(-81.3m)
1
= 2(-9.81m/s2)t2
+ (12.4m/s)t
for time. Get t = -3.0 s and t = 5.53 s. Keep the positive answer .
(a) Now find final velocity from
Vy = (-9.8m/s2)(5.53s)
+ (12.4m/s)
23
= -41.8m/s.
~
The total time the pot is visible is 0.54 s; the pot is visible for 0.27 s on the way down.
We'll define the initial position as the highest point and make our measurementsfrom there. Then
Yo= 0 and VOy= 0. Define tl to be the time at which the falling pot passesthe top of the window
Yl, then t2 = tl + 0.27 s is the time the pot passesthe bottom of the window y2 = yl -1.1 m. We
have two equations we can write, both based on Eq. 2-30,
Yl
=
I
Yo + VOytl -29tl'
Yl
=
I
(O) + (O)tl -29tl'
2
2
and
yl
1
-2gt2'
2
y2
=
m
1
= {0) + {0)t2 -29{tl
+ 0.278)2,
-1.1
Yo +
VOyt2
Isolate Yl in this last equation and then set the two expressionsequal to each other so that we can
solve for tl,
--
1
2
--
g t2
1
1
2
g t2
1
=
=
o =
1.
1.1
m-
) 2
29(tl
+
0.27
s
,
1 m- ~g(t~ + [0.548]tl + 0.07382),
1.1 m
~g([0.54
8]tl
+
0.073
82).
This last line can be directly solved to yield tl = 0.28 s as the time when the falling pot passesthe
top ofthe window. Use this value in the first equation above and we can find Yl = -~(9.8 m/s2)(0.28
S)2= -0.38 m. The negative sign is becausethe top of the window is beneath the highest point, so
the pot must have risen to 0.38 m above the top of the window.
P2-1 (a) The net shift is v(22m)2 + (17m)2) = 28m.
(b) The vertical displacement is (17m) sin(52°) = 13m.
P2-2
Wheel "rolls" through halí oí a turn, or 1rr = 1.41 m. The vertical displacement is 2r =
0.90 m. The net displacement is
The angle is
(} = tan-l(O.90m)/(1.41m)
= 33°.
~
We align the coordinate system so that the origin corresponds to the starting position of
the fl.y and that all positions inside the room are given by positive coordinates.
(a) The displacement vector can just be written,
~r = {10 ft)i + (12 ft)j + (14 ft)k.
(b) The magnitude of the displacement vector is 16.r¡ = y102 + 122 + 142 ft= 21 ft.
'>A
(c) The straight line distance between two points is the shortest possible distance, so the length
of the path taken by the fiy must be greater than or equal to 21 ft.
(d) If the fiy walks it will need to cross two faces. The shortest path will be the diagonal across
these two faces. If the lengths of sides of the room are ll, l2, and l3, then the diagonallength across
two faceswill be given by
v (ll + l2)2 + l~,
where we want to choosethe li from the set of 10 ft, 12 ft, and 14 ft that will minimize the length.
The minimum distance is when II = 10 ft, l2 = 12 ft, and l3 = 14. Then the minimal distance the
fly would walk is 26 ft.
P2-4
Choose vector a to lie on the x axis. Then a = ai and b = bxi + byj where bx = b cos (} and
by = bsin(}. The sum then has components
T", = a + bcos{} and Ty = bsin{}.
Then
r2
=
(a+
bcosO)
2
+
(bsinO)
2
,
a2+ 2abcosO+ b2.
P2-5
(a) Average speed is total distance divided by total time. Then
(b) Average speed is total distance divided by total time. Then
vav =
P2-6
(d/2)/(35.0
(d/2)
mi/hr)
+ {d/2)
+ (d/2)/(55.0
mi/hr)
= 42.8
.
ml/hr.
(a) We'll do jU8t one together. How about t = 2.08?
x = (3.0 m/s)(2.0 s) +
-4.0m/s2){2.0S)2
and
25
+ {1.Om/s3){2.0s)3
= -2.0m.
~
(a) Assume the bird has no size, the trains have some separation, and the bird is just
leaving one of the trains. The bird will be able to fiy from one train to the other be/orethe two
trains collide, regardlessof how close together the trains are. After doing so, the bird is now on the
other train, the trains are still separated, so once again the bird can fiy between the trains before
they collide. This processcan be repeated every time the bird touches one of the trains, so the bird
will make an infinite number of trips between the trains.
(b) The trains collide in the middle; therefore the trains collide after (51 km)/(34 km/hr) = 1.5
hr. The bird was fiying with constant speedthis entire time, so the distance fiown by the bird is (58
km/hr)(1.5 hr) = 87 km.
P2-8
(a) Start with a perfect square:
(v¡
o,
-V2)2
V~ + V~
2vlv2,
(V~ + V~)t¡t2
d~ + d~ + (v~ + V~)t¡t2
(V~t¡
+ V~t2)(t¡
v~t¡
+ t2)
+ V~t2
>
>
d~ + d~ + 2V¡V2t¡t2,
>
(d¡
2V¡V2t¡t2,
>
f;:+t;'
v¡t¡ + V2t2
d1 + d2
V1 d1 + v2d2
+ d2)2,
d¡ + d2
>
t¡ + t2
d1 + d2
Actually,
it only works ir d¡ + d2 > O!
(b) Ir v¡ = V2.
P2-9
(a) The average velocity during the time interval is vav = 6.x/6.t,
vav = (A+B(3s)3)
=(A+B(2s)3)
= (1.50 cm/s3)(19s3)/(ls)
or
= 28.5 cm/s.
(38) -(28)
(b)
(c)
(d)
(e)
v =
v =
v =
The
dx/dt =
dx/dt =
dx/dt =
midway
3Bt2 =
3Bt2 =
3Bt2 =
position
3(1.50
3(1.50
3(1.50
is (Xf
cm/s3)(2s)2 = 18 cm/s.
cm/s3)(3s)2 = 40.5 cm/s.
cm/s3)(2.5s)2 = 28.1 cm/s.
+ xi)/2, or
Xmid = A + B[(3s)3 + (2s)3)]/2 = A + (17.5s3)B.
This occurs when t = ~.
The instantaneous velocity at this point is
v = dx/dt = 3Bt2 = 3(1.50 cm/s3)(~)2
P2-10
Consider
the figure
below.
x
x
-Po;; 1
(a)
= 30.3 cm/s.
""'-
t
t
(b)
(c)
26
(d)
~
(a) The average velocity is displacement divided by change in time,
Vav
=
(2.0
m/s3)(2.0
8)3-
(2.0
m/s3){1.0
(2.0 s) -(1.0
8)3'
=
14.0
s)
1.0
m
= 14.0m/s.
s
The averageacceleration is the change in velocity. So we need an expression for the velocity, which
is the time derivative of the position,
From this we find averageacceleration
18.0 m/s
1.0 s
=
18.0
m/s2.
(b) The instantaneous velocities can be íound directly from v = (6.0 m/s2)t2, so v(2.0 s) = 24.0
m/s and v(1.0 s) = 6.0 m/s. We can get an expression íor the instantaneous acceleration by taking
the time derivative oí the velocity
Then the instantaneous accelerations are a(2.0 s) = 24.0 m/s2 and a(1.0 s) = 12.0 m/s2
(C) Since the motion is monotonic we expect the average quantities to be somewherebetween
the instantaneous values at the endpoints of the time interval. Indeed, that is the case.
x(m)
o
2
4
6
8
10
o
4
2
8 10
x =
1
-at2
2
+
Vit
=
1
-at2
2
2
4
6
8 10
t(s)
Start with Vf = Vi + at, but Vf = O, so Vi = -at,
so t = p¡x¡a
Converting,
o
t(s)
t(s)
P2-13
6
-at2
then
=
1
--at2
2
= J=2(19.2 ft)/(-32 ft/S2) = 1.10s. Then Vi = -(-32 ft/s2)(1.10s) = 35.2ft/s.
35.2 ft/s(1/5280
mi/ft)(3600
27
s/h) = 24 mi/h.
P2-14 (b) The averagespeed during while traveling the 160 m is
vav = (33.0m/s + 54.0m/s)/2
The time to travel the 160 m is t
(a) The acceleration is
(160
m)/(43.5m/8)
= 43.5m/s.
=;= 3.688.
2(160m)
2(33.0 m/8 }
= 5.69m/s2,
(3.688}2 -(3.688}
(c) The time required to get up to a speed of 33 mis is
t = v/a = (33.0m/8)/(5.69m/82)= 5.808.
( d ) The distance
moved from start
is
P2-15 (a) The distance traveled during the reaction time happens at constant speed;treac= d/v =
(15m)/(20m/s) = 0.75s.
(b) The braking distance is proportional to the speed squared (look at the numbers!) and in
this caseis dbrake= v2/(20m/s2). Then dbrake= (25m/s)2/(20m/s2) = 31.25m. The reaction time
distance is dreac= (25m/s)(0.75s) = 18.75m. The stopping distance is then 50 m.
P2-16 (a) For the car Xc = act2/2. For the truck Xt = Vtt. Set both Xi to the same value, and
then substitute the time from the truck expression:
x
act2/2
= ac(x/vt)2/2,
or
x = 2Vt2/ac =2(9.5m/s)2/(2.2m/s)
= 82m.
(b) The speed oí the car will be given by Vc = act, or
Vc = act = acx/vt = (2.2 m/s)(82 m)/(9.5m/s)
= 19m/s.
~
The runner covered a distance d1 in a time interval tl during the acceleration phase and
a distance d2 in a time interval t2 during the constant speed phase. Since the runner started from
rest we know that the constant speed is given by v = atl, where a is the runner's acceleration.
The distance covered during the acceleration phase is given by
dl
=
1
2
2atl.
The distance covered during the constant speed phase can also be found from
d2 = Vt2 = atlt2.
We want to use these two expressions,along with d¡ + d2 = 100 m and t2 = (12.2 s) -t¡,
100 m
=
1 2
d¡ + d2 = 2at¡ + at¡(12.2 S-t¡),
=
1 2
-2at¡+
=
-(1.40 m/s2)t~+ (34.2m/s)t¡.
a( 12.2s)t¡,
28
to get
This 185texpressionis quadratic in tl, and is solved to give tl = 3.40s or tl = 21.0s. Since the race
only 185ted12.2 s we can ignore the second answer.
(b) The distance traveled during the acceleration ph85e is then
1
2at~
d¡ =
= {1.40
m/s2){3.40s)2
= 16.2 m.
P2-18
(a) The ball will return to the ground with the same speed it was launched. Then the total
time of flight is given by
t = {Vf -Vi)/g
= ( -25m/s
-25
m/s)/(9.8m/s2)
= 5.1 s.
(b) For small quantities we can think in terms of derivatives, so
8t = (8vr -8vj)/g,
orT=2t:/g.
P2-19
Use y = -gt2/2, but only keep the absolute value. Then Y50 = (9.8m/s2)(0.05s)2/2
1.2 cm; YI00 = (9.8m/s2)(0.10s)2/2
= 4.9 cm; Y150 = (9.8m/s2)(0.15s)2/2
= 11 cm; Y200
(9.8m/s2)(0.20s)2/2
= 20 cm; Y250= (9.8m/s2)(0.25s)2/2
= 31 cm.
P2-20
The truck wi}} move 12 m in (12 m)/(55 km/h) = 0.785 s. The app}e wi}} fa}} y = -gt2 /2 =
-(9.81m/s2)(0.785s)2/2
= -3.02 m.
~
The rocket travels a distance d1 = !at~ = !(20 m/s2)(60 S)2= 36,000 m during the acceleration phase; the rocket velocity at the end of the acceleration phase is v = at = (20 m/s2)(60 s) =
1200m/s. The secondhalf of the trajectory can be found from Eqs. 2-29 and 2-30, with Yo= 36,000
m and VOy= 1200 m/s.
(a) The highest point of the trajectory occurs when Vy = 0, so
vy
(O)
122
=
VOy -gt,
=
{1200
mis)
-{9.8
mis2)t,
s
This time is used to find the height to which the rocket rises,
y
=
1
Yo + VOyt -2gt
2
,
1
-2(9.8
(36000 m) + (1200 m/8)(1228)
m/82)(122
8)2 = 110000 m.
(b) The easiest way to find the total time of flight is to solve Eq. 2-30 for the time when the
rocket has returned to the ground. Then
y
(O)
=
1
Yo + VOyt -2gt
2
,
-
This quadratic expression has two solutions for t; one is negative so we don't need to worry about
it, the other is t = 270s. This is the free-fall part of the problem, to find the total time we need to
add on the 60 secondsof acceleratedmotion. The total time is then 330 seconds.
')(\
P2-22 (a) The time required for the player to "fall" from the highest point a distance ofy = 15 cm
is V2Y79j the total time spent in the top 15 cm is twice this, or 2V2Y79 = 2V2(0.15 m)/(9.81 m/s) =
0.350s.
(b The time required for the player to "fall" from the highest point a distance of 76 cm is
2(0.76m)/(9.81m/s) = 0.394s, the time required for the player to fall from the highest point a
distance of (76 -15 = 61) cm is V2(0.61m)/g = 0.353s. The time required to fall the bottom 15
cm is the difference, or 0.041 s. The time spent in the bottom 15 cm is twice this, or 0.081 s.
P2-23 (a) The averagespeed between A and B is vav = (v + v/2)/2 = 3v/4. We can also write
vav = (3.0m)/~t = 3v/4. Finally, v/2 = v -g~t.
Rearranging, v/2 = g~t. Combining all of the
above,
v -
(
2-9
~
)
or
V2
=
{8.0
m)9.
3v
Then v = v(8.0m)(9.8:iñ:/s2)
= 8.85m/s.
(b) The time to the highest point above B is v /2 = gt, so the distance above B is
2
y=-~t2+~t=-~(~).
Then y = (8.85 m/s)2/(8(9.8m/s2))
+~(~)=~.
= 1.00m.
P2-24 (a) The time in free fall is t = F'iiiTii=
V-2(-145m)/(9.81m/s2)=
5.44s.
(b) The speed at the bottom is v = -gt = -(9.81m/s2)(5.44s) = -53.4m/s.
(c) The time for deceleration is given by v = -25gt, or t = -( -53.4m/s)/(25 x 9.81 /S2) =
0.218s. The distance through which deceleration occurred is
25g
y = -t2
2
~
Find
+ vt = (123m/s2)(0.218s)2 + (-53.4m/s)(0.218s) = -5.80m.
the time she fell from Eq. 2-30,
(O ft) = (144 ft) + (o)t -~(32
ft/S2)t2,
which is a simple quadratic with solutions t = :1:3.0s. Only the positive solution is of interest. Use
this time in Eq. 2-29 to find her speed when she hit the ventilator box,
Vy = (0) -(32 ft/s2)(3.0 s) = -96 ft/s.
This becomesthe initial velocity for the deceleration motion, so her averagespeedduring deceleration
is given by Eq. 2-27,
1
1
Vav,y= 2 (Vy + VOy)= 2 ((O) + ( -96 ft/s)) = -48 ft/s.
This averagespeed,usedwith the distance of 18 in (1.5 ft), can be usedto find the time of deceleration
Vav,y= ~y/ ~t,
and putting numbers into the expressiongives ~t = 0.031 s. We actually used ~y = -1.5 ft, where
the negative sign indicated that she was still moving downward. Finally, we use this in Eq. 2-26 to
find the acceleration,
(O) = ( -96 ft/s) + a(0.031 s),
which gives a = +3100 ft/S2.
through by 1 = g/(32 ft/S2).
In terms oí g this is a = 97g, which can be íound by multiplying
30
P2-26 Let the speed of the disk when it comes into contact with the ground be Vlj then the
averagespeed during the deceleration processis vl/2j so the time taken for the deceleration process
is tl = 2d/Vl, where d = -2mm. Butd is also give by d = at~/2 + Vltl, so
d=
1009
2d
)
~
2
+VI
2
or v~ = -200gd. The negative signs are necessary!.
The disk was dropped from a height h = -y and it first came into contact with the ground when
it had a speed of Vl. Then the averagespeed is vl/2, and we can repeat most of the above (except
a = -g instead of 100g), and then the time to fall is t2 = 2y/Vl,
or v~ = -2gy. The negative signs are necessary!.
Equating, y = 100d = 100(-2mm) = -0.2m, so h = O.2m. Note that although 100g's sounds
like plenty, you still shouldn't be dropping your hard disk drive!
P2-27
Measure from the feet! Jim is 2.8 cm tall in the photo, so 1 cm on the photo is 60.7
cm in real-life. Then Jim has fallen a distance Yl = -3.04m while Clare has fallen a distance
Y2 = -5.77 m. Clare jumped first, and the time she has been falling is t2j Jim jumped seconds, the
time he has been falling is tl = t2 -6.t.
Then Y2 = -gt~/2 and Yl = -gt~/2, or t2 = piii;Tii
=
..¡-2(-5.77m)/(9.81m/s2)
= 1.08s and tl = pi;j;Jij
= ..¡-2(3.04m)/(9.81m/s2)
= 0.79s. So
Jim waited 0.29 s.
P2-28 (a) Assuming she starts from rest and has a speed of Vl when she opens her chute, then
her averagespeedwhile falling freely is vl/2, and the time taken to fall yl = -52.0m is tl = 2yl Vl.
Her speed Vl is given by Vl = -gtl, or v~ = -2gyl. Then Vl = --2(9.81
m/s2)( -52.0m) =
-31.9m/s. We must use the negative answer, becauseshe fall down! The time in the air is then
tl = ,pii¡J¡j
= V-2(52.0m)/(9.81 m/s2) = 3.26s.
Her final speed is V2 = -2.90m/s, so the time for the decelera:tionis t2 = (V2 -vl)/a,
where
a = 2.10m/s2. Then t2 = (-2.90m/s --31.9m/s)/(2.10m/s2)
= 13.8s.
Finally, the total time of fiight is t = tl + t2 = 3.26s + 13.8s = 17.1s.
(b) The distance fallen during the deceleration phase is
9 2 + Vlt2 = -. ~(13.8s)2
y2 = --t2
2
2
+ (-31.9
mjs)(13.8s)
= -240m.
The total distance fallen is y = Yl + y2 = -52.0 m -240 m = -292 m. It is negative because she was
falling down.
P2-29 Let the speed of the bearing be Vl at the top of the windows and V2 at the bottom.
These speeds are related by V2 = Vl -gt12, where t12 = 0.125s is the time between when the
bearing is at the top of the window and at the bottom of the window. The average speed is
vav = (Vl + v2)/2 = Vl -gt12/2. The distance traveled in the time t12 is y12 = -1.20m, so
y12 = Vavt12
2
Vltl2
and expressionthat can be solved for Vl to yield
31
-gtl2/2,
Now that we know Vl we can find the height ofthe building abovethe top ofthe window. The time
the object has fallen to get to the top ofthe window is tl = -Vl/g = -(-8.99m/s)/(9.81m/s2)
=
0.916m.
The total time for falling is then (0.916s) + (0.125s) + (1.0s) ;= 2.04s. Note that we remembered
to divide the last time by two! The total distance from the top of the building to the bottom is then
y = -gt2/2
P2-30
= -(9.81m/s2)(2.04s)2/2
= 20.4 m.
Each hall falls from a highest point through a distance of 2.0m in
t =
V-2(2.0m)/(9.8m/82)
=
0.6398.
The time each ball spends in the air is twice this, or 1.28s. The frequency of tossesper ball is the
reciprocal, f = l/T = 0.781s-l. There are five ball, so we multiply this by 5, but there are two
hands, so we then divide that by 2. The tossesper hand per second then requires a factor 5/2, and
the tossesper hand per minute is 60 times this, or 117.
~
Assume each hand can toss n objects per second. Let 7"be the amount of time that any
one object is in the air. Then 2n7"is the number of objects that are in the air at any time, where
the "2" comes from the fact that (most?) jugglers have two hands. We'll estimate n, but 7"can be
found from Eq. 2-30 for an object which falls a distance h from rest:
o = h + (O)t-~gt2,
solving, t = .JihTg. But T is twice this, becausethe object had to go up before it coulcl come clown.
So the number of objects that can be jugglecl is
4n .j2h[¡j
We estímate n = 2 tosses/second. So the maxímum number of objects one could juggle to a heíght
h would be
3.6yh/meters.
P2-32 (a) We needto look up the height of the leaning tower to solve this! If the height is h = 56 m,
then the time taken to fall a distance h = -Yl is tl = ,piij;yij
= V-2(-56m)/(9.81m/s2)
= 3.4s.
The secondobject, however, has only fallen a a time t2 = tl -~t = 3.3 s, so the distance the second
object falls is y2 = -gt~/2 = -(9.81m/s2)(3.3s)2/2 = 53.4. The difference is yl -y2 = 2.9m.
(b) If the vertical separation is ~y = 0.01m, then we can approach this problem in terms of
differentials,
t5y = at t5t,
so t5t = (0.01 m)/[((9.81
m/s2)(3.4s)]
= 3 x 10-4 s.
P2-33 Use symmetry, and focus on the path from the highest point downward. Then ~tu = 2tu ,
where tu is the time from the highest point to the upper level. A similar expression exists for
the lower level, but replace U with L. The distance from the highest point to the upper level is
Yu = -gtt /2 = -g(~tu /2)2/2. The distance from the highest point to the lower level is YL =
-gtl/2 = -g(~tL/2)2/2.
Now H = Yu -YL = -g~tt/8
--g~tl/8,
which can be written as
9 = ~ti 8H
-~tt
32
.
~
E3-1 The Earth orbits the sun with a speed of 29.8 km/s. The distance to Pluto is 5900xl06 km.
The time it would take the Earth to reach the orbit of Pluto is
t =
(5900xlO6
km)/(29.8
km/s)
= 2.0xlO8s,
or 6.3 years!
E3-2 (a) a = F/m = (3.8N)/(5.5kg) = O.69m/82,
(b) t = Vf/a = (5.2 m/8)/(0.69m/82) =7.58.
(c) x = at2/2 = (0.69m/82)(7.58)2/2 = 20m.
~
2-27
Assuming constant acceleration we can find the average speed during the interval from Eq.
1
1
vav,x = 2 (vx +vOx) = 2 ((5.8x 106 m/s) + (0)) = 2.9x 106 m/s.
From this we can find the time spent accelerating from Eq. 2-22, since ~x = vav,x~t. Putting in
the numbers ~t = 5.17x 10-9s. The acceleration is then
ax-
-.~vx
-~t
(5.8x 106 m/s) -(0)
- 11
-.X
1015ms./
2
-{5.l7xlO-9s)
The net force on the electron is from Eq. 3-5,
LFx
= max = (9.llxlO-31kg)(l.lxlO15m/s2)
= l.OxlO-15N.
E3-4
The average speed while decelerating is vav = 0.7 x 107m/s. The time of deceleration is
t = x/vav = (1.0x10-14m)/(0.7x107m/s)
= 1.4x10-21S. The deceleration is a = ~v/t = (-1.4x
107m/s)/(1.4x10-21 S) = -1.0x1028m/s2. The force is F = ma = (1.67x10-27kg)(1.0x1028m/s2)
=
17N.
~
The net force on the sled is 92 N-90
directions. Then
ax
-¿Fx
m
N= 2 N; subtract because the forces are in opposite
-(2N)
I
~
=
8.0x
lO-2m/s2,
E3-6
53 km/hr is 14.7m/s. The average speed while decelerating is v..v = 7.4m/s. The time
of deceleration is t = x/v..v = (0.65m)/(7.4m/s)
= 8.8x 10-2s. The deceleration is a = f).v/t =
(-14.7m/s)/(8.8x10-2s)
= -17x102m/s2.
The force is F = ma = (39kg)(1.7x102m/s2) = 6600N.
E3-7
Vertical acceleration is a = F/m = (4.5 x 10-15N)/(9.11 x 10-31kg) = 4.9 x 1015m/s2. The
electron moves horizontally 33 mm in a time t = x/vx = (0.033m)/(1.2x107m/s)
= 2.8x10-9s.
The vertical distance deflected is y = at2/2 = (4.9 x 1015m/s2)(2.8 X 10-9 s)2/2 = 1.9 x 10-2m.
E3-8
(a) a = F/m = (29N)/(930kg)
= 3.1x10-2m/s2.
(b) x = at2/2 = (3.1 x 10-2 m/s2)(86400s)2 /2 = 1.2x 108 m.
(c) v = at = (3.1 x 10-2 m/s2)(86400s) = 2700m/s.
33
~
Write the expression for the motion of the first object as ¿::;Fx = mlalx and that of the
secondobject as ¿ Fx = m2a2x. In both casesthere is onlyone force, F, on the object, so ¿Fx = F.
We will solve these for the mass as ml = F / al and m2 = F / a2.Since al > a2 we can conclude that
m2 > ml
(a) The acceleration of and object with mass m2 -ml under the influence of a single force of
magnitude F would be
F
a==-m2 -ml
F
F/a2 -F/al
1
1/(3.30m/s2)
-1/(12.0m/s2)
which has a numerical value of a = 4.55 m/s2.
(b) Similarly, the acceleration of an object of mass m2 +ml
magnitude F would be
a---1/a2
1
,
under the influence of a force of
1
+ r/al
-1/(3.30m/s2)
+ 1/(12.0m/s2)
,
which is the same as part (a) except for the sign change. Then a = 2.59 m/s2.
E3-10
(a) The required acceleration is a = v/t
O.lc/t.
The
required
force
is F = ma
= O.lmc/t.
Then
F = O.l(l200xlO3kg)(3.00xlO8m/s)/(2.59xlO5S)
=
l.4xlO8N,
F = 0.1(1200x103kg)(3.00x108m/s)/(5.18x106s)
= ~.9x106N,
and
(b) The distance traveled during the acceleration phase is Xl = at~/2, the time required to
travel the remaining distance is t2 = X2/V where X2 = d- Xl. d is 5 light-months, or d = (3.00 x
108m/s)(1.30x107s) = 3.90x1015m. Then
t = t¡ + t2 = t¡ + d-Xl
V
=t¡
+
2d-at~
2v
=tl
+
2d-vti
.
2v
If t¡ is 3 days, then
t = (2.59xlO5s)
+ 2(3.90xlO15m)
-(3.00xlO7m/s)(2.59xlO5s)
2(3.00 x 107m/s)
=
1.30x108s
=
4.12
yr,
=
1.33x108s
=
4.20
yr,
if t¡ is 2 months, then
t = {5.18 x 1068) +
2(3.90 x 1015m) -(3.00
x 107m/8)(5.18 x 1068)
2(3.00 x lO7ffi7S)
~
(a) The net force on the second block is given by
LFx
= m2a2x = (3.8kg)(2.6m/s2) = 9.9N.
There is only one (relevant) force on the block, the force of block 1 on block 2.
(b) There is only one (relevant) force on block 1, the force of block 2 on block 1. By Newton's
third law this force has a magnitude of 9.9, N. Then Newton's second law gives EFx = -9.9
N= mlalx = (4.6 kg)alx. So alx = -2.2 m/s2 at the instant that a2x = 2.6 m/s2.
E3-12
(a) W = (5.00 lb)(4.448N/lb)
= 22.2N; m = W/g = (22.2N)/(9.81m/s2)
= 2.26kg.
(b) W = (240 lb)(4.448N/lb)
= 1070N; m = W/g = (1070N)/(9.81m/s2)
= 109kg.
(c) W = (3600 lb)(4.448N/lb)
= 16000N; m = W/g = (16000N)/(9.81m/s2)
= 1630kg.
34
E3-13 (a) W = (1420.00 lb) (4.448N/lb) = 6320Nj m = W/g = (6320N)/(9.81 m/s2) = 644kg.
(b) m = 412kg; W = mg = (412kg)(9.81m/s2) = 4040N.
E3-14
(a) W = mg = (75.0kg)(9.81m/s2)
= 736N.
(b) W = mg = (75.0kg)(3.72m/s2)
= 279N.
(c) W = mg = (75.0kg)(Om/s2) = ON.
(d) The mass is 75.0 kg at alllocations.
~
If 9 = 9.81 m/s2, then m = W/g = (26.0 N)/(9.81 m/s2) = 2.65 kg.
(a) Apply W = mg again, but now 9 = 4.60 m/s2, so at this point W = (2.65 kg)(4.60 m/s2) =
12.2 N.
(b) If there is no gravitational force, there is no weight, because 9 = 0. There is still mass,
however, and that mass is still 2.65 kg.
E3-16
Upward force balances weight, so F = W = mg = (l2000kg)(9.8lm/s2)
E3-17
Mass is m = W/g; net force is F=
F
E3-18
a = ~v/~t
l.29x lO5N.
~
ma, or F = Wa/g.
(3900 lb)(13 ft/s2)/(32
= (450 m/s)/(l.82s)
= 247m/s2.
¿ Fx = 2(1.4 x 105 N) = max. Then m
w = mg = (l.22x
lO5kg)(9.8l
= l.2x lO5N.
Then
ft/s2) = 1600 lb.
Net force is F = ma
(523kg)(247 m/s2) =
1.22 x 105 kg and
m!s2)
= l.20x
lO6N.
E3-20 Do part (b) first; there must be a 10 lb force to support the mass. Now do part (a), but
cover up the left hand side of both pictures. If you can't telJ which picture is which, then they must
both be 10 lb!
E3-21 (b) Average speed during deceleration is 40 km/h, or 11 m/s. The time taken to stop the
car is then t = X/vav = (61 m)/(11 m/s) = 5.6s.
(a) The deceleration is a = Av/~t = (22m/s)/(5.6s) = 3.9m/s2. The braking force is F =
ma = Wa/g = (13,000N)(3.9m/s2)/(9.81m/s2) = 5200N.
(d) The deceleration is same;the time to stop the car is then ~t = ~v/a = (11 m/s)/(3.9m/s2) =
2.8s.
(c) The distance traveled during stopping is x= vavt = (5.6m/s)(2.8s) = 16m.
E3-22
Assume acceleration of free fall is 9.81 m/s2 at the altitude of the meteor. The net force is
Fnet = ma = (0.25kg)(9.2m/s2)
= 2.30N. The weight is W = mg = (0.25kg)(9.81m/s2)
= 2.45N.
The retarding force is Fnet -W = (2.3N) -(2.45N)
= -0.15N.
~
(a) Fincl the time cluring the "jump clown" phase from Eq. 2-30.
1
(O m) = (0.48 m) + (o)t -2(9.8
m/s2)t2,
which is a simple quadratic with solutions t = :!:0.31 s. Use this time in Eq. 2-29 to find his speed
when he hit ground,
v y = (O) -(9.8
m/s2)(0.31
35
s) =
-3.1
m/s.
This becomesthe initial velocity for the deceleration motion, so his averagespeedduring deceleration
is given by Eq. 2-27,
1
1
Vav,y= 2 (vy + VOy)= 2 ((O)+ (-3.1 mis)) = -1.6 mis.
This average speed, used with the distance of -2.2 cm (-0.022 m), can be used to find the time of
deceleration
Vav,y= 6.y/6.t,
and putting numbers into the expressiongives 6.t = 0.014 s. Finally, we use this in Eq. 2-26 to find
the acceleration,
(O) = ( -3.1
mis)
+ a(0.014
s),
which gives a = 220 m/s2.
(b) The average net force on the man is
E3-24 The averagespeed of the salmon while decelerating is 4.6 ft/s. The time required to stop
the salmon is then t = X/Vav = {0.38 ft)/{4.6 ft/s) = 8.3 x 10-2s. The deceleration of the salmon
is a = Av/At = {9.2 ft/s)/{8.2-2s) = 110 ft/s2. The force on the salmon is then F = Wa/g =
{19 lb){110 ft/s2)/{32 ft/s2) = 65 lb.
~
From appendix G we find 11b = 4.448 N; so the weight is (100 lb)(4.448 N/11b) = 445
N; similarly the cord will break if it pulls upward on the object with a force greater than 387 N.
The mass of the object is m = W/g = (445 N)/(9.8 m/s2) = 45 kg.
There are two vertical forceson the 45 kg object, an upward force from the cord Foc (which has a
maximum value of387 N) and a downward force from gravity FoG. Then EFy = Foc-FoG = (387
N) -(445 N) = -58 N. Since the net force is negative, the object must be accelerating downward
according to
E3-26
(a) Constant speed means no acceleration, hence no net forcej this means the weight is
balanced by the force from the scale, so the scale reads 65 N .
(b) Net force on mass is Fnet = ma = Wa/g = (65N)(-2.4m/s2)/(9.81m/s2)
= -16N.. Since
the weight is 65 N, the scale must be exerting a force of ( -16 N) -I -65 N) = 49 N .
E3-27
The magnitude of the net force is W- R = (1600kg)(9.81 m/s2) -(3700N)
= 12000N.
The acceleration is then a = F/m = (12000N)/(1600kg)
= 7.5m/s2. The time to fall is
t = V~
= V2(-72m)/(-7.5m/s2)
The final speed is v = at = (-7.5m/s2)(4.4s)
E3-28
ft is
= 33m/s.
= 4.4s.
Get better brakes, eh?
The average speed during the acceleration is 140 ft/s. The time for the plane to travel 300
t = X/Vav
= (300
ft)/(140
ft/8)=
2.148.
The acceleration is then
a = ~v/~t
= (280 ft/s)/(2.14s)
= 130 ft/S2.
The net force on the plane is F = ma = Wa/g = (52000 Ib)(130 ft/s2)/(32 ft/s2) = 2.1x105Ib.
The force exerted by the catapult is then 2.1 x 105lb -2.4 x 104 lb = 1.86 x 105 lb.
36
E3-29
(a) The acceleration of a hovering rocket is O, so the net force is zero; hence the thrust must
equal the weight. Then T = W = mg = (51000kg)(9.81m/s2) = 5.0x105N.
(b) If the rocket accelerates upward then the net force is F = ma = (51000kg)(18m/s2)
=
9.2x 105 N. Now Fnet = T- W, so T = 9.2x 105 N + 5.0x 105N = 1.42x 106N.
E3-30
(a) Net force on parachute + person system is Fnet = ma = (77kg+5.2kg)(-2.5s2)
=
-210N. The weight ofthe system is W = mg = (77kg+5.2kg)(9.81s2)
= 810N. IfP is the upward
force of the air on the system (parachute) then P = F net + W = ( -210 N) + (810 N) = 600 N .
(b) The net force on the parachute is Fnet = ma = (5.2kg)(-2.5s2)
= -13N. The weight ofthe
parachute is W = mg = (5.2 kg)(9.81 m/s2) = 51 N. If D is the downward force of the person on the
parachute then D = -Fnet -W + P = -(-13N)
-(51N)
+ 600N = 560N.
~
(a) The total mass of the helicopter+car system is 19,500 kg; and the only other force
acting on the system is the force of gravity, which is
w = mg = (l9,500kg)(9.8m/s2)
= l.9lxlOsN.
The force of gravity is directed down, so the net force on the system is ¿ Fy = FBA -(1.91 x 105
N). The net force can also be found from Newton's second law: ¿ Fy = may = (19,500 kg)(1.4
m/s2) = 2.7x104 N. Equate the two expressionsfor the net force, FBA -(1.91x105 N) = 2.7x104
N, and solve; FBA = 2.2x105 N.
(b) Repeat the above steps except: (1) the system will consist only ofthe car, and (2) the upward
force on the car comesfrom the supporting cable only F cc .The weight of the car is W = mg = (4500
kg)(9.8 m/s2) = 4.4 x 104 N. The net force is ¿ Fy = Fcc -( 4.4 x 104 N), it can also be written as
¿Fy = may = (4500 kg)(1.4 m/s2) = 6300 N. Equating, Fcc = 50,000 N.
P3-1
(a) The acceleration is a = F/m = (2.7xlO-5N)/(280kg)
ment (from the original trajectory) is
y = at2/2 = (9.64x lO-8m/s2)(2.4s)2
(b) The acceleration is a = F/m
(from the original trajectory) is
y = at2/2
lO-5m/s2)(2.4s)2
The displace-
/2 = 2.8x lO-7m.
= (2.7xlO-5N)/(2.lkg)
= (l.3x
= 9.64xlO-8m/s2.
= l.3xlO-5m/s2.
The displacement
/2 = 3. 7x lO-5m.
P3-2 (a) The acceleration ofthe sled is a = F/m = (5.2N)/(8.4kg) = 0.62m/s2.
(b) The acceleration ofthe girl is a = F/m = (5.2N)/(40kg) =0.13m/s2.
(c) The distance traveled by girl is Xl = alt2/2; the distance traveled by the sled is X2 = a2t2/2.
The two meet when Xl + X2 = 15m. This happens when (al + a2)t2 = 30 m. They then meet when
t = ..¡(30m)/(0.13m/s2 + 0.62m/s2) = 6.3s. The girl moves Xl = (0.13m/s2)(6.3s)2/2 = 2.6m.
~
(a) Start with block one. It starts from rest, accelerating through a distance of 16 m in a
time of 4.2 s. Applying Eq. 2-28,
x
-16 m
=
=
1
2
Xo + vOxt + 2axt ,
(O) + (0)(4.2
find the acceleration to be ax = -1.8 m/s2,
37
s) + ~ax(4.2
8)2,
Now for the secondblock. The acceleration of the second block is identical to the first for much
the same reason that all objects fall with approximately the same acceleration.
(b) The initial and final velocities are related by a sign, then Vx = -VOx and Eq. 2-26 becomes
Vx
-VOx
-2vox
='
VOx
+
axt,
=
VOx
+
axt,
=
(-1.8
m/s2)(4.2
s).
which gives an initial velocity of Vo",= 3.8 m/s.
(c) Half of the time is spent coming down from the highest point, so the time to "fall" is 2.1 s.
The distance traveled is found from Eq. 2-28,
1
x = (O)+ (0)(2.1s)+ 2(-1.8m/s2)(2.1sf
= -4.0 m.
P3-4
(a) The weight ofthe engine is W = mg = (1400kg)(9.81m/s2) = 1.37x104N.
supports 1/3 of this, then the force on a bolt is 4600 N .
(b) If engine accelerates up at 2.60m/s2, then net force on the engine is
F net = ma = (1400kg)(2.60m/s
Ifeach bolt
z ) = 3.64x 103 N.
The upward force from the bolts must then be
B = Fnet + W = (3.64x103N)
+ (1.37x104N)
= 1.73x104N.
The force per bolt is one third this, or 5800 N .
P3-5 (a) If craft descendswith constant speed then net force is zero, so thrust balances weight.
The weight is then 3260 N .
(b) If the thrust is 2200 N the net force is 2200N -3260N = -1060N. The mass is then
m = F/a = (-1060N)/(-0.390m/s2)
= 2720kg.
(c) The acceleration due to gravity is 9 = W/m = (3260N)/(2720kg) = 1.20m/s2.
P3-6 The weight is origina1ly Mg. The net force is origina1ly -Ma. The upward force is then
origina1ly B = Mg -Ma. The goal is for a net force of (M -m)a and a weight (M- m)g. Then
(M-m)a
or m = 2Ma/(a
~
= B-
(M-m)g
= Mg-Ma-
Mg+mg
=mg -Ma,
+ g).
( a) Consider
all three carts as one system.
Then
¿ Fx
=
mtotalax,
6.5N
=
(3.1kg+2.4kg+1.2kg)ax,
0.97 m/s2
=
ax.
(b) Now chooseyour system so that it only contains the third car. Then
¿Fx
= F23 = m3ax = (1.2kg)(0.97m/s2).
The unknown can be solved to give F23 = 1.2N directed to the right.
(c) Consider a system involving the second and thirdcarts. Then LFx
applied to the system gives
F12 = (m2 +m3)ax
= (2.4kg+
.qR
1.2 kg)(0.97m/s2)
= F12, so Newton's law
=3.5N.
P3-8
(a) F =ma = (45.2kg + 22.8kg + 34.3kg)(1.32m/s2) = 135N.
(b) Consider only m3. Then F = ma = (34.3kg)(1.32m/s2)
= 45.3N.
(c) Consider m2 and m3. Then F = ma = (22.8kg + 34.3kg)(1.32m/s2)
= 75.4N.
P3-9
(c) The net force on each link is the same, Fnet = ma = (0.100kg)(2.50m/s2)
= 0.250N.
(a) The weight ofeach link is W = mg = (0.100kg)(9.81m/s2)
= 0.981N. On each link (except
the top or bottom link) there is a weight, an upward force from the link above, and a downward force
fromthe linkbelow. Then Fnet = U -D-W.
Then U = Fnet+W+D
= (0.250N)+(0.981N)+D
=
1.231 N + D. For the bottom link D = o. For the bottom link, U = 1.23N. For the link above,
U = 1.23N+ 1.23N = 2.46N. For the link above, U = 1.23N+2.46N=
3.69N. For the link above,
U = 1.23N + 3.69N = 4.92N.
(b) For the top link, the upward force is U = 1.23N + 4.92N = 6.15N.
P3-10
(a) The acceleration ofthe two blocks is a = F/(ml
the force of contact, and is
p = m2a = Fm2/(ml
+m2)
= (3.2N)(1.2kg)/(2.3kg+
(b) The acceleration of the two blocks is a =F/(ml
force of contact, and is
p = m la=
Fm~/{m;+
+m2) The net force on block 2 is from
1.2kg) = 1.lN.
+'m2) The net force on block I is from the
m2) = {3.2N)(2.3kg)/{2.3kg
+1.2kg)
=2.1N.
Not a third law pair, eh?
~
(a) Treat the system as including both the block and the rope, so that the mass of
the system is M + m. There is one (relevant) force which acts on the system, so LFx = P. Then
Newton's secondlaw would be written as P = (M +m)ax. Solve this for ax and get ax = P/(M +m).
(b) Now consider only the block. The horizontal force doesn't act on the block; instead, there is
the force of the rope on the block. We'll assumethat force has a magnitude R, and this iS:the only
(relevant) force on the block, so E Fx = R for the net fo:rceon the block.. In this case Newton's
secondlaw would be written R = Max. Yes, ax is the same in p~ (a) and (b); the acceleration of
the block is the same as the acceleration oÍ the block + rope. Substituting in the results from part
(a) we find
R=M+m
Mp
39
.
E4-1 (a) The time to passbetweenthe plates is t = x/vx = (2.3 cm)/(9.6x108 cm/s) = 2.4x10-9s.
(b) The vertical displacement of the beam is then y = ayt2/2 == (9.4 x 1016cm/s2)(2.4 X
10-9s)2/2 = 0.27 cm.
(c) The velocity components are Vx = 9.6 x 108 cm/s and Vy = ayt = (9.4 x 1016cm/s2)(2.4 X
10-9s) = 2.3x108cm/s.
E4-2
~
a = ~v! ~t = -(6.30i
(a) The velocity
-8.42j)(m!s)!(3
s) = ( -2.10i + 2.81j)(m!s2).
is given by
dr
dt
v
= ~ (Al) + ~ (Bfj)
=
+ ~ (ctk)
,
(O) + 2Btj + ck.
(b) The acceleration is given by
dv
-
~ ( 2Btj) + ~ ( ck ) ,
-
(O)+ 2Bj + (0).
dt
-a
(c) Nothing exciting happens in the x direction, so we will focus on the yz plane. The trajectory
in this plane is a parabola.
E4-4
(a) Maximum
x is when Vz = O. Since Vz = azt + vz,o, Vz = 0 when t = -vz,o/az
=
-(3.6m/s)/(
-1.2m/s2)
= 3.0s.
:
(b) Since Vz = 0 we have IvJ = Iv!ll. But v!I = a!lt + v!I,O = -(1.4m/s)(3.0s)
+ (Q) = -4.2m/s.
Then Ivl = 4.2m/s.
(c) r = at2/2 + vot, so
r=
[-(0.6m/s2)i
-(0.7m/s2)j](3.0S)2
+ [(3.6m/s)i](3.0s)
E4-5
F = Fl + F2 = (3.7N)j + (4.3N)i.
E4-6
or
{a) The acceleration is a = F /m = {2.2 m/s2)j.
= (5.4m)i -(6.3m)j.
Then a = F/'m = (0.71 m/s2)j + (0.83m/s2)i.
The yelocity after 15 seconds is v = at + Yo,
v = [(2.2m/s2)j](15s)+ [(42m/s)i] = (42m/s)i + (33m/s)j.
(b) r = 8t2/2 + vot, so
r = [(1.1 m/82)]](15
8)2 + [(42 m/8)i](158)
~
= (630m)i
+ (250 m)].
The block h8Ba weight W = mg = (5.1 kg)(9.8 m/s2) = 50 N.
(a) Initially P = 12 N, so Py = (12N) sin(25°) = 5.1 N and Px = (12N) cos(25°) = 11 N. Since
the upward component is less than the weight, the block doesn't leave the floor, and a normal force
will be present which will make E Fy = 0. There is on1y one contribution to the horizontal force,
so E Fx = Px. Newton's second law then gives ax = Px/m = (11 N)/(5.1 kg) = 2.2 m/s2.
(b) As P is incre8Bed,so is Py; eventually Py will be large enough to overcomethe weight of the
block. This happens just after Py = W = 50 N, which occurs when P = Py/sin(} = 120 N.
(c) Repeat part (a), except now P = 120 N. Then Px = 110 N, and the acceleration ofthe block
is ax = Px/m = 22 m/s2.
40
E4-8
(a) The block has weight W = mg = (96.0kg)(9.81m/s2)
= 942N. Px = (450N)cos(38°) =
355N; Py = (450N)sin(38°)
= 277N. Since Py < W the crate stays on the fioor and there is a
normal force N = W -Py.
The net force in the x direction is Fx = Px- (125N) = 230N. The
acceleration is ax = Fx/m = (230N)/(96.0kg)
= 2.40m/s2.
(b) The block has mass m = W/g = (96.0N)/(9.81m/s2)
= 9.79kg. Px = (450N)cos(38°) =
355N; Py = (450N) sin(38°) = 277N. Since Py > W the crate lifts off of the fioor! The net
force in the x direction is Fx = Px- (125N) = 230N. The x acceleration is ax = Fx/m =
(230N)/(9.79kg)
= 23.5m/s2. The net force in the y direction is Fy = Py -W = 181N. The y
acceleration is ay = Fy/m = (181N)/(9.79kg)
= 18.5m/s2. Wow.
E4-9
Let y be perpendicular
and x be parallel to the incline. Then p = 4600 N ;
Px = (4600 N) cos(27°)
= 4100N;
Py = (4600 N) sin(27°) = 2090 N.
The weight ofthe car is W = mg = (1200kg)(9.81m/s2)
Wx = (11800N)
sin(18°)
= 11800Nj
= 3650 N;
Wy = (11800N) cos(18°) = 11200N.
Since Wy > Py the car stays on the incline. The net force in the x direction is Fx = Px- Wx = 450N.
The acceleration in the x direction is ax = Fx/m = (450N)/(1200kg)
= O.375m/s2. The distance
traveled in 7.5s is x = axt2/2 = (0.375m/s2)(7.5s)2/2
= 10.5m.
E4-10
Constant speed means zero acceleration, so net force is zero. Let y be perpendicular and x
be parallel to the incline. The weight is W = mg = (110kg)(9.81 m/s2) = 1080N; Wx = W sin(34°);
Wy = Wcos(34°).
The push F has components Fx = Fcos(34°) and Fy = -Fsin(34°).
The
y components will balance after a normal force is introduced; the x components will balance if
Fx = Wx, or F = Wtan(34°) = (1080N)tan(34°)
= 730N.
~
If the x axi8 i8 parallel to the river and the y axi8 i8 perpendicular , then a = 0.121 mi 82.
The net force on the barge i8
LF
= ma = (9500kg)(0.12im/s2) = 110oiN.
The force exerted on the barge by the horse has components in both the x and y direction. If
p = 7900 N is the magnitude of the pull and (} = 18° is the direction, then p = p cos(}i + p sin (}j =
(7500i + 2400j) N.
Let the force exerted on the barge by the water be F: = Fw,xi + Fw,yj. Then E Fx = (7500
N) + Fw,x and E Fy = (2400 N) + Fw,y. But we already found E F, so
Fx = 1100 N
=
7500 N + Fw,x,
Fy = 0
=
2400 N + Fw,yo
Solving, Fw,x = -6400 N and Fw,y = -2400 N. The magnitude is found by Fw = VF;',x + F;',y
6800 N .
41
=
~
E4-12 (a) Let y be perpendicular and x be parallel to the direction of motion of the plane.
Then Wx = mgsine; Wy = mgcose; m = W/g. The plane is accelerating in the x direction, so
ax = 2.62m/s2; the net force is in the x direction, where Fx = max. But Fx = T- Wx, so
T"=
Fx
+
Wx
=
W
ax
+
Wsin{}
=
(7.93
x
4
10
N)
9
(b) There is no motion in the y direction, so
L = Wy = (7.93xlO4N)cos(27°)
~
(a) The hall rolled
off horizontally
so VOy = O. Then
y
( -4.23
= 7.07xlO4N.
=
ft)
VOyt-
1
2gt
2
(O)t-
~(32.2
,
ft/S2)t2,
which can be solved to yield t = 0.514 s.
(b) The initial velocity in the x direction can be found from x = vOxt; rearranging, VOx= x/t =
(5.11 ft)/(0.514 s) = 9.94 ft/s. Since there is no y component to the velocity, then the initial speed
is Vo= 9.94 ft/s.
E4-14
The electron travels for a time t = x/vx.
y = -gt2/2 in that time. Then
9
Y=-2
x
2
=-
The electron "falls" vertically through a distance
(9.81 m/s2)
2
v"'
(1.0 m)
(3.0x107m/s)
)2
=
-5.5
x 10-15
m.
E4-15
(a) The dart "falls" vertically through a distance y = -gt2/2 = -(9.81 m/s2)(0.19s)2 /2 =
-0.18m.
(b) The dart travels horizontally x = v:z:t = (10m/s)(0.19s) = 1.9m.
E4-16
The initial
velocity
components
Vx,Q
=
are
(15m/s)
cos(200)
=
14m/s
and
Vy,Q= -(15 m/s) sin(200) = -5.1 m/s.
(a) The horizontal displacement is x = vxt = {14m/s)(2.3s)
(b) The vertical displacement is
y = -gt2/2
~
+ Vy,ot = -(9.81
m/s2)(2.3s)2
/2 +( -5.1
= 32m.
m/s)(2.3s)
= -38m.
Find the time in terms of the the initial y component of the velocity:
vy
=
VOy -gt,
(O)
=
VOy -gt,
t
=
VOy/g.
42
Use this time to find the highest point:
2
Finally, we know the initial y component oí the velocity írom Eq. 2-6, SOYrnax= (vosin<t>o)2/2g.
E4-18
The horizpntal displacement is x = v:¡;t. The vertical displacement is y = -gt2/2. Combining, y = -g(x/v:¡;)2/2.
The edge of the nth step is located at y = -nw, x = nw, where w = 2/3 ft.
If Iyl > nw when x = nw then the ball hasn't hit the step. Solving,
<
g(nw/vx)2/2
nw,
2,
gnw/v;
2v;/(gw)
= 2(5.0
ft/s)2/[(32
ft/s2)(2/3
ft)]
= 2.34.
Then the balllands on the third step.
E4-19 (a) Start from the observation point 9.1 m above the ground. The ball will reach the
highest point when Vy = O,this will happen at a time t after the observation such that t = vy,olg =
(6.1 m/s)/(9.81 m/s2) = 0.62s. The vertical displacement (from the ground) will be
y=
-gt2/2+Vy,ot+yo
= -(9.81m/s2)(0.62s)2/2+
(6.1 m/s)(0.62 s) + (9.1 m) = llm.
(b) The time for the ball to return to the ground from the highest point is t = ,j'iii;;:;Tii =
V2(11 m)/(9.81 m7s2) = 1.5s. The total time of flight is twice this, or 3.0 s. The horizontal distance
traveled is x = vxt = (7.6m/s)(3.0s) = 23m.
(c) The velocity of the ball just prior to hitting the ground is
= at + yo = (-9.81 m/s2)j(1.5s)+ (7.6m/s)i = 7.6m/si -15m/suj.
The magnitude is V7.62 + 152(m/s) = 17m/s. The direction is
(} = arctan(-15/7.6}
= -63°.
E4-20 Focus on the time it takes the ball to get to the plate, assuming it traveled in a straight
line. The ball has a "horizontal" velocity of 135 ft/s. Then t = x/vx = (60.5 ft)/(135 ft/s) = 0.448s.
The ball will "fall" a vertical distance of y = -gt2/2 = -(32 ft/s2)(0.448 s)2/2 = -3.2 ft. That's in
the strike zone.
E4-21
Since R cx:1/9 one can write R2/R1 = 91/92, or
6.R = R2 -Rl
= Rl
1-
!!:!.)
92
=
(8.09
4~
m)
or
t2 = [(13 ft) tan(55°)
or t = 0.9868.
Then
-(3
ft)][2/(32m/82)]
v = (13 ft)/(co8(55°)(0.9868))
= 0.97382,
= 23ft/8.
E4-23 Vx = x/t = (150 ft)/(4.50s) = 33.3 ft/s. The time to the highest point is half the hang
time, or 2.25s. The vertical speedwhen the ball hits the ground is Vy = -gt = -(32 ft/s2)(2.25s) =
72.0 ft/s. Then the initial vertical velocity is 72.0 ft/s. The magnitude of the initial velocity is
v722 + 332(ft/s) = 79 ft/s. The direction is
e = arctan(72/33) = 65°.
E4-24
(a) The magnitude of the initial velocity of the projectile
was in the air for a time t where
x
x
t = ;;: = -;;c-¡;sO. -(2300ft)
is v = 264 ft/s.
~
,,~ ~
9.88
-(264
ft/s)
The projectile
, ";,!C,, tf,c;
cos( -27°)
(b) The height of the plane was -y where
-y = gt2/2 -Vy,ot = (32 ft/s)(9.8 S)2/2-
(264 ft/s) sin( -27°)(9.8s)
= 2700 ft.
~
Define the point the ballleaves the racquet as r = o.
(a) The initial conditions are given as VOx= 23.6 m/s and VOy= 0. The time it takes for the ball
to reach the horizontallocation of the net is found from
x
(12 m)
0.51
s
Find how far the ball has moved horizontally
1
y = VOyt-2gt
2
=
vOxt,
=
{23.6 m/s)t,
=
t,
in this time:
1
= (0)(0.51 s) -2(9.8
2
2
m/s )(0.51 s) = -1.3 m.
Did the ball clear the net? The ball started 2.37 m above the ground and "fell" through a distance
of 1.3 m by the time it arrived at the net. So it is still1.1 m above the ground and 0.2 m above the
net.
(b) The initial conditions are now given by VOx= (23.6 m/s)(cos[-5.0°]) = 23.5 m/s and VOy=
(23.6 m/s)(sin[-5.0°]) = -2.1 m/s. Now find the time to reach the net just as done in part (a):
t = x/vox = (12.0 m)/(23.5 m/s) = 0.51 s.
Find the vertical position of the ball when it arrives at the net:
1
1
2
,
y = VOyt--gt2 = (-2.1 m/s)(0.51 s) --(9.8 m/s )(0.51 s)2 = -2.3 m.
2
2
Did the ball clear the net? Not this time; it started 2.37 m above the ground and then passedthe
net 2.3 m lower, or only 0.07 m above the ground.
44
E4-26
The initial speed of the ball is given by v = J9R = J (32 ft/s2)(350 ft) = 106 ft/s.
The
time of flight from the batter to the wall is
x/vx
= (320
ft)/[(106
ft/8)
co8(45°)]
The height of the ball at that time is given by y = Yo + VOyt -~gt2,
y = (4 ft) + (106 ft/s)sin(45°)(4.3s)
-(16
= 4.38.
or
ft/s2)(4.3s)2
= 31 ft,
clearing the fence by 7 feet.
E4-27 The balllands x = (358 ft) + (39 ft) cos(28°) = 392 ft from the initial position. The ball
lands y = (39 ft) sin(28°) -(4.60 ft) = 14 ft above the initial position. The horizontal equation
is (392 ft) = vcos(}tj the vertical equation is (14 ft) = -(g/2)t2 + vsin(}t. Rearrange the vertical
equation and then divide by the horizontal equation to get
14 ft + (g/2)t2
,---"-;
,
=
tan(},
or
t2 = [(392 ft) tan(52°) -(14
ft)][2/(32m/82)]
or t = 5.528. Then v = (392 ft)/(co8(52°)(5.528))
= 30.582,
= 115 ft/8.
E4-28 Since ball is traveling at 45° when it returns to the same level from which it was thrown
for maximum range, then we can assumeit actually traveled ~ 1.6 m. farther than it would have
had it been launched from the ground level. This won't make a big difference, but that means that
R = 60.0m -1.6 m = 58.4m. If Vo is initial speed of ball thrown directly up, the ball rises to the
highest point in a time t = vo/g, and that point is Ymax= gt2/2 = v5/(2g) above the launch point.
But v5 = gR, so Ymax= R/2 = (58.4m)/2 = 29.2m. To this we add the 1.60m point of releaseto
get 30.8m.
~
The net force on the pebble is zero, so E Fy = o. There are only two forces on the
pebble, the force of gravity W and the force of the water on the pebble Fpw. These point in
opposite directions, so 0 = Fpw -W.
But W = mg = (0.150 kg)(9.81 m/s2) = 1.47 N. Since
Fpw = W in this problem, the force of the water on the pebble must also be 1.47 N.
E4-30
~
?~~~
Terminal speed is when drag force equal weight, or mg = bvT2, Then VT = Jiñii7b.
Eq.
4-22
is
Vy(t) = VT ( 1- e-bt/m)
,
where we have used Eq. 4-24 to substitute for the terminal speed. We want to solve this equation
for time when Vy(t) = vT/2, so
~VT
= VT (1-
~ =
(1-
e-bt/m)
e-bt/m),
e -bt/m
bt/m
-.!
-2
= -ln(1/2)
t=~ln2
45
,
E4-32
The terminal
m = 41Tpr3/3, so
b=
mg
VT
speed is 7m/s for a raindrop with r = 0.15 cm. The mass of this drop is
=
47r(l.O x lO-3kg/cm3)(O.l5
-:-
cm)3(9.Sl m/s2)
..
= 2.0 x 10-5kg/s.
3(7m/s)
E4-33 (a) The speed of the train is v = 9.58m/s. The drag force on one car is f = 243(9.58)N =
2330N. The total drag force is 23(2330N) = 5.36x104N. The net force exerted on the cars (treated
as a single entity) is F = ma = 23(48.6x103kg)(0.182m/s2) = 2.03x105N. The pull ofthe locomotive
is then p = 2.03x 105N+ 5.36x 104N = 2.57x 105N.
(b) If the locomotive is pulling the cars at constant speed up an incline then it must exert a
force on the cars equal to the sum of the drag force and the parallel component of the weight. The
drag force is the same in each case, so the parallel component of the weight is WII = W sin (} =
2.03x105N = ma, where a is the acceleration from part (a). Then
(} = arcsin(a/g) = arcsin[(0.182m/s2)/(9.81m/s2)] = 1.06°,
E4-34
(a) a = v2fr = (2.18 x 106mfs)2f(5.29 x 10-11m) = 8.98 x 1022mfs2.
(b) F = ma = (9.11 x 10-31kg)(8.98 x 1022mfs2) = 8.18x 10-8N, toward the center.
~
(a) v = ~
= v (5.2 m)(6.8)(9.8 m/s2) = 19 m/s.
(b) Use the fact that one revolution corresponds to a length of 27rr:
19~
s
(
1 rev
27!"(5.2 m) )
( ~1 min )
= 35~min
.
E4-36 (a) v = 27rr/T = 27r(15m)/(12s) = 7.85m/s. Then a = v2fr = (7.85m/s)2/(15m) =
4.11m/s2, directed toward center, which is down.
(b) Same arithmetic as in (a); direction is still toward center, which is now up.
(c) The magnitude of the net force in both (a) and (b) is F = ma = (75kg)(4.11 m/s2) = 310N.
The weight of the person is the same in both parts: W = mg = (75kg)(9.81m/s2) = 740N. At
the top the net force is down, the weight is down, so the Ferris wheel is pushing up with a force of
p = W -F = (740N) -(310N) = 430N. At the bottom the net force is up, the weight is down, so
the Ferris wheel is pushing up with a force of p = W + F = (740N) + (310N) = 1050N.
E4-37
(a) v = 21rr/T = 21r(20xlO3m)/(l.Os)
(b) a = v2/r = (l.26xlO5m/s)2/(20xlO3m)
E4-38
106m)
(b)
on the
245N.
= l.26xlO5m/s.
= 7.9xlO5m/s2,
(a) v = 27rr/T = 27r(6.37x 106m)/(86400s) = 463m/s. a = v2/r = (463m/s)2/(6.37x
= 0.034 m/s2.
The net force on the object is F = ma = (25.0kg)(0.034m/s2)
= 0.85N. There are two forces
object: a force up from the scale (S), and the weight down, W = mg = (25.0 kg)(9.80 m/s2) =
Then S = F + W = 245N + 0.85N = 246N.
~
Let Ax = 15 m be the length; tw = 90 s, the time to walk the stalled Escalatorj t8 = 60
s, the time to ride the moving Escalatorj and tm, the time to walk up the moving Escalator.
The walking speed of the person relative to a fixed Escalator is Vwe= Ax/tw; the speed of the
Escalator relative to the ground is Veg= Ax/t8j and the speed of the walking person relative to the
46
ground on a moving Escalator is Vwg = 6.x/tmo But these three speeds are related by Vwg = Vwe+Veg.
Combine all the above:
Vwg
~x
tm
Vwe
+
¡lx
= -+-,
tw
1
-+tw
tm
Veg,
¡lx
1ts
1
ts
Putting in the numbers, tm = 36 s.
E4-40 Let Vw be the walking speed, Vs be the sidewalk speed, and Vm = Vw+ Vs be Mary's speed
while walking on the moving sidewalk. All three cover the same distance x, so Vi = x/ti, where i is
one of w, s, or m. Put this into the Mary equation, and
l/tm = l/tw + l/ts = 1/(1508) + 1/(708) = 1/488.
E4-41 If it takes longer to fly westward then the speed of the plane (relative to the ground)
westward must be less than the speed of the plane eastward. We conclude that the jet-stream must
be blowing east. The speed of the plane relative to the ground is Ve= Vp + Vj when going east and
Vw = Vp -Vj when going west. In either case the distance is the same, so x = Viti, where i is e or
w. Since tw -te is given, we can write
x
x
tw-te=-=x
vp-'/Jj
vp+Vj
2v. J
.
Vp2-Vj2
Solve the quadratic if you want, but since Vj « Vp we can neglect it in the denominator and
Vj = vp2(0.83 h)/(2x)
= (600 mi/h)2(0.417
h)/(2700 mi) = 56 mi/hr.
E4-42 The horizontal component of the rain drop's velocity is 28m/s. Since v"' = v sin(}, v =
(28 m/s)/ sin(64°) = 31 m/s.
~
(a) The position of the bolt relative to the elevator is Ybe,the position of the bolt relative
to the shaft is yb8,and the position of the elevator relative to the shaft is Ye8.Zero all three positions
at t = Oj at this time VO,b8
= VO,e8
= 8.0 ft/s.
The three equations describing the positions are
ybs
Yes
y be
+
res
1 2
2gt ,
=
VO,bst-
=
1
2
VO,est + 2at
,
=
rbs,
where a = 4.0 m/s2 is the upward acceleration of the elevator. Rearrange the last equation and
solve for Ybe;get ybe= -1 (g + a)t2 , where advantage was taken of the fact that the initial velocities
are the same.
Then
t = v -2ybe/(g + a) = V -2( -9.0 ft)/(32
ft/S2 + 4 ft/S2) = 0.71 s
(b) Use the expression for yb8 to find how the bolt moved relative to the shaft:
1
ybs = VO,bst -2gt
2
= (8.0 ft)(0.71
1
s) -2(32
A7
ft/s
2
)(0.71 s)
2
= -2.4
ft.
E4-44 The speed of the plane relative to the ground is Vpg= (810 km)/(1.9 h) = 426 km/h. The
velocity components ofthe plane relative to the air are VN = (480 km/h)cos(21°) = 448 km/h and
VE = (480 km/h)sin(21°) = 172 km/h. The wind must be blowing with a component of 172 km/h
to the west and a component of 448 -426 = 22 km/h to the south.
E4-45
(a) Let
,
i point
east
andj
point
north.
The
velocityofthe
torpedo
is v = (50 km/h)isin(}+
,
(50 km/h)jcos(}. The initial coordinates of the battleship are then ro = (4.0 km)isin(20°) +
(4.0 km)jcos(20°) = (1.37 km)i+ (3.76 km)j. The final position ofthe battleship is r= (1.37 km+
24 km/ht)i + (3.76 km)j, where t is the time of impact. The final position of the torpedo is the
same, so
[(50 km/h)isine
+ (50 km/h)]cosejt
= (1.37 km + 24 km/ht)i
+ (3.76 km)],
or
[(50 km/h) sinO]t -24
km/ht = 1.37 km
and
[(50 km/h)cos{}]t = 3.76 km.
Dividing the top equation by the bottom and rearranging,
50sin{} -24 = 18.2cos{}.
Use any trick you want to solve this. I used Maple and found {} = 46.8°.
(b) The time to impact is then t = 3.76 km/[(50 km/h) cos(46.8O)1=0.110 h, or 6.6 minutes.
~
Let rA be the position of particle of particle A, and rB be the position of particle B. The
equations for the motion of the two particles are then
rA
=
rO,A + vt,
=
dj + vti;
...1-2 =
rB
2at ,
=
A collision will occur if there is atime
~a(sinei + cosej)t2.
when rA = rB. Then
dj + vti = ~a(sinei
+ cosej)t2,
but this is really two equations: d = !at2cose and vt = !at2sine.
Solve the second one for t and get t = 2v / (a sin e) .Substitute
then rearrange,
d
=
d
=
sin2 (}
=
1 -COS2 (}
=
'2
o =
cos
COS2
() +
2v2
ad
48
(} ,
cosO -
that into the first equation, and
This last expression is quadratic in cosO. It simplifies the solution if we define b = 2v/(ad) = 2(3.0
m/s)2/([0.4 m/s2][30 m]) = 1.5, then
Then cos () = 0.5 and () = 60° .
P4-2 (a) The acceleration of the ball is a = (1.20m/s2)i -(9.81 m/s2)j. Since a is constant the
trajectory is given by r = at2/2, since Vo = 0 and we choosero = 0. This is a straight line trajectory,
with a direction given by a. Then
(J = arctan(9.81/1.20)
= 83.0°,
and R = (39.0m)/tan(83.00)
= 4.79m. It will be useful to find H = (39.0m)/sin(83.00)
= 39.3m.
(b) The magnitude ofthe acceleration ofthe ball is a = v9.812 + 1.202(m/s2) = 9.88m/s2. The
time for the ball to travel down the hypotenuse of the figure is then t = ~-39.3m)/(9.88m/s2)
=
2.82s.
( c ) The magnitude of the speed of the ball at the bottom will then be
v = at = (9.88 m/s2)(2.82s)
= 27.9m/s.
P4-3
(a) The rocket thrust is T = (61.2 kN) cos(58.00)i + (61.2 kN) sin(58.00)i
51.9 kNj.~ The net force on the rocket is the F = T + W, or
F = 32.4 k Ni + 51.9 kNj-
(3030 kg)(9.81
m/s2)j
= 32.4 kNi +
= 32.4 kNi + 22.2 kNj.
The acceleration (until rocket cut-off) is this net force divided by the mass, or
a = 10.7m/s2¡+ 7.33m/s2j.
The position at rocket cut-off is given by
r
=
=
at2/2 = (10.7m/s2i+ 7.33m/s2j)(48.0s)2
/2,
1.23x 104mi+ 8.44x 103mj.
The altitude at rocket cut-off is then 8.44 km.
(b) The velocity at rocket cut-off is
v = at = (10.7m/s2i+ 7.33m/s2j)(48.0
s) = 514m/si + 352m/sj,
this becomesthe initial velocity for the "free fal1" part of the journey. The rocket will hit the ground
after t seconds,where t is the solution to
o = -(9.81
m/s2)t2 /2 + (352 m/s)t
+ 8.44 x 103 m.
The solution is t = 90.7s. The rocket lands a horizontal distance of x = vxt = (514m/s)(90.7s) =
4.66xl04 m beyond the rocket cut-off; the total horizontal distance covered by the rocket is 46.6 km+
12.3 km = 58.9 km.
49
~
P4-4
(a) The horizontal speed of the hall is Vx = 135 ft/s. It takes
x/vx
= {30.0
ft)/(135
ft/8)
= 0.2228
to travel the 30 feet horizontally, whether the first 30 feet, the last 30 feet, or 30 feet somewherein
the middle.
(b) The ball "falls" y = -gt2/2 = -(32 ft/s2)(0.222s)2/2 = -0.789 ft while traveling the first
30 feet.
(c) The ball "falls" a total of y = -gt2/2 = -(32 ft/s2)(0;444s)2/2 = -3.15 ft while traveling
the first 60 feet, so during the last 30 feet it must have fallen (-3.15 ft) -(-0.789 ft) = -2.36 ft.
(d) The distance fallen becauseof acceleration is not linear in time; the distance moved horizontally is linear in time.
P4-5
(a) The initial velocity of the hall has components
18.8mjs
v:¡;,o =(25.3m/s)cos(42.00)
and
Vy,O= (25.3m/s)sin(42.0~)
The ball is in the air for t =
(b) y = -gt2/2 + Vy,ot =
(c) Vx = vx,o = 18.8m/s.
(d) Since Vy > O the ball
= 16.9m/s.
x/vx = (21.8m)/(18.8m/s)
= 1.16s before it hits the wall.
-(4.91 m/s2)(1.16s)2 + (16.9m/s)(1.16s) = 13.0m.
Vy = -gt + Vy,O= -(9.81 m/s2)(1.16s) + (16.9m/s) = 5.52m/s.
is still heading up.
P4-6 (a) Theinitial vertical velocity is Vy,a= Vasin<t>a.The time to the highest point is t = Vy,a/g.
The highest point is H = gt2/2. Combining,
H
= y( Vo sin </>o/y )2/2
= V~ sin2 4>O/ (2y ) .
The range is R = (v~/g)sin2<t>o = 2(v~/g)sin<t>ocos<t>o. Since tan(} = H/(R/2),
2.
-Vo
Slll 2,¡,'!'0 ! g
1
-2(v5!g)
sin </>oCOS</>o= 2 tan </>o.
2H
tan(}
we have
=
R
(b) (} = arctan(O.5tan45°)
= 26.6°,
~
The components of the initial velocity are given by VOx= Vocos(} = 56 ft/s and VOy
vosin(} = 106 ft/s where we used Vo=120 ft/s and (} = 62°.
(a) To find h we need only find out the vertical position of the stone when t = 5.5 s.
(106 ft/8)(5.5
8)-
~(32 ft/81{5.5
8)2 = 99 ft.
(b) Look at this as a vector problem:
v
Vo + at,
( voxi + voyj) -gjt,
voxi + (VOy-gt)j,
(56 ft/s)i+
( (106 ft/s -(32
(56 ft/s)i+
(-70.0 ft/s)j.
50
ft/s2)(5.5
s) ) j,
The magnitude of this vector gives the speed when t = 5.5 s; v = V562 + ( -70)2 ft/s= 90 ft/s.
(c) Highest point occurs when Vy = 0. Solving Eq. 4-9(b) for time; Vy = 0 = VOy-gt
(106 ft/s) -(32 ft/s2)t; t = 3.31 s. Use this time in Eq. 4-10(b),
1
y = VOyt -29t2
1
s) -2(32
= (106 ft/s)(3.31
ft/s2)(3.31
=
8)2 = 176 ft.
P4-8
(a) Since R = (v5/g)sin 2</>0,it is suflicient to prove that sin2</>o is the same for both
sin 2(45° + a) and sin 2(45° -a).
sin 2(45° :1::a) = sin(90° :1::2a) = cos(:1::2a) = cos(2a).
Since the :f: dropped out, the two quaJ1tities are equal.
(b) <1>0
= (1/2) arcsin(Rg/v5) = (1/2) arcsin«20.0m)(9.81
choice is 90° -6.3° = 83.7° .
m/s2)/(30.0m/s)2)
= 6.3°. The other
~
To score the ball must passthe horizontal distance of 50 m with an altitude of no lessthan
3.44 m. The initial velocity components are VOx= Vocos() and VOy= Vosin () where Vo = 25 m/s,
and () is the unknown.
The time to the goal post is t = x/vOx = x/(vo cos()).
The vertical motion is given by
VOyt -2
y
1gt2
= ( 'IlOsin (})
y = x tan(} - ~
(
x
;;;;-oos-9
2
)
-2
1g
(
x
;;;;-oos-9
)
2
(tan2 (j + I) ,
2vo
which can be combined with numbers and constraints to give
(3.44 m)
5
(9.8 m/s2)(50
(50 m) tan(} -2(25
m/s)2 m)2 (tan 2 (} + 1) ,
3.44
5
50 tan (} -20
0
5
-20tan2
(tan2 (} + 1) ,
(} + 50tan(} -23
Solve, and tan (} = 1.25:!: 0.65, so the allowed kicking angles are between (} = 31° and (} = 62°,
P4-10
(a) The height of the projectile at the highest point is H = Lsin(j. The amount of time
before the projectile hits the ground is t = ,j'iiiTii
= v2Lsin(j/g.
The horizontal distance covered
by the projectile in this time is x = vxt = vv2Lsin(j/g.
The horizontal distance to the projectile
when it is at the highest point is x' = Lcos(j. The projectile lands at
D = x -x'
(b) The projectile
will
pass overhead
= vV2Lsin(j/g
ir D > O.
51
-Lcos(j.
P4-11 V2 =V; +V;. For a projectile Vx is constant, so we need only evaluate ~(V;)/dt2. The first
derivative is 2vy dvy/ dt = -2vyg. The derivative of this (the secondderivative) is -2g dvy/ dt = 2g2.
P4-12
1f1 is a maximum when r2 is a maximum.
r
2
2
+
r2 = X2 + y2, or
2
( -gt
/2
+
2
=
(vx,ot)
Vy,ot)
,
=
(votcOS<t>O)2+(votsip<t>o -gt2/2)2
=
v5t2 -vogt3
,
sin <t>o
+ g2t4 /4.
We want to look for the condition which will allow dr2/ dt to vanish. Since
dr2 / dt = 2v~t
-3vOgt2
sin <t>O+ g2t3
we can focus on the quadratic discriminant, b2 -4ac, which is
9
22
voY
.2,¡,
8
Slll
'l'a
22
-voY,
a quantity which will only be greater than zero ir 9 sin2 <t>o> 8. The critica! angle is then
<l>c =
arcsin(
J879)
=
70.5°,
f-p;¡:i3l
There is a downward force on the balloon of 10.8 kN from gravity and an upward force
of 10.3 kN from the buoyant force of the air. The resultant of these two forces is 500 N down, but
since the balloon is descendingat constant speedso the net force onthe balloon must be zero. This
is possible becausethere is a drag force on the balloon of D = bv2, this force is directed upward.
The magnitude must be 500 N, so the constant b is
b=
(500 N)
=
141
kg/m.
(1.88m/s)2
If the crew drops 26.5 kg of ballast they are "lightening"
{26.5kg){9.81
m/s2)
the balloon by
= 260 N.
This reduced the weight, but not the buoyant force, so the drag force at constant speed will now be
500N- 260N = 240N.
The new constant downward speed will be
v = .JD7b=
P4-14
v(240N)X141kg7m)
= 1.30m/s.
The constant b is
b = (500N)/(1.88m/s)
The drag force after "lightening"
v=
Dlb
= 266N .s/m.
the load will still be 240 N. The new downward speed will be
= (240 N)/(266N
.s/m)
= O.902m/s.
P4-15 (a) Initially vo= O, so D = 0, the only force is the weight, so a =-g.
(b) After some time the acceleration is zero, then W = D, or bvT2 = mg, or VT = v'ñi97b.
(c) When v = vT/2 the drag force is D = bvT2/4 = mg/4, so the net force is F = D- W =
-3mg/4. The acceleration is the a = -3g/4.
52
P4-16 (a) The net force on the barge is F = -D = -bv, this results in a differential equation
mdv/dt = -bv, which can be written as
dv/v
=
-(b/m)dt,
=
-(b/m)
JdV/V
ln(vf/vJ = -bt/m.
Then t = (m/b) ln(vi/vf).
(b) t= I(970kg)/(68N
~
.8/m)] ln(32/8.3)
(a) The acceleration
ay =
= idt
( ~ b (1-
which can be simplified as ay = ge-bt/m.
exponent can be expanded to give
dt,
= 198.
is the time derivative
~dt
J
of the velocity,
e-bt/m
))
= ~~e-bt/m
b m
,
For large t this expression approaches O; for small t the
= g -VTt,
bt'
ay ~g
1-m
P4-18
(a) We have v y = vT(l -e-bt/m)
oí the solution íor P4-17 to give
from Eq. 4-22; this can be substituted
into the last line
Y95=VT(t-;).
We can also rearrange
Eq. 4-22 to get t = -(m/b)
ln(l-
Vy/VT),
SO
y95 = VT2/y
But Vy/VT ~ 0.95, so
y95 = VT2/ 9 ( -ln(0.05)
(b) y95 = (42m/s)2/(9.81m/s2)(2.05)
-0.95)
VT2/g
(In
20
-19/20)
= 370 m.
P4-19
(a) Convert units first. v = 86.1 m/s, a = 0.05(9.81 m/s2) = 0.491 m/s2.
radius is r = v2/a = (86.1m/s)/(0.491m/s2)
= 15 km.
(b) v = .¡ar = ..¡(0.491 m/s2)(940m) = 21.5m/s. That's 77 km/hr.
53
The minimum
P4-20
(a) The position is given by r= Rsin(¡)ti+R(l-cos(¡)t)j,
where (¡) = 27r/(20s) = 0.314s-1
and R = 3.0m. When t = 5.0s r = (3.0m)i + (3.0m)jj when t = 7.5s r = (2.1m)i + (5.1m)j;
when t = 10 s r = (6.0 m)j. These vectors have magnitude 4.3 m, 5.5 m and 6.0 m, respectively. The
vectors have direction 45°, 68° and 90° respectively.
(b) ~r = ( -3.0 m)i + (3.0 m)j, which has magnitude 4.3 m and direction 135° .
(c) vav = ~r/~t
= (4.3m)/(5.0s) = 0.86m/s. The direction is the same as ~r.
(d) The velocity is given by v = &cOS(¡)ti+&sin(¡)tj.
v = (-0.94m/s)i.
At t = 5.0s v = (0.94m/s)j;
(e) The acceleration is given by a = -&2
at t = lOs a = ( -0.30m/s2)j.
COS(¡)tj. At t = 5.0s a =( -0.30m/s2)ij
sin(¡)ti + &2
at t = lOs
~
Start fram where the stane lands; in arder ta get the~e the stane fell thraugh a vertical
distance af 1.9 m while maving 11 m harizantally. Then
1
y = -2gt2
which
can
be written
as t =
F2Y
V yO
Putting in the numbers, t = 0.62s is the time of flight from the moment the string breaks. From
this time find the horizontal velocity,
P4-22
(a) The path traced out by her feet ha.s circumference Cl = 27rrcos50°, where r is the
radius of the earth; the path traced out by her head ha.s circumference C2 = 27r(r + h) cos 50°, where
h is her height. The difference is Ac = 27rh cos 50° = 27r(1.6 m) cos 50° = 6.46 m.
(b) a = v2/r = (27rr /T)2 /r = 47r2r/T2. Then Aa = 47r2Ar /T2. Note that Ar = h cos (J! Then
Aa = 47r2(1.6 m) cos 50° /(86400 S)2 = 5.44 x 10-9m/s2.
~
(a) A cycloid
looks something
like this:
,
/
,
~~~)
(b) The position of the particle is given by
r = (RsinúJt+ úJRt)i+ (RcosúJt+ R)j.
The maximum value of y occurs whenever coswt = 1. The minimum
coswt = -1. At either of those times sinwt = O.
The velocity is the derivative of the displacement vector ,
v = (&coswt+wR)i+
I -&sinúJt)j.
When y is a maximum the velocity simplifies to
v = (2f.¡)R)i+ (O)].
54
value of y occurs whenever
When y is a minimum the velocity simplifies to
v = (0)1+ (O)].
The acceleration is the derivative of the velocity vector ,
a = ( -&;2 sin(¡)t)i + ( -&;2 COS(¡)t)j.
When y is a maximum the acceleration simplifies to
a = (0)1+ { -&2)1.
When y is a minimum the acceleration simplifies to
a = (0)1+ (Rr.JJ2)j,
P4-24
(a) The speed of the car is Vc = 15,3m/s. The snow appears to fall with an angle (}
arctan(15,3/7.8) = 63°,
(b)'The'apparent
speed is V(15.3)2 + (7.8)2(m/s) = 17.2m/s.
P4-25 (a) The decimal angles are 89.994250° and 89.994278°. The earth moves in the orbit
around the sun with a speed of v = 2.98x 104m/s (Appendix C). The speed of light is then between
c = (2.98 x 104m/s) tan(89.994250°) = 2.97 x 108m/s and c = (2.98 x 104m/s) tan(89.994278°) =
2.98x 108m/s. This method is highly sensitive to rounding. Calculating the orbital speed from the
radius and period of the Earth 's orbit willlikely result in different answers!
P4-26 (a) Total distance is 21,so to = 21/v.
(b) Assume wind blows east. Time to travel out is tl = l/(v + u), time to travel back is
t2 = l/(v -u). Total time is sum, or
If wind blows west the times reverse,but the result is otherwise the same.
(c) Assume wind blows i1orth. The airplane will still have a speed of v relative to the wind, but
it will need to fiy with a heading away from east. The speed of the plane relative to the ground will
be Vv2 -u2. This will be the speed even when it fiies west, so
tN =
2[
.¡;;.-=-t;'
-to
-~.
(d) If u > v the wind sweeps the plane along in one general direction only; it can never By back.
Sort of like a black hole event horizon.
~
The velocity of the police car with respect to the ground is v pg = -76km/h¡. The velocity
of the motorist with respect the ground is Vmg = -62 km/hj.
The velocity of the motorist with respect to the police car is given by solving
--Vmg
so Vmp = 76km/hi -62
km/hj.
Vmp =
=
Vmp
+
Vpg,
This velocity has magnitude
v (76km/h)2
+ ( -62
55
km/h)2
= 98 km/h.
The direction is
f) = arctan( -62 km/h)/(76km/h)
= -39°,
but that is relative to i. We want to know the direction relative to the line oí sight. The line oí sight
is
a = arctan(57m)/(41m)
= -54°
relative to i, so the answer must be 15°.
P4-28 (a) The velocity of the plane with respect to the air is vpa; the velocity of the air with
respect to the ground is Vag, the velocity of the plane with respect to the ground is Vpg. Then
Vpg = vpa + Vag. This can be represented by a triangle; since the sides are given we can find the
angle between Vag and Vpg (points north) using the cosine law
(} = arccos
(\135)2
-(135)2
-(70)2
-2(135)(70)
(b) The direction of v pa can also be found using the cosine law,
(J =
arccos
(70)2 -(135)2
-(135)2
-2(135)(135)
56
30°.
~
There are three forces which act on the charged sphere-- an electric force, FE, the force of
gravity, W, and the tension in the string, T. All arranged as shown in the figure on the right below.
(a) Write the vectors so that they geometrically show that the sum is zero, as in the figure
on the left below. Now W = mg = (2.8 x 10-4 kg)(9.8 m/s2) = 2.7 x 10-3 N. The magnitude
of the electric force can be found from the tangent relationship, so FE = W tan (} = (2.7 x 10-3
N) tan(33°) = 1.8 x 10-3 N.
\
w
(a)
(b)
(b) The tension can be found from the cosine relation, so
T = W/cos(}
= (2;7 x lO-3N)/cos(33°)
=;: 3.2 x lO-3N.
E5-2
(a) The net force on the elevator is F = ma = Wa/g = (6200 Ib)(3.8 ft/s2)/(32 ft/S2) =
740 lb. Positive means up. There are two force on the elevator: a weight W down and a tension
from the cable T up. Then F = T- W or T = F + W = (740 lb) + (6200 lb) = 6940 lb.
(b) If the elevator acceleration is down then F = -740 lb; consequently T = F + W = ( -740 lb) +
(6200 lb) = 5460 lb.
E5-3 (a) The tension Tis up, the weight W is down, and the net force F is in the direction of the
acceleration (up). Then F =T -W. But F = ma and W = mg, so
m = T/(a + y) = (89N)/[(2.4m/s2) + (9.8m/s2)]= 7.3kg.
(b) T = 89N. The direction of velocity is unimportant. In both (a) and (b) the acceleration is
up.
E5-4 The averagespeedof the elevator during deceleration is vav = 6.0m/s. The time to stop the
elevator is then t = (42.0m)/(6.0m/s) = 7.0s. The deceleration is then a = (12.0m/s)/(7.0s) =
1.7m/s2. Since the elevator is moving downward but slowing down, then the acceleration is up,
which will be positive.
The net force on the elevator is F = ma; this is equal to the tension T minus the weight W.
Then
T = F + W = ma + mg = (l600kg)[(l.7m/s2) + (9.8m/s2)] = l.8x lO4N.
~
(a) The magnitude o~,the man's acceleration is given by
a= m2 -ml
m2+ml'
I
"
and is directed down. The time which elapseswhile he falls is found by solving y = VOyt+ ~ayt2,
or, with numbers, (-12 m) = (o)t + ~( -0.2g)t2 which has the Bolutions t = :1::3.5s. The velocity
with which he hits the ground is then v = VOy+ ayt = (O) + ( -0.2g)(3.5 s) = -6.9 m/s.
57
(b) Reducing the speed can be accomplished by reducing the acceleration. We can't change Eq.
5-4 without also changing one of the assumptions that went into it. Since the man is hoping to
reduce the speed with which he hits the ground, it makes sensethat he might want to climb up the
rope.
E5-6 (a) Although it might be the monkey which does the work, the upward force to lift him still
comesfrom the tension in the rope. The minimum tension to lift the log is T = Wl = mlg. The net
force on the monkey is T- W m = mma. The acceleration of the monkey is then
a = (m¡ -mm)g/mm
= [(15kg)
-(11
kg)](9.8m/s2)/(11
kg) = 3.6m/s2.
(b) Atwood 's machine!
a = (mi -mm)g/(ml
+ mm) = [(15 kg) -(11
kg)](9.8
m/s2)/[(15
kg) + (11 kg)] = 1.5 m/s.
(c) Atwood's machine!
T = 2mlmm9/(ml
+ mm) = 2(15kg)(11kg)(9.8m/s2)/[(15kg)
+ (llkg)]
= 120N.
E5- 7 The weight of eachcar has two compOnent8:a component parallel to the cable8WII = W 8in (}
and a component normal to the cable8WJ... The normal component i8 "balanced" by the 8upporting
cable. The parallel component act8 with the pull cable.
In order to acceleratea car up the incline there mU8t be a net force up with magnitude F = ma.
Then F = T above-Tbelow -WII' or
~T = ma + mgsin(J = (2800 kg)[(0.81 m/s2) + (9.8m/s2) sin(35°)}=
1.8x 104N.
E5-8
The tension in the cable is T, the weight ofthe man + platform system is W = mg, and the
net force on the man + platform system is F = ma = Wa/g = T- W. Then
T
~
Wa/g
+ W = W(a/g
+ 1) = (180 lb + 43 lb)[(1.2
ft/s2)/(32
See Sample Problem 5-8. We need only apply the (unlabeled!)
JLs
=
tan
to find the egg angle. In this case (} = tan-l(O.O4)
ft/S2) + 1] = 2311b.
equation
(}
= 2.3°,
E5-10 (a) The maximum force offriction is F = JJ.sN.Ifthe rear wheelssupport halfofthe weight
of the automobile then N = W/2 = mg/2. The maximum acceleration is then
a = F/m
{b)
a
T
{0.56){9.8mls
2
)12
=
2.7mls
2
= JJ.sN/m = JJ.s9/2.
.
E5-11 The maximum force of friction is F = J1,sN.Since there is no motion in the y direction the
magnitude of the normal force must equal the weight, N = W = mg. The maximum acceleration is
then
a = F/m = JLsN/m = JLsg= (0.95)(9.8m/s2)
E5-12
There is no motion in the vertical direction,
(470N)I[(9.8mls2)(79kg)]
= 0.61.
58
= 9.3m/s2.
so N = W = mg.
Then JLk = FIN
=
~
A 75 kg mass has a weight of W = (75kg)(9.8m/s2)
each end of the bar must be 368 N. Then
= 735 N, so the force of friction on
E5-14 (a) There is no motion in the vertical direction, so N = W = mg.
To get the box moving you must overcome static friction and push with a force of p ~ JLsN=
(0.41)(240N) = 98N.
(b) To keep the box moving at constant speed you must push with a force equal to the kinetic
friction, p = JLkN= (0.32)(240N) = 77N.
(c) If you push with a force of 98 N on a box that experiencesa (kinetic) friction of 77 N, then
the net force on the box is 21 N. The box will accelerate at
a = F/m = Fg/W
= (21 N)(9.8m/s2)/(240N)
= O.86m/s2.
E5-15
(a) The maximum braking force is F = J1sN. There is nomotion in the vertical direction,
so N = W = mg. Then F = J1smg = (0.62)(1500kg)(9.8m/s2)
= 9100N.
(b) Although we still use F = J1sN, N # W on an incline! The weight has two components;
one which is parallel to the surface and the other which is perpendicular .Since there is no motion
perpendicular to the surface we must have N = W.l = W cos ().Then
F=
JLsmg cosO = (0.62)(1500
kg)(9.8
m/s2)
cos(8;6°)
E5-16
JLs = tan (} iB the condition for an object to Bit without
(} = arctan(O.55} = 29°. The angle Bhould bereduced by 13°.
= 9000 N .
Blipping on an incline.
Then
~
(a) The force of static friction is less than .UsN, where N is the normal force. Since the
crat,eisn't moving up or down, ¿Fy =O = N- W. So in this case N = W = mg = (136 kg)(9.81
m/s2) = 1330 N. The force of static friction is less than or equal to (0.37)(1330N) = 492Nj moving
the crate will require a force greater than or equal to 492 N .
(b) The second worker could lift upward with a force L, reducing the normal force, and hence
reducing the force of friction. If the first worker can move the block with a 412 N force, then
412 ?; .UsN. Solving for N, the normal force needs to be less than 1110 N. The crate doesn't move
off the table, so then N + L = W, or L = W -N = (1330 N) -(1110 N) = 220 N.
(c) Or the second worker can help by adding a push sothat the total force of both workers is
equal to 492 N. If the first worker pusheswith a force of 412 N, the secondwould need to push with
a force of 80 N.
E5-18
The coefficient of static friction is J1s = tan(28.0°) = 0.532. The acceleration is a =
2(2.53m)/(3.92s)2 = .329m/s2. We will need to insert a negative sign since this is downward.
The weight has two components: a component parallel to the plane, WII = mg sin f}j and a
component perpendicular to the plane, W-L = mg cos f}.There is no motion perpendicular to the
plane, so N = W -L. The kinetic friction is then f = JLkN = JLkmgcosf}. The net force parallel to
the plane is F = ma = f -WII = JLkmgcosf} -mgsinf}.
Solving this for JLk,
Jl,k
-
(a + y sin (})/(y COS(}),
[( -0.329
m/s2)
+ (9.81 m/s2) sin(28.0°)]/[((9.81
59
m/s2) cos(28.0°)}
= 0.494.
E5-19
The acceleration is a = -2d/t2,
where d = 203m is the distance down the slope and t is
the time to make the run.
The weight has two components: a component parallel to the incline, WII = mgsin(}; and a
component perpendicular to the incline, W.l = mg cos (}.There is no motion perpendicular tot he
plane, so N = W.l. The kinetic friction is then f = J.tkN = J.tkmgcos(}. The net force parallel to
the plane is F = ma = f -WII = J.tkmgcos(} -mgsin(}.
Solving this for J.tk,
J.tk
=
(a + 9 sin 9)/(g
(gsin9
cos 9),
-2d/t2)/(gcos9).
If t = 61 s, then
JLk
=
JLk
=
if t = 42 s, then
E5-20
(a) If the block slides down with constant velocity then a = O and J.tk = tanO. Not only
that, but the force of kinetic friction must be equal to the parallel component of the weight, f = WII.
If the block is projected up the ramp then the net force is now 2WII = 2mgsinO. The deceleration
is a = 2gsinO; the block will travel a time t = vo/a before stopping, and travel a distance
d = -at2/2
+ vot = -a(vo/a)2/2
+ vo(vo/a) = v5/(2a) = v5/(4gsinO)
before stopping.
(b) SÍnce Jl,k< Jl,s,the incline is not steep enough to get the block moving again once it stops.
~
Let al be acceleration down frictionless incline of length l, and tl the time taken. The a2
is acceleration down "rough" incline, and t2 = 2tl is the time taken. Then
l =
1
2
2altl
and
l =
1
2a2
( 2tl ) 2 .
Equate and find al/a2 = 4.
There are two force which act on the ice when it sits on the frictionless incline. The normal force
acts perpendicular to the surface, so it doesn't contribute any components parallel to the surface.
The force of gravity has a component parallel to the surface, given by
WII = mgsin
and a component perpendicular
O,
to the surface given by
W.1.
=
mgcos(}.
The acceleration clown the frictionless ramp is then
al = ~
m
= gsinO.
When friction is present the force of kinetic friction is fk = JLkN j since the ice doesn 't move
perpendicular to the surface we also have N = W.l ; and finally the acceleration down the ramp is
a2 = Wll-fk
m
= g(sin9 -JLcos9).
60
~
Previously we found the ratio of al/a2, so we now have
sin (}
=
4sin8
sin 33°
=
4 sin 33° -4JL cos 33° ,
=
0.49.
11
-4JLcOS 8,
E5-22 (a) The static friction between A and the table must be equal to the weight of block B to
keep A from sliding. This means mBg = JLs(mA+ mc)g, or mc = mB/JLs-mA = (2.6kg)/(0.18)(4.4kg) = 10kg.
(b) There is no up/down motion for block A, so f = JLkN = JLkmAg.The net force on the system
containing blocks A and B is F = WB -f = mBg -JLkmAg; the acceleration of this system is then
a = gmB -JLkmA ={~.m/s2)(2.6kg)
mA + mB
-(0.1~)(4.4kg)
= 2.7m/s2.
(2.6kg) +(4.4kg)
~
There are four forces on the blockthe force of gravity, W = mg; the normal force,
N; the horizontal push, P, and the force of friction, f. Choose the coordinate system so that
components are either parallel (x-axis) to the plane or perpendicular (y-axis) to it. (J = 39°. Refer
to the figure below.
?1
The magnitudes of the x components of the forces are W x = W sin (), p x = p cos () and f; the
magnitudes of the y components of the forces are W y = W cos (), p y = p sin ().
(a) We consider the first the case of the block moving up the ramp; then f is directed down.
Newton 's second law for each set of components then reads as
Px -f
¿Fx
¿Fy
-
N-
-Wx
Py -Wy
= Pcos()
= N-
-f
Psin()
-Wsin()
-Wcos()
= max,
= may
Then the secandequatian is easy ta salve far N
N = Psin(} + Wcos(} = (46 N) sin(39°) + (4.8 kg)(9.8 m/s2) cos(39°) = 66 N.
The force offriction is found from f = JLkN = (0.33)(66 N) = 22 N. This is directed down the incline
while the block is moving up. We can now find the acceleration in the x direction.
max
=
=
Pcos(}-f-Wsin(},
(46N)cos(39°) -(22N)
-(4.8kg)(9.8m/s2)sin(39°)
= -16N.
So the block is slowing down, with an acceleration of magnitude 3.3 m/s2.
(b) The block has an initial speed ofvox = 4,3 m/s; it will rise until it stops; so we can use
Vy = O= VOy+ayt to find the time to the highest point. Then t = (Vy-voy)/ay = -( -4.3 m/s)/(3.3
m/s2 = 1.3 s. Now that we know the time we can use the other kinematic relation to find the
distance
y = VOyt + 4ayt2
~
(4.3
m/s)(1.3
s) + ~(-3.3
61
m/s2)(¡.3
8)2 = 2.8 m
(c) When the block gets to the top it might slide back down. But in order to do so the frictional
force, which is now directed up the ramp, must be sufficiently small so that f + p x .$:Wx. Solving
for f we find f .$:Wx -Px or, using our numbers from above, f.$: -6 N. Is this possible? No, so
the block will not slide back down the ramp, even if the ramp werefrictionless, while the horizontal
force is applied.
E5-24 (a) The horizontal force needs to overcome the maximum static friction, so P ~ .U8N~
.U8mg~ (0.52)(12kg)(9.8 m/s2) = 61 N .
(b) If the force acts upward from the horizontal then there are two components: a horizontal
component Px = Pcos(J and a vertical component Py = Psin(J. The normal force is now given by
W = Py + N; consequently the maximum force of static friction is now .U8N= .U8(mg-Psin(J).
The block will move only if Pcos(J ~ .U8(mg-Psin(J), or
(c) If the force acts downward from the horizontal then e = -62°,
so
E5-25
(a) If the tension acts upward from the horizontal then there are two components: a horizontal component T x = T cos () and a vertical component T y = T sin ().The normal force is now given
by W = Ty + N; consequently the maximum force of static friction is now JLsN = JLs(W -Tsin()).
The crate will move only if T cos () ?; JLs(W -T
sin ()) , or
(b) Once the crate starts to move then the net force on the crate is F = Tx -f.
The acceleration
is then
a
=
.JL¡TcosfJ
W
(2
(3
-Jlk(W
fJ)],
ft/S2
) ){(70
150
-Tsin
Ib)cos(17°)
-(0.35)¡(150
lb)
-(70
Ib)sin(17°)]},
lb
4.6 ft/S2,
E5-26
If the tension acts upward from the horizontal then there are two components: a horizontal
component T x = T cos (} and a vertical component T y = T sin (}.The normal force is now given by
W = Ty +N; consequently the maximum force ofstatic friction is now J.tsN = J.ts(W -Tsin(}).
The
crate will move only if T cos (} ?: J.ts(W -T
w
sin (}), or
:::; Tcos(}/JLs
+ Tsin(}.
We want the maximum, so we find dW/d(},
dW/dO
= -(T/JLs)
sinO + TcosO,
which equals zero when IJ.s= tan(J. For this problem (J = arctan(O.35) = 19°, so
w
~ (1220N)cos(19°)/(0.35)
+ (1220N)sin(19°)
62
= 3690 N.
E5-27
The three force on the know above A must add to zero, Construct a vector diagram:
TA + TB + Td = O, where Td refers to the diagonal rope. TA and TB must be related by TA =
TB tan(}, where (} = 41°,
T
There is no up/down motion of block B, so N = WB and f = JJ.sWB. Since block B is at rest
f = TB. Since block A is at rest W A = TA. Then
WA = WB(JLstan())
= (712N)(0.25)tan(41°)
= 155N.
E5-28
(a) Block 2 doesn't move up/down, so N = W2 = m2g and the force of friction on block 2
is f = J1,km2g.Block 1 is on a frictionless incline; only the component of the weight parallel to the
surface contributes to the motjon, and WII = mlg sin (}.There are two relevant forces on the two
mass system. The effective net force is the of magnitude WII -f,
so the acceleration is
a=g
m¡ sin () -¡¡'km2
=
The blocks accelerate clown the ramp.
(b) The net force on block 2 is F = m2a = T -f.
T = m2a + /lkm29
= {2.30 kg)[{1.25
1.25
m/s2.
The tension in the cable is then
m/s2)
+ {0.47){9.81
m/s2)]
= 13.5 N.
~
This problem is similar to Sample Problem 5- 7, except now there is friction which can act
on block E. The relevant equations are now for block E
-mBgcoslJ
=
O
and
T-
mBgsin(} ::I:f = mBa,
where the sign in front of f depends on the direction in which block B is moving. If the block is
moving up the ramp then friction is directed down the ramp, and we would use the negative sign.
If the block is moving down the ramp then friction will be directed up the ramp, and then we will
use the positive sign. Finally, if the block is stationary then friction we be in such a direction as to
make a = O.
For block A the relevant equation is
mAg -T
=,: mAa.
Combine the first two equations with f = JJ.N to get
T-
mEgsin
(} ::!:: J1,mEgcOS(}
63
=
mEa.
Take some care when interpreting friction for the static case, since the static value of .u yields the
maximum possible static friction force, which is not necessarily the actual static frictional force.
Combine this last equation with the block A equation,
mAg -mAa
mA
-mEgsin(J
mB sin() = (13.2kg)
:!:: ¡¡,mEgcos(J
-(42.6
kg)(0.669)
= mEa,
=
15.3kg
where the negative sign indicates that block B would slide downhill if there were no friction.
H the blocks are originally at rest we need to consider static friction, so the last term can be as
large as
f1,mBCOS(}
= (.56)(42.6kg)(0.743) = 17.7kg.
Since this quantity is larger than the first static friction would be large enough to stop the blocks
from accelerating if they are at rest.
(b) If block B is moving up the ramp we use the negative sign, and the acceleration is
~
~4.08m/s2.
where the negative sign means down the ramp. The block, originally moving up the ramp, will
slow down and stop. Once it stops the static friction takes over and the results of part (a) become
relevant.
(c) If block B is moving down the ramp we use the positive sign, and the acceleration is
=-1.30m/s2.
where the negative sign means clown the ramp. This means that if the block is moving clown the
ramp it will continue to move clown the ramp, faster ancl faster.
E5-30 The weight can be resolved into a component parallel to the incline, WII = Wsin(} and a
component that is perpendicular , W.L = W cos(}.There are two normal forces on the crate, one from
each side of the trough. By symmetry we expect them to have equal magnitudes; since they both
act per:pendicular to their respective surfaces we expect them to be perpendicular to each other .
They must add to equal the perpendicular component of the weight. Since they are at right angles
and equal in magnitude, this yields N2 + N2 = W.L2, or N= W.L/V2.
Each surface contributes a frictional force f = J1,kN= J1,k
W.L/ V2; the total frictional force is
then twice this, or V2J1,k
W.L.The net force on the crate is F = W sin (} -V2J1,k W cos(} down the
ramp. The acceleration is then
a
=
g(sinO
-V2'Uk
64
cosO).
E5-31 The normal force between the top slab and the bottom slab is N = W t = mtg. The force
of friction between the top and the bottom slab is f ~ .uN = .umtg. We don't yet know if the slabs
slip relative to each other, so we don't yet know what kind of friction to consider.
The acceleration of the top slab is
at =
(110N)/(9.7kg)
-J1,(9.8m/s2)
=
11.3m/s2
-J1,(9.8m/s2).
The acceleration of the bottom slab is
ab = JL(9.8m/s2)(9.7,kg)/(42kg)
= JL(2.3m/s2).
Can these two be equal? Only if f1,?; 0.93. Since the static coefficient is less than this, the block
slide. Then at = 7.6m/s2 and ab = 0.S7m/s2.
E5-32
(a) Convert the speed to ft/s: v = 88 ft/s. The acceleration is
a = v2jr
(b) a = 310 ft/S29/(32
~
= (88 ftjs)2j(25
ft) = 310 ftjs2.
ft/S2) = 9.79.
(a) The force required to keep the car in the turn is F = mv2 Ir = Wv2 I(rg),
F =
(10700N)(13.4m/s)2/[(61.0m)(9.81m/s2)]
(b) The coefficient of friction required is 1Ls=F/W
E5-34
or
= 3210N.
= (3210N)/(10700N)
= 0.300.
(a) The proper banking angle is given by
(} = arctan
V2
"Rg
=
arctan
(16.7m/s)2
(150m)(9.8m/s2)
=
11°.
(b) If the road is not banked then the force required to keep the car in the turn is F = mv2/r =
Wv2/(Rg) and the required coefficient of friction is
JLs = F/W
=
V2
Rg
E5-35
(a) This conical pendulum makes an angle (J = arcsin(0.25/1.4)
The pebble has a speed of
v = ~
= v(O.25m)(9.8m/s2)tan(10o)
= 10° with the vertical.
= O.66m/s.
(b) a = v2/r = (0.66m/s)2/(0.25m) = 1.7m/s2.
(c} T = mg/cosO = (0.053kg)(9.8m/s2)/cos(100) = O.53N.
E5-36 Ignoring air friction (there must be a forward component to the friction!), we have a normal force upward which is equal to the weight: N = mg = (85kg)(9.8m/s2) = 833N. There
is a sideways component to the friction which is equal tot eh centripetal force, F = mv2/r =
(85kg)(8. 7m/s)2 /(25m) = 257N. The magnitudeof the net force of the road on the person is
F = V(833N)2 + (257N)2 = 870N,
and the direction is (J = arctan(257/833)
= 17° off of vertical.
65
E5-37
(a) The speed is v = 27rrf = 27r(5.3x10-11m)(6.6x1015/s)
= 2.2x106m/s.
(b) The acceleration is a = v2/r = (2.2 x 106m/s)2 /(5.3x 10-11m) = 9.1 x 1022m/s2.
(c) The net force is F = ma = (9.1 x 10-31 kg)(9.1 x 1022m/s2) = 8.3x 10-8N.
E5-38
The bask~t has speed v = 27rr/t. The basket experiences a frictional force F = mv2/r
m(27rr/t)2/r
= 47r2mr/t2. The coefficient of static friction is Jl,s= F/N = F/W. Combining,
JLs
=
47r2r
=
47r2(4.6m)
.
gt2
~
There are two forces on the hanging cylinder: the force of the cord pulling up T and the
force of gravity W = Mg. The cylinder is at rest, so these two forces must balance, or T = W.
There are three forces on the disk, but only the force of the cord on the disk T is relevant here, since
there is no friction or vertical motion.
The disk undergoescircular motion, so T = mv2/r .We want to solve this for v and then express
the answer in terms of m, M, r, and G.
v
=
I!J.=
{!1!.
E5-40
(a) The frictional force stopping the car is F = JLsN = JLsmg. The deceleration of the car
is then a = JLsg. If the car is moving at v = 13.3 m/s then the average speed while decelerating
is half this, or vav = 6.7 m/s. The time required to stop is t = x/Vav = (21m)/(6.7m/s)
= 3.1s.
The deceleration is a = (13.3m/s)/(3.1s)
= 4.3 m/s2. The coefficient of friction is JLs = a/g =
(4.3m/s2)/(9.8m/s2)
= 0.44.
(b) The acceleration is the same as in part ( a) ., so r = v2/a = (13.3m/s)2/(4.3m/s2)
= 41m.
~
There are three forces to consider: the normal force of the road on the car N; the force of
gravity on the car W; and the frictional force on the car f. The acceleration of the car in circular
motion is toward the center of the circle; this means the net force on the car is horizontal, toward
the center. We will arrange our coordinate system so that r is horizontal and z is vertical. Then the
componentsofthe normal force are Nr = Nsin(} and Nz = Ncos(}; the components ofthe frictional
force are fr = f cos(} and f z = f sin(}.
The direction of the friction depends on the speed of the car .The figure below shows the two
force diagrams.
¡1
1:1
y
7
w
The turn is designed for 95 km/hr , at this speed a car should require no friction to stay on the
road. Using Eq. 5-17 we find that the banking angle is given by
tan(}b = ~ =
(26 m/s)2
= 0.33,
rg
(210 m)(9.8 m/s2)
66
~
for a bank angle of ()b= 18°.
(a) On the rainy day traffic is moving at 14 m/s. This is slower than the rated speed, so any
frictional force must be directed up the incline. Newton 's second law is then
¿Fr
Nr -fr
¿Fz
=
= Nsin() -fcos()
N z + f z -w
=
mv2
r
= N cos () + f sin () -mg
= O.
We can substitute f = J1,8Nto find the minimum value of J1,8which will keep the cars from slipping.
There will then be two equations and two unknowns, J1,8and N. Solving for N ,
= mv 2 and N(COS()+J1,8Sin()) =mg.
N(sin()-J1,8COS())
r
Combining,
Rearrange,
gr sin (J-V2 cos(J
J1,8
= gr COS
(J+ V2sin (J.
We know all the numbers. Put them in and we'll find J1,8
= 0.22
(b) Now the frictional force will point the other way, so Newton's second law is now
LFr
=
mv2
Nr+fr=Nsin(J+fcos(J=
r
LFz
=
Nz
-fz
-w
=
Ncos{}
-fsin{}
-mg
=
o.
The bottom equation can be rearranged to show that
mg
N=
cos
o -J.l;s
,
sin
o
This can be combined with the top equation to give
sinO
+
J.l;s cosO
~-,
-=
',,¡,g
.
cos f} -JL8 S1llf}
n /'V2
r
We can solve this final expression for v using all our previous numbers and get v = 35 m/s. That's
about 130 km/hr .
E5-42 (a) The net force on the person at the top of the Ferris wheel is mv2/r = W -N t, pointing
down. The net force on the bottom is still mv2/ r , but this quantity now equals N b -W and is point
up. Then Nb = 2W -Nt = 2(150 lb) -(125 lb) = 175 lb.
(b) Doubling the speedwould quadruple the net force, so the new scalereading at the top would
be (150 lb) -4[(150 lb) -(125 lb)] = 50 lb. Wee!
E5-43 The net force on the object when it is not sliding is F = mv2/r; the speed of the object is
v = 27rrf (f is rotational frequency here), so F = 47r2mrf2. The coefficient of friction is then at
least Jls = F /W = 47r2rf2 / g. If the object stays put when the table rotates at 331 rev/min then
JLs?: 47r2(0.13 m) (33.3/60 18)2/(9.8 m/s2) = 0.16.
If the object slips when the table rotates at 45.0 rev /min then
JLs::; 47r2(0.13m)(45.0/60/s)2/(9.8m/s2)
67
= 0.30.
E5-44
E5-45
(a) Assume that frigate bird fiies as we1l as a pilot. Then this is a banked highway problem.
The speed of the bird is given by V2 = gr tan (J, Butthere is also vt = 27rr, so 27rV2= gvt tan (J, or
E5-46
(a) The radius ofthe turn is r = v(33m)2 -(18m)2
= 28m. The speed ofthe plane is v =
27rrf = 27r(28m)(4.4/60/s)
= 13m/s. The acceleration is a = v2/r = (13m/s)2/(28m)
= 6.0m/s2.
(b) The tension has two components: Tx = Tcos(} and Ty = Tsin(}.
In this case (} =
arcsin(18/33) = 33°. All of the centripetal force is provided for by Tx, so
T=
(0.75kg)(6.0m/s2)/cos(33°)
= 5.4N.
( c ) The lift is balanced by the weight and Ty. The lift is then
Ty+W=
(5.4N)sin(33°)+(O.75kg)(9.8m/s2)
=
ION.
E5-47
(IJ.) The acceleration is a = v2/r = 47r2r/t2 = 47r2(6.37x 106m)/(8.64x 104s)2 = 3.37x
10-2m/s2. The centripetal force on the standard kilogram is F = ma = (1.00kg)(3.37x10-2m/s2)
=
0.0337 N .
(b) The tension in the balance would be T = W -F =(9.80N) -(0.0337N)
= 9.77N.
E5-48
(a) v = 4(0.179m/s4)(7.18s)3 -2(2.08 m/s2)(7.18s) = 235m/s.
(b) a = 12(0.179m/s4)(7.18s)2 -2(2.08m/s2)
= 107m/s2.
(c) F = ma = (2.17kg)(107m/s2) = 232N.
~
The force only has an x component, so we can use Eq. 5-19 to find the velocity.
Integrating,
1 'I
t -271"t2
,
)
Now put this expression into Eq. 5-20 to find the position a.s a functio~ of time
v x = Vo + ao
t
x = Xo + 1
t .
Vx dt = 11
Integrating,
T2
x = 'IJoT+ do (/~T2 -
ao-.
,2
Now We can put t = T into the expression
~"' =,va
3
for v.
+ aa ('1;'-
1
~T2)
68
= va + aaT/2.
P5-1 (a) There are two forces which accelerateblock I: the tension, T, and the parallel component
ofthe weight, W",l = mlgsin()lo Assuming the block acceleratesto the right,
m la
=
mlgsin
01 -T.
There are two forces which accelerate block 2: the tension, T, and the parallel component of the
weight, WII,2 = m2g sin (}2. Assuming the block 1 accelerates to the right, block 2 must also accelerate
to the right, and
m2a = T- m2gsin(}2.
Add
these
two
equations,
(m¡
+ m2)a
= m¡gsin(J¡
-m2gsin
(J2,
and then rearrange:
a=
m19 sin 01 -m29
sin 02
m¡ + m2
Or, take the two net force equations, divide each side by the mass, and set them equal to each other:
9sin
Rearrange,
T
{}1 -T/m1
=
T/m2
-gsin{}2°
1,
1
= 9sin 01+ gsinO2,
m¡ m2)
and then rearrange again:
T=
.
. (}
mlm29 (SIn
1 +SIn (}2.)
ml + m2
(b) The negative sign we get in the answer means that block 1 accelerates up the ramp.
(c) No acceleration happens when m2 = (3.70kg)sin(28°)/sin(42°) = 2.60kg. If m2 is more
massivethan this ml will accelerateup the plane; if m2 is less massive than this ml will accelerate
down the plane.
P5-2
(a) Since the pulley is massless, F = 2T. The largest value of T that will allow block 2 to
remain on the fioorisT
~ W2 = m2g. So F ~ 2(1.9kg)(9.8m/s2)
= 37N.
(b) T = F/2 = (110N)/2 = 55N.
(c) The net force on block 1 is T- W1 = (55N) -(1.2kg)(9.8m/s2)
= 43N. This will result in
an acceleration of a = (43N)/(1.2kg)
= 36m/s2.
~
As the string is pulled the two m8SSeS
will move together so that the configuration willlook
like the figure below. The point where the force is applied to the string is m8Ss1ess,
so ¿ F = O at
that point. We can take advantage of this fact and the figure below to find the tension in the cords,
F/2 = Tcos(}. The factor of 1/2 occurs becauseonly 1/2 ofF is contained in the right triangle that
h8ST 8Sthe hypotenuse. From the figure we can find the x component of the force on one m8Ssto
be Tx = Tsin(}. Combining,
Tx = ~ sin{}
F
2 cose = 2 tan{}.
69
But the tangent
is equal to
tan
(} =
x
-VL2 -X2
Opposite
Adjacent
And now we have the answer in the book.
!
What happens when x = L ? Well, ax is infinite according to this expression. Since that could
only happen if the tension in the string were infinite, then there must be some other physics that
we had previously ignored.
P5-4 (a) The force of static friction can be as large as f ~ JLsN= (0.60)(12 lb) = 7.2 lb. That is
more than enough to hold the block up.
(b) The force of static friction is actually only large enough to hold up the block: f = 5.0 lb.
The magnitude of the force of the wall on the block is then Fbw = V(5.0)2 + (12.0)2lb = 13 lb.
P5-5
(a) The weight has two components: normal to the incline, W.L = mg cos (} and parallel to the
incline, WII = mg sin (}.There is no motion perpendicular to the incline, so N = W.L = mg cos (}.The
force of friction on the block is then f = JLN = JLmgcos (}, where we use whichever JLis appropriate.
The net force on the block is F -f -WII = F :!: JLmgcos (} -mg sin (}.
To hold the block in place we use JLsand friction will point up the ramp so the :!: is +, and
F = (7.96 kg)(9.81
mjs2)[sin(22.0°)
-(0.25)
cos(22.0°)]
= 11.2N.
(b) To find the minimum force to begin sliding the block up the ramp we still have static friction,
but now friction points down, so
F = (7.96kg)(9.81m/s2)[sin(22.00)
+ (0.25)cos(22.00)]
=47.4N.
( c ) To keep the block sliding up at constant speed we have kinetic friction, so
F = (7.96 kg}(9.81
m/s2}[sin(22.0°}
+ (0.15} cos(22.0°}]
= 40.1 N.
P5-6
The sand will slide if the cone makes an angle greater than (} where .u8 = tan (}. But
tan(} = h/R or h = Rtan(}. The volume ofthe cone is then
Ah/3
= 7rR2h/3
= 7rR3tan(}/3
= 7rJ1,sR3/3.
~
There are four forces on the broom: the force of gravity W = mg; the normal force of the
floor N; the force offriction f; and the applied force from the person p (the book calls it F). Then
LFx
=
P", -f
LFy
=
N-
= PsinO
Py -w
-f
= ma""
= N-PcosO-mg
Salve the secandequatian far N,
N = PcosO
70
+ mg.
= may = O
(a) If the mop slides at constant
Psin(j
We can solve this for P (which
speed f = JLkN. Then
-f
= Psin(j
-J.l;k (PCOS(j + mg) = O.
was called F in the book);
P=
J.l;mg
sin (} -/Lk COS(} .
This is the force required to push the broom at constant speed.
(b) Note that p becomes negative (or infinite) ifsin(} .$:/LkCOS(}. This occurs when tan(}c .$:/Lk.
If this happens the mop stops moving, to get it started again you must overcome the static friction,
but this is impossible if tan (}o .$:/Ls
P5-8
(a) The condition to slide is J.l;s~ tanO. In this case, (0.63) > tan(24°) = 0.445.
(b) The normal force on the slab is N = W l. = mg cos O. There are three forces parallel to the
surface: friction, f = J.l;sN= J.l;smgcosOj the parallel component of the weight, WII = mgsinO, and
the force F. The block will slide if these don 't balance, or
F > J.l;smgcosO-mgsinO
= (1.8x 107kg)(9.8m/s2)[(0.63)
cos(24°) -sin(24°)]
= 3.0x 107N.
~
To hold up the smaller block the frictional force between the larger block and smaller block
must be as large as the weight of the smaller block. This can be written as f = mg. The normal
force of the larger block on the smaller block is N, and the frictional force is given by f ~ JLsN. So
the smaller block won't fall if mg ~ JLsN.
There is only one horizontal force on the large block, which is the normal force of the small block
on the large block. Newton 's third law says this force has a magnitude N, so the acceleration of the
large block is N = M a.
There is only one horizontal force on the two block system, the force F. So the acceleration of
this system is given by F = (M + m)a. The two accelerations are equal, otherwise the blocks won't
stick together. Equating, then, gives N/M = F/(M + m).
We can combine this last expressionwith mg ~ JLsN and get
mg ~ JLsF~
M
or
F?:
g(M
+m)m
J1sM
P5-10 The normal force on the ith block is Ni = migcos(); the force of friction on the ith block
is then fi = J1,imigcos().The parallel component of the weight on the ith block is WII,i = migsin().
(a) The netiorce on the system is
F = }:= mig(sin 0- J.LicosO).
i
Then
a
=
(9.81
{1.65 kg) + {3.22 kg)
3.46m/s2,
(b) The net force on the lower mass is m2a = WII,2 -!2T = (9.81 m/s2)(3.22kg)(sin29.5°
-O.127cos29.5°)
71
T, so the tension is
-(3.22
kg)(3.46m/s2)
= 0.922 N.
~
(c) The acceleration will stay the same, since the system is still the same. Reversing the order
of the massescan only result in a reversing of the tension: it is still 0.992 N, but is now negative,
meaning compression.
~
The rope wraps around the dowel and there is a contribution to the frictional force ~f
from each small segment of the rope where it touches the dowel. There is also a normal force ~N at
each point where the contact occurs. We can find ~N much the same way that we solve the circular
motion problem.
In the figure on the left below we seethat we can form a triangle with long side T and short side
~N. In the figure on the right below we see a triangle with long side r and short side r~(}. These
triangles are similar, so r~(}/r = ~N/T.
<J-~~
N
rL\e
~
íi::~
/
r
F'",
,1
\
Now ~f
""'"
J
"
= JL~N and T(()) + ~f
\
~ T(() + ~()).
\
I
r
"", ,
Combining,
and taking the limit as ~() -O,
dT=df
J
= Id(}
~~
JLT
integrating both sides of this expression,
J
~~
Ji,T
1
-lnT¡T2
Ji,
Tl
=
7r,
T2
In this case Tl is the weight and T2 is the downward force.
P5-12 Answer this out of order!
(b) The maximum static friction between the blocks is 12.0 N; the maximum acceleration of the
top block is then a = F/m = (12.0N)/(4.40kg) = 2.73m/s2.
(a) The net force on a system of two blocks that will accelerate them at 2.73m/s2 is F =
(4.40kg + 5.50kg)(2.73m/s2) = 27.0N.
(c) The coefficient of friction is JLs= F/N = F/mg = (12.0N)/[(4.40kg)(9.81m/s2)] = 0.278.
P5-13
The speed is v = 23.6m/s.
(a) The average speed while stopping is half the initial speed, or vav = 11.8 m/s. The time to stop
is t = (62m)/(11.8m/s)
= 5.25s. The rate of deceleration is a = (23.6m/s)/(5.25s)
= 4.50m/s2.
The stopping force is F = ma; this is related to the frictional force by F = JLsmg, so JLs= a/ 9 =
(4.50m/s2)/(9.81 m/s2) = 0.46.
72
(b) Thrning,
a = v2/r
Then Jl,s = a/g = (8.98m/s2)/(9.81
= (23.6 m/s)2/(62
m/s2)
m) = 8.98 m/s2,
= 0.92.
P5-14 (a) The net force on car as it travels at the top of a circular hill is F net = mv2/r = W -N;
in this casewe are told N = W/2, so Fnet = W/2 = (16000N)/2 = 8000N. When the car travels
through the bottom valley the net force at the bottom is F net = mv2/r = N -W.
Since the
magnitude of v, r, and hence F net is the same in both cases,
N = W/2 + W = 3W/2 = 3(16000N)/2 = 24000N
at the bottom of the valley.
(b) You leave the hill when N = O, or
v = .¡r:g ~ J(250m)(9.81 m/s2) = 50m/s.
(c) At this speed Fnet = W, so at the bottom of the valley N= 2W = 32000N.
P5-15
(a) v = 27rr/t = 27r(0.052m)(3/3.3s) = 0.30m/s.
(b) a = v2/r = (0.30m/s)2/(0.052m)
= 1.7m/s2, toward center.
(c) F = ma = (0.0017kg)(1.7m/s2) = 2.9xI0-3N.
(d) If the coin can be as far awayas r before slipping, then
J1,s= F:/mg = (27rr/t)2/(rg)
= 47r2r/(t2g) = 47r2(0.12m)/[(3/3.3s)2(9.8m/s2)]
= 0.59.
P5-16 (a) Whether you as8umeconstant 8peedor con8tant energy, the ten8ion in the 8tring mu8t
be the greate8t at the bottom of the circle, 80 that'8 where the 8tring will break.
(b) The net force on the 8tone at the bottom i8 T- W = mv2/r. Then
v = Vrg[T/W
-iJ=
J (2.9 ft)(32
ft~s2)[(9.2
lb)/(0.82
lb) -1];;:=
31 ft/s.
~
(a) In order to keep the ball moving in a circle there must be a net centripetal force Fc
directed horizontally toward the rod. There are only three forces which act on the ball: the force of
gravity, W = mg = (1.34kg)(9.81 m/s2) = 13.1N; the tension in the top string T1 = 35;ON, and the
tension in the bottom string, T2.
The components of the force from the tension In the top string are
T1,x = (35.0N) cos30° = 30.3N and T1,1/= (35.0N) sin30° = 17.5N.
The vertical components do balance, so
T!,y + T2,y = W,
or T2,11~(13.1N) -(17.5N)
= -4;4N. From this we can find the tensionin the bottom string,
T2 = T2,y/sin(-300) = 8.8N.
(b) The net force on the object wil1 be the sum of the two horizontal components,
Fc = (30.3N) +(8.8N)cos30°
= 37.9N.
(c) The speed will be found from
~
= ,¡F;;:¡m
v(37.9m)(1.70m)
,
sin 60° /(1.34kg)
73
= 6.45m/s.
P5-18
The net force on the cube is F = mv2 /r .The speed is 27rr(U. (Note that we are using (U in
a non-standard way!) Then F = 47r2mr(U2. There are three forces to consider: the normal force of
the funnel on the cube N; the force of gravity on the cube W; and the frictional force on the cube
f. The acceleration of the cube in circular motion is toward the center of the circle; this means the
net force on the cube is horizontal, toward the center. We will arrange our coordinate system so
that r is horizontal and z is vertical. Then the components of the normal force are N r = N sin (}
and N z = N cos (}; the components of the frictional force are f r = f cos (} and f z = f sin (}.
The direction of the friction depends on the speed of the cube; it will point up if (U is small and
down if (U is large.
(a) If(U is small, Newton's second law is
LFr
LFz
=
Nr -Ir
=
Nz + fz -w
= N sin {} -Icos{}
= NcosB
= 411"2mr(.;2,
+fsinB
-mg
= o.
We can substitute f = JLsN. Solving for N
N ( cos o + 11.8
sin O) = mg.
Combining,
sin () -JLs cos ()
47r2r(¿)2 = 9:cos (} + ¡¡;:s¡;;-¡¡ .
Rearrange,
(.;=
1
2;V~
!gsin(}-JLscos(}
cos(}+JLssin(}'
This is the minimum value.
(b) Now the frictional force will point the other way, so Newton's second law is now
LFr
LFz
=
Nr + Ir = N sin {} + Icos{} = 47r2mr(¡)2,
-
Nz -fz
-w
= Ncos()
-fsin()
-mg
= o.
We've swapped + and -signs, so
(U=
is the
maximum
.!.-
1:?: sin (} + /1.8COS(}
27r V ~ cos (} -JLs sin (}
value.
P5-19 (a) The radial distance from the axis of rotation at a latitude L is RcosL. The speed
in the circle is then v = 27rRcosL/T. The net force on a hanging object is F = mv2/(RcosL) =
47r2mRcosL/T2. This net force is notdirected toward the center ofthe earth, but is instead directed
toward the axis of rotation. It makes an angle L with the Earth 's vertical. The tension in the cable
must then have two components: one which is vertical (compared to the Earth) and the other which
is horizontal. If the cable makes an angle f) with the vertical, then 111= Tsinf) and T.L = Tcosf).
Then TII = FII and W -T.L = F.L. Written with a little more detail,
Tsin{}
= 47r2mRcosL
sinL/T2
~ T{},
and
Tcos(} = 47r2mRcOS2L/T2 + mg ~ T.
74
But 47r2R COS2
L/T2 « y, so it can be ignored in the last equation compared to y, and T ~ my.
Then from the first equation,
9 = 211"2
Rsin 2L/(gT2).
(b) This is a maximum
when sin2L
is a maximum,
which
(} = 27r2(6.37x 106m)/[(9.8m/s2)(86400S)2]
happens
when L = 45°. Then
= 1.7x 10-3 rad.
(c) The deflection at both the equator and the poles would be zero.
P5-20 a = (Fo/m)e-t/T. Then v = J~ adt = (FoT/m)e-t/T,
(a) When t = T v = (FoT/m)e-l = O.368(FoT/m).
(b) When t = T x = (FoT2/m)e-l = O.368(FoT2/m).
75
and x = J~ vdt = (FoT2/m)e-t/T.
E6-1 (a) VI = (m2/mI)V2 = (2650kg/816kg)(16.0 km/h) = 52.0 km/h.
(b) VI = (m2/mI)v2 = (9080kg/816kg)(16.0 km/h) = 178 km/h.
E6-2
Pi = (2000kg)(40 km/h)j = 8.00x 104kg .km/bj. Pf = (2000kg)(50 km/h)i = 1.00x 105kg .
km/hi. ~P = Pf -Pi = 1.00 x 105kg .km/hi -8.00 x 104kg .km/bj.
~P = V(~Px)2 + (~py)2 =
1.28 x 105kg .km/h. The direction is 38.7° south of east.
~
The figure below shows the initial and final momentum vectors arranged to geometrically
show Pf -Pi = ~p. We can use the cosine law to find the length of ~p.
""'
-pi
,,
~
y
~p
--
---
e
/
/
pf
The angle a = 42° + 42°, Pi = m v = (4.88kg)(31.4m/s)
~p is
6.p =
(153kg.m/s)2
+ (153kg.m/s)2
-2(153kg.m/s)2
= 153kg.m/s.
cos(84°)
Then the magnitude oí
= 205kg.m/s,
directed up from the plate. By symmetry it must be perpendicular .
E6-4
The change in momentum is ~p = -mv = -(2300kg)(15m/s)
= -3.5xl04kg
average force is F = ~p/ ~t = ( -3.5 x 104kg .m/s)/(0.54s)
= -6.5 x 104N.
E6-5
(a) The change in momentum i8 ~p = (-mv)
-2mv/~t.
(b) F = -2(0.14 kg)(7.8m/8)/(3.9x
-mvj
.m/s.
the average force i8 F = ~p/~t
The
=
10-38) = 560N.
E6-6
(a) J = ~p = (O.O46kg)(52.2m/s) -O = 2.4N .s.
(b) The impulse imparted to the club is opposite that imparted to the ball.
(c) F = ~p/ ~t = (2.4N .s)/(1.20x 10-3s) = 2000N.
~
Choose the coordinate system so that the ball is only moving along the x axis, with away
from the batter as positive. Then Pfx = mvfx = (0.150kg)(61.5m/s) = 9.23 kg.m/s and Pix =
mvix = (0.150kg)( -41.6 m/s) = -6.24 kg.m/s. The impu1seis given by Jx = Pfx -Pix = 15.47
kg.m/s. We can find the averageforce by application of Eq. 6-7:
F, av,:!:
=
Jx
~t
76
E6-8
The magnitude of the impulse is J = Ft5t = ( -984 N)(O.O270s) = -26.6N
.s. Then Pf =
Pi + ~P, so
The hall moves hackward!
E6-9
The change in momentum ofthe ball is ~p = (mv) -( -mv) = 2mv = 2(O.O58kg)(32m/s) =
3.7 kg .m/ s. The impulse is the area under a force -time graph; for the trapezoid in the figure this
area is J = Fmax(2ms + 6ms)/2 = (4ms)Fmax. Then Fmax = (3.7kg .m/s)/(4ms)
= 930N.
E6-10 The final speed of each object is given by Vi = J/mi, where i refers to which object (as
opposed to "initial"). The object are going in different directions, so the relative speed will be the
sum. Then
Vrel = V¡ + V2 = (300 N. s)[1/(1200 kg) + 1/(1800 kg)]
O.42m/s.
~
Use Simpson 's rule.
Jx
Then the area is given by
1
:r. -h(/o +4!1 + 2!2 +4!3 + ...+4!13+!14)
3
1
-(0.2ms)
3
(200
+
4.800
+
,
2 .1200...N)
which gives Jx = 4.28 kg.m/s.
Sincethe impulse is the changein momentum, and the ball started from rest, Pfx = Jx+Pix = 4.28
kg.m/s. The final velocity is then found from Vx = px/m = 8.6 m/s.
E6-12 (a) The averagespeedduring the the time the hand is in contact with the board is half of the
initial speed,or vav = 4.8m/s. The time of contact is then t = y/vav = (O.O28m)/(4.8m/s) = 5.8ms.
(b) The impulse given to the board is the same as the magnitude in the changein momentum of
the hand, or J = (0.54kg)(9.5m/s) = 5.1 N. s. Then Fav = (5.1 N. s)/(5.8ms) = 880N.
E6-13 Ap = J = FAt = (3000N)(65.0s) = l.95x lO5N .s. The direction of the thrust relative to
the velocity doesn't matter in this exercise.
E6-14 (a) p = mv = (2.14x 10-3kg)(483m/s) = 1.03kg .m/s.
(b) The impulse imparted to the wall in one second is ten times the above momentum, or
J = 10.3kg .m/s. The averageforce is then F av = (10.3kg .m/s)/(1.0s) = 10.3N.
(c) The averageforce for each individual particle is F av = (1.03kg. m/s)/(1.25x10-3 s) = 830N.
~
A transverse direction means at right angles,so the thrusters have imparted a momentum
sufficient to direct the spacecraft 100+3400 = 3500 km to the side of the original path. The spacecraft
is half-way through the six-month journey, so it has three months to move the 3500 km to the side.
This correspondsto a transverse speedof v = (3500x103m)/(90x86400s) = 0.45m/s. The required
time for the rocket to fire is ~t = (5400kg)(0.45m/s)/(1200N) = 2.0s.
E6-16
Total initial momentum is zero, so
Vm=-
fJs =
~~
m m
-
.!!!:!!iL~
lIs
=
-,
mmg
77
~
~
Conservation
of momentum:
Pf,m
mmVf,m
+
Vf,c
+
Pf,c
mcVf,c
=
Pi,m
=
mmVi,m
+pi,c,
+
~~!,m
-Vi,c
mcVi,c,
-mmVf,m
mc
=
~vc
{75.2
kg){~.33
mis)
-{75.2
kg){O)
(38.6 kg)
4.54m/s.
The answer is positive; the cart speed increases.
E6-18
Conservation
of momentum:
Pf,m +Pf,c
m m(Vf,c -Vrel) + mcVf,c
(m m + mc)Vf,c -1nmVrel
Avc
E6-19
Conservation of momentum.
Pf,m + Pf,c
m m (Vf,c -Vrel)
+ mcVf,c
(m m + mc)Vf,c -mmVrel
=
=
=
=
=
Pi,m+Pi,c,
(m m + mc)Vi,C,
(m m + mc)Vi,c,
mmVrel/(mm + mc),
WVrel/(W+ W).
Let m refer to motor and c refer to command module:
=
=
Pi,m+Pi,c,
(m m + mc)Vi,c,
=
(mm+mc)Vi,c,
=
.mmVrel+ (mm+ mc)Vi,C
Vf,c
(m m + mc)
4mc(125
km/h}+
(4mc + mc)(3860
km/h)
(4mc + mc)
E6-20
Conservation of momentum.
mlVl
,f +
m2V2
=
,f
Vl,f
The block on the left is 1, the other is 2.
=
m¡V¡,i + m2V2,i,
m2
V¡,i + -(V2
m¡
,i -V2 ,f),
=
(5.5m/s) + ~[(2.5m/S)
-(4.9m/s)],
1.9m/s,
E6-21
Conservation oí momentum. The block on the left is I, the other is 2.
mi
V1,f
+
m2V2,f
Vl,f
=
mlVl,i + m2V2.i,
=
Vl
-m2
i +
m
,
=
1
(5.5m/s)
-5.6
( V2 ..i
-V2
f
)
,
(2.4 kg)
+ -¡i:¡j"kg)[(-2.5m/s)
m/s.
78
-(4.9m/s)],
=
3960
km/h.
E6-22
Assume a completely inelastic collision. Can the Earth I and the meteorite 2. Then
That 's 2 mm/y!
~
Canservatian
af mamentum
Pf
=
Pi,
Pf,bl + Pf,bu
=
Pi,bl + Pi,bu,
+ mbuVf,bu
=
mblvi,bl
=
(715
mblVf,bl
(715
which
has
solution
g)Vf,bl
Vf
bl
+ (5.18
=
is used ta salve the prablem:
g)(428
mis)
+ mbuVi,bu.
g)(O)
+ (5.18
g)(672
mis),
1.77m/s.
,
E6-24
The y component oí the initial momentum is zero; thereíore the magnitudes oí the y components oí the two particles must be equal after the collision. Then
mQvQ
sin()Q
Va
E6-25
=
movo
=
movo
sin (}o,
sin (}o
mQvQ
sin(}Q
=
.(16
,
u)(1.20x105m/s)~~
=
4.15
x 105m/s.
The total momentum is
¡3 =
=
(2.0kg)[(15m/s)i + (30m/s)jj + (3.0kg)[(-10m/s)i + (5m/s)jj,
75kg. m/sj.
The final velocity of B is
VBf
=
1
(p-mAVAf
--
),
mB
E6-26
Assume electron travels in +x direction while neutrino travels in +y direction.
of momentum requires that
Conservation
p = -{1.2 x 10-22kg .mis)i -{6.4 x 10-23kg .mis)j
be the momentum oí the nucleus after the decay. This has a magnitude oí p = 1.4 x 10-22kg .mis
and be directed 152° from the electron.
79
E6-27
What
we know:
Pl,i
=
(1.50x 105kg)(6.20m/s)i
= 9.30x 105kg .m/si,
P2,i
=
(2. 78x 105kg)(4.30m/s)j
= 1.20x 106kg .m/sj,
P2,f
=
(2.78 x 105kg)(5.10 m/s)[sin(18°)i
=
4.38 x 105kg .m/si
+ cos(18°)j],
+ 1.35 x 106kg .m/sj.
Conservation of momentum then requires
Pl,f
=
(9.30x105kg.m/si)-(4.38x105kg.m/si)
=
4.92 x 105kg .m/s i -1.50 x 105kg .m/s i.
+(1.20 x 106kg .m/sj)
-(1.35
x 106kg .m/sj),
This corresponds to a velocity of
Vl,f = 3.28m/si
-1.00m/sj,
which has a magnitude oí 3.43 mis directed 17° to the right.
E6-28
Vf = -2.1
m/s.
~
We want ta salve Eq. 6-24 far m2 given that Vl,f = O and Vl,i = -V2,i.
substitutians
Making these
so m2 = 100 g.
E6-30
(a)
Rearrange
Eq.
6-27:
E6-31
Rearrange Eq. 6-27:
Vli '- Vlf
m2 = ml ,
= (2.0
Vli
E6-32
+
Vlf
I~Vli + Vli/4
= 1.2kg.
I'll multiply all momentum equations by g, then I can use weight directly without converting
to mass.
(a) Vf = [(31.8 T)(5.20 ft/s) + (24.2 T)(2.90 ft/s)]/(31.8 T + 24.2 T) = 4.21 ft/s.
(b) Evaluate:
80
~
Let the initial momentum of the first object be Pl,i = mVl,i, that of the second object be
P2,i = mV2,i, and that of the combined final object be Pf = 2mVf. Then
-+
Pl,i
+
-+
P2,¡;;:
-+
Pf,
implies that we can find a triangle with sides of length Pl,i, P2,i, and Pf. These lengths are
Pl,i
P2,i
=
mvj,
=
mvj,
Pf
2mvf
= 2mvj/2
= mvj,
so this is an equilateral triangle. This meansthe angle between the initial velocities is 120°.
E6-34 We need to change to the center of mass system. Since both particles have the same
mass, the conservation of momentum problem is effectively the same as a (vector) conservation of
velocity problem. Since one of the particles is originally at rest, the center of mass moves with
speed Vcm= Vli/2. In the figure below the center of mass velocities are primed; the transformation
velocity is Vt.
v t
Note that since Vt = v' li = v' 2i = v' lf = v' 2f the entire problem can be inscribed in a rhombus.
The diagonals of the rhombus are the directions of Vlf and V2fj note that the diagonals of a rhombus
are necessarilyat right angles!
(a) The target proton moves off at 90° to the direction the incident proton moves after the
collision, or 26° away from the incident pr9tons original direction.
(b) The y components ofthe final momenta must be equal, so v2fsin(26°) = Vlf8in(64°), or V2f =
Vlf tan(64°). The x components must add to the original momentum, so (514m/s) = V2fcos(26°) +
Vlfcos(64°), or
Vlf
(514m/s)/{tan(64°)
cos(26°) + cos(64°)} = 225m/s,
and
V2f = (225m/s)tan(64°)
E6-35
v cm = {(3.16 kg)(15.6m/s)+(2.84
= 461m/s.
kg)(-12.2m/s)}/{(3.16kg)+(2.84
itive means to the left.
81
kg)} = 2.44m/s, pos-
~
The force is the changein momentum over change in time; the momentum is the mass time
velocity, so
dp
mdv
m
F=
-=
At
-=dvAt
At
=2u.u
,
since J.Lis the mass per unit time.
P6-2
(a) The initial momentum is Pi = (1420kg)(5.28m/s)j
= 7500kg .m/sj.
After making the
right hand turn the final momentum is Pf = 7500 kg .m/s ¡. The impulse is j = 7500 kg .m/s ¡ 7500kg .m/sj, which has magnitude J = 10600kg .m/s.
(b) During the collision the impulse is j = 0-7500kg.m/s¡.
(c) The average force is F = J/t = (10600kg .m/s)/(4.60s)
(d) The average force is F = J/t = (7500kg .m/s)/(0.350s)
P6-3
The magnitude is J = 7500kg.m/s.
= 2300N.
= 21400N.
(a) Only the component ofthe momentum which is perpendicular
.1 = ~p
(b) F = -J/t
= -2(0.325
kg)(6.22
mis) sin(33°)j
= -2.20
to the wall changes. Then
kg .misj.
A
= -( -2.20kg .m/sj)/(O.O104s) = 212N.
P6-4 The change in momentum ofone bullet is ~p = 2mv = 2(O.OO30kg)(500m/s)= 3.0kg.m/s.
The averageforce is the total impulse in one minute divided by one minute, or
F av = lOO(3.0kg
.m/s)/(60s)
= 5.0N.
P6-5 (a) The volume of a hailstone is V = 47rr3/3 = 47r(0.5cm)3/3 = 0.524 cm3. The mass of a
hailstone is m = pV = (9.2 x 10-4kg/cm3)(0.524 cm3) = 4.8x 10-4kg.
(b) The change in momentum of one hai1stonewhen it hits the ground is
~p = (4.8 x 10-4kg)(25
m/s)
= 1.2 x 10-2kg
.m/s.
The hailstones fall at 25 m/s, which means that in one secondthe hailstones in a column 25 m high
hit the ground. Over an area of 10mx 20m then there would be (25m)(10m)(20m) = 500m3 worth
of hailstones, or 6.00x 105 hailstones per second striking the surface. Then
Fav = 6.00xlO5(l.2xlO-2kg
.m/s)/(ls)
= 7200 N.
P6-6 Assume the links are not connected once the top link is released. Consider the link that
starts h above the table; it falls a distance h in a time t = J2h79 and hits the table with a speed
v = gt = ..¡2Tí¡j.When the link hits the table h of the chain is already on the table, and L- h is yet
to come. The linear mass density of the chain is M / L, so when this link strikes the table the mass is
hitting the table at a rate dm/dt = (M/L)v = (M/L)..¡2Tí¡j. The averageforce required to stop the
falling link is then vdm/dt = (M/L)2hg = 2(M/L)hg. But the weight ofthe chain that is already
on the table is (M/L)hg, so the net force on the table is the sum ofthese two terms, or F = 3W.
~
The weight of the marbles in the box after a time t is mgRt because Rt is the number of
marbles in the box.
The marbles fa11 a distance h from rest; the time required to fal1 this distance is t = .j2ií7Y,
the speed of the marbles when they strike the box is v = gt = .¡2g7i". The momentum each marble
imparts on the box is then m.¡2g7i". If the marbles strike at a rate R then the force required to stop
them is Rm.¡2g7i".
82
The reading on the scale is then
w = mR( J29h
+ gt).
This will give a numerical result of
(4.60x 10-3kg)(115 S-I ) ( .¡2(9.81 m/s2)(9.62 m) + (9.81 m/s2)(6.50s) ) = 41.0N.
P6-8
(a) v = (108kg)(9.74m/s)/(108kg+
1930kg) = 0.516m/s.
(b) Label the person as object 1 and the car as object 2. Then mlVl+m2V2
and Vl = V2 + 0.520 m/s. Combining,
V2 = [1050 kg .m/s -(0.520
= (108kg)(9.74m/s)
m/s)(108 kg)1/(108 kg + 1930 kg) = 0.488 m/s.
P6-9 (a) It takes a time tl = J2h79 to fall h = 6.5 ft. An object will be moving at a speed
Vl = gtl = V2fig after falling this distance. If there is an inelastic collision with the pile then the
two will move together with a speed of V2= MV1/(M + m) after the collision.
Ifthe pile then stops within d = 1.5 inches, then the time ofstopping is given by t2 = d/(V2/2) =
2d/V2.
For inelastic collisions this correspondsto an averageforce of
F -..= .(M + m)v2 -(M
t2
+ m)v~ -M2v~
=2d
2(M+m)d
(gM)2
g(M+m)
~.
do
Note that we multiply through by 9 to get weights. The numerical result is F av = 130 t.
(b) For an elastic collision V2 = 2Mvl/(M + m); the time of stopping is still expressed by
t2 = 2d/V2, but we now know F av instead of d. Then
F
-mv2
av-
t2
-mv~
=-==-
4Mmv~
2d
(M+m)d
-2(gM)(gm)
g(M+m)
~
d'
or
d=
2(gM)(gm)~
,
g(M+m)
Fav
which has a numerical result of d = 0.51 inches.
But wait! The weight, which just had an elastic collision, "bounced" off of the pile, and then hit
it again. This drives the pile deeper into the earth. The weight hits the pile a second time with a
speedof V3= (M -m)/(M +m)vl; the pile will (in this secondelastic collision) then have a speedof
V4= 2M(M + m)v3 = [(M -m)/(M + m)]v2. In other words, we have an infinite seriesof distances
traveled by the pile, and if a = [(M -m)/(M
+ m)] = 0.71, the depth driven by the pile is
df = d(l + a2 + a4 + a6
)
=
l-a2
d
,
or d = 1.03.
The catjumps offofsled A; conservationofmomentum requires that MVA,l +m(VA,l +Vc) =
P6-10
O,
or
VA,l = -mvc/(m
+ M) = -(3.63kg)(3.05m/s)/(22.7kg
+ 3.63kg) = -0.420m/s.
The cat lands on sled B; conservation of momentum requires VB,l = m(VA,l + vc)/(m + M).
cat jumps off of sled B; conservation of momentum is now
MVB,2
+ m(VB,2
-Vc)
83
= m(VA.l
+ Vc),
The
or
VB,2 = 2mvc/(m + M) = (3.63kg)[(-0.420m/s)
+ 2(3.05m/s)]/(22.7kg
+ 3.63kg) = O.783m/s.
The cat then lands on cart A; conservation of momentum requires that (M + m)vA,2 = -MVB,2,
VA,2 = -(22.7kg)(0.783m/s)/(22.7kg+3.63kg)
= -0.675m/s.
~
We align the coordinate system so that west is +x and south is +y.
contributes the following to the initial momentum
A:
B
or
The each car
(2720 lb/g)(38.5 mi/h)i = 1.05x 105 lb. mi/h/gi,
:
(3640 lb/g)(58.0 mi/h)j = 2.11x 105 lb. mi/h/gj.
These become the components of the final momentum. The direction is then
2.11x 105 lb. mi/h/g
o
(} = arctan 1.05x 105 lb. mi/h/ 9 = 63.5 ,
south of west. The magnitude is the square root of the sum of the squares,
2.36x 105 lb. mi/h/ g,
and we divide this by the mass (6360 lb/g) to get the final speed after the collision: 37.1 mi/h.
P6-12 (a) Hall A must carry offamomentumofp = mBvi-mBv/2j,
whichwould be in adirection
f) = arctan(-0.5/1) = 27° from the original direction of E, or 117° from the final direction.
(b) No.
P6-13 (a) We assumeall balls have a mass m. The collision imparts a "sideways" momentum to
the cue ball ofm(3.50m/s) sin(65°) = m(3.17m/s). The other ball must have an equal, but opposite
"sideways" momentum, so -m(3.17m/s) = m(6.75m/s)sin(}, or (} = -28.0°.
(b) The final "forward" momentum is
m(3.50m/s) cos(65°) + m(6.75m/s) cos(-28°) = m(7.44m/s),
so the initial speed of the cue ball would have been 7.44m/s.
P6-14
Assuming M » m, Eq. 6-25 becomes
V2f = 2Vli -Vli
~
(a)
We
= 2(13 km/s) -(-12
get
g)
V2,f = (220 2(220
g) + (46.0
g) (45.0m/s)
(b) Doubling
the m88S of the clubhead
the mass of the clubhead
V2,f =
= 74.4m/s.
we get
2L440 g)
I .-~
U2,f = (440 g) + {46.0 g) t45.0mjs)
{c) Tripling
km/s) = 38 km/s.
r- .--'~ 1= dl.5m¡s.
we get
g)
(66üg)2{660
+ (46.0
g) t45.0m¡s)
= 84.1m/s.
Although the heavier club helps some, the maximum speed to get out of the ball will be less than
twice the speed of the club.
84
The balls are a, b, and c from left to right.
P6-16
There will always be at least two collisiollS.
After the first collision between a and b one has
Vb,l
=
va
and
Va,l
=
O.
After the first collision between b and c one has
Vc,l
= 2mvo/(m+
M)
and
Vb,2 = (m-
M)vo/(m
+ M).
(a) Ifm ;::::M then ball b continue to move to the right (or stops) and there are no more collisions.
(b) If m < M then ball b bounces back and strikes ball a which was at rest. Then
Va,2 = (m -M)vo/(m
+ M)
and
Vb,3 = O.
P6-17 All three balls are identical in mass and radii? Then balls 2 and 3 will move off at 30° to
the initial direction of the first ball. By symmetry we expect balls 2 and 3 to have the same speed.
The problem now is to define an elastic three body collision. It is no longer the case that the
balls bounce off with the same speed in the center of mass. One can't even treat the problem as two
separate collisions, one right after the other. No amount of momentum conservation laws will help
solve the problem; we need some additional physics, but at this point in the text we don't have it.
P6-18
and
The original speed is Vo in the lab frame. Let a be the angle of deflection in the cm ffame
v~ be the
A
initial
velocity
A
in the
cm frame.
Then
the
velocity
after
the
A
collision
in the
A
cm frame
is v~cosa i + v~ sin aj and the velocity in the lab frame is (v~cosa + v )i + v~sin aj, where v is the
speedof the cm frame. The deflection angle in the lab frame is
() = arctan[(v~ sina)/(v~ cosa + v)],
but
v =
mlVO/(ml
+
m2)
andv~
=
Vo -v
so v~ =
m2VO/(ml
(}= arctan[(m2 sina)/(m2
+
m2)
and
cosa + ml)].
(c) (} is a maximum when (cosa + ml/m2)/sina
is a minimum, which happens when cosa =
-ml/m2 if ml S m2. Then [(m2 sina)/(m2 cosa + ml)] can have any value between -00 and 00,
so (} can be between O and 7r.
(a) Ifml > m2 then (cosa+ml/m2)/sina
is a minimum when cosa = -m2/ml, then
[(m2 sina)/(m2
If tan(} = m2/ vm~ -m~
cosa + ml)] = m2/Vm~
-m~.
then m¡ is like a hypotenuse and m2 the opposite side. Then
(b) We need to change to the center of m8SSsystem. Since both particles have the same m8Ss,
the conservation of momentum problem is effectively the same 8Sa (vector) conservation of velocity
problem. Since one of the particles is originally at rest, the center of m8SSmoves with speed
Vcm = Vli/2. In the figure below the center of m8Ss velocities are primed; the transformation
velocity is Vt.
85
Vt
Note that since Vt = v' li = v' 2i = v' lf = v' 2f the entire problem can be inscribed in a rhombus.
The diagonals of the rhombus are the directions of Vlf and V2fj note that the diagonals of a rhombus
are necessarilyat right angles!
~
( a) The speed of the bullet after leaving the first block but before entering the second can
be determined by momentum conservation.
Pf,bl
mblVf,bl
(l.78kg)(l.48m/s)+(3.54x
+
Pf
=
Pi,
+ Pf,bu
=
Pi,bl
=
mblVi,bl
=
(l.78kg)(0)+(3.54x
mbuVf,bu
lO-3kg)(l.48m/s)
+ Pi,bu,
+
mbuVi.bu,
lO-3kg)Vi,bu,
which has solution Vi,bl = 746m/s.
(b) We do the same steps again, except applied to the first block,
mblVf,bl
(l.22kg)(0.63mis)+
Pf
=
Pi,
Pf,bl + Pf,bu
=
Pi,bl
+ mbuVf,bu
=
mblVi,bl
(3.54 x lO-3kg)(746 mis)
=
+Pi,bu,
+
mbuVi,bu,
(l.22kg)(0)+(3.54x
lO-3kg)Vi,bu,
which has solution Vi,bl = 963 m/s.
P6-20 The acceleration of the block down the ramp is al = gsin(22°). The ramp has a length of
d = h/sin(22°), so it takes a time tl = J2d7a;- = J2h79/sin(22°) to reach the bottom. The speed
when it reachesthe bottom is Vl = altl = V2g7i:.Notice that it is independent of the angle!
The collision is inelastic, so the two stick together and move with an initial speed of V2 =
mlVl/(ml + m2). They slide a distance x before stopping; the averagespeed while decelerating is
vav = v2/2, so the stopping time is t2= 2X/V2 and the deceleration is a2 = V2/t2 = v~/(2x). Ifthe
retarding force is f = (ml + m2)a2, then f = JLk(ml + m2)g. Glue it aIl together and
86
P6-21 (a) For an object with initial speedv and deceleration -a which travels a distance x before
stopping, the time t to stop is t = v/a, the average speed while stopping is v/2, and d = at2/2.
Combining, v = V2ax. The deceleration in this case is given by a = Jl,kg.
Then just after the collision
VA
=
V2(0.130)(9.81m/s2)(8.20m)=4.57m/s,
while
VE = V2(0.130)(9.81 m/s2)(6.10m)
(b) Va = [(1100 kg)(4.57m/s)
+ (1400kg)(3.94m/s)]/(1400kg)
87
= 3.94m/s,
= 7.53m/s.
E7-1
Xcm = {7.36x lO22kg){3.82 x lO8m)/{7.36x
than the radius of the Earth.
E7-2
Ir the particles
are l apart
lO22kg + 5.98x lO34kg) = 4.64x lO6m. This is less
then
Xl = mll(ml
+ m2)
is the distance from particle 1 to the center oí mass and
X2 = m2l(ml
+ m2)
is the distance from particle 2 to the center of mass. Divide the top equation by the bottom and
Xl/X2
~
The center of mass velocity
"cm
=
mlVl
mi
.(2210
= ml/m2.
is given by Eq. 7-1,
+m2V2
+ m2
kg)(105
,
km/h)
+ (2080
kg)(43.5
km/h)
=
75.2
km/h.
(2210 kg) + (2080 kg)
E7-4
They will meet at the center of mass, so
Xcm = (65kg)(9.7m)/(65kg+42kg)
= 5.9m.
E7-5
(a) No external forces, center of mass doesn't move.
(b) The collide at the center of mass,
Xcm = (4.29 kg)(1.64m)/(4.29kg
E7-6
+ 1.43kg) = 1.23m.
The range of the center of ma.ss is
R = v~sin2(}/g
= (466m/s)2sin(2
x 57.4°)/(9.8lm/s2)
= 2.0lxlO4m.
Half lands directly underneath the highest point, or 1.00x 104m. The other piece must land at x,
such that
2.01 x 104 m = (1.00 x 104 m + x)/2;
then x = 3.02 x 104 m.
~
The center of mass of the boat + dog doesn't move becausethere are no external forces on
the system. Define the coordinate system so that distances are measured from the shore, so toward
the shore is in the negative x direction. The changein position of the center of mass is given by
~Xcm
=
'
md~Xd
+ mb~Xb
=
O,
-1-
Both ~Xd and ~Xb are measured with respect to the shore; we are given ~Xdb = -8.50 ft, the
displacement of the dog with respect to the boat. But
~Xd
=
~Xdb
88
+
~Xb.
~
Since we want to find out about the dog, we'll substitute for the boat's displacementt
O = ~~~b
(~Xd -~Xdb)
md + mb
Rearrange and solve for ~Xd. Use W = my and multiply the top and bottom of the expressionby
y. Then
6.90 ft.
E1-8 Richardhag too much time on his hands.
The center of mags of the system is Xcm away from the center of the boat. Switching seats is
effectively the same thing ag rotating the canoe through 180°, so the center of mags of the system
hag moved through a distance of 2xcm = 0.412m. Then Xcm= 0.206m. Then
Xcm= (Ml-
ml)/(M + m + mc) = 0.206m,
where 1 = 1.47m, M = 78.4kg, mc = 31.6kg, and m is Judy's mags. Rearrange,
Mlm = .-=
(M + mc)Xcm
(78.4kg)(1.47m) ~ (78.4kg+ 31.6kg) (0.206m )
+Xcm
(1.47m)
+ (0.206
55.2kg.
m)
E7-9 It takes the man t = (18.2m)/(2.08m/s) = 8.75s to walk to the front of the boat. During
this time the center ofmass ofthe system has moved forward x = (4.16m/s)(8.75s) = 36.4m. But
in walking forward to the front of the boat the man shifted the center of mass by a distance of
(84.4kg)(18.2m)/(84.4kg + 425kg) = 3.02m, so the boat only traveled 36.4m -3.02m = 33.4m.
E7-10
Do each coordinate separately.
Xcm
=
(3kg)(O) + (8kg)(lm)
+ (4kg)(2m)
07m
(3kg) + (8kg) + (4kg)
and
E7-12 The velocity components of the center of mass at t = 1.42s are Vcm,x = 7.3m/s and
Vcm,y= (10.0m/s) -(9.81m/s)(1.42s) = -3.93m/s. Then the velocity components ofthe "other"
piece are
V2,x =
[(9.6kg)(7.3m/s)
-(6.5kg)(11.4m:/s)]/(3.1
kg)
= -1.30m/s.
and
V2,y =
[(9.6kg)(-3.9m/s)
-(6.5kg)(-4.6m/s)]/(3.1kg)
89
= -2.4m/s.
~
The center of mass should lie on the perpendicular bisector of the rod of mass 3M. We
can view the system as having two parts: the heavy rod of mass 3M and the two light rods each of
mass M. The two light rods have a center of mass at the center of the square.
Both of these center of massesare located along the vertical line of symmetry for the object.
The center of mass of the heavy bar is at Yh,cm= O, while the combinedcenter of mass of the two
light bars is at Yl,cm= L/2, where down is positive. The center of mass of the system is then at
y cm =
2Myl,cm
':' .r
+
.~
3Myh,cm
.r
-
~
=L/5.
5
E7-14
The two slabs have the same volume and have mass mi = PiV.
located at
-
{5 5
-.cm
) {7.85
{7.85
g/cm3)
3
g/cm
-{2.70
The center oí mass is
g/cm3)
-
-.cm2 68
3
) + {2.70
g/cm
)
from the boundary inside the iron; it is centered in the y and z directions.
E7-15
Treat the íour sides oí the box as one thing oí mass 4m with a mass located l/2 above the
base. Then the center oí mass is
Zcm = ([/2) (4m)/(4m
+ m) = 2[/5 = 2(0.4m)/5
= O.16m,
Xcm = y cm = O.2m.
E7-16 One piece moves off with momentum m(31.4m/s)i, another moves off with momentum
2m(31.4m/s)j. The third piece must then have momentum -m(31.4m/s)i -2m(31.4m/s)j
and
velocity -(1/3)(31.4m/s)i -2/3(31.4m/s)j
= -10.5m/si -20.9m/sj.
The magnitude of V3 is
23.4m/s and direction 63.3° away from the lighter piece.
E7-17
It will take an impulse of (84.7kg)(3.87mis)
= 328kg.mis to stop the animal. This would
come from firing n bullets where n = (328 kg .mis)i[(O.O126 kg)(975 mis) = 27.
E7-18
Conservation of momentum for firing one cannon ball of mags m with muzzle speed Vc
forward out of a cannon on a trolley of original total mags M moving forward with original speed
Vo is
Mvo = (M -m)Vl
+ m(vc + Vl) = MVl + mvc,
where Vl is the speed of the trolley after the cannonball is fired. Then to stop the trolley we require
n cannonballs be fired so that
n = (Mvo)/(mvc)
= [(3500 kg)(45m/s)]/[(65kg)(625m/s)]
= 3.88,
so n = 4.
~
Label the velocities of the various containers as Vk where k is an integer between one and
twelve. The mass of each container is m. The subscript "g" refers to the goo; the subscript k refers
to the kth container.
The total momentum before the collision is given by
...~
p
=
L..,
...
mVk,
+ mgVg,i = 12mvcont"cm+ mgVg,i'
k
90
We are told, however, that the initial velocity of the center of mass of the containers is at rest, so
the initial momentum simplifies to p = mgVg,i, and has a magnitude of 4000 kg.m/s.
(a) Then
p
(4000kg.m/s)
=
= 3.2m/s.
12m+mg
12(100.0kg) + (50kg)
(b) It doesn't matter if the cord breaks, we'll get the same answer for the motion of the center
of mass.
v~~
=
E7-20
(a) F = (3270 m/s)(480 kg/s) = l.57x lO6N.
(b) m = 2.55x lO5kg -(480 kg/s)(250s) = l.35x lO5kg.
(C) Eq. 7-32:
Vf = ( -3270 m/s) ln(1.35 x 105kg/2.55 x 105kg) = 2080 m/s.
E7-22
Eq.
7-32
rearranged:
Mf
~e e-l~tJ/tJrell
:;::¡ e-(22.6m/s)/(1230m/s)
=
0.982.
Mi
The fraction of the initial mass which is discarded is 0.0182.
E7-23 The loaded rocket has a weight of (1.11xl05kg)(9.81 m/s2) = 1.09xl06N; the thrust must be
at least this large to get the rocket offthe ground. Then v 2::(1.09xl06N)/(820kg/s) = 1.33xl03m/8
is the minimum exhaust speed.
E7-24 The acceleration down the incline is (9.8 m/s2) sin(26°) = 4.3 m/s2. It will take t =
v2(93m)/(4.3m/s2) = 6.6s. The sand doesn't affect the problem, so long as it only "leaks" out.
~
We'll use Eq. 7-4 to solve this problem, but since we are given weights instead of mass
we'll multiply the top and bottom by 9 like we did in Exercise 7-7. Then
-mlVl=Vcm
+ m2V29 = W1Vl + W2V2
Wl + W2
9
m¡ + m2
Now for the numbers
P7-1
(a) The balloon moves clown so that the center of mass is stationary;
o = MVb +mvm = MVb +m(v + Vb),
or Vb = -mv/(m + M).
(b) When the man stops so does the balloon.
91
P7-2
(a) The center oí ma.ss is midway between them.
(b) Mea.sure from the heavier ma.ss.
Xcm = (0.0560 m)(0.816kg)/(1.700kg)
= 0.0269 m,
which is 1.12 mm closer to the heavier mass than in part (a).
(c) Think Atwood 's machine. The acceleration of the two massesis
a = 2Am9/(ml
+ m2) = 2(0.034kg)9/(1.700kg)
= 0.04009,
the heavier going clown while the lighter moves up. The acceleration of the center of mass is
acm = (am¡
-am2)/(m¡
+ m2) = (O.O400g)2(O.O34kg)g/(1.700kg)
0.001609.
~
This is a glorified Atwood's machine problem. The total mass on the right side is the mass
per unit length times the length, mr = >.x; similarly the mass on the left is given by ml = >.(L -x) .
Then
a=
m2
-m¡
m2 + ml
g-
).x-
).(L-
x)
2x
g
,
.,-
AX -¡- A~1.1-X)
=
,
-L
1.1 g
,-
-
which salves the prablem. The acceleratian is in the rlirectian ar the sirle ar length X ir X > L/2.
P7-4 (a) Assume the car is massless.Then moving the cannonballs is moving the center of mass,
unless the cannonballs don't move but instead the car does. How far? L.
(b) Once the cannonballs stop moving so does the rail car.
~
By symmetry, the center of mass of the empty storage tank should be in the very center,
along the axis at a height Yt,cm= H /2. We can pretend that the entire mass of the tank, mt = M ,
is located at this point.
The center of massof the gasolineis also, by symmetry, located along the axis at half the height of
the gasoline, Yg,cm= x /2. The mass, if the tank were filled to a height H, is m; assuming a uniform
density for the gasoline, the mass present when the level of gas reachesa height x is mg = mx / H .
(a) The center of mass of the entire system is at the center of the cylinder when the tank is full
and when the tank is empty. When the tank is half full the center of mass is below the center. So as
the tank changesfrom full to empty the center of mass drops, reachessome lowest point, and then
rises back to the center of the tank.
(b) The center of mass of the entire system is found from
mx2+MH2
2mx+2MH.
Take the derivative:
~
dx
=
m
(mx2
+
2xMH
-MH2)
(mx+MH)2
Set this equal to zero to find the minimum; this means we want the numerator to vanish, or mx2 +
2xMHMH2 = O. Then
x=
-M+vM2+mM
~
Cc
m
92
H.
The center oí masswill
be located along symmetry axis. Call this the x axis. Then
Xcm =
:b J xdm,
4R
3;-'
P1-1
(a) The components of the shell velocity with respect to the cannon are
v~ = (556m/s)cos(39.00) = 432m/s and v~ = (556m/s)sin(39.00) = 350m/s.
The vertical component with respect to the ground is the same, Vy = v~' but the horizontal component is found from conservation of momentum:
M(vx
so Vx = (1400kg)(432m/s)(70.0kg+
(b) The direction
P7-8
-v~)
1400kg)
is (} = arctan(350/411)
v = (2870 kg)(252 m/s)/(2870kg
+ m(vx)
= 411m/s.
= O,
The resulting
speed is v = 540m/s.
= 40.4°.
+ 917kg) = 191 m/s.
P7-9 It takes (1.5m/s)(20kg) = 30N to accelerate the luggage to the speed of the belt. The
people when taking the luggage off will (on average) a1soneed to exert a 30 N force to remove it;
this force (becauseof friction) will be exerted on the belt. So the belt requires 60 N of additional
force.
P7-10
(a) The thrust must be at least equal to the weight, so
dmldt = (5860kg)(9.81m/s2)/(1170m/s) = 49.1kg/s.
(b) The net force on the rocket will need to be F = (5860kg)(18.3m/s2) = 107000N. Add this
to the weight to find the thrust, so
dmldt = [107000N+ (5860kg)(9.81m/s2)]/(1170m/s) = 141kg/s
~
Consider Eq. 7-31. We want the barges to continue at constant speed, so the left hand
side of that equation vanishes. Then
We are told that the frictional force is independent ofthe weight, since the speeddoesn't changethe
frictional force should be constant and equal in magnitude to the force exerted by the engine before
the shoveling happens. Then ¿ F ext is equal to the additional force required from the engines. We'll
call it P.
The relative speed of the coal to the faster moving cart has magnitude: 21.2- 9.65 = 11.6
km/h= 3.22 m/s. The mass flux is 15.4 kg/s, so p = (3.22 m/s)(15.4kg/s) = 49.6 N. The faster
moving cart will need to increase the engine force by 49.6 N. The slower cart won't need to do
anything, because the coal left the slower barge with a relative speed of zero according to our
approximation.
93
P7-12 (a) Nothing is ejected from the string, so Vrel= O. Then Eq. 7-31 reducesto mdv/dt = F ext.
(b) Since F ext is from the weight of the hanging string, and the fraction that is hanging is y / L ,
F ext = mgy/ L. The equation of motion is then ~y/dt2 = gy/ L.
(c) Take first derivative:
~dt
-2(
Yo
J97L)
( e.¡g¡¡;t
-e-ViiTit)
,
and then secondderivative,
d2y
dt2
Substitute
into equation
of motion.
= ~ ( V97i)2
( eat
It works!
Note that
2
94
+e-at
).
when t = O we have y = Yo.
~
An n-dimensional object can be oriented by stating the position of n different carefully
chosenpoints Pi inside the body. Since each point has n coordinates, one might think there are n2
coordinates required to completely specify the position of the body. But if the body is rigid then
the distances between the points are fixed. There is a distance dij for every pair of points Pi and
Pj. For each distance dij we need one fewer coordinate to specify the position of the body. There
are n(n -1)/2 ways to connect n objects in pairs, so n2 -n(n -1)/2 = n(n + 1)/2 is the number
of coordinates required.
E8-2
(1 rev/min)(27r
rad/rev)/(60
s/min) = 0.105 rad/s.
E8-3
(a) úJ = a + 3bt2 -4ct3.
(b) a = 6bt -12t2.
E8-4
(a) The radius is r = (2.3 x 1041y)(3.0 x 108m/s) = 6.9 x 1012m .y/s.
revolution is t = (27r6.9 x 1012m .y/s)/(250 x 103m/s) = 1.7x 108y.
(b) The Sun has made 4.5 x 109y/1.7 x 108y = 26 revolutions.
~
(a)
The time to make one
Integrate.
(Alz = "'o + lt
( 4at3 -3bt2)
dt = (Alo+ at4 -bt3
(b) Integrate, again.
[t
[t
A(} = lo (LIzdt = lo
((LIO + at4 -bt3)
I
dt = (LIot + sat5
I
-¡bt4
E8-6 (a) (1 rev/min)(27r rad/rev)/(60 s/min) = 0.105 rad/s.
(b) (1 rev/h)(27r rad/rev)/(3600 s/h) = 1.75xI0-3 rad/s.
(c) (1/12 rev/h)(27r rad/rev)/(3600 s/h) = 1.45x 10-3 rad/s.
E8-7
85 mi/h = 125 ft/8. The hall take8 t = (60 ft)/(125
make8 (30 rev/8)(0.488) = 14 revolution8 in that time.
E8-8
It take8 t = V2(10iñ)/(9.81m/82)
= 1.438 to falll0
then iJ) = (2.5)(27r rad)/(1.438) = 11 rad/8.
ft/8)
= 0.488 to reach the plate.
It
m. The average angular velocity is
~
(a) Since there are eight spokes, this means the wheel can make no more than 1/8 of a
revolution while the arrow traverses the plane of the wheel. The wheel rotates at 2.5 rev/s; it
makes one revolution every 1/2.5 = 0.4 s; so the arrow must pass through the wheel in less than
0.4/8 = 0.05 s.
The arrow is 0.24 m long, and it must move at least one arrow length in 0.05 s. The corresponding
minimum speed is (0.24 m)/(0.05 s) = 4.8 m/s.
(b) It does not matter where you aim, becausethe wheel is rigid. It is the angle thiough which
the spokeshave turned, not the distance, which matters here.
95
E8-10 We look for the times when the Sun, the Earth, and the other planet are collinear in some
specified order.
Since the outer planets revolve around the Sun more slowly than Earth, after one year the Earth
has returned to the original position, but the outer planet has completed less than one revolution.
The Earth will then "catch up" with the outer planet beforethe planet has completed a revolution.
If (}E is the angle through which Earth moved and (}p is the angle through which the planet moved,
then (}E = (}p + 211'
, since the Earth completed one more revolution than the planet.
If IJJp is the angular velocity of the planet, then the angle through which it moves during the
time Ts (the time for the planet to line up with the Earth). Then
(}E =
c.JETs =
c.JE =
(}p + 27r,
c.JpTs+ 27r,
c.Jp+27r/Ts
The angular velocity of a planet is w = 27r/T , where T is the period of revolution.
into the last equation above yields
l/TE
= l/Tp
Substituting
this
+ l/Ts.
~
We look for the times when the Sun, the Earth, and the other planet are collinear in some
specified order .
Since the inner planets revolve around the Sun more quickly than Earth, after one year theEarth
has returned to the original position, but the inner planet has completed more than one revolution.
The inner planet must then have "caught-up" with the Earth before the Earth has completed a
revolution. If (}E is the angle through which Earth moved and (}p is the angle through which the
planet moved, then (}p = (}E + 27r, since the inner planet completed one more revolution than the
Earth.
96
If IA)pis the angular velocity of the planet, then the angle through which it moves during the
time Ts (the time for the planet to line up with the Earth). Then
(}p =
IA)pTs =
(}E+27r,
IA)ETs+ 27r,
IA)p =
IA)E+ 27r/Ts
The angular velocity of a planet is lA)= 27r/T , where T is the period of revolution. Substituting this
into the last equation above yields
l/Tp
= l/TE + l/Ts.
E8-12
(a) a = (-78 rev/min)/(0.533
min) = -150 rev/min2.
(b) Average angular speecl while slowing clown is 39 rev/min,
21 rev.
so (39 rev/min)(0.533
min) =
E8-13
(a) a = (2880 rev/min -1170 rev/min)/(0.210
min) = 8140 rev/min2.
(b) Average angular speed while accelerating is 2030 rev/min, so (2030 rev/min)(0.210
425 rev.
min) =
E8-14
Find area under curye.
~(5 min+2.5
~
(a) iJ.JOz= 25.2
rad/s;
min)(3000 rey/min)
iJ.Jz= O; t
= l.l3xlO4
rey.
19.7 s; and Qz and <Pare unknown. From Eq. 8-6,
(.,)z =
(.,)Oz+ azt,
(O)
=
(25.2 rad/s)
az
=
-1.28
(b) We use Eq. 8- 7 to find the angle through
+ az(19. 7 s),
rad/s2
which
97
the wheel rotates.
(c)
<P =
248
rad~
=
39.5
rey.
21r rau
E8-16
(a) a = (225 rev/min -315 rev/min)/(1.00
min) = -90.0 rev/min2
(b) t= (0- 225 rev/min)/( -90.0 rev/min2) = 2.50 min.
(c) (-90.0 rev/min2)(2.50min)2/2
+ (225 rev/min)(2.50 min) = 281 rev.
E8-17 (a) The average angular speed was (90 rev)/(15s) = 6.0 rev/s. The angular speed at the
beginning of the interval was then 2(6.0 rev/s) -(10 rev/s) = 2.0 rev/s.
(b) The angular acceleration was (10 rev/s-2.0 rev/s)/(15s) = 0.533 rev/s2. The time required
to get the wheel to 2.0 rev/s was t = (2.0 rev/s)/(0.533 rev/s2) = 3.8s.
E8-18
(a) The wheel will rotate through an angle <I>where
t/> =
(563
cm)/(8.14
cm/2)
= 138 rad.
~
(a) We are given I/>= 42.3 rev= 266 rad, (AlOz= 1.44 rad/s, and (Alz= O. Assuming a
uniform deceleration, the average angular velocity during the interval is
1
lA)av
,z = 2 (lA)oz+ IA)z)= 0.72 rad/ s.
Then the time taken for deceleration is given by <t>
= lA)av
,zt, so t = 369 s.
(b) The angular acceleration can be found from Eq. 8-6,
IAJz
(O)
Ctz
=
(AJOz+ azt,
=
(1.44
rad/s)
+ az(369
=
-3.9
x 10-3
rad/s2.
s),
(c) We'll solve Eq. 8-7 for t,
<1> =
(133 rad)
1
2
<1>0+ IJJozt + ?Qzt
,
;,;, {0) + {1.44 rad/s)t
o =
-133+
{1.44s-1)t
1
+ 2{ -3.9
-{-1.95
x 10-3 rad/s2)t2,
x 10-3s-2)f.
Solving this quadratic expression yields two answers: t = 108 s and
=
630
s.
E8-20
The angular acceleration is a = (4.96 rad/s)/(2.33s)
= 2.13 rad/s2. The angle through
which the wheel turned while accelerating is <t>= (2.13 rad/s2)(23.0s)2/2
= 563 rad. The angular
speed at this time is (¿)= (2.13 rad/s2)(23.0s) = 49.0 rad/s. The wheel spins through an additional
angle of (49.0 rad/s)(46s -23s) = 1130 rad, for a total angle of 1690 rad.
E8-21
/JJ= (14.6 m/s)/(110m)
= 0.133 rad/s.
E8-22
The linear acceleration is (25m/s -12m/s)/(6.2s)
a = (2.1 m/s2)/(0.75m/2)
= 5.6 rad/s.
98
=2.1 m/s2, The angular acceleration is
~
(a) The angular speed is given by VT = i.l)r. So i.I) = vT/r
km) = 8.91 rad/hr. That's the same thing as 2.48x10-3 rad/s.
(b) aR = i.l)2r = (8.91 rad/h)2(3220 km) = 256000 km/h2, or
aR = 256000
km/h2(1/3600
h/s)2(1000
m/km)
= (28,700 km/hr)/(3220
= 19.8m/s2.
( c) If the speed is constant then the tangential acceleration is zero, regardless of the shape of the
trajectory!
E8-24
The bar needs to make
1.50 cm)(12.0 turns/cm)
Thi8 will happen i8 (18 rev)/(237
rev/min)
= 18 turns.
= 4.568.
E8-25 (a) The angular speed is úJ= (21rrad)/(86400s) = 7.27x10-5 rad/s.
(b) The distance from the polar axis is r = (6.37x106m) cos(40°) = 4.88x106m. The linear speed
is then v = (7.27x 10-5 rad/s)(4.88x 106m) = 355m/s.
(c) The angular speed is the same as part (a). The distance from the polar axis is r = (6.37x
106m)cos(O°)= 6.37x106m. The linear speedis then v = (7.27x10-5 rad/s)(6.37x106m) = 463m/s.
E8-26
(a) aT = (14.2 rad/s2)(O.O283m) = O.402m/s2.
(b) F\111speed is iA)= 289 rad/s. aR = (289 rad/s)2(O.O283m) = 2360m/s2.
(c) It takes
t = (289 rad/8)/(14.2 rad/82) = 20.48
to get up to full 8peed. Then x = (0.402m/82)(20.48)2/2
point on the rim mOVe8.
= 83.6m i8 the di8tance through which a
~
(a) The pilot seesthe propeller rotate, no more. So the tip ofthe propeller is moving with
a tangential velocity of VT = wr = (2000 rev/min)(27r rad/rev)(1.5 m) = 18900 m/min. This is the
same thing as 315 m/s.
(b) The observer on the ground sees this tangential motion and sees the forward motion of
the plane. These two velocity components are perpendicular , so the magnitude of the sum is
.¡(315m/s)2 + (133m/s)2 = 342m/s.
E8-28
aT = aR when
ra
= r(¿)2 = r(at)2,
E8-29
(a) aR = rúJ2 = ra:2t2.
(b) arr = ra:.
(c) Since aR = aTtan(57.00),
or t =
Jl/(0.236
t = ..¡tan(57.00)/a:.
rad/82)
= 2.068.
Then
0.77 rad = 44.1°.
E8-30 (a) The tangential speed of the edge of the wheel relative axle is v = 27m/s. úJ =
(27m/s)/(0.38m) = 71 rad/s.
(b) The average angular speed while slowing is 71 rad/s/2, the time required to stop is then
t = (30 x 27rrad)/(71 rad/s/2) = 5.3s. The angular acceleration is then a = (-71 rad/s)/(5.3s) =
-13 rad/s.
(c) The car moves forward (27m/s/2)(5.3s) = 72m.
99
E8-31 Ves, the speedwould be wrong. The angular velocity ofthe small wheel would be 1.1)
: Vt/rs,
but the reported velocity would be v = l.l)rl = Vtrl/rs. This would be in error by a fraction
~v
= (72 cm) -1 = 0.16.
Vt
E8-32
~
(a) Square both equations and then add them:
X2 + y2 = (RcoS('¡)t)2 + (Rsin(,¡)t)2 = R2,
which is the equation for a circle of radius R.
(b) Vx = -RJoIJsin(¿)t
= -(¿)y; Vy = RJoIJcOS(¿)t
= (¿)x. Square and add, v = (¿)R.The direction is
tangent to the circle.
(b) ax = -RJoIJ2cOS(¿)t
= -(¿)2X; ay = -RJoIJ2sin(¿)t
= -(¿)2y. Square and add, a = (¿)2R. The
direction is toward the center.
~
(a) The object is "slowing down",
because
it
is
-A
rotating
about
V = ¡;j X R = (14.3 rad/s)k
v = (-26.2m/s)i.
the
z
axis
and
A
so a. = ( -2.66
we
are
given
A
x [(1.83 m)j + (1.26 m)k].
the
rad/s2)k.
direction
of
We know the direction
¡;j.
Then
A
from
A
Eq.
8-19,
But only the cross term k x j survives, so
(b) We find the acceleration from Eq. 8-21,
a
E8-34
=
Q x R +r..J x v,
=
( -2.66 rad/s2)k x [(1.83 m)j + (1.26 m)k] + (14.3 rad/s)k
=
( 4.87 m/s2)i + ( -375 m/s2)j.
(a) F = -2miiJ x v = -2Tn¡,}vcos8, where 8 is the latitude.
F = 2(12kg)(27r
rad/86400s)(35m/s)
cos(45°)
x ( -26.2 m/s)i,
Then
= 0.043 N,
and is directed west.
(b) Reversing the velocity will reversethe direction, so east.
(c) No. The Coriolis force pushes it to the west on the way up and gives it a westerly velocity;
on the way down the Coriolis force slows down the westerly motion, but does not push it back east.
The object lands to the west of the starting point.
P8-1
(a) (¡.J= (4.0 rad/s) -(6.0 rad/s2)t+ (3.0 rad/s)t2. Then (¡.J(2.0s) = 4.0 rad/s and (¡.J(4.0s)=
28.0 rad/s.
(b) a&v = (28.0 rad/s -4.0 rad/s)/(4.0s -2.0s) = 12 rad/s2.
(C) a = -(6.0 rad/s2) + (6.0 rad/s)t. Then a(2.0s) = 6.0 rad/s2 and a(4.0s) = 18.0 rad/s2.
P8-2
If the wheel really does moye counterclockwise at 4.0 rey /min, then it turns through
(4.0 rev/min)/[(60
s/min)(24
frames/s)] = 2.78x10-3
rev/frame.
This means that a spoke has moved 2.78 x 10-3 rev. There are 16 spokes each located 1/16 of a
revolution around the wheel. If instead of moving counterclockwise the wheel was instead moving
clockwise so that a different spoke had moved 1/16 rev -2.78 x 10-3 rev = 0.0597 rev, then the
same effect would be present. The wheel then would be turning clockwise with a speed of YJ =
(0.0597 rev)(60 s/min)(24 frames/s) = 86 rev/min.
100
P8-3
(a) In the diagram below the Earth is shown at two locations a day apart. The Earth rotates
clockwise in this figure.
Note that the Earth rotates through 27rrad in order to be correctly oriented for a complete
sidereal day, but becausethe Earth has moved in the orbit it needs to go farther through an angle
(} in order to complete a solar day. By the time the Earth has gone all of the way around the sun
the total angle (} will be 27rrad, which means that there was one more sidereal day than solar day.
(b) There are (365.25 d)(24.000 h/d) = 8.7660x 103 hours in a year with 265.25 solar days. But
there are 366.25 sidereal days, so each one has a length of 8.7660x 103/366.25= 23.934 hours, or 23
hours and 56 minutes and 4 seconds.
P8-4
(a) The period is time per complete rotation, so (,¡J= 21r/T.
(b) a = ~(,¡J/~t, so
a =
~
= -2.30xlO-9rad/s2.
(3.16x 1078)
(c) t = (21r/0.033s)/(2.30x10-9rad/s2)
(d) 21r/To = 21r/T -at, or
(0.0338)2
= 8.28x101os, or 2600 years.
=
0.024
s.
~
The final angular velocity during the acceleration pha8e is (.cJz= azt = (3.0 rad/s)(4.0
s) = 12.0 rad/s. Since both the acceleration and deceleration pha8es are uniform with endpoints
(.cJz
= 0, the average angular velocity for both pha8es is the same, and given by half of the maximum:
(.cJav,z
= 6.0 rad/s.
lO1
The angle through which the wheel turns is then
<I>= ú)av,zt = (6.0 rad/s)(4.1 s) = 24.6 rad.
The time is the total for both phases.
(a) The first student seesthe wheel rotate through the smallest angle less than one revolution;
this student would have no idea that the disk had rotated more than once. Since the disk moved
through 3.92 revolutions, the first student will either assumethe disk moved forward through 0.92
revolutions or backward through 0.08 revolutions.
(b) According to whom? We've already answeredfrom the perspective of the secondstudent.
P8-6
(a)
(b)
(c)
(d)
!.A)= (0.652 rad/s2)t and a = (0.652 rad/s2).
!.A)= (0.652 rad/s2)(5.60s) = 3.65 rad/s
VT = !.A)r= (3.65rad/s)(10.4m)
= 38m/s.
aT = ar = (0.652 rad/s2)(10.4m) = 6.78m/s2.
aR = !.A)2r= (3.65 rad/s)2(10.4m) = 139m/s2.
P8-7 {a) (¿)= {27rrad)/{3.16x 107s) = 1.99x10-7 rad/s.
{b) VT = (¿)R= {1.99x 10-7 rad/s){1.50 x 1011m)= 2.99x 104m/s.
{c) aR = (¿)2R= {1.99x10-7 rad/s)2{1.50x 1011m)= 5.94x10-3m/s2.
P8-8 (a) a = ( -156 rev/min)/(2.2 x 60 min) = -1.18 rev/min2.
(b) The average angular speed while slowing down is 78 rev /min, so the wheel turns through
(78 rev/min)(2.2 x 60 min) = 10300revolutions.
(c) aT = (27rrad/rev)(-1.18 rev/min2)(0.524m) = -3.89 m/min2. That's the same as -1.08x
10-3m/s2.
(d) aR = (27rrad/rev)(72.5 rev/min)2(0.524m) = 1.73 x 104 m/min2. That's the same as
4.81m/s2. This is so much larger than the aT term that the magnitude of the total linear acceleration is simply 4.81m/s2.
~
(a) There are 500 teeth (and 500 spacesbetween these teeth); so disk rotates 27r/500 rad
between the outgoing light pulse and the incoming light pulse. The light traveled 1000 m, so the
elapsedtime is t = (1000m)/(3 x 108m/s) = 3.33x 10-6s.
Then the angular speed of the disk is (¡)z= 4>/t = 1.26x 10-2 rad)/(3.33 x 10-6s) = 3800 rad/s.
(b) The linear speed of a point on the edge of the would be
VT = (¿)R= (3800 rad/s)(0.05 m) = 190m/s.
P8-10 The linear acceleration of the belt is a = aAr A. The angular acceleration of C i8 ac =
a/rc = aA(rA/rc). The time required for C to get up to speed is
t = (27rrad/rev)(100 rev/min)(1/60 min/8) = 16.4s.
(1;60 rad/s2)(10.0/25.0)
P8-11
{a) The final angular speed is LLIo= {130 cm/s)/{5.80 cm) = 22.4 rad/s.
{b) The recording area is 7I"{Ro2-Ri2), the recorded track has a length 1 and width w, so
1=
71"[{5.80
cm)2-{2.50 cm2)J= 5.cm.
38 x 105
{1.60 x 10-4 cm)
(c) Playing time is t = (5.38x 105 cm)/(130 cm/s) = 41408, or 69 minutes.
102
P8-12
The angular position is given by 4>= arctan(vt/b).
(¿}--b2
The derivative (Maple!) is
vb
+ v2t2 ,
and is directed up. Take the derivative again,
a=
but
is clirectecl
2bv3t
(b2 + v2t2)2
,
clown.
~
(a) Let the rocket sled move along the line x = b. The observer is at the origin and
seesthe rocket move with a constant angular speed, so the angle made with the x axis increases
according to (} = ú.lt. The observer, rocket, and starting point form a right triangle; the position y
of the rocket is the opposite side of this triangle, so
tan(} = y/b implies y = b/tanr..Jt.
We want to take the derivative oí this with respect to time and get
v(t) = i.J)b/cos2(i.J)t).
(b) The speed becomesinfinite (which is clearly unphysical) when t = 7r/2(AJo
103
{a) First, F = {5.0 N)i.
f'
[yFz -zF1I]i
+ [zF:I; -xFz]j
[y(O) -(0)(0)]i
[-yF:1;]k
+ [xF1I -yF:1;]k,
+ [(O)F:I; -x(o)Jj
= -(3.0
m)(5.0
N)k
+ [x(o) -yF:1;]k,
= -(15.0
N. m)k.
(b) Now F = (5.0 N)]. Ignoring all zero terms,
f = [xFy]k
= (2.0 m)(5.0
N)k
= (10 N. m)k.
(c) Finally, F = ( -5.0 N)i.
T=
[-yFx]k
= -{3.0
m){-5.0
N)k
= {15.0
N.
m)k.
E9-2
(a) Everything is in the plane of the page, so the net torque will either be directed normal
to the page. Let out be positive, then the net torque is T = rlF1 sin (}1 -r2F2 sin (}2.
(b) T = (1.30m)(4.20N) sin(75.00) -(2.15 m)(4.90N) sin(58.00) = -3.66N .m.
E9-4
Everything is in the plane of the page, so the net torque will either be directed normal to
the page. Let out be positive, then the net torque is
T = (8.0m)(10N)sin(45°)
-(4.0m)(16N)sin(900)
+ (3.0m)(19N)sin(200)
=
12N .m.
~
Since r and s lie in the xy plane, then t = r x s must be perpendicular to that plane, and
can only point along the z axis.
The angle between rand sis 320°- 85° = 235°. So Itl = r8lsin(JI = (4.5)(7.3)lsin(235°)1 = 27.
Now for the direction of t. The smaller rotation to bring r into s is through a counterclockwise
rotation; the right hand rule would then show that the cross product points along the positive z
direction.
E9-6
a = {3.20)[cos{63.00)j + sin{63.00)k] and b = {1.40)[cos{48.00)i + sin{48.00)k].
=
axb
Then
(3.20) cos(63.0°)(1.40) sin(48.0°)¡
+(3.20) sin(63.0°)(1.40) cos( 48.0°)]
-(3.20) cos(63.0°)(1.40)cos(48.0°)k
1.51¡ + 2.67] -1.36k.
E9-7
b x a has magnitude absin<1>and points in the negative z direction. It is then perpendicular
to a, so c has magnitude a2bsin<1>. The direction of c is perpendicular to a but lies in the plane
containing vectors a and b. Then it makes an angle 7r/2 -<I> with b.
(a) In unit vector notation,
c
[( -3)( -3) -( -2)(1)]¡
+ [(1)(4) -(2)(
-3)]]
+ [(2)( -2) -( -3)(4)]k,
11¡ + 10] + 8k.
(b) Evaluate arcsin[li
x bl/(ab)], finding magnitudes with the Pythagoras relationship:
<I> =
arcsin
(16.8)/[(3.74)(5.39)]
104
=
56°.
~
This exercise is a three dimensional
f
=
[yF% -zFy]¡
=
[( -2.0
m)(4.3
+[(1.5
m)( -2.4
=
E9-10
{b)
E9-11
N) -(1.6
N) -(
N)]¡ + [(1.6 m)(3.5
m)(3.5
nothing
is zero.
N) -(1.5
m)(4.3
N)]j
N)]k,
(-2.6N)k,
A
then f = (0.85 m)(2.6 N)j -(
A
then
(a) The rotational
f
=
(-0.36m)(-2.6N)i
A
-0.36
m)(2.6 N)k
A
-(0.54m)(-2.6N)j
A
A
= 2.2 N. mj + 0.94 N. mk.
A
=
0.93N
.mi
A
+
1.4N
.mj.
inertia about an axis through the origin is
I=mr2
= (0.025kg)(0.74m)2
(b) a = (0.74 m)(22 N) sin(50°)/(1.4
E9-12
9-1, except
+ [xFy -yFx]k,
m)( -2.4
-2.0
of Ex.
[-4.8N.m]¡+[-0.85N.m]j+[3.4N.m]k.
-A
(a) F = (2.6 N)i,
-A
F =
+ [zFx -xF%]j
generalization
= 1.4x10-2kg.m2.
x 10-2kg .m2) = 890 rad/s.
{a) lo = {0.052kg){0.27m)2+{0.035kg){0.45m)2+{0.024kg){0.65m)2
= 2.1x10-2kg.m2.
{b) The center of mass is located at
-{0.052kg){0.27m)
Xcm -{0.052
+kg){0.035kg){0.45m)
+ {0.024kg){0.65m)
+ {0.035 kg) + {0.024
kg)
Applying the parallel axis theorem yields lcm = 2.1x10-2kg.m2
= 0.41 m.
-{0.11 kg){0.41 m)2 = 2.5x10-3kg.
m2.
~
(a) Rotational
We can use Eq. 9-10:
I = Lmnr~
inertia is additive so long as we consider the inertia about the same a;xis.
= (0.075 kg)(0.42m)2 + (O.O30kg)(0.65m)2 = O.O26kg.m2.
(b) No change.
E9-14
f = [(0.42m)(2.5N) -(0.65m)(3.6N)]k
= -1.29N .mk.
Using the result from E9-13,
a. = ( -1.29N .m k)/(0.026kg.m2)
= 50 rad/s2k. That's clockwise if viewed from above.
E9-15
(a) F = mi.lJ2r = (110kg)(33.5 rad/s)2(3.90m) = 4.81x105N.
(b) The angular acceleration is a = (33.5 rad/s)/(6.70s)
= 5.00 rad/s2. The rotational inertia
about the axis of rotation is I = (110 kg)(7 .80 m)2 /3 = 2.23 x 103kg .m2. T = I a = (2.23 x 103kg .
m2)(5.00 rad/s2) = 1.12x 104N .m.
E9-16
We can add the inertia.s for the three rods together ,
~
The diagonal distance from the axis through the center of mass and the axis through the
edge is h = v(a/2)2 + (b/2)2, so
I = Icm + Mh2 = bM
(a2 + b2) + M ((a/2)2 + (b/2)2) = (b
Simplifying, I = ~M (a2 + b2).
105
+ ¡) M (a2 + b2) .
E9-18 I = Icm + Mh2 = (0.56kg)(1.0m)2/12 + (0.56)(0.30m)2 = 9.7x 10-2kg .m2.
~
For particle one /1 = mr2 = mL2; for particle two 12 = mr2 = m(2L)2 = 4mL2.
rotational inertia of the rod is Irod = l(2M)(2L)2
= ~ML2. Add the three inertias:
(
8
=
5m+
The
,
3M~
E9-20
la) I = M R2 /2 = M(R/ J2)2.
(b) Let I be the rotational inertia. Assuming that k is the radius of a hoop with an equivalent
rotational inertia, then I = M k2 , or k = JI7M.
E9-21 Note the mistakes in the equation in the part (b) of the exercisetext.
(a) mn = M/N.
(b) Each piece has a thickness t = L / N, the distance from the end to the nth piece is Xn =
(n- 1/2)t = (n- 1/2)L/N. The axis of rotation is the center, so the distance from the center is
rn = Xn -L/2 = nL/N -(1 + 1/2N)L.
(c) The rotational inertia is
N
T
L
n=l
mnr~,
ML2
N
N3.L(n-1/2-N)2,
n=l
ML2
N
N3
.L
n=l
(n2 -(2N
+i}
6
N3
~
+ (N
+ 1/2)2)
,
,
( .N(N + 1)(2N
ML2.
+ l)n
ML2," ,~
2N3 -::r -2N3
n
N3
I
...'
-~2N+
j
1)~
2
+ (N+
1/2)2N
+N3
ML2/3.
E9-22
F = (46 N)(2.6
cm)/(13
cm) = 9.2N.
E9-23 Tower topples when center of gravity is no longer above base. Assuming center of gravity
is located at the very center of the tower, then when the tower leans 7.0 m then the tower falls. This
is 2.5 m farther than the present.
(b) f) = arcsin(7.0m/55m) = 7.3°.
E9-24 If the torque from the force is sufficient to lift edge the cube then the cube will tip. The
net torque about the edge which stays in contact with the ground will be T = Fd -mgd/2 if F is
sufficiently large. Then F ;:::mg /2 is the minimum force which will causethe cube to tip.
The minimum force to get the cube to slide is F ;:::JLsmg= (0.46)mg. The cube will slide first.
106
~
The ladder slips if the force of static friction required to keep the ladder up exceedsJ.t8N.
Equations 9-31 give us the normal force in terms of the massesof the ladder and the firefighter,
N = (m + M)g, and is independent of the location of the firefighter on the ladder. Also from Eq.
9-31 is the relationship between the force from the wall and the force of friction; the condition at
which slipping occurs is Fw ?; J.t8(m+ M)g.
Now go straight to Eq. 9-32. The a/2 in the secondterm is the location of the firefighter, who in
the example was halfway between the base of the ladder and the top of the ladder. In the exercise
we don't know where the firefighter is, so we'll replace a/2 with x. Then
-Fwh+Mgx+-=O
mga
3
is an expression for rotational equilibrium. Substitute in the condition of Fw when slipping just
starts, and we get
-(JLs(m + M)g) h + Mgx +
=o.
mga
3
Salve
this
far
x,
(9.3
m)
I
-0:- (45 kg)(7.6
,-- .\
m) = 6.6 m
This is the horizontal distance; the fraction of the total length along the ladder is then given by
x/a = (6.6 m)/(7.6 m) = 0.87. The firefighter can climb (0.87)(12m) = 10.4m up the ladder.
E9-26 (a) The net torque about the rear axle is (1360kg)(9.8m/s2)(3.05m-l.78m)-Ff(3.05m)
=
O, which hassolution Ff = 5.55x 103N. Each of the front tires support half of this, or 2.77x 103N.
(b) The net torque about the front axle is (1360kg)(9.8m/s2)(1.78m) -Ff(3.05m) = 0, which
has solution F f = 7.78x 103N. Each of the front tires support half of this, or 3.89x 103N.
E9-27
The net torque on the bridge about the end closest to the person is
(160 lb)L/4
+ (600 lb)L/2
-FfL
= 0,
which has a solution for the supporting force on the far end of F f = 340 lb.
The net force on the bridge is (160 lb)L/4+ (600 lb)L/2 -(340 lb) -F c = 0, so the force on the
close end of the bridge is F c = 420 lb.
E9-28
The net torque on the board about the left end is
Fr(1.55m)
-(142N)(2.24m)
-(582
N)(4.48m)
= O,
which has a solution for the supporting force for the right pedestal of F r = 1890N. The force on the
board from the pedestal is up, so the force on the pedestal from the board is down (compression).
The net force on the board is Fl+(1890N) -(142N) -(582 N) = 0, so the force from the pedestal
on the left is Fl = -1170N. The negative sign means up, so the pedestal is under tension.
~
We can assumethat both the force F and the force of gravity W act on the center of the
wheel. Then the wheel will just start to lift when
w x r + F x r = o,
or
Wsin() = F cos(),
107
where (} is the angle between the vertical (pointing down) and the line between the center of the
wheel and the point of contact with the step. The use of the sine on the left is a straightforward
application of Eq. 9-2. Why the cosine on the right? Because
sin(90°
-e)
=
cose.
Then F = W tan f}.We can express the angle f} in terms of trig functions, h, and r. r cos f} is the
vertical distance from the center of the wheel to the top of the step, or r -h. Then
Finally by combining the above we get
=w~
r-h
E9-30 (a) Assume that each of the two support points for the square sign experience the same
tension, equal to half of the weight of the sign. The net torque on the rod about an axis through
the hinge is
(52.3kg/2)(9.81m/s2)(0.95m) + (52.3kg/2)(9.81m/s2)(2.88m) -(2.88m)TsinO
= O,
where T is the tension in the cable and f} is the angle between the cable and the rod. The angle can
be found from f} = arctan(4.12m/2.88m)
= 55.00, so T = 416N.
(b) There are two components to the tension, one which is vertical, ( 416 N) sin(55.0°) = 341 N ,
and another which is horizontal, (416N) cos(55.0°) = 239N. The horizontal force exerted by the wall
must then be 239N. The net vertical force on the rod is F+ (341N) -(52.3kg/2)(9.81m/s2)
= 0,
which has solution F = 172 N as the vertical upward force of the wall on the rod.
E9-31
(a) The net torque on the rod about an axis through the hinge is
T = W(L/2)
cos(54.0°)
-TLsin(153.0°)
= O.
or T = (52.7 lb/2)(sin 54.0° / sin 153.0°) = 47.0 lb.
(b) The vertical upward force of the wire on the rod is Ty = Tcos(27.0°). The vertical upward
force of the wall on the rod is Py = W -Tcos(27.0°),
where W is the weight of the rod. Then
Py =
(52.71b)
-(47.0
lb)cos(27.0°)
= 10.81b
The horizontal force from the wall is balanced by the horizontal force from the wire.
(47.0 lb) sin(27.0°) = 21.3 lb.
Then
Px
=
E9-32
Ií the ladder is not slipping then the torque about an axis through the point oí contact with
the ground is
r = (W L/2) cos() -Nh/sin()
= O,
where N is the normal force ofthe edge on the ladder. Then N = WLcos() sin()/(2h).
N has two componentsj one which is vertically up, N y = N cos (), and another which is horizontal,
N x = N sin (). The horizontal force must be offset by the static friction.
The normal force on the ladder from the ground is given by
Ng
= W -NcosB
= W[l-
108
LCOS2 B sinB/(2h)].
The force of static friction can be as large as f = .UsN g, so
WLcos(j
sin2(j/(2h)
JLs = W[l -L COS2(j sin (j /(2h)]
L
=
cos
()
'2h -Lcos2
Put in the numbers
E9-33
and (j = 68.0°.
Then
sin2
()
(} sin(}"
.Us = 0.407.
Let out be positive. The net torque about the axis is then
T = (0.118m)(5.88N)
The rotational
-(0.118m)(4.13m)
-(O.O493m)(2.12N)
= O.102N
.m.
inertia of the disk is I = (1.92 kg)(0.118 m)2/2 = 1.34 x 10-2kg .m2.
(0.102N .m)/(1.34x
Then a =
10-2kg .m2) = 7.61 rad/s2.
E9-34
(a) I = Tia = (960 N .m)I(6.23 radls2) = 154kg .m2.
(b) m = (312)Ilr2 = (1.5)(154kg .m2)1(1.88m)2 = 65.4kg.
~
(a)
The
angular
~celeration
is
~(.,;
6.20
a = "Xt" -0.22
rad/s
-
=
s
28.2
rad/s2
{b) From Eq. 9-11, 'T = la = {12.0kg. m2){28.2 rad/s2) = 338N.m.
E9-36
The angular acceleration is a = 2(7r/2 rad)/(30s)2
= 3.5xlO-3
rad/s2. The required force
is then
F = 7"/r = Ia/r
Don't
= (8.7xlO4kg
.m2)(3.5xlO-3
rad/s2)/(2.4m)
= l27N.
let the door slam...
E9-37
The torque is T = r F , the angular acceleration is a = T/ I = r F / I. The angular velocity is
1.1.1
= lt
a dt =
rAt2
+
21
rBt3
3I'
so when t = 3.60 s,
ú.i = (9.88x10-2m)(0.496Nfs)(3.60s)2
2(1.14 x 10-3kg .m2)
+ (9.88x10-2m)(0.305Nfs2)(3.60s)3
3(1.14 x 10-3kg .m2)
E9-38
(a) a = 2()lt2.
(b) a = aR = 2()Rlt2.
(c) T1 and T2 are not equal. Instead, (Tl -T2)R
= la.
For the hanging block Mg -T1
Then
TI = Mg -2MR(}/t2,
and
T2 = Mg -2MR(}/t2
109
= 690 radfs.
-2(I/R)(}/t2.
= Ma.
~
Apply a kinematic equation from chapter 2 to find the acceleration:
y
ay
=
1
2
VOyt+ 2ayt ,
=
~t2 = 2(0.165
m) = 0 .m0586
(5.11
S)2
/
2
S
Closely follow the approach in Sample Problem 9-10. For the heavier block, ml = 0.512 kg, and
Newton 's second law gives
mlg -Tl
= mlay,
where ay is positive and down. For the lighter block, m2 = 0.463 kg, and Newton's secondlaw gives
T2 -m2g = m2ay,
where ay is positive and up. We do know that Tl > T2; the net force on the pulley creates a torque
which results in the pulley rotating toward the heavier mass. That net force is Tl -T2; so the
rotational form of Newton's second law gives
(TI -T2)
R = Iaz
= IaT/R,
where R = 0.049 m is the radius of the pulley and aT is the tangential acceleration. But this
acceleration is equal to ay, because everythingboth blocks and the pulley- are moving together.
We then have three equations and three unknowns. We'll add the first two together ,
mlg -Tl
+ T2 -m2g
=
mla1l + m2ay,
T1 -T2
=
(g-
ay)ml -(g
and then combine this with the third equation by substituting
(g -ay)ml
JL-l
ay
ml-
-(g
.!. + ;
+ ay)m2,
for T1 -T2,
+ ay)m2
=
Iay/R2,
m2 ] R2
=
I.
ay.
Now for the numbers:
(
{9.81
{0.0586
m/s2)
m/s2)
-1
(0.512 kg) -
=
7.23 kg,
=
O.O174kg.m2,
E9-40
The wheel turns with an initial angular speed of ú)o = 88.0 rad/s. The average speed while
decelerating is Ú)av= ú)o/2. The wheel stops turning in a time t = <t>/Ú)av
= 2<t>/ú)o.The deceleration
is then a = -ú)o/t = -ú)~/(2<t».
The rotational inertia is I = M R2 /2, so the torque required to stop the disk is T = I a =
-M R2ú)~/ (4<t>
) .The force of friction on the disk is f = JLN, so T = Rf. Then
JL= ~
= .(1.40kg)(~(88.0
4N<t>
4(130N)(17.6
rad/s)2 = 0.272.
rad)
E9-41
(a) The automobile h88 an initial speed of Vo = 21.8 m/s.
(¡)o= (21.8m/s)/(0.385m)
= 56.6 rad/s.
(b) The average speed while decelerating is (¡)av = (¡)o/2. ,
The angular speed is then
The wheel stops turning in a time
t = 4>/(¡)av= 24>/(¡)o.The deceleration is then
a = -(¡)0/t = -(¡)~/(24» = -(56.6 rad/s)2/[2(180
(c) The automobile traveled x = <Pr= (180 rad)(0.385m)
110
rad)] = -8.90 rad/s.
= 69.3m.
E9-42
(a) The angular acceleration is derived in Sample Problem 9-13,
The initial rotational speed was Wo= (1.30m/s)/(3.2x10-3m)
ú) = ú)o + at
= (406
rad/s)
+ (39.1
rad/s2)(0.945s)
= 406 rad/s. Then
= 443 rad/s,
which is the same as 70.5 rev/s.
~
(a) Assuming a perfect hinge at B, the only two vertical forces on the tire will be the
normal force from the belt and the force of gravity. Then N = W = mg, or N = (15.0 kg)(9.8
m/s2) = 147 N.
While the tire skids we have kinetic friction, so f = Jl.kN = (0.600)(147 N) = 88.2 N. The force
of gravity and and the pull from the holding rod AB both act at the axis of rotation, so can't
contribute to the net torque. The normal force acts at a point which is parallel to the displacement
from the axis of rotation, so it doesn't contribute to the torque either (becausethe cross product
would vanish); so the only contribution to the torque is from the frictional force.
The frictional force is perpendicular to the radial vector, so the magnitude of the torque is just
T = rf = (0.300 m)(88.2 N) = 26.5 N.m. This means the angular acceleration will be a = T/I =
(26.5 N.m)/(0.750 kg.m2) =35.3 rad/s2.
When ",-,R= VT = 12.0 m/s the tire is no longer slipping. We solve for "'-'and get "'-'= 40 rad/s.
Now we solve "'-'= "'-'o+ at for the time. The wheel started from rest, so t = 1.13 s.
(b) The length of the skid is x = vt = (12.0 m/s)(1.13 s) = 13.6 m long.
~
The problem of sliding down the ramp hBBbeen solved (see Sample Problem 5-8); the
critical angle (}8is given by tan(}8 = JL8.
The problem of tipping is actually not that much harder: an object tips when the center of
gravity is no longer over the bBBe.The important angle for tipping is shown in the figure below; we
can find that by trigonometry to be
o
tan(}t = A =
so (Jt= 34°,
111
(a) If J1s= 0.60 then (}s = 31° and the crate slides.
(b) If J1s= 0.70 then (}s = 35° and the crate tips before sliding; it tips at 34°.
P9-2 (a) The total force up on the chain needs to be equal to the total force downj the force
down is W. Assuming the tension at the end points is T then T sin () is the upward component, so
T = W/(2sin()).
(b) There is a horizontal component to the tension T cos() at the wallj this must be the tension
at the horizontal point at the bottom of the cable. Then Tbottom= W/(2tan()).
P9-3 (a) The rope exerts a force on the sphere which has horizontal Tsin(} and vertical Tcos(}
components, where (} = arctan(r/L). The weight ofthe sphere is balanced by the upward force from
the rope, so T cos(} = W. But cos(} = L / .¡;;.-:¡:7;'l, so T = W Vl + r2 / L2 .
(b) The wall pushesoutward against the sphere equal to the inward push on the sphere from the
rope, or p = Tsin(} = Wtan(} = Wr/L.
P9-4
'll:eat
and the two
2F = 2W /3.
lifting is T =
x = 3L/4.
the problem as having two forces: the man at one end lifting with force F = W /3
men acting together a distance x away from the first man and lifting with a force
Then the torque about an axis through the end of the beam where the first man is
2xW/3 -WL/2,
where L is the length ofthe beam. This expression equal zero when
~
(a) We can solve this problem with Eq. 9-32 after a few modifications. We'll assumethe
center of mass of the ladder is at the center, then the third term of Eq. 9-32 is mga/2. The cleaner
didn't climb half-way, he climbed 3.10/5.12 = 60.5% of the way, 80 the second term of Eq. 9-32
becomesMga(0.605). h, L, and a are related by L2 = a2+h2, so h = V(5.12 m)2 -(2.45 m)2 = 4.5
m. Then, putting the correction into Eq. 9-32,
Fw
=
.1
(4:5"tñ)
l(74.6
+ (10.3 kg)(9.81
kg)(9.81
m/s
m/s2)(2.45
2
)(2.45 m)(0.605)
m)/2]
,
,
269 N
(b) The vertical component of the force of the ground on the ground is the sum of the weight of
the window cleaner and the weight of the ladder, or 833 N .
112
The horizontal component is equal in magnitude to the force of the ladder on the window. Then
the net force of the ground on the ladder has magnitude
V(269N)2 + (833N)2 = 875N
and direction
I}
= arctan(833j269)
= 72° above the horizontal.
P9-6 (a) There are no upward forces other than the normal force on the bottom ball, so the force
exerted on the bottom ball by the container is 2W.
(c) The bottom ball must exert a force on the top ball which has a vertical component equal to the
weight of the top ball. Then W = N sin (Jor the force of contact between the balls is N = W / sin (J.
(b) The force of contact between the balls has a horizontal component p = N cos(J= W / tan (J,
this must also be the force of the walls on the balls.
P9-7
(a)
There
are
three
forces
on
the
ball:
-W,
weight
the
normal
force
from
the
lower
plane
N1,
and the normal force from the upper plane Ñ 2. The force from the lower plane has components
N1,x = -Nl sin(}l and Nl,y = Nl COS(}l. The force from the upper plane has components N2,x =
N2 sin(}2 and N2,y = -N2 COS(}2.Then N1 sin(}l = N2 sin(}2 and N1 COS(}l= W + N2 COS(}2.
Solving for N2 by dividing one expression by the other,
cos (}1
-=.+sin (}1
W
Cos (}2
N2 Sin(}2
-,
sin
(}2
or
N2
=
sin (}1 sin (}2
sin 02
W sin 01
sin(02 -01)
.
Then solve for Nl,
NI
=
w
Sin((}2
sin
(}2
-(}1)
,
(b) Friction changes everything.
P9-8
(a) The net torque about a line through A is
r = w x .:.-T L sin () = O,
so T = Wx/(Lsin()).
(b) The horizontal force on the pin is equal in magnitude to the horizontal component of the
tension: Tcos() = Wx/(Ltan()). The vertical component balances the weight: W- Wx/L.
(c) x = (520N)(2.75m)sin(32.00)/(315N) = 2.41m.
P9-9 (a) As long as the center of gravity of an object (even if combined) is above the base, then
the object will not tip.
Stack the bricks from the top down. The center of gravity of the top brick is L /2 from the edge
of the top brick. This top brick can be offset nor more than L /2 from the one beneath. The center
of gravity of the top two bricks is located at
Xcm = [(L/2) + (L)]/2 = 3L/4.
113
These top two bricks can be offset no more than L /4 from the brick beneath. The center of gravity
of the top three bricks is located at
Xcm = [(L/2) + 2(L)]/3
= 5L/6.
These top three bricks can be offset no more than LI6 from the brick beneath. The total offset is
then LI2 + LI4 + LI6 = llL112.
(b) Actually, we never need to know the location of the center of gravity; we now realize that
each brick is located with an offset LI(2n) with the brick beneath, where n is the number of the
brick counting from the top. The series is then of the form
(L/2)[(l/l)
+ (1/2) + (1/3) + (1/4) +
a series (harmonic, for those of you who care) which does not converge.
(c ) The center of gravity would be half way between the ends of the two ex:treme bricks. This
would be at N L/n; the pile will topple when this value exceeds L, or when N = n.
P9-10
(a) For a planar object which lies in the x -y plane, I", = J x2dm and Iy = J y2dm. Then
I", + ly = J(X2 + y2)dm = J r2 dm. But this is the rotational inertia about the z axis, sent r is the
distance from the z axis.
(b) Since the rotational inertia about one diameter (I",) should be the same as the rotational
inertia about anyother (ly) then I", = ly and I", =Iz/2 = MR2/4.
~
Problem 9-10 says that Ix + Iy = Iz for any thin, flat object which lies only in the
x -y plane.lt doesn't matter in which direction the x and y axes are chosen, so long as they are
perpendicular .We can then orient our square as in either of the pictures below:
y
--{>
x
By symmetry Ix = 11Iin either picture. Consequently, Ix = 11I= Iz/2 for either picture. It is
the same square, so Iz is the same for both picture. Then Ix is also the same for both orientations.
P9-12 Let Mo be the mass of the plate beforethe holes are cut out. Then M1 = Mo(al L)2 is the
mass of the part cut out of each hole and M = Mo -9M1 is the mass of the plate. The rotational
inertia (about an axis perpendicular to the plane through the center of the square) for the large
uncut square is MoL2/6 and for each smaller cut out is M1a2/6.
From the large uncut square'sinertia we need to remove M1a2/6 for the center cut-out, M1a2/6+
M1(LI3)2 for each of the four edge cut-outs, and M1a2/6 + M1(.J2LI3)2 for each of the comer
sections.
114
~
Then
MoL2
-9
-4
-Mla2
MlL2
-g-4
6
MoL2
-4~
-3~
6
2M1L2
,
6
9
.1
.
2L2
~
(a) From Eq. 9-15, I = J r2dm about some axis of rotation when r is measuredfrom that
axis. If we consider the x axis as our axis of rotation, then r = ~
, since the distance to the
x axis dependsonly on the y and z coordinates. We have similar equations for the y and z axes, so
Ix
-
I"
=
J
(y2
+"Z2)
dm,
lz
These three equations can be added together to give
I x + I y + lz = 2
J (x
2
+ y 2 + z 2 ) dm,
so if we now define r to be measured from the origin (which is not the definition used above), then
we end up with the answer in the text.
(b) The right hand side of the equation is integrated over the entire body, regardless of how
the axes are defined. So the integral should be the same, no matter how the coordinate system is
rotated.
P9-14
(a) Since the shell is spherically symmetric I", = Iy = Iz, so I", =
(2R2/3) J dm = 2M R2 /3.
(b) Since the solid ball is spherically symmetric I", = Iy = Iz, so
P9-15
(a) A simple
ratio
will suffice:
M
= 7rR2
dm
27rr
(b)
dI
(c)
I
P9-16
=
=
r2dm
=
(2Mr3/R2)dr.
JOR(2Mr3/R2)dr
(a)
Another
or dm =
dr
=
simple
MR2/2.
ratio
dm
will
suffice:
M
{4/3)7rR3
or dm =
~
(b)
dI
=
r2dm/2
=
[3M(R2
-Z2)2/8R3]dz.
115
2Mr
-R2
.dr.
(2/3) J r2dm =
( c ) There are a few steps to do here:
{R
l-R
=
I
3M(R2 -Z2)2
8R3
dz,
3M
{R
4R3 lo (R4 -2R2Z2
+ z4)dz,
~
The rotational acceleration will be given by az = ¿r/I.
The torque about the pivot comesfrorn the force of gravity on each block. This forces will both
originally be at right angles to the pivot arm, so the net torque will be ¿ r = mgL2 -mgLl , where
clockwise as seenon the page is positive.
The rotational inertia about the pivot is given by I = ¿mnr~ = m(L~ + L~). So we can now
find the rotational acceleration,
a
=
~
I
=
mgL2
m(L~
-mgL1
+ L~)
-L2
-g~
-L1
=
8.66
rad/s2.
The linear acceleration is the tangential acceleration, aT = aR. For the left block, aT = 1.73 m/s2;
for the right block aT = 6.93 m/s2.
P9-18 (a) The force of friction on the hub is JLkMg. The torque is T = JLkMga. The angular
acceleration has magnitude a = T/I = JLkga/k2. The time it takes to stop the wheel will be
t = ú)o/a = ú)ok2/(JLkga).
(b) The averagerotational speed while slowing is ú)o/2. The angle through which it turns while
slowing is ú)ot/2 radians, or ú)ot/(47r) = ú)~k2/(47rJLkga)
P9-19
(a) Consider a diíf.erential ring of the disk. The torque on the ring because of friction is
The net torque is then
T =
(b) The rotational
J dT =
acceleration
rR 2JLkM gr2
JO -¡¡;¡---dr
has magnitude
2
= 3JLkMgR.
a = T / I = ~ JLkg/ R. Then
t = ú)o/a
=
it will take a time
3&0
4JLkg
to
stop.
P9-20
We need only show that the two objects have the same acceleration.
Consider first the hoop. There is a force WII = W sin (} = mg sin (} pulling it down the ramp
and a frictional force f pulling it up the ramp. The frictional force is just large enough to cause
a torque that will allow the hoop to roll without slipping. This means a = aR; consequently,
fR = al = aI/R. In this case I = mR2.
The acceleration down the plane is
ma = mgsin(}
-f
= mgsin(}
-maI/R2
116
= mgsin(}
-ma.
Then a = 9 sin (} /2. The mass and radius are irrelevant!
For a block sliding with friction there are also two forces: WII = W sin (} = mg sin (} and f =
.Ukmgcos (}. Then the acceleration down the plane will be given by
a
=
gsin{}
-Jl,kgCOS{},
which will be equal to that or the hoop ir
= 21 tan(}.
~
This problem is equivalent to Sample Problem 9-11, except that we have a sphere instead
of a cylinder. We'll have the same two equations for Newton 's second law,
MgsinO-
f = Macm and N -MgcosO
Newton's second law for rotation willlook
= O.
like
-fR
= Icma.
The conditions for accelerating without slipping are acm = aR, rearrange the rotational equation to
get
f
=
-~
.-R2
=
-Icm(-acm)
R
,
and then
.Icm(acm)
Mgsm()
-R2
and solve for acm. For fun, let's write the rotational
= Macm,
inertia as 1 = {3M R2, where {3 = 2/5 for the
sphere. Then, upon some mild rearranging, we get
sin f)
acm = gU"!i
For the sphere, acm = 5/7gsin().
(a) If acm = 0.133g, then sin() = 7/5(0.133) = 0.186, and () = 10.7°.
(b) A frictionless block has no rotational propertiesj in this case {3 = 0! Then acm = 9 sin () =
0.186g.
P9-22
(a) There are three forces on the cylinder: gravity W and the tension from each cable T.
The downward acceleration of the cylinder is then given by ma = W -2T .
The ropes unwind according to Q = a/ R, but Q = T / I and I = mR2/2. Then
a = TR/I
Combining the above, 4T = W -2T,
(b) a = 4(mg/6)/m = 2g/3.
~
= (2TR)R/(mR2/2)
= 4T/m.
or T = W/6.
The force of friction required to keep the cylinder rolling is given by
f = ~Mgsin(};
the normal force is giveIi to be N = M 9 cos (}; so the coefficient of static friction is given by
f
1
JLB~ N = 3 tan(}.
117
P9-24
a = F / M, since F is the net force on the disk. The torque about the center of mass is F R,
so the disk has an angular acceleration of
a=
FR =
FR
=
MW72
2F
MR'
~9-25
This problem is equivalent to Sample Problem 9-11, except that we have an unknown rolling
object. We'll have the same two equations for Newton 's second law,
Mgsin{}
-f
= Macm
and N -Mgcos{}
= O.
Newton's second law for rotation willlook like
-IR
= Icma.
The conditions for accelerating without slipping are acm= aR, rearrange the rotational equation to
get
f
and
= -~
R
= -Icm(-acm)
R2
,
then
Mgsinf}
= Macm
-R2~
and solve for acm. Write the rotational inertia as I = ,8MR2, where ,8 = 2/5 for a sphere, fJ =!=1/2
for a cylinder, and ,8 = 1 for a hoop. Then, upon some mild rearranging, we get
sin f}
acm = g"f+ft
Note that a is largest when ,8 is smallest; consequently the cylinder wins. Neither M nor R entered
into the final equation.
118
ElO-l
l = rp = mvr = {13.7x 10-3kg){380 m/s){0.12m)
ElO-2
(a) r= mrx V, or
-A
1 = m(yvz
-zvy)i
+ m(zvx
-x'!!z)j
= 0.62 kg .m2/s.
A
+ m(xvy
-yvx)k.
A
(b) Ir v and r exist only in the xy plane then z = Vz = O, so only the u k term survives.
dr
dt
di-.
=
m-
xv+mrx
=
m v x v+mix
dt
-.dv
-
dt ,
a.
Now the cross product of a vector with itself is zero, so the first term vanishes. But in the exercise
we are t-.oldthe particle has constant velocity, so a = O, and consequently the secondterm vanishes.
Hence, 1 is constant for a single particle if v is constant.
EI0-4
(a) L = ¿ li; li = rimipi. Putting the numbers in for each planet and then summing (I
won't bore you withthe arithmetic details) yields L = 3.15x 1043kg.m2/s.
(b) Jupiter has l = 1.94x 1043kg.m2/s, which is 61.6% of the total.
E 10-5 I = mvr = m(27rr/T)r = 27r(84.3kg)(6.37xl06m)2
/(86400s)= 2.49x lOllkg .m2/s.
ElO-6
(a) Substitute
L
=
and expand:
L(rcm
~
LI
+ ~) x (miVcm + p~),
(mircm
Mrcm
,
X Vcm + rcm X Pi +miri
x vcm + rcm x (Lp~)
""
-
X Vcm
+ (Lmi~)
+ ri""
-'
X Pi
),
x Vcm + L~
x p~.
(b) But L: p~ = O and L: mi~ = O, becausethese two quantities are in the center of momentum
and center of mass. Then
ElO-7
(a)
Substitute
and
expand:
-I = mi-¡¡¡
d~
Pi
dri
= mi-¡¡¡
-- -miVcmo
-mi-¡¡¡-drcm = Pi
(b) Substitute and expand:
di:'
dt
The first term vanished becausev~ is parallel to p~.
119
( c ) Substitute
and expand:
di'
dt
""""""'
L...riX-
=
=
,d{Pi -miV
dt
cm)
Lr:
x (mi~
-mi8cm),
L~
x miai + (Lmi~)
x 8cm)
The second term vanishes because of the definition of the center of mass. Then
dL' = Lr:
x Fi,
dt
-where F i is the net force on the ith particle. The force F i may include both internal and external
components. If there is an internal component, say between the ith and jth particles, then the
torques from these two third law components will cancel out. Consequently,
di'
dt
ElO-8
(a)
'\:"""'-= L-ITi
-= Texto
Integrate.
f
f' dt = f
*dt
= f
dL = ~L.
(b) If I is fixed, ~L = I~(Li. Not only that,
fTdt
= fFrdt
= rfFdt
= rFav~t,
where we use the definition of average that depends on time.
~
(a) T~t = ~r. The disk starts from rest, so ~r = r -lo
ourselves with the magnitudes, so
= I. We need only concern
l = Al = TAt = {15.8 N.m){O.O33s) = O.521kg.m2/s.
(b) (¿)= 1/1 = (0.521 kg.m2/s)/(1.22x
10-3kg.m2)
= 427 rad/s.
EI0-I0
(a) Let Vo be the initial speed; the average speed while slowing to a stop is vo/2; the time
required to stop is t = 2x/vo; the acceleration is a = -vo/t = -v3/(2x).
Then
a = -(43.3m/s)2/[2(225m/s)]
= -4.17m/s2.
(b) a = a/r = (-4.17m/s2)/(0.247m)
= -16.9 rad/s2.
(c) T = la = (0.155kg .m2)(-16.9 rad/s2) = -2.62N .m.
EI0-ll
Let ri = Z+ ~. From the figure, Pl = -P2 and r'¡ = -~.
L
=
r1 + t = r1 X Pl + r2 X P2,
=
(rl--
-r2
=
(rl""
-r2
2 ""
X Pl.
-rl
""
)
)
Then
X Pl,
X Pl,
Since r'¡ and Pl both lie in the xy plane then L must be along the z axis.
120
ElO-l2
Expand:
L
=
¿:: ri = ¿:: ri X Pi,
'""'
L
'""'
miri
X Vi = L
¿::mi[(ri
.ri)¡;J -(ri
'""'
L
2
mi
[ri
miri
...
X (W X ri
.¡;J)i!)i],
2A
¡;J -(Zi
)
A
W )k
-(ZiXiW)i
A
-(ZiYiW
)j],
but if the body is symmetric about the z axis then the last two terms vanish, leaving
~
An impulse of 12.8 N.s will change the linear momentum by 12.8 N.s; the stick starts
from rest, so the final momentum must be 12.8 N.s. Sincep = mv, we then can find v = p/m = (12.8
N.s)/(4.42 kg) = 2.90 m/s.
Impulse is a vector, given by f Fdt. We can take the cross product of the impulse with the
displacement vector r (measured from the axis of rotation to the point where the force is applied)
and get
r x f Fdt ~ f r x Fdt,
The two sidesof the aboveexpressionare only equal if r has a constant magnitude and direction. This
won't be true, but if the force is of sufficiently short duration then it hopefully won't change much.
The right hand side is an integral over a torque, and will equal the change in angular momentum of
the stick.
The exercisestates that the force is perpendicular to the stick, then Irx FI = rF, and the "torque
impulse" is then (0.464 m)(12.8 N.s) = 5.94 kg.m/s. This "torque impulse" is equal to the change
in the angular momentum, but the stick started from rest, so the final angular momentum of the
stick is 5.94 kg.m/s.
But how fast is it rotating? We can use Fig. 9-15 to find the rotational inertia about the center
ofthe stick: I = ~ML2 = ~(4.42 kg)(1.23 m)2 = 0.557 kg.m2. The angular velocity ofthe stick is
w = l/I = (5.94 kg.m/s)/(0.557 kg.m2) = 10.7 rad/s.
EI0-14 The point of rotation is the point of contact with the plane; the torque about that point
is T = rmgsin(). The angular momentum is I(J.J,so T = la. In this case I = mr2/2 + mr2, the
secondterm from the parallel axis theorem. Then
a = ra = rT/I
ElO-l5
From Exercise
= mr2gsin(}/(3mr2/2)
8 we can immediately
Il(U)l
-U)o)/rl
= ~gsin(}.
3
write
= 12(U)2 -O)/r2,
but we also have rl(¿)l = -r2(¿)2. Then
1.1}
2- --2
rlr2I11.1}o
r~I2 -r2Il
121
.
EI0-16
(a) ~1..)/1..)= (l/Tl -1/T2)/(1/Tl)
= -(T2 -T1)/T2
= -~T/T,
which in this case is
-(6.0x 10-3s)(8.64x 104s) = -6.9X 10-8.
(b) Assuming conservation of angular momentum, ~I/ I = -~1..) /I..). Then the fractional change
would be 6.9 x 10-8.
~
The ratatianal
inertia
af a salid sphere is I = ~MR2;
=
=
Li
Ii¡-:;i
sa as the sun callapses
Lf,
Ifr:;Jf,
2
-1
MR
.2¡-:;.
-~MRf2r:;Jf,
-5
= Rf 2úJ¡.
1
5
Ri2¡-:;i
The angular frequency is inversely proportional to the period of rotation, so
Ti R?
Tf
Rf2
= {3.6 x 10
4
mm)
.
( {6.37Xl06m) )
{6.96 x 108m)
2 = 3.0 mm.
.
ElO-l8
The final angular velocity of the train with respect to the tracks is (Altt = Rv.
conservation of angular momentum implies
The
o = MR2úJ+ mR2(úJtt + úJ),
or
-m v
"' = (m + M)R"
ElO-l9
Thia ia much like a center of masa problem.
o = Ip</>p+ Im«/>mp + </>p),
or
4>mp
(Ip
-~-.
+
Im)t/>p
(12.6kg
{2.47x
Im
.m2)(25°)
lO-3kg
= 1.28x1050,
.m2)
That's 354 rotations!
ElO-20
(A)f = (Ii/lf)úJi
= [(6.13kg
.m2)/(1.97kg
.m2)](1.22
~
We have two disks which are originally
there are no external torques. We can write
-11,i + 12,i
1lQl,i
+ I2Q2,i
=
=
rev/s)
= 3.80 rev/s.
not in contact which then come into contact;
f1,f+t,f,
11¡;jl,f + 12¡;j2,fo
The final angular velocities of the two disks will be equal, so the above equation can be simplified
and rearranged to yield
ElO-22
l.L = lcos{}
= mvrcos{}
= mvh.
122
~
ElO-23
(a) (¿}f = (11/12)(¿}j, 11 = (3.66 kg)(0.363m)2
(¿}f = [(0.482kg
.m2)/(2.88kg
= O.482kg .m2.
.m2)](57.7
rad/s)
Then
= 9.66 rad/s,
with the same rotational sense as the original wheel.
(b) Same answer, since friction is an internal force internal here.
ElO-24
(a) Assume the merry-go-round
is a disk. Then conservation of angular momentum yields
I
(2mmR2 + mgR2)(¡J + (m rR2)(v/R)
= O,
or
ElO-25
Conservation
of angular
momentum:
(mmk 2 + mgR 2)!4)= mgR 2 (v/R),
so
úJ
ElO-26
=
(44.3 kg)(2.92m/s)/(1.22~)
(176kg)(0.916m)2
+ (44.3kg)(1.22m)2
= 0.496
rad/s.
Use Eq. 10-22:
ú)p
=
Mgr
IUJ
=
{0.492 kg){9,81 m/s2){3.88 x 10-2m)
{5.12 x 10-4kg .m2){27r28.6 rad/s)
= 2.04 rad/s
= 0.324 rev /s.
~
The relevant precessionexpression is Eq. 10-22.
The rotational inertia will be a sum of the contributions from both the disk and the axle, but
the radius of the axle is probably very small compared to the disk, probably as small as 0.5 cm.
Since I is proportional to the radius squared, we expect contributions from the axle to be less than
(1/100)2 of the value for the disk. For the disk only we use
Now for (¡),
I..; =
975
21[ rad
rev/min
1 rey
1 min
=
60
102
rad/s.
s
Then L = Ir.,; = 13.8 kg.m2/s.
Back to Eq. 10-22,
(Up
=
Mgr
=
{1.27
kg){9.81
13.8 kgm/s2){0.0610m)
.m2/s
= 0.0551
L
The time for one precessionis
t=
lrev
!.l)p
27r rad
(0.0551 rad/s)
123
=
114
s.
rad/s.
PI0-l
(a)
Positive z is out of the page.
..A
1 =
rmvsinOk
(b) f = rFsinOk
=
(2.91
A
m)(2.13kg)(4.18m)
sin(147°)k
A
=
14.1
kg
.m2/8k.
= (2.91 m)(1.88N) sin(26°)k = 2.40N .mk.
PI0-2 Regardlessof where the origin is located one can orient the coordinate system so that the
two paths lie in the xy plane and are both parallel to the y axis. The one of the particles travels
along the path x = vt, y = a, z = b; the momentum of this particle is Pl = mvi. The other
particle will then travel along a path x = c- vt, y = a + d, z = b; the momentum of this particle is
P2 = -mvi. The angular momentum of the first particle is
t = mvbj -mvak,
while that oí the second is
~ = -mvbj
+ mv(a
+ d)k,
so the total is 11 + t = mvdk.
~
Assume that the cue stick strikes the ball horizontally with a force of constant magnitude
F for a time ~t. Then the magnitude of the change in linear momentum of the ball is given by
F ~t = ~p = p, since the initial momentum is zero.
If the force is applied a distance x above the center of the ball, then the magnitude of the torque
about a horizontal aJdsthrough the center of the ball is 7"= xF. The change in angular momentum
of the ball is given by 7"~t = ~l = l, since initially the ball is not rotating.
For the ball to roll without slipping we need v = (J.lR.We can start with this:
Then x = I/mR is the condition
I = ~mR2, so x = ~R.
v
=
¡.)R,
E.
m
F~t
--.
m
F
--.
m
=
!R
I'
T~tR
I
xFR
I.
for rolling without
sliding from the start.
For a solid sphere,
PI0-4
The change in momentum of the block is M(V2 -Vl), this is equal to the magnitude the
impulse delivered to the cylinder. According to EI0-8 we can write M(V2 -vl)R = I(¿)f. But in the
end the box isn't slipping, so (¿)f= V2/R. Then
MV2
-MVl
=
(I/R
2
)V2,
or
V2 = vl/(l
+ I/MR2).
PI0-5 Assume that the cue stick strikes the ball horizontally with a force of constant magnitude
F for a time At. Then the magnitude of the change in linear momentum of the ball is given by
FAt = Ap = p, since the initial momentum is zero. Consequently, FAt = mvo.
If the force is applied a distance h above the center of the ball, then the magnitude of the torque
about a horizontal axis through the center of the ball is T = hF. The change in angular momentum
ofthe ball is given by TAt = Al = lo, since initially the ball is not rotating. Consequently,the initial
angular momentum of the ball is lo = hmvo = Iú)o.
124
The hall originally slips while moving, hut eventually it rolls. When it has hegun to roll without
slipping we have v = &. Applying the results from ElO-8,
m(v
-vo)R
+1(¡.;
-¡.;0)
=
O,
or
then, if v = 9vo/7,
P10-6
(a) Refer to the previous answer. We now want v = 1.1)
= O, so
m(v
-vo)R
2
+ 5mR
2V
R -hmvo
= O,
becomes
-voR -hvo
or h = -R.
That'll
= O,
scratch the felt.
(b) Assuming only a horizontal force then
~
We assumethe bowling ball is solid, so the rotational inertia will be I = (2/5)M R2 (see
Figure 9-15).
The normal force on the bowling ball will be N = Mg, where M is the mass ofthe bowling ball.
The kinetic friction on the bowling ball is Ff = J1,kN= J1,kMg.The magnitude of the net torque on
the bowling ball while skidding is then T = J1,kM
gR.
Originally the angular momentum of the ball is zeroj the final angular momentum will have
magnitude 1 = I ú) = I v I R, where v is the final translational speed of the ball.
(a) The time requires for the ball to stop skidding is the time required to change the angular
momentum to 1, so
Al
At = -=
T
(2j5)M ..
R2V
.ukMgR
j R
-
2v
5.ukg
.
Since we don't know v, we can't solve this for At. But the same time through which the angular
momentum of the ball is increasing the linear momentum of the ball is decreasing, so we also have
Ap = Mv -Mvo
a 1:= --"'--F¡
-J.tkMg
Combining,
~t
=
va -v
¡¡'k9
,
va -5¡¡,k9~t/2
¡.tkg
125
Vo -v
= -"'-Ji,k9
~
2/Lkg~t
=
~t
=
2vo -5JLkg6.t,
2vo
7JLkg'
2(8.50
-=
mis )
1.18s.
7(0.210)(9.81 m/s2)
(d) Use the expression for angular momentum and torque,
v = 5Jlkg~t/2
= 5(0.210)(9.81 m/s2)(1.18s)/2
(b) The acceleration of the ball is F / M = -Jlg .The
X
=
=
1 2
-at +vot,
2
1
--(0.210)(9.81m/s2)(1.18s)2
2
= 6.08m/s.
distance traveled is then given by
+ (8.50m/s)(1.18s)
= 8.6m,
(c) The angular acceleration i8 T/I = 5JLkg/(2R). Then
(}
=
1 2
-at + (,¡)ot,
2
=
~(0.210)(9.81 m(82) (1.188)2 = 32.6 rad = 5.19 revolutions.
4(0.11 m)
PI0-8
(a) 1 = llA)o = (1/2)MR21A)0.
(b) The initial speed is vo = &o. The chip decelerates in a time t =
the chip travels with an average speed of vo/2 through a distance of
vo
--=
y=Vavt=
Vo
vo/g,
and
during
this
time
R21A)2
-.
2 9
2g
( c) Loosing the chip won 't change the angular velocity of the wheel.
PlO-9
Since L =J(J) = 27rI/T and L is constant, then I <XT. But I <XR2, so R2 <XTand
6.T
2R6.R
T
R2
26.R
=~"
R
Then
AT
PI0-I0
Originally
=
the rotational
(86400
s)
"(6.37
2(30m)
x lO6m)
~0.8s.
inertia was
Ii
=
2
-MR
2
=
:g7r
.PoR
5
.
5
15
The average density can be found from Appendix C. Now the rotational
If
=
87r
l5(Pl
-P2)R1
5
87r
+ l5P2R
5
inertia is
,
where Pl is the density of the core, R1 is the radius of the core, and P2 is the density of the mantle.
Since the angular momentum is constant we have AT /T = Al/ I. Then
AT
T
so the day is getting
Pl -P2 ~
Po
R5
+ ~ -1
Po
= 10.3 -4c50 35705
5,52
longer .
1?¡::
4.50
63705 T ~
-I
r --~
= -U.127,
~
~
The cockroach initially h8B an angular speed of (Llc,i = -v/r.
The rotational inertia of
the cockroach about the axis of the turntable is I c = mR2. Then conservation of angular momentum
lci+18i
,
, =
I c(.1c,i+ 18(.18,i =
-mR2v/r
+ 1(.1 =
lcf+18f,
,
,
I c(.1c,f+ 18(.18,f
,
(mR2 + 1)(.1f,
1(.1-mvR
1.I.
PI0-12
(a) The skaters move in a circle of radius R = (2.92m)/2 = 1.46m centered midway
between the skaters. The angular velocity of the system will be (,¡}i= v/R = (1.38m/s)/(1.46m) =
0.945 rad/s.
,
2MRi2
úJ&
=
-úJ:
-(1.46m)2
--
fO-~4;t)
rM/R)
=
~-12
rM/R.
~
(a) Apply Eq. 11-2, W = Fscos4> = (190N)(3.3m)cos(22°) = 5803.
(b) The force of gravity is perpendicular to the displacement of the crate, so there is no work
done by the force of gravity.
(c) The normal force is perpendicular to the displacement of the crate, so there is no work done
by the normal force.
Ell-2
(a) The force required is F = ma = (106kg)(1.97m/s2)
= 209N. The object moves with
an average velocity vav = vo/2 in a time t = vo/a through a distance x = vavt = v8/(2a). So
x = (5l.3m/s)2/[2(l.97m/s2)]
The work done is W = Fx = (-209
(b) The force required
N)(668m)
= 668 m.
= l.40xlO5J.
is
F = ma = (106kg)(4.82m/s2)
x = (5l.3m/s)2/[2(4.82m/s2)]
= 273m. Theworkdone
= 511N.
is W = Fx = (-5llN)(273m)
= l.40xlO5J.
Ell-3
(a) W = Fx = (120N)(3.6m) = 430J.
(b) W = Fxcos(} = mgxcos(} = (25kg)(9.8m/s2)(3.6m) cos(117°) = 4O0N.
(c) W = Fxcos(}, but (} = 90°, so W = 0.
Ell-4
The worker pushes with a force P; this force has components Px = Pcos(} and Py = Psin(},
where (} = -32.0°. The normal force of the ground on the crate is N = mg- Py, so the force of
friction is f = Jl,kN = Jl,k(mg -Py). The crate moves at constant speed, so Px = f. Then
Pcos()
=
p
=
JLk(mg-
Psin
()),
JLkmg
cos
(}
+
Jl,k
siDJj
.
The work done on the crate is
w =
Ell-5
The components of the weight are WII = mgsin(} and W ~ = mgcos(}. The push p has
components PII = Pcos(} and P~ = Psin(}.
The normal force on the trunk is N = W ~ + P.l so the force of friction is f = .uk(mg cos (} +
Psin(}). The push required to move the trunk up at constant speed is then found by noting that
~I = WII + f .
Then
p = mg(tan(J
1 -ILk
(a)-The
w
work done by the applied
=
Pxcos(J
=
{52.3
+ ILk)
tan (J
force is
kg){9.81m/s2)[sin{28.0°)
1-
{0.19)
+ {0.19)
tan{28.0°)
cos{28.0°)]{5.95m)
=
21
(b) The work done by the force of gravity is
w = mgxcos{(} + 900) = {52.3 kg){9.81 m/s2){5.95m)
12R
cos{118°) = -1430J.
60J
.
Ell-6
O= arcsin(O.902m/1.62 m) = 33.8°.
The components of the weight are WII = mgsinO and W.L = mgcosO.
The normal force on the ice is N = W.L so the force of friction is f = JLkmgcosO. The push
required to allow the ice to slide down at constant speed is then found by noting that p = WII -f .
Then p = mg( sin O-JLk cosO).
(a) p = (47.2kg)(9.81m/s2)[sin(33.8°) -(0.110)cos(33.8°)1 = 215N.
(b) The work done by the applied force is W = Px = (215N)(-1.62m) = -348J.
(c) The work done by the force of gravity is
W = mgxcos(90° -0) = (47.2kg)(9.81 m/s2)(1.62m) cos(56.2°) = 417J.
~
Equation 11-5 describes how to find the dot product of two vectors from components,
a.b
=
axbx+ayby+azbz,
=
(3)(2) + (3)(1) + (3)(3) = 18.
Equation 11-3 can be used to find the angle between the vectors,
a
=
yf(3)2 + (3)2 + (3)2 = 5.19,
b
=
yf(2)2 + (1)2 + (3)2 = 3.74.
Now use Eq. 11-3,
a.b
cos<1>=
=
(18)
(5.19)(3.74)
= 0.927,
ab
and then 1/>= 22.0° .
Ell-8
a. b = (12)(5.8) cos(55°) = 40.
Ell-9
¡. s= (4.5)(7.3) cos(320° -85°)
EII-I0
= -19.
(a) Add the components individually:
r = (5 + 2 + 4)i + (4- 2 + 3)j + ( -6-
3 + 2)k = 11i + 5j -7k.
(b) (} = arccos(-7 /Vl12 + 52 + 72) = 120°.
(c) (} = arccos(a .b/Vlib), or
(} =
(5)(-2)
+ (4)(~) + (-6)(3)
= 124°.
.¡(52 + 42 + 62)(22 + 22 + 32)
~
There are two forces on the woman, the force of gravity directed down and the normal
force of the floor directed up. These will be effectively equal, so N = W = mg. Consequently,the
57 kg woman must exert a force of F = (57kg)(9.8m/s2) = 560N to propel herself up the stairs.
From the !jeferenceframe of the woman the stairs are moving down, and she is exerting a force
down, so the work done by the woman is given by
W = Fs = (560 N)(4.5 m) = 2500 J,
this work is positive becausethe force is in the same direction as the displacement.
The averagepower supplied by the woman is given by Eq. 11-7,
p = W/t = (2500 J)/(3.5 s) = 710 W.
129
Ell-12
p = W/t = mgy/t = (100 x 667 N) (152 m)/(55.0s)
Ell-13
p = Fv = (110N)(0.22m/s)
Ell-14
F = P/v,
~
but the units
= 1.84x 105W.
= 24 W.
are awkward.
p = Fv = (720 N)(26
m/s)
= 19000 W; in horsepower,
p = 19000W(1/745.7
hp/W)
=
25 hp.
Ell-16
Change to metric units! Then p = 4920W, and the flow rate is Q = 13.9 L/s.
density of water is approx.imately 1.00 kg/L , so the mass flow rate is R = 13.9 kg/s.
~
(a) Start by converting kilowatt-hours
1 kW.
The
to Joules:
h = (1000 W)(3600
s) = 3.6x 106 J.
The gas required to
0.026 gal
30 mi'
=
0.78
mi.
1 gal,
(b) At 55 mi/h, it will take
0.78 mi
lhr
55
= 0.014h = 51 s.
mi
The rate of energy expenditure is then (3.6 x 106 J) / (51 s) = 71000 W.
Ell-18
The linear speed is v = 27r(O.207m)(2.53 rev/s) = 3.29m/s. The frictional force is f =
JlkN = (0.32)(180N) = 57.6N. The power developed is p = Fv = (57.6N)(3.29m) = 19ÓW.
Ell-19
The net force required will be (1380kg -1220kg)(9.81m/s2)
= 1570N. The work is
W = Fy; the power output is p = W/t = (1570N)(54.5 m)/(43.0s) = 1990W, or p = 2.67 hp.
Ell-20
(a) The momentum change of the ejected material in one second is
~p = (70.2kg)(497m/s
-184m/s)
+ (2.92 kg)(497m/s)
The thrust is then F = ~p/ ~t = 2.34 x 104N .
(b) The power is p = Fv = (2.34x104N)(184m/s)
130
= 2.34x 104kg .m/s.
= 4.31x106W.
That's 5780 hp.
~
The acceleration on the object as a function oí position is given by
20 m/s2
a=
8m'
The work done on the object is given by Eq. 11-14,
X,
W = 18 Fzdx = 18 (10 kg).20.xm/s2
8m
dx
=
800
J.
Work is area between the curve and the line F = O. Then
Ell-22
~
(a) For
a spring,
F ~ -kx,
and
~F
= -k~x.
With no force on the spring,
This
is the
amount
less
than ~x the=
40
-c
-(-110
N) = of-omthe 7spring
~mm
k
=mark,
-I ~o)
(6500
so
the N/m)
position
m
with
no
force
on
it
is 23
mm.
(b)
~x
=
-10
mm
compared
~F
The
weight
of
the
last
object
to
the
100
=
-k~x
=
is
110N
-65N
Ell-24
(a) W = ~k(xr -Xi2)
(b) W = ~(1500N/m)[(1.52x
N
picture,
so
-(6500N/m)(-0.010m)
=
=
65N.
45N.
= ~(1500N/m)(7.60x
10-3m)2 = 4.33x 10-23.
10-2m)2 -(7.60 x 10-3m)2 = 1.30x 10-13.
E 11-25 Start with Eq. 11-20,and let Fx = 0 while Fy = -mg:
W = 1! (Fxdx + Fydy) = -mg1!
Ell~26
(a) Fo = mv5/ro = (0.675kg)(10.0m/s)2/(0.500m)
(b) Angular momentum is conserved, so v = vo(ro/r).
mv5r5/r3 = Fo(ro/r)3. The work done is
dy = -mgh.
= 135N.
The force then varies as F = mv2/r
-(0.500m)-2)
W=JFodr=
~
The kinetic
energy of the electron
4.2 eV
is
1.60 x 10-19 J'
1
=
6.7
x
10-19
J.
eV
Then
v=
~=
V m
y
f2{6.7xlO-19J)
(9.lxlO-31kg)
131
=
1.2 x 106m/s.
= 60.03.
Ell-28
(a) K = ~(110kg)(8.1m/s)2
(b) K = ~(4.2x10-3kg)(950m/s)2
(c) m = 91,400 tons(907.2 kg/ton)
= 3600J.
= 1900J.
= 8.29 x 107kg;
v = 32.0 knots(1.688
K
=
1
2(8.29x
ft/s/knot)(0.3048
lO7kg)(l6.5m/s)2
m/ft)
=
= 16.5m/s.
l.l3x
lOlOJ.
Ell-29
(b) ~K = W = Fx = (1.67x10-27kg)(3.60x1015m/s2)(O.O350m)
= 2.10x10-13J.
2.10x 10-13J/(1.60x 10-19J/eV) = 1.31 x 106eV.
(a) Kf = 2.10x 10-13J + !(1.67x 10-27kg)(2.40x 107m/s2) = 6.91 x 10-13J. Then
Vf = ,¡;¡K¡m
= V2(6.91 x 10-13J)/(1.67x10-27kg)
That's
= 2.88x 107m/s.
Ell-30
Work is negative ir kinetic energy is decreasing. This happens only in region CD.
work is zero in region BC. Otherwise it is positive.
~
The
(a) Find the velocity of the particle by taking the time derivative of the position:
v = ~
= (3.0m/8) -(8.0m/82)t
+ (3.0m/83)f.
Find v at two times: t = O and t = 48.
v(O)
v(4)
=
=
(3.0m/s) -(8.0m/s2)(O)
(3.0m/s) -(8.0
+ (3.0m/s3)(O?
m/s2)(4.0s)
+ (3.0 m/s3)(4.0S)2 = 19.0m/s
The initial kinetic energy is Ki = ~(2.80kg)(3.0m/s)2
Kf = ~(2.80kg)(19.0m/s)2
= 505J.
The work done by the force is given by Eq. 11-24,
w = Kf
-Ki
= 3.0m/s,
= 505J -13J
= 13J, while the final kinetic energy is
= 492J.
(b) This question is agking for the instantaneous power when t = 3.0s. p = Fv, so first find a;
a=
dv
:!: -(8.0
m/s2)
+ (6.0 m/s3)t.
dt
Then the power is given by p = mav, and when = 3 s this gives
p = mav = (2.80kg)(10m/s2)(6m/s)
Ell-32
w = ¿lK
= -Ki.
= 168W.
Then
1
W = -2(5.98x
1024kg)(29.8
x 103m/s)2
Ell-33
(a) K = !(l600kg)(20m/s)2 = 3.2x lO5J.
(b) p = W/t = (3.2xlO5J)/(33s) = 9.7xlO3W.
(C) p = Fv = mav = (l600kg)(20m/s/33.0s)(20m/s)
= 2.66x 1033J.
= l.9xlO4W.
Ell-34
(a) I = 1.40 x 104u .pm2(1.66 x 10-27kg/mboxu) = 2;32 x 10-47kg .m2.
(b) K = ~I(¿)2 = ~(2.32x10-47kg. m2)(4.30x 1012 rad/s)2 = 2.14x10-22J. That's 1.34 meV.
132
Ell-35
The translational
II 2~ 2
2 úJ -3
K t. Th en
kinetic
is Kt
=
~mv2,
the rotational
2(5.30x10-46kg
10-26kg)
3(1.94x
.m2) (500m/s)
~
úJ
energy
kinetic
energy
is Kr
= 6.75 x 1012 rad/s.
Y3lV
Ell-36
Kr = ~Iú)2 = ~(5l2kg)(0.976m)2(624
rad/8)2 = 4.75xl07J.
(b) t = W/P = (4.75 x l07J)/(8l30W)
= 58408, or 97.4 minute8.
~
From Eq.
11-29, Ki = !Ic.Ji2.
The object
is a hoop, so I = MR2.
1
1
Ki = 2MR2c.J2 = 2(31.4kg)(1.21m)2(29.6rad/s)2
Then
= 2.01 x 1043.
Finally, the averagepower required to stop the wheel is
w =
t
p=
Kf-Ki
t
Ell-38
The wheels are connected by a belt, 80 r A(¿}A
(a) Ir lA = lB then
IA
IR
(b) If instead KA
KB
=
18
(a) LV= 21r/T,
K
=
-
=~-
úJB
1
úJA
-,
3
or
lAJA =
3IAJB.
then
1A
Ell-39
IA/(.;A
rBIAJB,
2KA/úJA2
úJB2
2KB/úJB2
:.JA2
1
-9'
so
1
-Iw2
2
=
-47r2 (5.98
x lO24kg)(6.37
5
x lO6m)2
= 2.57x 1029J
(86,4008)2
(b} t = (2.57x 1029J}/(6.17x 1012W} = 4.17x 10168, or 1.3 billion year8.
Ell-40
(a) The velocities relative to the center oí mass are mIVI = m2V2j combine with VI +V2 =
910.0 mis and get
(290.0kg)Vl
=
(150.0kg)(910m/s
Vi),
or
Vl = (150kg)(910m/s)/(290kg
+ 150kg) = 310m/s
and V2 = 600m/s. The rocket case was sent back, so Vc = 7600m/s -310m/s
payload capsule was sent forward, so Vp = 7600 m/s + 600 m/s = 8200 m/s.
(b) Before;
After
1
Ki = 2(290kg
+ 150kg)(7600m/s)2= 1.271x 10loJ.
,
Kf
The
= 7290m/s.
"extra"
1
= 2(290 kg)(7290
1
rn/s)2 + 2(150 kg)(8200
energy carne frorn the spring.
133
rn/s)2
= 1.275 x 1010J.
The
~
Let the mass of the freight car be M and the initial speed be Vi. Let the mass of the
caboosebe m and the final speed of the coupled cars be Vf. The cabooseis originally at rest, so the
expressionof momentum conservation is
MVi = MVf + mvf = (M + m)vf
The decreasein kinetic energy is given by
K;-Kf
=
1
1
2
(
2
2Mvi
(M
1
2
-2Mvf
Vi
2
+
1
2mVf
2
-(M+m)Vf
2
)
Ell-42
Since the body splits into two parts with equal mass then the velocity gained by one is
identical to the velocity "lost" by the other. The initial kinetic energy is
1
Ki = 2(8.0kg)(2.Om/s)2 = 16J.
The final kinetic energy is 16 J greater than this, so
Kf
=
1
1
32J= 2(4.0kg)(2.Om/s+v)2+2(4.0kg)(2.0m/s-v)2,
=
~(8.0kg)[(2.0m/s)2 + v2],
so 16.0J = (4.0kg)v2. Then v = 2.0m/s; one chunk comesto a rest while the other moves off at a
speed of 4.0 m/s.
Ell-43
The initial velocity of the neutron is vo¡, the final velocity is Vlj. By momentum conservation the final momentum of the deuteron is mn(VO¡ -Vlj).
Then mdv2 = mn,;;lf+"ij'f.
There is also conservation of kinetic energy:
1
2
1
2
1
2
2mnVo = 2mnVl + 2mdV2.
Roun(iing the numbers slightly we have md = 2mn, then 4v~ = v3 + v~ is the momentum expression
and v3 = v~ + 2v~ is the energy expression. Combining,
2 2 - 2
Vo -Vl
2
(
2
2
+ Vo + Vl ) ,
or v~ = v3/3. So the neutron is left with 1/3 of its original kinetic energy.
134
Ell-44
(a) The third particle must have a momentum
P3
=
-(16.7x
10-27kg)(6.22x
106m/s)i + (8.35x10-27kg)(7.85
=
( -1.04i + O.655j) x 10-19kg .m/s.
(b) The kinetic energy can also be written
x 106m/s)j
as K = ~mv2 = ~m(p/m)2
= p2/2m.
Then the
kinetic energy appearing in this process is
K
=
1
1
2(16. 7x 10-27kg)(6.22 x 106m/s)2 + 2(8.35 x 10-27kg)(7.85x
+2(11. 7x 10-27kg)(1.23x
10-19kg .m/s}2 =
106m/s)2
23x lO-12J.
This is the same 88 7.66 MeV.
~
Change
your
units!
Then
w
(4.5 eV)(1.6x10-19
J/eV)
F = -;- =
(3.4 x 10-9 m)
= 2.1 x 10-10 N.
Pll-2
(a) If the acceleration is -g/4 the the net force on the block is -Mg/4,
the cord must be T = 3Mg/4.
(a) The work done by the cord is W = F .s= (3Mg/4)(-d)
= -(3/4)Mgd.
(b) The work done by gravity is W = F .s= ( -Mg)(-d)
= Mgd.
so the tension in
Pll-3
(a) There are four cords which are attached to the bottom load L. Each supports a tension
F, so to lift the load 4F = (840 lb) + (20.0 lb), or F = 215 lb.
(b) The work done against gravity is W = F .s = (840 lb) (12.0 ft ) = 10100 ft .lb.
(c) To lift the load 12 feet each segment of the cord must be shortened by 12 ftj there are four
segments, so the end of the cord must be pulled through a distance of 48.0 ft.
(d) The work done by the applied force is W = F .s= (215lb)(48.0 ft) = 10300 ft .lb.
Pll-4
The incline has a height h where h = W/mg = (680J)/[(75kg)(9.81
m/s2)]. The work
required to lift the block is the same regardless of the path, so the length of the incline l is l =
W/F = (680J)/(320N).
The angle of the incline is then
F
O = arcsin -lh = arcsin-mg
~
".".c
(320N)
""- .-=
= ar(;!;lIl
(75 kg)(9-:81
25.8°.
m782)
Pll-5
(a) In 12 minutes the horse moves x = (5280 ft/mi)(6.20 mi/h)(0.200 h) = 6550 ft. The
work done by the horse in that time is W = F .i = (42.0 lb) ( 6550 ft ) cos(27 .o°) = 2.45 x 105 ft .lb.
(b) The power output of the horse is
p = ~
105 ft~
(720s)
=-
..¡ ,- ,-
= J40 tt .lb/S550
1 hp
ft .lb/s
=
0.p.
618 h
Pll-6
In this problem (} = arctan(28.2/39.4) = 35.5°.
The weight of the block h8S components WII = mg sin (} and W 1. = mg cos (}.The force of friction
on the block is f = J.tkN = J.tkmgcos(}. The tension in the rope must then be
T = mg(sin (} + J.tkcos (})
in order to move the block up the ramp. The power supplied by the winch will be p = Tv, so
p = (1380kg)(9.81 m/s2)[sin(35.5°)
+ (0.41) cos(35.5°)](1.34m/s)
135
= 1.66x 104W.
~
If the power is constant then the force on the car is given by F = Plv. But the force is
related to the acceleration by F = ma and to the speed by F = m~ for motion in one dimension.
Then
F
-
dv
m-
=
dt
dxdv
m-dt dx
=
dv
=
mvdx
lv mv2dv
1
-m
3
3
v
P
-,
v
P
-,
v
~
v
P
-,
v
-
ix
=
Px.
Pdx,
We can rearrange this final expression to get vas a function of x, v = (3xP/m)1/3.
Pll-8
(a) If the drag is D = bv2, then the force required to move the plane forward at constant
speed is F = D = bv2, so the power required is p = Fv = bv3.
(b) p ocV3, SOif the speed increasesto 125% then p increasesby a factor of 1.253= 1.953, or
increasesby 95.3%.
Pll-9
(a) p = mgh/t,
p =
(100
but m/t is the persons per minute times the average mass, so
people/min)(75.0kg)(9.81
m/s2)(8.20
m)
=
1.01 x 104W.
(b) In 9.50 s the Escalator has moved (0.620m/s)(9.50s) = 5.89m; so the Escalator has "lifted"
the man through a distance of (5.89m)(8.20m/13.3m) = 3.63m. The man did the rest himself.
The work done by the Escalator is then W = (83.5kg)(9.81m/s2)(3.63m) = 2970J.
(c)Yes, becausethe point of contact is moving in a direction with at least some non-zero component of the force. The power is
p = (83.5 m/s2)(9.81m/s2)(0.620m/s)(8.20
m/13.3m)
= 313W.
(d ) If there is a force of contact between the man and the Escalator then the Escalator is doing
work on the mano
PII-I0
(a) dP/dv = ab -3av2, so Pmax occurs when 3V2 = b, or v = Vb73.
(b) F = P/v, so dF/dv = -2v, which means F is a maximum when v = O.
(c) No; P = 0, but F = ab.
~
(b)
Integrate,
w=
13xo F .d8=
o
or
W
,;
O
F;
:.-Xo
¡
3XO
j
:x
O
3Foxo/2.
136
-Xo)dx
=
Foxo
~ -3)
,
Pll-12
(a) Simpson's rule gives
1
W = 3 [(10N)+
4(2.4N) + (0.8N)] (1.0m) = 6.8J.
(b) W = JFds = J(A/x2)dx
= -A/x,
W = (9N .m2)(1/lm
-1/3m)
=6J.
~
The work required
to stretch
evaluating this between the limits of integration
the spring
1
xf
W
=
gives
from Xj to Xf is given by
3
k x dx =
k
¡Xf
4
k
-¡Xj
4
.
Xi
We then want to find the work required to stretch from x = l to x = 2l, so
k
WI-+21
4
k
) 4
=
¡(2l)
-¡(l
,
=
16~l4
4 -~l44
=
k 4
15¡1 = 15Wo.
,
Pll-14
(a) The spring extension is 8l = ~
-lo. The force from one spring has magnitude
k8l, but only the x component contributes to the problem, so
F=2k
is the force required to move the point.
The work required is the integral, W
w
=
kx2
I;
F dx,
which
-2klo.¡¡:¡--+-:;2
is
+
2kl~
Note that it does reduce to the expected behavior for x » lo.
(b) Binomial expansion of square root gives
,
1 X2
.¡¡:¡--+-;;; = lo 1+
1 X4
,
,..
2 l2O
8 l4O
SOthe first term in the above expansion cancels with the last term in W; the second term cancels
with the first term in W, leaving
1 X4
W=¡k"l-g-'
Pll-15
Number the springs clockwise from the top of the picture.
spring are
Fj.
F2
Fa
F4
=
=
=
=
k(lo -..¡X2
k(lo -..¡(lo
k(lo -..¡X2
k(lo -..¡(lo
137
+ (lo -y)2),
-;)2 +y2),
+ (lo + y)2),
+ x)Z + y2).
Then the four forces on each
The directions are much harder to work out, but for small x and y we can assumethat
F;
=
k(lo -..¡x2 + (10- y)2)j,
F~
=
k(lo -..¡(10-
F~
=
k(lo -..¡x2 + (lo + y)2)j,
F~
=
k(lo -..¡(lo
x)2 + y2)i,
+ x)2 + y2)i.
Then
Pll-16
(a) Kj = !(1100kg)(12.8m/s)2 = 9.0x 104J. Removing 51 kJ leaves39 kJ behind, so
or 30 km/h.
(b) 39 kJ, 88 W88 found above.
~
Let M be the mass of the man and m be the mass of the boy. Let VM be the original
speed of the man and Vm be the original speed of the boy. Then
1
2
-mvm
2
and
1
) 2
2M(VM
Combine
these two expressions
vt
+
1.0m/s
=
1
2mvm.
2
and solve for v M ,
=
o
=
The last line can be solved as a quadratic, and v M = (1.0mis)
first equation to find the speed of the boy,
vL
2VM
138
::1::(1.41 mis).
Now we use the very
Pll-18
(a) The work done by gravity on the projectile 88 it is raised up to 140 m is W =
-mgy = -(0.550kg)(9.81 m/s2)(140m) = -755J. Then the kinetic energy at the highest point is
1550J- 755J = 795J. Since the projectile must be moving horizontally at the highest point, the
horizontal velocity is Vx = V2(795J)/(0.550kg) = 53.8m/s.
(b) The magnitude of the velocity at launch is v = V2(1550J)/(0.550kg) = 75.1m/s. Then
Vy = V(75.1m/s)2 -(53.8m/s)2 = 52.4m/s.
(c) The kinetic energy at that point is !(0.550kg)[(65.0m/s)2 + (53.8m/sf] = 1960J. Since it
h88 extra kinetic energy, it must be below the launch point, and gravity h88 done 410 J of work on
it. That means itis y = (410J)/[(0.550kg)(9.81m/s2)] = 76.0m below the launch point.
Pll-19
(a) K = !mv2 = !(8.38 x 1011kg)(3.0 X 104m/s)2 = 3.77 x 102oJ. In terms of TNT,
K = 9.0 x 104 megatons.
(b) The diameter will be :r¡8.98 x 104 = 45 km.
PII-20
(a) W 9 = -(0.263kg)(9.81m/s2)(-0.118m)
= 0.304J.
(b) Ws = -~(252N/m)(-0.118m)2
= -1.75J.
(c ) The kinetic energy just before hitting the block would be 1.75 J -0.304 J = 1.45 J. The speed
is then v = V2(1.45J)/(0.263kg)
= 3.32m/s.
( d) Doubling the speed quadruples the initial kinetic energy to 5.78 J. The compression will then
be given by
-5.78J
=
1
-2(252N/m)y2
-(0.263kg)(9.81m/s2)y,
with solution y = 0.225 m.
~
(a) We can solve this with
w =
a trick
lx
oí integration.
F dx,
{X
dt
{t dx
lo maxdt dx = max lo &dt
max lt
v x dt = max lt
at dt,
1
2
-ma2t2
x
.
B88ically, we changed the variable of integration from x to t, and then used the fact the the acceleration W88constant so Vx = VOx+ axt. The object started at rest so VOx= O, and we are given in
the problem that Vf = atf. Combining,
(b) Instantaneous power will be the derivative of this, so
2
p= ~
=m(*)
139
t.
Pll-22
(a) a = (-39.0 rev/s)(27r rad/rev)/(32.0s)
= -7.66 rad/s2.
(b) The total rotational inertia of the system about the axis of rotation is
I = (6.40kg)(1.20m)2/12
+ 2(1.06kg)(1.20m/2)2
= 1.53kg
.m2.
The torque is then T = (1.53kg .m2)(7.66 rad/s2) = 11.7N .m.
(c) K = !(1.53kg .m2)(245 rad/s)2 =4.59x 104J.
(d) (} = fJ)avt = (39.0 rev/s/2)(32.0s) = 624 rev.
(e) Only the loss in kinetic energy is independent of the behavior óf the frictional torque.
Pll-23
The wheel turn with angular speed úJ = v/r, where r is the radius of the wheel and v the
speed of the car. The total rotational kinetic energy in the four wheels is then
The translational kinetic energy is K t = ~ (1040 kg)V2, so the fraction of the total which is due to
the rotation of the wheels is
--11.3
= 0.0213
or 2.13%.
520 + 11.3
Pll-24
(a) Conservation of angular momentum:
3.80 rev/s.
(b) K r O( II..)2 O( l2 / I, so
I..)f = (6.13kg .m2/1.97kg
Kf/Ki
.m2)
= Ii/lf
= (6.13kg
.m2)/(1.97kg
.m2)(1.22 rev/s) =
= 3.11.
~
We did the first part of the solution in Ex. 10-21. The initial kinetic energy is (again,
ignoring the shaft ) ,
since the secondwheel was originally at rest. The final kinetic energy is
since the two wheels moved M one. Then
Kj-Kf
!11ii:i1,i2
-!(11
+
- 1 .2
-1 1 1(.,;
2
,I
Kj
1 -(11
+
12)ii:if2
,
12)ii:if2
II¡;jl,i2
1-
II
~'
where in the last line we substituted from the results of Ex. 10-21.
Using the numbers from Ex. 10-21,
Ki-Kf
79.2%.
Kj
140
~
Pll-26
See the solution
to PlO-ll.
while
according to PlO-ll,
úlf
=
I(¡) -mvR
I +mR2
.
Then
K f-2-1
(II.JJ-mvR)2
I+mR2
.
Finally,
~K
Pll-27
See the solution
(a) Ki = ~Ii(J)i2, so
=
to PlO-l2.
Ki = ~ (2(51.2kg)(1.46m)2)
(0.945 rad/s)2 = 97.5J.
(b) Kr = ! (2(51.2kg)(0.470m)2) (9.12 rad/s)2 = 941 J. Th~~nergy
do while pulling themselves closer together .
Pll-28
K = ~mv2 = ~p2 = ~p.
Kf
=
=
~K
In all three cases ~p = (3000N)(65.0s)
comes from the work they
p. Then
2m
2k (pi2 + 2pi .~p
1 .(2pi .~p
+ (~p)2) ,
+ (~p)2) .
2m
= 1.95x105N .s and Pi = (2500kg)(300m/s)
= 7.50x105kg .
m/s.
(a) If the thrust is backward (pushing rocket forward),
~
K
-+2(7.50x105kg.m/s)(1.95x105N.s)+(1.95x105N.s)2=
-2(2500
+ 661107J
.x
kg)
.
(b) If the thru8t i8 forward,
~K
= .-2(7.50 x lO5kg .m/8)(l.95x
lO5N .8)+S~.~x
lO5N: 8)2,
2(2500 kg)
(c ) If the thru8t i8 8ideway8 the fir8t term vani8he8,
6.K = +(1:95 x 105N .8)2
2(2500 kg)
141
= 7.6lxlO6J.
-5.09
x lO7J.
~
There's
nothing
to integrate
W
where in the last line we factored
here! Start
1
2mVf
=
K f -K
=
~m (Vf2 -Vi2)
=
1
2m ( Vf -Vi)
the difference
i =
with
the work-energy
2
1
-2mVi
2
theorem
,
,
( Vf + Vi) ,
of two squares.
Continuing,
1
2 (mVf -m Vi) (Vf+ Vi) ,
w
1
2(Ap)(Vf+Vi),
but ~p = J, the impulse. That finishes this problem.
Pll-30
Let M be the mass of the helicopter. It will take a force Mg to keep the helicopter
airborne. This force comes from pushing the air down at a rate ~m/ ~t with a speed of v; so
Mg = v~m/~t.
The blades sweep out a cylinder of cross sectional area A, so ~m/~t = pAv.
The force is then Mg = pAv2; the speed that the air must be pushed down is v = ylMg7pA. The
minimum power is then
p = Fv
= Mgy
{ii;PA
= 2.49xlO5W.
Pll-31
(a) Inelastic collision, so Vf = mVi/(m
(b) K = ~mv2 = p2/2m, so
+ M).
~K
= ~m-l/(m+M)
Kj
ljm
Pll-32
Inelastic
collision,
=~
m+M'
so
The 1088in kinetic energy i8
~K
(1.88 kg)(10.3~
2
(1.88kg
+
4.92kg)(5.21
m/s)2
-2
This change is because of work done on the spring, so
x = v'2(33.7 J)/(1120N/m)
Pll-33
Pf,B = Pi,A + Pi,B -Pf,A,
Pf,B
=
so
[(2.0kg)(15m/s)
+[(2.0kg)(30m/s)
(12kg .m/s)i
= 0.245 m
+ (3,Okg)(-10m/s)
+ (3.0kg)(5.0m/s)
+ (15kg .m/s)j.
142
-(2.0kg)(-6.0m/s)]i
-(2.0
kg)(30m/s)]j,
=
33.7
J.
~
Then Vf,B = (4.0m/s)i
~K
=
+ (5.0m/s)j.
Since K = T(v~ + v~), the change in kinetic energy is
(2.0kg)[(-6.0m/s)2
+ (30m/s)2 -(15m/s)2
2
+ (5m/s)2
-(-10m/s)2
-
+ (3.0kg)[(4.0m/s)2
-(30m/s)2]
-(5.0m/s)2]
2
-315 J .
Pll-34
For the observer on the train the acceleration of the particle is a, the distance traveled is
Xt = !at2, so the work done as measured by thetrain is Wt = maxt = !a2t2. The final speed of
the particle as measured by the train is Vt = at, so the kinetic energy as measured by the train is
K = !mv2 = !m(at)2. The particle started from rest, so ~Kt = W t.
For the observer on the ground the acceleration of the particle is a, the distance traveled is
Xg = !at2 + ut, so the work done as measured by the ground is W 9 = maxg = !a2t2 + maut.
The final speed of the particle as measured by the ground is Vg = at + u, so the kinetic energy as
measuredby the ground is
1
1
1
1
Kg = 2mV2 = 2m(at + U)2 = 2a2t2 + maut + 2mu2.
But the initial kinetic energy as measuredby the ground is ~mu2, so W 9 = LlKg.
Pll-35
(a) Ki = ~mlVl,i2.
(b) After collision Vf = mlVl,i/(ml
+ m2), SO
2
m¡V¡,i
m¡
+
1
= -ml
c
Vl,i ~
ml
2
m2
ml
+
m2
(c) The fraction lost was
ml +m2
ml +m2
(d) Note that Vcm= Vf. The initial kinetic energy of the system is
1
I 2
1
I 2
K i = 2ml v l,i + 2m2V 2,i .
The final kinetic energy is zero (they stick together!), so the fraction lost is 1. The amount lost,
however, is the same.
Pll-36
Only consider the first two collisions, the one between m and m', and then the one between
m' and M.
Momentum conservation applied to the first collision means the speed of m' will be between
v' = mvo/(m+m') (completely inelastic)and v' = 2mvo/(m+m') (completelyelastic). Momentum
conservationapplied to the secondcollision meansthe speedof M will be betweenV = m'v' /(m' +M)
and V = 2m' v' / (m' + M) .The largest kinetic energy for M will occur when it is moving the fastest,
so
2
2 ' ,
4 ,
v' =
mvo
and V =
m+m'
m v
m'+M
-m
mvo
(m+m')(m'+M)
We want to maximize Vas a function of m' , so take the derivative:
dV
-=
dm'
4mvo(mM -m'2)
(m'+M)2(m+m')2.
This vanishes when m' = JTñM .
143
~
E12-1
(a) Integrate.
U(x)
(b)
W
= U(x)
-U(x
+ d),
= -{X
G~dX
Joo
Ir d «
(:;; -m
I
Start
) = Gm¡ m2;¡;;-:¡:"ijj"'
d
I
x then x(x + d) ~ X2, SO
w
~
x
so
W = Gm¡m2
E12-2
= -G~,
with
Eq.
~
Gmlm2
~d.
12-6.
=
U(x) -U(Xo)
Finishing the integration,
U(x)
Ir
we
choose
Xo
= U(Xo) + ~
00 and U(XO) = O we would
U(x)
( e-f:Jx~ -e-f:JX2)
be left with
= -a
-Bx2
2{je
E12-4
~K
= -~U
so ~K
= mg~y.
The power output is then
p = (58%).(lOOOkg/m3)(73,800m3)(9.8lm/s2)(96.3m)
= 6.74xlOsW.
(608)
E12-5
~U = -~K,
so !kX2 = !mv2,
k = ~2
X2
Then
= (2.38kg)(lO.OxlO3/s)
(1.47 m)2
= l.lOxlOBN/m.
Wow.
E12-6
~U 9 + ~U s = O, since K = 0 when the man jumps and when the man stops.
~Us = -mg~y = (220 lb)(40.4 ft) = 8900 ft .lb.
Kf + Uf
1
2
2mVf + mgyf
~Vf2
2
+ g( -r
)
Ki+Ui,
1
2
-m
2
Vi
+
mgyi,
=
Rearranging,
Vf = V -29( -r)
=
-2(9.81 m/s2)( -0.236 m)
144
2.15 m/s.
Then
E12-8
(a) K = ~mv2 = !(2.40kg)(150m/s)2
= 2.70x 104J.
(b) Assuming that the ground is zero, U = mgy = (2.40kg)(9.81m/s2)(125m)
(c) Kf = Ki + Ui since Uf =O. Then
I
-In
Vf -y
2 S2.
(2. 70x
70x 104J)
104J) + (2.94
(2.94 x 103J)
103J) -
~-
= 2.94x103J.
/ s.
-m 158
(2.40 kg)
Only (a) and (b) depend on the mass.
E12-9 (a) Since ~y = O, then ~U = 0 and ~K = o. Consequently, at B, v = vo.
(b) At C Kc = KA + UA -UC, or
1
2 1
2
h
2mvc = 2mvo + mgh -mg2'
or
VB=~=VVO2+9h.
(c) At
D KD
=KA
+UA-
UD,
or
1
-m
2
VD
2
~mvo2 + mgh -mg(O),
or
VE
=
v'VO2
+
2gh.
E12-10
From the slope ofthe graph, k = (0.4N)/(O.O4m)
(a) AK = -AU, so ~mVf2 = ~kXi2, or
= 10N/m.
(10N/m)
Vf = VI ~(O.O550m)
(b) AK=
~
-AU,
so ~mVf2 = ~k(Xi2 -Xf2),
= 2.82m/s.
or
(a) The force constant ofthe spring is
k = F/x = mg/x = (7.94kg)(9.81 m/s2)/(0.102m)
= 764N/m.
(b) The potential energy stored in the spring is given by Eq. 12-8,
U = ~kX2 = ~(764N/m)(0.286m
2
2
+ O.102m)2 = 57.5J.
(c) Conservation of energy,
Kf + Uf =
1
2
1
2
=
2mvf + mgyf + 2kxf
1
1
2(0)2 + mgh + 2k(0)2
-
Ki+Ui.
1
-mv.
2
2
1
+
1
2
1 + -kx.
2
1 .
m gy .
1
2
1
2
2(0)
+ mg(O) + 2kxi .
Rearranging,
k
h = ~Xi2
.I
= 2(7.94kg)(9.81m/s2)
(764N/m)
145
(0.388m)2
= 0.738 m.
E12-12
The annual mass of water is m = (1000kg/m3)(8 x 1012m2)(0.75m).
The change in
potential en~rgy each year is then AU == -mgy, where y = -500 m. The power available is then
E12-13
(a) From kinematics,
(b)U=mgysoK=Uo-U=mg(h-y).
E12-14
then
so K
=
~mg2t2
and
U = Úo -K
,
= mgh
-~mg2t2.
The potential energy is the same in both cases. Consequently,mgE~YE
y M = (2.05
~
v = ~gt,
The working
m -
10m)(9.81m/s2)/(1.67m/s2)
is identical
to Ex.
mgM~yM,
and
+ 1.10m = 6.68m.
12-11,
Kf + Uf
1
2
1
2
2mVf + mgyf + 2kxf
.
= K.1+U 1,
1
2
= 2mV¡
+ mgy¡
1
2
+ 2kX¡ ,
=
so
The distance up the incline is given by a trig relation,
d = k/sinO
= (1.92m)/sin(27°)
= 4.23m.
E12-16
The vertical position of the pendulum is y = -lcos(}, where (} is measured from the
downward vertical and I is the length of the string. The total mechanical energy of the pendulum is
1
E = 2mVb2
if we set U = O at the bottom of the path and Vb is the speed at the bottom.
U = mg(l +y).
(a) K = E- U = ~mvb2 -mgl(l
-cose). Then
v = v(8.12m/s)2
(b) U = E -K,
-2(9.81
m/s2)(3.82kg)(1-
In
this
case
cos58,OO) = 5.54m/s.
but at highest point K = O. Then
(J
= 83.1°.
arccos
(c) E = ~(1.33kg)(8.12m/s)2
= 43.83.
E12-17
The equilibrium position
~(ky)y = ~mgy. So 2~Us = -~Ug.
is when F = ky = mg.
146
Then
~U 9 =
-mgy
and ~U s =
E12-18 Let the spring get compresseda distance x. If the object fell from a height h = 0.436m,
then conservation of energy gives !kx2 = mg(x + h). Solving for x,
x = f
:!:: J (f)
2 + 2fh
only the positive answer is of interest, so
x =
{2.14(1860N/m)
kg){9.81 m/s2)
(
{2.14
:!:
kg){9.81
m/s2)
)
2 +
2 {2.14
kg){9.81
{1860N/m)
m/s2)
{0.436
m)
=
0.111
m.
{1860N/m)
~
The horizontal distance traveled by the marble is R = vtf, where tf is the time of flight
and v is the speedof the marble when it leavesthe gun. We find that speedusing energy conservation
principles applied to the spring just before it is releasedand just after the marble leavesthe gun.
=
Kj+Uj
1
O + 2kx2
=
Kj = O because the marble isn't moving originally,
compressed. Substituting
Kf
+ Uf,
I
2
-m
2
v
+
O.
and Uf = O because the spring is no longer
R into this,
1
-kx2
2
1
=
-m
2
.~)2.
We have two values for the compression,Xl and X2, and two ranges, R1 and R2. We can put both
pairs into the above equation and get two expressionsj if we divide one expression by the other we
get
~
(~). = (~)
2
We can easily take the square root of both sides, then
~-&
x¡
-R¡
.
R1 was Bobby's try, and was equal to 2.20 -0.27 = 1.93m. Xl = 1.1 cm was his compression. If
Rhoda wants to score, she wants R2 = 2.2 m, then
2.2m 1.1 cm = 1.25 cm.
X2
=
1.93m
E12-20 Conservation of energy- U1 + K1 = U2 + K2- but U1 = mgh, K1 = O, and U2 = 0, so
K2 = !mv2 = mgh at the bottom of the swing.
The net force on Tarzan at the bottom of the swing is F = mv2/r , but this net force is equal to
the tension T minus the weight W = mg. Then 2mgh/r = T -mg. Rearranging,
=
This isn't enaugh ta break the vine, but it is clase.
147
241
lb.
E12-21 Let point 1 be the start position of the first mass, point 2 be the collision point, and point
3 be the highest point in the swing after the collision. Then U1 = K2, or ~mlv12 = mlgd, where Vl
is the speed of ml just before it collides with m2. Then Vl = yI2gd.
After the collision the speedofboth objects is, by momentum conservation, V2 = mlvl/(ml +m2).
Then, by energy conservation, U3 = K~, or ~(ml +m2)v22 = (ml +m2)gy, where y is the height
to which the combined massesrise.
Combining,
:¿
2(ml
2g
E12-22
~K
= -~U,
m12v12
=
V22
y=
mi
=
mi
+m2)2g
+
d.
m2
so
1
1
2mV2+ 2IiJ)2= mgh,
whereI = ! M R2 and iJ)= v/ R. Combining,
1
1
2mv2
+ 4Mv2
= mgh,
so
v =
~
y
=
2;;;: + M
V
/4(0.0487
kg)(9.81m/s2)(0.540m)
2(0.0487
kg) + (0.396 kg)
=
1.45m/s.
~
There are three contributions to the kinetic energy: rotational kinetic energy of the shell
(Ks), rotational kinetic energy of the pulley (Kp), and translational kinetic energy of the block
( K b ) .The conservation of energy statement is then
;
Ksi+K
p
,
i+Kbi+Ui
=
,
,
(O) + (0) + (0) + (0)
=
Ksf+K
p ,f+Kbf+Uf,
,
,
1
2 1
2
1
2
2Is(J.Js +2'p(J.Jp +2mVb
+mgy.
Finally, y = -h and
(.;sR=
(.;pT
=
Vb.
Combine all of this together, and our energy conservation statement willlook like this:
which can be fairly easily rearranged into
E12-24
The angular speed of the flywheel and the speed of the car are related by
k=~-1 v
(1490 rad/s)
mis)
-{24.0
=
62.1
radim.
The height oí the slope is h = (1500m) sin(5.000) = 131m. The rotational inertia oí the Hywheel is
I =
~
(194N)
2 (ij:8i""iñ7B')
(0.54 m) 2 = 2.88 kg .m2.
148
~
(a) Energy is conserved as the car moves down the slope: Ui = Kf, or
m gh =
1
-mv2
2
1
+
-I(¿)2
2
=
1
-mv2
2
+
1
-Ik2v2
2
,
or
v = V I m2mgh
+ Ik2
.1 (822kg)
2(822
kg)(9.81 .~2)(62.1
m/s2)(131
= V
+ (2.88kg
m)
rad/m)2
= 13.3m/s,
or 47.9 m/s.
(b) The average speed down the slope is 13.3m/s/2=
6.65m/s. The time to get to the bottom
is t = (1500m)/(6.65m/s)
= 226s. The angular acceleration ofthe disk is
a = ~ = (13.3'm/s)(62.1
t
(2268)
(c)
p = T(Q = Ia(Q,
p =
rad/m)
= 3.65 rad/s2.
so
(2.88kg
.m2)(3.65
rad/s2)(13.3m/s)(62.1
rad/m)
E12-25
(a) Far the salid sphere I = ~mr2; if it ralls withaut
energy means K i = U f .Then
1
2mv2
1
+ 2
)(
(
2
Smr2
V
)2
-;:
= 8680W.
slipping w = v/r;
canservatian af
= mgh.
or
E12-26
Conservation of energy means ~mv2 + ~I(¿)2 = m9h.
h = 3v2/49, so
1
-m
2
2
or
I = 2r 2
(
3
4m
(.; =
v/r
and we are told
3V2
1 V2
-1'-2
r2
v
But
mg49'
1
-2m
)
=
1
2mr
,2
which cauld be a salid disk ar cylinder .
~
We agsumethe cannon ball is solid, so the rotational inertia will be I = (2/5)M R2
The normal force on the cannon ball will be N = Mg, where M is the magsof the bowling ball.
The kinetic friction on the cannon ball is Ff = JLkN = JLkMg. The magnitude of the net torque on
the bowling ball while skidding is then T = JLkMgR.
Originally the angular momentum of the cannon ball is zero; the final angular momentum will
have magnitude l = I (¡)= I v / R, where v is the final translational speed of the ball.
The time requires for the cannon ball to stop skidding is the time required to changethe angular
momentum to l, so
~t=
~l
T
=
{2/5)M
R2V / R
.,-"N";=.
2v
5J1,k9
Since we don't know At, we can't solve this for v. But the same time through which the angular
momentum of the ball is increB8ingthe linear momentum of the ball is decreB8ing,so we also have
¡\t=
.Ap -Mv-Mvo
-=
-vo-v
-JLkMg
-l'J
/1kg
Combining,
2v
=
5Jl,kg
2v
go
,-.
~
~
-I
-2
r-,--r-,--r-,--1
I
I
I
-v
Jl,kg
5(vo -v),
=
v::;:
2
va
5vo/7
4
I
I
L-.J--LI
I
I
I
r-,-I
I
L-.JI
I
I
-L-.J--I
I
I
I
I
-,--r-,--1
I
I
L-.J--L-.J--I
I
I
I
I
I
I
O
23456
I
I
I
I
I
I
I
I
I
I
I
I
,-.,
'-"2
"';;:
~1
3
o
o
1
2
3
4
5
6
x (m)
E12-28
E12-29 (a) F = -~U/~x
= -[(-17 J) -(-3J)]/[(4m)
-(lm)]
= 4.7N.
(b) The total energy is !(2.0kg)(-2.0m/s)2 + (-7J), or -3J. The particle is constrained to
move between x = 1 m and x = 14m.
(c) When x!= 7m K = (-3J) -(-17 J) = 14J. The speed is v = ..¡2(14J)/(2.0kg) = 3.7m/s.
E12-30
Energy is conserved, so
1
2 1 2
2mvo = 2mv + mgy,
or
v = .¡;¡=-2;;
,
which depends only on y.
~
(a) We can find Fx and Fy from the appropriate
Fx
-=
Fy
=
{)U
derivatives
of the potential,
-kx,
ax
au
=-ky.
ay
The force at point (x, y) is then
F = Fxi + Fyj = -kxi
-kyj.
(b) Since the force vector points directly toward the origin there is no angular component, and
F9 = O. Then Fr = -kr where r is the distance from the origin.
(c) A spring which is attached to a point; the spring is free to rotate, perhaps?
150
E12-33
We'll jUBt do the paths, showing only non-zero terms.
Path 1: W = J~(-k2a)dy)
= -k2ab.
Path 2: W = Joa(-k1b) dx) = -k1ab.
Path 3: W = (cos<1>
sin<1»JOd(-k1 -k2)rdr
These three are only equal if k1 = k2.
P12-2
= -(kl
The hall just reaches the top, so K2 = O.
y2(mgL)/m=
+ k2)ab/2.
Then
K1
=
U2
-tJ1
=
mgL,
so
Vi
J29L.
P12-3 Measure distances along the incline by x, where x = O is measured from the maximally
compressedspring. The vertical position of the mass is given by xsin(}. For the spring k =
(268N)/(0.0233m) = 1.15x 104N/m. The total energy of the system is
1
2(l.l5xlO4N/m)(O.O548m)2
= l7.3J.
(a) The block needs to have moved a vertical distance xsin(32.00),
where
l7.3J = (3.l8kg)(9.8lm/s2)xsin(32.00),
or x = 1.05m.
(b) When the block hits the top of the spring the gravitational
AU = -(3.18kg)(9.81
potential energy has changed by
m/s2)(1.05 m -O.O548m) sin(32.00) = 16.5, J;
hence thespeed is v = .¡2(16.5 J)/(3:18 kg) = 3.22 m/s.
151
P12-4
The potential energy associated with the hanging part is
¡
O
10
M
U=
Mg
-gydy=
L
-L/4
2
2LY
-L/4
=- MgL
32 ,
sothe work required is W = MgL132.
~
(a) Considering
points
p and Q we have
Kp+Up
(O) + mg(5R)
4mgR
=
KQ + UQ,
=
~mv2 + mg(R),
2
1
2
-mv,
2
=
v.
There are two forces on the block, the normal force from the track,
N=
=
mv2
m(8gR)
,
R
=8mg,
R
and the force of gravity W = mg. They are orthogonal so
Fnet = v(8mg)2
+ (mg)2 = V65mg
and the angle from the horizontal by
tan
-mg
8mg
(} =
k:
-.:!
8'
or (} = 7.13° below the horizontal.
(b) If the block barely makes it over the top of the track then the speed at the top of the loop
(point S, perhaps?) is just fast enough so that the centripetal force is equal in magnitude to the
weight,
mvs2/R=mg.
Assume the block wag releaged from point T. The energy conservation problem is then
KT+UT
=
(O) + mgyT
=
K s + Us.
1
2
2mvs + mgys,
yT
S,
1
2(R) + m(2R),
5R/2.
P12-6
The wedge slides to the left, the block to the right. Conservation of momentum requires
M Vw + mvb,x = O. The block is constrained to move on the surface of the wedge, so
tan a =
Vb,y
Vb
,
,:¡:-Vw
or
Vb,y = Vb,x tana(l
152
+ m/M).
Conservation of energy requires
1
2 1
2
2mVb +2Mvw
= mgh.
Combining,
= mgh,
( tan2
(sin2 a(M
+ m)2
(M2
a(l
+ M2
+ mM
+ m/M)2
COS2 a + mMcos2
+ mMsin2
((M
) Vb,x2
=
2gh,
a) Vb,x2
=
2M2ghcoS2
a,
2M2ghcos2
a,
2M2ghcoS2
a,
+ 1 + ~
a + m2 sin2 a) Vb,x2
+ m)(M
+ msin2
a))
Vb,x2
=
=
or
vw=
P12-7
U(x) = -J Fx dx = -Ax2/2 -Bx3/3.
(a) U = -( -3.00N/m)(2.26m)2
/2- ( -5.00N/m2)(2.26m)3
(b) There are two points to consider:
U1
=
-( -3.00N/m)(4.91
U2
=
-(-3.00N/m)(1.77m)2/2-
K1
m)2 /2-
1
2(1.18kg)(4.13m/s)2
/3 = 26.9J.
( -5.00 N/m2)(4.91 m)3/3 = 233J,
(-5.00N/m2}(1.77m)3/3
= 13.9J,
= 10.1 J.
P12-8
Assume that Uo = Ko = o. Then conservation of energy requires K = -U;
v = ,fiijFilj
.
(a) v = V2(9.81m/s2)(1.20m)
= 4.85m/s.
(b) v =
2(9.81m/s2)(1.20m -0.45m0.45m) = 2.43m/s.
consequently,
~
consequently,
Assume that Uo = Ko = o. Then conservation of energy requires K = -U;
v = ,fiijFii'j
.If the ball barely swings around the top of the peg then the speed at the top of the
loop is just fast enough so that the centripetal force is equal in magnitude to the weight,
mv2/R = mg.
The energy conservation
problem
is then
mv2 =
mg(L -d)
=
d
=
2mg(L- 2(L -d))
2mg(2d- L),
3L/5.
153
= 2mg(2d -L)
P12-10
The speed at the top and the speed at the bottom are related by
1
2
1
2
2mVb = 2mVt + 2mgR.
The magnitude of the net force is F = mv2 / R, the tension at the top is
Tt =mvt2/R-mg,
while tension at the bottom is
Tb = mvb 2/ R + mg,
The
difference
is
~T
= 2mg + m(Vb2 -Vt2)/R
= 2mg + 4mg = 6mg.
~
Let the angle (} be meBBuredfrom the horizontal to the point on the hemisphere where
the boy is located. There are then two components to the force of gravity- a component tangent
to the hemisphere, WII = mgcos(}, and a component directed radially toward the center of the
hemisphere, WJ. = mgsin(}.
While the boy is in contact with the hemisphere the motion is circular so
mv2/R
= W.L -N.
When the hoy leaves the surface we have mv2/R
= Wl., or mv2 = mgRsin(J.
Now for energy
conservation,
1
-mv2
2
K+U
-
Ko + Uo,
+
=
~m(Of
2
mgy
+ mgR,
~gRsin(} + mgy =
1 +y =
2Y
R,
y
P12-12
(a) To be in contact at the top requires mVt2 / R = mg. The speed at the bottom would
be given by energy conservation .,
1
2 1
2
2mVb = 2mVt + 2mgR,
SOVb = ~
is the speed at the bottom that will allow the object to make it around the circle
without loosing contact.
(b) The particle willlose contact with the track if mv2 / R ::; mg sin (J.Energy conservation gives
~mv~ = ~mv2 + mgR(l
2
2
+ sin(J)
for points above the half-way point. Then the condition for "sticking"
or, if Va = O.775vm,
5(0.775)2-
2 ~ 3 sin(},
or (} = arcsin(I/3).
154
to the track is
P12-13
The rotational
inertia
is
Conservation oí energy is
so (JJ =
,/9Ii7(4I¡
.
P12-14 The rotational speed of the sphere is (J)= v/r; the rotational kinetic energy is Kr =
!I(J)2 = kmv2.
(a) For the marble to stay on the track mv2/ R = mg at the top of the track. Then the marble
needsto be releasedfrom a point
or h = R/2 + R/5 + 2R = 2.7 R.
(b) Energy conservation gives
1
1
6mgR = 2mv2
+ Smv2
+ mgR,
or mv2 / R = 50mg /7. This corresponds to the horizontal force acting on the marble.
P12-15
!mv5
+ mgy = o, where y is the distance
Vo = v'=2g¡;
beneath
the rim,
or y = -r
cos Oo. Then
= \!2grcosOo.
P12-16
(a) For El the atoms will eventually move apart completely.
(b) For E2 the moving atom will bounce back and forth between a closest point and a farthest
point.
(c) U ~ -1.2x 10-19J.
(d) K = El -U ~ 2.2xl0-19J.
(e) Find the slope of the curve, so
which would point toward the larger mass.
~
The function needs to fall off at infinity in both directions; an exponential envelop~
would work, but it will need to have an -X2 term to force the potential to zero on both sides. So we
propose something of the form
U(x) = P(x)e-{JX2
where P(x) is a polynomial in x and f3 is a positive constant.
We proposed the polynomial because we need a symmetric function which has two zeroes. A
quadratic of the form ax2 -Uo would work, it has two zeroes, a minimum at x = O, and is symmetric.
So our trial function is
U(x) = (ax2 -UO) e-í3"'2.
1!'í!'í
~
This function should have three extrema. Take the derivative, and then we'll set it equal to zero,
dU
dx
= 2axe-f30:2
-2
(ax2
-UO)
{3xe-f3:¡:2,
Setting this equal to zero leavestwo possibilities,
x
2a-2(ax2-Uo){3
The first equation is trivial,
=
o,
=
0.
the second is easily rearranged to give
x=:i:
~
y---¡¡a-
These are the points ::I::Xl. We can, if we wanted, try to find Q and {3 from the picture, but you
might notice we have one equation, U(Xl) = U1 and two unknowns. It really isn't very illuminating
to take this problem much farther, but we could.
(b) The force is the derivative of the potential; this expression was found above.
(c) As long as the energy is le88than the two peaks, then the motion would be oscillatory, trapped
in the well.
P12-18
(a) F = -{)U/{)r,
or
F = -Uo
ro
-+r2
1
e-r/ro
r
(b) Evaluate the force at the four points:
1
F(ro)
-2(Uo/ro)e-
F(2ro)
F(4ro)
F(10ro)
,
-(3/4)(Uo/rO)e-2,
=
=
-(5/16)(Uo/ro)e-4,
-(11/100)
(Uo/ro)e-l0.
The ratios are then
F('
F(
F(11
{3/8)e-l
4ro)
=
IFI ~ro) =
Oro)
/FI ~ro)
{11/200)e-9
2ro)
/FI ~ro)
{5/32)e-3
156
= 0.14,
= 7.8 x 10-3,
= 6.8 x 10-6.
~
If the projectile had not experienced air drag it would have risen to a height y2, but
becauseof air drag 68 kJ of mechanical energy was dissipated so it only rose to a height yl. In
either case the initial velocity, and hence initial kinetic energy, was the same; and the velocity at
the highest point was zero. Then W = ~U, so the potential energy would have been 68 kJ greater,
and
Lly = LlU/mg
is haw much higher
it wauld
= (68xl03J)/(9.4kg)(9:81m/s2)
have gane withaut
= 740 m
air frictian.
E13-2 (a) The road incline is (J= arctan(0.08) = 4.57°. The frictional forces are the same; the car
is now moving with a vertical upward speedof (15m/s) sin(4.57°) = 1.20m/s. The additional power
required to drive up the hill is then ~p = mgvy = (1700kg)(9.81m/s2)(1.20m/s) = 20000W. The
total power required is 36000W.
(b) The car will "coast" if the power generated by rolling downhill is equal to 16000W, or
Vy = (16000W)/[(1700kg)(9.81m/s2)] = 0.959m/s,
clown. Then the incline is
() = arcsin(O.959m/s/15m/s)
This corresponds
E13-3
Apply
to a downward
energy
grade of tan(3.67°)
= 3.67°.
= 6.4 %.
conservation:
1
1
2mv2
+ muy + 2ky2
= O,
so
v = J-2(9.81m/s2)(-O.O84m)
-(262
N/m)(-O.O84m)2/(1.25kg)
E13-4
The car climbs a vertical distance of (225m)sin(10°)
change in energy of the car is then
1 (16400 N)
~E = -2¡¡¡:s¡;ñ7S')(31.4m/s)
2
= O.41m/s.
= 39.1m in coming to a stop. The
+ (16400N)(39.1m)
= -1.83x10
5
J.
~
(a) Applying conservation ofenergy to the points where the hall was dropped and where
it entered the oil,
1
2
2mvf +mgyf
1
2vf2 +g(O)
Vf
=
1
2
2mvi +mgyi,
=
1
2(0)2 +gyi,
=
~2gyi,
=
v2(9.81m/s2)(0.76m)
= 3.9m/s.
(b) The change in internal energy of the ball + oil can be found by considering the points where
the ball was released and where the ball reached the bottom of the container .
~E
=
Kf+Uf-Kj-Uj,
1
2
1
2mVf + mgyf -2m(O)
1
2(l2.2x
2
lO-3kg)(l.48m/s)2
-mgyj,
-(l2.2x
-O.l43J
157
lO-3kg)(9.8lm/s2)(
-0.55m -O.76m),
E13-6
(a) Ui = (25.3kg)(9.81m/s2)(12.2m)
(b) Kf = !(25.3kg)(5.56m/s)2
= 391 J.
(c) ~Eint = 3030J -391 J = 2640J.
E13-7
= 3030J.
(a) At atmospheric entry the kinetic energy is
K
The gravitational
potential
energy is
u = (7.9x lO4kg)(9.8m!s2)(l.6x
lOsm) = l.2x lO11J.
The total energy is 2.6xlO12J.
(b) At touch clown the kinetic energy is
3.8x lO8J.
E13-8
~E/~t
= (68 kg)(9.8m/s2)(59m/s)
= 39000J/s.
~
Let m be the mass of the water under consideration.
energy "lost" which appears as kinetic energy is
Then the percentage of the potential
Kf-Ki
Ui-Uf'
Then
Kf-Kj
=
Uj-Uf
1
2m
(Vf 2 -Vi
Vr
-Vi2
-2gAy
(13m/s)2
2
)/
(mgyi
-mgyf)
,
-(3.2m/s)2
--2(9.81m/s2)(-15m)'
=
54%.
The rest oí the energy would have been converted to sound and thermal energy.
E13-10
The change in energy is
1
~E = 2(524kg)(62.6m/s)2 -(524 kg)(9.81 m/s2)(292m)
E13-11
Uf = Kj -(34.6J).
h =
4.74x 105J.
Then
! {7.81
mLs)2
2 {9.81
m/s2)
-
which means the distance along the incline is (2.28m)/sin(33.00) = 4.19m.
158
E13-12
(a) Kf
= Ui -Uf,
so
Vf = J2(9.81m/s2)[(862m)
-(741
m)] = 48.7m/s.
rhat 's a quick 175 km/h; but the speed at the bottom of the valley is 40% of the speed of sound!
(b)~E=Uf-Ui,SO
~E=
(54.4
kg)(9.8lm/s2)[(862m)
-(74lm)]
= -6.46xlO4J;
which means the internal energy of the snow and skis increased by 6.46 x 104J .
~
The final potential energy is 15% less than the initial kinetic plus potential energy of the
ball, so
=Uf.
O.85{Ki+U;
But
Ui = Uf,
SO Ki
= O.15Uf/0.85,
and
then
E13-14 Focus on the potential energy. After the nth bounce the ball will have a potential energy
at the top of the bounce of Un = O.9Un-l. Since U <Xh, one can write hn = (0.9)nho. Solving for n,
n = log(hn/ho)/
log(O.9) = log(3 ft/6
ft)/
log(O.9) = 6.58,
which must be rounded up to
E13-15 Let m be the mass of the ball and M be the mass of the block.
The kinetic energy of the ball just before colliding with the block is given by K1 = UO, so
Vl = V2(9.81 m/s2)(0.687m) = 3.67m/s.
Momentum is conserved, so if V2 and V3 are velocities of the ball and block after the collision
then mVl = mV2+ MV3. Kinetic energy is not conserved,instead
Combine the energy and momentum expressions to eliminate V3:
2
mvl
Mv~
which can be formed into a quadratic. The solution for V2 is
2m
V2
=
~
V2(M2
2(M+m)
-mM)
VI
=
(0.600
~
95) m/s.
The corresponding solutions for V3 are then found from the momentum expression to be V3 =
0.981m/s and V3 = 0.219. Since it is unlikely that the ba11passedthrough the block we can toss
out the secondset of answers.
E13-16
Ef = K f + U f = 3mgh,
or
Vf = ..¡2(9.81 mis2)2(0.18
159
m) = 2.66 mis.
~
We can find the kinetic energy of the center of mass of the woman when her feet leave
the ground by considering energy conservation and her highest point. Then
1
-m
2
2
Vi
+
mgyi
1
-m
2
Vi
=
1
2
2mVf + mgyf ,
=
mgAy,
=
{55.0 kg){9.81 m/s2){1.20m -O.90m)
162J.
(a) During the jumping phase her potential energy changed by
~u = mg~y = (55.0 kg)(9.81m/s2)(0.50m)
= 270J
while she was moving up. Then
Fext
E13-18
(b) The ice skater needsto lose ~(116kg)(3.24mjs)2
= 609J of internal energy.
(a) The average force exerted on the rail is F = (609 J)j(0.340m) = 1790 N.
E13-19
12.6 km/h is equal to 3.50m/s; the initial kinetic energy of the car is
1
2(2340kg)(3.50m/s)2= l.43xlO4J.
(a) The force exerted on the car is F = (l.43x lO4J)!(0.64m)
(b) The change increase in interna! energy of the car is
(2.24 x 104N)(0.640 m -0.083 m) = 1.25 x 104J.
~Eint
E13-20
= 2.24x lO4N.
Note that v~ = v~2- 2"~ ."cm + Vcm2,Then
K
=
~
-I
L...mnVn
V cm+
n
Kint
-
~
L,,¡
mn
-I
V n
L~mn
n
v cm2
vcm+Kcm.
The middle term vanishesbecauseof the definition of velocities relative to the center of mass.
160
~
~
Momentum conservation requires mvo = mv+ MV, where the sign indicates the direction.
We are assuming one dimensional collisions. Energy conservation requires
1
2
1
2
1
2mvo = 2mv
+ 2MV
2
+ E.
Combining,
J.
2
-mvo
2
Mv~
1
=
-m
2
=
MV2
1
(m
-M
2
-voM
+m(VO
-V)2
2
v
+
m
)2
-v
M
+E
,
+ 2(M/m)E.
Arrange this as a quadratic in v,
(M
+ m) V2 -(2mvo)
v +
(2(M/m)E
+ mv5
-Mv5)
= o.
This will only have real solutions ir the discriminant (b2-4ac) is greater than or equal to zero. Then
(2mvo)2 ~ 4 (M + m) (2(M/m)E
+ mv5 -Mv5)
is the condition for the minimum Va. Solving the equality condition,
4m2v~ = 4(M + m) (2(M/m)E
or M2v8 = 2(M+m)(M/m)E.
+ (m -M)v~)
,
One last rearrangement, and Va = v2(M+m)E/(mM).
~
(a) The initial kinetic energy will equal the potential energy at the highest point plus the
amount oí energy which is dissipated because oí air drag.
mgh + fh
h
=
=
(b) The final kinetic energy when the stone lands will be equal to the initial kinetic energy minus
twice the energy dissipated on the way up, so
1
-m
2
2
v
-
V2
v
P13-2 The object starts with U = (0.234kg)(9.81 m/s2)(1.05 m) = 2.41J. It will move back and
forth across the flat portion (2.41J)/(0.688J) = 3.50 times, which means it will come to a rest at
the center of the flat part while attempting one last right to left journey.
161
P13-3 (a) The work done on the block block becauseof friction is
{0.210){2.41 kg){9.81 m/s2){1.83 m) = 9.09 J .
The energy dissipated because of friction is {9.09J)/0.83 = 11.0J.
The speed of the block just after the bullet comes toa rest is
3.02 m/s.
(b) The initial speed of the bullet is
Vo
=
M+m,
m
P13-4 The energy stored In the spring after compression is ~(193N/m)(0.0416m)2 = 0.167J.
Since 117 mJ was dissipated by friction, the kinetic energy of the block before colliding with the
spring was 0.284J. The speed of the block was then
v = y'2(0.284 J)/(1.34 kg) = 0.651m/s.
P13-5
(a) Using Newton's second law, F = ma, so for circular motion around the proton
e2
= F = k-;:2'
mv2
r
The kinetic energy of the electron in an orbit is then
K
1
=; -mv2
2
=
1
(1
The change in kinetic energy is
6.K
=
-ke
2
2
1 e2
-k-.
2
r
--r2
1
)
rl
(b) The potential energy difference is
1
r2 ke2
~u
( c ) The total
= --;¡-dr
rl
= -ke2
r
(1
--r2
1
)
rl
energy change is
'1
--.r2
1
rl
P13-6 (a) The initial energy of the system is (4000 lb)(12ft) = 48,000 ft .lb. The safety device
removes (1000 lb )(12ft) = 12,000 ft , lb before the elevator hits the spring, so the elevator has a
kinetic energy of 36, 000 ft .lb when it hits the spring. The speed of the elevator when it hits the
spring is
v
= .12(36,000 ft .lb)(32.0 ft/s1 = 240 ft/
V
(4000 lb)
.s.
(b) Assuming the safety clamp remains in effect while the elevator is in contact with the spring
then the distance compressedwill be found from
36,000 ft .lb = ~(10, 000 Ib/ft)y2-
162
(4000 lb)y + (1000 lb)y.
This is a quadratic expression in y which can be simplified to look like
5y2 -3y
-36
= O,
which has solutions y = {0.3 :f: 2.7) ft. Only y = 3.00 ft makes sense here.
{ c ) The distance through which the elevator will bounce back up is found from
33,000 ft = (4000 lb)y -(1000
lb)y,
where y is measuredfrom the most compressedpoint of the spring. Then y = 11 ft, or the elevator
bounces back up 8 feet.
(d) The elevator will bounce until it has traveled a total distance so that the safety device
dissipates all of the original energy, or 48 ft.
~
The net force on the top block while it is being pulled is
11.0N -F¡
= 11.0N -(0.35)(2.5kg)(9.81m/s2)
= 2.42 N.
This means it is accelerating at (2.42N)/(2.5kg) = O.968m/s2. That acceleration willlast a time
t = V2(0.30m)/(0.968m/s2) = O.787s. The speed of the top block after the force stops pulling is
then (0.968m/s2)(0.787s) = O.762m/s. The force on the bottom block is Ff, so the acceleration of
the bottom block is
(0.35)(2.5kg)(9.81 m/s2)/(10.0kg) = O.858m/s2,
and the speed after the force stops pulling on the top block is (0.858m/s2)(0.787s) = O.675m/s.
(a) W = Fs = (11.0N)(0.30m) = 3.3J of energy were delivered to the system, but after the
force stops pulling only
were present as kinetic energy. So 0.296J is "missing" and would be now present as internal energy.
(b) The impulse received by the two block system is then J = (11.0N)(0.787s) = 8.66N.s. This
impulse causesa changein momentum, so the speedof the two block system after the external force
stops pulling and both blocks move as one is (8.66N.g)(12.5kg) = 0.693m/s. The final kinetic energy
is
1
-(12.5
2
this means that
P13-8
0.0023
kg)(0.693m/s)2
are dissipated.
Hmm.
163
=
3.0023;
El4-1
E14-2 Consider the force from the Sun and the force from the Earth. Fs/FE = MSrE2/MErs2,
since everything else cancels out in the expression. We want the ratio to be one; we are also
constrained becauserE + rS = R is the distance from the Sun to the Earth. Then
ME (R -rE)2
R-rE
TE
=
MSrE2,
=
~
VM;
~rE,
=
{1.50 x 1011m)/
1
I (1.99
+y
x 1030kg)
(5.98
= 2.6 x lO8m.
x 10:¿4)
~
The masses of each object are ml = 20.0 kg and m2 = 7.0 kg; the distance between the
centers of the two objects is 15 + 3 = 18 m.
The magnitude of the force from Newton's law of gravitation is then
F = Gmlm2
-(6.67x
10-11N .m2/kg2)(20.0kg)(7.0kg)
r2
~
= 2.9x 10-11N.
.18 m):&
The force of gravity on an object near the surface of the earth is given by
GMm
(re + y)2 .,
F-
where M is the mass of the Earth, m is the mass of the object, re is the radius of the Earth, and
y is the height above the surface of the Earth. Expand the expression since y « re. We'l1 use a
Taylor expansion, where F(re + y) ~ F(re) + yaF/arej
GMm
-GMm
Since we are interested in the difference between the force at the top and the bottom, we really want
~F
= 2y ~
3
= 2JL~
re
where in the last part we substituted
re
2
= 2JLW
re
re
,
for the weight, which is the same as the force of gravity,
w
=
GMm
-=--.
2
re
Finally,
L\F
= 2(411
m)/(6.37x
106m)(120
164
lb)
= 0.015
lb.
E14-6
9 cx: l/r2,
so 91/92 = r~/r~ .Then
r2 = V(9.8lm/s2)/(7.35m/s2)(6.37x
lO6m) = 7.36xlO6m.
That's 990 kilometers above the surface of the Earth.
E14-7
(a) a = GM/r2,
or
(6.67 x 10-11 N. m2 /kg2)(1.99
l-
(10.0 x 103m)2
17. =
x 103Okg)
= 1.33x 1012m/s2.
(b) v = ..¡¡ax = v'2(1.33x 1012m/s2)(1.2m/s) = 1.79x 106m/s.
E14-8
(a) 90 = GM/r2,
or
go =
(b)
W m =
We(9m/ge)
(6.67x
lO-llN
.m2 /kg2)(7.36x
(l.74x
lO6m)2
lO22kg)
= l.62m/s2.
so
w m = (100 N)(1.62
m/s2/9.81
m/s2)
= 16.5 N.
(c) Invert 9 = GM/r2;
r = /GM1ii
=
(6.67x
lO-llN
.m2 /kg2)(5.98
x lO24kg)/(l.62m/s2)
= l.57x
lO7m.
That's 2.46 Earth radii, or 1.46 Earth radii above the surface of the Earth.
~
The object fell through y = -10.0mj
the time required to fall would then be
t = v=2ii7ii = V-2(-10.0m)/(9.81m/82) = 1.438.
We are interested in the error, that means taking the total derivative of y = -!gt2,
óy
8y = O so -~8gt
= g8t,
which
=
I
-2ógt2
can be rearranged
<5t =
and getting
-gtót.
as
-~
29
The percentage
(0.05%)(1.43s)
error in t needs to be 8t/t
= 0.7 ms.
= 0.1%/2
= 0.05%.
The absolute error is then 8t =
E14-10 Treat mass which is inside a spherical shell as being located at the center of that shell.
Ignore any contributions from shells farther away from the center than the point in question.
(a) F = G(M1 + M2)m/a2.
(b) F = G(Ml)m/b2.
(c)F=O.
165
~
For a sphere oí uniíorm
density
and radius
R > r,
M
M(r)
1;;:3
3
-FR3'
where M is the total mass.
The force of gravity on the object of mass m is then
GMm r3
-;:2w
F=
=
GMmr
R3
.
9 is the free-fall acceleration of the object, and is the gravitational
GMr
9= ~
force divided by the mass, so
GM r
GM R-D
= W
R = w
-:n--.
Since R is the distance from the center to the surface, and D is the distance of the object beneath
the surface, then r = R -D is the distance from the center to the object. The first fraction is the
free-fall acceleration on the surface, so
R-D
=98
l-Q
R
E14-12
The work required to move the object is GMsm/r,
But ir this equals mc2 we can write
mc2
R
where r is the gravitational
radius.
= GMsm/r,
GMS/C2.
For the sun, r = (6.67 x 10-11 N. m2 /kg2)(1.99 x 103Okg)/(3.00 x 108m/s)2 ;
2.1 x 10-6 Rs.
E14-13
1.47 x 103m. That's
The distance from the center is
= (80000)(3.00x 108m/8)(3.16x 1018)= 7.6x 102om.
The mass of the galaxy is
M = (l.4x
The
escape
velocity
v = ,f'iGMF
El4-14
so
lOll)(l.99x
lO30kg) = 2.8x lO41kg.
is
= y2(6.67X
lO-llN
.m2 /kg2)(2.8x
lO41kg)/(7.6x
Staying in a circular orbit requires the centripetal force be equal to the gravitational
mVorb 2 Ir = GMmlr
or mVorb2 = GMmlr.
kinetic
lO20m) = 2.2x lO5m/s.
But
-GMmlr
is the gravitational
2,
potential
energy
mvesc2/2
= GMm/r
which has solution vesc = y'2Vorbo
166
force,
= mVorb2,
energy;
to escape one requires
a
~
(a) Near the surface of the Earth the total energy is
but
CM
RE2'
g=
so the
total
energy
is
GMEm
,
RE
GMEm
RE-
,
RE
RE
This is a positive number, so the rocket will escape.
(b) Far from earth there is no gravitational potential energy, so
1
-mv2
2-
= GME
GMEm
R
2
m
RE
E
=
gmRE
,
RE
with solution v = V29RE.
E14-16
The rotational
acceleration of the sun is related to the galactic acceleration of free fall by
47r2mr/T2
= GNm2/r2,
where N is the number of "sun" sized stars of ma.ss m, r is the size of the galaxy, T is period of
revolution of the sun. Then
N =
47r2r3
GmT2
~
Energy
47r2(2.2 x 1020m)3
=
(6.67x 10-11N .m2 /kg2)(2.0x
conservation
is Ki + Ui = Kf+
= 5.1xl010.
1030kg)(7.9x 1015S)2
Uf, but at the highest
point
Kf
Uf
GMEm
R
1
R
1
R
R
The distance above the Earth's surface is 2.19x 107 m -6.37x
167
106m = 1.55 x 106 m.
= O, so
El4-18
(a) Free-fall acceleration is 9 = GM/r2. Escapespeedis v = ,fiGMr;;.
Then v = J2gr =
J2(1.30m/s2)(1.569xl06m) = 2.02xl03m/s.
(b) Uf = Ki + Ui. But U/m = -gor~/r, so
1
2.09x 106m.
That 's 523 km above the surface.
(c) Kf = Uj -Ufo But U/m = -gor~/r,
v =
so
2(l.30m/s2)(l.569xlO6m)2[l/(l.569xlO6m)
-l/(2.569xlO6m)]
= l260m/s.
(d) M = gr2/G, or
M
= (l.30m/s2)(l.569xlO6m)2/(6.67XlO-IIN.m2/kg2)
= 4.8xlO22kg.
= 3.34xlO7m/s.
(b)
Apply
~K
= -~U.
Then
mv2
= Gm2(1/r2
-l/rl),
SO
=
5.49
x 107 m/s.
El4-20
Call the particles 1 and 2. Then conservation of momentum requires the particle to have
the same momentum of the same magnitude, p = mVl = MV2. The momentum of the particles is
given by
Then Vrel = IVll + IV21 is equal to
Vrel
= mMv2G/d(m+M)
=
mM ..¡2G/d(m + M)
..¡2G(m
+ M)/d
El4-21
The maximum speed is mv2 = Gm2 / d, or v = ,fGm1d.
E14-22
Tl/r~
=
Ti/r~,
O!
T2 = T1(r2/rl)3/2
= (1.00
168
y)(1.52)3/2
= 1.87 y.
~
for M-
We can use Eq. 14-23 to find the mass of Mars; all we need to do is rearrange to solve
M = ~
=
47r2(9.4x106m)3
GT2
(6.67xl0-11N.m2/kg2)(2.75x104S)2
E14-24
Use GM/r2
M
El4-25
Tl/rr
= 6.5x1023k
9
.
= 47r2r/T2, so M = 47r2r3/GT2, and
=
47r2(3.82xl08m)3
(6.67xlO-llN.m2/kg2)(27.3
=
X
5.93xl024k
.
864008)2
9
= Ti /r~, or
T2 = T1(r2/rl)3/2
= (1.00 month)(1/2)3/2
= 0.354 month.
El4-26
Geosynchronous orbit W88 found in Sample Problem 14-8 to be 4.22 x 107m. The latitude
is given by
(} = arccos(6.37x106m/4.22x107m)
= 81.3°.
47r2
{6.67x
lO-llN
(l.74xlO6m)3
.m2 /kg2){7.36
= 4.24xlO7s2.
x lO22kg)
So T = 6.5 x 1038.
(a) The 8peed of the 8atellite i8 the circumference divided by the period, or
v = .27rr
~ 27r(1.74x 106m) = 1.68x103m/8.
-T
E14-28
-(6.5xlO3s)
The total energy is -GMm/2a.
1
-m
2
v
Then
2
--
GMm
GMm
~'
-
r
so
V2 =
CM
(
~
-.!.
r
E14-29
a
ra = a(l + e), so from Ex. 14-28,
Va=VGM(~-~);
rp = a(l -e),
so from Ex. 14-28,
Dividing one expression by the other,
-!2/(1-e)-1
Vp -vay
2/(1 + e) -1
= {3.72
km/s)y
169
12/0.121
27i":"88""=""l = 58.3
km/s.
~
El4-30
(a)
Convert.
3.156x
1078
2
AU
3
l.496xlOllm,
y
which is G = 39.49AU3/Ms2 .y2.
(b) Here is a hint: 47r2 = 39.48. Kepler's law then looks like
MS2
T2=
.y2
r3
M'
AU3,
~
Kepler's third law states T2 cx:r3, where r is the mean distance from the Sun and T is
the period of revolution. Newton was in a position to find the acceleration of the Moon toward the
Earth by assuming the Moon moved in a circular orbit, since ac = v2/ r = 47r2r/T2 .But this means
that, becauseof Kepler's law, ac cx:r/T2 cx:1/r2.
E14-32
( a) The force of attraction
between
the two bodies is
F=
GMm
(r+R)2"
The centripetal acceleration for the body of mass m is
CM
T¡.;2
¡;;-:¡:Rj'i'
CM
=
(¿}2
r3(1
+
47r2
T2
R/r)2
3
(
,
2
GMr1+R/r).
(b) Note that r = Md/(m+M)
and R = md/(m+M).
Then R/r = m/M, so the correction is
(1 + 5.94x 1024/1.99x 1030)2= 1.000006for the Earth/Sun 8y8tem and 1.025 for the Earth/Moon
8y8tem.
El4-33
(a) Use the results of Exercise 14-32. The center of mass is located a distance r =
2md/(m + 2m) = 2d/3 from the star of mass m and a distance R = d/3 from the star of mass 2m.
The period of revolution is then given by
2
47r2
T2
=
)
3
=
-d
G(2m)
3
47r2
-,
3Gm
(b) Use Lm = mr2(.¡), then
Lm =
LM
mr2
=
M:R2
m(2d/3)2
=2.
(2m)(d/3)2
(c) Use K = IúJ2/2 = mr2úJ2/2. Then
Km
KM
-
mr2
MR2
-
m(2d/3)2
(211i)(d/3)2
170
=2.
d3,
El4-34
Since we don't know which direction the orbit will be, we will assumethat the satellite
on the surface of the Earth starts with zero kinetic energy. Then Ei = Ui.
~U = Uf -Ui to get the satellite up to the specified altitude. ~K = Kf = -Uf/2. We want
to know if ~U -~K
is positive (more energy to get it up) or negative (more energy to put it in
orbit). Then we are interested in
The "break-even" point is when rf = 3ri/2 = 3(6400 km)/2
Earth.
(a) More energy to put it in orbit.
(b) Same energy for both.
(c) More energy to get it up.
~
9600 km, which is 3200 km above the
(a) The approximate force of gravity on a 2000 kg pickup truck on Eros will be
(b) Use
v =
~
=
V-;:-
~u
=
{6.67x lO-llN
-c-
y
(a) u = -GMm/r.
El4-36
1.(6067x 10-11N om2/kg2)(5.0 x 1015kg)
=
6.9m/s.
(7000 m)
The variation is then
.m2 /kg2){l.99
x lO30kg){5.98 x lO24kg)
= O, so I~KI
= 1.78x 1032J.
1
1
(1.47 x 1011m)
{1.52 x 1011m)
78 x 1032 J.
(b) ~K
+ ~U
= ~E
(c) ~E = 0.
(d) Since ~l = 0 and l = mvr,
Vp -Va
= Vp
we have
1-~
ra
But Vp ~ vav = 271"(1.5x 1011m)i(3.16
=
=Vp
X 107 s) = 2.98 x 104mis.
Then
3.29x
10
-2
Vp.
~v = 981 mis.
El4-37
Draw a triangle. The angle made by Chicago, Earth center, satellite is 47.5°.The distance
from Earth center to satellite is 4.22xl07 m. The distance from Earth center to Chicago is 6.37xl06m.
Applying the cosine law we find the distance from Chicago to the satellite is
(4.22x 107m)2+ (6.37x 106m)2-2(4.22x 107m)(6.37x 106m)cos(47.5°) = 3.82x 107m.
Applying the sine law we find the angle rilade by Earth center, Chicago, satellite to be
(
arcsin
{4.22X107m)
.
81]
{3.82 x 107 m)
That's 36° above the horizontal.
171
47.5°)
= 126°.
El4-38
(a) The new orbit is an ellipse with eccentricity given by r = a'(l + e). Then
e= r/a' -1 = (6.64x106m)/(6.52x106m) -1 = 0.0184.
The distance at P' is given by rp¡ = a'(l -e).
Upl = Up~
1 -e
The potential energy at P' is
= 2(-9.76x1010J)~
= -2.03x1011J.
1 -0.0184
The kinetic energy at P' is then
Kpl
That
would
= (-9.94xlOlOJ)
mean v = V2(l.O4x
-(-2.03xlOllJ)
lOllJ)/(3250kg)
= l.O4xlOllJ.
= 8000m/s.
(b) The average speed is
v=
27r(6.52 x lO6m) .
(--
.-'I
=
7820 m
/ s.
~
(a) The Starshine satellite was approximately 275 km above the surface of the Earth on
1 January 2000. We can find the orbital period from Eq. 14-23,
.41T2 .
=
T2
CM
r3,
471"2
{6.67x
lO-llN
(6.65 x 106m)3 = 2.91 x 10782,
.m2 /kg2){5.98x
lO24kg)
so T = 5.39 x 1038.
(b) Equation 14-25 give8 the total energy of the 8y8tem of a 8atellite of mas8 m in a circular orbit
of radiu8 r around a 8tationary body of mas8 M » m,
E=-GMm
2r
We want the rate of change of this with respect to time, so
dE
GMm dr
dt -2r2
dt
We can estimate the value of dr / dt from the diagram. I'll chooseFebruary 1 and December 1 as my
two referencepoints.
~r
dr
'""'-=
'""'
dt
t=to
~t
-(240~(300
km)
~
-I
km/day
(62 days)
The rate of energy 1088i8 then
dE
dt
=
~~)0-11N.m2/kg2)~24kg)(39kg)
-1000m
=
2(6.65 x lO6m)2
8.64
x
104
-2.0J/s.
s
~
The object on the top experiencesa force down from gravity W1 and a force down from
the tension in the rope T. The object on the bottom experiencesa force down from gravity W2 and
a force up from the tension in the rope.
In either case,the magnitude of Wi is
J.v:. =
~
GMm
~
r.
~
172
or
GMm
GMm
T =
.1
1 'I
2
,2-2
,rl
r21
GMm r~ -r~
-;¡r2 r2 .
1 2
Now rl ~ r2 ~ R in the denominator, but r2 = rl + I, so r~ -r~
T~
~ 2RI in the numerator.
Then
GMml
R3
Pl4-2
For a planet of uniform density, g = GM/r2
= G(47rpr3/3)/r2
=
47rGpr/3.
Then
ir p is
doubled while r is halved we find that g will be unchanged.
Pl4-3
(a) F = GMm/r2,
(b) The acceleration of
relative acceleration is then
is independent of m relative
Pl4-4
(a) 9 = GM/r2,
a = F/m = GM/r.
the Earth toward the center of mass is aE = F/M = Gm/r2.
The
GM/r + Gm/r = G(m + M)/r.
Only if M » m can we assume that a
to the Earth.
8g = -(2GM/r3)8r.
In this case 8r = h and M = 47rpr3/3. Then
8W = m 8g = 87rGpmh/3.
(b) 6.W/W = 6.g/g = 2h/r. Then an error of one part in a million will occur when h is one
part in two million of r, or 3.2 meters.
~
(a) The magnitude of the gravitational
FA=
force from the Moon on a particle at A is
GMm
¡;:-=Rj2'
173
To simplify assume R « r and then substitute
( r -R)
=
FT = GMm~2(r)2
~
( d ) Repeat
part
~ r. The force difference simplifies to
( c ) except we want r + R instead
2GMmR
r3
oí r -R.
Then
=
FA-Fc
To simplify assume R « r and then substitute
v
rT
=
(r + R) ~ r. The force difference simplifies to
GM m ~=-2GMmR
2
r (r ) 2
r3
The negative sign indicates that this "apparent" force points away from the moon, not toward it.
(e) Consider the directions: the water is effectively attracted to the moon when closer, but
repelled when farther.
Pl4-6
F net = mrl.l.Js2,
where I.l.Js
is the rotational speed of the ship. But since the ship is moving
relative to the earth with a speed v, we can write I.l.Js
= l.I.J::1::v/r,
where the sign is positive ifthe ship
is sailing east. Then F net = mr(1.I.J
::I::v/r)2.
The scale measuresa force W which is given by mg -F net, or
W = mg -mr(1.I.J::I::v/r)2.
Note that Wo = m(g -TIA)2). Then
w =
~
~
(a) a = GM/r2
Pl4-7
we
-r(,¡J2. (,¡Jis the rotational speed ofthe Earth. Since Frank observes a = 9/2
have
9/2
=
GM/r2
-r(.c)2,
r2
=
(2GM
-2r3(.c)2)/9,
r
=
J2(GM~(.c)2)/9
Note that
CM
= {6.67x
lO-11N
.m2 /kg2){5.98
x lO24kg) = 3.99x lO14m3/82
while
r3úJ2 = {6.37x
lO6m)3{27r/864008)2
174
= l.37x
lO12m3/82.
Consequently, r3(¿}2can be treated as a perturbation
ro
=
2[(3.99x 1014m3/s2) -(6.37x
rl
=
2[(3.99xl014m3/s2)
of GM bear the Earth. Solving iteratively,
106m)3(27r/86400s)2]/(9.81
m/s2) = 9.00x 106m,
-(9.00xl06m)3(27r/86400s)2]/(9.81m/s2)
= 8.98xl06m,
which is close enough for me. Then h = 8.98 x 106m -6.37 x 106m = 2610 km.
(b)~E=Ef-Ei=UfI2-Ui.
Pl4-8
Then
(a) Equate centripetal force with the force of gravity.
47r2mr
GMm
=
T2'
-;¡;;¡47r2
--.
~
T =
(b) T = V37r/(6.7xlO-llN
G(4/3)7rr3p
r3
rE
VGP
.m2/kg2)(3.0xl03kg/m3)
= 68008.
Pl4-9
(a) One can find 8g by pretending the Earth is not there, but the material in the hole is.
Concentrate on the vertical component of the resulting force of attraction. Then
8g =
GMd
-.-
r2 r'
where r is the straight line distance from the prospector to the center of the hole and M is the mass
of material that would fill the hole. A few substitutions later ,
8g =
47rGpR3d
3 ( v'iJ'""+X'
)3
or
(0.800)(d2
+ 2.25 x lO4m2) = d2,
which has solution d = 300 m. Then
R3 =
3(14.0x10-5m/s2)(300m)2
41T(6.7x 10'-11N .m2 /kg2)(2800 kg/m3) ,
so R = 250 m. The top of the cave is then 300 m -250 m = 50 m beneath the surface.
(b) All of the formulae stay the same except replace p with the difference between rock and water .
d doesn 't change, but R will now be given by
R3 = 41T(6.7
3(14.0~10-5m/s2)(300m)2
x 10-11 N. m2 /kg2)(1800 kg/m3) ,
so R = 292 m, and then the cave is located 300 m -292 m = 7 m beneath the surface.
175
Pl4-10
9 = GM/r2, where M is the mass enclosed in within the sphere of radius r.
dg = (G/r2)dM -2(GM/r3)dr,
so that 9 is locally constant if dM/dr = 2M/r. Expanding,
47rr2Pl
=
87rr2p/3,
Pl
=
2p/3.
Then
~
The force of gravity on the small sphere of mass m is equal to the force of gravity from
a solid lead sphere minus the force which would have been contributed by the smaller lead sphere
which would have filled the hole. So we need to know about the size and mass of the lead which was
removed to make the hole.
The density of the lead is given by
M
p = 1;:R3
3
The hole has a radius of R/2, so if the density is constant the mass of the hole will be
Mh = pV =
( R) 3
(FR3M ) 37r
4
2
=
M
8
The "hole" is closer to the small sphere; the center of the hole is d -R/2
whole lead sphere minus the force of the "hole" lead sphere is
GMm
(a) Use v = r.,;VR2 -r2,
Pl4-12
T
away. The force of the
G(M/8)m
where r.,;= VGME/R3.
Then
=
rT
rO dr
ro dr
Jo dt = JR d;:7dt = JR -;-'
=
rO
dr
JRr.,;~'
7r
2iJ)
Knowing that
we can find T = 1260 s = 21 min.
(b) Same time, 21 minutes. To do a complete journey would require four times this, or 27r/(JJ.
That's 84 minutes!
(c ) The answers are the same.
P14-13
(a) 9 = GM/r2
and M = 1.93x 1024kg + 4.01 x 1024kg + 3.94x 1022kg = 5.98x 1024kg so
9 = (6.67x 10-11N .m2 /kg2)(5.98x
1024kg)/(6.37x 106m)2 = 9.83m/s2.
(b) Now M = 1.93 x 1024kg + 4.01 x 1024kg = 5.94 x 1024kg so
9 = (6.67x 10-11N .m2 /kg2)(5.94x
1024kg)/(6.345x 106m)2 = 9.84m/s2.
(c) For a uniform body, 9 = 47rGpr/3 = GMr/R3,
9 = (6.67x 10-11N .m2 /kg2)(5.98x
so
1024kg)(6.345x 106m)/(6.37x
176
106m)3 = 9.79m/s2.
P14-14 (a) Use g = GM/r2, then
9 = {6.67x
10-11N
.m2 /kg2){1.93x
1024kg)/{3.490x
106m)2
= 106 m/s2.
The variation with depth is linear if core has uniform density.
(b) In the mantle we have 9 = G ( M c + M) / r2 , where M is the amount of the mass of the mantle
which is enclosed in the sphere of radius r. The density of the core is
The density of the mantle is harder to find,
We can pretend that the core is made up of a point mass at the center and the rest has a density
equal to that of the mantle. That point mass would be
Then
9 = GMpjr2
Find
+ 47rGpmrj3.
dg j dr , and set equal to zero. This happens
2M pfr3
when
= 471"Pmf3,
or r = 4.93 x lO6m. Then g = 9.29 m/s2. Since this is less than the value at the end points it must
be a minimum.
~
(a) We will use part of the hint, but we will integrate instead of assuming the bit about
gav; doing it this way will become important for later chapters. Consider a small horizontal slice of
the column of thickness dr. The weight of the material above the slice exerts a force F(r) on the
top of the slice; there is a force of gravity on the slice given by
dF=
GM(r) dm
r2
where M(r) is the mass contained in the sphere oí radius r,
4
-7rr
3
M(r)
3
po
Lastly, the mass oí the slice dm is related to the thickness and cross sectional area by dm = pA dr .
Then
~rdr
3
Integrate both sides of this expression. On the left the limits are O to F center,on the right the limits
are R to O; we need to throw in an extra negative sign because the force increases as r decreases.
Then
dF=
Divide
both
2
F =
-7rGAp2
3
sides by A to get the compressive
stress.
177
R2,
(b) Put in the numbers!
( c) Rearrange,
and then put in numbers;
R =
27r(6.67x
3(4.0xlO7N/m2)
lO-llN
.m2/kg2)(3000kg/m3)2
= l.8xlO5m.
Pl4-16
The two mass increments each exert a vertical and a horizontal force on the particle, but
the horizontal components will cancel. The vertical component is proportional to the sine of the
angle, so that
2Gm>..
y
Pl4-11
For any arbitrary point p the cross sectional area which is perpendicular to the axis
dA' = r2dn is not equal to the projection dA onto the surface of the sphere. It dependson the angle
that the axis makes with the normal, according to dA' = cos(}dA. Fortunately, the angle made at
point 1 is identical to the angle made at point 2, so we can write
dn1
dQ2,
dAl/r~
dA2/r~
But the mass of the shell contained in dA is proportional to dA, so
r~dml
Gmdml/r~
=
=
r~dm2,
Gmdm2/r~.
Consequently,the force on an object at point p is balanced by both cones.
(b) Evaluate f dn for the top and bottom halves of the sphere. Since every dn on the top is
balanced by one on the bottom, the net force is zero.
Pl4-18
Pl4-19
Assume that the small sphere is always between the two spheres. Then
w =
AUl
(6.67x
+ AU2,
10-llN
.m2 /kg2)(0.212kg)
[(7.16kg)
9.85 x 10-11 J.
178
-(2.53kg)]
I
1
~
Pl4-20
Note that ~mVesc2 = -UO. where Uo is the potential
Energy conservation gives
K
v
-
Ko + Uo.
1
2 1
2
2mvo -2mvesc ,
=
/~2
~
V vD -Vesc-.
energy at the burn-out
height.
2
~
(a) The force of one star on the other is given by F = Gm2/d2, where d is the distance
between the stars. The stars revolve around the center of mass, which is halfway between the stars
so r = d/2 is the radius of the orbit of the stars. If a is the centripetal acceleration of the stars, the
period of revolution is then
T=
~=
V -¡;-
~=
V -p
~.
V --a-;;;:-
The numerical value is
T =
.1
y {6.67x
167r2{1.12 x 1011m)3
10-11N
.m2 /kg2){3.22x
1030kg)
= 3.21 x 1078 = 1.02
y.
(b) The gravitational potential energy per kilogram midway between the stars is
-2~
= -2 (6.6~ x 10-~1 ~ .m2 /k~~)(~.22 x lO30kg)
~~
(1.12 x 1011m)
-3.84 x 109J /kg.
An object of mass M at the center between the stars would need (3.84x 109J /kg)M kinetic energy
to escape,this correspondsto a speed of
v
=
,fiKTM=
Pl4-22
(a) Each differential
on the particle,
V2(3.84xlOgJ/kg)=
8.76xlO4m/s.
mass segment on the ring contributes the same amount to the force
dF = ~::,
r2 r
where r2 = x2 + R2. Since the differential mass segmentsare all equal distance, the integration is
trivial, and the net force is
GMmx
F = (X2 + R2)3/2 ,
(b) The potential energy can be found by integrating
~u
= 10
{00 Fdx
= lo{00
with respect to x,
(.1:2~Mmx
+R2)3/2
d x- -GMm
~.
Then the particle of mass m will pass through the center of the ring with a speed v = .;;¡JS:ü7T;; =
V2GM7R.
179
~
(a)
Consider
the
following
diagram.
,
/
/
/
""'
I
..
\
\
,
The distance r is given by the cosine law to be
r2 = R2 + R2 -2R2
cos()
2R2(1-
cos()).
The force between two particles is then F = Gm2/r2. Each particle has a symmetric partner, so
only the force component directed toward the center contributes. If we call this the R component
we have
FR = Fcosa
= Fcos(90°
-0/2)
Fsin(O/2).
Combining,
Fnet
When there are only 9 particles the angles are in steps of 40°j the (}i are then 40°, 80°, 120°, 160°,
200°, 240°, 280°, and 320°. With a little patience you will find
'\:""'
= 6.649655,
~ij2)
L..t
1
-COS
().
.~
~
using these angles. Then F net = 3.32Gm2 / R2.
(b) The rotational period of the ring would need to be
v ~3.32Gm
.
Pl4-24
The potential energy ofthe system is U = -Gm2 Ir. The kinetic energy is mv2. The total
energy is E = -Gm2 Id. Then
dr
dt
= 2VGm(1/r -l/d),
180
so the time to come together
is
{o
dr
T = Jd 2vGm(1/r-
~{1
= Y 4Gm
l/d)
Jo
~
Y l=-xl
[d3
dx=
¡vam'
Pl4-25
(a) E = U/2 for each satellite, so the total mechanical energy is -GMm/r.
(b) Now there is no K, so the total mechanical energy is simply U = -2GMm/r.
2 is becausethere are two satellites.
(c) The wreckagefalls vertically onto the Earth.
Pl4-26
is
Let ra = a(l + e) and rp = a(l-
The factor of
e). Then ra + re = 2a and ra -rp = 2ae. So the answer
2(0.0167)(1.50
x 1011m) = 5.01 x 109m,
or 7.20 solar radii.
P14-27
Pl4-28
The net force on an orbiting star is
M
-;:2 + m4r2
F=Gm
This is the centripetal force, and is equal to 47r2mr/T2. Combining,
47r2
=
G
47=3(4M
+
m),
T2
so
T
=
47rVr3/[G(4M
+
m)].
P14-29 (a) v = ,¡a¡;¡¡r,
v =
so
(6.67x lO-llN
.m2 /kg2)(5.98x
(b) T = 27r(7.0lxlO6m)/(7.54xlO3m/s)
(c) Originally Ea = U /2, or
(6.67x
E = -2(7.01
10-11N
lO24kg)/(7.0l
x lO6m) = 7.54x lO3m/s.
= 5.84xlO3s.
.m2 /kg2)(5.98x
x 106m)
1024kg)(220kg)
After 1500 orbits the energy is now -6.25 x 109J -(1500)(1.40
distance from the Earth is then
r = -2(
(6.67x
¡0-llN
.m2 /kg2)(5.98
x ¡024kg)(220kg)
-6.46
x ¡ogJ)
The altitude is now 6790 -6370 = 420 km.
(d) F = (1.40x105J)/(27r7.01x106m)
= 3.2x10-3N.
(e) No.
181
= -6.25x
109J.
x 105J) = -6.46 x 109J. The new
= 6. 79x
¡O6m.
P14-30 Let the satel1ite S be directly overhead at some time. The magnitudE)of the speed is
equal to that of a geosynchronoussatel1ite T whose orbit is not inclined, but since there are both
paral1eland perpendicular componentsto the motion of S it wil1 appear to move north while "losing
ground" compared to T. Eventually, though, it must pass overhead again in 12 hours. When S is as
far north as it wil1 go (6 hours) it has a velocity which is paral1el to T, but it is located in a region
where the required speed to appear fixed is slower. Hence, it wil1 appear to be "gaining ground"
against the background stars. Consequently,the motion against the background stars appears to be
a figure 8.
P14-31 The net force of gravity on one star becauseof the other two is
2GM2
L2
The stars orbit about a point r = L/2cos{300)
,
-.-
)
2GM2
-cos
Mv2
v
2
from any star. The orbita! speed is then found from
0
M
cos(300)
(30
F=
[, /2 cos( 30° )
r
or
v
=
L2
,fGM7L.
Pl4-32
A parabolic path wi11eventually escape;this means that the speed of the comet at any
distance is the escapespeedfor that distance, or v = ,f2GMF. The angular momentum is constant,
and is equal to
~l = m VArA = m.'¡2GMrA.
For a parabolic path, r = 2rA/(1 + cosO). Combining with Eq. 14-21 and the equation before that
one we get
dO -~(1
+ COSO)2.
"dt -4rA2
The time required is the integral
T=
Note that vrA3jGM
~
VGM
f7r/2
lo
dO
(1+cosO)2
~
=VGM
()
2
3
is equal to lj27r years. Then thetime for the comet to move is
~
There are three forces on loose matter(of mass mo) sitting on the moon: the force of
gravity toward the moon, Fm = Gmmo/a2, the force of gravity toward the planet, FM = GMmo/(ra) 2, and the normal force N of the moon pushing the loosematter away from the center of the moon.
The net force on this loose matter is FM + N -Fm, this value is exactlyequal to the centripetal
force necessaryto keep the loose matter moving in a uniform circle. The period of revolution of the
loose matter is identical to that of the moon,
T = 2 7r ,¡;:¡;JG¡;¡
,
but since the loose matter is actually revolving at a radial distance r -a
182
the centripetal force is
Only if the normal force is zero can the loose matter can lift off, and this will happen when p c =
PM -Pm, or
M(r
-a)
r3
M
¡;:--::--aj'
-
Ma2
-
m
-~
-m(r
,
-a)2
a2(r -a)2
Ma2(r
-3r2a3
-a)3
+ 3ra4 -a4
=
=
Mr3a2
~
Let r = ax, then x is dimensionless; let fJ = m/M,
simplifies to
-mr3(r
( -r5
thenfJ
-a)2,
+ 2r4a -r3a2)
is dimensionless. The expression then
-3x2 + 3x -1 = {3(-X5 + 2x4 -X3).
If we assumethan x is very large (r » a) then only the largest term on eachside survives. This means
3x2 ~ {3x5, or X = (3/{3)1/3. In that case, r = a(3M/m)1/3. For the Earth's moon rc = 1.lx107m,
which is only 4,500 km away from the surface of the Earth. It is somewhat interesting to note that
the radius r is actually independent of both a and m if the moon has a uniform density!
183
~
The pressure
in the syringe
( 42.3
p=
E15-2
m =
The total
kg/m3)+(0.25x10-3m3)(1000
The weight is (18.7kg)(9.8m/s2)
F
=
AAp,
= 4.29x lO5Pa.
kg/m3)+(0.4x10-3m3)(800
kg/m3)
= 1.87 kg.
= 18N.
so
F = {3.43 m){2.08
E15-4
N)
7r(l.l2 x lO-2m/ 8)2
mass oí Huid is
(0.5x10-3m3)(2600
E15-3
is
B~V/V
=
-~p;
m){l.OO
V
~p
=
L3;
=
atm-
~V
~
0.962 atm){1.01
L2~L/3.
(140 x 109Pa)
x 105pa/
atm)
= 2.74 x 104N.
Then
~
=
3(0.85
2.74x
lO9Pa.
m)
~
There is an inward force F 1 pushing the lid closed from the pressureof the air outside the
boxj there is an outward force F2 pushing the lid open from the pressure of the air inside the box.
To lift the lid we need to exert an additional outward force F3 to get a net force of zero.
The magnitude of the inward force is F1 = p outA, where A is the area of the lid and p out is
the pressure outside the box. The magnitude of the outward force F2 is F2 = PinA. We are told
F3 = 108 lb. Combining,
=
Fl -F3,
PinA
=
PautA
Pin
=
Paut -F3/A,
F2
so Pin = (15 lb/in2
E15-6
(108 lb)/(12
in2) = 6.0 lb/in2.
h = Ap/ pg, so
h=
(0.05 atm)(1.01 x 105P~~)
~p
E15-8
6.p = (1024kg/m3)(9.81m/s2)(118m)
~
=
=
0.52
m.
(1000 kg/m3)([8lli/s2)
E15-7
then
-F3,
(1060kg/m3)(9.81
m/s2)(1.83m)
90 x 104Pa.
=
=
19x 106Pa.Add this to Poi the total pressureis
1.29 x 106Pa.
The pressure
P2 -Pl
differential
= -pg(y2
-Yl)
assuming
we don 't have a sewage pump:
= (926kg/m3)(9.8lm/s2)(8.l6m
We need to overcomethis pressure difference with the pump.
E15-10
{a) p = {1.00 atm)e-5.00/8.55 = 0.557 atm.
{b) h = {8.55 km) ln{1.00/0.500) = 5.93 km.
184
-2.08 m) = 5.52xlO4Pa.
E15-11 The mercury will rise a distance a on one side and fall a distance a on the other so that
the difference in mercury height will be 2a. Since the massesof the "excess" mercury and the water
will be proportional, we have 2apm = dpw, so
E15-12
(a) The pressure (due to the water alone) at the bottom oí the pool is
p =
(62.45
lb/ft3)(8.0
ft)
= 500 lb/ft2.
The force on the bottom is
F = (500 Ib/ft2)(80
ft)(30
ft) = 1.2 x 106 lb.
The averagepressure on the sirle is half the pressure on the bottom, so
F =
(250
lb/ft2)(80
ft)(8.0
ft)
= 1.6 x 105 lb.
The averagepressure on the end is half the pressure on the bottom, so
F = (250 lb/ft2)(30
ft)(8.0
ft) = 6.0x 104 lb.
(b) No, since that additional pressure acts on both sides.
~
(a) Equation 15-8 can be used to find the height y2 of the atmosphere if the density is
constant. The pressureat the top of the atmosphere would be P2 = O, and the height of the bottom
Yl would be zero. Then
y2 = {1.01 x 105pa)/ [{1.21 kg/m3){9.81 m/s2)] = 8.51 x 103m.
{b) We have to go back to Eq. 15-7 for an atmosphere which has a density which varies linearly
with altitude. Linear variation of density means
p=po
1--
Y'
Ymax
)
Substitute this into Eq. 15-7,
P2 -Pl
=
dy,
In this case we have Ymax = 2pl / (pg ) , SOthe answer is twice that in part ( a) , or 17 km.
E15-14
~p = (1000kg/m3)(9.8m/s2)(112m)
(1.1 x 106Pa)(1.22m)(0.590m) = 7.9x 105N.
= 1.1 x 106Pa. The force required is then F
185
E15-15 (a) Choose any infinitesimally small spherical region where equal volumes ofthe two fluids
are in contact. The denserfluid will have the larger mass. We can treat the system as being a sphere
of uniform mass with a hemisphere of additional mass being superimposed in the region of higher
density. The orientation of this hemisphereis the only variable when calculating the potential energy.
The center of mass of this hemisphere will be a low as possible only when the surface is horizontal.
So all two-fluid interfaces will be horizontal.
(b) If there exists a region where the interface is not horizontal then there will be two different
values for ~p = pgh, depending on the path taken. This means that there will be a horizontal
pressure gradient, and the fluid will flow along that gradient until the horizontal pressure gradient
is equalized.
E15-16 The mags of liquid originally in the first vessel is ml = pAh1; the center of gravity is
at hl/2, so the potential energy of the liquid in the first vessel is originally U1 = pgAh~/2. A
similar expression exists for the liquid in the second vessel. Since the two vesselshave the same
cross sectional area the final height in both containers will be hf = (h1 + h2)/2. The final potential
energy of the liquid in eachcontainer will be U f = pgA(h1 + h2)2/8. The work done by gravity is
then
U1 + U2 -2Uf,
pg~
[2h~
+
2h~
-(h~
+
2h1h2
+
h~)]
4
~(hl
4
-h2)2,
~
There are three force on the block: gravity (W = mg), a buoyant force Bo = mwg, and a
tension To. When the container is at rest all three forces balance, so Bo -W -To = o. The tension
in this case is To = (mw -m)g.
When the container accelerates upward we now have B -W -T = ma. Note that neither the
tension northe buoyant force stay the same; the buoyant force increases according to B = mw(g+a).
The new tension is then
T = mw(g
E15-18
(a) F1/d'f=
+ a) -mg
-ma
= (mw
-m)(g
+ a) = To(l+
a/g).
F2/d~, so
F2 = (18.6 kN)(3.72
cm)2/(51.3
cm)2 = 97.8 N.
(b) F2h2 = F1h1, so
h2
=
(1.65m)(18.6
kN)/(97.8N)
=
314m.
E15-19
(a) 35.6 kN; the boat doesn't get heavier or lighter just because it is in different water!
(b) Yes.
=
E15-20
(a) P2 = Pl(V1/V2) = (1000kg/m3)(0.646) = 646kg/m3.
(b) P2 = Pl(V1/V2) = (646 kg/m3)(0.918)-1 = 704kg/í:n3.
186
-8.51
x
10-2m3.
~
The can has a volume of 1200 cm3, so it can displace that much water.
provide a buoyant force of
This would
B = pVg = (998kg/m3)(1200 x 10-6m3)(9.81 m/s2) = 11.7N.
This force can then support a total mass of (11.7N)/(9.81m/s2) = 1.20kg. If 130 9 belong to the
can, then the can will be able to carry 1.07 kg of lead.
3
E15-22
P2 = Pl{V1/V2)
E15-23
Let the object have a mass m. The buoyant force of the air on the object is then
= {0.98
g/cm3){2/3)-1
=
1.47 g/cm
Bo = ~mg.
Po
There iB alBoa buoyant force on the brasB,equal to
Bb
=
Pa
-mg.
Pb
The fractional error in the weighing is then
Bo -Bb
mg
E15-24
The volume
-(0.00122/cm3)
-(3.4
-(0.0012
g/cm3)
g/C~3)
(8.0
g/cm
= 2.0x
10-4
)
oí iron is
Vi = (6l3ÜN)/(9.8lm/s2)(787ükg/m3)
= 7.94xlü-2m3.
The buoyant force of water is 6130 N -3970 N = 2160 N. This corresponds to a volume of
v w = (2160 N)/(9.81
The volume
E15-25
Ft
of air is then 2.20 x 10-1m3
m/s2)(1000
-7.94
kg/m3)
x 10-2m3
= 2.20 x 10-1m3.
= 1.41 x 10-1m3.
(a) The pressure on the top surface is p = Po + pgL/2.
=
(Po + pgL/2)L2,
[(1.01 x 105Pa) + (944 kg/m3)(9.81 m/s2)(0.608m)/2]
(b) The pressure on the bottom surface is p = Po + 3pgL/2.
Fb
=
The downward force is
(0.608m)2 = 3.84x 104N,
The upward force is
(po + 3pgL/2)L2,
[(1.01
x 105Pa)
+ 3(944kg/m3)(9.81
m/s2)(0.608m)/2]
(c) The tension in the wire is given by T= W+ Ft -Fb,
T = { 4450 N) + {3.84 x 104N) -{
(0.608m)2
or
4.05 x 104N) = 2350 N .
(d) The buoyant force is
B = L3 pg =
(0.6083)(944
kg/m3)(9.81
187
m/s2)
= 2080 N .
= 4.05x
104N.
~
E15-26
The fish has the ( average) density of water if
mf
Pw=Vc+Va
or
mf
va=
-VC'
Pw
We want the fraction V a/(V c + Va), so
Va
1
Vc
-Pw~
Vc+Va
mf
1-
Pw/ Pc = 1-
(1.024
g/cm3)/(1.08
g/cm3)
= 5.19 x 10-2.
~
There are three force on the dirigible: gravity (W = mgg), a buoyant force B = mag,
and a tension T. Since these forces must balance we have T = B -w.
The masses are related to
the densities, so we can write
T = (Pa -Pg)Vg
E15-28
= (l.2lkg/m3
-O.796kg/m3)(l.l7xlO6m3)(9.8lm/s2)
= 4.75xlO6N.
~m = ~pV, so
~m
= [(0.160kg/m3)
-(0.0810
kg/m3))(5000m3)
= 395kg.
E15-29 The volume of one log is 7r(1.05/2 ft)2(5.80 ft) = 5.02 ft3. The weight of the log is
(47.3 lb/ft3)(5.02 ft3) = 237 lb. Each log if completely submerged will displace a weight of water (62.4 lb/ft3)(5.02 ft3) = 313 lb. So each log can support at most 313 lb- 237 lb = 76 lb. The
three children have a total weight of 247 lb, so that will require 3.25 logs. Round up to four.
E15-30
(a) The ice will hold up the automobile if
Pw >
ma+mi
Vi
ma
At +Pi.
Then
~
If there were no water vapor pressure above the barometer then the height of the water
would be yl = p/(pg), where p = Po is the atmospheric pressure. If there is water vapor where there
should be a vacuum, then p is the difference, and we would have y2 = (Po -Pv ) / (pg ) .The relative
error is
(Yl
-Y2)/Yl
=
[Po/(pg) -(Po -Pv )/(pg)] / [Po/(pg)] ,
pv/po = (3l69Pa)/(l.OlxlO5Pa)
= 3.14%.
E15-32
p =
E15-33
h = (90)(l.Ol x lO5pa)/(8.60 m/s2)(l.36
E15-34
~U = 2(4.5x lO-2N/m)47r(2.l
(1.01
x 105pa)/(9.81
m/s2)(14m)
= 740kg/m3.
x lO4kg/m3) = 78 m.
x lO-2m)2 = 5.0x lO-4J.
188
~
The force required is just the surface tension times the circumference of the circular
patch. Then
F
E15-36
ilU
=
(0./072N/m)27r(O.l2m)
=
= 2(2.5xlÜ-2N/m)47r(l.4xlÜ-2m)2
5.43x
lO-2N.
= l.23xlÜ-4J.
P15-1 (a) One can replace the two hemisphereswith an open flat end with two hemisphereswith
a closed flat end. Then the area of relevanceis the area of the flat end, or 7rR2.The net force from
the pressure difference is 6.pA = 6.p7rR2;this much force must be applied to pull the hemispheres
apart.
(b) F = 7r(O.9)(1.01x 105Pa)(0.305m)2= 2.6x 104N.
~
( a) The resultant
force on the wall will be
F
=
!!Pdxdy,
= 11 -pgy)Wdy,
= pgD2W/2.
(b) The torque will is given by T = F(D-y)
T =
(the distance is from the bottom) so ifwe generalize,
11pydxdy,
=
!(-pg(D-y))yWdy,
=
pgD3W/6.
( c ) Dividing to find the location of the equivalent resultant force,
d = T/F = (pgD3W/6)/(pgD2W/2)
= D/3,
this distance being measured from the bottom.
P15-4
p = pgy = pg(3.6 m); the force on the bottom is F = pA = pg(3.6 m)7r(O.60 m)2 = 1.2967rpg.
The volume of liquid is
v = (1.8m)
The weight
is W = pg(2.037m3).
[7r(0.60m)
The ratio
= 2.037m3
is 2.000.
P15-5
The pressure at b is pc(3.2xl04m)+pmY.
Set these quantities equal to each other:
Pc(3.8 x 104m + d) + Pm(Y -d)
+4.6x10-4m2]
The pressure at a is pc(3.8xl04m+d)+pm(Y-d).
=
Pc(3.2 x 104m) + PmY,
Pc(6 x 103m + d)
=
Pmd,
d
=
Pc(6x 103m)/(pm -Pc),
(2900kg/m3)(6x
189
103m)/(400kg/m3)
= 4.35 x 104m.
P15-6
(a) The pressure (difference) at a depth y is Ap = pgy. Since {1= m/V,
then
m AV
Ap
Ap ~ -Vv
= PsB.
Then
(b)
~p/
Ps
Psgy/B,
~p/
~
so
p ~
(1000
kg/m3)(9.8m/s2)(4200m)/(2.2x
(a) Use Eq. 15-10, p = (Poi Po)p, then Eq.
10gpa)
15-13 willlook
= 1.9%.
like
(Poi Po)p = (POI po)POe-h/a.
(b) The upward velocity of the rocket as a function of time is given by v = art. The height of
the rocket above the ground is given by y = !art2. Combining,
Put this into the expressionfor drag, along with the equation for density variation with altitude;
D = CApv2
= CApoe-Y/a2yaro
Now take the derivative with respect to y,
dD/dy
= ( -1/a)CApoe-Y/a(2yar)
+ CApoe-Y/a(2ar).
This will vanish when y = a, regardless of the acceleration aro
P15-8
(a) Consider a slice of cross section A a depth h beneath the surface. The net force on the
Huid above the slice will be
Fnet = ma = phAg,
Since the weight of the Huid above the slice is
W=mg
=phAg,
then the upward force on the bottom of the Huid at the slice must be
w + Fnet = phA(g
+ a),
so the pressure is p = F/A = ph(g + a).
(b) Change a to -a.
(e) The pressure is zero (ignores atmospherie eontributions.
P15-9 (a) Consider a portion of the liquid at the surface. The net force on this portion is F =
ma = mai. The force of gravity on-.this yort~on is W A= ~mgj. There must then be a buoyant
force on the portion with direction B = F -W = m(ai + gj). The buoyant force makes an angle
(} = arctan(a/g) with the vertical. The buoyant force must be perpendicular to the surface of the
fluid; there are no pressure-relatedforces which are parallel to the surface. Consequently,the surface
must make an angle (} = arctan(a/g) with the horizontal.
(b) It will still vary as pghj the derivation on page 334 is still valid for vertical displacements.
190
~
P15-10
dp/dr
= -pg,
but
now
g = 47rGpr/3.
Then
fP
lo
dp
=
4
-37rGp2
p
=
2
37rGp2
fr
lR
(R2
rdr,
-r2)
~
We can start with Eq. 15-11, except that we'll write our distance interms
if y. Into this we can substitute our expression for g,
9 =
of r instead
R2
go2.
r
Substituting, then integrating,
dp
~dr
Po
,
p
goPoR2 dr
=
dp
p
dp
1~
In
-
p
p
Po
If k = 9OPoR2/PO,
then
p
=
poek(l/r-l/R)
P15-12
(a) The net force on a small volrime of the fluid is dF = r(Ll2dm directed toward the
center. For radial displacements, then, dF/dr = ~r(Ll2dm/dr or dp/dr = -r(Ll2p.
(b) Integrating
outward,
r
1
p = Pc + lo fJ'I.J2rdr = Pc + 2pr2(A.!2.
(c) Do part (d) first.
(d) It will still vary as pgh; the derivation on page 334 is still valid for vertical displacements.
(c) The pressure anywhere in the liquid is then given by
1
p = Po + 2pr2iJ)2-pgy,
where Po is the pressure on the surface, y is measured from the bottom of the paraboloid, and r is
measuredfrom the center. The surface is defined by p = Po, so
1
-pr2iJ)2 -pgy = O,
2
or
y
=
r2úJ2/29.
P15-13
The total m88S of the shell is m = Pw7rdo3/3, orit
in the shell is m = pj7r(do3 -dj3)/3,
so
d .3
l
-Pi
-~
-Pw
d
wouldn't barely float. The m88S of iron
3
O ,
Pí
so
(1000kg/m3){0.587m)
di = 3 (7870kg/m3)(7870kg/m3)
191
= 0.561 m.
P15-14 The wood will displace a volume ofwater equal to (3.67kg)/(594kg/m3)(0.883) = 5.45x
10-3m3 in either case. That corresponds to a mass of (1000kg/m3)(5.45 x 10-3m3) = 5.45kg that
can be supported.
(a) The mass of lead is 5.45kg -3.67kg = 1.78kg.
(b) When the lead is submergedbeneath the water it displaceswater, which affects the "apparent"
massof the lead. The true weight of the lead is mg, the buoyant force is (Pw/ Pl)mg , so the apparent
weight is (1- pw/Pl)mg. This means the apparent mass of the submerged lead is (1- pw/Pl)m.
This apparent mass is 1.78 kg, so the true mass is
(11400kg/m3)
m = (11400kg/m3)
-(¡¡j¡j¡¡kg)(1.78kg)
~
We
initially
= 1.95kg.
have
! -Po
4 -Pmercury
When water is poured over the object the simple relation no longer works.
Once the water is over the object there are two buoyant forces: one from mercury, F1, and one
from the water, F2. Following a derivation which is similar to Sample Problem 15-3, we have
F1 = Pl V1g and F2 = P2V2g
where Pl is the density of mercury, V1 the volume of the object which is in the mercury, P2 is the
density of water, and V2 is the volume of the object which is in the water. We also have
F1 + F2 = PoVog and V1+ V2 = Vo
as expressionsfor the net force on the object (zero) and the total volume of the object. Combining
these four expressions,
Pl Vl + P2V2 = PoV o,
or
=
PI VI + P2 (Vo -VI)
(PI -P2)
VI
=
VI
=
PoVo,
(Po-p2)V(
Po-P2
Vo
Pl -P2
The left hand side is the fraction that is submerged in the mercury, so we just need to substitute
our result for the density of the material from the beginning to solve the problem. The fraction
submerged after adding water is then
V1
Vo
=
Po-P2
Pl
Pl/4-
-P2
P2
P15-16 (a) The car floats if it displaces a mass of water equal to the mass of the car. Then
V = (1820kg)/(1000kg/m3) = 1.82m3.
(b) The car has a total volume of 4.87m3+0.750m3+0.810m3 = 6.43m3. It will sink ifthe total
mass inside the car (car + water) is then (6.43m3)(1000kg/m3) = 6430kg. So the massofthe water
in the car is 6430kg-1820kg = 4610kg when it sinks. That's a volume of (4610kg)/(1000kg/m3) =
4.16m3.
192
P15-17 When the beaker is halffilled with water it has a total massexactly equal to the maximum
amount of water it can displace. The total mass of the beaker is the mass of the beaker plus the
mass of the water inside the beaker. Then
pw(mg/pg
+ Vb) = mg + PwVb/2,
where mg/ Pg is the volume oí the glass which makes up the beaker. Rearrange,
Pg
P15-18
=
mg/pw
mg-Vb/2
--
(0.390 kg)
(0.390 kg)/(1000 kg/m3) -(5.00
=
2790kg/m3.
x 10-4 m3)/2
(a) If each atoro is a cube then the cube has a side of length
l = {/(6.64x 10-27 kg)/(145kg/m3) = 3.58x 10-1Om.
Then the atomic surface density is l-2 = (3.58x 10-10m)-2 = 7.8x 1018/m2.
(b) The bond surface density is twice the atomic surface density. Show this by drawing a square
array of atoms and then joining each adjacent pair with a bond. You will need twice as many bonds
as there are atoms. Then the energy per bond is
(3.5 x lO-4N/m)
2(7.8 x 1018/m2)(1.6 x 10-19 J/eV)
= 1.4x
10-4
eV.
~
Pretend the bubble consists of two hemispheres. The force from surface tension holding
the hemispheres together is F = 2'YL = 47rr'Y. The "extra" factor of two occurs because each
hemispherehas a circumference which "touches" the boundary that is held together by the surface
tension of the liquid. The pressuredifference between the inside and outside is ~p = F / A, where A
is the area of the flat side of one of the hemispheres,so ~p = (47rr'Y)/(7rr2)= 4'Y/r .
P15-20 Use the results of Problem 15-19. To get a numerical answer you need to know the surface
tension; try 'Y = 2.5x 10-2N/m. The initial pressure inside the bubble is Pi = Po + 4'Y/ri. The final
pressure inside the bell jar is P = Pf -4'Y / rf .The initial and final pressure inside the bubble are
related by piri3 = Pfrf3. Now for numbers:
Pi = {1.00x105Pa)
+4{2.5x10-2N/m)/{1.0x10-3m)
Pf = {1.0 x 10-3m/1.0
x 10-2m)3{1.001
p = {1.001 x 102Pa) -4{2.5x
= 1.001 x 105Pa.
x 105Pa) = 1.001 x 102Pa.
10-2N/m)/{1.0x
10-2m)
= 90.1Pa.
P15-21 The force on the liquid in the space between the rod and the cylinder is F = 'YL =
27r'Y(R+ r). This force can support a mass of water m = F/g. This mass has a volume V = m/p.
The cross sectional area is 7r(R2-r2), so the height h to which the water rises is
h
=
27r"((R
pg7r(R2
+ r)
-r2)
=
2(72.8x
pg(R
2"(
-r)
,
lO-3N/m)
(1000 kg/m3)(9.81 m/s2)( 4.0 x 10-3m)
193
= 3.71x 10-3m.
P15-22
(a) Refer to Problem 15-19. The initial pressure difference is
4(2.6x lO-2N/m)/(3.20x
lO-2m) = 3.25Pa.
(b) The final pressure difference is
4(2.6x lO-2N/m)/(5.80x
lO-2m) = l.79Pa.
(c) The work done against the atmosphere is p~V, or
(d) The work done in stretching the bubble surface is 'YAA, or
(2.60x
10-2N/m)47r[(5.80x
10-2m)2
-(3.20
194
x 10-2m)2]
= 7.65x 10-43.
E16-1
R = Av = 7r~v/4
and V = Rt, so
t =
E16-2
Then
Alvl
= A2v2, Al
4(1600
m3)
7r(0.345m)2(2.62m)
=
65308
= 7rd~/4 for the hose, and A2 = N7r4
V2
for the sprinkler,
where N = 24.
(0.75 in)2
=
~
We'll assume that each river has a rectangular cross section, despite what the picture
implies. The cross section area of the two streams is then
Al = (8.2m)(3.4m) = 28m2 and A2 = (6.8m)(3.2m) = 22m2.
The volume flow rate in the first stream is
Rl = A1vl
= (28m2)(2.3m/s)
= 64m3/s,
while the volume flow rate in the second stream is
R2 = A2v2
= (22m2)(2.6m/s)=
57m3/s.
The amount of fluid in the stream/river system is conserved,so
R3 = Rl + R2 = (64m3/s) + (57m3/s) = 121m3/s.
where R3 is the volume flow rate in the river. Then
D3 = R3/(V3W3) = (121m3/s)/[(10.7m)(2.9m/s)]
=3.9m.
E16-4 The speed of the water is originally zero so both the kinetic and potential energy is zero.
When it leaves the pipe at the top it has a kinetic energy of ~{5.30m/s)2 = 14.0J/kg and a
potential energy of {9.81m/s2){2.90m) = 28.4J/kg. The water is flowing out at a volume rate of
R = {5.30m/s)7r{9.70 x 10-3m)2 = 1.57x 10-3 m3/s. The mass rate is pR = {1000kg/m3){1.57 x
10-3m3/s) = 1.57kg/s.
The power supplied by the pump is {42.8J/kg){1.57kg/s) = 67.2W.
~
There are 8500 km2 which collects an average oí (0.75) (0.48 m/y), where the 0.75 reflects
the íact that 1/4 oí the water evaporates, so
R =
(8500(103
m)2]
(0. 75) (0.48 m/y)
Then the speed of the water in the river is
v = R/A = (97m3 /s)/[(21 m)(4.3m)]
E16-7
(a) ~p = pg(yl -Y2) + p(V~ -v~)/2.
~p = (62.4 lb/ft3)(572
(b) A2v2 = A1vl,
ft) + [(62.4 lb/ft3)/(32
= 1.1 m 18.
Then
ft/s2)][(1.33
ftjS)2-
(31.0 ft/s)2]/2
SO
A2 =
(7.60
ft2)(1.33
ft/s)/(31.0
195
ft/s)
= 0.326
ft2,
= 3.48x1041b/ft2.
E16-8
(a)
A2v2
= A1vl,
V2 =
(b)
dp
= p(v~
so
(2.76m/s)[(0.255m)2
-(O.O480m)2j/(0.255m)2
2.66m/s.
-v~)/2,
~p = (1000 kg/m3)[(2.66m}s)2
~
=
(b) We will do part
R =
-(2.76m/s)2J/2
= -271 Pa
(b) first.
(100 m2)(1.6
1 y
m/y)
=
5.1
x 10-6
m3/s.
365 x 24 x 60 x 608
(b) The speed ofthe flow R through a hole of cross sectional area a will be v = R/a. p = Po+pgh,
where h = 2.0 m is the depth of the hole. Bernoulli's equation can be applied to find the speed of
the water as it travels a horizontal stream line out the hole,
1
2
po+2PV
=P,
where we drop any terms which are either zero or the same on both sides. Then
v = ..¡2(p -Po)/p
= J29h = ..¡2(9.81 m/s2)(2.0m)
Finally, a = (5.1x10-6m3/s)/(6.3m/s)
= 6.3m/s.
= 8.1 X 10-7m2, or about 0.81 mm2.
E16-10
{a) V2 = {Al/A2)Vl = {4.20 cm2){5.18m/s)/{7.60
{b) Use Bernoulli's equation:
P2 + pgy2
cm2) = 2.86m/s.
1
2
+ 2PV2 = Pl + pgyl
1
2
+ 2PVl'
Then
P2
=
2.56 x lO5Pa.
E16-11
(a) The wind speed is (110 km/h)(1000
difference is then
~p
m/km)/(3600
1
=
-(1.2kg/m3)(30.6m/s)2
2
=
s/h) = 30.6m/s.
562Pa.
(b) The lifting force would be F = (562Pa)(93m2) = 52000N.
E16-12
The pressure difference
is
1
~p = 2(1.23kg/m3)(28.0m/s)2
= 482N.
The net force is then F
(482 N)(4.26m)(5.26m)
= lO800N.
196
The
pressure
~
The lower pipe has a radius rl = 2.52 cm, and a cross sectional area of Al = 7rr~.The
speed of the fluid flow at this point is Vl. The higher pipe has a radius of r2 = 6.14 cm, a cross
sectional area of A2 = 7rr~, and a fluid speed of V2. Then
A
-
A
lVl-
2-
2
2V20rrlvl-r2v2.
Set yl = O for the lower pipe. The problem specifies that the pressures in the two pipes are the
same, so
1 2
= Po+ 2PV2+ pgy2,
1 2 + gy2,
2V2
=
We can combine the results of the equation of continuity with this and get
v~
V2
V~ (1-
v~ + 2gy2,
=
(Vlr~/r~)2
=
2gy2,
=
2gy2/ (1-
rt/r~)
V2
Then
1
=
+ 2gy2,
rf/ri)
.
1
v~ = 2(9.81m/s2)(11.5m)/ (1- (0.0252m)4/(0.0614m)4) = 232m2/s2
The volume fiow rate in the bottom (and top) pipe is
R = 1I"r~Vl= 11"(0.0252
m)2(15.2m/s) = 0.0303m3/s.
E16-14
(a) As instructed,
1 2
Po+ 2PVl + pgyl -Po
=
1 2
+ 2PV3+ pgy3,
~V~+ g(y3-Yl),
But y3 -yl = -h, so V3 = J29h.
(b) h above the hole. Just reverse your streamline!
(c) It won't come out as fast and it won't rise as high.
~
Sea level will be defined as y = O, and at that point the Huid is assumed to be at rest.
Then
1 2
Po+ 2PVl + pgyl
=
1 2
Po + 2PV2+ pgy2,
=
1 2
2V2 + gy2,
where y2 = -200 m. Then
V2 = ,p¡gij;
= V-2(9.81m/s2)(-200m)
197
= 63m/s.
E16-16
Assume
streamlined
flow, then
1 2
= P2 + 2PV2
+ pgy2,
(Pl
-p2)/
p +
g(yl
1 2
=
-Y2)
2
.
-V2
Then upon rearranging
V2 =
..¡2 [{2.1){1.01
x 105Pa)/{1000kg/m3)
+ {9.81
m/s2){53.0m)]
= 38.3m/s.
E16-17 (a) Points 1 and 3 are both at atmospheric pressure, and both will move at the same
speed. But since they are at different heights, Bernoulli's equation will be violated.
(b) The flow isn't steady.
E16-18 The atmospheric pressuredifference betweenthe two sideswill be ~p = ~PaV2.The height
difference in the U-tube is given by ~p = Pwgh. Then
~
(a) There are three forces on the plug. The force from the pressure of the water, F1 =
PIA, the force from the pressure of the air, F2 = P2A, and the force of friction, F3. These three
forces must balance, so F3 = F1 -F2, or F3 = PIA -P2A. But Pl -P2 is the pressure difference
between the surface and the water 6.15 m below the surface, so
F3
=
L\.p A = -pgyA,
-(998
kg/m3)(9.81
m/s2)( -6.15
m)7r(O.O215 m)2,
87.4N
(b) To find the volume of water which flows out in three hours we need to know the volume
flow rate, and for that we need both the cross section area of the hole and the speed of the flow.
The speed of the flow can be found by an application of Bernoulli's equation. We'll consider the
horizontal motion only- a point just inside the hole, and a point just outside the hole. These points
are at the same level, so
1 2
Pl + 2PVl + pgyl =
Pl
I 2
P2 + 2PV2+ pgy2,
-P2
I
2
+ 2PV2.
Combine this with the resuIts of PascaI's principIe above, and
The volume oí water which flows out in three hours is
v = Rt = (11.0m/8)1r(0.0215m)2(3
E16-20
x 36008) = 173m3.
Apply Eq. 16-12:
Vi =
y2(9.81
m/s2)
(0.262
m)(810
kg/m3)
198
/ (1.03
kg/m3)
= 63.6 m/s.
~
We'll agsume that the central column of air down the pipe exerts minimal force on the
card when it is deflected to the sides. Then
1 2
Pl + 2PVl + pgyl
=
1 2
P2 + 2PV2 + pgy2,
Pl
=
1 2
P2 + 2PV2.
The resultant upward force on the card is the area of the card times the pressure difference, or
1
= 2pAv2.
F = (Pl -P2)A
E16-22 If the air blows uniformly over the surf8{:eof the plate then there can be no torque about
any axis through the center of mass of the plate. Since the weight also doesn't introduce a torque,
then the hinge can't exert a force on the plate, becauseany such force would produce an unbalanced
torque. Consequently mg = ApA. Ap = pv2/2, so
'fv = ~
=
(l.:i~-:J!~fl~~{~~
7t~~k)2 = 3O.9m/s.
E16-23 Consider a streamline which passesabove the wing and a streamline which passesbeneath
the wing. Far from the wing thetwo streamlines are close together, move with zero relative velocity,
and are effectively at the same pressure. So we can pretend they are actually one streamline. Then,
since the altitude difference between the two points above and below the wing (on this new, single
streamline) is so small we can write
The lift force is then
L
E16-24
=
~pA
=
~pA(Vt2
-Vu2)
(a) From Exercise 16-23,
L= ~(1.17kg/m3)(2)(12.5 m2) {(49.8m/s)2 -(38.2m/s)2]
;;;;1.49x 104N.
The mass of the plane must be m =L/g = (1.49x 104N)/(9.81 m/s2) = i520 kg.
(b) The lift is directed straight up.
(c) The lift is directed 15° off thevertical toward the rear of the plane.
(d) The lift is directed 15° off the vertical toward the front otthe plane.
~
The larger pipe has a radius rl = 12.7 cm, and a cross sectiona1area of Al = 1rr~.The
speed ofthe fluid flow at this point is Vl. The smaller pipe has a radius of r2 = 5.65 cm, a cross
sectional area of A2 = 1rr~, and a fluid speed of V2. Then
A1vl = A2V2or r~vl = r~v2.
Now Bernoulli's equation. The two pipes are at the same level, so Yl = y2. Then
1 2
Pl + 2PVl + pgyl
=
1 2
Pl + 2PVl =
199
1 2
P2 + 2PV2+ pgy2,
1 2
P2 + 2PV2.
Combining this with the results from the equation of continuity,
I
2
Pl + 2PVl
v~
2
vIl
v~
(
4
rI
)
-~
v~
=
2
p
(P2
(1-
-Pl)
rt/r~)
.
It may laak a mess, but we can salve it ta find v¡
R = Av = 7r(O.127m)2(1.41
mis)
= 7.14x lO-3m3is.
That's about 711iters/second.
E16-26 The lines are parallel and equally spaced,so the velocity is everywhere the same. We can
transform to a referenceframe where the liquid appears to be at rest, so the Pascal's equation would
apply, and p + pgy would be a constant. Hence,
Po =p+pgy+
1
2PV2
is the same for all streamlines.
E16-27
(a) The "particles" of fluid in a whirlpool would obey conservation of angular momentum,
meaning a particle off mass m would have l = mvr be constant, so the speed of the fluid as a function
of radial distance from the center would be given by k = vr, where k is some constant representing
the angular momentum per mass. Then v = k/r .
(b) Since v = k/r and v = 27rr/T, the period would be T cx:r2.
(c) Kepler's third law says T cx:r3/2.
E16-28
~
pipe.
Rc
=
2000.
Then
(a) The volume flux is given; from that we can find the average speed of the fluid in the
But
200
Reynold's number from Eq. 16-22 is then
R=
pDv
(13600 kgim3)(0.0376 m)(8.03 x 10-4 mis)
{1.55 x 10-3N .s/m2)
=
265.
This is well below the critical value of 2000.
(b) Poiseuille's Law, Eq. 16-20, can be used to find the pressure difference between the ends of
the pipe. But first, note that the mass fiux dm/ dt is equal to the volume rate times the density
when the density is constant. Then p dV/ dt = dm/ dt, and Poiseuille's Law can be written as
óp=
8r¡L
--
dV
7rR4
dt
P16-1 The volume oí water which needsto fiow out oí the bay is
v = (6l00m)(5200m)(3m)
= 9.5x lO7m3
during a 6.25 hour (22500s) period. The averagespeed through the channel must be
P16-2
(a) The speed of the fluid through either hole is v = .¡2¡¡fi. The mass flux through a hole
is Q = pAv, so PIA1 = P2A2. Then Pl/P2 = A2/Al = 2.
(b) R = Av, so Rl/R2 = Al/A2 = 1/2.
(c) If R1 = R2 then A1..j2¡¡n¡ = A2J29h2, Then
h2/hl = (Al/A2) 2 = (1/2) 2 = 1/4.
So
h2
=
h1/4.
~
(a) Apply Torricelli's law (Exercise 16-14): v = V297i:.The speedv is a horizontal velocity,
and serves as the initial horizontal velocity of the fluid "projectile" after it leavesthe tank. There
is no initial vertical velocity.
This fluid "projectile" falls through a vertical distance H- h before splashing against the ground.
The equation governing the time t for it to fall is
-(H-
1
-2gt
h) =
2
,
Solve this for the time, and t = v'2(H -h)/g. The equation which governs the horizontal distance
traveled during the fall is x = vxt, but Vx = v and we just found t, so
(b) How many values of h willlead to a distance of x? We need to invert the expression, and
we'll start by squaring both sides
X2 = 4k(H-
k) = 4kH
-4k2,
and then solving the resulting quadratic expression for h,
h = 4!f~i16H2
-16x2
= ~ ( H :!:: ,¡¡¡¡--::-;;-;-)
8
201
~
For values of x between O and H there are two real solutions, if x = H there is one real solution,
and if x > H there are no real solutions.
If h1 is a solution, then we can write hl = (H + ~)/2,
negative. Then h2 = (H + ~)/2 is also a solution, and
h2 =
(H
+ 2h1 -2H)/2
where ~ = 2h1 -H
could be positive or
= h1 -H
is also a solution.
(c ) The farthest distance is x = H, and this happens when h = H /2, as we can see from the
previous section.
P16-4 (a) Apply Torricelli's law (Exercise 16-14): v = V2g(d + h2), assuming that the liquid
remains in contact with the walls of the tube until it exits at the bottom.
(b) The speed of the fl.uid in the tube is everywhere the same. Then the pressure difference at
various points are only functions of height. The fl.uid exits at C, and assuming that it remains in
contact with the walls of the tube the pressure difference is given by ~p = p(h1 + d + h2), so the
pressureat B is
p = Po -P(hl
+ d + h2).
(C) The lowest possible pressure at B is zero. A8Sume the flow rate is so slow that Pascal's
principle applies. Then the max:imum height is given by O = Po + pgh1, or
h1 = {1.01 x 105pa)/[{9.81
m/s2){1000
kg/m3)]
=
10.3 m.
P16-5 (a) The momentum per kilogram of the fluid in the smaller pipe is Vl. The momentum per
kilogram of the fluid in the larger pipe is V2. The change in momentum per kilogram is V2 -Vl.
There are pa2V2kilograms per second flowing past any point, so the change in momentum per
secondis pa2V2(V2-Vl). The change in momentum per second is related to the net force according
to F = ~p/ ~t, so F = Pa2V2(V2
-Vl). But F ~ ~p/a2, SOPl -P2 ~ Pv2(V2-Vl).
(b) Applying the streamline equation,
=
P2 + pgy2
1
2
+ 2PV2'
P2 -Pl
(c) This question asks for the 1088of pressure beyondthat which would occur from a gradual1y
widened pipe. Then we want
Ap
=
1
2
2
2P(Vl -V2) -Pv2(Vl
1
2
2
2P(vJ -V2) -PV2Vl
1
2
2P(Vl
-2V1V2
-V2
),
2
+ PV2'
2
+ V2) =
1
2P(Vl
-V2)
2
.
P16-6 The juice leavesthe jug with a speed v = ..¡2"iiij, where y is the height of the juice in the
jug. If A is the cross sectional area of the base of the jug and a the cross sectional area of the hole,
then the juice flows out the hole with a rate dV/dt = va = a..¡2"iiij, which means the level of jug
varies as dy/dt = -(a/A)..¡2"iiij. Rearrange and integrate,
202
~
.¡Y)
~ ~ fif)
J29at/A.
= t
( Jh -..jfj)
When y = 14h/15 we have t = 12.0s. Then the part in the parenthesis on the left is 3.539x 102s.
The time to empty completely is then 354 seconds,or 5 minutes and 54 seconds. But we want the
remaining time, which is 12 secondsless than this.
~
The greatest possible value for v will be the value over the wing which results in an air
pressure of zero. If the air at the leading edge is stagnant (not moving) and has a pressure of Po,
then Bernoulli 's equation gives
1
2
po=2PV,
or v = ,j'ip;;Tii
= V2(1.01xl05pa)/(1.2kg/m3)
= 410m/s.
This value is only slightly larger than
the speedoí sound; they are related becausesound wavesinvolve the movement oí air particles which
"shove" other air particles out oí the way.
P16-8
Bernoulli's
equation
together
Pl
+
with
continuity
1
2
P2 + 2PV2'
1
2
?PVl
-
Pl-P2
gives
1 (
2
2P V2 -VI'
1
(A
)
2
I
2P
2
,
2
2
A2VI
2
-VI
,
But Pl -P2 =«(J' -(J )gh. Note that we are not assuming (J is negligible compared to (J'. Combining,
Vl = A2
"
P16-9
J~(P'(A -p)ghA )
p
2 1
2
2
.
(a) Bernoulli's equation together with continuity gives
Pl
Then
Vi
=
f2(2.12)(1.01x 105Pa) = 4.45m/s
V (1000kg/m3)(21.6)
,
and then V2= (4.75)(4.45m/s) = 21.2m/s.
(b) R = 7r(2.60x10-2m)2(4.45m/s) = 9;45x 10-3 m3/s.
203
P16-10
(a) For Fig. 16-13 the velocity is constant, or v = vi. ds = idx + Jdy. Then
f v. ds = v f dx = O,
because § dx = o.
(b)ForFig.16-16thevelocityisv=(k/r)r.
dS=rdr+Ord4>.
f
v. ds = v f
Then
dr = O,
because f dr = O.
Pl6-11
(a) For an element of the fluid of mass dm the net force as it moves around the circle is
dF = (v2/r)dm. dm/dV = p and dV = Adr and dF/A = dp. Then dp/dr = pv2/r.
(b) From Bernoulli's equation p + pv2/2 is a constant. Then
dp
~+Pv~
dv
=O,
or v/r + dv/dr = O, or d(vr) = 0. Consequently vr is a constant.
(c) The velocity is v = (k/r)r. dS= rdr + Ord<t>.Then
f v. dS= v f dr = O,
because .f dr = O. This means the fiow is irrotational.
P16-12
F/A
F/A
= 7Jv/D,
= (4.0x
so
lO19N
.s/m2)(O.O48m/3.l6x
lO7S)/(l.9x
lO5m)
= 3.2x
lO5Pa.
~
A fiow will be irrotational if and only if § v. ds = O for all possible paths. It is fairly
easy to construct a rectangular path which is parallel to the fiow on the top and bottom sides, but
perpendicular on the left and right sides. Then only the top and bottom paths contribute to the
integral. v is constant for either path (but not the same), so the magnitude v will come out of the
integral sign. Since the lengths of the two paths are the same but v is different the two terms don't
cancel, so the fiow is not irrotational.
p 16-14
( a) The area of a cylinder is A = 27rrL .The velocity gradient if dv / dr .Then the retarding
force on the cylinder is F = -1](27rrL)dv/dr.
(b) The force pushing a cylinder through is F' = A~p = 7rr2~p.
(C) Equate, rearrange, and integrate:
Then
v=
~(R2
41)L
204
-r2).
~
~
The volume flux (called R¡ to distinguish it from the radius R) through an annular ring
of radius r and width 8r is
tSRf = tSAv = 21TrtSrV,
where v is a function of r given by Eq. 16-18. The mass flux is the volume flux times the density,
so the total mass flux is
dm
dt
=
dr,
8r¡L.
P16-16 The pressure difference in the tube is ~p = 4')'/r, where r is the (changing) radius of the
bubble. The mass flux through the tube is
dm
dt
R is the radius
of the tube.
dm = pdV,
1
4p7rR4'Y
8;¡¡;;:--'
and dV = 47rr2dr.
r2 r3 dr
=
rl
4
4
rl -r2
¡
Then
t
R 4 '"'1
O
8-;;"Ldt,
pR4'"'1
-;¡;¡¡;
t ,
Then
t=
2(1.80 x 10-5N
.s/m2)(0.112
(0.54 x 10-3m)4(2.50
m)
[(38.2 x 10-3m)4
x 10-2N/m)
205
-(21.6
x 10-3m)4J
= 3630 s.
~
For a perfect spring IFI = klxl. x = O.157m when 3.94kg is suspendedfrom it. There
would be two forces on the object- the force of gravity, W = mg, and the force of the spring, F.
These two force must balance, so mg = kx or
k- --= mg
x
(3.94 kg)(9.81 m/s2) = 0.246N /m.
.-
(0.157m)
Now that we know k, the spring constant, we can find the period of oscillations from Eq. 17-8,
E17-2
(a) T = 0.484s.
(b) f = l/T = 1/(0.484s) = 2.07s-1.
(C) (¿)= 27rf = 13.0 rad/s.
(d) k = mw2 = (0.512kg)(13.0 rad/s)2= 86.5N/m.
(e) Vm = (¿)Xm= (13.0 rad/s)(0.347m) = 4.51 m/s.
(f) Fm = mam = (0.512kg)(13.0 rad/s)2(0.347m) = 30.0N.
E17-3
am = (27rf)2xmo
Then
f =
V(9.81
m!s2)/(1.20x
10-6~)/(27r)
= 455 Hz.
E17-4
(a) "" = (271")/(0.6458) = 9.74 rad/8. k = 7n(,.)2= (5.22kg)(9.74 rad/8)2 = 495N/m.
(b) Xm = vm/"" = (0.153m/8)/(9.74rad/8)
= 1.57x10-2m.
(c) f = 1/(0.6458) = 1.55 Hz.
~
(a) The amplitude is halfofthe distance between the extremes ofthe motion, so A = (2.00
mm)/2 = 1.00 mm.
(b) The maximum blade speed is given by Vm = ú)xm. The blade oscillates with a frequency of
120 Hz, so ú) = 27rf = 27r(120s-1) = 754 rad/s, and then Vm = (754 rad/s)(0.001m) = 0.754m/s.
(c) Similarly, am = ú)2Xm'am = (754 rad/s)2(0.001m) = 568m/s2.
E17-6
(a) k = mw2 =
(b) f = /k7m/21T =
2 = l.25xlOSN/m
830kg/4)/21T = 2.63/s.
E17-7
(a) x = (6.12m)cos[(8.38 rad/s)(1.90s) + 1.92 rad] = 3.27m.
(b) v = -(6.12m)(8.38/s)sin[(8.38
rad/s)(1.90s) + 1.92 rad] = 43.4m/s.
(c) a = -(6.12m)(8.38/s)2cos[(8.38
rad/s)(1.90s) + 1.92 rad]= -229m/s2.
(d) f = (8.38 rad/s)/27r = 1.33/s.
(e)T=I/f=O.750s.
~
If the driye wheel rotates
at 193 rey /min
W = (193 rev/min)(27r
then v m =WXm
= (20.2 rad/s)(0.3825
then
rad/rev)(1/60
m) = 7.73m/s.
206
s/min)
= 20.2 rad/s,
E17-10
k = (0.325kg)(9.81m/s2)/(1.80x10-2
m) = 177N/m.
=
0.691
s.
E17-11
For the tides (¿}= 27r/(12.5 h). Half the maximum occurs when COS(¿}t
= 1/2, or (¿}t= 7r/3.
Then t = (12.5 h)/6 = 2.08 h.
E17-12
The two will separate if the (maximum) acceleration exceeds g.
(a) Since (AJ= 27r/T = 27r/(1.18s) = 5.32 rad/s the maximum amplitude is
Xm = (9.81m/s2)/(5.32
(b) In this case (AJ= ..¡(9.81m/s2)/(O.O512m)
E17-13
(a) ax/x
= -",2.
rad/s)2 = O.347m.
= 13.8 rad/s. Then f = (13.8 rad/s)/27r
2.20/8.
Then
w = V-(-123m/s)/(0.112m)
= 33.1 rad/s,
so f = (33.1 rad/s)/21r = 5.27/s.
(b) m = k/'-'l2 = (456N/m)/(33.1
rad/s)2 = O.416kg.
(c) x = XmCOS'-'lt;V = -Xm'-'lSin'-'lt. Combining,
X2 + (v/'-'l)2 = Xm2 COS2
'-'lt + Xm2 sin2 '-'lt = Xm2.
Consequently,
X m = /(0.112m)2
+ (-13.6m/s)2/(33.1
rad/s)2
= 0.426 m.
E17-14 Xl = XmCOS(¿}t,
X2 = XmCOS((¿}t
+ <f¡). The crossing happens when Xl = xm/2, or when
(¿}t= 7r/3 (and other values!). The san1econstraint happens for X2, except that it is moving in the
other direction. The closest value is (¿}t+ <f¡= 27r/3, or <f¡= 7r/3.
~
(a) The net force on the three cars is zero before the cable breaks. There are three forces
on the cars: the weight, W, a normal force, N, and the upward force from the cable, F. Then
F = WsinfJ
= 3mgsinfJ.
The frequency oí oscillation oí the remaining two cars after the bottom car is releasedis
Numerically, the frequency is
(b) Each car contributes equally to the stretching of the cable, so one car causesthe cable to
stretch 14.2/3 = 4.73 cm. The amplitude is then 4.73 cm.
207
E17-16 Let the height of one side over the equilibrium position be x. The net restoring force
on the liquid is 2pAxg, where A is the cross sectional area of the tube and 9 is the acceleration of
free-fall. This corresponds to a spring constant of k = 2pAg. The mass of the fluid is m = pAL.
The period of oscillation is
T = 21T.fi
~
(a) There are two forces on
We'll 888ume the log is cylindrical.
If
cross sectional area of the log, then V
7nw = PwV is the m888 of the displaced
= 1Tfij.
the log. The weight, W = mg, and the buoyant force B.
x is the length of the log beneath the surface and A the
= Ax is the volume of the displaced water. Furthermore,
water and B = 7nwg is then the buoyant force on the log.
Combining,
B = PwAgx,
where Pw is the density of water. This certainly looks similar to an elastic spring force law, with
k = PwAg. We would then expect the motion to be simple harmonic.
(b) The period of the oscillation would be
T~27rfi~27r~,
where m is the total mass of the log and lead. We are told the log is in equilibrium when x = L = 2.56
m. This would give us the weight of the log, since W = B is the condition for the log to float. Then
m=~=~=pAL.
9
9
From this we can write the period of the motion as
T=27r
~
=
3.21
s.
yp:Ag
E17-18
(a) k = 2(1.18J)/(O.O984m)2 = 244N/m.
(b) m = 2(1.18J)/(1.22m/s)2
= 1.59kg.
(c) f = [(1.22m/s)/(O.O984m)]/(27r)
= 1.97/s.
E17-19
(a) Equate the kinetic energy of the object just after it leaves the slingshot with the
potential energy of the stretched slingshot.
k=
mv2
=
(0.130
~
kg)(11.2
(1:53
(b) N = (6.97x 106 N/m)(1.53m)/(220N)
x 103m/s)2
= 6.97x lO6N/m.
m)2
= 4.85 x 104 people.
E17-20
(a) E = kxm2/2, U = kx2/2 = k(xm/2)2/2 = E/4.
is 25% potential and 75% kinetic.
(b) If U = E/2 then kx2/2 = kxm2/4, or x = xm/~.
208
K = E -U
= 3E/4.
The the energy
~
(a)
am
=
úJ2Xm
SO
~
~
úJ=
Xm
= I (7,93x 103m/s2) =
y (1.86x 10-3m)
2.06
x
103
rad/s
The period oí the motion is then
T=
(b) The maximum
211"
= 3.05x 10-38.
(¿}
speed of the particle
is found
by
v m = I.UXm = (2.06 x 103 rad/s)(1.86
x 10-3 m) = 3.83 m/s.
(c) The mechanical energy is given by Eq. 17-15, except that we will focus on when Vx = vm,
because then x = O and
E11-22
(a) f = Jk7ffi/27r = i,~~~!1m)/(5.13kg)/27r
(b) Ui = (988N/m)(0.535m)
/2 = 141 J.
(c) Ki = (5.13kg)(11.2m s 2 2 = 322J.
( d) Xm = V2E7k
E17-23
=
2(322 J + 141 J) / (988 N /m) = 0.968 m.
(a) iAJ= v'(538N/m)/(1.26kg)
= 20.7 rad/s.
X m = v'(0.263m)2
(b) 4>= arctan
= 2.21/s.
-(-3.72m/s)/[(20.7
+ (3.72m/s)2/(20.7
rad/s)(0.263
m)]}
rad/s)2
= 0.319m.
= 34.3°.
E17-24 Before doing anything else apply conservation of momentum. If Vo is the speed of the
bullet just before hitting the block and Vl is the speed of the bullet/block system just after the two
begin moving as one, then Vl = mvo/(m+M), where m is the mass ofthe bullet and M is the mass
of the block.
For this system úJ= .¡k/(m + M).
(a) The total energy of the oscillation is ~(m+ M)v~, so the amplitude is
= m voy 11
k¡;ñ:¡:-M)"
The
numerical
value
is
1
(500Njm)(O.O50kg + 4.00kg) = O.167m.
Xm = (O.O50kg)(150mjs)
(b) The fraction of the energy is
(m+M)v~
mv2
O
E17-25
--
m+M
m
2
m
m
--
~m+M
-
m+M
L = (9.82 m/s2)(1.00s/27r)2
= 0.249 m.
209
(0.050 kg)
= 1.23x 10-2.
( 0.050 kg + 4.00 kg)
E17-26
T = (1808)/(72.0).
Then
2
27r(72.0)
y=
9.66 m/so
{1.53 m)
-¡"i8¡¡;;¡--
~
We are interested in the value oí ()m which will make the second term 2% oí the first
term. We want to solve
which has solution
or (}m= 33°.
(b) How large is the third term at this angle?
32
.4
-sm
2242
(1
(jm
32
-=
2
--sm
22
)
.2 (jm :2
9
-=
2
22
(
-0.02
4
)2
or 0.0009, which is very small.
E17-28
Since T cx:.¡iTg
we have
~
Let the period of the clock in Paris be T1. In a day of length D1 = 24 hours it will
undergo n = D/Tl oscillations. In Cayennethe period iB T2. n oscillations should occur in 24 hours,
but since the clock runs slow, D2 iB 24 hours + 2.5 minutes elapse. So
T2 = D2jn = (D2jD1)T1
= [(1442.5 min)j(1440.0
Since the ratio of the periods is {T2/Tl)
= ,¡-(ij;Jg:;j,
(a) Take the differential
= 1.0017Tl.
the 92 in Cayenne is
92 = 91{Tl/T2)2 = {9.81m/s2)/{1.0017)2
E17-30
min)]Tl
= 9.78m/s2.
of
2
47r2
~
g=
(10m)
T
X 105
m
=
T2
so 8g = ( -87r2 X 105 m/T3)8T.
The relative error is then
Note that T is not the period here, it is the time for 100 oscillations!
~
8g
uT
---29 -T.
If óg/g = 0.1% then óT/T = 0.05%.
(b) For g ~ 10m/s2 we have
T
Then
8T
~
(0.0005)
(987
s)
~
300
~
27r(100)v(10m)/(10m/82)
=
InS.
210
6288.
E17-31
T = 27rV(17.3m)/(9.81
m/82)
= 8.348.
E17-32 The spring will extend until the force from the spring balancesthe weight, or when Mg =
kh. The frequency of this system is then
~
The frequency
of oscillation
is
where d is the distance from the pivot about which the hoop oscillates and the center of massof the
hoop.
The rotational inertia I is about an axis through the pivot, so we apply the parallel axis theorem.
Then
I = Md2 +Icm
= Md2 +Mr2.
But d is r, since the pivot point is on the rim of the hoop. So I = 2M ~ , and the frequency is
1 {"j;ii¡.f
1 {9
1 I (9.81 m/s2)
f = 2;V -zM(ii = 2;V 2d = 2;V 2(0.653m) = 0.436 Hz.
(b) Note the above expression looks like the simple pendulum equation if we replace 2d with l.
Then the equivalent length of the simple pendulum is 2(0.653m) = 1.31m.
E17-34
Apply
Eq.
17-21:
T2~
I=
E17-35
Then
=
~ = (0.192N .m)/(0.850
T
7.70
xlO-2kg.m2.
47r2
=
27r.jf/K
rad) = 0.226 N .m.
=
27rV(O.834kg
I = ~(95.2kg)(0.148m)2
.m2)/(0.226N
.m)=
12.1
= O.834kg .m2.
s.
E11-36
x is d in Eq. 17-29. Since the hole is driIIed off center we appIy the principIe axis theorem
to find the rotational inertia:
Then
1
-ML2+Mx2
12
=
T2Mgx
1
l2(1.00m)2 + X2 =
(8.33x 10-2m2)-(1.55 m)x + X2 = o.
This has solutions x = 1.49m and x = 0.0557 m. Use the latter.
211
~
~
For a stick of length L which can pivot about the end, I
of such a stick is located d = L /2 away from the end.
The frequency of oscillation of such a stick is
kML2.
The center oí mags
f
f
=
f
=
1!39
2;Vu'
This means that f is proportional to Ji7L, regardless of the mass or density of the stick. The ratio
of the frequency of two such sticks is then h/ h = .¡t;;7L;,
which in our case gives
f2 = hv'L;¡¡;;:
E17-38
(,
\
,
The rotational
= hV(L1)/(2L1/3)
= 1.22h.
inertia of the pipe section about the cylindrical
M
Icm = 2 [r~ +r~]
(a) The total rotational
axis is
M
= 2 [(0.102m)2 + (0.1084m)2] = (1.11xl0-2m2)M
inertia about the pivot axis is
I = 2Icm + M(O.102m)2 + M(O.3188m)2 = (0.134m2)M.
The period of oscillation is
(b) The rotational
inertia of the pipe section about a diameter is
M
M
Icm = 4 [r~ + r~] = 4 [(O.lO2m)2 + (O.lO84m)2] = (5.54x lO-3m2)M
The total rotational
inertia about the pivot axis is now
I = M(I.11 x 10-2m2) + M(O.102m)2 + M(5.54x
10-3m2) + M(O.3188m)2 = (0.129m2)M
The period of oscillation is
T = 27r.1
V M(9.81 (0.129m2)M
m/s2)(0.2104
The percentage difference with part {a) i8 {0.038)/{1.608)
m)
=
1.57 s.
= 1.9%.
E17-39
E17-40
E17-41
(a) Since effectively x = y, the path is a diagonalline.
(b) The path will be an ellipse which is symmetric about the line x = y.
(c) Since cos((A)t+ 900) = -sin((A)t), the path is a circle.
212
E17-42 (a)
(b) Take two time derivatives and multiply by m,
F = -mALV2 ( i cos(;)t + 9j cos3(;)t)
(c)U=-JF.dr,so
u
=
~ mA
2 úJ2 ( COS2 úJt +
9 COS2 3<..Jt )
2
(d) K = ~mv2,
so
K
=
~mA2w2
2
(sin2wt+9sin23wt)
;
And then E = K + U = 5mA2¡.;2.
(e) Yes; the period is 21!"!¡.;.
~
The w which describes the angular velocity in uniform circular motion is effectively the
same w which describes the angular frequency of the corresponding simple harmonic motion. Since
w = Vkliñ, we can find the effective force constant k from knowledge of the Moon 's mass and the
period of revolution.
The moon orbits with a period of T, so
27r
T
I.AJ=
27r
.
(27.3 x 24 x 36008) = 2.66
~~-x 10-61l'ad/8.
--.~
-
Thi8 can be used to find the value of the effective force constant k from
k = Tn(LI2= (7.36 x l022kg)(2.66 x lO-6rad/8)2
E17-44
= 5.2lxlOllN/m.
(a) We want to know when e-bt/2m = 113, or
t = :~ln3
=
b
2(1.52kg)
ln3 = 14.78
(0.227 kg/s)
(b) The (angular) frequency is
=
2.31
rad/s.
The number of oscillations is then
(14.7s)(2.31 rad/s)/27r = 5.40
~
The first
dx
dt
=
derivative
Xm(
oí Eq.
-b/2m)e-bt/2m
-Xme-bt/2m
The second derivative
~
=
is quite
-xm(
7-39 is
COS((..1't
«(b/2m)
+
cos((..1't
<1» +
+
<1» +
Xme-bt/2m(
(..I' sin((..1't
-(..I')
+
sin((..1't
+
<1»,
<1»)
a bit messier;
-b/2m)e-bt/2m
((b/2m)
cos((J.}'t + 4» + (J.}'sin((J.}'t + 4»)
dX2
-Xme-bt/2m
Xme-bt/2m
( (b/2m)(
-(J.}') sin((J.}'t + 4» + ((J.}')2COS((J.}'t+ 4») ,
( ((J.}'b/m) sin((J.}'t + 4» + (b2/4m2
213
-(J.}'2) cos((J.}'t + 4» )
Substitute these three expressionsinto Eq. 17-38. There are, however, some fairly obvious simplifications. Every one of the terms above has a factor of xm, and every term above has a factor of
e-bt/2m, SOsimultaneously with the substitution we will cancel out those factors. Then Eq. 17-38
becomes
m [(w'b/m) sin(w't + 4»+ (b2/4m2-W'2) COS(w't
+ 4»]
-b [(b/2m)cos(w't+ 4»+ w' sin(w't +4»] + k cos(w't+ 4» =
{)
Now we collect terms with cosine and terms with sine,
(-.1'b--.I'b) sin(-.1't +<I» + (mb2/4m2 --.1'2 -b2 /2m + k) cos(-.1't + <I»= o.
The coefficient for the sine termis identically zero; furthermore, becausethe cosine term must then
vanish regardlessof the value of t, the coefficient for the sine term must also vanish. Then
(mb2/4m2 -m,{..;'2 -b2/2m + k) = Q1
or
12
k
úJ -m
If this condition
is met, then Eq.
b2
4m2 .
17-39 is indeed a solution
of Eq.
17-38.
E17-46 (a) Four complete cycles requires a time t4 = 87r/úJ'. The amplitude decays to 3/4 the
original value in this time, so 0.75 = e-bt4/2m, or
87rb
ln( 4/3) = 21n(¿}'
It is probably reasonableat this time to assumethat b/2m is small compared to ú.iso that ú.i' ~ ú.i.
We'll do it the hard way anyway. Then
",'2
~
m
-
( )
=
2
~
2m
k
m
=
Numerically, then,
= O.112kg/s.
(b)
E17-47
(a) Use Eqs. 17-43 and 17-44. At resonance 1.1)"= 1.1),80
G = JbV
and then X m = F m/lJ¡.¡.
(b) Vm = (¡)Xm = F m/b.
214
= bIo,;,
E17-48
We need the first two derivatives of
F m-COS(¡)t-{3
( II
X=-
)
G
The derivatives are easy enough to find,
dx
-=
~I
dt
-úJ")
sin(úJ"t
-{3),
G
and
d2x
dt2 We'll
substitute
this into Eq.
.
17-42,
m
Then we'll cancel out a.smuch a.swe can and collect the sine and cosine terms,
(k -m«(Ll")2) COS«(LlI't
-{J) -(búJ")sin«(Ll"t -.B) = GCOS(Ll"t.
We can write the left hand side of this equation in the form
A cos al cos a2 -A
sin al sin a2,
if we let a2 = YJ'it -/3 and choose A and al correctly. The best choice is
Acosal
A SIn
.J.. al
=
k-m(YJ")2,
=
(A.II)1 ,
and then taking advantage of the fact that sin2 + COS2= 1,
A2 = (k-
m(v;")2)2 + (/)(.;")2,
which looks like Eq. 17-44! But then we can apply the cosine angle addition formula, and
Acos(al
+ v;"t -{3) = G cosv;"t.
This expression needs to be true for all time. This means that A = G and al = {3.
and
d2x
--
dt2
We'll
substitute
this into Eq. 17-42,
m
215
Then we'll cancel out as much as we can and collect the sine and cosine terms,
(k -m(úJ"')2)
COS(úJ"'t -/3)
-(bú1"')
sin(úJ"'t
-/3)
= G cosúJ"t.
We can write the left hanrl sirle of this equation in the form
Acosal
cosa2 -Asinal
sina2,
if we let a2 = (.J"'t -/3 and choose A and al correctly. The best choice is
A cos al
A sin al
=
=
k -m(w"')2,
~II/ ,
and then taking advantage of the fact that sin2 + COS2= I,
A2 = (k -m(f.AJ"')2)2 + (bI.J.J"')2,
which looks like Eq. 17-44! But then we can apply the cosine angle addition formula, and
A COS(Ql+ ú)"'t -;8)=
G COSú)"t.
This expression needs to be true for all time. This means that A = G and al + ",,"'t -.8 = ",,"t and
al = .8 and ",,"' = ",,".
E17-50 Actually, Eq. 17-39 is not a solution to Eq. 17-42 by itself, this is a wording mistake in
the exercise. Instead, Eq. 17-39 can be added to any solution of Eq. 17-42 and the result will sti11
be a solution.
Let Xn be any solution to Eq. 17-42 (such as Eq. 17-43.) Let Xh be given by Eq. 17-39. Then
x = X n + Xho
Take the first two time derivatives oí this expression.
m
+ k (Xn + Xh) = F mCOSiJ)"t.
+b
Rearrange and regroup.
Consider the second term on the left. The parenthetical expression is just Eq. 17-38, the damped
harmonic oscillator equation. It is given in the text (and proved in Ex. 17-45) the Xh is a solution,
so this term is identically zero. What remains is Eq. 17-42; and we took as a given that Xn was a
solution.
(b) The "add-on" solution of Xh representsthe transient motion that will die away with time.
216
~
~
The time between "bumps" is the solution to
vt
t
x,
3600 s
=
1 hr
= 0.886s
The angular frequency is
21T
--T
= 7.09 rad/s
This is the driving frequency, and the problem states that at this frequency the up-down bounce
oscillation is at a maximum. This occurs when the driving frequency is approximately equal to the
natural frequency of oscillation. The force constant for the car is k, and this is related to the natural
angular frequency by
k =
2
rTUJ.J =
w
-úJ
2
,
9
where W = (2200 + 4 x 180) lb=
2920 lb is the weight
of the car and occupants.
Then
When the four people get out of the car there is less downward force on the car springs. The
important relationship is
6.F = k6.x.
In this case6.F = 720 lb, the weight of the four people who got out of the car.6.x is the distance
the car will rise when the people get out. So
Ax=
= (720 lb) = 0.157 ft ~ 2 in.
AF
4590 lb/ft
k
E17-52
The derivative is easy enough to find,
dx
Pm
-I
-
dt
-(¿)") sin((¿)"t -¡3),
G
The velocity amplitude is
v m
=
Fm
II
-úJ,
G
~Vm2(w"2
Pm
-w2)2
Pm
V(m,(JJ" -k/(¡)")2
+ b2w'12 ,
+ b2 .
Note that this is exactly a maximum when iJ.J"= iJ.J.
E17-53
The
reduced
mags
is
m = (1.13kg)(3.24kg)/(1.12kg+ 3.24kg)= O.840kg.
The period of oscillation is
T = 27rJ(0.840kg)/(252N/m)
217
= 0.3638
E11-54
E17-55
Start by multiplying
then
add
2ml
+ m2).
=
K
and
the kinetic energy expression by (ml + m2)/(ml
m2Vl V2 -2ml
m2Vl V21
=
2(
)
m¡
+
((m¡v¡
+
m2V2)2
+
m¡m2(V¡
-V2)2)
m2
But m¡ V2 + m2V2 = O by conservation
of momentum,
so
(mlm2)
K
2(m¡
+ m2)
m
2(V¡
-V2)2.
(V¡ -V2)2
P17-1
The roass ofone silver atoro is {O.lO8kg)/{6.02xlO23)
constant is
k = (l.79xlO-25kg)47r2(lO.OxlO12/S)2
,
= l.79xlO-25kg.
= 7.07xlO2N/m.
P17-2
(a) Rearrange Eq. 17-8 except replace m with the total mass, or m+M.
T2/(47r2), or
M
=
The effective spring
(k/47r2)T2
Then (M +m)/k
=
-m.
(b) When M = O we have
m = [(605.6N/m)/(47r2)](0.90149 S)2= 12.467kg.
(c) M = [(605.6N/m)/(47r2)](2.08832 S)2-(12.467kg) = 54.432kg.
~
The max.imum
static
friction
is F¡ :::; JLsN. Then
F¡ = JLsN = JLsW = JLsmg
is the maximum available force to accelerate the upper block. So the maximum acceleration is
F¡
am = --J.Lsg
m
The maximum
possible
amplitude
of the oscillation
xm=
is then given by
.am (,)2
J1s9
k/(m+M)'
where in the last part we substituted the total mass of the two blocks because both blocks are
oscillating. Now we put in numbers, and find
0.119
218
m.
P17-4
(a) Equilibrium occurs when F = O, or b/r3 = a/r2. This happens when r = b/a.
(b) dF/dr = 2a/r3 -3b/r4.
At r = b/a this becomes
dF/dr
= 2a4/b3 -3a4
fb3 = ~a4/b3,
which correspondsto a force constant of a4/b3.
(C) T= 27rvmlk = 27rYmb3/a2, where m is the reduced mass.
P17-5
Each spring helps to restore the block. The net force on the block is then of magnitude
F1 + F2 = k1x + k2x = (k1 + k2)X = kx. We can then write the frequency as
f=~
/I=
27r V ~
With a little algebra,
f =
1-~
27ry ~'
/ 1 k1 .1
k2
y ¡;2 ;;:¡ + ¡;2 ;;:¡ ,
.fii+fi.
P17-6 The tension in the two spring is the same, so k1xl = k2x2, where Xi is the extension of the
ith spring. The total extension is Xl + X2, SOthe effective spring constant of the combination is
F
-====
X
F
X1 +X2
xdF
1
+ X2/
F
1/k1
1
+ 1/k2
The period is theD
T-
.!.- /I.!.- I~
-211" V:;;;: -211" y (k1 + k2)m
With a little algebra,
f
1 .1 klk2
2;y (kl + k2)m,
1 1
i
2;y m/kl + m/k2,
1 11
2;y 1/ú.i~+¡/ú.i~'
1~
2ú.i2
I
2
2;
~'
hf2
v'7ri7i'
219
k1k2
k1 + k2 ,
~
(a) When a spring is stretched the tension is the same everywhere in the spring. The
stretching, however, is distributed over the entire length of the spring, so that the relative amount
of stretch is proportional to the length of the spring under consideration. Half a spring, half the
extension. But k = -F/x, so half the extension means twice the spring constant.
In short, cutting the spring in half will create two stiffer springs with twice the spring constant,
so k = 7.20 N/cm for each spring.
(b) The two spring halves now support a mass M. We can view this as each spring is holding
one-half of the total mass, so in effect
f=
lr--;;-
:2-;YM72
or, solving for M ,
M=
2k
=
2(720N/m) S-I
47r2(2.S7
)2
=
4.43kg.
4;2"12
P17 -8 Treat the spring a being composedof N masslessspringlets each with a point mass ms/ N
at the end. The spring constant for each springlet will be kN. An expression for the conservation
of energy is then
m
2v
N
ms ~
2
2
N
Vk ~
2
+ 2N L..;vn + 2 L..;Xn = E .
1
1
Since the spring stretches proportionally along the length then we conclude that each springlet
compresses the same amount, and then Xn = A/ N sin f.¡)tcould describe the change in length of each
springlet. The energy conservation expression becomes
v = A(,¡)COS
(,¡)t.The hard part to sort out is the vn, since the displacements for all springlets to one
side of the nth must be added to get the net displacement. Then
A
Vn = nN(,¡)cos(,¡)t,
and the energy expressionbecomes
m
-+
m
N
8~
2
2:N3L..n2
I
Replace the sum with an integral, then
and the energy expressionbecomes
~ (m+
This will only be constant if
or
T
=
27r.¡(m
+
ms/3)/k.
220
P17-9
(a) Apply conservation of energy. When x = Xm v = O, so
1
2kxm2
1
1
2m[v(0)r + 2k[x(0)r,
=
m
k[v(O)J2 +[x(O)12,
Xm2 =
xm =
(b) When
t =Ox(O)
V[v(0)/tIJ]2 + [X(O)]2.
= XmCOSt/> and v(O) = -wxmsint/>,
v(O)
=--
wx(o)
so
sint/> = tant/>.
cos~
P17-10
~
Conservation
of momentum
for the bullet
block
collision
gives m v = ( m + M)Vf
or
m
Vf = ~V.
This Vf will be equal to the maximum oscillation speedVmoThe angular frequency for the oscillation
is given by
1.1)
= {;;h
.
Then the amplitude for the oscillation is
v m
Xm=
(.;
P17-12
(a) W = Fs, or mg = kx, so x = mg/k.
(b) F = ma, but F = W -Fs = mg -kx, and since ma = md2x/dt2,
f1l!x
md"i2" + kx = mg.
The solution can be verified by direct substitution.
(c) Just look at the answer!
(d) dE/dt is
P17-13 The initial energy stored in the spring is kxm2/2. When the cylinder pa.ssesthrough the
equilibrium point it ha.sa translational velocity Vm and a rotational velocity IJ)r= vm/ R, where R
is the radius of the cylinder. The total kinetic energy at the equilibrium point is
Then the kinetic energy is 2/3 translational and 1/3 rotational. The total energy of the system is
221
(a) Kt = (2/3)(8.40J) = 5.60J.
(b) Kr = (1/3)(8.40J) = 2.80J.
(c ) The energy expression is
1
which leads to a standard
expression
2
3m'
-I
2
1
+
v
-k
2
2
for the period
2
x
with
=
E
3M/2
,
replacing
m. Then T = 21T,f3M7'ik.
P17-14
(a) Integrate the potential energy expression over one complete period and then divide by
the time for one period:
1
¡
xm
-2
2xm
~kx2dx
=
1 1k 3
-2xm
3 Xm ,
-xm
1 X 2
-k
6
m .
This is one-third the total energy. The averagekinetic energy must be two-thirds the total energy,
or {1/3)kxm2.
P17-15 The rotational inertia is
'1
2(0.144m)2
I=~MR2+Md2=M
2
+
(0.102m)2
= (2.08x 10-2m2)M.
The period of oscillation is
=
27r
(2.08
M(9.81
x 10-2m2)M
m/82)(0.102m)
=
0.9068.
T=2,,~
P17-16
(a) The rotational
inertia of the pendulum about the pivot is
(b) The center of mass location is
(c) The period of oscillation is
T = 211"
(0.291
kg .m2)/(0.272kg
+0.488kg)(9.81
222
m/82)(0.496m)
= 1.768.
~
(a) The rotational inertia oí a stick about an axis through a point which is a distance d
from the center oí mass is given by the parallel axis theorem,
1
I = I
+ md2 = -.::.-mL2
cm
"'
+
md2
.
The period of oscillation is given by Eq. 17-28,
= 27rV
f L2 + 12d2
12gd
(b) We want to find the minimum period, so we need to take the derivative of T with respect to
d. It 'lllook weird, but
dT
12¿;;
dd =7r V12gd3(L2
This will vanish when 12~
P17-18
is
-L2
+ 12d2) .
~ L2, or when d = L/..jf2.
The energy stored in the spring is given by kx2/2, the kinetic energy oí the rotating wheel
~(M R2) (~) 2,
where v is the tangential velocity of the point of attachment of the spring to the wheel. If x =
Xm sinú)t, then v = Xmú)COSÚ)t,and the energy will only be constant if
(¿) 2-
k
r2
-
MW"
P17-19 The method of solution is identical to the approach for the simple pendulum on page 381
exceptreplace the tension with the normal force of the bowl on the particle. The effective pendulum
with have a length R.
P17-20
given by
Let x be the distance from the center of mass to the first pivot point. Then the pertod is
T=21T
Salve this far x by expressing
~.
y-¡;¡-¡¡;-
the abave equatian
(
~
471"2
as a quadratic:
) x=I+Mx2,
or
MgT2'\
x+I=O
,~)
There are two solutions. One correspondsto the first location, the other the secondlocation. Adding
the two solutions together will yield L; in this casethe discriminant of the quadratic will drop out,
leaving
Mx2-
L = XI + X2 = ,MgT2
~
Then
g
=
471"2L/T2,
223
-~
-47r2"
P17-21
In this problem
I
=
{2.50
{0.210 mf
kg)
+ {0.760m + O.210m)2
) = 2.41kg/ .m2.
2
The center of mass is at the center of the disk.
(a) T = 27r (2.41kg/ .m2)/(2.50kg)(9.81m/s2)(0.760m + O.210m) = 2.00s.
(b) Replace Mgd with Mgd + K.and 2.00s with 1.50s. Then
K.= 47r2(2
(.41 kg/ .m2)
)
2
-(2.50kg)(9.81m/s2)(0.760m+O.210m)=18.5N.m/rad.
1.50s
P11-22 The net force on the bob is toward the center of the cÍrcle, and h88 magnitude F net =
mv2/ R. This net force comesfrom the horizontal component of the tension. There is also a vertical
component of the tension of magnitude mg. The tension then h88 magnitude
T = J(mg)2 + (mv2/R)2 = mJg2 +v4/R2.
It is this tension which is important in finding the restoring force in Eq. 17-22; in effect we want to
replace 9 with J g2 + v4/ R2 in Eq. 17-24. The frequency will then be
1
f=2;V
I
L
Jg2+V4/R2.
~
(a) Consider an object of mass m at a point p on the axis of the ring. It experiencesa
gravitational force of attraction to all points on the ring; by symmetry, however, the net force is not
directed toward the circumference of the ring, but instead along the axis of the ring. There is then
a factor of cos(} which will be thrown in to the mix.
The di8tance from p to any point on the ring is r = V R2 + Z2, and (} is the angle between the
axis on the line which connects p and any point on the circumference. Con8equently,
cosO = z/r,
and then the net force on the star of mass m at p is
GMm
GMmz
F = -;:r¡-- cos(} =
r3
GMmz
= (R2 + Z2)3/2 .
(b) If z « R we can apply the binomial expansion to the denominator,
(R2 + Z2)-3/2
= R-3
1+ (~))-3/2
~R-3
-~
and
(~)2)
Keeping terms only linear in z we have
GMm
F = -w-z,
which corresponds to a spring constant k = GMm/ R3. The frequency of oscillation is then
f = ~/(27r)
= ,¡¡¡¡;¡¡w
/(27r).
(c) Using some numbers from the Milky Way galaxy,
f =
(7x 10-11N .m2 /kg2)(2x
1043kg)/(6x 1019m)3/(27r) = 1 x 10-14 Hz.
224
P17-24 (a) The acceleration of the center of mass (point C) is a = F/M. The torque about an
axis through the center of mass is tau = F R/2, since O is R/2 away from the center of mass. The
angular acceleration of the disk is then
a = T/I = (FR/2)/(MR2/2)
= F/(MR).
Note that the angular acceleration will tend to rotate the disk anti-clockwise. The tangential component to the angular acceleration at p is aT = -aR = -F / M; this is exactly the opposite of the
linear acceleration, so p will not (initially) accelerate.
(b) There is no net force at p .
P17-25
The value for k is closest to
k ~
(2000kg/4)(9.8lm/s2)/(O.lOm)
=
4.9xlO4N/m.
One complete oscillation requires a time t¡ = 27r/(¿)'.The amplitude decays to 1/2 the original
value in this time, so 0.5 = e-btl/2m, or
ln(2)
=
~
2m(,¡)"
It is not reasonable at this time to assume that b/2m is small compared to YJso that YJ' ~ YJ. Then
2
úJ
~
m
-
'2
(2m)
=
2
~
k
m
2
=
=
Then the value for b is
= 1100kg/s.
b=
P17-26
a = ~x/dt2
= -AúJ2cOSúJt. Substituting
-mA(L}2
COS(L}t + kA3
into the non-linear equation,
COS3(L}t = FCOS(L}dt.
Now let f.A)d= 3ú.I. Then
kA 3 COS3(¿}t -mAw2
Expanrl
the right
cos (¿}t = F cos &.Jt
hanrl sirle aB COS3(¿}t = 4 COS3(¿}t -3
kA3 COS3 (¿}t -mA(¿}2
cos (¿}t, then
COS(¿}t = F(4
This will only work if 4F = kA3 and 3F = mA¡,.;2,
4mA(,¡)2 = kA3, so A <X(,¡)and then F <X(,¡)3<X (,¡)d3,
225
COS3 (¿}t -3
COS(¿}t)
Dividing one condition by the other means
P17-27
(a) Divide the top and the bottom by m2. Then
and in the limit as m2 -00
+
mi
=
m¡m2
m¡
(ml/m2)
m2
the value of (ml/m2)
mlm2
lim
m2-+00 m¡ + m2
226
-O,
+
so
l'
E18-1
(a) f = v!A = (243m!8)!(0.0327m)
= 7.43x 103 Hz.
(b) T = 1! f = 1.35x 10-48.
E18-2
(a) f = (12)/(308) = 0.40 Hz.
(b) v = (15m)/(5.08) = 3.0m/8.
(c) >. = v/f = (3.0m/8)/(0.40 Hz) = 7.5m.
~
(a) The time for a particular point to move from maximum displacement to zero displacement is one-quarter of a period; the point must then go to maximum negative displacement,
zero displacement, and finally maximum positive displacement to complete a cycle. So the period is
4(178 ms) = 712 ms.
(b) The frequency is f = l/T= 1/(712x10-3s) = 1.40 Hz.
(c) The wave-speedis v = f>. = (1.40 Hz)(1.38m) = 1.93m/s.
Use Eq. 18-9, except let f = l/T:
E18-4
x
y =
(0.0213
m)
sin
27r
(0.114m)
-(385
Hz)t
= (0.0213
m) sin [(55.1
rad/m)x
-(2420
rad/s)tJ
~
The dimensions for tension are [F] = [M][L]/[T]2 where M stands for m8SS,L for length, T
for time, and F stands for force. The dimensions forlinear m8SSdensity are [M]/[L]. The dimensions
for velocity are [L]/[T].
Inserting this into the expression v = Fa / JLb,
[L]
=
ffl
There are three equations here. One for time, -1 = -2aj one for length, 1 = a + b; and one for
mass, O = a- b. We need to satisfy all three equations. The first is fairly quickj a = 1/2. Either of
the other equations can be used to show that b = 1/2.
E18-6
(b)
(c)
(d)
(e)
(a) Ym = 2.30mm.
f = (588 rad/s)/(27r rad) = 93.6 Hz.
v = (588 rad/s)/(1822 rad/m) = O.323m/s.
A = (27r rad)/(1822 rad/m) = 3.45mm.
Uy = YmW= (2.30mm)(588 rad/s) = 1.35m/s.
E18-7
(b)
(c)
(d)
(e)
(f)
(a) Ym = 0.060m.
.>..= (27r rad)/(2.07r rad/m) = 1.0m.
f = (4.07r rad/s)/(27r rad) = 2.0 Hz.
v = (4.07r rad/s)/(2.07r rad/m) = 2.0m/s.
Since the second term is positive the wave is moving in the -x direction.
Uy = YmW= (0.060m)(4.07r rad/s) = 0.75m/s.
E18-8
v = JF7jL
= V(487N)/[(O.O625kg)/(2.15m)]
= 129m/s.
227
~
We'll
first
find the linear
mass density
by rearranging
Eq.
18-19,
F
Ji,=V2
Since this is the same string, we expect that changing the tension will not significantly
linear mass density. Then for the two different instances,
F1
.=2
~
change the
F2
V2
We want to know the new tension, so
= 135N
E18-10
First v = (317 rad/s)/(23.8
rad/m)
= 13.32 m/s. Then
Ji,= F/V2 = (16.3N)/(13.32m/s)2
= O.O919kg/m.
E18-11
(a) Ym = O.O5m.
(b) >. = (0.55m) -O.15m)
= O.40m.
(c)v = JF7Jt =
(3.6N)/(O.O25kg/m)
= 12m/s.
(d) T = ¡If = >.Iv = (0.40m)/(12m/s)
= 3.33xl0-2s.
(e) Uy = YmúJ= 27rymlT = 27r(O.O5m)/(3.33x 10-2s) = 9.4m/s.
(f) k = (27r rad)/(O.40m) = 5.07r rad/m; úJ = kv = (5.07r rad/m)(12m/s)
= 607r rad/s.
The
phase angle can be found from
(0.04 m) = (0.05 m) sin( <I>
),
or 4>= 0.93 rad. Then
y =
El8-12
(0.05 m) sin[(5.07r
rad/m)x
+ (607r rad/s)t
+ (0.93
rad)].
(a) The tensions in the two strings are equal, so F = (0.511 kg)(9.81 m/s2)/2
= 2.506 N.
The wave speed in string 1 is
v = JF7j-¡ = V(2.506N)/(3.31x
10-3kg/m)
= 27.5m/s,
while the wave speed in string 2 is
v = ~
(b) We have ,¡F;7¡;;
= v'(2.506N)/(4.87xl0-3kg/m)
= ,¡F;7¡;;,
or Fl/ JLl = F2/ JL2. But Fi = Mig, so Ml/ JLl = M2/ JL2. Using
M=M1+M2,
MI
=
M-M1
=
M
11.1
~+
J1,1
MI
MI
and
M2
=
/L2
11.2
{0.511kg)
,
/L2
/(
1
1
-+- I /1.2,
/1.1
--/1.2
M
=
kg)
{4.87 {0.511
x 10-3kg/m)
=
0.207 kg
-{0.207kg)
= 22.7m/s.
I
= 0.304kg.
228
~
18-19,
We need to know the wave speed before we do anything else. This is found from Eq.
Our equations are then Xl = vt = (162 m/s)t, and X2 = v(t+29.6 ms) = (162 m/s)(t+29.6 ms) =
(162m/s)t+4.80m.
We can add these two expressions together to solve for the time t at which the
pulses meet,
lO.3m = XI + X2 = (l62m/s)t + (l62m/s)t + 4.80m = (324m/s)t + 4.80m.
which haB solution
X2 = 7.55m.
E18-14
= 0.01708.The
two pulses meet at XI = (162m/s)(O.O170s)
(a) [)y/[)r = (A/r)k cos(kr -r.,;t) -(A/r2)
8
28y
(3;r
(3;
= Ak cos(kr
-l.IJt)
sin(kr -r.,;t). Multiply this by r2, and then find
.
sm(kr
-Ak2r
= 2.75m, or
-l.IJt)
-Ak
cos(kr
-l.IJt).
Simplify, and then divide by r2 to get
lo
--rr2 or
20Y
=
-(Ak2
Ir)
sin(kr
-(LIt).
or
Now find {}2y/8t2 = -Aú.l2 sin(kr -(,¡)t). But since 1/V2 = k2/(,¡)2,the two sides are equal.
(b) [length]2.
E18-15
The liner mass density is JL= (0.263kg)/(2.72m)
v = y(36.1 N)/(9.669x 10-:¿kg/m) = 19.32m/s.
Pav = !jJ,(AJ2ym2v,SO
(0} =
~I
V (9.669
= 9.669xlO-2kg/m.
2(85.5W)
x 10-2kgim)(7.
=
70 x 10-3mr
(19.32
mis
1243
The wave speed is
radis.
)
Then f = (1243 rad/s)/27r = 199 Hz.
E18-16 (a) If the medium absorbs no energy then the power flow through any closed surface
which contains the source must be constant. Since for a cylindrical surface the area grows as r, then
intensity must fall of as 1/ r .
(b) Intensity is proportional to the amplitude squared, so the amplitude must fall off as 1/ .¡r .
~
The intensity is the averagepower per unit area (Eq. 18-33); as you get farther from the
source the intensity falls ofI becausethe perpendicular area increases. At same distance r from the
saurce the total possible area is the area of a spherical shell of radius r, so intensity as a function of
distance from the saurce would be
Pav
I=
47rr2
229
We are given two intensities: /1 = 1.13W /m2 at a distance rlj 12 = 2.41 W /m2 at a distance
r2 = rl -5.30 m. Since the average power of the source is the same in both cases we can equate
these two values as
=
47rr~11 =
47rr~12,
47r(rl -d)212,
47rr~11
where d = 5.30m, and then salve far rl. Daing this we find a quadratic expressianwhich is
r~I1
=
o =
o =
o =
The solutions to this are rl = 16.8m and rl = 3.15m; but since the person walked 5.3 m toward
the lamp we will assumethey started at least that far away. Then the power output from the light
is
p = 47rr~I1
=47r(16.8m)2(1.13W
/m2)
= 4.01 x 103W.
E18-18
Energy density is energy per volume, or u = U /V. A wave front of cross sectional area A
sweeps out a volume of V = Al when it travels a distance l. The wave front travels that distance
l in a time t = l/v. The energy flow per time is the power, or p = U /t. Combine this with the
definition of intensity, I = p / A, and
I=~
~
u = uV = uAI
=uv.
At
At
At
A=
Refer to Eq. 18-40, where the amplitude of the combined wave is
2ym cos(L\<t>/2),
where y m is the amplitude of the combining waves. Then
cos(~<P/2)
which
has solution
E18-20
=
(1.65ym)/(2ym)
= 0.825,
A<1>= 68.8° .
Consider only the point x = O. The displacement y at that point is given by
y = Yrnl sin(ú.lt)
This can be written
+ Yrn2 sin(ú.lt
+ 7r/2)
= Yrnl sin(ú.lt)
+ Yrn2 COS(ú.lt).
as
y = Ym(Al
where Ai = Ymi/Ymo But
if Ym is judiciously
sinú)t + A2 cosú)t),
chosen, Al = cos{3 and A2 = sin {3, so that
y = y m sin(c.lt+
/3).
Since we then require A~ + A~ = I, we must have
ym =
V(3.20
cm)2
+ (4.19
230
cm)2
= 5.27
cm.
E18-21
The easiest approach is to use a phasor representation oí the waves.
Write the phasor components as
=
=
XI
yI
X2
y2
ymlcosl/>l!
yml sinl/>l!
-ym2cosl/>2,
= ym2sinl/>2!
and then use the cosine law to find the magnitude of the resultant.
The phase angle can be found from the arcsine of the opposite over the hypotenuse.
E18-22 (a) The diagrams for all times except t = 15ms should show two distinct pulses, first
moving closer together, then moving farther apart. The pulses don not flip over when passing each
other. The t = 15ms diagram, however, should simply be a flat line.
E18-23
Use a program such as Maple or Mathematica
to plot this.
E18-24 Use a program such as Maple or Mathematica to plot this.
~
(a) The wave speed can be found from Eq. 18-19; we need to know the linear mass
density, which is JL= m/L = (0.122kg)/(8.36m)
= O.O146kg/m. The wave speed is then given by
(96.7N)
v = y [F~ = y I (O.O146kg/m)
= 81.4m/s.
(b) The longest possible standing wave will be twice the length of the string; so). = 2L = 16.7m.
(c) The frequency of the wave is found from Eq. 18-13, v = f)..
E18-26
(a) v = V(152N)I(7.16x
10-3kglm) = 146mls.
(b) >..= (213)(0.894m) = 0.596 m.
(c) f = vi>.. = (146 mls)I(O.596m) = 245 Hz.
E18-27
(a) y = -3.9 cm.
(b) y = (0.15m) sin[(0.79 rad/m)x + (13 rad/s)t].
(c) y = 2(0.15m)sin[(0.79 rad/m)(2.3m)]cos[(13
rad/s)(0.16 s)] = -0.14m.
E18-28
(a) The amplitude is halí oí 0.520 cm, or 2.60mm. The speed is
v =
(137
rad/s)/(1.14
rad/cm)
= 1.20m/s.
(b) The nodes are (7rrad)/(1.14 rad/cm) = 2.76 cm apart.
( c ) The velocity of a particle on the string at position x and time t is the derivative of the wave
equation with respect to time, or
Uy = -(0.520
cm)(137
rad/s)
sin[(1.14
rad/cm)(1.47
231
cm)] sin[(137
rad/s)(1.36s)]
= -0.582m/s.
~
(a) We are given the wave frequency and the wave-speed, the wavelength is found from
Eq. 18-13,
v
>. = 7 = (388m/s)
(622 Hz)
The standing
wave has four loops, so from Eq.
= O.624m
18-45
is the length of the string.
(b) We can just write it clown,
y =
(1.90
mm)
sin[(27r
/0.624m)x]
E18-30
(a) In = nv/2L = (1)(250m/s)/2(0.150m)
(b) >..= v/1 = (348m/s)/(833 Hz) = O.418m.
E18-31
v = vfFf¡¡
= ,fFL7rñ.
Then In = nv/2L
cos[(27r622
S-I )t].
= 833 Hz.
= n.¡FT4mL,
h = (1)J(236N)/4(0.107kg)(9.88m)7.47
so
Hz,
and !2 = 2h = 14.9 Hz while !3 = 3h = 22.4 Hz.
E18-32
(a) v = JF7jl = v'FL?m = ..¡(122N)(1.48m)/(8.62x
10-3kg) = 145m/s.
(b) >..1= 2(1.48m) = 2.96m; >..2= 1.48m.
(c) h = (145m/s)/(2.96m)
= 49.0 Hz; f2 = (145m/s)/(1.48m)
= 98.0 Hz.
~
Although the tied end of the string forces it to be a node, the fact that the other end
is loose means that it should be an anti-node. The discussion of Section 18-10 indicated that the
spacing between nodes is always >./2. Since anti-nodes occur between nodes, we can expect that the
distance between a node and the nearest anti-node is >./4.
The longest possible wavelength will have one node at the tied end, an anti-node at the loose
end, and no other nodes or anti-nodes. In this case >./4 = 120 cm, or >.= 480 cm.
The next longest wavelength will have a node somewherein the middle region of the string. But
this means that there must be an anti-node between this new node and the node at the tied end
of the string. Moving from left to right, we then have an anti-node at the loose end, a node, and
anti-node, and finally a node at the tied end. There are four points, each separated by >./4, so the
wavelength would be given by 3>./4 = 120 cm, or >.= 160 cm.
To progress to the next wavelength we will add another node, and another anti-node. This will
add another two lengths of >./4 that need to be fit onto the string; hence 5>./4 = 120 cm, or >.= 100
cm.
In the figure below we have sketched the first three standing waves.
232
E18-34 (a) Note that fn = nh. Then fn+l -fn = h. Since there is no resonant frequency
betweenthese two then they must differ by 1, and consequently h = (420 Hz) -(315 Hz) = 105 Hz.
(b) v = f>. = (105 Hz)[2(0.756m)] = 159m/s.
P18-1 (a) >. = v/f and k = 360° />..
x = (55°)A/(36üO)
= 55(353m/s)/360(493
O.109m.
Hz)
(b) (J) = 360° f , so
f/J = LAJt= (360O)(493
P18-2
Hz)(1,12
x 10-38)
= 199°,
(¿)= (27r rad)(548 Hz) = 3440 radls; >. = vi f and then
k = (27r rad)/[(326m/s)/(548
Finally, y = (1.12x 10-2m) sin[(10.6 rad/m)x
Hz)]
= 10.6
rad/m.
+ (3440 rad/s)t].
~
( a) This problem really isn 't as bad as it might look. The tensile stress S is tension per
unit cross sectional area, so
F
S =
-or
A
F = SA.
We already know that linear mass density is JL= m/ L, where L is the length of the wire. Substituting
into Eq. 18-19,
[i
~
r---s---
V=y¡=ym¡Ly;;;¡¡AL).
But AL is the volume of the wire, so the denominator is just the mass density p.
(b) The maximum speed of the transverse wave will be
v
{8 -{(720 x 106Pa) = 300m/s.
y p -Y" (7800kg/m3)
P18-4
{a) f = (,)/211"= {4.08 rad/s)/{211" rad) = 0.649 Hz.
(b) >. = v/f = {0.826m/s)/{0.649 Hz) = 1.27m.
{c) k = {211"rad)/{1.27m) = 4.95 rad/m, so
y =
(5.12
cm) sin[(4.95
rad/mx
-(4.08
rad/s)t
+ 4>],
where <I>is determined by (4.95 rad/m)(9.60 x 10-2m) + <I>= (1.16 rad), or <I>= 0.685 rad.
(d) F = JLV2= (0.386kg/m)(0.826m/s)2
= 0.263m/s.
P18-5 We want to show that dy/dx = Uy/v. The easy way, although not mathematically rigorous:
dy
dx
P18-6
The maximum value for Uy occurs when the cosine function in Eq. 18-14 returns unity.
Consequently, um/Ym = úJ.
233
[E:f!J
(a) The linear mass density changesas the rubber band is stretched! In this case,
m .
JL= L +¿lL
The tension in the rubber band is given by F = kAL. Substituting this into Eq. 18-19,
v=f¡¡=J~.
(b) We want to know the time it will take to travel the length of the rubber band, so
L+AL
L+AL
v=ort=
t
v
Into this we will substitute our expression for wave speed
.
t=(L+~L)/~.
=~
k~L
We have to possibilities to consider: either ~L « L or ~L » L. In either case we are only
interested in the part of the expressionwith L + ~L; whichever term is much largerthan the other
will be the only significant part.
Then if ~L « L we get L + ~L ~ L and
t= V'(m(L+~L)~ ,
kAL
V~,
~
so that t is proportional to 1/JXL".
But if AL » L we get L + AL ~ AL and
-rm
~~
v~
-yk'
so that t is constant.
Pl8-8
(a) The tension in the rope at some point is a function of the weight of the cable beneath
it. Ifthe bottom ofthe rope is y = O, then the weight beneath some point y is W = y(m/L)g. The
speedofthe wave at that point is v = VT/(m/L)=
vy(M/L)g/(m/L)=
.¡¡¡y.
(b) dy/dt = .¡¡¡y, so
dt
dy
=
m'
t =
fL
lo
iy
.
=2JL79.
"m
(c) No.
P18-9
(a) M = jJl,dx,
so
M =
¡
L
kxdx
o
Then k = 2M~
(b) v = vFIJL = VF7kZ,
dt
=
then
.¡ki1F
dx ,
234
1
= -kL2,
2
P18-10 Take a cue from pressureand surface tension. In the rotating non-inertial referenceframe
for which the hoop appears to be at rest there is an effective force per unit length acting to push on
each part of the loop directly away from the center. This force per unit length has magnitude
~F
~L
V2
= (~m/~L)-;:
V2
= JL-;:'
There must be a tension T in the string to hold the loop tQgether. Imagine the loop to be replaced
with two semicircular loops. Each semicircular loop has a diameter part; the force tending to pull
offthe diameter section is (~F/~L)2r = 2JlV2.There are two connections to the diameter section,
so the tension in the string must be half the force on the diameter section, or T = JlV2.
The wave speed is Vw = JT7¡i = v.
Note that the wave on the string can travel in either direction relative to an inertial observer.
One wave will appear to be fixed in space; the other will move around the string with twice the
speed of the string.
~
If we assumethat Handel wanted his violins to play in tune with the other instruments
then all we need to do is find an instrument from Handel's time that will accurately keep pitch over
a period of several hundred years. Most instruments won't keep pitch for even a few days becauseof
temperature and humidity changes;some (like the piccolo?) can't even play in tune for more than
a few notes! But if someonefound a tuning fork...
Since the length of the string doesn't change, and we are using a string with the same mass
density, the only choice is to changethe tension. But f cx:v cx:.JT , so the percentage change in the
tension of the string is
Tf-Ti
=
ff2
Ti
= I(440
-fi2
HZ)2
, .~--(422.5
~ --,
"
HZ)2
= 8.46 %.
fi2
P18-12
P18-13 (a) The point sourcesemit spherical wavesj the solution to the appropriate wave equation
is found in Ex. 18-14:
A
y~ = -sin(kri -(..;t).
ri
If ri is sufficiently large compared to A, and rl ~ r2, then let rl = r- 8r and r2 = r -8r;
A
-+-~
rl
A
2A
r2
r
with an error of order (r5r/r)2. So ignore it.
Then
yl + y2
Ym
(b) A maximum
multiple
(minimum)
of 7r (a half odd-integer
~
~ [sin(krl
r
-úJt)
=
~
-úJt)
=
2A
k
-;:-COS2(rl
occurs
multiple
sin(kr
+ sin(kr2
cos ~(rl
235
,
-r2),
-!2).
when the operand
of 7r)
-úJt)]
of the cosine,
k(rl
-r2)/2
is an integer
P18-14
The direct wave travels a distance d from S to D. The wave which refiects off the original
layer travels a distance V d2 + 4H2 between S and D. The wave which refiects off the layer after
it has risen a distance h travels a distance vld2 + 4(H + h)2. Waves will interfere constructively if
there is a difference of an integer number of wavelengths between the two path lengths. In other
words originally we have
v ([2 + 4H2 -d
and later we have destructive
interference
= n>..,
so
Vd2 +4(H
+ h)2-
d = (n+
1/2»...
We don 't know n, but we can subtract the top equation from the bottom and get
Vd2 + 4(H + h)2 -V
~
The
wavelength
d2 + 4H2 = AI2
is
>..= v/ f = (3.00x 108m/s)/(13.0x
106Hz) = 23.1 m.
The direct wave travels a distance d from S to D. The wave which reflects off the originallayer
travels a distance v d2 + 4H2 between S and D. The wave which reflects off the layer one minute
later travels a distance Vd2 + 4(H + h)2. Waves will interfere constructively if there is a difference
of an integer number of wavelengths between the two path lengths. In other words originally we
have
vd2
+
+
4(H
4H2-
d =
nlA,
and then one minute later we have
vd2
+
h)2
-d
=
n2>..
We don't know either ni or n2, but we do know the difference is 6, so we can subtract the top
equation from the bottom and get
..¡d2
+4(H
+
h)2-
..¡d2
+
4H2
=
6>.
Between the second and the third lines we factored out d2 + 4H2; that last line is from the binomial
expansion theorem. We put this into the previous expression, and
.¡d2+4(H+h)2-.¡d2+4H2
/
4H
"\
4H
vál
236
+ 4H2
h
=
6>.,
=
6>.,
=
6>..
Now what were we doing? We were trying to find the speed at which the layer is moving. We know
H, d, and >.; we can then find h,
h
=
6(23.1
m)
4(510
x 103m)
v' (230 x 103mr + 4(510 x 103m)2 = 71.0 m.
The layer is then moving at v = (71.0m)/(60s)
P18-16
The equation
of the standing
= 1.18m/s.
wave is
y = 2ym sin k X cos (J)t.
The transverse speed of a point on the string is the derivative of this, or
Uy =
-2ymúJsink
x sinúJt,
this has a maximum value when (AJt-7r /2 is a integer multiple of 7r.The
Um
=
2ym(AJsin
maximum value is
kx.
Each mass element on the string dm then has a maximum kinetic energy
{dm/2)um2
dKm
= Ym2(AJ2 sin2 k x dm.
Using dm = Jl.dx, and integrating over one loop from k x = O to k x = 7r, we get
Km
= Ym2(.,)2JL/2k
= 21T2ym2 fJLv.
P18-17 (a) For 100% reflection the amplitudes of the incident and reflected wave are equal, or
Ai = Ar, which puts a zero in the denominator of the equation for SWR. If there is no reflection,
Ar = 0 leaving the expressionfor SWR to reduce to Ai/Ai = 1.
(b) Pr/Pi = A~/A~. Do the algebra:
Ai +Ar
=
SWR,
=
=
SWR(Ai -Ar),
Ai(SWR -1),
=
(SWR- l)/(SWR + 1).
Ai-Ar
Ai+Ar
Ar(SWR
+ I)
Ar/Ai
Square this, and multiply by 100.
P18-18
Mea.sure with a ruler; I get 2Amax = 1.1 cm and 2Amin = 0.5 cm.
(a) SWR = (1.1/0.5) = 2.2
(b) (2.2 -1)2/(2.2 + 1)2 = 0.14%.
yi
Yt
Yr
=
Asink1(x
=
B sin
=
Csink1(x
-Vlt),
k2(X
-V2t),
+
Vlt),
where the subscripts i, t, and r refer to the incident, transmitted, and reflected wavesrespectively.
237
~
Apply the principle of superposition. Just to the left of the knot the wave has amplitude yi + Yr
while just to the right of the knot the wave has amplitude yt. These two amplitudes must line up
at the knot for a1l times t, or the knot wi1l come undone. Remember the knot is at x = O, so
Asin
k1 ( -Vlt)
-A
+ C sink1
sin k1vlt
Yi +Yr
=
yt!
( +Vlt)
=
Bsin
=
-Bsin
+ Csink1vlt
k2(-V2t),
k2v2t
We know that k1vl = k2v2 = (.o),so the three sin functions are all equivalent, and can be canceled.
This leavesA = B + C.
(b) We need to match more than the displacement, we need to match the slope just on either
side of the knot. In that casewe need to take the derivative of
yi + Yr = Yt
with respect to x, and then set x
o. First we take the derivative,
d
-(Yi
dx
k1A
cos k1 (X -Vlt)
+
k1 C cos
+
k1 (X +
Yr)
d
=
Vlt)
;¡;
=
(Yt)
,
k2Bcosk2(x
-V2t),
and then we set x = O and simplify,
klA cos kl ( -Vlt)
+ kl C cos k1 ( +Vlt)
-
k2B cos k2( -V2t),
+ klCcosklvlt
=
k2B cos k2v2t.
klAcosklvlt
This last expressionsimplifies like the one in part (a) to give
k1(A+
C) = k2B
We can combine this with A = B + C to solve for C ,
k1(A + O)
=
O(k1 + k2)
-
k2(A-
O),
A(k2 -k1),
A~
k1 + k2.
O
If k2 < k1 C will be negative; this means the reflected wave will be inverted.
P18-20
P18-21
Find
the wavelength
from
A = 2(0.924m)/4 = 0.462m.
Find
the
wavespeed
from
v
Find
the
tension
IA = (60.0 Hz)(0.462m)
= 27.7m/s.
from
F=¡¡,V2
= (0.0442
kg)(27.7m/s)2/(0.924m)
238
36.7N.
P18-22 (a) The frequency of vibration f is the same for both the aluminum and steel wires; they
don not, however, need to vibrate in the same mode. The speed of waves in the aluminum is Vl,
that in the steel is V2. The aluminum vibrates in a mode given by nl = 2L1f /Vl, the steel vibrates
in a mode given by n2 = 2L2f /V2. Both nl and n2 need be integers, so the ratio must be a rational
fraction. Note that the ratio is independent of f, so that Ll and L2 must be chosen correctly for
this problem to work at all!
This ratio is
(b) There are three nodes in the aluminum and six in the steel. But one of those nodes is shared,
and two are on the ends of the wire. The answer is then six.
2?!:1
E19-1
(a) v = fA = (25 Hz)(0.24m) = 6.0m/s.
(b) k = (27r rad)/(0.24m) = 26 rad/m; (¿)= (27r rad)(25 Hz) = 160 rad/s. The wave equation is
8 = (3.0x 10-3m) sin[(26 rad/m)x
+ (160 rad/s)tl
E19-2
(a) [~P]m = 1.48Pa.
(b) f = (3347r rad/s)/(27r rad) = 167 Hz.
(c) >. = (27r rad)/(1.077r rad/m) = 1.87m.
(d) v = (167 Hz)(1.87m) = 312m/s.
~
(a) The wavelength is given by >..= v/f = (343m/s)/(4.50xlO6Hz
= 7.62xlO-5m.
(b) The wavelength is given by >..= v/f = (l500m/s)/(4.50xlO6Hz
= 3.33xlO-4m.
Note: There is a typo; the mean free path should have been measured in "JLm" instead of
E19-4
"pm".
>.min
l.Ox lO-6m; fmax = (343m/s)/(l.OxlO-6m)
E19-5
(a) A = (240m/s)/(4.2x
E19-6
(a) The speed of sound is
= 3.4x lO8Hz.
lO9Hz) = 5.7x lO-8m.
v = (331 m/s)(6.21 x 10-4 mi/m) = 0.206 mi/s.
In five seconds the sound travels (0.206 mi/s)(5.0s)
(b) Count seconds and divide by 3.
= 1.03 mi, which is 3% too large.
~
Marching at 120 pacesper minute means that you move a foot every half a second. The
soldiers in the back are moving the wrong foot, which means they are moving the correct foot half
a secondlater than they should. If the speed of sound is 343m/s, then the column of soldiers must
be (343m/s)(0.5s) = 172m long.
E19-8 It takes (300m)/(343m/s) = 0.87s for the concert goer to hear the music after it has passed
the microphone. It takes (5.0x106m)/(3.0x108m/s) = 0.017s for the radio listener to hear the music
after it has passedthe microphone. The radio listener hears the music first, 0.85s before the concert
goer.
E19-9
x/vp = tp and x/vs = ts; subtracting
x = ~t/[l/vs
E19-10
-l/vp]
and rearranging,
= (1808)/[1/(4.5
Use Eq. 19-8, Sm = [6.p]m/kB,
km/s) -1/(8.2
km/s)] = 1800 km.
and Eq. 19-14, v = VJirp;;.
Then
[6.p]m = kBsm = kv2pOSm = 27rfvposmo
Insert into Eq. 19-18, and
= 27r2Pvf2sm2,
~
If the source emits equally in all directions the intensity at a distance r is given by the
average power divided by the surface area of a sphere of radius r centered on the source.
The power output of the source can then be found from
p = [A = I(47rr2) = (197xl0-6W/m2)47r(42.5m)2
240
= 4.47W.
E19-12
Use the results of Exercise 19-10.
(1.13x 10-6W /m2)
27r2(1.21kg/m3)(343 m/s)(3i3 HZ)2 = 3.75x lO-8m.
Bm=
E19-13
U = IAt = (1.60x 10-2W /m2)(4. 70x 10-4m2)(3600s)
E19-14
Invert
Eq.
19-21:
11/ 12 = 10(1.00dB)/10
~
= 2.71 x 10-2W.
= 1.26.
(a) Relative sound level is given by Eq. 19-21,
SLl
-SL2
=
10 lag
I
-2
I2
ar
I'
-2
I2
=
10(8Ll-8L2)/lO
,
so if ~SL = 30 then /1/ 12 = 1030/10 = 1000.
(b) Intensity is proportional to pressure amplitude squared according to Eq. 19-19j so
~Pm,l/ ~Pm,2 = ,ffJI;;
E19-16
= v'1OOO= 32.
We know where her ears hurt, so we know the intensity at that point. The power output
is then
p = 47r(1.3m)2(1.0W/m2)
= ~lW.
This is less than the advertised power .
E19-17
Use the results oí Exercise 18-18, I = uv. The intensity is
I = (5200W)/47r(4820m)2
= 1.78x 10-5W /m2,
so the energy density is
u =
E19-18
12 = 2/1,
since
I/v=
(l.78xlO-5W/m2)/(343m/s)=
I cx 1/r2
then
r~ = 2r~.
D
D(h-l)
D
~
5.l9xlO-8J/m3.
Then
=
yf2(D -51.4 m),
=
yf2(51.4 m),
176m.
=
The sound level is given by Eq. 19-20,
SL
=
lOlog
I
"1;:;
where lo is the threshold intensity of 10-12 W /m2. Intensity is given by Eq. 19-19,
I=~
2pv
If we assumethe maximum possible pressure amplitude is equal to one atmosphere, then
I = ~
=
2pv
(1.01x 105Pa)2
2(1.21kg/m3)(343m/s)
241
= 1.22x 107W/m2.
The sound level would then be
I
1.22 x 107W /m2
SL = 101ogT
= 101og
2
.Lo
(10-12 W /m )
= 191 dB
E19-20
Let one person speak with an intensity 11. N people would have an intensity NI1.
ratio is N, so by inverting Eq. 19-21,
The
N = 1Ü(15dB)/10 = 31.6,
so 32 people would be required.
E19-21
Let one leafrustle
is N, so by Eq. 19-21,
with an intensity 11. N leaves would have an intensity NI1. The ratio
SLN
= (8.4 dB) + 10 log{2. 71 x 105) = 63 dB.
E19-22 Ignoring the finite time means that we can assume the sound waves travels vertically,
which considerably simplifies the algebra.
The intensity ratio can be found by inverting Eq. 19-21,
/1/ 12 = 10(30dB)/10
= 1000.
But intensity is proportional to the inverse distance squared, so 11/12 = (r2/r1)2, or
r2 = (115m),¡-(i"{Xjjj
= 3640 m.
~
A minimum will be heard at the detector if the path length difference between the
straight path and the path through the curved tube is half of a wavelength. Both paths involve a
straight section from the sourceto the start of the curved tube, and then from the end of the curved
tube to the detector. Since it is the path difference that matters, we'll only focus on the part of
the path between the start of the curved tube and the end of the curved tube. The length of the
straight path is one diameter, or 2r. The length of the curved tube is half a circumference, or 7rr.
The difference is (7r-2)r. This difference is equal to half a wavelength, so
(7r-2)r
=
>./2,
>.
r =
=
(42.0
27r -4
E19-24
cm)
211"
=
18.4
cm.
-4
The path length difference here is
V(3.75m)2
+ (2.12m)2-
(3.75 m) = 0.5578 m.
(a) A minimum will occur ifthis is equal to a halfinteger number ofwavelengths, or (n-l/2».
0.5578m. This will occur when
f
=
(n
(343m/s)
-1/2)¡¡¡:5578m)
=
(n-
1/2)(615
=
Hz).
(b) A maximum will occur ir this is equal to an integer number or wavelengths, or n>. = 0.5578 m.
This will occur when
(343m/s)
f = n(O.5578
m) = n(615Hz).
242
E19-25
The path length difference here is
v(24.4m
+ 6.10m)2 + (15.2m)2 -v(24.4m)2
+ (15.2m)2 = 5.33m.
A maximum will occur ir this is equal to an integer number or wavelengths,or n>. = 5.33m. This
will occur when
f = n(343m/s)/(5.33m)
The two lowest frequencies
= n(64.4
Hz)
are then 64.4 Hz and 129 Hz.
E19-26 The wavelength is >.= (343m/s)/(300 Hz) = 1.143m. This meansthat the sound maxima
will be half of this, or 0.572m apart. Directly in the center the path length difference is zero, but
since the waves are out of phase, this will be a minimum. The maxima should be located on either
side of this, a distance (0.572m)/2 = 0.286m from the center. There will then be maxima located
each 0.572m farther along.
E19-27
(a) h = v/2L and !2 = v/2(L -~L).
Then
1-~=
~ -!2
or ~L = L(ll/r).
(b) The answers are ~L = (0.80m)(1 -5/6)
= O.133m; ~L = (0.80m)(1 -4/5)
~L = (0.80m)(13/4) = O.200m; and ~L=
(0.80m)(1 -2/3) = O.267m.
E19-28
= O.160m;
The wavelength is twice the distance between the nodes in this case, so >. = 7.68 cm. The
frequency is
f = (1520
m/s)/(7.68
x 10-2m)
= 1.98 x 104Hz.
~
The well is a tube open at one end and closed at the other; Eq. 19-28 describes the
allowed frequenciesof the resonant modes. The lowest frequency is when n = 1, so h = v/4L. We
know h; to find the depth of the we11,L, we need to know the speed of sound.
We should usethe information provided, instead of looking up the speedof sound, becausemaybe
the we11is fi11edwith some kind of strange gas.
Then, from Eq. 19-14,
v =
{fi = I {1.41 x 105 Pa)
y p
y {1.21 kg/m3)
= 341 m/s.
The depth of the well is then
L = v/(4f¡)
E19-30
= (341m/s)/[4(7.20
Hz)] = 11.8m.
(a) The resonant frequencies of the pipe are given by In = nv/2L,
In
= n(343m/s)/2(0.457m)
= n(375
or
Hz).
The lowest frequency in the specified range is f3 = 1130 Hz; the other allowed frequencies in the
specified range are f 4 = 1500 Hz, and f5 = 1880 Hz.
243
~
The maximum reflected frequencies will be the ones that undergo constructive interference, which means the path length difference will be an integer multiple of a wavelength. A
wavefront will strike a terrace wall and part will reflect, the other part will travel on to the next
terrace, and then reflect. Since part of the wave had to travel to the next terrace and back, the path
length difference will be 2 x 0.914m = 1.83m.
If the speed of sound is v = 343m/s, the lowest frequency wave which undergoes constructive
interference will be
v
f = A = (3~3m/s)
(1.83 m) .= 187 Hz
Any integer multiple of this frequency will also undergo constructive interference, and will also be
heard. The ear and brain, however, will most likely interpret the complex mix of frequenciesas a
single tone of frequency 187 Hz.
E19-32 Assume there is no frequency between these two that is amplified. Then one of these
frequenciesis fn = nv/2L, and the other is fn+l = (n + 1)v/2L. Subtracting the larger from the
smaller, ~f = v/2L, or
L = v/2~f
= (343m/s)/2(138
Hz -135
Hz) = 57.2 m.
E19-33
(a) v = 2Lf = 2(0.22 m) (920 Hz) = 405m/s.
(b) F = V2J.t= (405 m/s)2(820x 10-6kg)/(0.220m)
= 611N.
E19-34
f ()( v, and v ()( JF,
so f ()( JF.
Doubling
f requires
F increase by a factor
of 4.
~
The speed of a wave on the string is the same, regardlessof where you put your finger,
so f>. is a constant. The string will vibrate (mostly) in the lowest harmonic, so that >.= 2L, where
L is the length of the part of the string that is allowed to vibrate. Then
/2>.2
2/2L2
=
=
L2
So you need to place your finger 5 cm from the end.
E19-36
The open organ pipe has a length
Lo = v/2f¡
= (343m/s)/2(291
Hz) = 0.589 m.
The secondharmonic ofthe open pipe has frequency 2!1; this is the first overtone ofthe closedpipe,
so the closed pipe has a length
Lc = (3)v/4(2f¡)
(3)(343m/s)/4(2)(291 Hz) = 0.442m.
~
The unknown frequency is either 3 Hz higher or lower than the standard fork. A small
piece of wax placed on the fork of this unknown frequency tuning fork will result in a lower frequency
because f cx Jk1iñ.
If the beat frequency decreases then the two tuning forks are getting closer
in frequency, so the frequency of the first tuning fork must be above the frequency of the standard
fork. Hence, 387 Hz.
244
E19-38
If the 8tring i8 too taut then the frequency is too high, or f = (440 + 4)Hz.
T = 1/ f = 1/(444 Hz) = 2.25 x 10-38.
Then
E19-39 One of the tuning forks need to have a frequency 1 Hz different from another. Assume
then one is at 501 Hz. The next fork can be played against the first or the second, so it could have
a frequency of 503 Hz to pick up the 2 and 3 Hz beats. The next one needsto pick up the 5, 7, and
8 Hz beats, and 508 Hz will do the trick. There are other choices.
E19-40
I = vi>. = (5.5 mls)I(2.3m) = 2.39 Hz. Then
I' = I(v + vo)lv = (2.39 Hz)(5.5 mls + 3.3mls)I(5.5mls)
E19-42
= 3.8 Hz.
Solve Eq. 19-44 for vs;
v s = (v+vo)!/!'
-v
= (343m/s+2.63m/s)(1602
Hz)/(1590 Hz) -(343m/s)
E19-43 Vs = (14.7 Rad/8)(0.712m) = 10.5m/8.
(a) The low frequency heard i8
f' = (538 Hz)(343m/8)/(343m/8 + 10.5m/8) = 522 Hz.
(a) The high frequency heard i8
f' = (538 Hz)(343m/8)/(343m/8 -10.5m/8)
E19-44
= 555 Hz,
Salve Eq. 19-44 far Vs;
Vs = V -vi
i I' = (343 mis) -(343
mis) (440 Hz)i(444
E19-45
E19-46
The approaching
E19-47
The departing
car "hears"
intruder
"hears"
The beat ftequency is 28.3 kHz -28.14
kHz = 160 Hz.
245
Hz) = 3.1 mis.
= 5.24m/s.
E19-48 (a) I' = (1000 Hz)(330m/s)(330 m/s + 10.0m/s) = 971 Hz.
(b) I' = (1000 Hz)(330r:n/s)(330 m/s -10.0m/s) = 1030 Hz.
(c) 1030 Hz -971 Hz = 59 Hz.
~
(a) The frequency
I'
=
"heard"
f ~
f -Vs
by the wall is
(343m/s)+(O)
= (438 Hz)(343m/s) -(19.3m/s)
= 464 Hz
(b) The wall then reflects a frequency of 464 Hz back to the trumpet player. Sticking with Eq.
19-44, the source is now at rest while the observer moving,
= (464 Hz)~-+(19.3m/s)
f'=f~
E19-50
The body
part
-"
=
490
Hz
(343m/s) -(O)
v-vs
"hears"
V+Vb
f'=f
v
This sound is reflected back to the detector which then "hears"
I" = I'~
= I.!!---+vb
v -Vb
v -Vb
Rearranging,
I"
Vb/V
so
v ~
2(lxlO-3m/s)/(l.3xlO-6)
E19-51
~
-I"
..,.,1 ~f
-I
+
I
..'.'2¡'
l500m/s.
The wall "hears"
I' = I ~
= (39.2
v -v
(343m/s)+
kHz).
s
(343m/s)
(O)
-= 40.21
kHz
-(8.58m/s)
This sound is refl.ected back at the same frequency, so the bat "hears"
I.
f'=f~
P19-1 (a) tair
)~
= (40.21 kHz)S343m/s)
L/Vair
+ (8.58m/s)
,
v -.,v s
=
41.2
kHz.
(343m/s) -(O)
and tm = L/v,
so the difference
6.t = L(l/vair
is
-l/v)
(b) Rearrange the above result, and
L
(0.120s)/[1/(343m/s)
-1/(6420m/s)]
43.5 m.
P19-2
The 8tone falls for a time tl where y = gt~/2 i8 the depth of the well. Note y i8 po8itive in
thi8 equation. The 8ound travel8 back in a time t~where v = y/t2 i8 the 8peed of 8ound in the well.
t1 + t2 = 3.008,80
2y = g(3.00s
-t2)2
=g[(9.00s2)
-(6.00s)y/v
+ y2/V2],
or, using 9 = 9.81 m/s2 and v = 343 m/s,
y2 -(2.555
x 105 m)y + (1.039 x 107 m2) = 0, .
which has a positive solution y = 40.7 m.
246
12 = I1(rl/r2)2
= (962J1W /m2)(6.11 m/28.5m)2
= 44.2J1W /m2.
(C) We'll do this part first. The pressure amplitude is found from Eq. 19-19,
¡)..Pm= .¡2pvj
=
2(1.21 kg/m3)(343 m/s)(962 x 10-6W /m2) = 0.894 Pa.
(b) The displacement amplitude is found from Eq. 19-8,
8m = ¡)..Pm/(kB),
where k = 27rf /v is the wave number. From Eq. 19-14 w know that B = pv2, SO
¡)..Pm
s m-
P19-4
27rfpv
(0.894Pa)
-27r(2090
= 1.64x 10-7m.
Hz)(1.21kg/m3)(343m/s)
(a) If the intensities are equal, then A Pm <X.¡-¡;ü, so
J~Pm]water -(998kg/m3)(1482m/~)
(1.2 kg/m3)(343 m/s)
[¡j.pm]air
{b) If the pressure
amplitudes
Iwater
= 59.9.
are equal, then I cx:cx:1! pv , SO
-
(1.2 kg!m3)(343m!s)
= 2.78xl0-'4,
(998 kgim3)(1482 mis)
Iair
P19-5
The energy is dissipated on a cylindrical surface which grows in area as r, so the intensity is
proportional to l/r. The amplitude is proportional to the square root ofthe intensity, so Sm CXl/vr.
P19-6 (a) The first position correspondsto maximum destructive interference, so the wavesare half
a wavelength out of phase;the secondposition correspondsto maximum constructive interference, so
the wavesare in phase. Shifting the tube has in effect added half a wavelength to the path through
B. But each segment is added, so
A = (2)(2)(1,65
cm)
= 6.60
cm,
and f = (343m/s)/(6.60 cm) = 5200 Hz.
(b) Imin CX(Sl-S2)2, Imax CX(Sl +S2)2, then dividing one expression by tbe other and rearranging
we find
~=
S2
~+~=
~-~
V90+VfP=2
V90-VfP
P19-7
(a) I = P/47rr2 = (31.6W)/47r(194m)2 = 6.68x10-5W/m2.
(b) P = IA = (6.68x10-5W/m2)(75.2x10-6m2)
= 5.02x10-9W.
( c) U = Pt = (5.02 x 10-9W) (25.0 min) (60.0 s/min) = 7.53 /l.J.
P19-8 Note that the reverberation time is logarithmically related to the intensity, but linearly
related to the sound level. As such, the reverberation time is the amount of time for the sound level
to decreaseby
~SL = 101og(10-6) = 60 dB.
Then
t = (87 dB)(2.6s)/(60
247
dB)
= 3.88
~
What the device is doing is taking all of the energy which strikes a large surface area and
concentrating it into a small surface area. It doesn't succeedjonly 12% of the energy is concentrated.
We can think, however, in terms of power: 12% of the average power which strikes the parabolic
refiector is transmitted into the tube.
Ifthe sound intensity onthe refiector is 11, then the averagepower is P1 = I1A1 = 117rr~,where
rl is the radius of the refiector. The averagepower in the tube will be P2 = O.12P1,so the intensity
in the tube will be
T
-~
-0.12l17rr~
.L2-
-
-2
-.12
A2
O
12l
7rr2
1
r2
Since the lowest audible sound has an intensity of lo = 10-12 W /m2, we can set l2 = lo as the
condition for "hearing" the whisperer through the apparatus. The minimum sound intensity at the
parabolic reflector is
T
2
.LO
r2
I1 = Q"l2
-;:¡.
Now for the whisperers. Intensity falls off a.s 1/d2, where d is the distance from the source. We
are told that when d = 1.0 m the sound level is 20 dB; this sound level ha.s an intensity of
I =101020/10
=
10010
Then at a distance d from the source the intensity must be
11
=
100.l
(1
0-
m)2
d2
This would be the intensity "picked-up" by the parabolic reflector. Combining this with the condition
for being able to hear the whisperers through the apparatus, we have
or, upon some rearranging,
d = (.Jf2m)!:!.
= 346m
= (.Jf2m)~
r2
(0.005 m)
P19-10
(a) A di8placement nade; at the center the particle8 have nawhere ta ga.
(b) Thi8 8y8tem8 act8 like a pipe which i8 cla8ed at ane end.
(c) vVB7p, 8a
T = 4(O.OO9)(6.96x
lOSm).¡(l.Ox
lOlOkg/m3)/(l.33x
lO22Pa)
= 228.
P19-11 The cork filing collect at pressure antinodes when standing waves are present, and the
antinodes are each half a wavelength apart. Then v = j.>..= j(2d).
P19-12
(a) f = v/4L = (343 m/s)/4(1.18m)
(b) F = Ji,V2= Ji,f2>.2, or
= 72.7 Hz.
F = (9.57xl0-3kg/0.332m)(72.7
Hz)2[2(0.332m)]2
248
= 67.1 N.
~
In this problem the string is observed to resonate at 880 Hz and then again at 1320 Hz,
so the two corresponding values of n must differ by 1. We can then write two equations
~
2L
and solve these for v. It is somewhat easier to first salve for n. Rearranging both equations, we get
(880
Hz)
n
v
=2L
Combining these two equations we get
(880 Hz)
=
~.
Now that we know n we can find v,
v = 2(0.300m)~
= 264m/s
2
And, finally, we are in a position to find the tension, since
F = JLV2= (0.652xl0-3kg/m)(264m/s)2
= 45.4 N.
P19-14
(a) There are five choices for the first fork, and four for the second. That gives 20 pairs.
But order doesn't matter, so we need divide that by two to get a maximum of 10 possible beat
frequencies.
(b) If the forks are ordered to have equal differences (say, 400 Hz, 410 Hz, 420 Hz, 430 Hz, and
440 Hz) then there will actually be only 4 beat frequencies.
P19-15
v = (2.25 x lO8m/s)/ sin(58.00) = 2.65 x lO8m/s.
P19-16 (a) h = (442 Hz)(343m/s)/(343m/s
-31.3m/s)
!2 = (442 Hz)(343m/s)/(343m/s
so ~f = 81 Hz.
(b) h = (442 Hz)(343m/s
-31.3 m/s)/(343m/s)
!2 = (442 Hz)(343m/s+
= 486 Hz, while
+ 31.3m/s) = 405 Hz,
= 402 Hz, while
31.3 m/s)/(343m/s)
= 482 Hz,
so A! = 80 Hz.
~
The sonic boom that you hear is not from the sound given off by the plane when it is
overhead, it is from the sound given off before the plane was overhead. So this problem isn't as
simple as distance equals velocity x time. It is very useful to sketch a picture.
249
~
-
I
o
We can find the angle (j from the figure, we'll get Eq. 19-45, so
sin{} = !- = S330m/s)
v..
I(396 mis)
= 0.833
or (} = 56.4°
Note that v8 is the speed of the source, not the speed of sound!
Unfortunately t = 12s is not the time between when the sonic boom leavesthe plane and when
it arrives at the observer. It is the time between when the plane is overhead and when the sonic
boom arrives at the observer. That 's why there are so many marks and variables on the figure. Xl
is the distance from where the sonic boom which is heard by the observer is emitted to the point
directly overhead; X2 is the distance from the point which is directly overhead to the point where
the plane is when the sonic boom is heard by the observer. We do have X2 = v8(12.0s). This length
forms one side ofa right triangle HSO, the opposite side of this triangle is the side HO, which is
the height of the plane above the ground, so
h = x2tan(}
P19-18
= (343m/s)(12.0s)tan(56.4°)
= 7150m.
(a) The target "hears':
f'=fB~'
This sound is refl.ected back to the detector
f
-
v
which
1
'-
v
then
I
r-
"hears"
,
.v+v
-8,
v-v
u-v'
(b) Rearranging,
V/v = Ir -!8
Ir+18
~ -.' ir -is
2 18
where we have assumed that the source frequency and the reflected frequency are almost identical,
so that when added Ir + 18 ~ 218.
I' = f ~
= {1030 Hz) {5470 km/h)
v -v s
{5470 km/h)
250
+ {94.6 km/h)
-{20.2
km/h)
= 1050 Hz
(b) The reflected signal has a frequency equal to that of the signal received by the second sub
originally. Applying Eq. 19-44 again,
I' = I ~
v-vs
P19-20
{5470 km/h)
{20.2 km/h)
= {1050 Hz)(5470
km/h) +
-{94.6
km/h)
In this case v s = 75.2 km/h-
I'
30.5 km/h = 12.4m/s.
(989 Hz)(1482m/s)(1482
=
1070
Hz
Then
997 Hz.
m/s -12.4m/s)
P19-21
There is no relative motion between the source and observer, so there is no frequency shift
regardless of the wind direction.
P19-22
(a) v s = 34.2m/s and Va = 34.2m/s, so
l'
=
(525
Hz)(343m/s
+ 34.2m/s)/(343m/s
-34.2m/s)
(b) v s = 34.2m/s + 15.3m/s = 49.5m/s and Va = 34.2m/s -15.3m/s
I'
= (525
Hz)(343
(c) v s = 34.2m/s -15.3m/s
mis
+ 18.9 mis)i(343
mis
-49.5
mis)
641 Hz.
18.9
mis,
so
= 647 Hz.
= 18.9m/s and Va = 34.2m/s + 15.3m/s = 49.5m/s, so
I' = (525 Hz)(343m/s
+ 49.5 m/s)/(343m/s
251
-18.9m/s)
= 636 Hz.
E20-1
(a) t= x/v = (0.20 m)/(0.941)(3.00x 108m/8) = 7.1 x 10-108.
(b) y = -gt2/2 = -(9.81 m/82)(7.1 x 10-108)2/2 = 2.5x 10-18m.
E20-2
~
E20-4
L = LoV1-
L = LoVl-
Solve
~t
U2/C2 = (2.86m)v1-
U2/C2=
= ~to/Vl-
(0.999987)2
(1.68m)vl-
U2/C2
for
= 1.46 cm.
(0.632)2 = 1.30m.
u:
=
~
2.97x
lO8m/s.
We can apply ~x = v~t to find the time the particle existed before it decayed. Then
=
3.53
x
10-12
s.
The proper lifetime of the particle is
E20-6
Apply Eq. 20-12:
E20-7
(a) L = Lo..¡lu2/c2 = (130m)..¡1(0.740)2 = 87.4m
(b) ~t = L/v = (87.4 m)/(0.740)(3.00 x 108m/8) = 3.94x 10-78.
E20-8 ~t = ~to/ ..¡1 -U2 / c2 = (26 ns) ..¡1 -(0.99)2
= 184 ns. Then
L = v~t = (0.99) (3.00 x 108m/8)(184x 10-98) = 55m.
E20-9 (a) Vg = 2v = (7.91 + 7.91) km/s = 15.82 km/s.
(b) A relativistic treatment yields Vr = 2v/(1 + v2/c2). The fractional error is
E20-10
Invert Eq. 20-15 to get /3 = V1 -1/'Y2.
(a) /3 = V1 -1/(1.01)2
= 0.140.
(b) /3 = V1 -1/(10.0)2
= 0.995.
(c) /3 = V1 -1/(100)2
= 0.99995.
(d) /3 = V1 -1/(1000)2
= 0.9999995.
252
~
The distance traveled by the particle is (6.0 y)c; the time required for the particle to
travel this distance is 8.0 years. Then the speed of the particle is
v=
~x
(6.0 y)c
3
= (&OY)
= ¡c.
~t
The speed parameter ,8 is given by
v
/3=-=1-c
E20-12
'"Y= 1/ VI -(0.950)2
x'
t'
E20-13
=
=
(3.20)[(1.00 x 105m) -(0.950)(3.00
-4.
x 108m/s)(2.00 x 10-48)] = 1.38x 105m,
(3.20) [(2.00 x 10-48) -(1.00x105m)(0.950)/(3.00x108m/8)]
x'
=
(1.081)[(3.20x
t'
=
(1.081)[(2.50
(b) 'Y = 1/ v'1 -(0.380)2
t'
3
c
= 3.20. Then
(a) 'Y = 1/ .¡1 -(0.380)2
x'
2c
=
=
= -3.73x10-4s.
= 1.081. Then
108m) -(0.380)(3.00x
s) -(3.20
108m/s)(2.50s)]
x 108m)(0.380)/(3.00
x 108m/s)]
= 3.78x
107m,
= 2.26 s.
= 1.081. Then
(1.081)[(3.20x108m) -(-0.380)(3.00x 108m/8)(2.508)) = 6.54x108m,
(1.081)[(2.508) -(3.20 x 108m)(-0.380)/(3.00 x 108m/8)) = 3.148.
E20-14
E20-15
(a) v~ = ( -u)/(1 -O) and v~ = cV1 -U2/C2.
(b) (V~)2 + (V~)2 = U2 + C2 -U2 = (:2.
E20-16
v' = {0.787c + O.612c)f[1 + {0.787){0.612)] = O.944c.
~
( a) The first part is easyj we appear to be moving away from A at the same speed as A
appears to be moving away from us: 0.347 c.
(b) Using the velocity transformation formula, Eq. 20-18,
E20-18
v' = (0.788c -0.413c)/[1
E20-19
(a) 7 = 1/ VI -(0.8)2 = 5/3.
,
v"'
+ (0. 788)( -0.413)]
Vx
=
/(1-
,
vtJ
Then v' = cV( -4/5)2
=
12
= 25c,
-3(0.8c)
uVy/c2)
v -u
1- yUVy/C2
= 0.556c.
-5[1
-
(0)
-(O)]
-(0.8c)
1 -(0)
-4
--5c.
:
+ (12/25)2 = 0.933c directed (J = arctan( -12/20)
253
= 31° East of South.
(b)
'Y =
1/ .¡1
-(0.8)2
= 5/3.
=
I
v x
1-
Vx -u
uvx/&
-
(O)
-(
1
vy
=
I
v y
1(1-
Then v' = cy(4/5)2
-3(0.8c)
-5[1
4
-0.8c)
=+5c,
-(O)
-
-(0)]
12
25c.
uvx/&)
+ (12/25)2 = O.933c directed (J = arctan(20/12)
= 59° West of North.
E20-20
This exercise should occur in Section 20-9.
(a) v = 21r(6.37x106m)c/(ls)(3.00x108m/s)
= O.133c.
(b) K = {'Y -1)mc2 = (1/ .¡1(0.133F -1)(511 keV) = 4.58 keV.
(c) Kc = mv2/2 = mc2(v2/c2)/2 = (511 keV)(0.133)2/2 = 4.52 keV. The percent error is
(4.52E20-21
~L
~L
E20-22
= L' -Lo
= -1.31%.
so
= 2(6.370xlO6m)(l-
y
-{29.8
x 103m/s)2 /{3.00 x 108m/s)2) = 6.29 x 10-2m.
(a) ~L/Lo = 1- L' /Lo so
~L
{1-
VI -{522
(b) We want to 8o1ve~t -~t'
which has solution ~t
The length
m/s)2 /{3.00 x 108m/s)2) =
51 x 10-12.
= 1 JL8,or
1 Jl,S= ~t(l
~
4.58)/(4.58)
-1/
..¡1-
(522 m/s)2 /(3.00 x 108m/s)2),
6.61 x 1058. That'8 7.64 day8.
of the ship as measured
L = LoVl-V2/C2
in the "certain"
= (358m)vl-
reference
frame is
(0.728)2 = 245 m.
In a time ~t the ship will move a distance Xl = Vl~t while the micrometeorite will move a distance
X2 = V2~t; since they are moving toward each other then the micrometeorite will pass then ship
when Xl + X2 =L. Then
At = L/(Vl
+ V2) = (245m)/[(0.728
+ 0.817)(3.00x 108m/8)] = 5.29 x 10-18.
This answer is the time measured in the "certain"
the time as measured on the ship,
reference frame. We can use Eq. 20-21 to find
~t = ~t' + U~X'/C2 = (5.29x 10-7 s) + (0.728c)(116m)/c2
Vl-
U2/C2
= 1.23x 10-6s.
V1 -(0.728)2
E20-24
(a) "1 = 1/ ..¡1- (0.622)2 = 1.28.
(b) ~t = (183 m)/(0.622)(3.00x 108m/s) = 9.81 x io-7s. On the clock, however,
~t' = ~t/'Y
{9.81 x 10-7 s)/{1.28) = 7.66 x 10-7 s.
254
~
E20-25
(a) ~t = (26.0 ly)/(0.988)(1.00 ly/y) = 26.3 y.
(b) The signal takes 26 years to return, so 26 + 26.3 = 52.3 years.
(c) ~t' = (26.3 Y)Vl -(0.988)2 = 4.06 y.
E20-26 (a) '1 = (1000 y)(l y) = 1000;
v = cV1 -1/'12
~ c(l -1/2y)
= 0.9999995c
(b) No.
E20-27
{5.6lx
E20-28
p2 = m2c2 = m2v2/(1 -V2/C2),
lO29 MeV/c2)c/{3.00xlO8m/s)
=
l.87xlO21
MeV/c.
so 2V2/C2 = 1, or v = V2c.
~
The magnitude of the momentum of a relativistic
the velocity is given by Eq. 20-23,
mv
particle in terms of the magnitude of
p=~.
The speed parameter, {3, is what we are looking for, so we need to rearrange the above expression
for the quantity v/c.
p/c
mv/c
=
.,f1=""V'TC'J'
~
c
mc
p
mc
m,8
=
~'
=
~
=
VI/
{3
{32 "'- I.
p
Rearranging,
mc2
=
VI/
=
1
:82 -i,
-
(:]2 -1,
1
{3'
{3
(a) For the electron,
{3 =
(12;5 MeV /c)c
= 0.999.
..¡(0.511 MeV /c2)2c4 + (12.5 MeV /C)2C2
:,a=
(12.5 MeV /c)c
= 0.0133.
..¡(938 MeV /C2)2C4+ (12.5 MeV /c)2c2
(b) For the proton,
255
E20-30 K = mc2('Y-1), so 'Y = 1 + K/mc2. /3 = V1-1/'Y2.
(a) 'Y = 1 + (1.0 keV)/(511 keV) = 1.00196. /3 = 0.0625c.
(b) 'Y = 1 + (1.0 MeV)/(0.511 MeV) = 2.96. /3 = 0.941c.
(c) 'Y = 1 + (1.0 GeV)/(0.511 MeV) = 1960. /3 = 0.99999987c.
We rearrange this to salve for /3 :::; v / c,
2
It is actually
much ea.sier to find 1', since
I
-,
"1 -..¡1-
so K = "Ymc2-mc2
V2/C2
implies
1=
K+mc2
mc2
(a) For the electron,
=
0.9988,
and
(b) For the proton,
(938 MeV /C2)C2
fJ=yl-
(10 MeV) + (938 MeV /c2)c2
2
= 0.0145,
and
(b) For the alpha particle,
=0.73,
and
E20-32
'Y = 1/ V1 -(0.99)2 = 7.089.
(a) E = 'Ymc2 = (7.089)(938.3 MeV) = 6650 MeV.
(938.3 MeV /c2)(0.99c)(7.089) = 6580 MeV /c.
(b) E = 'Ymc2 = (7.089)(0.511 MeV) = 3.62 MeV.
(0.511 MeV /c2)(0.99c)(7.089) = 3.59 MeV /c.
256
K = E -mc2
= 5710 MeV.
p =
mv"Y
=
K = E -mc2
= 3.11 MeV.
p =
mv')'
=
E20-33
~m/~t
= (1.2x1041W)/(3.0x108m/s)2
~
~t
E20-34
E-
(a) If K
= 1.33x1024kg/s,
= (1.33x1024kg/s)(3.16x107s/y)
(1.99 x 103°kg/sun)
which is
= 21.1
mc2 = 2mc2, then E = 3mc2, SO7 = 3, and
v = cv'l
-1/72
= cv'l
-1/(3)2
= O.943c.
(b) If E = 2mc2, then 7 = 2, and
v = cV1 -1/72
mc2
.--mc2
K
\11We want
to expand
the
1 -.82
= cV1 -1/(2)2
= O.866c.
( (1-
=
mc2
+
-mc2f34
8
.82) -1/2
v2/c2
part
for
small
.8,
(1 -{32) -1/2
Inserting this into the kinetic energy expression,
K
But /3
=
1
-mc2f32
2
3
+
V/C,80
1
3 V4
K = -mv2 + -m
+ ...
2
8 C2
(b) We want to know when the error because of neglecting the second ( and higher) terms is 1%;
or
0.01
This will happen
E20-36
when v/c = V(0.01)4/3)
Kc = (lOOOkg)(20m/s)2/2
Kr
~
3
Start
V4
1
= (-m-)/(-mv
8
C2
2
) =
2
3
-4
(V ) 2
c
= 0.115.
= 2.0xlOsJ.
The relativistic
calculation is slightly harder:
=
(lOOOkg)(3xlO8m/s)2(l/..¡l-(20m/s)2/(3xlO8m/s)2-l),
Rj
(lOOOkg) [~(20m/S)2
=
2.0x lOsJ + 6.7x lO-lOJ.
with
+ ~(20m/s)4/(3XlO8m/s)2
+ ...] ,
Eq. 20-34 in the form
E2
(pc)
2 +
(mc2)2
The rest energy is mc2, and if the total energy is three times this then E = 3mc2, so
(3mc2)2
=
(pc)2 + (mc2)2,
8(mc2)2
=
(pC)2,
h
mc
-p.
257
E20-38
The initial
kinetic
energy is
2mv2
2v
1
Ki=2m
-(1+v2/c2)2'
1+v2/c2
The final kinetic energy is
E20-39
This exercise is much more involved than the previous one!
The initial kinetic energy is
mc2
=
Kj
2
F(~)2/:;
,
-mc
mc2(l
+
~2/C2)2
v2/c2)
2
-4V2/C2
-mc
m(c2 + V2)
m(c2 -V2)
1- v2/c2 -1v2/c2
,
,
2mv2
1-v2/c2.
The final kinetic energy is
Kf
=
m¿
2
2--2mc,
V;-(V~)2/-:;
2
mc2
1-?/-?\In
I..
y1mc2
2
1
-021-2\-2mc2,
(V2/C2)(2
-v
2
/ c 2-
2~-2~
1-v2/c2
-V2/C2)
2mc2,
1-v2/c2
,
2mv2
l-v2/c2'
E20-40
For a particle with mass, 'Y = K/mr?
+ 1. For the electron, 'Y = {0.40)/{0.511) + 1 = 1.78.
For the proton, 'Y = {lO)/{938) + 1 = 1.066.
For the photon, pc = E. For a particle with mass, pc = .¡{K
pc =
[{0.40 MeV) + {0.511 MeV)]2 -{0.511
+ mc2)2 -m2c4.
MeV)2 = 0.754 MeV.
For the proton,
pc = V[(10 MeV) + (938 MeV)]2 -(938
(a)
(b)
(c)
(d)
MeV)2 = 137 MeV.
Only photons move at the speed of light, so it is moving the fastest.
The proton, since it has smallest value for 'Y.
The proton has the greatest momentum.
The photon has the least.
258
For the electron,
~
Work
is change in energy, so
w
=
mc2/vl-
(Vf/C)2-
mc2/vl-
(Vi/C)2,
(a) Plug in the numbers,
w
= {0.511
MeV){1/v1
~{0.19)2-1/Vl-{0.18)2)
= 0.996
keV.
(b) Plug in the numbers,
w
E20-42
= {0:511
E = 2¡moc2
MeV)(11
= mc2,
..¡1-(0.99)2
-11
..¡1 -(0.98)2)
=
1.05 MeV.
SO
m = 2lmo = 2(1.30mg)/ VI -(0.580)2 = 3.19mg.
E20-43
(a) Energy conservation requires Ek = 2E1r,or mkc2 = 2'"Ym1rC2,
Then
'Y = (498 MeV)12(140 MeV) = 1.78
This corresponds to a speed of v = cV1 -11(1.78)2 = O.827c.
(b) 'Y = (498 MeV + 325 MeV)I(498 MeV) = 1.65, so v = cV1-11(1.65)2
(c) The lab frame velocities are then
(0.795)
+( -0.827)
VI, = ~+
(O.795)(
-0.827) c = -0.0934c,
+ (0.827)
, = 1(0.795)
V2
+ (O.795)(0.827)
c = 0.979c,
The corresponding kinetic energiesare
K1
=
K1
(140
MeV)(1/v1-
= (140
MeV)(1/
(-0.0934)2-1)
VI
-(0.979)2-
= 0.614
1) = 548 MeV
E20-44
P20-1
(a) ')' = 2, so v = V1 -1/(2)2
= O.866c.
(b)')'=2.
P20-2
(a) Classically, v' = (0.620c) + (0.470c) = 1.0gc. Relativistically,
(b) Cla.ssically,v' = (0.620c) + (-0.470c) = 0.150c. Relativistically,
+ ( -0.470c)
v , = (0.620c)
1 + (0.620)(
-0.470) = 0.211c.
259
MeV
= O.795c.
P20-3
(a) '"Y= 1/ V1 -(0.247)2
= 1.032. Use the equations from Table 20-2.
~t = (1.032)[(0) -(0.247)(30.4x103m)/(3.00x108m/s)]
= -2.58x10-5s.
(b) The red flash appears to go first.
P20-4
Once again, the "pico" should have been a J1,.
"'I'= 1/ VI -(0.60)2 = 1.25. Use the equations from Table 20-2.
At = (1.25)[(4.0xI0-6s)
-(0.60)(3.0xI03m)/(3.00xI08m/s)]
= -2.5xI0-6s.
~
We can choose our coordinate system so that u is directed along the x axis without
loss of generality. Then, according to Table 20-2,
~x'
=
')'(~x -u~t),
~y'
=
~y,
~z'
=
~z,
c~t'
=
')'(c~t -u~x/c).
any
Square these expressions,
(~X')2
=
'"Y2(~X -u~t)2
(~y')2
=
(~y)2,
(~Z')2
C2(~t')2
= '"Y2((~X)2 -2u(~x)(~t)
-(~Z)2,
= '"Y2(C~t-U~X/C)2
+ (~t)2) ,
= '"Y2(C2(~t)2 -2u(~t)(~x)
+ U2(~X)2/C2) .
We'll add the first three equations and then subtract the fourth. The left hand side is the equal to
(~X')2 + (~y')2 + (~Z')2 -C2(~t')2,
while the right hand side will equal
'"Y2((~X)2 + U2(~t)2 -C2(~t)2
-U2/C2(~X)2)
+ (~y)2 + (~Z)2,
which can be rearranged as
'"Y2(1'"Y2(1-
u2/c2) (~X)2 + '"Y2( U2 -c2)
(~t)2 + (~y)2 + (~Z)2,
U2/C2) (~X)2 + (~y)2 + (~Z)2 -C2'"Y2 (1-
But
u2/c2) (~t)2.
1
"(2= 1 -U2/C2
,
so the previous expression will simplify to
(~X)2 + (~y)2 + (~Z)2 -C2(~t)2.
P20-6 (a) Va:= [(0.780c)+ (0.240c)]![1+ (0.240)(0.780)]
= 0.859c.
(b) Va:= [(0) + (0.240c)]![1+ (0)] = 0.240c,while
Vy= (0.780c)vi -(0.240)2fl1 + (0)] = 0.757c.
Then v = V(0.240c)2+ (0.757c)2= 0.794c.
(b) v~ = [(0) -(0.240c)]![1 + (0)] = -0.240c, while
v~ = (0.780c)Vi~ (0.240)2![1+ (O)]=0.757c.
Then v' = V( -0.240c)2+ (0.757c)2= 0.794c.
260
~
If we look back at the boost equation we might notice that it looks very similar to the
rule for the tangent of the sum of two angles. It is exactly the same as the rule for the hyperbolic
tangent,
tanh(~l
+~2)
tanh al + tanh a2
=
1
h
+
tan
h
~1
tan
.
~2
This means that each boost of /3 = 0.5 is the same as a "hyperbolic" rotation of ~r where tanh ~r =
0.5. We need only add these rotations together until we get to ~f, where tanh~f = 0.999.
~f = 3.800, and ~R = 0.5493. We can fit (3.800)/(0.5493) = 6.92 boosts, but we need an integral
number, so there are seven boosts required. The final speed after these seven boosts will be 0.9991c.
P20-8
(a) Ir ~x'
= O, then ~x
= u~t,
or
u = (730 m)/(4.96x
(b) '"1= 1/v'1At'
P20-9
10-68) = 1.472x 108m/8 = 0.491c.
(0.491)2 = 1.148,
= (1.148)¡(4.96
x 10-68) -(0.491)
(730 m)/(3
x 108)] = 4.32 x 10-68.
Since the maximum value for u is c, then the minimum ~t is
At ?: (730m)/(3.00xlO8m/s)
= 2.43xlO-6s.
P20-10
(a) Yes.
(b) The speed will be very close to the speed of light, consequently"1 ~ (23,000)/(30)
Then
v = VI -1/,2
~ 1 -1/2,2
= 766.7.
= 1 -1/2(766.7)2 = O.99999915c.
P20-11 (a) 6.t' = (5.00¡¡,s)Vl- (0.6)2 = 4.00¡¡,s.
(b) Note: it takes time for the reading on the S' clock to be seen by the S clock. In this case,
6.tl + 6.t2 = 5.00¡¡,s,where 6.tl = x/u and 6.t2 = x/c. Solving for 6.tl,
and
~t~
= {3.125
.us)V1
-{0.6)2
= 2.50.us.
P20-12
The only change in the components of 6.r occur parallel to the boost. Then we can choose
the boost to be parallel to 6.r and then
Ar'
='Y[Ar
-u(O)]
=
'YAr
?
Ar,
since "Y?; 1.
~
(a) Start
with
Eq. 20-34,
E2
= (pC)2 + (mc2)2,
and substitute into this E = K + mc2 ,
K2
+ 2Kmc2
+ (mc2)2
261
=
(pc)2
+ (mc2)2.
We can rearrange this, and then
(b) As v/c;;..,. O we have K-+
becomes
K2 + 2K mc2
-
m
=
~mv2 and P-+
m
(pC)2,
(pc) 2 -K2
2Kc2
mv, the classicallimits.
Then the above expression
m2v2c2 -~m2v4
"""
But v / c -+ O, so this expression reduces to m = m in the classicallimit,
(c) We get
which is a good thing.
P20-14
Since E » mc2 the particle is ultra-relativistic
and v ~ c. 'Y = (135)/(0.1396) = 967.
Then the particle has a lab-life of At'=
(967)(35.0 x 10-9s) = 3.385 x 10-5s. The distance traveled
is
x = (3.00 x 108m/s)(3.385
x 10-58)
= 1.016x
104m,
so the pion decays 110 km above the Earth.
~
(a) A completely inelastic collision means the two particles, each of mass ml, stick
together after the collision, in e(fectbecoming a new particle of mass m2. We'll use the subscript 1
for moving particle of mass ml, the subscript O for the particle which is originally at rest, and the
subscript 2 for the new particle after the collision. We need to conserve momentum,
Pl +Po
"/lmlUl
+ (O)
=
P2,
=
'Y2m2U2,
and we need to conservetotal energy,
El
7lmlc
2
+
+mlc
Eo
2
-
E2,
=
'"Y2m2C
2
,
Divide the momentum equation by the energy equation and then
'YIUl
71
+ 1 = U2.
But Ul = cyl
-l/'Y~
, SO
U2
262
(b) Using the momentum equatio~,
m2
P20-16
(a) K = W = JFdx = J(dp/dt)dx = J(dx/dt)dp = Jvdp.
(b) dp = m'Ydv + mv(d'Y/dv)dv. Now use Maple or Mathematica to save time, and get
dp = (1-
mdv
mv2 dv
v2/c2)1/2 + c2(1 -v2/c2)3/2
.
Now integrate:
K
mv2
=
dv,
c2(1-
V2/C2)3/2
~.
P20-17 (a) Since E= K + mc2, then
Enew = 2E = 2mc2 + 2K = 2mc2(1 + K/mc2).
(b) Enew = 2(0.938 GeV) + 2(100 GeV) = 202 GeV.
(c) K = (100 GeV)/2 -(0.938 GeV) = 49.1 GeV.
P20-18
(a) Assume only one particle is formed. That particle can later decay, but it sets the
standard on energy and momentum conservation. The momentum of this one particle must equal
that of the incident proton, or
p2C2 = [(mc2 + K)2 -m2c4].
The initial
energy was K + 2mc2, so the mass of the "one"
M2c4 = [(K + 2mc2)2 -p2C2]
particle
is given by
= 2K mc2 + 4m2c4.
This is a measureof the available energyj the remaining energy is required to conservemomentum.
Then
Enew = VM2"¡;4 = 2mc2vl
263
+ K/2mc2.
P20-19
The initial momentum is m'"Yivio The final momentum is (M -m)'"Yfvfo Manipulating
momentum conservation equation,
=
m'YiVi
1
=
~
M-m
=
-;;¡;y;¡¡;M-m
1
+
=
m'Yi/3i
264
the
~
(a) We'll assumethat the new temperature scale is related to the Celsius scale by a linear
transformation; then Ts = mTc + b, where m and b are constants to be determined, Ts is the
temperature measurement in the "new" scale, and Tc is the temperature measurement in Celsius
degrees.
One of our known points is absolute zero;
We have two other
points,
Ts
=
mTc
(O)
=
m( -273.15°C)
the melting
+b,
and boiling
{Ts )bp
{Ts)mp
=
=
+ b.
points
for water ,
m(100°C)+ b,
m(O°C)+ b;
we can subtract the top equation from the bottom equation to get
(TS)bp
-(Ts)Smp
=
100
Com.
We are told this is 180 SO,so m = 1.8 SO/Co. Put this into the first equation and then find b,
b = 273.15°Cm = 491.67°S. The conversion is then
Ts = (1.8 SO/CO)Tc+
(491.67°S).
(b) The melting point for water is 491.67°S; the boiling point for water is 180 SO above this, or
671.67°S.
E21-2
TF = 9( -273.15 deg O)/5 + 32°F = -459.67°F .
E21-3 (a) We'1l assumethat the new temperature scale is related to the Celsius scale by a linear
transformationj then Ts = mTc + b, where m and b are constants to be determined, Ts is the
temperature measurement in the "new" scale, and Tc is the temperature measurement in Celsius
degrees.
One of our known points is absolute zero;
We have two other
points,
Ts
=
mTc
(O)
=
m( -273.15°C)
the melting
(Ts
(Ts
)hp
)mp
+b,
and boiling
=
=
+ b.
points
m(100°C)
+
for water ,
b,
m(Ooq)+,b;
we can subtract the top equation from the bottom equation to get
(TS)bp -(T8)Smp = 100 Com.
We are told this is 100 Qo, so m = 1.0 Qo/Co. Put this into the first equation and then find b,
b = 273.15°C = 273.15°Q. The conversion is then
Ts
= Tc
+ (273.15°8).
(b) The melting point for water is 273.15°Q; the boiling point for water is 100 Qo above this, or
373.15°Q.
(c) Kelvin Scale.
265
E21-4
(a) T = (9/5)(6000K -273.15) + 32 = 10000°F.
(b) T = (5/9)(98.6°F -32) = 37.0°C.
(c) T= (5/9)(-70°F
-32) = -57°C.
(d) T = (9/5)( -183°C) + 32 = -297°F.
(e) It depends on what you think is hot. My mom thinks
(5/9)(79°F -32) = 26°C.
E21-5
T = (9/5)(310 K -273.15)
79°F is too warm; that's
T =
+ 32 = 98.3°F, which is fine.
E21-6
(a) T = 2(5/9)(T -32), so -T/10 = -32, or T = 320°F.
(b) 2T = (5/9)(T -32), so 13T/5= -32, orT = -12.3°F,
~
If the temperature (in Kelvin) is directly proportional to the resistance then T = kR,
where k is a constant of proportionality.
We are given one point, T = 273.16 K when R = 90.35 n,
but that is okay; we only have one unknown, k. Then (273.16 K) = k(90.35 n) or k = 3.023 K/n.
If the resistance is measured to be R = 96.28 n, we have a temperature of
T = kR = {3.023 K/0){96.28
E21-8
n) = 291.1 K.
T = (510°C)/(0.028 Y)V , 80 T = (1.82 x 104°C/y)(0.0102
Y) = 186°C.
~
We must first find the equation which relates gain to temperature, and then find the gain
.at the specified temperature. If we let G be the gain we can write this linear relationship as
G=mT+b,
where m and b are constants to be determined. We have two known points:
(30.0)
(35.2)
=
m(20.0°
O) + b"
=
m(55.0°
O) + b.
Ir we subtract the top equation from the bottom we get 5.2 = m(35.0° C), or m = 1.49C-1. Put
this into either or the first two equations and
(30.0) = (0.149C-1)(20.0°
C) + b,
which has a solution b = 27.0
Now to find the gain when T = 28.0 °C:
G = mT + b = (0.149C-1)(28.0°
E21-10
P/Ptr = (373.15 K)/(273.16K)
C) + (27.0) = 31.2
= 1.366.
E21-11 100 cm ~is
1000 torr. PHe = (100 cm Hg)(373K)/(273.16K) = 136.550cm Hg. Nitrogen records a temperature whic};l is 0.2K higher, so PN = (100 cm Hg)(373.2K)/(273.16K) =
136.623cm Hg. The difference is 0.073 cm Hg.
E21-12
AL
E21-13
AL = (3.2xlO-6/Co)(20Ü
= (23 x 10-6/CO)(33
m)(15CO)
=
1.1 x 10-2m~
in)(60CO) = 3.8xlü-2in.
266
E21-14
L'
=((2.725cm)[l
+
(23xlü-6¡Co)(l28CO)]=
2.733
c
~
We want to focus on the temper'8.ture change, not the absolute temperature.
case, ~T = Tf -Ti = (42°C) -( -5.0°C) = 47Co.
Then
~L
E21-16
= (11 x 10-6 C-1 )(12.0 m)(47 C6;) = 6.2 x 10-3 m.
~A = 2aA~T,
~A
In this
so
=
2(9xlO-6/Co){2.0m)(3.0m)(30CO)
=
3.2x
lO-3m2.
~
(a) We'll apply Eq. 21-10. The surface area of a cube is six times the area of one face,
which is the edge length squared. So A = 6(0.332m)2 = 0.661m2. The temperature change is
~T = (75.0°C) -(20.0°C) = 55.0Co. Then t~e increase in surface area is
~A = 2aA~T = 2(19 x 10-6C-1)(0.661 m2)(55.0 CO) = 1.38 x 10-3 m2
(b) We'll now apply Eq. 21-11. The volume of the cube is the edge length cubed, so
V = (0.332m)3 = 0.0366m3.
and then from Eq. 21-11
A V = 2aV AT
= 3(19 x 10-6 C-1 )(0.0366 m3)(55.0CO)
= 1.15 x 10-4 m3,
is the change in volume of the cube,
E21-18
V' = V(l + 3a~T),
v'
so
(530 cm3)[1 + 3(29x
E21-19
(a) The slope is approximately
(b) The slope i~ zero.
E21-20
10-6/CO)(-172CO)]
= 522 cm: 3
,6 x 10-4/CO,
Ar = ({3/3)rAT, so
~r
(3.2x 10-5 /K)/3](6.37x
106m)(2700
K) = 1.8 x 105m.
~
We'll assumethat the steel ruler measureslength correctly at room temperature. Then
the 20.05 cm measurementof the rod is correct. But both the rod and the ruler will expand in the
oven, so the 20.11 cm measuremeptof the rod is not the actuallength of the rod in the oven. What
is the actuallength of the rod in the oven? We can only answer that after figuring out how the 20.11
cm mark on the ruler moveswhen the ruler expands.
Let L = 20.11 cm correspond to the ruler mark at room temperature. Then
AL = asteelLAT = (11 X lO-6C-1)(20.11 cm)(250CO) = 5.5 x 10-2 cm
is the shift in position of the mark as the ruler is raised to the higher temperature. Then the change
in length of the rod is not (20.11 cm) -(20.05 cm) ~ 0.06 cm, becausethe 20.11 cm mark is shifted
out. We need to add 0.055 cm to this; the rod chánged length by 0.115 cm.
The coefficient of thermal expansion for the rod is
AL
a = ~
L~T
=-
(0.115 cm)
.",
"
(20.05 cm)(250CO )
267
= 23 x 10-6 C-1,
E21-22
A = ab, A' = (a+
~a)(b+
6.A
~b) = ab+
=
a~b + b~a + ~a~b,
so
a~b + b~a + ~a~b,
A(~b/b
~
+ ~a/a
+ ~a~b/ab),
A(a~T+a~T),
2aA~T.
~
Solve this problem by assuming the solid is in the form of a cube.
If the length of one side of a cube is originally Lo, then the volume is originally Vo = L3. After
heating, the volume of the cube will be V = L3, where L = Lo + ~L.
Then
=
v
L3
,
(Lo
+
~L)3,
(Lo
+
aLo~T)3,
Lg(l
+
a~T)3.
As long as the quantity a~T is much less than one we can expand the last line in a binomial
expansion as
V ~ Vo(l
so,the
change
in volume
is ~V
+ 3a~T
+
),
~ 3aVo~T.
E21-24
{a) ~A/A = 2{0.18%) = {0.36%).
(b) ~L/L = 0.18%.
(c) ~V/V = 3{0.18%) = {0.54%).
{d) Zero.
{e) a = (0.0018)/{100CO) = 1.8x10-5/Co.
E21-25
p' -p
= m/V'-
m/V = m/(V
+ AV) -m/V
6.p = -(m/V)(6.V/V)
~ -mAV/V2.
Then
= -p.8AT.
E21-26
Use the results oí Exercise 21-25.
(a) AV/V = 3AL/L = 3(0.092%) = 0.276%. The change in density is
Llp/p
(b) a = /3/3=
~
(0.28%)/3(40CO)
= -LlV/V
=
-(0.276%)
= 2.3xlO-5/Co.
-0.28
Must be aluminum.
The diameter oí the rod as a function oí temperature
ds = ds,Q(l + QsLlT),
The diameter of the ring as a function of temperature
db =$: db,Q(l
is
+ ab~T).
268
is
~
We are interested in the temperature when the diameters are equal,
=
=
ds,o(l + asAT)
asds,oAT
-abdb,oAT
db,O(l+ ab~T),
db,O-ds,o,
db,O-ds,o
~T
asds,o -abdb,O
AT =
~~oooc~92-cm1
{2.992 cm) -{3.000
cm)
,
335 co .
The final temperature
E21-28
is then Tí = (25°) + 335 Co = 360°,
(a) ~L = ~L1 + ~L2 = (L1al + L2a2)~T.
~L
L~T
o:=
(b) Since L2 = L -L1
QlLl
+
--
alL1
The effective value for a is then
+ a2L2
we can write
Q2(L
=
-Ll)
aL
,
Ll
=
L~al
-a2
,
{0.524 m) {13 x 10-6) -{11
{19 x 10-6) -{11
x 10-6)
x 10-6) = 0.131 m.
The brass length is then 13.1 cm and the steel is 39.3 cm.
E21-29
At lOO°C the glass and mercury each have a volume Va. After cooling, the difference in
volume changes is given by
~ v = Vo(3ag -,8m)~T.
Since m = pV, the mass of mercury that needs to be added can be found by multiplying
the density of mercury. Then
~m = (0.89lkg)[3(9.0xlO-6/CO)
This is the additional
-(l.8xlO-4/co)](-l35CO)
though by
= O.Ol84kg.
amount required, so the total is now 909 g.
E21-30
(a) The rotational inertia is given by I = J r2dm; changing the temperature
r -r'
= r + ~r = r(l + a~T). Then
so ~I
(b)
= 2aI~T.
Since
L = IúJ, then
~w
O = úJ~I
~ -2(19x
+,I~úJ.
Rearranging,
10-6/co)(230
~úJ/úJ
rev/s)(170CO)
"'
269
= -~I/I
=
~ -1.5
rev/s.
-2a~T.
Then
requires
~
This problem is related to objects which expand when heated, but we never actually need
to calculate any temperature changes. We will, however, be interested in the change in rotational
inertia. Rotational inertia is directly proportional to the squareof the (appropriate ) linear dimension,
so
If/li = (rf/rJ2,
(a) If the bearings are frictionless then there are no external torques, so the angular momentum
is constant.
(b) I[the angular momentum is constant, then
Li
IiúJi
=
Lf,
=
If"'fo
We are interested in the percent change in the angular velocity, which is
(Llf
-(Lli
(Lli
=
I.
=~-1=
If
iAJf
iAJj
/
, 2
(~)
1
' 2
-1= (~)-
-1 = -0.36%.
(c) The rotational kinetic energy is proportional to Iw2 = (Iw)w = Lw, but L is constant, so
E21-32 (a) The period of a physical pendulum is given by Eq. 17-28. There are two variables
in the equation that depend on length. I, which is proportional to a length squared, and d, which
is proportional to a length. This means that the period have an overall dependence on length
proportional to vr .Taking the derivative,
1P
~ dP = --dr
2 r
~P
(b) ~P/P
= (O. 7x 10-6Co)(10CO)/2
At
E21-33
= {30
1
~ -Pa~T.
2
= 3.5 x 10-6.
After
x 24 x 60 x 60 s)(3.5
30 days the clock will
x 10-6)
be slow by
= 9.07 s.
Refer to the Exercise 21-32.
~p
=
(36008){l9x
lO-6Co)(
-20CO)/2
= 0.688.
E21-34
At 22°C the aluminum cup and glycerin each have a volume Va. After heating, the
difference in volume changes is given by
~
V
=
Vo(3aa
-.Bg)~T.
The amount that spills out is then
~v
= (110 cm3)[3(23x10-6/CO)
-(5.1
x 10-4/CO)](6CO)
= -0.29
cm3.
E21-35
At 20.0°C the gla.ss tube is filled with liquid to a volume Va. After heating, the difference
in volume changes is given by
~~V = Vo(3ag -/31)~T.
The cross sectional area oí the tube changes according to
~A = Ao2ag~T.
270
Consequently,the height of the liquid changesaccording to
~v
~V/VO
=
(ha + ~h)(Aa
~
ha~A
=
~A/Aa
+ ~A)
-haA,
+ Aa~h,
+ ~h/ha.
Then
~h
E21-36
=
(a) 13 =
(1.28m/2)[(1.1
(4.2xlÜ-5/CO)](l3CO)
x 10-5/co)
(dV/dT)/V.
IfpV
= nRT,
then
.8 = (nR/p)/V
(b) Kelvins.
(c) /3 ~ 1/(300/K)
pdV
= nRdT,
= 2.6xlü-4m.
so
= nR/pV =l/T.
= 3.3x 10-3/K.
E21-37
(a) V = (1 mol)(8.31 J/mol. K)(273 K)/(1.01 x 105Pa) = 2.25 x 10-2m3.
(b) (6.02x1023mol-l)/(2.25x104/cm3)
= 2.68~1019.
E21-38
n/V = p/kT,
n/V
~
(a)
Using
so
= (1.01 X 10-13Pa)/(1.38X
Eq.
10-23J/K)(295
K) = 25 part/cm;
3
21-17,
n
pV
RT
(b) Use the same expression again,
v=
nRT
E21-40
(a) n =pV/RT
(b) Tí = TiPíVí/PiVi,
E21-41
p
= (l.OlxlO5pa)(l.l3xlO-3m3)/(8.3lJ/mol.K)(3l5K)
so
Pi = (14.7+ 24.2) lb/in2 = 38.9 lb/in2.
Pf
=
Pf =piTfVi/TiVf,
(38.9 lb/in2)(299K)(988
in3)
;
=41.71b/m
= 4.36xlO-2mol.
SO
.2
.
(270 K)(1020 inv)
The gauge pressure is then (41.7 -' 14.7) lb/in2 = 27.0 lb/in2.
E21-42
Sínce p = F/A and F = mg, a reasonable estímate for the mass of the atmosphere ís
m = pA/g = {1.01x 105Pa)47r{6.37x106m)2/{9.81m/s2) = 5.25x1018kg.
271
E21-43
P = Po + pgh, where h is the depth. Then Pf = 1.01 x 105Pa and
Pi = {1.01 x 105Pa) + {998kg/m3){9.81
V f = ViPiTf/PfTi,
E21-44
m/s2){41.5m)
= 5.07x 105Pa.
so
The new pressure in the pipe is
Pf =PiVi/Vf=
(1.01x105Pa)(2)
= 2.02x105Pa.
The water pressure at some depth y is given by P = Po + pgy, SO
1J= .(2.02 x 105Pa) -ll.01
Then
the
water/air
interface
inside
(998
the
tube
kg/m3)(9.81
is at
a
x 105Pa) = 1n .':tm
depth
m/s2)
of
10.3
m;
~~.~ so
h
=
(10.3m)
+
(25.0m)/2
=
22.8m.
~
(a) The dimensions of A must be [time]-l, as can be seen with a quick inspection of the
equation. We would expect that A would depend on the surface area at the very least; however,
that means that it must also depend on some other factor to fix the dimensionality of A.
(b) Rearrange and integrate,
¡:o'
ln(~T/,
P21-2
d~T
~T
=
-1tAdt,
~To)
~T
-At
-
,
~Toe-At.
First find A.
Then find time to new temperature
difference.
t = ln(ó.To/ó.T)
t
This happens
97.8 -45
= 53 minutes
= ln[(29CO)/(~12°)]
(3.30 x 10-3/min)
= 97.8min
later .
P21-3
If we neglect the expansion of the tube then we can assuIhe the cross sectional area of the
tube is constant. Since V = Ah, we can assume that ~ V = A~h. Then since ~ V = ,BVo~T, we
can write ~h = ,Bho~T .
P21-4 For either container we can write piVi = niRTi. We are told that Vi and ni are constants.
Then 6.p = AT1 -BT2, where A and B are constants. When T1 = T2 6.p = O, so A = B. When
T1 = Ttr and T2 = Tb ~ have
(120
so A
1.202 m m Hg/K.
m m
Hg)=
A(373K
-273.16K),
Then
T = (90 m m Hg) + (1.202 m m Hg/K)(273.16~
= 348K.
(1.202 mm Hg/K)
Actually, we could have assumedA was negative, and then the answer would be 198K.
272
~
~
Start
with
a differential
form
for Eq. 21-8, dL/dT
{L dL
=
= aLa,
rearrange,
and integrate:
{T aLa dT,
JLo
JTo
T
=
L-Lo
Lo
f
adT,
JTO
L
P21-6
~L
=
Q;L~T,
Lo
so
1
-aL
~T
-~
~t
~
=
= (23 XlO-6
(96xlO-9m/8)
iCO)(L8 x lO-2m) = 0.23°C/8.
AL
At
(a) Consider the work that was done for Ex. 21-27. The length of rod a is
La = La,o(l
+ Qa6.T),
Lb = Lb,o(l
+ ab6.T).
while the length of rod b is
The difference is
La,o(l
La -Lb
La,o
which
will
be
a
constant
is
La
OQa
=
Lb
OQb
,
+ aa~T)
La,o -Lb,O
+ ab~T),
-Lb,Oab)~T,
or
,
Li,o
(b) We w~t
-Lb,o(l
~ Lb,O + (La,oaa
cx: l/aio
= 0.30 m so
k/aa
-k/ab
= 0.30 m,
where k is a constant of proportionality;
k = {0.30 m)/ (l/{llxlO-6/CO)
-l/{l9xlO-6/CO))
= 7.84xlO-6m/,Co.
The two lengths are
La = (7.84 x 10-6m/Coj/(11
x 10-6 jCO)=
0.713m
for steel and
Lb = (7.84xl0-6m/CO)/(19xl0-6/CO)
= O.413m
for brass.
P21-8 The fractional increasein length ofthe bar is ~L/Lo = a~T. The right triangle on the left
has base Lo/2, height x, and hypotenuse(Lo + ~L)/2. Then
x =
~V(LO
+ ~L)2-
With numbers,
273
L~ =
P21-9
We want to evaluate V = Vo(l + J /3 dT); the integral is the area under the graph; the
graph looks like a triangle, so the result is
V = Vo[l + (16CO)(O.OOO2/CO)/2]
= (1.0016)Vo.
The density is then
p = PO(Vo/V)
P21-10
= (1000kg/m3)/(1.0016)
= O.9984kg/m3.
At 0.00°C the glassbulb is filled with mercury to a volume Vo. After heating, the difference
in volume changesis given by
~V = Vo((3-3a)~T.
Since To = 0.0°C, then ~T = T, if it is measured in °C. The amount of mercury in the capillary is
~ V, and since the cross sectional area is fixed at A, then the length is L = ~ V / A, or
V
L = A((3 -3a)~T.
P21-11
Let a, b, and c correspond to aluminum, steel, and invar, respectively. Then
cosc =
a2 + b2 -¿
2a b
,-:
We can replace a with ao(l + aa6.T), and write similar expressions for b and c. Since ao = bo = co,
this can be simplified to
cosC =
Expand
this as a Taylor
(1 + aa~T)2 + (1 + ab~T}2 -:fJ + ac~
2(1 + aaAT)(l
series in terms
of AT,
+ ab~.l.)
and we find
1
1
cosC ~ 2 + 2 (aa + ab -2ac)
AT.
Now solve:
~T
2 cos(~9=-95°) -1
+ (11 x 10-6 /CO)--2
= (23 x-10-6JCO)
The final temperature
(O~7x 10-6/Co)
= 46.4°C.
is then 66.4°C.
P21-12 The bottom of the iron bar moves downward according to AL = aLAT. The center of
mass of the iron bar is located in the center; it moves downward half the distance. The mercury
expands in the glass upwards; subtracting off the distance the iron moves we get
Ah = {3hAT -AL
= ({3h-aL)AT.
The center of mass in the mercury is located in the center. If the center of mass of the system is to
remain constant we require
miAL/2 = mm(Ah -AL)/2;
or, since p = m V = mAy,
pjaL
= Pm({3h -2aL).
Solving for h,
h =
.~
c
~
(13.6
x 103kg/m3)(18
274
~
The volume of the block which is beneath the surface of the mercury displaces a mass
of mercury equal to the mass of the block. The mass of the block is independent of the temperature
but the volume of the displaced mercury changesaccording to
v m =
v m,o(l
+.BmAT).
This volume is equ8:1to the depth which the block sinks times the cross sectional area of the block
(which doeschangewith temperature) .Then
hshb2
= hs,Qhb,Q2(1
+ {3m~T),
where hs is the rlepth to which the block sinks anrl hb,Q= 20 cm is the length of the sirle of the
block. But
hb = hb,o(l
+ abAT),
so
l+f3m~T
hs = hs,o-(1
+ ab~T)2 .
Since the changes are small we can expand the right hand side using the binomial expansion; keeping
terms only in ~T we get
hs ~ hs,o(l + ({:Jm-2ab)~T),
which means the block will sin k a distance hs -hs,o given by
hs,o(/3m
-2ab)AT
= hs,o [ (1.8 x 10-4/CO)
-2(23
x 10-6/CO)J
(50 CO) = (6.7 x 10-3)hs,o.
In order to finish we need to know how much of the block was submerged in the first place. Since
the fraction submerged is equal to the ratio of the densities, we have
h8,0/hb,O
= Pb/Pm
= (2.7xl03kg/m3)/(1.36xl04kg/m3),
so hs,Q= 3.97 cm, and the change in depth is 0.27 mm.
P21-14
The area of glass expands according to ~Ag = 2agAg~T.
according to
~Ac
+ ~Ai
= 2(acAc
The are of Dumet wire expands
+ aiAi)~T.
We need these to be equal, so
QgAg
=
2
Qgrg
(
Qg
rc
~
2)
2
+
ri
:i
QcAc
-
-
Qc
2
(
2
rc
-Qc
2
r~2
=
+ QiAi,
(
2
)
-ri
rc
2
-ri
2
+
Qiri
)
,
2
+
Qiri
,
Qc -Qg
Qc -Qi
P21-15
P21-16
V2 = V1(Pl/P2)(T1/T2),
V2 = (3.47m3)[(76
SO
cm Hg)/(36
cm Hg)][(225K)/(295K)]
275
= 5;59m3.
~
Call the containers one and two so that Vi = 1.22 L and V2 = 3.18 L. Then the initial
number oí moles in the two containers are
nl,i
The
total
PiVl
R T 1.and
=
n2i
,
PiV2
=
~.
is
n ===
pi(V1 + V2)/(RTi).
Later the temperatures are changed and then the number of moles of gas in each container is
nl,f
~
RT1,f
=
and n2,f = ~.
RT2,f
The total is still n, so
+-
~(~
.T1,f
V2
pi(Vl
=
+ V2)
RT;
T2,f)
We can solve this for the final pressure, so long as we remember to convert all temperatures
Kelvins,
pi(VI
+ V2)
Pf=
(
Ti
VI
V2
-+T1,f
T2,f
to
)
or
=
74 atm.
P21-18 Originally nA = PAVA/RTA and nB = PBVB/RTB; VB = 4VA. Label the final state of
A as C and the final state of B as D. Aftermixing, nc = pcVA/RTA and nD = PDVB/RTB, but
Pc = PD and nA +nB = nc +nD. Then
PA/TA
+ 4PB/TB
= Pc (l/TA
+ 4/TB),
or
.(5x 105Pa)/(300K) + 4(1 x 105Pa)/(400K)
Pc=
=
1/(300K)
+ 4/(400
2.00
x 105Pa.
K)
P21-19 If the temperature is uniform then all that is necessaryis to substitute Po = nRT /V and
p = nRT /V; cancel RT from both sides, and then equate n/V with nv.
P21-20 Use the results of Problem 15-19. The initial pressureinside the bubble is Pi = Po+4'Ylri.
The final pressure inside the bell jar is zero, SOPf= 4'Ylrf. The initial and final pressure inside the
bubble are related by piri3 = Pfrf3 = 4'Yrf2. Now for numbers:
pj = (1.01
x 105Pa)
+ 4(2.5
x 10-2N/m)/(2.0
x 10-3m)
=
1.0105
and
I (1.0105
rf
=
~VI
x 105Pa)(2.0
4(2.5x
x 10-3m)3
lO=2N/m)
P21-21
P21-22
276
= 8.99x
lO-2m.
x 105Pa.
E22-1
(a) n = (2.56 g)/(l97 g/mol) = l.30x lO-2mol.
(b) N = (6.02xlO23mol-l)(l.30xlO-2mol)
= 7.83xlO21,
10-23J/K)(293
E22-2
I(a) N = pV/kT = (1.01 x 105pa)(1.00m3)/(1.38x
=41.5 mol. Then
(b) n = (2.50x1025)/(6.02~1023mol-l)
m = (41.5 mol)[75%(28
~
g/mol)
+ 25%(32
g/mol)]
K) = 2.50x
1025.
= 1.20kg.
(a) We first need to calculate the molar mass oí ammonia. This is
M = M{N)
+ 3M{H)
= {14.0 g/mol)
+ 3{1.01 g/mol)
= 17.0 g/mol
The number oí moles oí nitrogen present is
n = m/Mr
= (315 g)/(17.0 g/mol) = 18.5 mol.
The volume of the tan k is
v = nRT/p = (18.5 mol)(8.31 J/mol. K)(350K)/(1.35 x lO6Pa) = 3.99x lO-2m3.
(b) After the tank is checkedthe number of moles of g8$ in the tank is
n = pV/(RT) = (8.68x lO5Pa)(3.99x lO-2m3)/[(8.31 J/mol. K)(295K)] = 14.1 mol.
In that case,4.4 mol must have escaped;that correspondsto a mass of
m = nMr
E22-4
=
(4.4
(a) The volume per particle is v/N
mol)(17.0
g/mol)
= 74.8
g.
= kT/P,so
V/N = (1.38x 10-23 J/K)(285K)/(1.01
x 105Pa) = 3.89x 10-26m3.
The edge length is the cube root of this, or 3.39x 10-9m. The ratio is 11.3.
(b) The volume per particle is V/NA, so
V/NA = (18x10-6m3)/(6.02x
1023) = 2.99x10-29m3.
The edge length ls the cube root of this, or 3.10 x 10-1om. The ratio is 1.03.
E22-5
The volume per particle is V / N = kT / p , so
V/N = (1.38x 10~23J/K)(308K)/(1.22)(1.01
x 105Pa) = 3.45x 10-26m3.
The fraction actually occupied by the particle is
47r(0.710 x 10-10m)3/~
= 4.34 x 10-5.
(3.45 x lO-26m3 )
E22-6
The component of the momentum normal to the wall is
Py
{3.3 x lO-27kg){l;OX
lO3mis) cos{55°) = 1.89 x lO-24kg .mis.
The pressureexerted on the wall is
FP=A-
.(1.6 x 1023/8)2(1.89 x 10-24kg ~~18)
(2.0 x 10-4m2)
277
= 3.0 x 103Pa.
~
(a)
From
Eq.
22-9,
Vrms
=
f!
Then
p =
1.23
x
10-3
atm
=
124
Pa
and
p =
1.32
x
10-5
lkg
g/cm3
100 cm
1m
)
iOOüg
3
=
1.32
x
10-2
kg/m3.
Finally,
(b) The molar density of the gas is just n/V;
law as
~ = p ~
(1240Pa)
v
:nT
but this can be found quickly from the ideal gas
-(8.31J/mol.K)(317K)
(C) We were given the density,
which
~ ~..~ " ~~
is mass per volume,
(1.32 x 10-2 kg/m3)
-..
p
~=
= J. 71 y ln-l
(4.71 x 10-1 mol/m3)
=
Tnnl/Tn3
...~.f.~
.
so we could find the molar
28.0
mass from
g/mol.
But what gas is it? It could contain any atom lighter than silicon; trial and error is the way to go.
Someof my guessesinclude C2H4 (ethene), CO (carbon monoxide), and N2. There's no way to tell
which is correct at this point, in fact, the gas could be a mixt~re of all three.
E22-8
The density is p = m/V
= nMr/V,
p = (0.350 mol)(0.0280
'I'he
rms
speed
or
kg/mol)/7r(0.125m/2)2(0.560m)
= 1.43kg/m3.
is
E22-9 {a) N/V = p/kT = {1.01 x 105pa)/{1.38x 10-23J/K){273K) = 2.68x 1025/m3.
{b) Note that Eq. 22-11 is wrong; for the explanation read the last two paragraphs in the first
column on page 502. We need an extra factor of .j2, SO7rJ2= V/ .j2N >., so
,
d=
E22-10
.c
1/ V21T(2.68 x 1025/m3)(285 x 10,-9m) = 1.72 x 10-1Om.
(a) >. = V¡V2N7rd2,
so
(b) Particles effectively follow ballistic trajectories.
278
~
We have v = f>., where >. is the wavelength (which we will set equal to the mean free
path), and v is the speed of sound. The mean free path is, from Eq. 22-13,
>.=~
v'27rd2p
so
f=
E22-12
V27rd2pv
=
V27r(315
x 10-12m)2(1.02
kT
~~X
1.01 x 105Pa)(343
(1.38x 10-23J/K)(291
mis)
= 3.88x lOgHz.
K)
(a) p = (1.10x 10-6 mm Hg)(133 Pa/mm Hg) = 1.46x 10-4Pa. The particle density is
N/V = (1.46x 10-4Pa)/(1.38x 10-23J/K)(295 K) = 3.59x 1016/m3.
(b) The mean free path is
), = 1/ .¡(2) (3.59x 1016/m3)7r(2.20x 10-10m)2 = 130m.
E22-13
Note that
vav cx: VT,
while
>..cx: T. Then the collision
rate is proportional
to l/VT.
Then
E22-14
(a) vav = (65 ~m/s)/(10) = 6.5 km/s.
(b) Vrms = V(505 km/s)/-(-iO) = 7.1 km/s.
~
(a)
The
average
is
The mean-square value is
~(200s)2 + 2(500m/s)2 + 4(600 m/s)2
4+2+4
= 2.1 x 105m2/s2.
The root-mean-square value is the square root of this, or 458 m/s.
(b) I'll be lazy. Nine particles are not moving, and the tenth has a speed of 10 m/s. Then the
averagespeed is 1 m/s, and the root-mean-square speed is 3.16 m/s. Look, Vrmsis larger than vav!
(c) Can vrms=vav? Assume that the speeds are not all the same. Transform to a frame of
referencewhere vav = 0, then some of the individual speedsmust be greater than zero, and some
will be less than zero. Squaring these speedswill result in positive, non-zero, numbers; the mean
square will necessarilybe greater than zero, so Vrms> 0.
Only if all of the particles have the same speed will Vrrns=Vav.
E22-16
Use Eq. 22-20:
E22-17
Use Eq. 22-20:
279
E22-18
Eq. 22-14 is in the form N = Av2e-Bv2.
dN
= 2Ave-
B
v
Taking the derivative,
2
B 2
v,
-2ABv3e-
dv
and setting this equal to zero,
V2 = l/B
~
We want
= 2kT/m.
to integrate
vav
=
vdv,
The easiestway to attack this is first with a change of variables- let x = mv2/2kT , then kT dx =
mvdv. The limits of integration don't change, since .¡-00 = 00. Then
vav
(2;A;T
m ) 3/2
=
47r
=
2 -;;;;:-
roo 2kT
Jo
-;;;;-xe-x
kT
~dx,
( 2kT ) 1/2 roo
Jo xe-Qxdx
The factor of a that was introduced in the last line is a Feynman trick; we'll set it equal to one when
we are finished, so it won't change the result.
Feynman's trick looks like
d (e ax
{,ea da- J eApplying
this to our original
vav
dx = J
8~e
-axdx
problem,
=
280
=
f
( -x)e-axdx.
E22-20
We want to integrate
2
(V)av
=
1 roo
(
NJo
Nv)v
=
-47r
=
47r -v2e-mv
3/2
m
1
¡
N
00
2 dv,
(
N
o
¡
3/2
2 /2kT
27rkT
00
8kT
-xe
~m
¡
dv,
2/2kT V2 dv.
o
The eagiest way to attack this is first with a change of variablesy'2)CT7mdx = dv. The limits of integration don't change. Then
=
V2
27rkT
)
(
)
m
-v2e-mv
(X)
4
let X2 = mv2 /2kT , then
-x2
dX
O
Look up the integral; although you can solve it by first applying a Feynman trick (see solution to
Exercise 22-21) and then squaring the integral and changing to polar coordinates. I looked it up. I
found 3~ /8, so
E22-21
Apply Eq. 22-20:
Vrms = V3(1.38x
E22-22
Since
Vrms (X JT¡m,
10-23J/K)(287K)/(5:2x
10-17kg) = 1.5x lü-2m/s.
we have
T=
{299 K) {4/2)
= 598 K,
or 325°C.
E22-23
(a) The escape speed is found on page 310; v = 11.2 x 103m/s. For hydrogen,
T = (2)(1.67x 10-27kg)(11.2 x 103m/s)2 /3(1.38x 10-23J/K)
= 1.0x 104K.
For oxygen,
T = (32)(l.67x
lO-27kg)(ll.2x
lO3m/s)2 /3(l.38x
lO-23J/K)
= l.6x
lO5K.
(b) The escape speed is found on page 310; v = 2.38 x 103m/s. Por hydrogen,
T = (2)(1.67x 10-27kg)(2.38 x 103m/s)2 /3(1.38x 10-23J/K)
= 460K.
For oxygen,
T = (32)(1.67x 10-27kg)(2.38x
103m/s)2 /3(1.38x 10-23 J/K) = 7300K.
(c) There should be more oxygen than hydrogen.
E22-24
(a) vav = (70 km/s)/(22)
= 3.18 km/s.
(b) Vrms = V (250 km2/s2)/(22)
= 3.37 km/s.
(c) 3.0 km/s.
281
~
According
to the equation
directly
beneath
(¡) = v4>/L = (212 m/s}(0.0841
E22-26
Fig. 22-8,
rad}/(0.204m}
87.3 rad/s.
If Vp = Vrmsthen 2T2 = 3T1, or T2/Tl = 3/2.
E22-27
Read the last paragraph on the first column of page 505. The distribution
proportional to
2
/
B
v3e-mv
taking
the derivative
2kT
dN / dv and setting
dN
=
of speeds is
..
v3e-
v,
equal to zero yields
=
B v ..
3v2e-
4
-2Bv
B
e-
v,
..
dv
and setting this equal to zero,
V2 = 3/2B
= 3kT /m.
E22-28
(a) v = V3(8.31 J/mol. K)(4220K)/(0.07261
kg/mol)
(b) Half of the sum of the diameters, or 273 pm.
(c ) The mean free path of the germanium in the argon is
,\ = 1/V2(4.13x
= 1200m/s.
1025/m3)7r(273x 10-12m)2 = 7.31 x 10-8m.
The collision rate is
(l200m/s)/(7.3lxlO-8m)
~
The fraction
Change variables
of particles
according
that
~
~
Vii
(kT)3/2
= l.64xlOlO/s.
interests
¡
us is
0.03kT E1/2e-E/kT
0.01kT
to E / kT = x, so that
dE = kT dx. The integral
¡
-x1/2e-x
0.03
2
Vii
is then
dx.
0.01
Since the value of x is so small compa.red to 1 throughout
according to
e-x
dE.
~
1-
x for
the ra.nge of integration,
x «
we ca.n expa.nd
1.
The integral then simplifies to
2
~
J7r"
E22-30
Write
N(E)
¡
0.03
x1/2(1
-x)
dx =
0.01
= N(Ep
[
2
2
--x3/2
J7r"
3
2
5/2
]
0.03
= 3.09x 10-¡
-5x
0.01
+ E). Then
N(Ep
+ t:) R! N(Ep)
+ t:
+ ...
But the very definition of Ep implies that the first derivative is zero. Then the fraction of [particles
with energiesin the range Ep :I::O.O2kTis
2
v'ii
or 0.04,ff7'ieii
1
(kT)3/2
-(
kT
/ 2)
1/2 e -1/2
= 9.68 x 10-3.
282
(O.O2kT),
~
The volume correction is on page 508; we need first to find d. If we assume that the
particles in water are arranged in a cubic lattice (a bad guess, but we'll use it anyway), then 18
grams of water has a volume of 18x 10-6m3, and
d3 = {18 x10-6 m3)
{6.02 x 1023) = 3.0 x 10-29m3
is the valume allacated ta each water malecule. In this case d = 3.1 x 10-1om. Then
1
4
b = 2(6.02 x 1023)( 37r(3.1 x 10-10m)3) = 3.8 m3/mal.
E22-32
d3 = 3b/27rNA,
or
d =
3 3(32
x 10-6m3/~ol)
211"(6.02
= 2.9 x lO-10m.
x 1023/mol)
E22-33
a haS units of energy volume per square mole, which is the same as energy per mole times
volume per mole.
P22-1 Salve (1- x)(1.429) + x(1.250) = 1.293 far x. The result is x = 0.7598.
P22-2
~
The only thing that matters is the total number of moles of gas (2.5) and the number of
moles of the second gas (0.5). Since 1/5 of the total number of moles of gas is associatedwith the
secondgag, then 1/5 of the total pressure is associatedwith the second gas.
P22-4
Use Eq. 22-11 withthe
appropriate
~
inserted.
P22-5 (a) Since >. cx 1/J2 , we have
~=
dn
r&
V~
=
I (27.5x lO-8m)
= 1.67.
y (9.90x lO-8m)
(b) Since >..(X l/p, we have
= 49.5x lO-8m.
(c) Since ). <XT, we have
=
283
7.87x
10-8
m.
~
P22-6
We can assumethe molecule will collide with something. Then
1 = 1°CAe-crdr
=A/c,
so A = c. Ir the molecule has a mean free path or A, then
rce-cr dr = l/c,
>.=1°C
soA=c=l/A.
~
What is important here is the temperature; since the temperatures are the samethen the
averagekinetic energiesper particle are the same. Then
1
2ml(VrmS,I)
2
1
2
= 2m2(Vrms,2)
.
We are given in the problem that Vav,2= 2Vrms,l°According to Eqs. 22-18 and 22-20 we have
Vrms = ~
Combining
this with
the kinetic
= ~~
= ~va.v.
energy expression
m¡
above,
2
Vrms,2
=
m2
= 4.71.
Vrms,l
P22-8 (a) Assume that the speedsare not all the same. Transform to a frame of referencewhere
vav = O, then some of the individual speeds must be greater than zero, and some will be less
than zero. Squaring these speedswill result in positive, non-zero, numbers; the mean square will
necessarilybe greater than zero, so Vrms> 0.
(b) Only if all of the particles have the same speed will Vrms=Vav.
~
(a) We need to first find the number oí particles by integrating
N
1°C
N(v) dv,
1
°C
I
2
c
3
V dv = 3vo.
CV2 dv +
1vO
va
Invert, then C = 3N/v3.
(b) The average velocity is found from
'av =
1 foo
N lo
Using our result from above,
vav
=
284
N(v)vdv.
As expected, the averagespeed is less than the maximum speed. We can make a prediction about
the root mean square speed; it will be larger than the averagespeed (seeExercise 22-15 above) but
smaller than the maximum speed.
(c) The root-mean-square velocity is found from
Using our results from above,
2
Vrms
=
~
{VO
Nlo
2
v2dv,
vg
-;
Vo
3N
-v
{VO
v4dv
=
~ ~
v3 5
-3
-Svo
2
.
lo
Then, taking the square root,
Is .¡375 > 3/4? It had better be.
P22-10
P22-11
P22-12
P22-13
P22-14
~
The massofair displaced by 2180m3 is m = (1.22kg/m3)(2180m3) = 2660kg. The mass
of the balloon and basket is 249 kg and we want to lift 272 kg; this leavesa remainder of 2140 kg for
the massofthe air inside the balloon. This correspondsto (2140kg)/(0.0289kg/mol) = 7.4x104mol.
The temperature of the gas inside the balloon is then
T = (pV)/{nR)
= [{1.01 x 105Pa){2180 m3)]/[{7.4x
That's 85°C.
P22-16
P22-17
285
104mol){8.31 J/mol.
K) = 358K.
~
We apply Eq. 23-1,
H=kA~
~x
The rate at which heat flows out is given as a power per area (mW /m2), so the quantity given is
really H / A. Then the temperature difference is
~T
= !:!: ~x
A k
= (O 54 w /m2)
(33,000
(2.5 W /m.
m)
= 710K
K)
The heat flow is out, so that the temperature is higher at the base oí the crust. The temperature
there is then
710 + 10 = 720 °C.
E23-2
We apply Eq. 23-1
E23-3
(a) AT/Ax = (136 CO)/(0.249m) = 546CO/m.
(b) H = kAAT/Ax
= (401W /m. K)(1.80m2)(546CO /m) = 3.94x 105W.
(c) TH = (-12°C + 136CO) = 124°C. Then
T = (124°C)
-(546
CO /m)(O.ll
E23-4
(a) H= (0.040W/m.K)(1.8m2)(32CO)/(0.012m)
(b) Since k has increased by a factor of (0.60)/(0.04)
m)
= 63.9°C.
= 190W.
= 15 then H Bhould alBo increase by a
factor of 15.
~
There are three possible arrangements: a sheet of type 1 with a sheet of type 1; a sheet
of type 2 with a sheet of type 2; and a sheet of type 1 with a sheet of type 2. We can look back on
Sample Problem 23-1 to see how to start the problem; the heat flow will be
for substances of different typesj and
A~T
H1l
j L
=
(Ljkl)
+ (Ljk1)
for a double layer if substance 1. There is a similar expression for a double layer of substance 2.
For configuration ( a) we then have
while for configuration
(b) we have
((1/k1) + (1/k2))-1
klk2
= ((kl + k2)!(k2k2))-l =
kl
286
+
k2.
~
Then which is larger
2k1k2
(k1 + k2)/2 or
?
~.
If k1 » k2 then the expression become
kl/2 and 2k2,
so the first expressionis larger, and therefore configuration (b) has the lower heat flow. Notice that
we get the same result if k1 « k2!
E23-6
There's a typo in the exercise.
H = A,1.T / R; since the heat flows through one slab and then through the other, we can write
(T1 -T",)/ R1 = (T", -T2)/ R2. Rearranging,
T", = (T1R2 + T2R1)/(Rl
+ R2).
E23-7
Use the results of Exercise 23-6. At the interface between ice and water Tx = 0°C. Then
R1T2 + R2T1 = O, or klTl/ L1 + k2T2/ L2 = 0. Not only that, L1 + L2 = L, so
k1T1L2 + (L -L2)k2T2
=0,
so
L2
(1.42m)(1.67W /m. K)( -5.20°C)
=
(1.67W /m .K)(-5.20°C)
-(0.502W
/m .K)(3.98°C)
= 1.15m.
E23-8 ~T is the same in both cases. So is k. The top configuration has Ht = kA~T/(2L). The
bottom configuration has Hb = k(2A)~T/L. The ratio of Hb/Ht = 4, so heat flows through the
bottom configuration at 4 times the rate ofthe top. For the top configuration Ht = (10J)/(2 min) =
5J/min. Then Hb = 20J/min. It will take
t = (30J)/(20J/min)
= 1.5 min.
~
(a) This exercisehas a distraction: it asks about the heat fiow through the window, but
what you need to find first is the heat fiow through the air near the window. We are given the
temperature gradient both inside and outside the window. Inside,
a similar
expression
AT
(20°C) -(5°C)
~x
(0.08 m)
= 190CO/m;
exists for outside.
From Eq. 23-1 we find the heat flow through
H = kA.AT
---~x
the air;
= (O-O2R
w.., /m .K)(n
Rm
an ~o
/m' -1
\
1\~.~
...1)2(1
\~~~
~ I...J
-~.v ~ 'XT
...
The value that we 8J:rived at is the rate that heat flows through the air across an 8J:eathe size of
the window on either side of the window. This heat flow had to occur through the window as well,
so
H
=
1.8W
answersthe window question.
(b) Now that we know the rate that heat flows through the window, we are in a position to find
the temperature difference acrossthe window. Rearranging Eq. 32-1,
~T = H~x
-=
kA
so we were well justified
small.
(1.8W)(O.OO5m)
= () ()')~~o
(1.0 W /m. K)(0.6 m)2 -~.~~~
in our approximation
that the temperature
287
'"' ,
drop across the glass is very
E23-10
(a) W = +214J, done on means positive.
(b) Q = -293J, extracted from means negative.
(c) AEint = Q + W = ( -293J) + (+214J) = -79.0J.
E23-11
(a) AEint along any path between these two points is
~Eint = Q + W = (50J) + (-20J)
= 30J.
Then along ibf W = (30J) -(36J)
= -6J.
(b) Q = (-30J) -(+13J)
= -43J.
(c) Eint.f = Eint.i + ~Eint = (10J) + (30J) = 40J.
(d) ~Eintib = (22J) -(10J)
= 12J; while ~Eintbf = (40J) -(22J)
= 18J. There is no work
done on the path bf, so
Qbf = AEintbf
and Qib = Qibf -Qbf
E23-12
= (36J) -(18J)
-Wbf
= (18J)
-(O)
= 18J,
= 18J.
Q = mL = (0.10)(2.1 x 108 kg)(333x 103J/kg) = 7.0x 1012J.
~
We don 't need to know the outside temperature because the amount of heat energy
required is explicitly stated: 5.22 GJ. We just need to know how much water is required to transfer
this amount of heat energy. Use Eq. 23-11, and then
Q
(5.22 x 109 J)
~
= 4.45 x 104 kg.
m=-
c~T
(4iOOJ/kg.
K)(50.0°C -22.0°C)
This is the mass of the water, we want to know the volume, so we'll use the density, and then
E23-14
E23-15
Q = mL, so m = (50.4 x 103J)/(333x 103J/kg) = 0.151kg is the mount which freezes.
Then (0.258 kg) -(0.151 kg) = 0.107 kg is the amount which does not freeze.
E23-16
(a) w = mg~y;
if IQI = IWI, then
~T=
mg6.y
mc
~
There are three "things" in this problem: the copper bowl (b), the water (w), and the
copper cylinder (c). The total internal energy changesmust add up to zero, so
AEint,b
+ AEint,w
+ AEint,c
= O.
As in Sample Problem 23-3, no work is done on any object, so
Qb + Qw + Qc = O.
288
The heat transfers for these three objects are
Qb
=
mbCb(Tf,b
Qw
=
mwCw(Tf,w
Qc
=
mccc(Tf,c
-Ti,b),
-Ti,w)
+
Lvm2,
-Ti,c).
For the most part, this looks exactly like the presentation in Sample Problem 23-3; but there is an
extra term in the second line. This term refiects the extra heat required to vaporize m2 = 4.70 9 of
water at loo°C into steam loo°C.
Some of the initial temperatures are specified in the exercise: Ti ,b = Ti ,w = 21.0°C and T f ,b =
Tf,w
= Tf,c
=
loo°C.
( a) The heat transferred
to the water,
Qw =
then,
is
(0.223kg)(4190J/kg.K) ((100°C) -(21.0°C)) ,
+(2.26 x 106J /kg) (4.70x 10-3kg) ,
8.44 x 104J .
=
Thi8 an8Werdiffer8 from the back ofthe book. I think that they (or was it me) u8ed the latent heat
of fu8ion when they 8hould have U8edthe latent heat of vaporization!
(b) The heat tran8fered to the bowl, then, i8
Qw = (0.146 kg)(387 J/kg .K) ((100°0) -(21.0°C))
= 4.46 x 103 J.
( c) The heat transfered from the cylinder was transfered into the water and bowl, so
Qc =
-Qb
-Qw
= -{4.46
x lO3J)
-{8.44
x lO4J)
= -8.89
x lO4J.
The initial temperature of the cylinder is then given by
Ti
c =
Tí
,
c
,
Qc
-
mcCc
E23-18 The temperature of the silver must be raised to the melting point and then the heated
silver needsto be melted. The heat required is
Q = mL + mcLlT = (O.l30kg)[(lO5x
E23-19
lO3J/kg) + (236J/kg .K)(l235K
(a) Use Q = mc6.T, m = pV, and t = Q/P.
JmaCa
+
Pw
v
-289 K)] = 4.27x lO4J.
Then
wCw],6.T
p
~)(900J/kg.K)
+ (998kg/m3)(0.64x
1~(4190J/kg.K)](100~2~q2
(2400 W)
(b) Use Q = mL, m = pV, and t = Q/P.
t=
Then
PwVwLw
p
is the additional time required.
289
=
117
s.
E23-20
The heat given off by the steam will be
Qs = msLv + msCw(50CO).
The hear taken in by the ice will be
Qi = miLf
+ miCw(50 Co;
Equating,
ms
=
Lf
+
Cw(50
CO)
~
The linear dimensions of the ring and sphere change with the temperature
cording to
~dr
~ds
=
=
change ac-
ardr(Tf,r -Ti,r),
asds(Tf,s -Ti,s).
When the ring and sphere are at the same (final) temperature the ring and the sphere have the
same diameter. This means that
dr + 6.dr = ds + 6.ds
when Tf,s = Tf,r. We'll solve these expansion equations first, and then go back to the heat equations.
dr + 6.dr
dr (1 + ar(Tf,r -Ti,r))
=
ds + 6.ds,
=
ds (1 + as(Tf,s -Ti,s))
,
which can be rearranged to give
ardrTf,r -asdsTf,s
= ds (1-
asTi,s) -dr
(1-
arTi,r) ,
but since the final temperatures are the same,
Tf = .ds(1-
asTi,s) -dr
ardr
Putting
(1-
arTi,r)
-asds
in the numbers,
Tf =
= 34.1oC.
No work is done, so we only have the issue of heat flow, then
Qr + Qs = o.
Where "r" refers to the copper ring and "s" refers to the aluminum sphere. The heat equations are
Qr =
Qs =
mrCr(Tf -Ti,r),
msCs(Tf -Ti,s).
290
E23-22 The problem is compounded becausewe don't know if the final state is only water, only
ice, or a mixture of the two.
Consider first the water. Cooling it to 0°C would require the removal of
Qw = (0.200kg)(4190J/kg .K)(O°C -25°C) = -2.095x 104J.
Consider now the ice. Warming the ice to would require the addition of
Qi = (0.100kg)(2220J/kg .K)(O°C + 15°C) = 3.33x 103J.
The heat absorbed by the warming ice isn't enough to cool the water to freezing. However, the ice
can melt; and if it does it will require the addition of
Qim = (0.100kg)(333x 103J/kg) = 3.33x 104J.
This is far more than will be liberated by the cooling water, so the final temperature is o°C, and
consists of a mixture of ice and water.
(b) Consider now the ice. Warming the ice to would require the addition of
Qi = (0.050kg)(2220J/kg .K)(O°C + 15°C) = 1.665x 103J.
The heat absorbed by the warming ice isn't enough to cool the water to freezing. However, the ice
can melt; and if it does it will require the addition of
Qim = (0.050kg)(333x103J/kg) = 1.665x104J.
This is still not enough to cool the water to freezing. Hence, we need to solve
Qi + Qim + miCw(T -o°C) + mwCw(T -25°C) = 0,
which has solution
T = (4190J/kg.
K)(0.200kg)(25°C)
-(1.665x
103J) + (1.665x 104J)
=
(4190J/kg.
2.5°C.
K)(0.250kg)
E23-23
(a) c = (320J)/(0.0371 kg)(42.0°C -26.1°C) = 542 J/kg .K.
(b) n = m/M = (37.1 g)/(51.4 g/mol) = 0.722 mol.
(c) c = (542J/kg .K)(51.4x 10-3 kg/mol) = 27.9J/mol. K.
E23-24
(1) W = -p6.V = (15Pa)(4m3) = -60J for the horizontal path; no work is done during
the vertical pathj the net work done on the gas is -60 J .
(2) It is easiest to consider work as the (negative of) the area under the curvej then W =
-(15Pa + 5Pa)(4m3)/2 = -40J.
(3) No work is done during the vertical path; W = -p6.V = (5Pa)(4m3) = -20J for the
horizontal pathj the net work done on the gas is -20J.
~
Net work done on the gas is given by Eq. 23-15,
W=-IPdV.
But integrals are just the area under the curve; and that 's the easy way to solve this problem. In the
caseof closed paths, it becomesthe area inside the curve, with a clockwise sensegiving a positive
value for the integral.
The magnitude of the area is the same for either path, since it is a rectangle divided in half by
a square. The area of the rectangle is
(15x103Pa)(6m3) = 90x103J,
so the area of path 1 (counterclockwise) is -45 kJ; this means the work done on the gas is -(-45 kJ)
or 45 kJ. The work done on the gas for path 2 is the negative of this becausethe path is clockwise.
291
E23-26
During
the isothermal
expansion,
Wl = -nRT
In V2
Vi
-Pl Vlln P.!
P2
During cooling at constant pressure,
W2 = -P2¡lV
= -P2(Vl -V2)
= -P2Vl(1-Pl!P2)
= Vl(Pl-P2).
The work done is the sum, or
E23-27
During
the isothermal
expansion,
w
=
V2
-nRTln
.
=
-PI
V2
VIln
VI'
V¡
so
= l.l4xlO3J.
E23-28
(a) pV'Y is a constant, so
P2 = P1(V1/V2)'Y = (1.00 atm)((11)/(0.5
1))1.32=
2.50
atm;
T2 = T1(P2/P1)(V2/V1), SO
T2 = (273 K) .(2.50 at~
(0.5 I)
-=341K.
(1.00 atm) ~
(b)
V3 = V2(P2/Pl)(T3/T2),
SO
=0.40
( c ) During
the adiabatic
process,
(1.01 x 105Pa/atm)(1
x 10-3m3 /1)
[(
W12
=
2.5
atm
)(
0.5
1)
-(1.0
atm)(11)]
=
78.9
J.
.
(1.32) -1
During the cooling process,
W23 = -p~V = -(1.01 x 105pa/atm)(2.50
atm)(l
The net work done is W123 = 78.9J + 25.2J = 104.1 J.
292
x 10-3m3/1)[(0.4
I) -(0.5
J)] = 25.2J.
~
E23-30
Air is mostly diatomic (N2 and O2), so use T = 1.4.
(a) pV'Y is a constant, so
T2 = T1(P2/Pl)(V2/V1),
SO
or 96°C.
(b) The work required for delivering 1 liter of compressed air is
W12 = .(1.01 x 105Pa/atm)(lx10-3m3/1)[(2.3
atm)(1.0 I) -(1.0
atm)(1.0 1/0.552)] = 123J.
(1.40) -1
The number of liters per secondthat can be delivered is then
6.V/6.t
=
(230W)/(123J/l)
E23-31
Eint,rot = nRT = (1 mol)(8.31 J/mol.
E23-32
Eint,rot = ~nRT = (1.5)(1 mol)(8.31 J/mol.
~
(a) Invert
Eq.
=
K)(298K)
1.87
= 2480J.
K)(523K)
= 6520J.
32-20,
Tf = Ti~
PiVi
= (250 K) (l450x lO3Pa)(l.36m3)
(l22xlO3Pa)(lO.7m3)
-378
K,
which is the same as 105°C.
( c ) Ideal gas law, again:
n = [P~]/[RT]
= [(1450 x 103pa)(1.36m3)]/[(8.31
J/mal.
K)(378K)]
= 628 mal.
(d) From Eq. 23-24,
before the compressionand
2.96 x 106 J
after the compression.
(e) The ratio of the rms speeds will be proportional
to the square root of the ratio of the internal
energies,
V(1.96x 106J)/(2.96x 106J) = 0.813;
we can do this becausethe number of particles is the same before and after, hence the ratio of the
energiesper particle is the same as the ratio of the total energies.
293
E23-34
We can assume neon is an ideal gas. Then ~T = 2~Eint/3nR,
E23-35
At constant pressure, doubling the volume is the same as doubling the temperature.
Q = nCp~T
7
= {1.35 mol)2{8.31J/mol.
or
Then
K) {568 K -284 K) = 1.12x104J.
E23-36
(a) n = m/M = (12 g)/(28 g/mol) = 0.429 mol.
(b) This is a constant volume process, so
Q = nCvAT
= {0.429
Q = nCp~T
mol)~{8.31
= {4.34
J/mol.
K){125°C
J/mol.
mol){29.
-25°C)
K)(62.4K)
= 891J.
= 7880J.
(b) From Eq. 23-28,
5
5
Eint = 2nR~T
= 2(4.34
mol)(8.31J/mol. K)(62.4K) = 5630J.
(c) From Eq. 23-23,
Ktrans
=
3
2nR~T
=
5
2(4.34
K)/(4.00
mol)(8.31
J/mol.
E23-38
Cv = ~(8.31 J/mol.
E23-39
Each species will experience the same temperature
Q
K)(62.4K)
= 3380J.
g/mol) = 3120J/kg .K.
change, so
= Ql + Q2+ Q3,
= nlC1~T + n2C2~T+ n3C3~T,
Dividing this by n = n¡ + n2 + n3 and ~T will return the specific heat capacity oí the mixture, so
G -n¡
G¡ + n2G2 + n3G3
n¡ + n2 + n3
E23-40
W AB = O, since it is aconstant volume process, consequently, W = W AB+W ABC = -15J.
But around a closed path Q = -W, so Q = 15 J. Then
QCA
= Q -QAB
-QBC
= (15J)
Note that this heat is removedfrom the system!
294
-(20J)
-(OJ)
= -5J.
~
According to Eq. 23-25 (which is specific to ideal gases),
3
~Eint = 2nR~T,
and for an isothermal process ~T = O, so for an ideal gas ~Eint = 0. Consequently, Q + W = 0 for
an ideal gas which undergoes an isothermal process.
But we know W for an isotherm, Eq. 23-18 shows
Then finally
Vi
E23-42 Q is greatest for constant pressure processesand least for adiabatic. W is greatest (in
magnitude, it is negative for increasing volume processes)for constant pressure processesand least
for adiabatic. AEint is greatest for constant pressure (for which it is positive), and least for adiabatic
(for which is is negative).
E23-43
(a) For a monatomic gas, 'Y = 1.667. Fast process are often adiabatic, so
T2 = T1(Vl/V2)'Y-l
= (292 K)[(1)(1/10)]1.667-1
= 1360 K.
(b) For a diatomic gas, "Y= 1.4. Fast process are often adiabatic, so
T2 = T1(V1/V2)'Y-l = (292K)[(1)(1/10)]l04-1
= 733 K.
E23-44
This problem cannot be solved without making some assumptions about the type of process occurring on the two curved portions.
~
Ir the pressure and volume are both doubled along a straight line then the process can
be described by
p=!?!-V
V1
The final point involves the doubling of both the pressure and the volume, so according to the ideal
gas law, pV = nRT, the final temperature T2 will be four times the initial temperature T1.
Now for the exercises.
(a) The work done on the gas is
¡
W
=
-pdV
2
¡
=
I
2
(
-f!;..VdV
=
I
VI
-f!;..
2
~
VI
2
-~
2
2
)
We want to express our answer in terms oí TIo First we take advantage oí the íact that V2 = 2VI,
then
295
E23-46
The work done is the area enclosed by the path. If the pressure is measured in units of
10MPa, then the shape is a semi-circle, and the area is
W = (1T/2)(1.5)2(10MPa)(lx10-3m3)
The heat is given by Q = -W
= 3.53x104J.
= -3.53 x 104 J .
E23-48
Constant Volume
(a) Q = 3(3.15 mol)(8.31 J/mol. K)(52.0K) = 4080J.
(b) W = o.
( c ) ~Ermint = 3(3.15 mol) (8.31 J /mol .K) (52.0 K) = 4080 J .
Constant Pressure
(a) Q = 4(3.15 mol)(8.31J/mol.
K)(52.0K) = 5450J.
(b) W = -P~V = -nR~T
= -(3.15 mol)(8.31 J/mol. K)(52.0K)=
-1360J.
(c) ~Ermint = 3(3.15 mol)(8.31 J/mol. K)(52.0K) = 4080J.
Adiabatic
(a) Q = 0.
(b) W = (PfV f -Pi Vi)/('Y -1) = nR~T/('Y -1) = 3(3.15 mol)(8.31 J/mol. K)(52.0K)
(c) ~Ermint = 3(3.15 mol)(8.31 J/mol. K)(52.0K) = 4080J.
P23-1
(a) The temperaturedifference is
(5 Co/9 FO)(72°F --20°F)
= 51.1Co.
The rate of heat loss is
H = (1.0W /m. K)(1.4m2)(51.1
CO)/(3.0x 10-3m) = 2.4x 104W.
(b) Start by finding the R values.
Rg = (3.0x 10-3m)/(1.0W
/m. K) = 3.0x 10-3m2 .K/W,
Ra = (7.5x 10-2m)/(O.O26W /m. K) = 2.88m2 .K/W.
Then use Eq. 23-5,
Get double pane windowsl
296
= 4080J.
P23-2
(a) H = (428W /m. K)(4. 76x 10-4m2)(100 CO)/(1.17m) = 17.4 W.
(b) ~m/~t
= H/L = (17.4 W)/(333x103J/kg)
= 5.23x10-5kg/s, which is the same as 188 g/h.
~
Follow the example in Sample Problem 23-2. We start with Eq. 23-1:
H
-kA~
H
=
dr
,
2 dT
k( 47rr ) "dr" ,
j
T2
=
kdT,
Tl
=
k(T1
=
411"k(Tl
-T2),
411"k(Tl
-T2)rlr2
H
-T2),
.
r2 -rl
P23-4
(a) H = (54xlO-3W/m2)47r(6.37xlO6m)2
(b) Using the results of Problem 23-3,
AT
=
(2.8x lOI3W)(6.37x
47r(4.2W /m.
Since T2 = 0°C, we expect
P23-5
Since H
according to
= 2.8xlO13W.
lO6m -3.47x
K)(6.37x
lO6m)(3.47
lO6m)
x lO6m)
7.0x lO4co.
TI = 7.0xlO4co.
-kAdTjdx,
then Hdx
JHdx
= -aTdT.
=
H is a constant,
so integrate
both
side
-JaTdT,
HL =
H
P23-6 Assume the water is all at 0°C. The heat flow through the ice is then H
rate of ice formation is ~m/ ~t = H/ L. But ~m = pA~x, so
6.x
6.t
il. 7W /m.
(920 kg/m3)(333x
K)(10CO)
=
1.11
x 10-6m/s.
lO3J/kg)(0.05 m)
That's the same as 0.40 cm/h.
~
(a) Start
with
the heat equation:
Qt + Qi + Qw = O,
297
,1
=
H
PAL=
k~T
-P-j;;;'
kAAT/x;
the
~
where Qt is the heat from the tea, Qi is the heat from the ice when it melts, and Qw is the heat
from the water (which used to be ice). Then
mtCt(Tf
-Tt,i)
+ miLf
+ mwCw(Tf
-T
w,i)
= O,
which, since we have assumedall of the ice melts and the massesare all equal, can be solved for Tf
as
Ti
CtTt,i
=
+ CwT w,i -Lf
Ct+Cw
(4190J/kg. K)(90°C)+
(4190J/kg .K)(O°C)
-(333
x 103J/kg)
(4190J/kg .K) + (4190J/kg .K)
5.3°C.
(b) Once again, assumeall of the ice melted. Then we can do the same steps, and we get
TI
,--- CtTt.i + CwTw,i =--LE
Ct+Cw
,(~190J/kg
.K)(70°C)
+
(4190J/kg
.K)(O°C)
-(333~103J/kg)
(4190J/kg. K) + (4190J/kg .K)
-4.7°C.
So we must have guessedwrong when we assumed that all of the ice melted. The heat equation
then simplifies to
TntCt(Tf
-Tt,J
+ miLf
= O,
and then
mi
=
TntCt(Tt,i
-Tf)
Lf
(0.520
kg)(41~OJ/kg
.K)(90°C
-00)
(333 x lO3J/kg)
0.458 kg.
P23-8
c = Q/m~T
= H/(~m/~t)~T.
But ~m/~t
= p~V/~t.
Combining,
= 2.4x
(8.2 x lO-6m3/s)(0.85
(AV/At)pAT
P23-9
(a) n = NA/M,
x lO3kg/m3)(l5
lO3J/kg
.K.
CO)
so
(b) Eav = ~kT, so
P23-10
Qw + Qt = O, so
CtATt
+ mwCw(Tr
-TU
= O,
or
Ti=
.(0.3 kg)(4190J/kg
.m)(44.4°C)
+ (46.1 J/K)J44.4°C
(0.3kg)(4190J/kg.
298
m)
-15.0°C)
= 45.5°C.
~
We can use Eq. 23-10, but we will need to approximate
line is straight then we use c = mT + b. I approximate m from
m = .(14J /mol. K) -(3
(50ijK)
Then
J /mol .K)
-(200
c first. If we assume that the
= 3.67 x 10-2 J /mol.
K)
find b from those same data points,
b = (3 J /mol
.K)
-(3.67
x 10-2 J /mol) (200 K) = -4.34
J /mol
.K.
Then from Eq. 23-10,
h
-n
Tf cdT,
Q
Ti
=
nJTi{Tf
=
n
=
n (~(Tf2
-Ti2)
=
{0.45mal)
( {3.67
(mT+b)dT,
[2T
m
2 +bT
] Ti
Tf ,
,
+ {-4.34J/mol.
l.l5xlO3J.
P23-12
8Q = nC8T,
+ b(Tf
x 10-2
-Ti)
) ,
J /mal)
( (500
K)2
-{200
K)2)
2
K){500K -200K))
so
Q
n [(0.318 J/mol.
K2)T2/2
-(0.00109
J/mol.
K3)T3/3
-(0.628
n(645.8 J/mol).
Finally,
{645.8 J/mol){316
Q
P23-13
g)/{107.87 g/mol) = 189J.
TV"Y-l is a constant, so
T2 = (292 K)(1/1.28)(1.40)-1
P23-14
w=
-JpdV,
= 265 K
so
w =
J [ nRT
an2
]
-v-nb-V2
-nRTln(V
-nb)
dV,
~I
-nRT In Vf-nb
Vi-nb
299
-aff
2
J/mol.
goK
K)T]50K
,
~
We can use Eq. 23-10, but we will need to approximate
line is straight then we use c = mT + b. I approximate m from
m = .(14J /mol. K) -(3
(50ijK)
Then
J /mol .K)
-(200
c first. If we assume that the
= 3.67 x 10-2 J /mol.
K)
find b from those same data points,
b = (3 J /mol
.K)
-(3.67
x 10-2 J /mol) (200 K) = -4.34
J /mol
.K.
Then from Eq. 23-10,
h
-n
Tf cdT,
Q
Ti
=
nJTi{Tf
=
n
=
n (~(Tf2
-Ti2)
=
{0.45mal)
( {3.67
(mT+b)dT,
[2T
m
2 +bT
] Ti
Tf ,
,
+ {-4.34J/mol.
l.l5xlO3J.
P23-12
8Q = nC8T,
+ b(Tf
x 10-2
-Ti)
) ,
J /mal)
( (500
K)2
-{200
K)2)
2
K){500K -200K))
so
Q
n [(0.318 J/mol.
K2)T2/2
-(0.00109
J/mol.
K3)T3/3
-(0.628
n(645.8 J/mol).
Finally,
{645.8 J/mol){316
Q
P23-13
g)/{107.87 g/mol) = 189J.
TV"Y-l is a constant, so
T2 = (292 K)(1/1.28)(1.40)-1
P23-14
w=
-JpdV,
= 265 K
so
w =
J [ nRT
an2
]
-v-nb-V2
-nRTln(V
-nb)
dV,
~I
-nRT In Vf-nb
Vi-nb
299
-aff
2
J/mol.
goK
K)T]50K
,
P23-17
The speed of sound in the iodine gas is
v = fA = (1000 Hz)(2
x 0.0677m)
P23-18 w = -Q = mL = (333xlO3J/kg)(O.l22kg)
P23-19
(a)
Process AB
Q = ~(1.0 mol)(8.31 J/mol.
W=O.
~Eint = Q + W = 3740J.
Process BC
K)(300K)
= 135m/s.
= 4.06xlO4J.
= 3740J.
Q=O.
W = (PfV f -Pi Vi)/('Y -1) = nR~T/('Y -1) = (1.0 mol)(8.31 J/mol. K)( -145K)/(1.67
-1800 J
~Eint = Q + W = -1800J.
Process AB
Q = ~(1.0 mol)(8.31 J/mol. K)( -155K) = -3220J.
W = -P~V = -nR~T
= -(1.0 mol)(8.31J/mol.
K)(-155K)
= 1290J.
~Eint = Q + W = 1930J.
Cycle
Q = 520J; W = -510J
-1)
=
(rounding error!); ~Eint = 10J (rounding error!)
P23-20
P23-21
is
Pf = (16.0 atm)(50/250)1.40 = 1.68 atm. The work done by the gas during the expansion
w =
96.0J.
This process happens 4000 times per minute, but the actual time to complete the process is half of
the cycle, or 1/8000 of a minute. Then p = (96J)(8000)/(60s)
= 12.8x 103W.
301
~
.
~
For isothermal processesthe entropy expression is almost trivial, ~S = Q/T, where if Q
is positive (heat flow into system) the entropy increases.
Then Q = T~S = (405K)(46.2J/K) = 1.87xl04J.
E24-2 Entropy is a state variable and is path independent, so
(a) ~Sab,2 = ~Sab,l = +0.60J/K,
(b) ~Sba,2 = -~Sab,2 = -0.60J/K,
E24-3
(a) Heat onlyenters
along the top path, so
Qin = T6.S = (400 K)(0.6J/K
-O.lJ/K)
= 200J.
-0.63/K)
= -1253.
(b) Heat leaves only bottom path, so
Qout = T~S
= (250 K)(0.l
3/K
Since Q + W = O for a cyclic path,
w = -Q = -[(200J)
+ (-125J)]
= -75J.
E24-4
(b) For isothermal process, Q = -W,
~s
( c ) Entropy
~
= Q/T=
then
(l.69xlO4J)/(4l0K)
change is zero for reversible
adiabatic
= 4l.2J/K.
processes.
(a) We want to find the heat absorbed, so
Q = mcAT
= (1.22 kg)(387 J/mol.
K) ((105°C)
-(25.0°C))
(b) We want to find the entropy change, so, according to Eq. 24-1,
~s
The entropy change oí the copper block is then
E24-6
~s
= Q/T
= mL/T,
~s
=
so
(0.001
kg)( -333
x 103 J/kg)/(263K)
302
=
-1.27J/K
= 3.77x 104J.
~
E24-7 Use the first equation on page 551
E24-8
(a)
(b)
(c)
(d)
~S = Q/T c -Q/Th.
~S = (260J)(1/100K
~S = (260J)(1/200K
~S = (260J)(1/300K
~S = (260J)(1/360K
-1/400K)
-1/400K)
-1/400K)
-1/400K)
=
=
=
=
1.95J/K.
0.65J/K.
0.217 J/K.
0.0722J/K.
~
(a) If the rod is in a steady state we wouldn't expect the entropy of the rod to change.
Heat energy is flowing out of the hot reservoir into the rod, but this process happens at a fixed
temperature, so the entropy change in the hot reservoir is
ilSH
( -1200
=
J)
2!!
=
-2.98J/K.
( 403 K)
TH
The heat energy flows into the cold reservoir, so
L\Sc
=
=
2.!!
(1200 J)
= 4.04J/K.
(297K)
TH
The total change in entropy of the system is the sum of these two terms
~s
= ~SH
+ ~SC
= 1.06J/K.
(b) Since the rod is in a steady state, nothing is changing, not even the entropy.
E24-10
(a) Qc + Ql = O, so
mccc(T-
T c) + mlC¡(T
-Ti)
= O,
which can be solved for T to give
T=
~~!~)S387
J/kg
.K) (400 K) + (0.10kg)(129J/kg
.K) (200 K)
=
320
K.
(0.05kg)(387 Jjkg .K) + (0.10kg)(129Jjkg .K)
(b) Zero.
(c) ~S = mclnTf/Tj,
so
~
The total mass of ice and water is 2.04 kg. If eventually the ice and water have the same
mass, then the final state will have 1.02 kg of each. This means that 1.78kg -1.02 kg = 0.76kg of
water changed into ice.
(a) The change of water at o°C to ice at o°C is isothermal, so the entropy change is
~s
=
.2
r=
-mL
T
(0.76 kg)(333 x 103J/kg)
.
(273 K)
(b) The entropy change is now +927 J/K.
303
-927 J/K.
E24-12
(a) Qa + Qw = O, 80
maca(T-
T a) +
mwCw(T
-T
w)
=
O,
=
-24.4J/K.
(c) For the water,
= 27.5J/K.
(d) AS = (27.5J/K) + (-24.4J/K)
E24-13
(a) e = 1(b) W = Qh -Qc
= 3.1J/K.
(36.2 J/52.4J) = 0.309.
= (52.4J) -(36.2J)
= 16.2J.
E24-14
(a) Qh = (8.18 kJ)/(0.25) = 32.7 kJ, Qc = Qh -W = (32.7 kJ) -(8.18 kJ) = 24.5 kJ.
(b) Qh = (8.18 kJ)/(0.31) = 26.4 kJ, Qc = Qh -W = (26.4 kJ) -(8.18 kJ) = 18.2 kJ.
~
One hour's worth oí coal, when burned, will provide energy equal to
{382 x 103kg) {28.0 x 106J /kg) = 1.07 x 1013J .
In this hour , however, the plant only generates
(755 x lO6W)(3600s)
= 2.72xlO12J.
The efficiency is then
e = (2.72xl012J)/(1.07xl013J)
= 25.4%.
E24-16
We use the convention that all quantities are positive, regardless of direction.
Qi,A = 3Qi,B; and Qo,A = 2Qo,B. But W A = Qi,A -Qo,A, so
5WB
or, applying
= 3Qi,B
-2Qo,B,
WB = Qi,B -Qo,B,
5WB
3WB
WB/Qi,B
=
=
=
3Qi,B-2(Qi,B -WB),
Qi,B,
1/3 = eB.
Then
304
W A = 5W B;
E24-22
(a) The area of the cycle is ~V ~p = PoVo, so the work done by the gas is
w
= (l.Olxl05pa)(0.0225m3)
= 2270J.
(b) Let the temperature at a be Tao Then
Tb = Ta(Vb/Va)(Pb/Pa)
2Ta.
Let the temperature at c be Tc' Then
Tc = Ta(Vc/Va)(Pc/Pa)
Consequently, ~Tab = Ta and ~Tbc = 2Tao Putting
constant pressure heat expressions,
= 4Ta.
this information
into the constant volume and
and
Qbc
so that Qac = YPo Yo, or
13
Qac = 2{1.01
x 105Pa){0.0225
m3)
l.48x lO4J.
{c) e = {2270J)/{14800J)
= 0.153.
{d) ec = 1- {Ta/4Ta) = 0.75.
~
According
to Eq.
24-15,
Now we salve the question out oí order .
(b) The work required to run the freezer is
IWI = IQLI/K
= (185 kJ)/(5.70)
= 32.5 kJ.
(a) The freezer will discharge heat into the room equal to
IQLI + IWI = (185 kJ) + (32.5 kJ) = 218 kJ.
E24-24 (a) K = IQLlllwl = (5683)/(1533) = 3.71.
(b) IQHI = IQLI + IWI = (5683) + (1533) = 7213.
E24-25
K = TL/(TH -TL); IWI = IQLI/K = IQLI(TH/TL
(a) IWI = (10.0J)(300K/280K
-1) = O.714J.
(b) IWI = (10.0J)(300K/200K
-1) = 5.00J.
(c) IWI = (10.0J)(300K/100K
-1) = 20.0J.
(d) IWI = (10.0J)(300K/50K
-1) = 50.0J.
E24-26
306
-1).
~
We wi11 start with the 8Bsumption that the air conditioner Ís a Carnot refrigerator .
K = TL/(TH -TL);
IWI = IQLI/K = IQL1(TH/TL -1). For fun, 1'11convert temperature to the
abso1ute Rankine sca1e! Then
IQLI = (1.0J)/(555°R/5300R
E24-28
The best coefficient
of performance
-1)
21J.
is
Kc = (276 K)/(308K
-276
K) = 8.62.
The inventor claims they have a machine with
K = (20 kW -1.9kW)/(1.9
kW)
= 9.53.
Can 't be done.
E24-29 {a) e = 1- {258K/322K) = 0.199. IWI = {568J){0.199) = 113J.
{b) K = {258K)/{322K -258K) = 4.03. IWI = {1230J)/{4.03) = 305J.
E24-30
The temperatures are distractors!
IWI
=
IQHI-IQLI
=
IQHI-
KIWI,
so
JWI = IQHI/(l
Then p = (1.58 MJ)/(3600s)
E24-31
K = (260K)/(298K
+ K) = (7.6 MJ)/(l
= 440W.
-260
K) = 6.8.
E24-32
K = (0.85)(270K)/(299K-270K)
1.89x 105J of work. Then
IQLI = KIWI
~
+ 3.8) = 1.58 MJ.
= 7.91. In 15 minutes the motor can do (210W)(900s)
= (7.9l)(l.89xlO5J)
= l.50xlO6J.
The Ca;rnot engine has an efficiency
E = I -T2
T;=
IWI
TQJ"
The Carnot refrigerator has a coefficient of performance
K
=
T4
T;-::"T¡
=
IQ41
TWr'
Then we combine the last two expressions,and
~T3
-T4
-
IQ31-IWI =
IWI
307
~
IWI
-1.
=
Finally, combine them all,
T4
-
;¡;-::-T¡-
IQ1I
Tl
-T2
Now, we rearrange,
=
~
IQli
E24-34
(a)
Integrate:
ln N! ~ lN
ln x dx = N ln N -N
+ 1 ~ N ln N -N
.
(b) 91, 752, and about 615,000. You will need to use the Stirling approximation
double inequality to do the lBSt two:
.[i;nn+l/2e-n+l/(12n+l)
< n! < .[i;nn+l/2e-n+l/(12n)
extended to a
.
~
(a) For this problem we don't care how the particles are arranged inside a section, we
only care how they are divided up between the two sides.
Consequently,there is only one way to arrange the particles: you put them all on one side, and
you have no other choices. So the multiplicity in this case is one, or Wl = 1.
(b) Once the particles are allowed to mix we have more work in computing the multiplicity.
Using Eq. 24-19, we have
N!
N!
W2 = (N /2)!(N /2)! = ""("{Ñ7"iji)"
(c) The entropy oí a state oí multiplicity
w is given by Eq. 24-20,
s = klnw
For part (a), with a multiplicity
of 1, 81 = O. Now for part (b),
82 = kln
N!
=
klnN!
-2kln(N/2)!
¡¡N¡2¡ij'
and we need to expand each of those terms with Stirling's approximation.
Combining,
82
=
= k (NlnN
=
kNlnN-
=
kNln2
-N)
kN
-2k
-kNlnN
((N/2)
ln(N/2)
+ kNln2
-(N/2))
,
+ kN,
Finally, ~8 = 82 -81 = kN ln 2.
(d) The answer should be the same; it is a free expansion problem in both cases!
308
~
We want
to evaluate
~s
=
{Tf~
JT;
T
{Tf
JT;
l
,
nAT3
dT
T
,
Tf nAT2
dT ,
Ti
~
(Tf3
-Ti3)
.
Into this last expression,which is true for many substancesat sufficiently low temperatures, we
substitute the given numbers.
-(
4.8 mal) (3. 15 x 10-5 J /mal.
AS-
K4)
{
3
(10K)3
-(5.0
K)3)
=
4.41
x
10-'};
J
/K:
P24-2
P24-3
(a) Work is only done along path ab, where Wab = -p6.V
(b) 6.E Jbc = ~nR6.Tbc, with a little algebra,
= -3Po6.Vo. So Wabc= -3poVo.
3
3
3 -4)poVo = 6poVo.
~Eintbc= 2(nRTc
-nRTb) = 2(PcVc
-PbVb) = 2(8
6.Sbc
=
~nRln(Tc/Tb),
with
a little
algebra,
3
3
ASbc = 2nRln(pc/Pb) = 2nRln2.
(c) Both
P24-4
are zero for a cyclic
(a) For an isothermal
process.
process,
P2 = Pl (Vl/V2)
= Pl/3.
For an adiabatic process,
P3 = P1 (V1/V2)'Y
= P1 (1/3)1.4
= O.215p1.
For a constant volume process,
T3 = T2(P3/P2)
= Tl(O.215/0.333)
= O.646T1,
(b) The easiestones first: ~Eint12 = O, W23 = 0, Q31 = 0, ~S31 = 0. The next easier ones:
5
5
~Eint23 = 2nR~T23
= 2nR(O.646Tl
-T1) = -0.885PlVl,
Q23= ~Eint23-W23 = -0.885pl Vl,
~Eint31= -~Eint23 -~Eint12 = O.885pl
V1,
W31= ~Eint31-Q31 = O.885pl
V1.
309
Finally, some harder ones:
W12
-nRT11n(V2/V1)
= -P1 V11n(3)
Q12 = ~Eint12
-W12
=
-1.10p1
V1,
= 1.10p1 V1
And now, the hardest:
~Sl2
= Ql2/Tl
~S23 = -~Sl2
P24-5
Note that
TA = TB = Tc/4
= -1.10nR.
= TD.
Process
I: ABC
(a) QAB = -WAB = nRToln(VB/VA)
~(4poVo -PoVo) = 4.5poVo.
(b)
(c)
(d)
~SAC
-~S3l
= 1.10nR,
=PoVoln2.
QBC = ~nR(Tc-TB)
WAB=
-nRToln(VB/VA)
= -PoVoln2.
WBC = O.
Eint = ~nR(Tc
-TA)
= ~(PcVc -PAVA)
= ~(4poVo -PoVo)
~SAB = nRln(VB/VA)
= nRln2;
~SBC = ~nRln(Tc/TB)
= 4nR.
= ~(PCVC-PBVB)
= 4.5poVo.
= ~nRln4
= 3nRln2.
=
Then
Process
II: ADC
(a) QAD = -WAD
= nRToln(VD/VA)
= -PoVoln2.
QDC = ~nR(Tc
-TD)
= ~(pcVC PDVD) = ~(4poVo -PoVo) = 10poVo.
(b) WAB = -nRToln(VD/VA)
=PoVoln2.
WDC = -p~V
= -Po(2Vo -Vo/2)
= -~poVo.
(c) Eint = ~nR(Tc
-TA)
= ~(PcVc -PAVA)
= ~(4PoVo -PoVo) = 4.5poVo.
(d) ~SAD = nRln(VD/VA)
= -nRln2;
~SDC = ~nRln(Tc/TD)
= ~nRln4
= 5nRln2.
Then
~SAC = 4nR.
P24-6
The heat required to melt the ice is
Q
=
m(Cw~T23+ L + ci~T12),
(O.O126kg)[(4190J/kg. K)(15CO) + (333x 103J/kg) + (2220J/kg .K)(10CO)],
5270J.
The change in entropy of the ice is
~Si
=
m[Cw ln(T3/T2)
+ L/T2
(O.O126kg)[(4190J/kg.K)
+(2220J/kg.K)
+ Ci ln(T2/Tl)],
ln(288/273)
+ (333 x 103J/kg)/(273K),
ln(273/263)],
19.24J/K
The change in entropy ofthe lake is ~Sl = (-5270 J)/(288K) = 18.29J/K. The change in entropy
of the system is 0.95J/kg.
~
(a) This is a problem where the total internal energy of the two objects doesn't change,
but since no work is done during the process,we can start with the simpler expressionQl + Q2 = O.
The heat transfers by the two objects are
Ql
=
mlCl(Ti
-T1,i),
Q2
=
m2C2(T2
-T2,i).
Note that we don't call the final temperature Tf here, becausewe are not assuming that the two
objects are at equilibrium.
310
We combine these three equations,
m2c2(T2
-T2,i)
=
-m1c1(T1
m2c2T2
=
m2c2T2,i
=
m1c1
T2 i + -(T1
,
m2C2
T2
-T1,i),
+ m1c1(T1,i
-T1),
i -T1)
,
As object 1 "cools clown" , object 2 "heats up" , as expectecl.
(b) The entropy change of one object is given by
~S=
rTf~=mcln~,
JT;
T
Ti
and the total entropy change for the system will be the sum of the changesfor each object, so
~s
= mlclln
TTI + m2c21n
i,l
T2
Ti,2
Into the this last equation we need to substitute the expressionfor T2 in as a function of T1. There's
no new physics in doing this, just a mess of algebra.
(c) We want to evaluate d(AS)/dT1. To save on algebra we will work with the last expression,
remembering that T2 is a function, not a variable. Then
dT2
dTl
=
mlcl
We could consider writing T2 out in all of its glory, but what would it gain us? Nothing. There is
actually considerably more physics in the expression as written, because...
(d) ...we get a maximum for ~S when d(~S)/dTl = O, and this can only occur when T1 = T2
according to the expression.
P24-8
Tb = {10.4 x 1.01 x 105Pa){1.22 m3)/{2 mol){8.31 J/mol.
realistic? T a can be found after finding
Pc = Pb(~/Vc)'Y
= (10.4
x 1.01 x 105Pa)(1.22/9.13)1.67
K) = 7.71 x 104K. Maybe not so
= 3.64x
104Pa,
Then
Ta = Tb(pa/Pb)
= (7.7lx
lO4K)(3.64x
311
lO4/l.05X
lO6) = 2.67xlO3K.
Similarly,
Tc = Ta(Vc/Va)
(a) Heat is added during
3
Qab = 2nR(Tb
-Ta)
= (2.67x
103K)(9.13/1.22)
= 2.00x 104K.
process ab onlyj
3
= 2(2 mol)(8.31
J/mol.
K)(7.71
x 104K -2.67x
103K) = 1.85x 106J.
(b) Heat is removed during process ca only;
Qca = ~nR(Ta -Tc)
= ~ (2 mol)(8.31 J /mol .K)(2.67 x 103K -2.00 x 104K) = -0.721 x 106J .
(c) W = IQabl-IQcal = (1.85x106J) -(0.721 x 106J) = 1.13x106J.
(d) e = W/Qab = (1.13x106)/(1.85x106) = 0.611.
P24-9 The pV diagram for this processis Figure 23-21, except the cycle goes clockwise.
(a) Heat is input during the constant volume heating and the isothermal expansion. During
heating,
3
3 mol)(8.31 J/mol. K) (600K -300 K) = 3740J;
Ql = 2nR~T
= 2(1
During isothermal expansion,
Q2 = -W2 = nRTln(V
f/Vi)
= (1 mol)(8.31 J/mol.
K)(600K)
ln(2) = 3460J;
so Qin = 7200J.
(b) Work is only done during the second and third processes; we've a;lready solved the second,
W2 = -3460J;
W3 = -P~V
= PaVc -PaVa = nR(Tc -Ta)
So w = -970J.
(c) e = IWI/IQinl
P24-10
= (970 J)/(7200J)
(a) Tb = Ta(pb/Pa)
Td = Ta(Va/Vd)'Y-l
Pd = Pa(Va/Vd)'Y
Heat
in
occurs
cd, so Qo =
during
~nR~Tcd
-300K)
= 2490J;
= 3Ta;
Pc = Pb(Vb/Vc)'Y
(b)
K)(600K
= 0.13.
Tc = Tb(Vb/Vc)'Y-l
process
= (1 mol)(8.31 J/mol.
process
ab,
= 2.87nRTa.
= 3Ta(1/4)0.4
~3pa(1/4)1.4
=
= O.430pa;
= Ta(1/4)0.4
=Pa(1/4)1.4
so Qi
=
1.72Ta;
= O.574Ta;
= O.144pa.
~nR~Tab
=
5nRTa;
Heat
out
occurs
during
Then
e F 1- (2.87nRTa/5nRTa)= 0.426.
P24-11
(c) (VE/VA) = (PA/PE) = (0/0.5) = 2. The work done on the gas during the isothermal
compressionis
w = -nRTln(VB/VA)
= -(1 mol)(8.31J/mol.
K)(300K)
ln(2) = -1730J.
Since ó.Eint = O along an isotherm Qh = 1730 J .
The cycle has an efficiency of e = 1- (100/300) = 2/3. Then for the cycle,
w = eQh = (2/3)(1730J)
312
= 1150J.