oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Question 1, chap 1, sect 6.
part 1 of 1
10 points
Needing help, the secretary of the
United States Department of Agriculture
asked your teacher, “If a chicken-anda-half can lay an egg-and-a-half in a
day-and-a-half, how many days will it
take two chickens to lay twenty eggs?”
Please help your teacher select the correct
answer to the secretary’s question.
1. Two chickens will lay twenty eggs in six
days.
2. Two chickens will lay twenty eggs in nine
days.
3. Two chickens will lay twenty eggs in
twenty-one days.
4. Two chickens will lay twenty eggs in
twenty-four days.
5. Two chickens will lay twenty eggs in
twenty-two days.
6. Two chickens will lay twenty eggs in fifteen days. correct
7. Two chickens will lay twenty eggs in
twelve days.
8. Two chickens will lay twenty eggs in sixteen days.
9. Two chickens will lay twenty eggs in ten
days.
10. Two chickens will lay twenty eggs in
twenty days.
Explanation:
1
Basic Concept: The information given
in the question is the rate of egg production
in one instance and you must make this rate
compatible with another instance. The rate
of egg production is constant. The number of
eggs per chicken per day is a constant.
Solution: Since it takes a chicken-and-a-half
a day-and-a-half to lay an egg-and-a-half, it
will take one chicken one-and-a-half days to
lay one egg. Therefore, to lay twenty eggs it
will take two chickens fifteen days.
Alternative Method: Unit analysis is
basic to every physics problem and is central
to this problem. The rate of egg production is
the number of eggs produced per chicken per
day. In the given instance the rate is
{3/2 eggs}
{3/2 chickens} {3/2 days}
2
eggs
=
.
3 chickens · days
rate =
(1)
In the requested instance, the number of
chickens is (2 chickens) and the number of
eggs is (20 eggs). The number of days N is
to be determined. Therefore in the requested
instance, the rate is
rate =
{20 eggs}
.
{2 chickens} {N }
(2)
The rate is constant, so equating the rates
Eqs. (1) and (2), we have
20 eggs
2 eggs
=
3 chicken · days
2 chickens · N
Solving for the number of days N , we have
20 eggs 3 chickens · days
2 chickens
2 eggs
= 15 days .
N =
The correct answer: “Two chickens will lay
twenty eggs in fifteen days.”
Note: The early chicken catches the worm.
Question 2, chap 14, sect 3.
part 1 of 1
10 points
A weight (with a mass of 63 kg) is suspended from a point near the right-hand upper end of a uniform boom whose mass is
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
29 kg . Supporting this boom are a cable runing from this same point on the boom to a
point on the wall (the left-hand vertical coordinate at a height of 10 m) and a pivot (at
the origin of the coordinate axes) on the same
wall.
The acceleration of gravity is 9.8 m/s2 .
Boom and Weight
10
Using the figure
α = arctan
8
7
6
2
2
63 kg
0
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
Figure: Drawn to scale.
Calculate the tension in the cable.
Correct answer: 678.715 N (tolerance ± 1
%).
Explanation:
Let :
Ty x + Tx y−Wb xb − Ww x = 0 .
Ty
Tx
7
(x, y)
6 (0, y)
5
Ww
Mb
4
3
α
Wb
2
1
0
-1
-2
Mb = 29 kg ,
Mw = 63 kg ,
ℓ = 9 boom length ,
ℓ
xb = = 4.5 m ,
2
x = 8 m,
y = 6.5 m , and
h = 10 m .
The static equilibrium conditions are
X
Fx = 0 ,
X
Fy = 0 , and
X
τ = 0 , or
T
8
(6)
Force Scale is 77.7262 N/m
9
3
1
(0, h)
10
g
9k
Vertical Height (m)
4
(5)
x
6.5 m
= arctan
8m
◦
= 39.0939 ,
h−y
θ = arctan
x
10 m − 6.5 m
= arctan
8m
◦
= 23.6294 .
T
5
y
θ
Vertical Height (m)
9
2
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (m)
Let the cable make an angle θ with the
horizontal. Then using Eq. 6 and the figure,
we have
slope = tan θ =
Tx =
Ty
h−y
.
=
x
Tx
x
Ty .
h−y
(7)
(8)
(1)
(2)
(3)
(4)
Since
Ty
h−y
=p
,
T
(h − y)2 + x2
T (h − y)
Ty = p
,
(9)
(h − y)2 + x2
sin θ =
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
R
Using the pivot at the origin as the fulcrum,
and substituting Tx from Eq. 8 into the static
equilibrium condition for torque Eq. 4, we
have
3
r
Wb xb + Ww x = Ty x + Tx y
xy
Wb xb + Ww x = Ty x + Ty
h−y
y
Wb xb + Ww x = Ty 1 +
x
h−y
hx
.
Wb xb + Ww x = Ty
h−y
Ty =
(Wb xb + Ww x) (h − y)
,
hx
and substituting Ty from Eq. 9, we have
T (h − y)
p
p
(h − y)2 + x2
T
(h −
y)2
+ x2
=
(Wb x + Ww x) (h − y)
xh
=
(Wb xb + Ww x)
.
xh
p
(h − y)2 + x2
xh
= [(29 kg) (4.5 m) + (63 kg) (8 m)]
× (9.8 m/s2 )
p
(10 m − 6.5 m)2 + (8 m)2
×
(8 m) (10 m)
T = (Wb xb + Ww x)
= 678.715 N .
Question 3, chap 15, sect 3.
part 1 of 1
10 points
The cylindrical disk has mass m and outer
radius r with a radial mass distribution
(which may not be uniform) so that its mo7
ment of inertia is m r 2 .
9
The disk rolls (perpendicularly to the axis)
without slipping in a cylindrical trough, see
figure below.
Determine (for small displacements from
equilibrium) the period of harmonic oscillation which the disk undergoes.
s
13 R − r
1. T = 2 π
7
g
s
8 R−r
2. T = 2 π
5 g
s
9 R−r
3. T = 2 π
7 g
s
5 R−r
4. T = 2 π
3 g
s
3 R−r
5. T = 2 π
2 g
s
19 R − r
6. T = 2 π
10 g
s
17 R − r
7. T = 2 π
9
g
s
13 R − r
8. T = 2 π
9
g
s
15 R − r
9. T = 2 π
8
g
s
16 R − r
correct
10. T = 2 π
9
g
Explanation:
Basic Concepts: Let r be the radius of
the disk and R be the radius of the cylindrical trough. The disk is rolling without
slipping. Choose the point of contact as our
axis. Around this point, the rotational inertia
of the disk is, using parallel axis theorem,
I=
7
16
m r2 + m r2 =
m r2 .
9
7
(1)
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Solution: Let the angle of rotation around
this instantaneous contact point be φ and the
angle the center of the disk makes from the
center of the trough to the vertical be θ.
θ
r
R−
S is the common arc length
S = r (φ + θ)
S = Rθ
r
φ
θ
S
S
Note: The dotted curve is a
hypocycloid denoting the path of the
contact point at equilibrium as the
disk rolls back and forth in the cylindrical trough.
The arc length along the disk (which rolls
back and forth) must equal the arc length
along the cylindrical trough (both arc lengths
are labeled S in the figure).
Therefore, φ and θ are related by
r (φ + θ) = R θ , or
r φ = (R − r) θ , so
dφ
dθ
r
= (R − r)
,
dt
dt
from the rolling without slipping condition.
Now the torque equation around the point of
contact is
X
I
τ : m g r sin θ = −I
Substituting I from Eq. 1, we have
d2 θ
9
m g r2
+
θ=0
dt2
16 (R − r) m r 2
9
g
d2 θ
+
θ = 0.
2
dt
16 (R − r)
R
d2 φ
dt2
d2 φ
+ mgrθ = 0,
dt2
since sin θ ≈ θ. Substituting θ in place of φ,
we have
R − r d2 θ
+ mgrθ = 0
I
r
dt2
m g r2
d2 θ
+
θ=0.
dt2
I (R − r)
4
(2)
Equation 2 is the differential equation for simple harmonic motion. The coefficient of θ is
ω 2 . Therefore
s
9
g
ω=
, and
16 (R − r)
s
16 (R − r)
.
T = 2π
9
g
Alternative Solution: The kinetic energy,
2
2
dx
dφ
1
1
+ I
K= m
2
dt
2
dt
2
dθ
1
= m (R − r)2
2
dt
2
7
dφ
2
+
mr
18
dt
2
dθ
1
= m (R − r)2
2
dt
2
dθ
7
2
m (R − r)
+
18
dt
2
8
dθ
= m (R − r)2
,
9
dt
dθ
7
.
since Idisk = m r 2 , v = r ω, and ω =
9
dt
The potential energy is
U = mgh
= m g (R − r) (1 − cos θ)
1
≈ m g (R − r) θ 2 ,
2
1 2
θ . Energy is conserved,
since 1−cos θ ≈
2
therefore E = K +U = constant, and we have
2
8
2 dθ
E = m (R − r)
9
dt
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
dE
dt
d2 θ
dt2
1
+ m g (R − r) θ 2
2
= constant , so
16
d θ d2 θ
=
m (R − r)2
9
dt dt2
dθ
θ
+m g (R − r)
dt
= 0 . Therefore,
9
g
+
θ = 0.
(3)
16 (R − r)
Equations 2 and 3 are the same equation for
simple harmonic motion.
Correct answer: 0.466472 s (tolerance ± 1
%).
Explanation:
Let : R = 40 cm = 0.4 m ,
v0 = 0.8 m/s ,
m = 7 kg , and
µ = 0.07 .
From the perspective of the surface, let the
speed of the center of the disk be vsurf ace .
Using the frictional force f , we can determine
the acceleration
Question 4, chap 13, sect 2.
part 1 of 5
10 points
Assume: When the disk lands on the surface it does not bounce.
The disk has mass 7 kg and outer radius
40 cm with a radial mass distribution (which
may not be uniform) so that its moment of
2
inertia is m R2 .
3
A disk is given a hard kick (impulse) along
a horizontal surface at time t0 . The kicking
force acts along a horizontal line through the
disk’s center, so the disk acquires a linear velocity 0.8 m/s but no initial angular velocity.
The coefficient of friction between the disk
and the surface is 0.07 .
The kinetic friction force between the surface and the disk slows down its linear motion
while at the same time making the disk spin
on its axis at an accelerating rate. Eventually,
the disk’s rotation catches up with its linear
motion, and the disk begins to roll at time
trolling without slipping on the surface.
The acceleration of gravity is 9.8 m/s2 .
7 kg
40 cm , radius
0.8 m/s
2
I= m R2
3
µ = 0.07
How long t = trolling − t0 does it take for
the ball to roll without slipping?
5
f = µmg,
X
and
Fsurf ace = m a , or
m a = µ m g , so
a = µg.
(1)
Since
vsurf ace = v0 − a t , we have
= v0 − µ g t .
(2)
After pure rolling begins at trolling there is no
longer any frictional force and consequently
no acceleration. From the perspective of the
center of the disk, let the tangential velocity
of the rim of the disk be vdisk and the angular
velocity be ω ; the angular acceleration is
X
τ = I α , so
τ
α=
I
µmgR
=
2
m R2
3
3 µg
(3)
=
2 R
3 (0.07) (9.8 m/s2 )
=
2
(0.4 m)
= 2.5725 rad/s2 .
The time dependence of ω is
ω = αt
3 µg
t,
=
2 R
vdisk ≡ R ω
3
= µgt.
2
so
(4)
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
When the disk reaches pure rolling, the velocity from the perspective of the surface will be
the same as the velocity from the perspective
of the center of the disk; that is, there will
be no slipping. Setting the velocity vdisk from
Eq. 4 equal to vsurf ace from Eq. 2 gives
vdisk = vsurf ace
3
µ g t = v0 − µ g t , or
2
5
µ g t = v0 , so
2
2 v0
t=
5 µg
(0.8 m/s)
2
=
5 (0.07) (9.8 m/s2 )
(5)
= 0.466472 s .
Question 5, chap 13, sect 2.
part 2 of 5
10 points
Once the disk rolls without slipping, what
is its linear speed?
Correct answer: 0.48 m/s (tolerance ± 1 %).
How far ∆s does the ball slide until it begins
to roll without slipping?
Correct answer: 0.298542 m (tolerance ± 1
%).
Explanation:
The ball slides at a constant deceleration.
Using Eqs. 1 and 5, we have
1
∆s = v0 t − a t2
2
2
1
2 v0
2 v0
− µg
= v0
5 µg
2
5 µg
2
2
v
2
= 0
−
µ g 5 25
8 v02
(7)
=
25 µ g
8
(0.8 m/s)2
=
25 (0.07) (9.8 m/s2 )
= 0.298542 m .
Through what angle ∆θ does the disk rotate while sliding before it begins to roll without slipping?
Correct answer: 0.0445448 rev (tolerance ±
1 %).
(6)
or using Eqs. 4 and 5, we have
3
µgt
2
2 v0
3
= µg
2
5 µg
3
= v0 .
5
Question 6, chap 13, sect 2.
part 3 of 5
10 points
Question 7, chap 13, sect 2.
part 4 of 5
10 points
Explanation:
Using Eqs. 1 and 5, we have
vsurf ace = v0 − a t
= v0 − µ g t
2 v0
= v0 − µ g
5 µg
2
= v0 1 −
5
3
= v0
5
3
= (0.8 m/s)
5
= 0.48 m/s ,
6
vdisk =
(6)
Explanation:
Starting from rest, the disk spins at a constant angular acceleration. Using Eqs. 3 and
5, we have
1
∆θ = α t2
2
2
2 v0
1 3 µg
=
2 2 R
5 µg
2
3 v0
=
(8)
25 µ g R
(0.8 m/s)2
3
=
25 (0.07) (9.8 m/s2 ) (0.4 m)
= 0.279883 rad
= 16.0361 ◦ .
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
7
1
1
2
2
I ωrolling
+ m vrolling
2 h2
vrolling i2
1 2
m R2
=
2 3
R
1
2
+ m vrolling
2
1 2
1
2
2
=
m vrolling
+ m vrolling
2 3
2
2 3
2
1
v0
+1
= m
2
5
3
2 5
1
2 3
= m v0
2
5
3
3
= K0 , so
5
Kf
3
=
.
K0
5
The number of revolutions is
∆θ
∆θrev =
360◦
(16.0361 ◦ )
=
360◦
= 0.0445448 rev .
Kf =
Note: When comparing Part 3 to Part 4,
the ball slides more than it spins due to the
fact it is slipping on the surface.
∆s > R ∆θ
8 v02
3 v02
>
.
25 µ g
25 µ g
Question 8, chap 13, sect 2.
part 5 of 5
10 points
What is the ratio of the final kinetic energy (after pure rolling occures) to the initial
kinetic energy?
Kf
7
=
K0
12
Kf
7
2.
=
K0
13
Kf
6
=
3.
K0
11
Kf
9
=
4.
K0
11
Kf
8
=
5.
K0
15
Kf
5
6.
=
K0
7
Kf
4
=
7.
K0
7
Kf
2
=
8.
K0
3
Kf
3
9.
= correct
K0
5
Kf
7
=
10.
K0
11
Explanation:
1.
Question 9, chap 16, sect 99.
part 1 of 1
10 points
Amplitude (centimeter)
You are given f1 (x), a transverse wave that
moves on a string that ends and is FIXED in
place at x = 5 m. As the problem begins, the
wave is moving to the right at v = 1 m/s.
Using Eq. 6, vrolling =
K0 =
1
m v02 ,
2
3
v0 ,
5
and
v
3
2
1
0
-1
-2
-3
0
1
2
3 4 5 6 7 8 9 10
Distance (meter)
Hint: Consider the image of the wave reflected about the FIXED point x = 5 m in the
following diagram. The image will be moving to the left at v ′ = −1 m/s (in the opposite
direction from the real wave).
Amplitude (centimeter)
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
v
3
v′
3
2
2
1
1
0
4. 0
-1
-1
-2
-2
-3
0
1
3 4 5 6 7 8 9 10
Distance (meter)
What is the shape of the wave on the string
after 3 s?
2
3
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
correct
3
2
2
1
1
5. 0
1. 0
-1
-1
-2
-2
-3
8
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
3
3
2
2
1
1
6. 0
2. 0
-1
-1
-2
-2
-3
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
3
3
2
2
1
1
7. 0
3. 0
-1
-1
-2
-2
-3
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Initial time, t = 0 s
1
8. 0
-1
-2
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
3
2
1
9. 0
-1
-2
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
3
2
3
2
1
0
-1
-2
-3
3 4 5 6 7 8 9 10
Distance (meter)
After 3 s the positions of the two waves have
both moved 3 meters in opposite directions.
The resultant sum of the two waves is the
light gray line.
Superposition, at t = 3 s
3
Amplitude (centimeter)
2
Amplitude (centimeter)
3
-3
0
1
2
0
1
2
2
1
0
-1
-2
-3
1
10. 0
3 4 5 6 7 8
Distance (meter)
9 10
-2
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
Amplitude (centimeter)
Resultant, at t = 3 s
-1
-3
9
3
2
1
0
-1
-2
-3
0
1
2
3 4 5 6 7 8
Distance (meter)
9 10
Question 10, chap 9, sect 1.
part 1 of 2
10 points
Explanation:
The initial wave (real) on the string is represented with a dashed line and its reflected
wave (imaginary) is represented with a dotted
line.
Assume we find matter with negative mass;
i.e., m < 0. We want to examine the consequences of the gravitational force on negative
mass. Since the unit vector, r̂12 , is in the direction of the displacement vector, ~r12 , the
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
gravitational force exerted on mass m2 , by
mass m1 , is
The acceleration of m2 due to the gravitational force from m1 is
~ 21 = −G m1 m2 r̂12
F
2
r12
~a2 =
~
~
and, vice versa F21 = −F12 , for the gravitational force exerted on mass m1 , by mass,
m2 is
~ 12 = −G m1 m2 r̂21
F
2
r21
F21
r̂12
m1
m2
Notice: The reversal of the indices and the
negative sign indicating an attractive force,
when these two masses are both positive,
m > 0. Thus, the masses will accelerate as indicated in the diagram below. [Hint: Recall,
~ = m~a.]
F
a
a
m1
m2
If both masses are negative; i.e., m1 < 0
and m2 < 0, which diagram indicates the
relevant acceleration?
a
1.
2.
a
m1
m2
a
a
m1
m2
correct
3.
4.
a
a
m1
m2
a
m1
a
m2
Explanation:
Basic Concepts: From Newton’s second
Law, we have the force on mass m2 is
~ 21 = m2 ~a2 .
F
The sign (direction) of the acceleration of m2
is independent of the sign of mass m2 and
therefore depends only on the sign of mass
m1 !
10
~ 21
F
m1
= −G 2 r̂12 .
m2
r12
A mass of either sign will be attracted
to a positive mass and a mass of either
sign will be repelled from a negative mass.
a
a
m1
m2
Question 11, chap 9, sect 1.
part 2 of 2
10 points
If the masses have mixed signs, m1 > 0 and
m2 < 0; i.e., m1 has a positive mass and m2
has a negative mass, which diagram indicates
the relevant acceleration?
a
1.
a
m1
2.
a
3.
a
m2
a
m1
m2
a
m1
m2
correct
4.
a
m1
a
m2
Explanation:
A mass of either sign will be attracted to a
positive mass and a mass of either sign will be
repelled from a negative mass. Since m1 > 0
is positive, m2 is attracted to m1 and since
m2 < 0 is negative, m1 is repelled from m2 .
a
a
m1
m2
However, if the masses have opposite mixed
signs, m1 < 0 and m2 > 0, this diagram
indicates the relevant acceleration.
a
a
m1
m2
Comment: Will this mean that if there
is negative mass in the universe, the universe
will forever be expanding?
Since we (positive mass beings) are repelled
from negative mass, it is easy to understand
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
why we have not as yet detected matter with
negative mass. That is, all (either positive or
negative) mass will be repelled from negative
mass (accelerating away from) since the time
of the big bang.
Is there an experiment proposed which will
detect negative mass? Professor John Cramer
has proposed that just like the experiments
which have detected massive positive mass
objects in outer space by detecting a convex
lens effect due to light bending around the
gravitational field of these dark massive object, the same experiment could be used to
detect concave lens effect from large negative
mass objects.
Question 12, chap 9, sect 99.
part 1 of 2
10 points
Your instructor is on planet Krypton
which has three moons. We will call these
moons X, Y, and Z. The densities of the
three moons are related as
The force of gravity is
F =G
gK = G
2.
3.
B
A
Moon
θ
4.
5.
6.
L
Your instructor visually observes the three
moons. The visual acceptance angle of the
moons’ diameters are related as
θX = θY
θX = 2 θZ .
Newton’s 2nd law is
X
F = ma
MK
.
2
RK
(2)
To first order, compare the ratio of the
∆gY
differential gravitational force
, where
∆gX
∆g = gA − gB is the free fall acceleration
at points A and B on planet Krypton due
to the two moons Y and X, and where
∆g = gA − gB .
The figure is not drawn to scale. Consider
L ≫ RK .
RM
(1)
therefore
1.
RK
mM
r2
Comparing Newton’s 2nd law with the force
of gravity, we see that the gravitational acceleration g of an object at a distance r from a
mass M is
F
g=
,
m
ρY = 3 ρX
ρZ = ρX .
Krypton
11
7.
8.
9.
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
∆gY
∆gX
≈2
≈ 3 correct
≈
1
27
≈9
1
2
1
≈
3
≈
≈1
≈
1
9
≈ 27
10. None of these.
Explanation:
Let L be the distance from Krypton to one
of its moons (e.g., moon X) and RK be the
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
radius of the planet Krypton. The difference
∆gX is
∆gX = gA − gB
MX
MX
=G
−G
(3)
2
(L − RK )
(L + RK )2
Since L ≫ RK , we will make an approximation, which is clearer if we rewrite the the first
fraction [in equation (3)] as
1
1
1
(4)
= 2
2
2
(L − RK )
L
RK
1−
L
RK
The geometrical series, where x =
L
1
= 1 + x + x2 + x3 + ...
1−x
can be differentiated to yield
1
= 1 + 2 x + 3 x2 + ...
2
(1 − x)
and since x is small, we will only keep the
first two terms. Then equation (3) can be
rewritten as
G MX
∆gX ≈
L2
RK
RK
− 1+2 −
× 1+2
L
L
G MX
RK
≈
4
2
L
L
4 G MX RK
≈
L3
3
MX
RK
≈4
gK ,
(5)
MK
L
MK
since gK = G 2 .
RK
Therefore
3
RK
∆gX
MX
≈4
.
(6)
gK
MK
L
4
Using M ≡ ρ V ≡ π ρ R3, where M is the
3
sphere’s mass, and where V is the sphere’s
volume, we have
3
ρX RX3 RK
∆gX
≈4
3 L3
gK
ρK RK
1 ρX 3
≈
θ ,
2 ρK X
(7)
12
2 RX
where θX =
is the visual acceptance
L
angle of the moon’s diameter.
Using equation (7), the relationship between the differential gravitational forces at
points A and B on planet Krypton due to
the two moons Y and X, is
∆gY
gK
∆gX
gK
≈
1
2
1
2
ρY 3
θ
ρK Y
.
ρX 3
θX
ρK
(8)
Therefore
3
θY
ρY
ρX
θX
3
1
3
≈
1
1
≈ 3.
∆gY
≈
∆gX
(9)
Question 13, chap 9, sect 99.
part 2 of 2
10 points
Compare the ratio of the differential grav∆gZ
, where ∆g = gA − gB is
itational force
∆gX
the free fall acceleration at points A and B on
planet Krypton due to the two moons Z and
X, and where ∆g = gA − gB .
∆gZ
∆gX
∆gZ
2.
∆gX
∆gZ
3.
∆gX
∆gZ
4.
∆gX
1.
≈
1
2
≈4
≈
1
3
≈1
5. None of these.
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
∆gZ
1
≈
∆gX
4
1
∆gZ
7.
≈ correct
∆gX
8
∆gZ
≈2
8.
∆gX
∆gZ
9.
≈3
∆gX
∆gZ
10.
≈8
∆gX
Explanation:
Using equation (7), the relationship between the differential gravitational forces at
points A and B on planet Krypton due to
the two moons Z and X, is
6
1
2
≈
1
2
ρZ 3
θ
ρK Z
.
ρX 3
θX
ρK
acceleration (m/s2 )
6.
∆gZ
gK
∆gX
gK
13
5
4
3
2
1
0
−1
0
1
2
3
4 5
time (s)
6
7
8
9
What is the velocity at 1 s?
Correct answer: 7 m/s (tolerance ± 1 %).
Explanation:
In order to use the above graph, let
(8)
x0 = x0,1 = −18 m ,
v0 = v0,1 = 5 m/s ,
(t0 , a0 ) = (t0,1 , a0 ) = (0 s, 0 m/s2 ) ,
Therefore
3
θZ
ρZ
ρX
θX
3
1
1
≈
1
2
1
1
≈ 3 = .
2
8
∆gZ
≈
∆gX
(t1 , a1) = (t0,1 , a1,2 ) = (0 s, 2 m/s2 ) ,
(9)
Note: It is worth observing that the Sun
and Moon both visually appear to have nearly
the same visual acceptance angle from the
Earth θ. The differential gravitational effect
(which is partly responsible for the tides) of
the Moon is greater since it is more dense
that the sun, not necessarily because it is
closer (see Part 1).
The Earth is even more dense than the
Moon.
Question 14, chap 2, sect 4.
part 1 of 4
10 points
Consider the plot below describing the acceleration of a particle along a straight line
with an initial position of −18 m and an initial velocity of 5 m/s.
(t2 , a2) = (t2,3 , a1,2 ) = (3 s, 2 m/s2 ) ,
(t3 , a3) = (t2,3 , a3,4 ) = (3 s, 5 m/s2 ) ,
(t4 , a4) = (t4,5 , a3,4 ) = (9 s, 5 m/s2 ) ,
and
2
(t5 , a5 ) = (t4,5 , a5 ) = (9 s, 0 m/s ) .
Basic Concepts: The plot shows a curve
of acceleration versus time.
The change in velocity is the area (a1,2 t)
between the acceleration curve and the time
axis
v = v0,1 + a1,2 t ,
where the acceleration is constant.
Solution: With constant acceleration (a1,2 =
2 m/s2 ),
v = v0,1 + a1,2 t
(1)
= (5 m/s) + (2 m/s2 ) (1 s)
= 7 m/s .
Equations 1 and 3 are plotted below.
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
14
velocity (m/s) [× 0.1]
Correct answer: 36 m/s (tolerance ± 1 %).
5
Explanation:
The calculation is done in two parts, each
with constant acceleration (a1,2 = 2 m/s2 )
and (a3,4 = 5 m/s2 ).
4
3
2
v = v0,1 + a1,2 t2,3 + a3,4 [t − t2,3 ]
1
= (5 m/s) + (2 m/s ) (3 s)
+ (5 m/s2 ) [(8 s) − (3 s)]
= 36 m/s ,
0
−1
(3)
2
0
1
2
3
4 5
time (s)
6
7
8
9
where
Question 15, chap 2, sect 4.
part 2 of 4
10 points
v2,3 = v0,1 + a1,2 t2,3
= (5 m/s) + (2 m/s2 ) (3 s)
= 11 m/s .
What is the position at 1 s?
Correct answer: −12 m (tolerance ± 1 %).
Explanation:
Basic Concepts: The change in position
1
is the area (v0,1 t + a t2 ) between the veloc2
ity curve and the time axis
1 2
at .
2
With constant acceleration
x = x0,1 + v0,1 t +
Solution:
(a1,2 = 2 m/s2 ),
1
a t2
2 1,2
= (−18 m) + (5 m/s) (1 s)
1
+ (2 m/s2 ) (1 s)2
2
= −12 m .
position (m) [× 0.01]
x = x0,1 + v0,1 t +
(2)
Question 17, chap 2, sect 4.
part 4 of 4
10 points
What is the position at 8 s?
Correct answer: 123.5 m (tolerance ± 1 %).
Explanation:
The calculation is done in two parts, each
with constant acceleration (a1,2 = 2 m/s2 )
and (a3,4 = 5 m/s2 ).
1
a t2
(4)
2 1,2 2,3
1
+ v2,3 [t − t2,3 ] + a3,4 [t − t2,3 ]2
2
= (−18 m) + (5 m/s) (3 s)
1
+ (2 m/s2 ) (3 s)2
2
+ (11 m/s) [(8 s) − (3 s)]
1
+ (5 m/s2 ) [(8 s) − (3 s)]2
2
= 123.5 m ,
x = x0,1 + v0,1 t2,3 +
Equations 2 and 4 are plotted below.
2
1
0
−1
0
1
2
3
4 5
time (s)
6
7
Question 16, chap 2, sect 4.
part 3 of 4
10 points
What is the velocity at 8 s?
8
9
where
1
a1,2 t22,3
2
= (−18 m) + (5 m/s) (3 s)
1
+ (2 m/s2 ) (3 s)2
2
= 6 m.
x2,3 = x0,1 + v0,1 t2,3 +
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Hanging Gold Chain
10
A thin flexible gold chain of uniform linear
density has a mass of 12 g . It hangs between
two 10 cm high vertical walls (vertical axes)
which are a distance of 10 cm apart horizontally (x-axis), as shown in the figure below.
The acceleration of gravity is 9.8 m/s2 .
Hanging Gold Chain
10
8
Vertical Height (cm)
Question 18, chap 5, sect 5.
part 1 of 1
10 points
Vertical Height (cm)
9
7
6
5
3
2
8
1
7
0
5
4
3
2
1
0
0
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (cm)
Figure: Drawn to scale.
Find the magnitude of the horizontal force
the gold chain exerts on the left-hand wall.
Correct answer: 0.182152 mN (tolerance ±
12 %).
Fx
4
9
6
15
Fx
0
W
1
2 3 4 5 6 7 8 9 10
Horizontal Distance (cm)
Solution: The mass of the gold chain is
totally supported by the right-hand wall. The
vertical force W = m g at the right-hand wall
is the force of gravity on the chain. The
horizontal force Fx at the right-hand wall is
equal in magnitude to the horizontal force Fx
at the left-hand wall.
As well, the horizontal force Fx at the righthand wall is the vertical force W divided by
the slope (a tangent line to the chain’s curvature) at that point.
Using a ruler to draw a line tangent to the
curve at the right-hand wall (as shown in the
figure above), one can estimate an approximate slope, close to the slope using algebraic
equations, which is
3.5 cm
3.5 cm
=
10.0 − 4.5788
5.4212 cm
= 0.645613 .
The force of gravity is
Explanation:
Basic Concepts: At the right-hand point
of attachment, the tangent line defines the
W
.
slope of the chain; slope = tan θ =
Fx
W = m g = (12 g) (9.8 m/s2 ) = 117.6 mN .
W
3.5 cm
Since tan θ =
=
= 0.645613 ,
Fx
5.4212 cm
the horizontal force is
W
Fx =
tan θ
117.6 mN
=
0.645613
= 182.152 mN
= 0.182152 N .
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Mathematical Equations: The equation
for this hanging rope is a catenary
x
1 x/b
−x/b
y = cosh =
e +e
b
2
1
1
= ex/b + e−x/b , so
2
2
x
i
b= h
p
ln y + y 2 − 1
=
10 cm
h
i
p
ln (3.5 cm) + (3.5 cm)2 − 1
= 5.19522 cm , therefore x = 10 cm
dy
1 x/b
1 −x/b
=
e −
e
dx
2b
2b
at x = 10 cm , where b = 5.19522 cm
dy
= 0.645613 = tan θ ,
dx
6. 115.059 N
7. 97.3855 N
8. 80.4972 N
Explanation:
Drawing a diagram to scale of the vectors,
we have
F2
F
F1
θ
Scale: 10 N
as previously found using the figure.
Question 19, chap 3, sect 3.
part 1 of 2
10 points
~ 1 , and F
~ 2 are shown
Given: Two vectors F
below.
The magnitude and direction of these vectors are
F1
F2
θ1
θ2
= 62 N ,
= 76 N ,
= 339◦ ,
= 26◦ .
and
where
and
The angles are measured from the positive
x axis with the counter-clockwise angular direction as positive.
What is the magnitude of the resultant vec~ k, where F
~ =F
~1 + F
~2 ?
tor kF
where :
F1
θ1
F1x
F1y
F2
θ2
F2x
F2y
F
θ
Fx
Fy
= 62 N ,
= 339◦ ,
= 57.882 N ,
= −22.2188 N ,
= 76 N ,
= 26◦ ,
= 68.3083 N ,
= 33.3162 N , and
= 126.677 N ,
= 5.02576◦ ,
= 126.19 N ,
= 11.0974 N .
~ 1 and F
~2
The x components of the forces F
are
1. 107.488 N
2. 152.214 N
3. 126.677 N correct
4. 118.366 N
5. 121.5 N
16
F1x = F1 cos(339◦ ) = 57.882 N
F2x = F2 cos(26◦ ) = 68.3083 N .
And the y components are
F1y = F1 sin(339◦ ) = −22.2188 N
F2y = F2 sin(26◦ ) = 33.3162 N .
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
The x and y components of the resultant vec−1 11.0974 N
= tan
~ are
tor F
126.19 N
Fx = F1 x + F2 x
= (57.882 N) + (68.3083 N)
= 126.19 N
Fy = F1 y + F2 y
= (−22.2188 N) + (33.3162 N)
= 11.0974 N .
Hence the magnitude of the resultant vector
~ k is
kF
q
~ k = Fx2 + Fy2
kF
q
= (126.19 N)2 + (11.0974 N)2
= 126.677 N .
Question 20, chap 3, sect 3.
part 2 of 2
10 points
Note: Give the angle in degrees, use counterclockwise as the positive angular direction,
between the limits of −180◦ and +180◦ from
the positive x axis.
What is the direction of this resultant vec~?
tor F
= 5.02576◦ .
The “arctan” function is defined between limits −90◦ and +90◦ . Thus you must check to
see in which quadrant your resultant vector
lies. You should use the “atan2(y,x)” in your
calculator, if available, since it does not have
this ambiguity.
Question 21, chap 6, sect 1.
part 1 of 1
10 points
A force F acts to the right on a 5.38 kg
block. A 2.57 kg block is stacked on top of
the 5.38 kg block and can slide on it with
a coefficient of friction of 0.15 between the
blocks. The table has a coefficient of friction
of 0.16.
The acceleration of gravity is 9.8 m/s2 .
The system is in equilibrium.
µ1
3. 5.02576◦ correct
4. 136.551◦
5. −78.5869◦
2.57 kg
5.38 kg
1. 100.406◦
2. −27.8886◦
17
F
µ2
Find the force F required to accelerate the
5.38 kg block at 3.4 m/s2 .
Correct answer: 38.6089 N (tolerance ± 1
%).
Explanation:
6. −104.594◦
7. 52.7888◦
8. 19.9747◦
Explanation:
The angle is given by
Fy
θ = arctan
Fx
Given :
m1
m2
µ1
µ2
= 2.57 kg ,
= 5.38 kg ,
= 0.15 , and
= 0.16 .
Consider the free body diagrams below for
each mass
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
2T
T
Adding equations (1) and (2) yields
m1
(m1 + 4 m2 ) a = 4 F − 6 µ1 m1 g
− 4 µ2 (m1 + m2 ) g .
µ1 m1 g
µ1 m1 g
18
m2
F
Therefore
µ2 (m1 + m2 ) g
Basic Concepts : The acceleration of the
two masses will be different because of the
pulley system between them. The tensions in
the strings will be different.
Solution Let T be the tension in the string
attached to m2 on its left. Its acceleration will
be a and directed to the right. Because of the
pulley system between m1 and m2 , 2 T will be
the tension in the string attached to m1 on its
left, and the acceleration of that string, and
a
of m1 , will be
and will be directed to the
2
left.
For the mass m1 , 2 T acts to the left and
the motion to the left defines the frictional
force µ1 m1 g acting to the right, with the
a
acceleration directed to the left
2
a
= 2 T − µ1 m1 g .
Fnet1 = m1
2
(m1 + 4 m2 ) a 3 µ1 m1 g
+
4
2
+ µ2 (m1 + m2 ) g .
[2.57 kg + 4 (5.38 kg)] (3.4 m/s2 )
=
4
3 (0.15) (2.57 kg) (9.8 m/s2 )
+
2
+ (0.16) (2.57 kg + 5.38 kg) (9.8 m/s2 ) .
= 38.6089 N .
F =
Question 22, chap 13, sect 2.
part 1 of 1
10 points
A massless rope is wrapped around a uniform cylinder of radius R and mass M , as
shown in the figure.
Assume: The unwrapped portion of the
rope is vertical and the axis of the cylinder
remains horizontal.
Multiplying by 2,
m1 a = 4 T − 2 µ1 m1 g .
ω
(1)
For the mass m2 , notice first that the normal
force will be N = (m1 + m2 ) g , and since the
frictional force µ1 m1 g acted to the right on
m1 , it acts to the left on m2 .
For the block m2 , the force F acts to the
right, and the tension T and the two frictional
forces µ1 m1 g and µ2 (m1 + m2 ) g act to the
left, with the acceleration a directed to the
right
Fnet2 = m2 a
= F − T − µ1 m1 g − µ2 (m1 + m2 ) g .
Multiplying by 4
4 m2 a = 4 F − 4 T − 4 µ1 m1 g
− 4 µ2 (m1 + m2 ) g .
(2)
M
R
What is the linear acceleration of the cylinder?
g
4
3g
2. a =
5
1. a =
3. a = g
g
5
3g
5. a =
4
g
6. a =
2
4. a =
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
7. a = 2 g
2g
correct
3
2g
9. a =
5
g
10. a =
3
Explanation:
Choose positive direction as shown in the
figure.
8. a =
19
and the other side is in water of density
1000 kg/m3 .
The air is partially removed at the upper
part of the tube and the valve is closed. The
height of the water above its pool surface is
0.8 m . The height of the liquid above its pool
surface is 0.44 m . The difference in the heights
of the pool surfaces is 0.21 m .
⊗ Valve
0.8 m
T
0.44 m
α
a
0.21 m
Mg
R
From linear kinematics
Mg−T =Ma
and from rotational kinematics
test liquid
water
Figure: Not drawn to scale
Find the density of the liquid on the left.
Correct answer: 1818.18 kg/m3 (tolerance ±
1 %).
Explanation:
TR=Iα
where
a = Rα
and
1
M R2 .
2
Combine these equations to obtain
a
1
2
(M g − M a) R =
MR
2
R
1
Mg−Ma = Ma
2
3
g= a
2
2
a = g.
3
I=
Question 23, chap 18, sect 4.
part 1 of 1
10 points
One method of measuring the density of a
liquid is illustrated in the figure. One side
of the U-tube is in the liquid being tested,
Let : ρw = 1000 kg/m3 ,
hw = 0.8 m ,
h = 0.44 m , and
∆h = 0.21 m .
Note: The difference in the heights of the
pools 0.21 m does not matter since atmospheric pressure is nearly the same at both
pool heights.
The pressure at the upper surface of each
liquid is given by
P = Patm − ρw g hw = Patm − ρ g h.
Therefore,
hw
ρw
h
(0.8 m)
(1000 kg/m3 )
=
(0.44 m)
ρ=
= 1818.18 kg/m3 .
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
What is the angular frequency ω ′ of this
resultant wave?
Correct answer: 4.9 s−1 (tolerance ± 1 %).
Question 24, chap 16, sect 4.
part 1 of 5
10 points
Thanks to the trigonometric identity
θ1 − θ2
×
(0)
sin θ1 + sin θ2 = 2 cos
2
θ1 + θ2
× sin
,
2
the superposition
y1 = A sin(k x − ω t − φ1 ),
y2 = A sin(k x − ω t − φ2 ),
(2)
(3)
(of similar frequencies ω1 = ω2 = ω, wave
numbers k1 = k2 = k, and amplitudes A1 =
A2 = A) is itself a harmonic wave
y1+2 = A1+2 sin(k1+2 x−ω1+2 t−φ1+2 ). (4)
The following five questions concern the properties of this superposed wave (4) in terms
of the parameters of the original waves (2)
and (3):
A1 = A2 = A = 3.2 cm,
k1 = k2 = k = 3.9 m−1 ,
ω1 = ω2 = ω = 4.9 s ,
φ1 = 4.72 rad,
and φ2 = −0.241 rad.
⋆
⋆
⋆
Consider the time-plot of the resultant wave
y1+2 (t) at fixed position x = 0:
+A1+2
+A
0
−A
−A1+2
T
2T
y1 +y2
θ1 + θ2
φ + φ2
= kx − ωt −
,
2
2
φ − φ2
θ1 − θ2
=
,
2
2
and thus
sin(k x − ω t − φ1 ) +
+ sin(k x − ω t − φ2 ) =
φ1 − φ2
×
= 2 cos
2
φ + φ2
.
× sin k x − ω t −
2
Consequently, the superposition wave (1) can
be written as
−1
y2
θ1 = k x − ω t − φ1 ,
θ2 = k x − ω t − φ2 ,
(1)
of two harmonic waves
0
Explanation:
Applying the trigonometric identity (0) to
we have
y1+2 (x, t) = y1 (x, t) + y2 (x, t)
y1
20
y1+2 = A sin(k x − ω t − φ1 )
+ A sin(k x − ω t − φ2 )
θ1 − θ2
= 2 A cos
×
2
φ + φ2
× sin k x − ω t −
,
2
(5)
which is indeed a harmonic wave of the
form (4) where
ω1+2 = ω = 4.9 s−1 ,
k1+2 = k = 3.9 m−1 ,
φ1 − φ2
A1+2 = 2 A cos
2
= −5.05166 cm,
φ + φ2
= 2.2395 rad.
and φ1+2 =
2
(6)
(7)
(8)
(9)
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Sometimes (as now) eq. (8) yields a negative
amplitude of the superposition wave. In such
cases, we should redefine
A1+2 7→ −A1+2 ,
φ1+2 7→ φ1+2 + π. (10)
Such redefinition results leaves the waveform (4) unchanged, but it gives us a conventionally positive amplitude.
Generally,
θ
−
θ
1
2
A1+2 = 2 A × cos
(11)
2
φ + φ2
(12)
φ1+2 =
2
θ1 − θ2


> 0,
 0 if cos
2 +
θ1 − θ2


< 0.
 π if cos
2
This answers parts 1 through 4 of the problem.
Alternatively, we may solve part 1 by
looking at the graph
+5
Question 25, chap 16, sect 4.
part 2 of 5
10 points
Now consider the snapshot of the superposition wave at time t = 0:
+A1+2
+A
0
−A
−A1+2
0
y1
λ
y2
2λ
y1 +y2
What is the wavelength of this resultant
wave?
Correct answer: 1.61107 m (tolerance ± 1
%).
Explanation:
As explained in part 1, the superposition
wave has the same wave number
A (cm)
k1+2 = k1 = k2 = 3.9 m−1
(7)
as the two original waves, and therefore the
same wavelength
λ1+2 = λ1 = λ2 =
t (seconds)
1.28 s
y2
2T
y1 +y2
and noting the curves y1 (t) and y2 (t) have
equal periods
T1 = T2 =
2π
= 1.28 s
ω
and therefore their sum y1+2 (t) must have
exactly the same period
Alternatively, we may solve part 2 by
looking at the graph
+5
−5
T1+2 = T1 = T2
x (meters)
and hence the same frequency
ω1+2 = ω = 4.9 s−1 .
2π
= 1.61107 m. (12)
k
A (cm)
−5
y1
21
y1
(6)
1.61 m
y2
2λ
y1 +y2
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
and noting the two waveforms y1 (x) and y2 (x)
have are periodic with equal wavelengths
λ1 = λ2 =
2π
= 1.61107 m.
k
and therefore their superposition is also periodic with exactly the same wavelength
λ = λ1 = λ2 = 1.61107 m.
Correct answer: 4.68089 cm (tolerance ± 1
%).
Explanation:
The simplest solution is to evaluate the
separate displacements of the two waves
y1 = A sin(k x − ω t − φ1 )
h
= 3.2 cm sin (3.9 m−1 )(6 m)
−1
− (4.9 s
Question 26, chap 16, sect 4.
part 3 of 5
10 points
Explanation:
As explained in part 1,
φ1 + φ2
2
θ1 − θ2


> 0,
 0 if cos
2 +
θ1 − θ2


< 0,
 π if cos
2
= 5.38109 rad
≡ 5.38109 rad (mod 2π).
i
y2 = A sin(k x − ω t − φ2 )
h
= 3.2 cm sin (3.9 m−1 )(6 m)
− (4.9 s−1 )(6 s) − (−0.241 rad)
= 1.60163 cm,
i
and add them up:
φ1+2 =
y1+2 = y1 + y2 = 4.68089 cm.
(12)
Question 27, chap 16, sect 4.
part 4 of 5
10 points
What is the amplitude A1+2 of the resultant
wave?
Correct answer: 5.05166 cm (tolerance ± 1
%).
Explanation:
As explained in part 1,
φ
−
φ
1
2
A1+2 = 2 A × cos
2
= 5.05166 cm.
)(6 s) − (4.72 rad)
= 3.07926 cm,
What is the initial phase φ1+2 of the resultant wave? (Answer between 0 and 2π.)
Correct answer: 5.38109 rad (tolerance ± 1
%).
22
Alternatively, having evaluated the parameters (6–12) of the resultant wave, we may
substitute them into eq. (4) and directly evaluate
y1+1 = A1+2 sin k1+2 x − ω1+2 t − φ1+2
h
= (5.05166 cm) × sin (3.9 m−1 )(6 m)
i
− (4.9 s−1 )(6 s) − (5.38109 rad)
= 4.68089 cm.
Question 29, chap 18, sect 6.
part 1 of 1
10 points
(11)
Question 28, chap 16, sect 4.
part 5 of 5
10 points
What is the displacement of the resultant
wave at t = 6 s and x = 6 m?
Note: Patm used below is the atmospheric
pressure.
An incompressible, non-viscous liquid of
density ρ flows with speed vz into a pipe of diameter dz . The diameter of the pipe changes
to dy at its exit end. The elevation of the entrance is h above the elevation of the exit end
of the pipe. The pressure at the exit of the
pipe is Py .
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Pz
dz
23
Question 30, chap 17, sect 4.
part 1 of 2
10 points
Py
vz
vy
h
dy
Applying Bernoulli’s principle, what is the
pressure Pz at the entrance end of the pipe?
1
ρ (vz 2 − vy 2 )
2
1
2. Pz = Py − Patm + ρ g h + ρ (vy 2 − vz 2 )
2
1
3. Pz = −ρ g h + ρ (vy 2 − vz 2 )
2
1
4. Pz = Py − ρ g h + ρ (vy 2 − vz 2 ) cor2
rect
An ambulance is traveling north at
50.6 m/s, approaching a car that is also traveling north at 34.1 m/s. The ambulance driver
hears his siren at a frequency of 891 cycles/s.
The velocity of sound is 343 m/s.
50.6 m/s
34.1 m/s
Ambulance
Car
1. Pz = Py + ρ g h +
5. Pz = Py + ρ g h
6. Pz = ρ g h +
1
ρ (vz 2 − vy 2 )
2
7. Pz = Py − ρ g h
8. Pz = Py − Patm − ρ g h +
1
ρ (vz 2 − vy 2 )
2
1
ρ (vz 2 − vy 2 )
2
1
10. Pz = Py + ρ g h + ρ (vy 2 − vz 2 )
2
Explanation:
Applying Bernoulli’s principle to the fluid
flow at the entrance and exit of the pipe gives
9. Pz = Py − ρ g h +
1
1 2
ρ vz = Py + ρ g yy + ρ vy2
2
2
1
Pz = Py + ρ g (yy − yz ) + ρ (vy 2 − vz 2 ) .
2
We also have yy − yz = −h, since the entrance
height yz is greater than the exit height yy .
Therefore
Pz + ρ g y z +
Pz = Py − ρ g h +
1
ρ (vy 2 − vz 2 ) .
2
What is the wavelength at any position in
front of the ambulance for the sound from the
ambulance’s siren?
Correct answer: 0.328171 m (tolerance ± 1
%).
Explanation:
Let : vcar
vamb
vsound
f
= 34.1 m/s ,
= 50.6 m/s ,
= 343 m/s , and
= 891 cycles/s .
By the Doppler effect, the wavelength of the
sound created by a source with rest frequency
f and speed vsource is
λ=
vsound ± vsource
.
f
The wave speed relative to a moving observer
is
′
v = vsound ± vobserver
and the observed frequency is
′
v
f = .
λ
′
Note: The wavelength is specified in the
reference frame of the medium of propagation.
Sound waves always travel at a given speed
with respect to their medium of propagation.
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
Every observer (moving or at rest) will measure the same wavelength for the sound from
the siren, because length measurements will
not depend on the velocity of the measurer.
However, as sources/observers move through
the medium at different velocities, they see
the sound waves move past them at different
velocities. As a result, the number of wavefronts passing them in a given time interval
(i.e., the frequency of the sound) must change.
The wavelength of the sound emitted in
front of the ambulance is
′
vsound − vamb
λ =
f
343 m/s − 50.6 m/s
=
891 cycles/s
= 0.328171 m .
The negative sign arises because the ambulance driver is traveling in the same direction
as these sound waves and therefore perceives
them as being slower than sound waves emitted when the ambulance is at rest. This results in a smaller wavelength; intuitively, the
wavefronts are compressed together by the
motion of the siren.
=
At what frequency does the driver of the
car hear the ambulance’s siren?
Correct answer: 941.279 cycles/s (tolerance
± 1 %).
Explanation:
The car driver sees sound waves with a
wavelength λ overtaking him with a velocity
of vsound − vcar . the negative sign appears
because the car is traveling in the same direction as the sound, making the sound appear
slower. Therefore the car driver hears a sound
with a frequency
′
f =
vsound − vcar
.
′
λ
In terms of the original frequency,
′
f =
vsound − vcar
f
vsound − vamb
343 m/s − 34.1 m/s
(891 cycles/s)
343 m/s − 50.6 m/s
= 941.279 cycles/s .
Question 32, chap 17, sect 3.
part 1 of 2
10 points
The figure below represents a sound wave
in a hollow pipe with both ends open.
ℓ
Determine the wavelength of the sound
wave in this hollow pipe.
2ℓ
5
2ℓ
2. λ =
7
1. λ =
3. λ = ℓ
4. λ =
5. λ =
Question 31, chap 17, sect 4.
part 2 of 2
10 points
24
6. λ =
7. λ =
8. λ =
2ℓ
9
2ℓ
3
ℓ
correct
4
ℓ
3
ℓ
2
9. λ = 2 ℓ
Explanation:
Basic Concepts: When both ends of a
pipe are open, the wavelength is
λ=
2ℓ
,
N
where
N = 1, 2, 3, 4, · · · ,
(1)
where N is the number of nodes.
When one end of a pipe is closed and the
other end open, the wavelength is
λ=
4ℓ
, where
2N − 1
N = 1, 2, 3, · · · , (2)
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
since there is node at one end.
Solution: There are eight nodes (N = 8)
in the air column.
If the number of quarter wavelengths is
even, both ends have either nodes or antinodes; however, if the number of quarter
wavelengths is odd, one end has a node and
the other end has an anti-node.
λ
The number J of quarter wavelengths in
4
the length of the pipe ℓ is J = 16 , since
ℓ
λ
4
J =
or
λ=
4ℓ
4ℓ
=
.
J
16
The number of quarter wavelengths is even.
We will use Eq. 1 to solve for the number of
nodes.
2ℓ
,
N
2ℓ
=
8
ℓ
.
=
4
λ=
where
N =8
Question 33, chap 17, sect 3.
part 2 of 2
10 points
Consider another organ pipe which has one
end open and one end closed.
4ℓ
correct
15
4ℓ
7. λ =
7
6. λ =
8. λ = 4 ℓ
4ℓ
17
Explanation:
Solution: There are eight nodes (N = 8)
in the air column.
If the number of quarter wavelengths is
even, both ends have either nodes or antinodes; however, if the number of quarter
wavelengths is odd, one end has a node and
the other end has an anti-node.
λ
The number J of quarter wavelengths in
4
the length of the pipe ℓ is J = 15 , since
9. λ =
J =
1. λ =
2. λ =
3. λ =
4. λ =
5. λ =
4ℓ
3
4ℓ
11
4ℓ
13
4ℓ
9
4ℓ
5
ℓ
λ
4
or
λ=
4ℓ
4ℓ
=
.
J
15
The number of quarter wavelengths is odd.
We will use Eq. 2 to solve for the number of
nodes.
4ℓ
,
2N −1
4ℓ
=
2 (8) − 1
λ=
=
ℓ
Determine the wave length of the sound
wave in this hollow pipe.
25
where
N =8
4ℓ
.
15
Question 34, chap 9, sect 4.
part 1 of 1
10 points
When considering only the Sun, Earth,
Mercury, and Mars in a planetary system,
which statement is correct?
1. Mars goes around the Sun in the opposite
angular direction from Mercury.
2. The Mars and Mercury go around the
Earth. correct
3. Mars goes around the Earth but Mercury
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
doesn’t go around the Earth since its orbit is
smaller.
4. At night the Earth stops until morning
and then goes around the Sun during daytime
only.
5. The Sun, Earth, Mars, and Mercury all
go around each other with the same angular
momentum.
6. At night the Sun stops until morning and
then goes around the Earth during daytime
only.
Explanation:
Basic Concept: Kepler’s third law states
that the orbital period squared T 2 is porpotional to the semi-major axis cubed R3 ; i.e.,
T 2 = K R3 .
For convenience let us assume an imaginary
solar system and choose orbits for the planets
whose periods are integral multiples of each
other, for example
RM ercury = 1.9842511315 ≈ 2 ,
REarth = 5 ,
RM ars = 7.903700526 ≈ 8 ,
s
3
TM ercury
REarth
=
= 4,
TEarth
RM ercury
s
3
RM ars
TEarth
=
= 2.
TM ars
REarth
The upper diagram shows a Sun concetric
diagram of a solar system. This diagram
was proposed by Galileo Galilei (an Italian
scientist 1564-1642).
The lower diagram shows an Earth concentric diagram of the same solar system. This
diagram was proposed by Tycho Brahe (a
Danish astronomer 1546-1601).
Both diagrams mathematically describe the
planet’s orbits.
Figure:
In the Earth concentric (lower) diagram, the parametric
equations for the Sun’s (circular),
Mars’, and Mercury’s orbits are
xSun = REarth cos θ ,
ySun = REarth sin θ ,
θ
xM ars = RM ars cos − 5 cos θ ,
2
θ
yM ars = RM ars sin − 5 sin θ ,
2
xM ercury = RM ercury cos 4 θ − 5 cos θ ,
yM ercury = RM ercury sin 4 θ − 5 sin θ .
The solid and dashed line segments
in the lower diagram indicate time
periods equal to 60◦ arcs of the
planet Earth in the upper diagram.
Note: Since the coordinate transformation from the Sun’s concentric
26
oldfinal 01 – JYOTHINDRAN, VISHNU – Due: May 10 2007, 4:00 am
diagram to the Earth’s concentric diagram involves an angular transformation, the number of revolutions
Mercury makes around the Sun for
each of the Earth’s revolutions is 3
not 4, as one might nievely expect.
A coordinate system can be placed at the
center of mass of the solar system (upper, circular orbits) or a coordinate system can be
placed at the Earth (lower, either epicycloid
(Mars’ orbit) or hypocycloid (Mercury’s orbit).
That is, both the Sun and Earth rotate
around each other. Their motion is relative.
The center of mass of the solar system is
near the surface of the Sun; consequently, the
most convenient picture of the Solar System
is the upper diagram.
Note:
The teeth on gears are patterned on epitrochoids meshing with hypotrochoids. This notion was instigated by Tacho
Brahe’s work and has revolutionized automobile transmissions.
Question 35, chap 18, sect 6.
part 1 of 1
10 points
A jet of water squirts out horizontally from
a hole on the side of the tank as shown below.
h
1.9 m
0.51 m
Figure: Not drawn to scale.
If the hole has a diameter of 3.41 mm, what
27
is the height “h” of the water above the hole
in the tank?
Correct answer: 3.42237 cm (tolerance ± 1
%).
Explanation:
Given :
∆x = 0.51 m ,
∆y = −1.9 m , and
D = 3.41 mm .
Consider the motion of the water after leaving the tank. Vertically,
1
∆y = viy ∆t − g (∆t)2
2
1
= − g (∆t)2
2
since viy = 0 m/s, so
s
−2 ∆y
.
∆t =
g
Horizontally,
∆x = vx ∆t
since ax = 0 m/s , so
r
g
∆x
.
= ∆x
vx = r
−2 ∆y
−2 ∆y
g
If point 1 is at the top of the tank and point
2 at the level of the hole, with
P1 = P2 = Patm and v1 ≈ 0,
1
ρ g y1 = ρ vx 2 + p g y2
2
since v1 = vx .
1
ρ g (y1 − y2 ) = ρ vx 2
2
vx 2
(y1 − y2 ) =
2g
(∆x)2
h=
2 g (∆t)2
g
(∆x)2
=
2g
−2 ∆y
(∆x)2
=
−4 ∆y
100 cm
(0.51 m)2
=
−4 (−1.9 m)
1m
2
= 3.42237 cm .