Version 036 – Test 4 – swinney – (58535)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A cue ball initially moving at 2.8 m/s strikes
a stationary eight ball of the same size and
mass. After the collision, the cue ball’s final
speed is 1.6 m/s .
2.8 m/s
v1
θ
v
v2
This also means that ~v forms the hypotenuse of a right triangle, so
s
θ + φ = 90◦
m/
1. 6
Before
1
θ
cos θ =
φ
After
Find the cue ball’s angle θ with respect to
its original line of motion. Consider this to
be an elastic collision (ignoring friction and
rotational motion).
1. 63.9584
2. 60.0
3. 61.4281
4. 58.6677
5. 67.3801
6. 66.1238
7. 57.5485
8. 63.1528
9. 55.1501
10. 68.0945
v1
v
θ = arccos
v 1
v
= arccos
~ = −ı̂ F1 − ̂ F2 + k̂ F3
F
acts at the upper end of the rod.
What is the initial torque ~τ about the lower
end of the rod?
1. (ı̂ F3 + k̂ F1 ) ℓ correct
2. −(ı̂ F3 + ̂ F2 + k̂ F1 ) ℓ
Explanation:
4. (ı̂ F3 − k̂ F1 ) ℓ
5. (−ı̂ F1 + ̂ F2 − k̂ F3 ) ℓ
6. ı̂ F3 ℓ
7. −ı̂ F3 ℓ
8. (−ı̂ F3 + k̂ F1 ) ℓ
9. (ı̂ F1 + ̂ F2 + k̂ F3 ) ℓ
1
1
1
m v 2 = m v12 + m v22
2
2
2
2
2
2
v = v1 + v2
1.6 m/s
2.8 m/s
002 10.0 points
A rod is initially along the y axis and has
length ℓ. A force
3. Zero
KE is conserved, so
= 55.1501◦ .
Correct answer: 55.1501 ◦.
Let : v = 2.8 m/s and
v1 = 1.6 m/s .
and
10. (ı̂ F1 + k̂ F3 ) ℓ
Explanation:
Version 036 – Test 4 – swinney – (58535)
~r = ̂ ℓ, so the torque is
~
~τ = ~r × F
= (̂ ℓ) × (−ı̂ F1 − ̂ F2 + k̂ F3 )
= −(̂ × ı̂) ℓ F1 − (̂ × ̂) ℓ F2 + (̂ × k̂) ℓ F3
= k̂ ℓ F1 + ı̂ ℓ F3
003
10.0 points
2
by
I =
1
M R2
2
The magnitude of angular momentum is
L = Iω =
1
M R2 ω
2
The magnitude of the torque is given by
p
τ = R2 + d2 F sinθ
and the value of sinθ is given by
sinθ = √
A uniform solid disk with radius R and
mass M is mounted on a fixed axle passing
through its center (which is not shown). A
constant force F is applied as shown. At the
instant shown, the angular velocity of the disk
ω is in the −ẑ direction (where +x̂ is to the
right, +ŷ is up, and +ẑ is out of the page,
toward you). The length of the string is d.
The angular momentum, torque and kinetic
energy of the disk are respectively:
1
1
1. M R2 ωẑ, −RF ẑ, M R2 ω 2
2
4
1
1
2. M R2 ωẑ, RF ẑ, − M R2 ω 2
2
4
1
1
3. M R2 ωẑ, RF ẑ, M R2ω 2
2
4
1
1
4. M R2 ωẑ, −RF ẑ, − M R2ω 2
2
4
1
1
5. − M R2ωẑ, −RF ẑ, M R2ω 2 correct
2
4
1
1
6. − M R2ωẑ, −RF ẑ, − M R2 ω 2
2
4
1
1
7. − M R2ωẑ, RF ẑ, M R2 ω 2
2
4
1
1
8. − M R2ωẑ, RF ẑ, − M R2ω 2
2
4
Explanation:
The direction of both L and thtat of torque
are along the −ẑ direction.
The moment of inertia of the disk is given
R
R 2 + d2
Thus, the value of torque is
τ = RF
The kinetic energy is given by
K =
1
1
I ω 2 = M R2 ω 2
2
4
004 10.0 points
A uniform bar of mass M and length ℓ is
propped against a very slick vertical wall as
shown. The angle between the wall and the
upper end of the bar is θ. The force of static
friction between the upper end of the bar and
the wall is negligible, but the bar remains at
rest (in equilibrium).
~w
F
θ
~n
~fs
If we take the pivot at the point where the
barX
touches
the floor, which expression below
is
~τ , where x is along the floor and y
z
is along the wall?
−M g
1. ℓ
cos θ − Fw sin θ = 0
2
Mg
2. ℓ Fw sin θ −
cos θ = 0
2
Version 036 – Test 4 – swinney – (58535)
3. ℓ (−fs sin θ − n cos θ) = 0
Mg
4. ℓ
+ Fw = 0
2
Mg
sin θ = 0
5. ℓ Fw cos θ +
2
Mg
sin θ = 0 correct
6. ℓ Fw cos θ −
2
Mg
− Fw = 0
7. ℓ
2
Mg
8. ℓ Fw sin θ +
cos θ = 0
2
9. ℓ (fs sin θ − n cos θ) = 0
Mg
cos θ − Fw sin θ = 0
10. ℓ
2
Explanation:
Taking the pivot at the point where the bar
touches the floor, the torque into the paper is
M g ℓ sin θ
and the torque out of the paper is
2
Fw ℓ cos θ, so
X
Mg
sin θ = 0 .
~τz = ℓ Fw cos θ −
2
005 10.0 points
A uniform plank of length 2 L is balanced at
its center of mass. A mother with a mass of
50 kg sits at a distance L/5 from the center of
mass and is balanced with her child sitting at
the opposite end of the seesaw on the end of
the plank.
What is the mass of the child?
1. 20 kg
2. 10 kg correct
3. 5 kg
r1 = L/5 ,
r2 = L ,
and
In order for the system to be in static equilibrium, the sum of the torques must be zero,
r1 m1 g = r2 m2 g ,
so the mass of the child is
r1
1
m2 =
(50 kg) = 10 kg .
m1 =
r2
5
006 10.0 points
Using the Einstein model of a solid, what is
the change in entropy when adding 3 quanta
of energy to a system of 5 atoms that already
has 6 quanta of energy stored in it?
23 · 11
1. ∆S = kb ln
correct
12
23!
2. ∆S = kB ln
14! · 6!
24
3. ∆S = kB ln
5
23!
4. ∆S = kB ln
7! · 6!
5. ∆S = kB ln (10)
23!
6. ∆S = kB ln
9! · 6!
14!
7. ∆S = kB ln
9! · 7
10
8. ∆S = kB ln
7
23!
9. ∆S = kB ln
14! · 9!
23
10. ∆S = kB ln
9·6
Explanation:
Entropy is defined as S = kB ln Ω where Ω
is the number of possible microstates, which
is the number of ways to arrange q quanta in
n harmonic oscillators, i.e.
4. 15 kg
5. 25 kg
Explanation:
Let :
3
m1 = 50 kg ,
Ω=
(n − 1 + q)!
(n − 1)! · q!
Version 036 – Test 4 – swinney – (58535)
Here, n − 1 = 3 · 5 − 1 = 14 and initially
qi = 6, but in the final state qf = 9. Thus,
the change in entropy is
∆S = Sf − Si
(14 + 9)!
(14 + 6)!
= kb ln
− kb ln
14! · 9!
14! · 6!
 (14+9)! 
= kb ln  14!·9!
(14+6)!
14!·6!

23 · 22 · 21
= kb ln
9·8·7
23 · 11
= kb ln
3·4
23 · 11
= kb ln
12
4
gr
2R
gr
6. |a| =
r+R
gR
7. |a| =
r
2
R
8. |a| = g
r
g R2
9. |a| = 2
r + R2
2 g r2
10. |a| =
correct
2 r 2 + R2
Explanation:
We will take the upward direction as the
positive y direction .
5. |a| =
T
007 10.0 points
A yo-yo consists of two identical, uniform
disks whose total mass is M and radius R
connected between the disks by a shaft of
radius r and negligible mass. The moment of
1
inertia of the yo-yo is I = M R2 . See the
2
figure below.
y
θ
W =Mg
Looking at our free body diagram above,
we see that the net force on the yo-yo is
Fnet = T − M g = −M a .
R
Similarly, the net torque on the yo-yo about
the center of mass is
ω
r
What is the magnitude of the acceleration
|a| of the center of mass of the yo-yo as it is
falling downward with the string unwinding
and not slipping?
r
g r
1. |a| =
2 R
r !
R
2. |a| = g 1 +
r
2gr
3. |a| =
2r +R
r
4. |a| = g 1 +
R
τnet = −T r sin(90◦ )
= −T r
= I α,
To relate the rotational and translational
motion we have the fact that the string does
not slip as it unrolls from the yo-yo. This is
expressed in the no-slip condition, −a = r α .
Substituting for α, we have
T =
Ia
.
r2
We can now substitute the tension T into
our equation for the acceleration.
T = M g − M a = M (g − a) ,
so
Version 036 – Test 4 – swinney – (58535)
Ia
r2
Ia
g=
+a
2
M r
I
= 1+
a
M r2
"
2 #
1 R
a,
= 1+
2 r
M (g − a) =
a=
so
2 g r2
.
2 r 2 + R2
008
10.0 points
5
Explanation:
Since block A has more oscillators than
block B, A will have more quanta than B at
thermal equilibrium. This situation is similar
to the left side of the graph in text figure
12.26. As block B transfers quanta to block
A, block B’s entropy decreases while block A’s
entropy increases. But block A gains more
entropy than block B loses, so the system
increases in entropy overall, as required by
the second law of thermodynamics.
009 10.0 points
Two masses undergo a front-to-back collision.
The masses are m1 = m, with initial velocity
2v0 , and m2 = 4.7m, with initial velocity
v0 . Due to the collision, they stick together,
forming a compound system.
2v0
m1
Blocks A and B have 300 and 200 oscillators, respectively. The blocks are placed in
contact with each other in an insulated environment. At the instant the blocks touch,
block A has 10 quanta of energy while block
B has 90 quanta of energy. What will happen
some time after contact?
I. The entropy of block A will increase.
II. The entropy of block B will increase.
III. The entropy of the two block system will
increase.
1. I only
2. I,III only correct
3. II only
4. I,II,III
5. I,II only
6. III only
7. II,III only
8. None
v0
m2
If m = 0.83 kg and v0 = 4.6 m/s, find the
magnitude of the loss in kinetic energy after
the collision.
1. 7.24081
2. 36.5533
3. 32.6157
4. 61.4017
5. 55.3524
6. 28.3024
7. 21.375
8. 4.73882
9. 37.5497
10. 8.82867
Correct answer: 7.24081 J.
Explanation:
Basic Concepts:
Conservation of momentum:
pf = pi
Kinetic energy:
1
K = mv 2
2
Conservation of kinetic energy does not hold
for inelastic collisions. The initial kinetic energy of mass m1 is
1
1
K1i = m1 v12 = m(2v0 )2 = 2mv02
2
2
Version 036 – Test 4 – swinney – (58535)
and for mass m2 we find
1
1
K2i = m2 v22 = (4.7m)(v0 )2
2
2
so the total initial kinetic energy is
Ktot,i = K1 + K2 =
4.35mv02
What is the magnitude of the acceleration
downward of the suspended mass when the
system is released?
1.
Both masses are going the same way, pick
that direction as positive. The initial total
momentum of the system is
3.
ptot,i = m1 v1 + m2 v2
= m(2v0 ) + 4.7m(v0 ) = 6.7mv0
4.
Conservation of momentum tells us that
2.
5.
ptot,f = ptot,i = 6.7mv0
The relation between energy and momentum
is
1
1
p2
K = mv 2 =
(mv)2 =
2
2m
2m
Thus, since we know ptot,f = 6.7mv0 and the
mass of the compound system is 5.7m,
Ktot,f =
(6.7mv0 )2
22.445mv02
=
2(5.7m)
5.7
Finally, the loss in kinetic energy is
22.445
mv02
Ktot,i − Ktot,f = 4.35 −
5.7
= 7.24081 J
010 10.0 points
A mass m is held at rest on a horizontal
table. The friction coefficient between the
3
mass and the table is µk = . A massless
4
string connected to m passes horizontally over
a disk-shaped pulley that has radius R, mass
2 m, and inertia I = m R2 . The string wraps
3
around the pulley and a mass m hangs from
2
it.
6
6.
7.
8.
9.
10.
5
g.
7
5
g.
14
1
g.
2
2
g.
7
4
g.
5
3
g.
10
3
g.
7
2
g.
5
4
g.
7
3
g. correct
14
Explanation:
Let the direction from m toward the pulley
be positive. Let T1 be the tension in the
string between m and the pulley, and T2 be
the tension in the string between the pulley
and mass 3m. Then apply the momentum
principle to m and 3m
3
m g = ma
4
3
3
(2) m g − T2 = m a
2
2
(1) T1 −
Apply the angular momentum principle to
the pulley,
m
µk
dL
dt
=Iα
a
=I
R
τ=
3/2m
Version 036 – Test 4 – swinney – (58535)
a
which becomes R (T2 − T1 ) = m R2 ,
R
which can be written
(3) T2 − T1 = ma
Adding the three equations, (1)+(2)+(3),
we have
−3
3
3
m g + mg = (m + m + m) a
4
2
2
which is, after cancelling the common factor
m
3
7
g= a
4
2
Hence the acceleration is
3
g.
14
011 10.0 points
A spool (similar to a yo-yo) is pulled in three
ways as shown. There is sufficient friction for
rotation. Note that in for spool b the cord is
attached to the inner radius and the tension
force is aligned with the point of contact, as
indicated by the dashed line.
a
b
c
In (a) there is a clockwise torque about the
point of contact with the table, so the spool
rolls to the right.
In (b) the line of action extends through
the point of table contact, yielding no lever
arm and therefore no torque; with a force
component to the right, the spool slides to
the right without rolling.
In (c) the torque produces clockwise rotation so the spool rolls to the right.
012 10.0 points
A child is initially sitting near the outer
rim of a revolving merry-go-round. Suddenly,
the child moves towards the center of the
merry-go-round (while it is still revolving).
For the merry-go-round+child system, let the
symbols L and K refer to the magnitude of
the angular momentum (about the center of
the merry-go-round) and rotational kinetic
energy, respectively.
Consider the following statements:
Ia. L is constant
Ib. L increases
Ic. L decreases
IIa. K is constant
IIb. K increases
IIc. K decreases
Which of these statements are true?
1. Ib, IIc
In what direction will each spool move (in
the order spool a, spool b, spool c)?
2. Ia, IIb correct
1. None of these
3. Ia, IIa
2. right; left; left
4. Ib, IIb
3. right; right left
5. Ib, IIa
4. left; right; right
6. Ia, IIc
5. right; right; right correct
7. Ic, IIb
6. right; left; right
8. Ic, IIa
Explanation:
7
Version 036 – Test 4 – swinney – (58535)
9. Ic, IIc
Explanation:
Since there is no external torque acting on
the system, the angular momentum is conserved. The kinetic energy of the system is
given by
L2
K =
.
2I
The moment of inertia decreases when the
child moves closer to the center so the kinetic
energy increases. Thus, the correct answer is
Ia, IIb, or choice 2.
013 10.0 points
A person spins a tennis ball on a string in a
horizontal circle in the x −y plane as shown in
the image. The ball is moving clockwise when
viewed from above. Just when the string is
parallel with the y-axis, the ball is hit with a
force in the direction of the ball’s velocity.
z
Consider the ball as the system. The applied force is tangential to velocity so the
direction of the torque about the center of the
horizontal circle caused by this force is in the
−z direction. The change in angular momentum is always aligned with the net torque on
~ is also in the −z direction.
the system so ∆L
014 10.0 points
A sphere of radius R and mass M is held
on a ramp that makes an angle θ with the
horizontal by a string that pulls with force F
directed up the ramp and running over the
top of the sphere.
~F
R
Mg
~v
This causes a change in the ball’s angular
~ in the
momentum ∆L
1. +x direction
2. −z direction correct
3. The angular momentum does not change
(zero torque).
4. +y direction
θ
If the magnitude of this force is as small as
it can be and still hold the sphere at rest on
the ramp, what must be the value of µs , the
coefficient of static friction between sphere
and ramp?
1.
6. +z direction
7. −x direction
Explanation:
2
tan θ
2. tan 2 θ
3. 1.0 precisely
4.
1
2 tan θ
5. Zero; no static friction is needed in this
case.
tan θ
correct
2
sin θ
7.
tan θ
cos θ
8.
tan θ
6.
5. −y direction
fs
n
y
x
8
Version 036 – Test 4 – swinney – (58535)
9.
9
1
2
10. tan θ
φ
Explanation:
Let x be parallel to the ramp and y perpendicular to it.
X
b
b
ℓ
xcm
T
Fx = F + fs − M g sin θ = 0 and
X
Fy = n − M g cos θ = 0 .
Taking torques about the point where the
sphere touches the ramp,
X
If the mass of the beam is 8 kg, the tension
4
3
in the wire is 40 N, sin φ = and cos φ = ,
5
5
how far is the center of mass of the beam from
the hinge? The acceleration due to gravity is
10 m/s2 .
τz = (M g sin θ) R − 2 F R = 0
1. 0.375 ℓ
F =
2. Not enough information is given.
M g sin θ
2
to keep the sphere from rotating.
In order for F to be a minimum, fs must be
a maximum, so
3. 0.625 ℓ
4. 0.18 ℓ
5. 0.3 ℓ
fsmax = µs n = µs M g cos θ
and
6. 0.32 ℓ
7. 0.4 ℓ correct
F min = M g (sin θ − µs cos θ)
M g sin θ
= M g (sin θ − µs cos θ)
2
tan θ
.
µs =
2
015 10.0 points
A horizontal, nonuniform beam of mass M
and length ℓ is hinged to a vertical wall at
one side, and attached to a wire on the other
end. The bar is motionless and a wire exerts
a force T at an angle of φ with respect to the
vertical.
8. 0.666667 ℓ
9. 0.16 ℓ
10. 0.5 ℓ, of course
Explanation:
Consider the free-body diagram, with the
forces due to the wall not shown:
Version 036 – Test 4 – swinney – (58535)
5.
6.
φ
b
b
xcm
ℓ
T
m~g
7.
8.
9.
10.
With the z-axis pointing out of the paper,
the magnitude of the net torque through the
hinge is
τz = T ℓ cos φ − m g xcm = 0
m g xcm = T ℓ cos φ
T cos φ
ℓ
xcm =
mg
4
(40 N)
5
= 0.4 .
=
(8 kg)(10 m/s2 )
016 10.0 points
A bullet of mass 2 m moving with velocity
v strikes tangentially the edge of a spoked
wheel of radius R, and the bullet sticks to the
edge of the wheel. The spoked wheel has a
mass 7 m concentrated on its rim (neglect the
mass of the spokes). The wheel, initially at
rest, begins to rotate about its center, which
remains fixed on a frictionless axle.
What is the angular velocity of the spoked
wheel after the collision?
2
5
3
2.
8
2
3.
7
4
4.
9
1.
v
R
v
R
v
R
v
R
10
1 v
4 R
4 v
11 R
1 v
3 R
3 v
10 R
5 v
7 R
2 v
correct
9 R
Explanation:
The total angular momentum of the system is conserved since there is no external
force. The total angular momentum before
the collision is
Lbef ore = R (2 m) v
and after the collision
Laf ter = Lbullet + Lwheel
= (2 m) R2 ω + (7 m) R2 ω
= (9 m) R2 ω ,
so from conservation of angular momentum,
Lbef ore = Laf ter
R (2 m) v = (9 m) R2 ω
2 v
ω=
.
9 R
The total kinetic energy is not conserved because the collision is not elastic.
017 10.0 points
A uniform rod of mass 1.7 kg is 2 m long. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 2 m from the center of mass
of the rod. The rod is released from rest at
an initial angle of 69 ◦ with respect to the
horizontal, as shown.
Version 036 – Test 4 – swinney – (58535)
11
I = ICM + m d2 =
1
13
m ℓ2 + m ℓ2 =
m ℓ2 .
12
12
Since the rod is uniform, its center of mass
is located at ℓ . The weight m g acts at the
center of mass, so the magnitude of the torque
is
69◦
1.7 kg
O
2m
2m
What is the magnitude of the vertical component of the force of the pivot on the rod at
the instant the rod is in a horizontal position?
The acceleration due to gravity is 9.8 m/s2 .
The moment of inertia of the rod about its
1
center of mass is
m ℓ2 .
12
1. 2.71385
2. 0.678462
3. 2.11077
4. 1.96
5. 2.86462
6. 1.28154
7. 0.753846
8. 3.31692
9. 1.43231
10. 0.603077
Correct answer: 1.28154 N.
τ =r×F =Iα
13
m ℓ2 α = ℓ m g sin β = ℓ m g
12
12 g
α=
.
13 ℓ
The vertical component of the tangential
acceleration is
12
12 g
ℓ=
g
ay = α r =
13 ℓ
13
and the sum of the forces acting on the rod in
the vertical direction is
Fy − m g = −m ay ,
so the vertical reaction force Fy on the pivot
is
12
g
Fy = m (g − ay ) = m g −
g =m
13
13
2
9.8 m/s
= (1.7 kg)
13
= 1.28154 N .
Explanation:
Let : ℓ = 2 m ,
d = ℓ = 2 m,
θ = 69◦ ,
m = 1.7 kg , and
β = 90◦ .
018 10.0 points
A sticky ball collides with a rod that can rotate on a frictionless axle through the middle
of the rod, as shown. The axle is fixed to the
ground and the rod is initially at rest.
Fy
Fx
O
β
~p
mg
Consider the system of the sphere and the
rod, and let E be the mechanical energy of
~ the linear momentum of the
the system, P
Version 036 – Test 4 – swinney – (58535)
~ the angular momentum of the
system, and L
system.
What is conserved after the collision?
12
9. 38.5714
10. 30.9091
Correct answer: 38.5714 rad.
1. E only
Explanation:
~ only correct
2. L
Let : r = 0.28 m ,
∆vcm = 8 m/s , and
∆t = 2.7 s .
~ only
3. P
~ and P
~
4. L
The acceleration of the center of mass acm
and the angular acceleration of the wheel α
are related by α = acm /r.
~ P
~ , and E
5. L,
~ and E
6. P
1 ∆vcm
r ∆t
8 m/s
1
=
0.28 m 2.7 s
= 10.582 rad/s2 .
~ and E
7. L
α=
Explanation:
The mechanical energy of the system is not
conserved because this is an inelastic collision.
Likewise, the linear momentum is not conserved because the axle applies a net force on
the system. However, there is no net torque
on the system from the axle so the angular
momentum is conserved.
019 10.0 points
A bicycle wheel of radius 0.28 m rolls down
a hill without slipping. Its center of mass
velocity ~vcm increases at a constant rate from
0 to 8 m/s in 2.7 s.
~vcm
Through what angle does the wheel turn in
the 2.7 s?
1. 28.0645
2. 31.25
3. 28.9286
4. 33.871
5. 26.8966
6. 25.1613
7. 24.2188
8. 34.4828
Using the standard rotational kinematics
equation gives:
1 2
αt
2
1
= (10.582 rad/s2 ) (2.7 s)2
2
= 38.5714 rad .
θ=
020 10.0 points
The motor driving a grinding wheel with a
rotational inertia of 0.79 kg · m2 is switched
off when the wheel has a rotational speed
of 36 rad/s. After 12 s, the wheel has slowed
down to 28.8 rad/s due to a constant frictional
force.
What is the magnitude of the torque exerted by frictional force?
1. 0.109677
2. 0.255273
3. 0.29878
4. 0.474
5. 0.228167
6. 0.513333
7. 0.316522
8. 0.55
9. 0.526024
Version 036 – Test 4 – swinney – (58535)
10. 0.145753
Correct answer: 0.474 N. m.
Explanation:
We have
τ ∆t = ∆L = ∆(I ω) ,
so that
I |ω1 − ω0 |
∆t1
(0.79 kg · m2 ) (36 rad/s − 28.8 rad/s)
=
12 s
= 0.474 N · m .
|τ | =
13