# For all problems, use g=10m/s The correct answers are indicated in .

```For all problems, use g=10m/s2
The correct answers are indicated in blue italics.
16. (1 point) In which of the following solid objects of uniform density does the center of
mass NOT lie within the mass of the body?
A. A cube
B. A doughnut
C. A sphere
D. A cylinder
E. None of the above
A simple examination of the symmetry should be enough to give the answer to this one.
17. (1 point) An 80 kg man standing on a frictionless surface kicks forward a 0.04 kg
stone lying at his feet so that it acquires a velocity of +4.0 m/s. What velocity does the
man acquire as a result?
A. +0.002 m/s
B. +0.09 m/s
C. -0.002 m/s
D. -0.09 m/s
E. None of the above.
Momentum is conserved, and the initial momentum is zero, so
p f = 0 = mman vman + mstone vstone . Solving this for vman and plugging in the numbers gives
18. (1 point) Two masses of 1 kg are placed at x=4, y=5 and x=-4, y=5 and a 3 kg mass
is placed at the origin, as shown in the figure below. Where on the Y-axis will the center
of mass lie?
Y
1 kg (4,5)
1 kg (-4,5)
X
3 kg (0,0)
A. y=1
B. y=2
Physics 221 2005 Su Exam 2
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C. y=3
D. y=4
E. y=5
First take the CM of the two 1 kg masses. It appears like a mass of 2 kg at the point
(0,+5). Then, combining this with the 3 kg mass, we find that the overall CM is at y=2,
x=0.
19. (1 point) Two masses collide as shown in the figure below. Their masses and
velocities before and after the collision are as shown. Which of the following choices
describes the collision?
10 m/s
Before:
-3 m/s
2.0 kg
0.5 kg
After:
-10.8 m/s
0.5 kg
2.2 m/s
2.0 kg
A. Inelastic
B. Explosive
C. Elastic
D. Not enough information
The quick way to get the answer is to observe that the relative velocities are unchanged,
which is one of the characteristics of an elastic collision. Alternatively, you can calculate
the kinetic energies before and after the collision and see that they are the same.
20. (1 point) What is the velocity of the center of mass of the particles in question 19?
A. -0.4 m/s
B. 0.4 m/s
C. -1.0 m/s
D. 1.0 m/s
E. None of the above
This can be calculated from
v cm = p tot / M tot = (0.5kg )(10m / s ) + (2.0kg )(−3.0m / s ) /(0.5kg + 2.0kg )
So, vcm = (5.0kgm / s − 6.0kgm / s ) / 2.5kg = −0.4m / s
21. (1 point) Which of the following kinds of motion can an ideal rigid body NOT
execute?
A. Rotation
B. Acceleration of the center of mass in response to an external force
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C. expansion and contraction
D. None of the above
Expansion and contraction are changes in shape which are not a property of ideal rigid
bodies.
22. (1 point) Four balls of mass 2m, 2m, 3m, and 4m are located at the corners of a
square of side L as shown in the figure below. Which of the marked axes has the
smallest moment of inertia?
G
C
2m
F
3m
L
A
B
E 2m
4m
D
H
A. AB
B. CD
C. EF
D. GH
E. Both AB and CD
By calculating the moments of inertia for each axis we find
I AB = I CD = 114 mL2
I EF = 3mL2
I GH = 52 mL2
So, IGH is the smallest.
23. (1 point) A string is wound around a cylinder which is mounted horizontally on a
frictionless shaft. The radius of the cylinder is 0.40 m and its mass is 0.4 kg. A mass of 3
kg is tied to the rope and allowed to fall. What is the acceleration of the falling block?
A. 9.4 m/s2 upward
B. 9.4 m/s2 downward
C. 10.7 m/s2 upward
D. 10.7 m/s2 downward
E. None of the above
Use N’s 2nd law on the falling block to get ma=T-mg, where m is the mass of the block
(3.0 kg) and T is the tension in the string. Use the angular form of N’s 2nd on the rotating
Physics 221 2005 Su Exam 2
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1
a
MR 2 and α = because there is no slipping. From
2
R
the definition of torque we also have τ = − RT , where the minus sign comes from the
relative directions of R and T. Solve the two equations
T = m(a + g ) and
cylinder to get τ = Iα , where I =
1 MR 2 a
1
= − Ma
2 R R
2
for a and put in the numbers to get answer B.
T =−
24. (1 point) In the problem above, what is the total kinetic energy after the weight has
fallen 3.0 m from rest?
A. 77.4 kgm2/s2
B. 72.0 kgm2/s2
C. 90.0 kgm2/s2
D. 5.6 kgm2/s2
E. Insufficient information to solve
Since energy is conserved (there is no loss due to friction) the total kinetic energy is just
the change in potential energy, or mgh = (3.0kg )(10m / s 2 )(3.0m) = 9kgm / s 2 .
25. (1 point) Two gears are connected by a chain. The radii of the gears are 0.3m and
0.5m. If the smaller gear is rotating at a constant angular velocity of 15 rad/s, what is the
angular velocity of the larger gear?
E. None of the above
The linear speed must be the same at the point of contact of the two gears, so the angular
speed goes inversely with the radius. Hence
R
0.3m
ω2 = 1 ω1 =
R2
0.5m
26. (2 points) A uniform 2 kg plank of length 1.5 meters rests on two supports, A and B,
as shown below. Two upward forces are applied, a 7 Nt force at 40 cm from A and a 12
Nt force 1.2 m from A. At which point or points will the plank lift off the supports?
Physics 221 2005 Su Exam 2
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1.5 m
2.0 Kg
A
B
7 Nt
12 Nt
A. A
B. B
C. Both A and B
D. Neither A nor B
You must solve for the net torques about points A and B and see if either is enough to lift
the plank at the other end. We can neglect the normal forces because of the following: at
the pivot end the normal force contributes nothing to the torque and at the far end it
contributes nothing if the torque of the remaining forces is enough to lift the plank. We
must add the weight of the plank, which gives a downward force of 20N at a position of
0.75m. We use the usual convention of counterclockwise being positive. So the torques
τ A = (0.4m)(7 N ) − (0.75m)(20 N ) + (1.2m)(12 N ) = +0.2 Nm
τ B = (−0.3)(12 N ) + (0.75m)(20 N ) − (1.1m)(7 N ) = +3.7 Nm
In both cases we have a positive torque. For pivoting about A this is what we want to lift
the plank of B, but for pivoting about B a positive torque drives the plank harder against
A. Thus only B lifts up.
27. (2 points) A hoop of radius 3 m has a mass of 120 kg. It rolls along a horizontal floor
so that its center of mass has a speed of 0.20 m/s. How much work has to be done to stop
it?
A. 2.4 J
B. 3.6 J
C. 4.8 J
D. 9.6 J
E. None of the above.
1 2 1 2
v
I ω + mv . Since I = mR 2 and ω =
this becomes
2
2
R
W = mv 2 = (120kg )(0.2m / s ) 2 = 4.8 J . Note that the answer does not depend on the
Use W = ∆KE =
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28. (2 points) A 15 kg sign is supported by a wire at an angle of 30o, as shown in the
figure. The sign starts 0.5 m from the wall and has a length of 2.5 m. The point of support
against the wall cannot slip. What is the tension in the cable?
30o
2.5 m
0.5 m
A. 175 Nt
B. 300 Nt
C. 260 Nt
D. 150 Nt
E. Not enough information is given.
Drawing the free body diagram for this gives the figure below.
T
Fh
o
30
N
mg
Tsin30o
Tcos30o
Fh is the force parallel to the wall exerted by the pivot (or hinge) holding the bar to the
wall. N is the normal force exerted by the wall on the bar, mg is the weight of the sign,
and T is the tension in the wire, which is resolved into tits vertical and horizontal
components in the figure. If we choose our axis to be about the point of support, at the
left end of the bar, we only need the torque equation to solve the problem; the vertical
and horizontal force equation would give us N and Fh in addition, but we don’t need
these.
∑τ = 0 = −(15.0kg )(10m / s 2 )(1.75m) + T sin 30o (3.00m) , which solves to
T=175 Newtons.
29. (2 points) A disk of mass M and radius R spins on a shaft with an angular velocity of
30.0 rad/s. Suddenly, a second disk of mass M/2 and radius R/2 at rest is added to the
shaft. What is the angular velocity of the new system?
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E. None of the above
Use conservation of angular momentum, Li = L f , so I iωi = I f ω f and ω f =
I iωi
. Since
If
2
 M  R  9
I i = MR 2 and I f = MR 2 +    = MR 2 . So ω f = 89 ωi = 26.7 rad / s .
 2  2  8
30. (2 points) A block B is at rest on a frictionless surface, as shown in the figure below.
An identical block A is attached to one end of a string of length R. Block A is released
from the horizontal position and collides with B. The two blocks stick together after the
impact. What is the speed of the two blocks just after the impact?
A
R
B
A.
2gR
2 gR
2
C. 2gR
B.
D. 2 2gR
E. None of the above
Use conservation of energy to get the velocity of block A just before it hits and
conservation of momentum to get the combined velocity just after the impact. Since this
is an inelastic collision, you cannot equate the kinetic energy after the collision to the
change in potential energy.
Finding the velocity of block A from conservation of energy, vA = mgR .
1
1
Applying conservation of momentum, mv A = 2mv f , so v f = vA =
2 gR .
2
2
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