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CIRCLE
THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE
QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.
1. Tripling the
mass of the bob on a simple pendulum will cause a change in the
frequency of the pendulum swing by what
factor?
a.
0.330.
b.
1.00.
(4)
c.
3.00.
d.
9.00.
2. What condition
or conditions are necessary for static equilibrium?
a.
ΣFx = 0
b. ΣFx = 0, ΣFy
= 0, Σt = 0
(4)
c.
Σt = 0
d.
ΣFx = 0, ΣFy = 0
3. Two equal
forces are applied to a door at the doorknob.
The first force is applied perpendicular to the door; the second force
is applied at 30° to the plane of the door.
Which force exerts the greater torque?
a. The first applied perpendicular to the door.
b.
The second applied at an angle.
(4)
c.
Both exert equal non-zero torques.
d.
Both exert zero torques.
4. A solid
cylinder of mass 10 kg is pivoted about a frictionless axis thought the center
O. A rope wrapped around the outer
radius R1 = 1.0 m, exerts a force F1 = 5.0 N to the
right. A second rope wrapped around
another section of radius R2 = 0.50 m exerts a force F2 =
6.0 N downward. What is the angular acceleration of the cylinder?
a.
1.0 rad/s2
b.
0.60 rad/s2
(4)
c. 0.40 rad/s2
d.
0.80 rad/s2
5. A mass on a
spring undergoes SHM. When the mass is
at its maximum displacement from equilibrium, its instantaneous velocity
a. Is
maximum.
b. Is
less than maximum, but not zero.
(4)
c. Is zero.
d.
Cannot be determined from the information given.
2.
The moment of inertia of a rigid body
a. Depends on the axis of rotation.
b. Cannot be negative.
(4)
c. Depends on mass distribution.
d. All of the above.
7. A centrifuge
rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to
rest by a frictional torque of 
If the mass of the
rotor is 4.80 kg and it can be approximated as a solid cylinder of radius
0.0710 m,
a.
What
is the rotor’s moment of inertia?
I = ˝ MR2
= ˝ (4.80 kg)(0.0710 m)2 = 0.0121 kgm2
b.
What
is the angular acceleration of the rotor as it is brought to rest?
a
= St/I = (1.20 N-m)/(0.0121
kgm2) = 99.2 rad/s2
c. Through how
many revolutions will the rotor turn before coming to rest?
w2 =
w02 + 2 a (q - q0)
w0 = 10,300 rpm = 1079 rad/s
w
= 0
(q - q0) =
- w02 / (2 a) = (1079
rad/s)2/(-2 x 99.2 rad/s2) = 5868 rad = 934 rev
8. A
uniform disk turns at 
around a frictionless spindle.
A nonrotating rod, of the same mass as the disk and length equal to the disk’s
diameter, is dropped onto the freely spinning disk. They then both turn around
the spindle with their centers superposed. What is the angular frequency in 
of the combination?
Ii
= ˝ MR2
If
= ˝ MR2 + (1/12) ML2 = ˝ MR2 + (1/12) M(2R)2
= ˝ MR2 + (1/3) MR2 = (5/6) MR2
Ii wi = If wf
wf =
Ii
wi / If
wf =
(2.40 rev/s) (˝ MR2) /( 5/6 MR2) = 1.44 rev/s
9.
A 0.150-kg toy is undergoing simple harmonic motion on the end of a horizontal
spring with spring constant k = 300 N/m. The amplitude of the motion is 0.200
m.
a. Find the total mechanical energy of
the toy at any point of its motion.
E
= ˝ kA2 = ˝ (300 N/m) (0.200
m)2 = 6.00 J
c. Find the maximum speed attained by the
toy.
˝
m Vmax2 = ˝ kA2
m
Vmax2 = kA2
Vmax2
= (k/m)A2
Vmax
= ((300 N/m)/(0.150 kg))˝ (0.200 m) = 8.94
m/s
d. How many oscillations per second does
the toy undergo?
T
= 2 p (m/k)˝
T
= 2 p ((0.150 kg)/(300 N/m))˝
T
= 0.140 s
F
= 1/T = 7.11 Hz
e. What is the speed of the toy at 10.0
cm to the right from the equilibrium position?
V
= +- (k/m(A2 – x2)) ˝
V
= +- ((300 N/m)/(0.150 kg)((0.200 m)2 – (0.100 m)2))
˝ = +- 7.75 m/s
V
= 7.75 m/s
7. Pictured below is a very light wooden plank
with two masses, 10.0 kg each, on top of it.
Find the
reaction forces at points A and B.
NAx = NA cos (90o) =
0
NAy = NA sin (90o) =
NA
NBx = NB cos (90o) =
0
NBy =NB sin
(90o) = NB
w1x = w1 cos (-90o) =
0
w1y = w1 sin (-90o) =
- m1 g = - (10.0 kg)(9.81 m/s2)
= - 98.1 N
w2x = w2 cos (-90o) =
0
w2y = w2 sin (-90o) =
- m2 g = - (10.0 kg)(9.81 m/s2)
= - 98.1 N
SFx = 0 =>
0 = 0
SFy = 0 =>
NA + NB – 98.1 N – 98.1 N = 0
NA +
NB= 196 N
Assuming that the axis of rotation is at point A:
tNA =
0
tNB =
NB (3.00 m)
tw1 =
- (98.1 N) (0.750 m) = - 73.6 N-m
tsp =
- (98.1 N) (2.50 m) = - 245 N-m
St = 0
NB
(3.00 m) – 73.6 N-m – 245 N-m = 0
NB =
106 N
NA=
196 N – 106 N = 90.0 N