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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.
- Which one
of the following quantities is at maximum when an object in simple
harmonic motion is at its maximum displacement?
a. Velocity.
(8)
b. Acceleration.
c. Kinetic energy.
d. Frequency.
2. Consider a point on a bicycle wheel as the
wheel makes exactly four complete revolutions about a fixed axis. Compare the linear and
angular displacement of the point.
a. Both are zero.
b. Only the angular displacement is zero.
(8)
c.
Only the linear displacement is zero. (Dx = 0,
Dq = 8 p)
d. Neither is zero.
3.
A 40.0-kg boy is standing on the edge of a stationary 30.0-kg platform that is
free to rotate. The boy tries to walk around the platform in a counterclockwise
direction. As he does:
a. The platform doesn’t rotate.
b. The platform rotates in a clockwise
direction just fast enough so that the boy remains stationary relative to the
ground.
(8)
c. The platform rotates in a
clockwise direction while the boy goes around in counterclockwise direction
relative to the ground. (LP +
LB = 0)
d. Both go around with equal angular
velocities but in opposite directions.
4. A pendulum clock is set to run accurately at
sea level. It is then brought to the top of a high mountain, where it is found
to:
a. Run unchanged.
b. Run slow. As g
decreases, T increases.
(8)
c. Run fast.
d. Stop running.
5. The moment of inertia of a spinning body
about its spin axis depends on its:
a. Angular speed, shape, and mass.
b. Angular acceleration, mass, and axis
position.
(8)
c. Mass, shape, axis position, and
size.
d. Mass, size, shape, and speed.
6. A 64.0-kg person stands on a diving board
supported by two pillars, one at the end of the board, the other 1.10-m away.
The pillar at the end of the board exerts a downward force of 828 N. The board
has a length of 3.00 m and a mass of 10.0 kg.
- How far from that pillar is the
person standing?
tN2= 0
tN1
= + (828 N)(1.10 m) = 911 Nm
tMg = - (64.0 kg)(9.80 m/s2)(x)
tmg
= - (10.0 kg)(9.80 m/s2)(0.400 m) =
-39.2 Nm
St = 0
- (64.0 kg)(9.80 m/s2)(x) + 911 Nm – 39.2 Nm= 0 x = 1.39 m
= 1.39 m (from the second pillar or 2.49 m from
the first)
- Find the force exerted by the second
pillar.
(M g)x = M g cos (- 90o) = 0
(M g)y = M g sin (- 90o) = - (64kg) (9.80 m/s2) = - 627 N
(m g)x = m g cos (- 90o) = 0
(m g)y = m g sin (- 90o) = - (10.0kg) (9.80 m/s2) = - 98.0 N
(N1)x = N1 cos (-90o) = 0
(N1)y = N1 sin (-90o) = - 828 N
(N2)x = N2 cos (90o) = 0
(N2)y = N2 sin (90o) = N2
SFx = 0
SFy = 0 - 828 N - 627 N – 98.0 N + N2 = 0 N2 = 1553 N
7.A 500 g mass is undergoing simple harmonic oscillation that
is described by the following equation for its position x(t) from the equilibrium:
x(t) = (0.500 m) cos{
(6.00 p rad/s) t + (p rad)/6.00}
a. What is the amplitude A of the oscillations?
A
= 0.500 m
b. What is the angular w
frequency of the oscillations?
w = 6.00 p rad/s
c. What is the frequency f of the oscillations?
f = w/2p = (6.00 p rad/s)/(2 p rad) = 3.00 Hz
d. What is the period T of the oscillations?
T
= 1/f = 0.333 s
e. What is the fall…(oops!)
spring constant k?
k/m = w2 k = m w2 = (0.500 kg)(6.00 p
rad/s )2 = 178 N/m
f. What is the position of the oscillator
when t = 0 s?
x(t) = (0.500 m) cos{(p rad)/6.00}
= 0.433 m
g. At what time t > 0 is the oscillator first at maximum distance from the
equilibrium?
cos{ (6.00 p rad/s) t + (p rad)/6.00}
= 1
(6.00
p
rad/s) t + (p rad)/6.00
= 2 p n
n
= 1 t = 0.305
s (That corresponds actually to x = A, find the time for x = -A)
h. What is the maximum speed of the
oscillator?
Vmax = w A = (6.00 p rad/s) (0.500 m) = 9.42 m/s
i. What is the magnitude of the maximum
acceleration of the oscillator?
amax = w2 A = (6.00 p rad/s)2 (0.500 m) = 178
m/s
9.
A girl of mass 30.0 kg, initially running tangentially at a speed of 5.00 m/s,
jumps onto a disc-shaped merry-go-round initially at rest as shown below. The
mass of merry-go-round is 100 kg and its radius is 1.50 m. (Oh well, the moment of inertia of a solid disk is I =1/2 mR2).
- Find the
angular speed of the merry-go-round after the girl jumps on.
Li =
LG = m v R
Lf
= LG/MGR = IG/MGR w
= (mR2 + ˝
MR2) w
Li =
Lf
m v R = (mR2 + ˝ MR2) w
w = (m v R)/ (mR2 + ˝ MR2)
w
= 1.25 rad/s
- What is
the change in kinetic energy of the girl-merry-go-round system?
Ki = KG = ˝ m v2 = 375 J
Kf = KG/MGR =
˝ IG/MGR w2 = ˝ (mR2 + ˝ MR2) w2
= 141 J
Change
in K = - 234 J
- How would
you account for this change?
Kinetic energy was lost during the collision.
9.
A particle moving in a circle of radius 2.00 m begins at rest when t = 0 s and
attains speed of 10.0 m/s at the end of one revolution with a constant angular
acceleration. Mass of the particle is 1.00 kg.
- Determine
the angular speed after one revolution.
wf = Vf
/R = 5.00 rad/s
- Determine
the magnitude of angular acceleration and the time it takes to complete
the first revolution.
wf2 = wi2 + 2 a (qf - qi) qf - qi = 2 p
a = ( wf2
- wi2 )/ (2(qf - qi)) = 2.00 rad/s2
- How much
work was done on the particle during that time?
W
= Kf – Ki
= ˝ m V2 = 50.0 J