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AP Physics C
Spring, 2017
Circular-Rotational Motion Mock Exam
Name: Answer Key
Mr. Leonard
Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.
(22pts )
1. A stuntman drives a motorcycle through a vertical loop which has a radius of 2.5 m. The motorcycle’s
engine keeps the motorcycle moving at a constant speed as it drives through the loop.
22 pts
(a) (6 pts) Draw a free-body diagram of the motorcycle when bike is at the top of the loop.
Solution: Remember, the centripetal force is not a force in and of itself. Therefore, you should not
include it in a free-body diagram.
Fg
FN
(b) (8 pts) What is the minimum speed that the motorcycle must maintain to make it through the loop?
Solution: When the motorcycle reaches the top of the loop, both the gravitational force and the
normal force point towards the center of the circular path. Therefore Fg + FN = m v 2 /r. From
this formula we can see that there is a relationship between FN and v. In general, the faster the
motorcycle travels through the loop, the larger the normal force. Since there is a minimum value for
the normal force (FN = 0 N), there is likewise a minimum value for v. Setting FN equal to zero and
solving for v gives:
v2
r
v2
=
m
r
v2
=
r
= gr
√
= gr
m
=5
s
Fg = m
mg
g
v2
v
m
(c) (8 pts) Suppose the stuntman drives through the loop at 10 . If the motorcycle and the stuntman
s
have a combined mass of 150 kg, what was the Normal Force exerted by the loop on the motorcycle
when the bike reached the top of the loop?
Solution: Since Fg + FN = m v 2 /r,
22 pts
AP Physics C/Circular-Rotational Motion Mock Exam– Page 2 of 6 –Name: Answer Key
v2
r
v2
FN = m
− mg
r 2
v
−g
=m
r
= 130 N
Fg + FN = m
(22pts )
2. A coin is placed 20 cm from the center of a turntable. The turn table starts from rest and begins to
rad.
rotate with an angular acceleration of α = 12 2 . The coefficient of friction between the coin and the
s
turntable is 0.4.
(a) (4 pts) What is the tangential acceleration of the coin?
Solution: Recall that tangential acceleration is given by at = r α. Therefore, at = (0.2 m)(12rad/s2 ) =
2.4 m/s2 .
(b) (8 pts) What is the maximum centripetal acceleration the coin can withstand without sliding?
Solution: The force responsible for holding the coin on the turntable is the static frictional force.
Since there is a maximum static frictional force, there is a maximum acceleration. We can find the
magnitude of the maximum acceleration by setting the frictional force equal to m a.
22 pts
f = ma
µ
mg =
ma
a = µg
Because the turntable is speeding up as it spins, the coin’s acceleration can be broken into two
non-zero components: tangential and centripetal.
q Because these two components are orthogonal,
the magnitude of the coin’s total acceleration is a2t + a2c . Setting this equal to the maximum acceleration we found above, and solving for ac gives:
q
a2t + a2c = µ g
a2t + a2c = (µ g)2
a2c = (µ g)2 − a2t
q
ac = (µ g)2 − a2t )
m
= 3.2 2
s
(c) (4 pts) What is the maximum angular velocity the coin can withstand without sliding?
Solution: The centripetal acceleration is related to the angular velocity by ac = r ω 2 . So, ω =
p
ac /r = 4 rad./s.
(d) (6 pts) How long after the turntable started spinning was it before the coin started sliding?
Solution: Using α = ∆ω/∆t gives:
22 pts
AP Physics C/Circular-Rotational Motion Mock Exam– Page 3 of 6 –Name: Answer Key
∆ω
∆t
∆ω
∆t =
α
= 0.333 s
α=
(22pts )
3. A grinding wheel has a radius of 0.2 m. The wheel starts from rest when a motor begins rotating the
rad
wheel with an angular acceleration of 1.2 2 .
s
(a) (6 pts) What is the angular velocity of the turntable the instant it completes one full revolution?
22 pts
Solution: Using ωf2 = ω02 + 2 α ∆θ gives:
0
2
7
ω0 + 2 α ∆θ
=
rad
2
ωf = 2 1.2 2 (2π)
s
rad
ωf = 3.88
s
ωf2
(b) (4 pts) How fast is a point on the outer edge of the wheel moving the instant it completes one full
revolution?
Solution: Using v = ωr gives v = 0.777 m/s.
(c) (8 pts) What is the magnitude of the acceleration of a point on the outer edge of the wheel?
Solution: Recall that when looking at a point on a rotationally accelerating object, the acceleration
has two components: centripetal, and tangential. The centripetal acceleration is given by: ac = r ω 2 ,
while the tangential acceleration is given by at = r α. Since these two components are orthogonal,
the magnitude of the acceleration is given by:
q
a2c + a2t
q
2
= (rω 2 ) + (r α)2
p
= r ω 4 + α2
m
= 15.1 2
s
a=
(d) (4 pts) After the wheel completes one full rotation, the motor dies. It takes 10 seconds for the wheel
to come to rest. What was the angular acceleration of the wheel after the motor died?
rad
Solution: Using α = ∆ω/∆t, we see that α = −0.388 2 .
s
22 pts
AP Physics C/Circular-Rotational Motion Mock Exam– Page 4 of 6 –Name: Answer Key
(12pts )
4. A conical pendulum is formed by attaching a 500 gram ball to the end of a 1.0 m string. If the mass
moves in a circular path with a radius of 20 cm, how fast is the mass moving?
12 pts
1.0 m
0.2 m
Solution: There are two forces acting on the pendulum bob: gravity and tension.
T
θ
Fg
X
Because the bob does not move vertically up or down, we can conclude that
Fy = 0 N. Therefore,
the vertical component of the tension force must cancel the gravitational force: Ty = Fg = 4.9 N.
When the bob swings in a circular path as shown above, a component of the tension points towards the
center of the circular path. So, Tx = m v 2 /r. If we could find the x-component of the tension, we could
use this relationship to solve for v.
Since we know the vertical component of the tension, we can use the geometry of the problem to solve
for Tx . From the free-body diagram, we can see that tan θ = Tx /Ty . We can also see from the free-body
diagram that sin θ = 0.2; therefore, θ = 11.5◦ and tan θ = 0.204. Setting Tx /Ty = tan θ = 0.204 and
solving for Tx gives:
Tx
= 0.204
Ty
Tx = 0.204Ty
= 1.00
Setting this equal to m v 2 /r gives:
12 pts
AP Physics C/Circular-Rotational Motion Mock Exam– Page 5 of 6 –Name: Answer Key
m
v2
= Tx
r
r
v=
Tx r
s m
(1 N)(0.2 m)
0.5 kg
m
v = 0.632
s
v=
(8pts )
5. Two satellites, A and B, are in different circular orbits about the Earth. The orbital speed of the satellite
A is three times that of satellite B. Find the ratio TA /TB .
Solution: Since we are given the ratio of the orbital speeds and we are asked to find the ratio of the
orbital periods, the easiest way to solve this problem is using proportions. To do this, we must find a
formula which relates the period, T , to the velocity, v.
When a satellite is in a circular orbit around the Earth, the centripetal force required to keep the satellite
in a circular path is generated by gravitational force. Therefore:
8 pts
Fc = Fg
v2
GME
m
m =
2
r
r
GME
2
v =
r
2πr
. Since we want an expression
We can relate this to the period of the motion, by recalling that v =
T
relating v to T , we need to solve this equation for r and plug it inso the expression above. Solving for r
vT
gives: r =
. Plugging this into the expression above gives:
2π
v2 =
2
GME
r v = GME
v3 =
2π
vT
2πGME
T
Therefore T is proportional to v −3 . Therefore, if v increases by a factor of 3, T will decrease by a factor
of 33 = 27. Therefore:
1
TA
=
TB
27
8 pts
AP Physics C/Circular-Rotational Motion Mock Exam– Page 6 of 6 –Name: Answer Key
(6pts )
6. Planet X orbits the star Omega with an orbital period of 200 Earth days. Planet Y orbits Omega at a
distance that is four times larger than Planet X. How long is a year on Planet Y?
Solution: Since we are given RY /RX , and we are asked about the periods, we should use Kepler’s Third
Law. Recall that Kepler’s Third Law says:
6 pts
3
RY3
RX
=
2
TX
TY2
Or rearranging to solve for
TY
:
TX
3
R3
RX
= Y2
2
TX
TY
2 3
RY
TY
=
TX
RX
3/2
TY
RY
=
TX
RX
TY
=
TX
RX
RY
3/2
3/2
RX
RY
= 8 (200 days)
TY = TX
= 1600 days
6 pts