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AP Physics
Chapter 11 Review
Rotational Motion
1.
m
A car of mass 1000 kg moves with a speed of 50 sec
on a
circular track of radius 100 m. What is the magnitude of its
angular momentum relative to the center of the race track?
€
L = mvR
m
L = 1000Kg(50 sec
)(100m)
L = 5.0 ×10
€
2
6 Kg⋅m
sec
2
2.
A solid cylinder of radius R = 1.0 m and mass 10 kg
rotates about its axis. When its angular velocity is
10 rad
sec , its angular momentum is ?
L = Iω
(
)
L = 12 mR 2 ω
€
2
1
L = 2 (10Kg )(1m ) (10 rad
sec )
L = 50
€
Kg⋅m 2
sec
3
3. In the figure, a 1.5-kg weight swings in a vertical circle at the end
of a string having negligible weight. The string is 2 m long. If the
weight is released with zero initial velocity from a horizontal
position, its angular momentum at the lowest point of its path
relative to the center of the circle is approximately
From Conservation of Energy,
(
)
v = 2gh = (2) 9.8 secm 2 (2m) = 6.26 secm
L = mvR = (1.5Kg)(6.26 secm )(2m )
L =18.8
Kg⋅ m 2
sec
4
4.
A massless rope is wrapped around a uniform cylinder that
has radius R and mass M, as shown in the figure. Initially,
the unwrapped portion of the rope is vertical and the
cylinder is horizontal. The linear acceleration of the
cylinder is
5
#4
FNet = ma
Ia
Mg − 2 = Ma
R
Ia
Mg = Ma + 2
R

I 
Mg =  M + 2  a

R 
Mg
=a

I 
M + 2

R 
Mg
=
a
2
1


MR
M + 2 2 
R 

Mg
=a
1
(M + 2 M)
Mg
=a
3M
2
2
3
g=a
6
€
€
5. A merry-go-round of radius R
= 2.0 m has a moment
of
2
inertia I = 250Kg ⋅ m and is
rev
. A child
rotating at 10 min
whose mass is 25 kg jumps
onto the edge of the merrygo-round. The new angular
rev
speed
(in
) of the merry-gomin
€ round€is approximately
€
#5
Li = L f
Iiω i = I f ω f
I MGRω i = ( IMGR + IC )ω f
IMGRω i
=ωf
(IMGR + IC )
250Kg ⋅ m 2
rev
10
2 (
min ) = ω f
2
250Kg ⋅ m + 25Kg(2m)
rev
7.1 min
=ωf
7
6. Two blocks, m1 = 1kg and
m2 = 2kg, are connected by
a light string as shown in
the figure. If the radius of
the pulley is 1 meter and
its moment
of inertia is
2
5 Kg ⋅ m , the acceleration
of the system is ?
€
m2
m1
FNet = ma
Ia
m2 g − 2 − m1g = ( m1 + m2 )a
R
(m 2 − m1 ) g = a

I 
 m1 + m 2 + 2 

R 
(1Kg)
g= a

5Kg ⋅ m 2 
 3Kg +

2 
(1m) 

g
=a
8
8
€
€
7. A conical pendulum bob of mass
m, is set into motion in a circular
path in a horizontal plane as
shown in the figure. The angular
momentum of the bob about the
T cosθ vertical axis through the point P
is
P
θ
€
€
T sin€θ
mg
L = mvR
€
#7
€
€
9
#7
7
θ
θ
R = lsinθ
T cosθ
€
€
l
R
€
T sin€θ
mg
€ T = mg
cosθ
€
€
2
mv
T sin θ€ =
R
RT sin θ
=v
m
10
€
L = mvR
 RT sin θ 
L = m
( l sin θ )
m 



mg
sin θ 
 (l sin θ )
cosθ
( l sin θ )
L = m
m






2
gl
sin
θ
(
)

( l sin θ )
L=m

cosθ 


L = ml sin 2 θ
gl
cos θ
11
€
8. A thin rod of mass 2m and length l is struck at one end by
a ball of clay of mass m, moving with speed v as shown in
the figure. The ball sticks to the rod. After the collision, the
angular momentum of the clay-rod system about A, the
midpoint of the rod, is
90°
A
#8
ω
l
€
12
€
Find the angular velocity of the system after the
collision.
()
1 2m l 2 + m l
I = 12
( )
2
(
I = 16 ml 2 + 14 ml 2
2
5
I = 12 ml
)
2
Li = L f
l
2
5
mv = 12 ml ω f
2
v = 56 lω f
6v
=ωf
5l
13
L = Iω
L=
5 ml 2ω
12
L=
5 ml 2
12
L= 1
mlv
2
€
6v
5l
m
9. A particle of mass m = 0.10kg and speed v = 5.0 sec collides and
sticks to the end of a uniform solid cylinder of mass M = 1.0kg and
radius R = 20 cm. If the cylinder is initially at rest and is pivoted
about a frictionless axle through its center, what is the final
angular velocity?
€
Li = L f
[
]
mvR = 12 MR 2 + mR 2 ω
[
mvR
=
ω
1 MR 2 + mR 2
2
Kg⋅m 2
sec
0.1
=ω
2
0.024Kg ⋅ m
]
4.16 rad
=ω
sec
(0.1Kg)(5 secm )(0.2m )
=ω
2
2
1 (1Kg )( 0.2m) + 0.1Kg 0.2m
(
)( ) ]
[2
€
14
10. A skater extends her arms horizontally, holding a 5-kg mass
in each hand. She is rotating about a vertical axis with an
angular velocity of one revolution per second. If she drops
her hands to her sides, what will the final angular velocity
(in rev
sec ) be if her moment of inertia remains approximately
constant at 5 Kg˙m2, and the distance of the masses from the
axis changes from 1 m to 0.1 meters?
Li = L f
€
[
] [
]
IS + 2(5Kg)(1m 2 ) ω i = IS + 2(5Kg)(0.1m 2 ) ω f
15 rev
sec
= ωf
5.1
3 rev
sec = ω f
15
€
11. A solid cylinder
rolls down an
incline as shown
in the figure. The
l i n e a r
acceleration of its
center of mass is
θ
UTop = K + Kθ
2
1
2
mgh = mv + Iω
2
1
2
mgh = mv +
3
4
gh = v
#11
1
2
4
3
(
1 1
2 2
2
v 
mR ) 
 R
2
2
2
gh = v
16
h
d=
sin θ
4 gh = v
3
2
2
o
v = v + 2ad
4 gh = 2a h
3
sin θ
2g=a 1
3
sin θ
2 gsinθ = a
3
17
€
12. A solid sphere, spherical shell, solid cylinder and a
cylindrical shell all have the same mass, m and radius R. If
they are all released from rest at the same elevation and
roll without stopping, which reaches the bottom of an
inclined plane first?
UTop = K + Kθ 1
mgh = 12 mv 2 + 12 Iω 2
v
mgh = mv + ( mR ) 
R
1
2
θ
Ring
2
1
2
2
2
gh = v 2
gh = v
18
#12
UTop = K + Kθ 1
mgh = 12 mv 2 + 12 Iω 2
2
1
2
mgh = mv +
7
10
gh = v
10
7
1
2
(
2
5
v
mR ) 
 R
2
2
2
gh = v
θ
Sphere
€
15
€
UTop = K + Kθ 1
mgh = 12 mv 2 + 12 Iω 2
2
1
2
mgh = mv +
(
1 1
2 2
gh = 43 v 2
4
3
v
mR ) 
 R
2
2
1. Sphere
2. Cylinder
3. Ring
gh = v
cylinder
16
€
#13
xCofG
2m
0.3Kg
0.2Kg
mx 0.3Kg(2m)
∑
=€
= €
= 1.2m from€the 0.2Kg mass.
∑ m 0.5Kg
2
I = ∑ mR = (0.2Kg)(1.2m) + (0.3Kg)(0.8m)
2
I = (0.288 + 0.192)Kg ⋅ m
I = 0.48Kg⋅ m 2
2
2
13. Two particles (m1 = 0.20 kg,
m2 = 0.30 kg) are positioned
at the ends of a 2.0-m long
rod of negligible mass. What
is the moment of inertia of this
rigid body about an axis
perpendicular to the rod and
through the center of mass?
21
14. A uniform rod (m = 2.0 kg, l = 3 m) is free to rotate about
a frictionless pivot at one end. The rod is released from rest
in the horizontal position. What is the magnitude of the
angular acceleration of the rod at the instant it is 60° below
the horizontal?
τ = Iα
l 1 2
mgcos60 = 3 ml α
2
(
3 9.8 secm 2
)
3g
cos θ = α =
cosθ
2l
2( 3m)
rad
α = 2.45 sec
2
€
15. A uniform rod (m = 1.5 kg) is 2.0 meters long. The rod is
pivoted about a horizontal, frictionless pin through one end.
The rod is released from rest at an angle of 30° above the
horizontal. What is the angular acceleration of the rod at the
instant it is released? What is the angular speed the instant
the rod reaches the vertical position?
2m
€
θ
mg
€
22
€
#15a
τ = Iα
l
mg cos θ = ( 13 ml )α
2
2
1
g cos θ = 13 l α
2
3g
cos 30° = α
2l
( )
rad
6.37 sec
2 =α
€
23
#15b
UTop = K Bottom
l
4
3l 1 2
mg = 2 Iω
3l
h=
4
4
€
3l 1 1 2 2
l
mg = 2 3 ml ω
2
4
3 1 €2
g = 3l ω
€
2
9g
=ω
2l
(
)
( )
24
€
16.
Stars originate as large bodies of slowly
rotating gas. Because of gravity, these
clumps of gas slowly decrease in size. The
angular velocity of a star increases as it
shrinks because of
a. conservation of angular momentum
b. conservation of linear momentum
c. conservation of energy
d. the law of universal gravitation
e. conservation of mass
25
17. The object shown below has mass m and velocity v.
The direction of its angular momentum vector with
respect to an axis perpendicular to the page through
point O is:
v
O
Right Hand Rule!
Into the page
26
m
2m
O
r
2r
€
18. Two objects move in
parallel circles
around a
€
rotation axis O with equal
tangential speeds. r2 = r.
r1 = 2r. m2 = 2m; m1 = m.
Forces of equal magnitude
are applied opposite to
their velocities to stop
them. Which statement is
correct?
a. m1 will stop first because it
has the smaller mass.
b. m2 will stop first because it
has the smaller radius.
c. m1 will stop first because the
torque on it is greater.
d. m2 will stop first because it
has the smaller moment of
inertia.
e. both objects will stop at the
same time because the angular
accelerations are equal.
27
19. Which of the following are different physical
quantities but have the same units?
I. Work
II. Kinetic Energy
a. I and II only
b. I and III only
c. II and III only
III. Torque
d. I, II, and III
e. None of these
W = F ⋅ x = Nm = J
2
Kg
⋅
m
K = 12 mv 2 =
= Nm = J
2
sec
= F⊥ x = Nm
τ
28
20. A uniform thin rod of mass m and length l is suspended
form one end by a frictionless pivot so that it can swing
freely in the plane of the paper, as shown below. When
the rod is at rest it is struck by a clay ball of equal mass m
with an initial velocity vo at an angle of 60° with the
vertical. The clay ball strikes the rod at the free end and
sticks to it.
vo
€
60°
29
20.2
The linear vilocity v of the free end just after the
collision would be most nearly?
Li = L f
mvrsin φ = ( Irod + Imass )ω f
(
2
2
1
mv o sin60°l = 3 ml + ml
)ω
f
v 
4
v o sin60° = 3 l 
l
3 v sin60° = v
4 o
vo
60°
€
We could also
use the cos 30°!
26
21. The engine of a car generates a constant power P. If the
car starts from rest, and after t seconds its velocity is v, the
speed of the car at 2t would be:
dW
P=
and W = ΔK
dt
dW = Pt = 12 mv 2 − 0
€
€
P is constant, soooooo
t ∝v2
and v ∝ t
at 2t,
v2
=
v
2t
t
v2 =
2v
31
€
22. A merry-go-round
2
I
=
400Kg
⋅
m
;R = 2m)
(
is rotating at ωo with a
75 Kg person standing on
the edge of the platform.
The person then walks to
the
center of the
platform.
The new
angular velocity of the
merry-go-round is most
nearly?
(400Kg ⋅ m
2
Li = L f
(I
(I
mgr
+ IP )ω i = Imgrω f
mgr
+ IP )
Imgr
+ ( 75Kg)(2m)
400Kg ⋅ m
2
2
ωi = ω f
)ω =ω
i
f
7ω
4 i
=ωf
32
23. A disk with a moment of inertia relative to a vertical axis
through its center is dropped onto a disk having half its
moment of inertia which is rotating with an angular
velocity ωo. What is the final angular velocity of the
system?
Li = L f
1I
2
ω o = I + 12 I ω f
( ) (
1I
(2 )ω = ω
)
f
3I o
2
1ω =ω
f
3 o
33
€
24. Two blocks of masses m1 and m2 are connected by a light
cord that passes over a pulley of mass M, as shown. Block
m2 slides on a frictionless horizontal surface. The blocks
and pulley are initially at rest. When m2 is released, the
blocks accelerate and the pulley rotates. The total angular
momentum of the system of the two blocks and the pulley
relative to the axis of rotation of the pulley is:
a.
b.
c.
d.
e.
f.
g.
h.
proportional to the radius of the pulley.
proportional to the velocity of the blocks.
proportional to the length of the strings.
proportional to L1, the length of string from the pulley to m1.
proportional to L2, the length of string from the pulley to m2.
conserved, because the Earth doesn't move.
true for all of the above.
m2
true only to (a) and (b) above.
m1
€
€
€
θ = 0.4e
2t
What is the magnitude of the total
linear acceleration at t = 0 on a
point on the object that is 4 cm from
the axis?
dθ
2t
ω=
= 2(0.4 )e
dt
dω
2t
α=
= 4 (0.4 )e
dt
2( 0)
e =1
a = Rα
rad
a =1.6 sec
2 (0.04m )
a = 0.064 secm 2
rad
α = 4 (0.4 ) = 1.6 sec
2
35
€