Ch. 11 Review Angular Momentum solns PDF document

37 Pages·2012·0.21 MB·English

by  Cheryl Stock

AP Physics
Chapter 11 Review
Rotational Motion
1
. A car of mass 1000 kg moves with a speed of 50 m on a
sec
circular track of radius 100 m. What is the magnitude of its
angular momentum relative to the center of the race track?
€
L = mvR
L = 1000Kg(50 m )(100m)
sec
2
6 Kg⋅m
L = 5.0 × 10
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 ra d , its angular momentum is ?
sec
L = Iω
€ (1 2 )
L = mR ω
2
L = 1 (10Kg)(1m)2 (10 rad )
2 sec
2
Kg⋅m
L = 50
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 m (2m) = 6.26 m
2 sec
sec
L = mvR = (1.5Kg)(6.26 m )(2m)
sec
2
Kg⋅m
L =18.8
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
F = ma
Net Mg
= a
Ia
Mg − = Ma  1 MR2 
R2  M + 2 
2
R
 
Ia
Mg = Ma +
Mg
2
R
= a
(M + 1 M)
 I 
2
Mg =  M +  a
2
 R  Mg
= a
3
Mg M
= a 2
 I 
 M +  2
g = a
2
 R 
3
6
€
€
5. A merry-go-round of radius R
= 2.0 m has a moment of
L = L
2
inertia I = 2 5 0 K g ⋅ m and is
i f
rotating at 10 re v . A child
min
Iω = I ω
whose mass is 25 kg jumps
i i f f
onto the edge of the merry-
go-ro u n d . T h e n e w a n g u l a r I ω = (I + I )ω
MGR i MGR C f
speed (in r e v ) of the merry-go-
€
€ min
round is approximately I ω
MGR i =ω
f
(I + I )
MGR C
€
2
250Kg ⋅ m
(10 rev ) =ω
2 min f
250Kg ⋅ m2 + 25Kg(2m)
7.1 rev =ω
min f
#5
7
€
6. Two blocks, m = 1kg and
F = ma
1
Net
m = 2kg, are connected by
2
a light string as shown in Ia
m g − − m g = (m + m )a
the figure. If the radius of
2 2 1 1 2
R
the pulley is 1 meter and
its moment of inertia is (m − m )
5 K g ⋅ m 2 , the acceleration 2 1 g = a
 I 
of the system is ?
 m + m + 
1 2 2
 R 
(1Kg)
€
g = a
 
2
5Kg ⋅ m
 3Kg + 
 
2
(1m)
 
g
m
= a
2
m
8
1
8
€
€
€
7. A conical pendulum bob of mass
m, is set into motion in a circular
P
path in a horizontal plane as
θ
shown in the figure. The angular
momentum of the bob about the
T cosθ
vertical axis through the point P
€ is
€
T sinθ
€
L = mvR
mg
€
#7
€
9
€
#7
7
θ
l
θ
R = lsinθ
T cosθ
R
€
€
€
T sinθ
€ 2
€ mv
T sinθ =
R
mg
€
RT sinθ
mg
€ = v
T =
m
cosθ
€
10
€
€