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Physics II
Homework II
CJ
Chapter 9; 86, 89 10; 2, 6, 10, 17, 20, 41
9.86. IDENTIFY: Apply conservation of energy to the system of two blocks and the pulley.
SET UP: Let the potential energy of each block be zero at its initial position. The kinetic
energy of the system is the sum of the kinetic energies of each object. v  R , where v is the
common speed of the blocks and  is the angular velocity of the pulley.
EXECUTE: The amount of gravitational potential energy which has become kinetic energy
is K   4.00 kg  2.00 kg   9.80 m s 2   5.00 m   98.0 J. In terms of the common speed v of the
blocks, the kinetic energy of the system is
1
1 v
K  (m1  m2 )v 2  I  
2
2 R
1
(0.480 kg  m 2 )  2
K  v 2  4.00 kg  2.00 kg 
  v (12.4 kg).
2
(0.160 m) 2 
2
.
v
Solving for v gives
EVALUATE: If the pulley is massless, 98.0 J  12 (4.00 kg  2.00 kg)v 2 and
moment of inertia of the pulley reduces the final speed of the blocks.
98.0 J
 2.81 m s.
12.4 kg
v  5.72 m/s .
The
9.89. IDENTIFY: I  I1  I 2 . Apply conservation of energy to the system. The calculation is similar
to Example 9.9.
SET UP:

v
R1
for part (b) and  
v
R2
for part (c).
EXECUTE: (a) I  1 M1R12  1 M 2 R22  1 ((0.80 kg)(2.50 102 m)2  (1.60 kg)(5.00 102 m)2 )
2
2
2
I  2.25 103 kg  m 2 .
(b) The method of Example 9.9 yields
v
v
2 gh
1  ( I mR12 )
.
2(9.80 m s2 )(2.00 m)
 3.40 m s.
(1  ((2.25 103 kg  m2 ) (1.50 kg)(0.025 m)2 ))
The same calculation, with R2 instead of
R1
gives
v  4.95 m s.
EVALUATE: The final speed of the block is greater when the string is wrapped around the
larger disk. v  R , so when R  R2 the factor that relates v to  is larger. For R  R2 a larger
fraction of the total kinetic energy resides with the block. The total kinetic energy is the
same in both cases (equal to mgh), so when R  R2 the kinetic energy and speed of the block
are greater.
10.2. IDENTIFY:
SET UP:
  Fl with l  r sin  . Add the two torques to calculate the net torque.
Let counterclockwise torques be positive.
EXECUTE:
1   Fl
1 1  (8.00 N)(5.00 m)  40.0 N  m .
 2   F2l2  (12.0 N)(2.00 m)sin30.0°  12.0 N  m .
28.0 N  m ,
  
1
  2  28.0 N  m .
The net torque is
clockwise.
EVALUATE: Even though F1  F2 , the magnitude of 1 is greater than the magnitude of  2 ,
because F1 has a larger moment arm.
10.6. IDENTIFY: Use   Fl  rF sin  for the magnitude of the torque and the right-hand rule for
the direction.
SET UP:
r  0.250 m and   37°
In part (a),
EXECUTE: (a)   (17.0 N)(0.250 m)sin37°  2.56 N  m . The torque is counterclockwise.
(b) The torque is maximum when   90° and the force is perpendicular to the
wrench. This maximum torque is (17.0 N)(0.250 m)  4.25 N  m .
EVALUATE: If the force is directed along the handle then the torque is zero. The torque
increases as the angle between the force and the handle increases.
10.10. IDENTIFY: Apply  z  I z to the wheel. The acceleration a of a point on the cord and the
angular acceleration  of the wheel are related by a  R .
SET UP: Let the direction of rotation of the wheel be positive. The wheel has the shape of
a disk and I  12 MR 2 . The free-body diagram for the wheel is sketched in Figure 10.10a for a
horizontal pull and in Figure 10.10b for a vertical pull. P is the pull on the cord and F is the
force exerted on the wheel by the axle.
EXECUTE: (a)  z 
(b)
tan  
Fy
Fx

z
I

1
2
(40.0 N)(0.250 m)
 34.8 rad/s2 . a  R  (0.250 m)(34.8 rad/s 2 )  8.70 m/s 2 .
(9.20 kg)(0.250 m)2
Fx   P , Fy   Mg
Mg (9.20 kg)(9.80 m/s2 )

P
40.0 N
98.6 N and is directed at
the cord.
.
F  P 2  (Mg )2  (40.0 N)2  ([9.20 kg][9.80 m/s 2 ]) 2  98.6 N .
and   66.1° . The force exerted by the axle has magnitude
66.1° above
the horizontal, away from the direction of the pull on
(c) The pull exerts the same torque as in part (a), so the answers to part (a)
don’t change. In part (b), F  P  Mg and F  Mg  P  (9.20 kg)(9.80 m/s2 )  40.0 N  50.2 N .
The force exerted by the axle has magnitude 50.2 N and is upward.
EVALUATE: The weight of the wheel and the force exerted by the axle produce no torque
because they act at the axle.
Figure 10.10
10.17. IDENTIFY: Apply  z  I z to the post and  F = ma to the hanging mass. The acceleration
a of the mass has the same magnitude as the tangential acceleration atan  r of the point on
the post where the string is attached; r  1.75 m  0.500 m  1.25 m .
SET UP: The free-body diagrams for the post and mass are given in Figures 10.17a and b.
The post has I  13 ML2 , with M  15.0 kg and L  1.75 m .
EXECUTE: (a)
F
y
 may for
 z  I z for the post gives Tr   13 ML2  . a  r so  
the mass gives mg  T  ma . These two equations give
 ML2 
a
and T   2  a .
r
 3r 
mg  (m  ML2 /[3r 2 ])a and




m
5.00 kg
a
g 
(9.80 m/s 2 )  3.31 m/s 2 .
2
2 
2
2 
 m  ML /[3r ] 
 5.00 kg  [15.0 kg][1.75 m] / 3[1.25 m] 

a 3.31 m/s 2

 2.65 rad/s 2 .
r
1.25 m
(b) No. As the post rotates and the point where the string is attached moves in
an arc of a circle, the string is no longer perpendicular to the post. The torque due to
this tension changes and the acceleration due to this torque is not constant.
(c) From part (a), a  3.31 m/s2 . The acceleration of the mass is not constant. It
changes as  for the post changes.
EVALUATE: At the instant the cable breaks the tension in the string is less than the weight
of the mass because the mass accelerates downward and there is a net downward force on it.
Figure 10.17
10.20. IDENTIFY: Only gravity does work, so
2
 12 I cm 2 .
Ki  Ui  Kf  U f . K f  12 Mvcm
Wother  0
and conservation of energy gives
SET UP: Let yf  0 , so Uf  0 and yi  0.750 m . The hoop is released from rest so
vcm  R . For a hoop with an axis at its center, I cm  MR 2 .
EXECUTE: (a) Conservation of energy gives
MR 2 2  Mgyi .  
Ki  0 .
U i  K f . K f  12 MR 2 2  12 ( MR 2 ) 2  MR 2 2 ,
so
gyi
(9.80 m/s2 )(0.750 m)

 33.9 rad/s .
R
0.0800 m
(b) v  R  (0.0800 m)(33.9 rad/s)  2.71 m/s
EVALUATE: An object released from rest and falling in free-fall for 0.750 m attains a
speed of 2g (0.750 m)  3.83 m/s . The final speed of the hoop is less than this because some of
its energy is in kinetic energy of rotation. Or, equivalently, the upward tension causes the
magnitude of the net force of the hoop to be less than its weight.
10.41. IDENTIFY: Apply conservation of angular momentum to the motion of the skater.
SET UP: For a thin-walled hollow cylinder I  mR2 . For a slender rod rotating about an axis
through its center, I  121 ml 2 .
Li  Lf
EXECUTE:
so Iii  If f .
I i  0.40 kg  m  (8.0 kg)(1.8 m)2  2.56 kg  m2 . I f  0.40 kg  m 2  (8.0 kg)(0.25 m) 2  0.90 kg  m 2 .
2
 Ii
 If
f  
1
12

 2.56 kg  m 2 
(0.40 rev/s)=1.14 rev/s .
 i  
2 
 0.90 kg  m 

EVALUATE: K  12 I 2  12 L .  increases and L is constant, so K increases. The increase in
kinetic energy comes from the work done by the skater when he pulls in his hands.