Download Chapter 10 Homework and Practice Problems 10.1, 10.10, 10.17

Chapter 10 Homework and Practice Problems
10.1, 10.10, 10.17, 10.32, 10.37, 10.42, 10.63, 10.82
10.1.
Use Eq. (10.2) to calculate the magnitude of the torque and use the right-hand rule
illustrated in Figure 10.4 in the textbook to calculate the torque direction.
(a) SET UP: Consider Figure 10.1a.
EXECUTE:   Fl
l  rsin  (400 m)sin90
l  400 m
  (10.0 N)(4.00 m)  400 N  m
Figure 10.1a
This force tends to produce a counterclockwise rotation about the axis; by the right-hand
rule the vector  is directed out of the plane of the figure.
(b) SET UP: Consider Figure 10.1b.
EXECUTE:   Fl
l  rsin  (400 m)sin120
l  3464 m
  (10.0 N)(3.464 m)  346 N  m
Figure 10.1b
This force tends to produce a counterclockwise rotation about the axis; by the right-hand
rule the vector  is directed out of the plane of the figure.
(c) SET UP: Consider Figure 10.1c.
EXECUTE:   Fl
l  rsin  (400 m)sin30
l  200 m
  (10.0 N)(2.00 m)  200 N  m
Figure 10.1c
This force tends to produce a counterclockwise rotation about the axis; by the right-hand
rule the vector  is directed out of the plane of the figure.
(d) SET UP: Consider Figure 10.1d.
EXECUTE:   Fl
l  rsin  (2.00 m)sin60  1.732 m
  (10.0 N)(1.732 m)  173 N  m
Figure 10.1d
This force tends to produce a clockwise rotation about the axis; by the right-hand rule the
vector  is directed into the plane of the figure.
(e) SET UP: Consider Figure 10.1e.
EXECUTE:   Fl
r  0 so l  0 and   0
Figure 10.1e
(f) SET UP: Consider Figure 10.1f.
EXECUTE:   Fl
l  rsin,   180,
so
l 0
and   0
Figure 10.1f
The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action
of the force passes through the axis.
10.10.
The constant force produces a torque which gives a constant angular acceleration to the
disk and a linear acceleration to points on the disk.
ANALYZE:  z  I z applies to the disk, z2  02z  2 z (  0 ) because the angular
acceleration is constant. The acceleration components of the rim are atan  r and
arad  r 2 ,
2
2
 arad
.
and the magnitude of the acceleration is a  atan
EXECUTE: (a)  z  I z gives Fr  I z . For a uniform disk,
I
1
1
Fr (300 N)(0200 m)

 750 rad/s 2.
MR 2  (400 kg)(0200 m) 2  0800 kg  m 2.  z 
2
I
2
2
0800 kg  m
  0  0.200 rev  1.257 rad. 0 z  0, so z2  02z  2 z (  0 ) gives
z  2(750 rad/s2 )(1257 rad)  4342 rad/s. v  r  (0200 m)(4342 rad/s)  0868 m/s.
(b)
atan  r  (0200 m)(750 rad/s2 )  150 m/s 2.
2
2
 arad
 406 m/s2.
arad  r 2  (0200 m)(4342 rad/s)2  3771 m/s2. a  atan
The net acceleration is neither toward the center nor tangent to the disk.
10.17. Apply  F  ma to each box and  z  I z to the pulley. The magnitude a of the
acceleration of each box is related to the magnitude of the angular acceleration  of the
pulley by a  R .
ANALYZE: The free-body diagrams for each object are shown in Figure 10.17a–c. For
the pulley, R  0250 m and I  12 MR2. T1 and T2 are the tensions in the wire on either side
of the pulley. m1  120 kg, m2  500 kg and M  200 kg. F is the force that the axle exerts
on the pulley. For the pulley, let clockwise rotation be positive.
EXECUTE: (a)
 Fx  max
weight gives
a  R
and
and
for the 12.0 kg box gives
m2 g  T2  m2a.  z  I z
T2  T1  12 Ma.
T1  m1a.  Fy  ma y
for the pulley gives
for the 5.00 kg
(T2  T1) R 
Adding these three equations gives
 12 MR2 .
m2 g  (m1  m2  12 M )a




m2
500 kg
2
2
a 
g  
 (980 m/s )  272 m/s .
 m1  m2  1 M 
12

0
kg

5

00
kg

1

00
kg


2


T1  m1a  (120 kg)(272 m/s2 )  326 N. m2 g  T2  m2a
Then
gives
T2  m2 ( g  a)  (500 kg)(980 m/s  272 m/s )  354 N. The tension to the left of the
pulley is 32.6 N and below the pulley it is 35.4 N.
2
(b)
a  272 m/s2
2
(c) For the pulley,
 Fx  max
gives
Fx  T1  326 N
and
 Fy  ma y
gives
Fy  Mg  T2  (200 kg)(980 m/s2 )  354 N  550 N.
The equation
three objects.
m2 g  (m1  m2  12 M )a
says that the external force m2 g must accelerate all
10.32. The power output of the motor is related to the torque it produces and to its angular
velocity by P   zz , where  z must be in rad/s.
ANALYZE: The work output of the motor in
P
60.0 s
is
2
(900 kJ)  600 kJ,
3
so
600 kJ
 100 W. z  2500 rev/min  262 rad/s.
600 s
EXECUTE:
z 
P
z

100 W
 0382 N  m
262 rad/s
For a constant power output, the torque developed decreases when the rotation speed of
the motor increases.
10.37. (a) Use
L  mvr sin 
(Eq. (10.25)):
Consider Figure 10.37.
L  mvrsin 
(200 kg)(120 m/s)(800 m)sin1431
EXECUTE:
L  115 kg  m2 /s
Figure 10.37
To find the direction of L apply the right-hand rule by turning r into the direction of v
by pushing on it with the fingers of your right hand. Your thumb points into the page, in
the direction of L.
(b) By Eq. (10.26) the rate of change of the angular momentum of the rock equals the
torque of the net force acting on it.
EXECUTE:   mg (800 m) cos 369  125 kg  m2 /s2
To find the direction of  and hence of dL/dt , apply the right-hand rule by turning r into
the direction of the gravity force by pushing on it with the fingers of your right hand.
Your thumb points out of the page, in the direction of dL/dt.
L and dL/dt are in opposite directions, so L is decreasing. The gravity force is
accelerating the rock downward, toward the axis. Its horizontal velocity is constant but
the distance l is decreasing and hence L is decreasing.
10.42. ANALYZE: L is conserved if there is no net external torque.
Use conservation of angular momentum to find  at the new radius and use K  12 I 2 to
find the change in kinetic energy, which is equal to the work done on the block.
EXECUTE: (a) Yes, angular momentum is conserved. The moment arm for the tension
in the cord is zero so this force exerts no torque and there is no net torque on the block.
(b) L1  L2 so I11  I22. Block treated as a point mass, so
of the block from the hole.
I  mr 2,
where r is the distance
mr121  mr222
2
2
 r1 
 0300 m 
 1  
 (175 rad/s)  700 rad/s
 0150 m 
 r2 
2  
(c)
K1  12 I112  12 mr1212  12 mv12
v1  r11  (0300 m)(175 rad/s)  0525 m/s
K1  12 mv12  12 (00250 kg)(0525 m/s)2  000345 J
K2  12 mv22
v2  r22  (0150 m)(700 rad/s)  105 m/s
K2  12 mv22  12 (00250 kg)(105 m/s) 2  001378 J
K  K2  K1  001378 J  000345 J  00103 J
(d)
But
Wtot  K
Wtot  W , the
work done by the tension in the cord, so
W  00103 J.
Smaller r means smaller I. L  I is constant so  increases and K increases. The work
done by the tension is positive since it is directed inward and the block moves inward,
toward the hole.
10.63. ANALYZE: Use  z  I z to find the angular acceleration just after the ball falls off and
use conservation of energy to find the angular velocity of the bar as it swings through the
vertical position.
SET UP: The axis of rotation is at the axle. For this axis the bar has
1 m L2 ,
I  12
bar
where
and L  0800 m. Energy conservation gives K1  U1  K2  U2. The gravitational
potential energy of the bar doesn’t change. Let y1  0, so y2   L/2.
mbar  380 kg
EXECUTE: (a)  z  mball g (L/2) and
z 
z 
mball g ( L/2)
2
1 m L2  m
ball ( L/2)
12 bar
2

2
1 m L2  m
I  I ball  I bar  12
bar
ball ( L/2) .  z  I z

2g 
mball


L  mball  mbar /3 
gives
and

2(980 m/s ) 
250 kg
2

  163 rad/s .
0800 m  250 kg  [380 kg]/3 
(b) As the bar rotates, the moment arm for the weight of the ball decreases and the
angular acceleration of the bar decreases.
(c)
K1  U1  K2  U 2 . 0  K2  U 2.

mball gL
mball L2 /4  mbar L2 /12

1 (I
2 bar
 I ball ) 2   mball g ( L/2).


g
4mball
980 m/s2 
4[250 kg]




L  mball  mbar /3 
0800 m  250 kg  [380 kg]/3 
  570 rad/s.
As the bar swings through the vertical, the linear speed of the ball that is still attached to
the bar is v  (0400 m)(570 rad/s)  228 m/s. A point mass in free-fall acquires a speed of
2.80 m/s after falling 0.400 m; the ball on the bar acquires a speed less than this.
10.82. Apply conservation of energy to the motion of the ball as it rolls up the hill. After the
ball leaves the edge of the cliff it moves in projectile motion and constant acceleration
equations can be used.
(a) SET UP: Use conservation of energy to find the speed v2 of the ball just before it
leaves the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top
of the hill. Take y  0 at the bottom of the hill, so y1  0 and y2  280 m.
K1  U1  K2  U 2
EXECUTE:
1 mv 2
1
2
 12 I12  mgy2  12 mv22  12 I22
Rolling without slipping means   v/r and
7
mv12
10
7
 mgy2  10
mv22
v2  v12  10
gy2  1526 m/s
7
1
I 2
2
 12

2
mr 2
5
 (v/r)
2
 15 mv2.
SET UP: Consider the projectile motion of the ball, from just after it leaves the top of
the cliff until just before it lands. Take  y to be downward. Use the vertical motion to
find the time in the air:
v0 y  0, a y  980 m/s2 , y  y0  280 m, t  ?
EXECUTE: y  y0  v0 yt  12 a yt 2 gives t  239 s
During this time the ball travels horizontally
x  x0  v0 xt  (1526 m/s)(239 s)  365 m.
Just before it lands,
v y  v0 y  a yt  234 m/s
and
vx  v0 x  153 m/s
v  vx2  v 2y  280 m/s
(b) At the bottom of the hill,   v/r  (250 m/s)/r. The rotation rate doesn’t change while
the ball is in the air, after it leaves the top of the cliff, so just before it lands
  (153 m/s)/r. The total kinetic energy is the same at the bottom of the hill and just
before it lands, but just before it lands less of this energy is rotational kinetic energy, so
the translational kinetic energy is greater.