Download Chapter 8 (1, 3, 6, 7, 13, 19, 22, 39, 40, 44, 45, 52, 54, 56, 57, 63, 65

Chapter 8 (1, 3, 6, 7, 13, 19, 22, 39, 40, 44, 45, 52, 54, 56, 57, 63, 65 & 66)
8.1
Since the friction force is tangential to a point on the rim of the wheel, it is perpendicular to the radius line
connecting this point with the center of the wheel. The torque of this force about the axis through the
center of the wheel is then  = rf sin 90.0º = rf, and the friction force is
f 
8.3

r

76.0 N  m
 217 N
0.350 m
First resolve all of the forces shown in Figure P8.3 into components parallel to and perpendicular to the
beam as shown in the sketch below.
(a)
 O     25 N  cos 30  2.0 m    10 N  sin 20   4.0 m    30 N  m
or
(b)
 C     30 N  sin 45   2.0 m    10 N  sin 20  2.0 m    36 N  m
or
8.6
0 = 30 N  m counterclockwise
c = 30 N  m counterclockwise
The object is in both translational and rotational equilibrium. Thus, we may write:
Fx  0

Fy  0 
and
Fx  Rx  0
Fy  Ry  Fg  0
 O  0 


Fy  cos    Fx  sin    Fg  cos    0
2

8.7
Requiring that  = 0, using the shoulder joint at point O as a pivot, gives
   Ft sin 12.0   0.080 m    41.5 N   0.290 m   0 or Ft  724 N
Then Fy  0

 Fsy   724 N  sin 12.0  41.5 N = 0 , yielding Fsy  109 N
Fy  0 gives Fsx   724 N  cos12.0 = 0 , or Fsx  708 N
Therefore,
Fs 
8.13
Fsx2  Fsy2 
 708 N 2  109 N 2
 716 N
Requiring that xcg  mi xi mi  0 gives
 5.0 kg   0    3.0 kg   0    4.0 kg   3.0 m    8.0 kg  x
 5.0  3.0  4.0  8.0  kg
 0
or 8.0 x + 12 m = 0 which yields x = – 1.5 m
Also, requiring that ycg  mi yi mi  0 gives
 5.0 kg   0    3.0 kg   4.0 m    4.0 kg   0    8.0 kg  y
 5.0  3.0  4.0  8.0  kg
or 8.0y + 12 m = 0 yielding y = – 1.5 m
 0
Thus, the 8.0-kg object should be placed at coordinates (–1.5 m, –1.5m) .
8.19
Consider the torques about an axis perpendicular to the page through the left end of the rod.
  0  T 
100 N   3.00 m    500 N   4.00 m 
 6.00 m  cos 30.0
T  443 N
Fx  0  Rx  T sin 30.0   443 N  sin 30.0
Rx = 221 N toward the right
Fy  0  Ry  T cos 30.0  100 N  500 N  0
Ry = 600 N  (443 N) cos 30.0 = 217 N upward
8.22
(a)
See the diagram below
(b)
If x = 1.00 m, then
 left end  0    700 N  1.00 m    200 N   3.00 m 
  80.0 N   6.00 m    T sin 60.0°   6.00 m   0
giving T = 434 N.
Then, Fx  0  H  T cos 60.0°  0 , or H   343 N  cos 60.0°  172 N
and Fy  0  V  980 N +  343 N  sin 60.0°  0 , or V = 683 N.
(c)
When the wire is on the verge of breaking, T = 900 N and
  left end    700 N  xmax   200 N   3.00 m 
  80.0 N   6.00 m     900 N  sin 60.0°   6.00 m   0
which gives xmax = 5.14 m
8.39
I 
1
1
2
MR2   150 kg   1.50 m   169 kg  m 2
2
2
and
 
 f  i
t

 0.500
rev s  0   2  rad 


rad s2


 1 rev 
2 .00 s
2
Thus,   F  r  I  gives
F 
8.40
(a)
I

r
169 kg  m2   2

rad s2 

1.50 m
 177 N
It is necessary that the tensions T1 and T2 be different in
order to provide a net torque about the axis of the pulley
and produce and angular acceleration of the pulley.
Since intuition tells us that the system will accelerate in
the directions shown in the diagrams at the right
when m2 > m1, it is necessary that T2 > T1.
(b)
We adopt a sign convention for each object with the positive
direction being the indicated direction of the acceleration of that
object in the diagrams at the right. Then, apply Newton’s
second law to each object:
For m1 :
Fy  m1a

T1  m1g  m1a
or
T1  m1  g  a 
[1]
For m2 :
Fy  m2 a
For M :
  I 
 m2 g  T2  m2 a m2
 rT2  rT1  I 
or
T2  m2  g  a 
[2]
or
T2  T1  I  r
[3]
Substitute Equations [1] and [2], along with the relations I  Mr 2 2 and   a r , into Equation [3] to
obtain
m2  g  a   m1  g  a  
Mr 2  a 
Ma



2r  r 
2
or
M

 m1  m2  2  a   m2  m1  g
and
a 
(c)
 m2
 m1  g
m1  m2  M 2

 20.0 kg  10.0 kg   9.80 m s2 
20.0 kg  10.0 kg +  8.00 kg  2
 2.88 m s2




From Equation [1]: T1   10.0 kg  9.80 m s2  2.88 m s2  127 N .
From Equation [2]: T2   20.0 kg  9.80 m s2  2.88 m s2  138 N .
8.44
(a)
(b)
Hoop:
I  MR2   4.80 kg   0.230 m   0.254 kg  m2
Solid Cylinder:
I 
1
1
2
MR2   4.80 kg   0.230 m   0.127 kg  m 2
2
2
Solid Sphere:
I 
2
2
2
MR2   4.80 kg   0.230 m   0.102 kg  m 2
5
5
Thin, Spherical, Shell:
I 
2
2
2
MR2   4.80 kg   0.230 m   0.169 kg  m 2
3
3
2
When different objects of mass M and radius R roll
without slipping   a  R  down a ramp, the one with the largest
translational acceleration a will have the highest
translational speed at the bottom. To determine the
translational acceleration for the various objects,
consider the free-body diagram at the right:
Fx  Ma
  I


Mg sin   f  Ma
f R  I  a R
[1]
or f  Ia R2
[2]
Substitute Equation [2] into [1] to obtain
Mg sin   Ia R 2  Ma
or
a 
Mg sin 
M  I R2
Since M, R, g are  the same for all of the objects, we see that the translational acceleration (and
hence the
translational speed) increases as the moment of inertia decreases.
Thus, the proper rankings from highest to lowest by translational speed will be:
Solid sphere; solid cylinder; thin, spherical, shell; and hoop
(c)
When an object rolls down the ramp without slipping, the friction force does no work and
mechanical energy is conserved. Then, the total kinetic energy gained equals the gravitational
potential energy given up: KEr  KEt  PEg  Mgh and KEr  Mgh 
1
2
M v2 , where h is
the vertical drop of the ramp and  is the translational speed at the bottom. Since M, g, and h are the
same for all of the objects, the rotational kinetic energy decreases as the translational speed
increases. Using this fact, along with the result of Part (b), we rank the object’s final rotational
kinetic energies, from highest to lowest, as:
hoop; thin, spherical, shell; solid cylinder; and solid sphere
8.45
(a)
Treating the particles on the ends of the rod as point masses, the total moment of inertia of the
rotating system is I  I rod  I 3  I 4  mrod L2 12  m3 (L 2)2  m4 (L 2)2 . If the mass of the
rod can be ignored, this reduces to
 L
I  0   m3  m4   
 2
2
  3.00 kg  4.00 kg   0.500 m   1.75 kg  m 2
and the rotational kinetic energy is
2
KEr 
(b)
1 2
1
I 
1.75 kg  m 2
2
2

  2.50
rad s   5.47 J
2
If the rod has mass mrod  2.00 kg
I 
1
 2.00 kg  1.00 m 2  1.75 kg  m2  1.92 kg  m 2
12
and
KEr 
8.52
1 2
1
I 
1.92 kg  m 2
2
2

  2.50
rad s   6.00 J
2
As the bucket drops, it loses gravitational potential energy. The spool
gains rotational kinetic energy and the bucket gains translational kinetic
energy. Since the string does not slip on the spool,   r where r is the
radius of the spool. The moment of inertia of the spool is I 
1
2
Mr 2 ,
where M is the mass of the spool. Conservation of energy gives
 KE
t
 KEr  PEg

f

 KEt  KEr  PEg

i
1
1
m 2  I  2  mgy f  0  0  mgyi
2
2
or

1
11

2
m  r    Mr 2   2  mg yi  y f

2
22

This gives
 

2mg yi  y f

 m  21 M  r 2


2  3.00 kg  9.80 m s2
  4.00 m 
 3.00 kg+ 21  5.00 kg    0.600 m 
2
 10.9 rad s
8.54
8.56


(a)
L  I   MR2    2.40 kg   0.180 m 
(b)
1
1

2
L  I    MR 2     2.40 kg   0.180 m   35.0 rad s   1.36 kg  m 2 s
2

2
(c)
2
2

2
L  I    MR 2     2.40 kg   0.180 m   35.0 rad s   1.09 kg  m 2 s
5

5
(d)
2
2

2
L  I    MR 2     2.40 kg   0.180 m   35.0 rad s   1.81 kg  m 2 s
3

3
(a)
Yes, the bullet has angular momentum about an axis through the
2
 35.0
rad s   2.72 kg  m 2 s
hinges of the door before the collision. Consider the sketch at
the right, showing the bullet the instant before it hits the door. The
physical situation is identical to that of a point mass mg moving in
a circular path of radius r with tangential speed t = i. For that
situation the angular momentum is
 
Li  I i  i  mB r 2  i   mB r i
 r 


and this is also the angular momentum of the bullet about the axis
through the hinge at the instant just before impact.
(b)
No, mechanical energy is not conserved in the collision. The bullet embeds itself in the door with
the two
moving as a unit after impact. This is a perfectly inelastic collision
in which a significant amount of mechanic cal energy is converted to other forms, notably thermal energy.
(c)
Apply conservation of angular momentum with Li  mBri as discussed in part (a). After impact,
L f  I f  f   I door  Ibullet   f 

1
2

Mdoor L2  mBr 2  f where L = 1.00 m = the width of the
door and r = L – 10.0 cm = 0.900 m. Then,
L f  Li
f 

mBri
1
Mdoor L2  mBr 2
3



 0.005 kg   0.900 m  1.00  103
ms

1
18.0 kg  1.00 m 2   0.005 kg   0.900 m 2
3
yielding  f  0.749 rad s .
(d)
The kinetic energy of the door-bullet system immediately after impact is
KE f 
or
1
1 1
2
2
2
I f  2f    18.0 kg   1.00 m    0.005 kg   0.900 m    0.749 rad s 
2
2 3

KE f  1.68 J .
The kinetic energy (of the bullet) just before impact was
KEi 
8.57
1
1
mBi2   0.005 kg  1.00  103 m s
2
2

2
 2.50  10 3 J
Each mass moves in a circular path of radius r = 0.500 m/s about the center of the connecting rod. Their
angular speed is
 

r

5.00 m s
 10.0 m s
0.500 m
Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating
system is
L  I    m1r 2  m2r 2     4.00 kg  3.00 kg   0.500 m  10.0 rad s   17.5 J  s
2
8.63
The initial angular velocity of the puck is
i 
 t i
ri

0.800 m s
rad
 2 .00
0.400 m
s
Since the tension in the string does not exert a torque about the axis of revolution, the angular
momentum of the puck is conserved, or If f = Ii i.
Thus,
 Ii
 If
f  
2

 mri2 
 0.400 m 




 i
 2 i
 0.250 m   2 .00 rad s   5.12 rad s

 mrf 
The net work done on the puck is
Wnet  KE f  KEi 


1
1
1
m 2 2
 rf  f  ri2i2 
I f  2f  I i i2   mrf2  2f  mri2 i2  



2
2
2
2 


or
 0.120 kg  
Wnet 
2
2
2
2
2 
 0.250 m   5.12 rad s    0.400 m   2 .00 rad s  
This yields Wnet  5.99  102 J .
8.65
(a)
From conservation of angular momentum, If f = Ii i, so
 Ii 
 I1 
o
 i  
 I1  I 2 
 If 
f  
2
(b)
KE f 
 I

 I
1
1
1
  I1 
I f  2f   I1  I 2   1   o2   1   I1  o2   
 KEi
2
2
 I1  I 2 
 I1  I 2   2
  I1  I 2 
or
KE f
KEi

I1
I1  I 2
Since this is less than 1.0, kinetic energy was lost.
8.66
The initial angular velocity of the system is
rev   2 rad 
 0.40 rad s
s   1 rev 

 i   0.20

The total moment of inertia is given by
I  I man  I cylinder  mr 2 
1
1
2
M R2   80 kg r 2   25 kg  2 .0 m 
2
2
Initially, the man is at r  2 .0 m from the axis, and this gives I i  3.7  102 kg  m 2 . At the end,
when r = 1.0, the moment of inertia is I f  1.3  102 kg  m 2 .
(a)
From conservation of angular momentum, If f = Ii i, or
 Ii 
 3.7  102 kg  m 2 
 0.40 rad s   1.14  rad s  3.6 rad s
 i  
 1.3  102 kg  m 2 
 If 
f  
(b)
The change in kinetic energy is KE 
KE 
1
I
2 f
 2f  21 I f  i2 , or
2
1
rad 
1
rad 


1.3  102 kg  m 2  1.14 

3.7  102 kg  m 2  0.40 



2
s 
2
s 




or KE  5.4  102 J . The difference is the work done by the man as he walks inward.
2