Chapter 8
8
Rotational Equilibrium and
Rotational Dynamics
PROBLEM SOLUTIONS
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 
8.2

r

76.0 N  m
 217 N
0.350 m
The torque of the applied force is  = rF sin. Thus, if r = 0.330 m,  = 75.0º, and the torque has the maximum allowed value of max = 65.0 N  m, the applied force is
F 
8.3
 max
65.0 N  m

 204 N
r sin 
 0.330 m  sin 75.0 
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.
Page 8.1
Chapter 8
(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
8.4
0 = 30 N  m counterclockwise
c = 30 N  m counterclockwise


The lever arm is d  1.20  102 m cos 48.0 8.03  103 m , and the torque is
  Fd   80.0 N   8.03  103 m   0.642 N  m counterclockwise
8.5
(a)
  Fg   lever arm    mg    sin  


  3.0 kg  9.8 m s2   2.0 m  sin 5.0   5.1 N  m
(b)
The magnitude of the torque is proportional to the sin , where  is the angle between the direction of the
force and the line from the pivot to the point where the force acts. Note from the sketch that this is the same as the
an
gle the pendulum string makes with the vertical.
Since sin  increases as  increases, the torque also increases with the angle.
8.6
The object is in both translational and rotational equilibrium. Thus, we may write:
Fx  0 
Fx  Rx  0
Fy  0 
Fy  Ry  Fg  0
and
Page 8.2
Chapter 8
 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 sin12.0   0.080 m    41.5 N   0.290 m   0 or Ft  724 N
Then Fy  0

 Fsy   724 N  sin12.0  41.5 N = 0 , yielding Fsy  109 N
Fy  0 gives Fsx   724 N  cos12.0 = 0 , or Fsx  708 N
Therefore,
Fs 
8.8
(a)
Fsx2  Fsy2 
 708 N 2  109 N 2
 716 N
Since the beam is in equilibrium, we choose the center
as our pivot point and require that
 center  FSam  2.80 m   FJoe 1.80 m   0
or
FJoe  1.56FSam
[1]
Also,
Fy  0

FSam  FJoe  450 N [2]
Substitute Equation [1] into [2] to get the following:
FSam  1.56FSam  450 N
or
FSam 
Page 8.3
450 N
 176 N
2.56
Chapter 8
Then, Equation [1] yields FJoe  1.56 176 N   274 N .
(b)
If Sam moves closer to the center of the beam, his lever arm about the beam center decreases, so the force
FSam must increase to continue applying a clockwise torque capable of offsetting Joe’s counterclockwise
torque. At the same time, the force FJoe would decrease since the sum of the two upward forces equal the
magnitude of the downward gravitational force.
(c)
If Sam moves to the right of the center of the beam, his torque about the midpoint would then be counterclockwise. Joe would have to hold down on the beam in order to exert an offsetting clockwise torque.
8.9
Require that  = 0 about an axis through the elbow and perpendicular to the page. This gives


    2.00 kg  9.80 m s2   25.0 cm + 8.00 cm    FB cos 75.0  8.00 cm   0
or
FB 
8.10
 19.6 N   33.0 cm 
 8.00 cm  cos 75.0
 312 N
Since the bare meter stick balances at the
49.7 cm mark when placed on the fulcrum, the
center of gravity of the meter stick is located 49.7 cm
from the zero end. Thus, the entire weight of the meter
stick may be considered to be concentrated at this point.
The free-body diagram of the stick when it is balanced
with the 50.0-g mass attached at the 10.0 cm mark is as
given at the right.
Requiring that the sum of the torques about point O be zero yields
  50.0 g  g   39.2 cm  10.0 cm   M g  49.7 cm  39.2 cm   0
or
 39.2 cm  10.0 cm 
M   50.0 g  
 139 g
 49.7 cm  39.2 cm 
Page 8.4
Chapter 8
8.11
Consider the remaining plywood to consist of two
parts: A1 is a 4.00-ft-by-4.00-ft section with center of gravity
located at (2.00 ft, 2.00 ft), while A2 is a 2.00-ft-by-4.00-ft section with
center of gravity at (6.00 ft, 1.00 ft). Since the plywood is uniform, its
mass per area  is constant and the mass of a section having
area A is m = A. The center of gravity of the remaining
plywood has coordinates given by




 
  6.00 ft 





 


xcg 
16.0 ft 2  2.00 ft   8.00 ft 2
mi xi
 A1 x1   A2 x2


mi
 A1   A2
16.0 ft 2  8.00 ft 2
ycg 
16.0 ft 2  2.00 ft   8.00 ft 2 1.00 ft 
mi yi
 A1 y1   A2 y2


 1.67 ft
mi
 A1   A2
16.0 ft 2  8.00 ft 2
 3.33 ft
and
8.12
(a)


Mg   90.0 kg  9.80 m s2  882 N
(b)


mg   55.0 kg  9.80 m s2  539 N
The woman is at x = 0 when n1 is greatest. With this location of the woman, the counterclockwise torque
about the center of the beam is a maximum. Thus, n1 must be exerting its maximum clockwise torque about
the center to hold the beam in rotational equilibrium.
(c)
n1 = 0 As the woman walks to the right along the beam, she will eventually reach a point where the beam will
start to rotate clockwise about the rightmost pivot. At this point, the beam is starting to lift up off of the left
most pivot and the normal force exerted by that pivot will have diminished to zero.
(d)
When the beam is about to tip, n1 = 0 and Fy = 0, gives 0+n2 – Mg – mg = 0, or
n2  Mg  mg  882 N  539 N  1.42  103 N
Page 8.5
Chapter 8
(e)
 0 when the beam is about to tip (n = 0) gives
Requiring that   rightmost
1
pivot
  4.00 m  x  mg   4.00 m  3.00 m  Mg  0
or  mg  x  1.00 m Mg   4.00 m mg , and
x   1.00 m 
M
 4.00 m
m
Thus,
x   1.00 m 
(f)
 90.0 kg 
 55.0 kg 
 4.00 m  5.64 m
left  0 gives
When n1 = 0 and n2 = 1.42 × 103 N, requiring that  end


0   539 N  x   882 N   3.00 m   1.42  103 N  4.00 m   0
or
x 
3.03  103 N  m
 5.62 N
539 N
which, within limits of rounding errors, is the same as the answer to part (e).
8.13
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
or 8.0 x + 12 m = 0 which yields x = – 1.5 m
Also, requiring that ycg  mi yi mi  0 gives
Page 8.6
 0
Chapter 8
 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
 0
or 8.0y + 12 m = 0 yielding y = – 1.5 m
Thus, the 8.0-kg object should be placed at coordinates (–1.5 m, –1.5m) .
8.14
(a)
As the woman walks to the right along the beam,
she will eventually reach a point where the beam will start
to rotate clockwise about the rightmost pivot. At this point,
the beam is starting to lift up off of the leftmost pivot and the
normal force, n1, exerted by that pivot will have diminished to
zero.
Then, Fy  0

0  mg  Mg  n2  0 , or
n2   m  M  g
(b)
left  0 gives
When n1 = 0 and n2 = (m + M)g, requiring that  end
0   mg  x   Mg 
(c)
L
  mg  Mg 
2
 0
or
M

x  1 

m 
 M

L
 2m 
If the woman is to just reach the right end of the beam (x = L) when n1 = 0 and n2 = (m+M)g (i.e., the beam is
ready to tip), then the result from Part (b) requires that
M

L  1 

m 
8.15
 M

L or
 2m 

 1  M 2m 
1  M m 
 m  M 2
 
L
 m  M 
In each case, the distance from the bar to the center of mass of the body is given by
xcg 
marms xarms  mtorso xtorso  mthighs xthighs  mlegs xlegs
mi xi

mi
marms  mtorso  mthighs  mlegs
where the distance x for any body part is the distance from the bar to the center of gravity of that body part. In each
Page 8.7
Chapter 8
case, we shall take the positive direction for distances to run from the bar toward the location of the head.
Note that mi   6.87  33.57  14.07  7.54  kg  62.05 kg .
With the body positioned as shown in Figure P8.15b, the distances x for each body part is computed using the
sketch given below:
 
xarms   rcg
arms
xtorso 
 rcg
xthighs 
xlegs 
 
arms

arms
arms
 0.239 m

  0.548 m + 0.337 m  0.885 m
torso
torso
torso

 
 rcg
thighs
thighs
   0.548 + 0.601+ 0.151 m  1.30 m
 
 rcg
legs
   0.548 + 0.601+ 0.374 + 0.227  m  1.75 m
With these distances and the given masses we find
xcg 
 62.8 kg  m
  1.01 m
62.05 kg
With the body positioned as shown in Figure P8.15c, we use the following sketch to determine the distance x for
each body part:
 
xarms   rcg
arms
xtorso 
 rcg
arms
 0.239 m
 
torso
  0.548 m  0.337 m  0.211 m
Page 8.8
Chapter 8
xthighs 
xlegs 
arms
arms


torso
torso
 
 rcg

thighs
thighs
   0.548  0.601  0.151 m  0.204 m
 
 rcg
legs
   0.548  0.601  0.374  0.227  m  0.654 m
With these distances, the location (relative to the bar) of the center of gravity of the body is
xcg 
8.16
 0.924 kg  m
  0.015 m  0.015 m towards the head
62.05 kg
With the coordinate system shown below, the coordinates of the center of gravity of each body part may be computed:
xcg,arms  0
ycg,arms 
 
xcg,torso  rcg
xcg,thighs 
xcg,legs 
torso
torso
torso

arms
 
 rcg
arms
 0.309 m
 0.337 m
ycg,torso  0
 
thighs
ycg,thighs  rcg
 
thighs
thighs
cos 60.0 rcg
ycg,legs 
thighs
sin 60.0 0.324 m
 rcg
cos 60.0 0.676 m
 
legs
 1.02 m
sin 60.0 0.131 m
With these coordinates for individual body parts and the masses given in Problem 8.15, the coordinates of the center of mass for the entire body are found to be
xcg 
marms xcg,arms  mtorso xcg,torso  mthighs xcg,thighs  mlegs xcg,legs
marms  mtorso  mthighs  mlegs
and
Page 8.9

28.5 kg  m
 0.459 m
62.05 kg
Chapter 8
ycg 
8.17
marms ycg,arms  mtorso ycg,torso  mthighs ycg,thighs  mlegs ycg,legs
marms  mtorso  mthighs  mlegs

6.41 kg  m
 0.103 m
62.05 kg
The free-body diagram for the spine is shown below.
When the spine is in rotational equilibrium, the sum of the torques about the left end (point O) must be zero. Thus,
 2L 
 L
Ty 
  350 N      200 N   L   0

 3 
 2
Yielding Ty  T sin12.0  562 N .
The tension in the back muscle is then T 
562 N
= 2.71  103 N  2.71 kN .
sin 12.0
The spine is also in translational equilibrium, so Fx  0  Rx  Tx  0 and the compression force in the
spine is
Rx  Tx  T cos12.0=  2.71 kN  cos12.0  2.65 kN
8.18
In the free-body diagram of the foot
given at the right, note that the force
R (exerted on the foot by the tibia) has been
replaced by its horizontal and vertical
components. Employing both conditions of
equilibrium (using point O as the pivot point)
gives the following three equations:
Page 8.10
Chapter 8
Fx  0  R sin15.0  T sin   0
or
R 
T sin 
sin 15.0 
[1]
Fy  0  700 N  R cos15.0  T cos   0
[2]
  700 N  18.0 cm  cos    T  25.0 cm  18.0 cm   0
O  0 
or
T = (1 800 N) cos 
[3]
Substituting Equation [3] into Equation [1] gives
 1 800 N 
R  
sin  cos 
 sin 15.0  
[4]
Substituting Equations [3] and [4] into Equation [2] yields
 1 800 N 
2
 tan 15.0   sin  cos    1 800 N  cos   700 N
which reduces to: sin  cos  = (tan 15.0º) cos2  + 0.104 2.
Squaring this result and using the identity sin2   1  cos2  gives
 tan2 15.0   1 cos4    2 tan 15.0   0.104 2   1 cos2   0.104 2   0
2
In this last result, let u = cos2  and evaluate the constants to obtain the quadratic equation
1.071 8 u2   0.944 2  u   0.010 9  0
The quadratic formula yields the solutions u = 0.869 3 and u = 1.011 7.
Thus,
  cos1


0.869 3  21.2 or   cos1


0.011 7  83.8
Page 8.11
Chapter 8
We ignore the second solution since it is physically impossible for the human foot to stand with
the sole inclined at 83.8° to the floor. We are the left with:  = 21.2º.
Equation [3] then yields
T  1 800 N  cos 21.2  1.68  103 N
and Equation [1] gives
R 
8.19
1.68  103 N  sin 21.2
sin 15.0
 2.34  103 N
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
8.20
Consider the torques about an axis perpendicular to the page through the left end of the scaffold.
  0  T1  0    700 N  1.00 m    200 N  1.50 m   T2  3.00 m   0
From which, T2 = 333 N.
Page 8.12
Chapter 8
Then, from Fy = 0, we have
T1+T2 – 700 N – 200 N = 0
or
T1 = 900 N – T2 = 900 N – 333 N = 567 N
8.21
Consider the torques about an axis
perpendicular to the page and through
the left end of the plank.
 = 0 gives
  700 N   0.500 m    294 N  1.00 m    T1 sin 40.0    2 .00 m   0
or T1 = 501 N.
Then, Fx  0 gives  T3  T1 cos 40.0°  0 , or
T3   501 N  cos 40.0  384 N
From
Fy  0 , T2  994 N  T1 sin 40.0  0 ,
or T2  994 N   501 N  sin 40.0  672 N .
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
Page 8.13
Chapter 8
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
8.23
The required dimensions are as follows:
d1   4.00 m  cos 50.0°  2 .57 m
d2  d cos 50.0°   0.643 d
d3   8.00 m  sin 50.0  6.13 m
Fy  0 yields F1  200 N  800 N = 0 , or F1 = 1.00  103 N.
When the ladder is on the verge of slipping,
f   fs max  s n  s F1
or


f   0.600  1.00  103 N  600 N
Then, Fx = 0 gives F2 = 600 N to the left.
Finally, using an axis perpendicular to the page and through the lower end of the ladder  = 0, gives
–(200 N)(2.57 m) – (800 N)(0.643)d+(600 N)(6.13 m) = 0
or
d 
 3.68  103  550 
0.643  800 N 
Nm
 6.15 m when the ladder is ready to slip
Page 8.14
Chapter 8
8.24
(a)
(b)
The point of intersection of two unknown forces is always a good choice as the pivot point in a torque calculation. Doing this eliminates these two unknowns from the calculation (since they have zero lever arms about
the chosen pivot) and makes it .easier to solve the resulting equilibrium equation .
L

0  0  mg  cos    T  L sin    0
2

(c)
   hinge  0
(d)
Solving the above result for the tension in the cable gives
T 

 mg 2  L cos 
L sin 

mg
2 tan 
or
T 
 16.0 kg   9.80
m s2
2 tan 30.0


 136 N
Fy  mg  0
(f)
Solving the above results for the components of the hinge force gives
(g)
and
and


Fy  mg  16.0 kg  9.80 m s2  157 N
Attaching the cable higher up would allow the cable to bear some of the weight, thereby reducing the stress
on the hinge. It would also reduce the tension in the cable.
8.25

Fx  0
Fx = T = 136 N
Fx  T  0
Fy  0
(e)
Consider the free-body diagram of the
material making up the center point in
Page 8.15
Chapter 8
the rope given at the right. Since this
material is in equilibrium, it is necessary
to have Fx = 0 and Fy = 0, giving
Fx  0: T2 sin   T1 sin   0
or T2 = T1, meaning that the rope has a uniform tension T throughout its length.
T cos   T cos   475 N  0 where cos  
Fy  0:
0.500 m
 6.00 m 2   0.500 m 2
and the tension in the rope (force applied to the car) is
 475 N 
475 N
T 

2 cos
8.26
 6.00 m 2   0.500 m 2
2  0.500 m 
 2.86  103 N  2.86 kN
(a)
(b)
  lower  0
or
T 
(c)
Fx  0

Fy  0

end

L

0  0  mg  cos    T  L sin    0
2

mg  cos  
mg

cot 
2  sin  
2
 T  s n  0
n  mg  0
or
T  s n
[1]
n  mg
[2]
Substitute Equation [2] into [1] to obtain T = smg.
Page 8.16
Chapter 8
(d)
Equate the results of parts (b) and (c) to obtain s = cot /2
This result is valid only at the critical angle  where the beam is on the verge of slipping
(i.e., where fs = (fs)max is valid.)
(e)
At angles below the critical angle (where fs   fs max is valid), the beam will slip. At larger angles, the
static friction force is reduced below the maximum value, and it is no longer appropriate to use  s in the calculation.
8.27
Consider the torques about an axis perpendicular
to the page and through the point where the force
T acts on the jawbone.
  0  (50.0 N) (7.50 cm)  R (3.50 cm)  0 , which
yields R  107 N .
Then, Fy  0    50.0 N  + T  107 N  0 , or T  157 N .
8.28
Observe that the cable is perpendicular to the boom. Then, using  = 0 for an axis perpendicular to the page and
through the lower end of the boom gives
L

3
  1.20 kN   cos 65°   T 
2

4

L    2.00 kN   L cos 65°   0

or .T = 1.47 kN
From Fx  0 , H  T cos 25°  1.33 kN to the right
and Fy = 0 gives
V = 3.20 kN – T sin 25 = 2.58 kN upward
8.29
First, we resolve all forces into components parallel to and
perpendicular to the tibia, as shown. Note that  = 40.0 and
Page 8.17
Chapter 8
wy = (30.0 N) sin 40.0 = 19.3 N
Fy = (12.5 N) sin 40.0 = 8.03 N
and
Ty = T sin 25.0
Using  = 0 for an axis perpendicular to the page and
through the upper end of the tibia gives
d
d
 T sin 25.0°  5  19.3 N  2   8.03 N  d
 0
or T = 209 N.
8.30
When x = xmin , the rod is on the verge of slipping, so
f   fs max  s n  0.50 n
From Fx  0 , n  T cos 37°  0, or n  0.80 T . Thus,
f  0.50  0.80 T   0.40 T
From Fy  0 , f  T sin 37  2 w  0, or
0.40 T  0.60 T  2 w  0 , giving T  2 w
Using  = 0 for an axis perpendicular to the page and through the left end of the beam gives
w  xmin  w  2.0 m    2 w  sin 37°   4.0 m   0 , which reduces to xmin = 2.8 m.
8.31
The moment of inertia for rotations about an axis is I  mi ri2 , where ri is the distance mass mi is from that axis.
(a)
For rotation about the x-axis,
I x   3.00 kg   3.00 m    2 .00 kg   3.00 m 
2
2
  2.00 kg   3.00 m    4.00 kg   3.00 m   99.0 kg  m 2
2
2
Page 8.18
Chapter 8
(b)
When rotating about the y-axis,
I y   3.00 kg   2 .00 m    2 .00 kg   2 .00 m 
2
2
  2.00 kg   2.00 m    4.00 kg   2.00 m   44.0 kg  m 2
2
(c)
2
For rotations about an axis perpendicular to the page through point O, the distance ri for each mass is
ri 
 2 .00 m 2   3.00 m 2

13.0 m
Thus,


IO   3.00  2.00  2.00  4.00  kg  13.0 m2  143 kg  m2
8.32
The required torque in each case is  = I. Thus,
 x  Ix    99.0 kg  m2   1.50 rad s2   149 N  m
 y  I y    44.0 kg  m2 1.50 rad s2   66.0 N  m
and
O  IO   143 kg  m2 1.50 rad s2   215 N  m
8.33
 net
 0.330 m   250 N   87.8 kg  m 2
rF sin 90




0.940 rad s2
(a)
 net  I 
(b)
For a solid cylinder, I = Mr2/2, so
M 
(c)

I 

2 87.8 kg  m2
2I

r2
 0.330 m 2

 1.61  103 kg
  0   t  0   0.940 rad s2   5.00 s   4.70 rad s
Page 8.19
Chapter 8
8.34


(a)
I  2Idisk  Icylinder  2 MR2 2  mr 2 2 or I  MR2  mr 2 2
(b)
g = 0 Since the line of action of the gravitational force passes through the rotation axis, it has zero lever arm
about this axis and zero torque.
(c)
The torque due to the tension force is positive. Imagine gripping the cylinder with your right hand so
your fingers on the front side of the cylinder point upward in the direction of the tension force. The thumb of
your right hand then points toward the left (positive direction) along the rotation axis. Because   I  , the
torque and angular acceleration have the same direction. Thus, a positive torque produces a
positive angular acceleration. When released, the center of mass of the yoyo drops downward, in the
negative direction. The translational acceleration is negative.
(d)
Since, with the chosen sign convention, the translational acceleration is negative when the angular acceleration is positive, we must include a negative sign in the proportionality between these two quantities. Thus, we
write: a = r or  = –a/r
(e)
Translation:
Fy  mtotal a
(f)
T   2M  m  g   2M  m  a
[1]
Rotational:
  I 
(g)


rT sin 90  I 
or
rT  I 
[2]
Substitute the results of (d) and (a) into Equation [2] to obtain

mr 2  a

 a r 
T  I   I
   MR 2 

r
 r 
2  r 2

  R2
m
T    M     as
2
  r 

or
Substituting Equation [3] into [1] yields
2
  M  R r   m 2  a   2 M  m  g   2 M  m  a


Page 8.20
or
a 
  2M  m  g
2 M  M  R r   3m 2
2
[3]
Chapter 8
8.35

  2  2.00 kg   1.00 kg  9.80 m s2

(h)
a 
(i)
From Equation [1], T   2M  m   g  a    5.00 kg  9.80 m s2  2.72 m s2  35.4 N .
( j)
y   0  t  at 2 2
(a)
Consider the free-body diagrams of the cylinder and
2  2.00 kg    2.00 kg  10.0 4.00   3 1.00 kg  2
2
 2.72 m s2


2  y 
t 
a
2  1.00 m 

2.72 m s2

 0.857 s
man given at the right. Note that we shall adopt a sign
convention with clockwise and downward as the positive
directions. Thus, both a and  are positive in the indicated
directions and a = r. We apply the appropriate form of Newton’s
second law to each diagram to obtain the following:
Rotation of Cylinder:   I 

rT sin 90°  I , or T  I  r,
so
T 
11
  a
Mr 2   


r
r 2
and
1
Ma
2
T 
[1]
Translation of man:
Fy  ma

mg  T  ma
Equating Equations [1] and [2] gives
a 
(b)

T  m  g  a
or
1
2
Ma  m  g  a  , or

 75.0 kg  9.80 m s2
mg

 3.92 m s2
m M 2
75.0 kg+  225 kg 2 
From a = r, we have
Page 8.21
[2]
Chapter 8
a
3.92 m s2

 9.80 rad s2
r
0.400 m
 
(c)
As the rope leaves the cylinder, the mass of the cylinder decreases, thereby decreasing the moment of inertia.
At the same time, the weight of the rope leaving the cylinder would increase the downward force acting tangential to the cylinder, and hence increase the torque exerted on the cylinder. Both of these effects will cause
the acceleration of the system to increase with time. (The increase would be slight in this case, given the
large mass of the cylinder.)
8.36
The angular acceleration is   ( f  i ) t   (i t ) since  f  0
Thus, the torque is   I    (I i t ) . But, the torque is also  = – fr, so the magnitude of the required friction force is
f


12 kg  m 2  50 rev min   2  rad   1 min 
I i

 1 rev   60 s   21 N
r  t 
 0.50 m   6.0 s 
Therefore, the coefficient of friction is
k 
8.37
8.38
f
21 N

 0.30
n
70 N
(a)
  F  r   0.800 N   30.0 m   24.0 N  m
(b)
 
(c)
a t  r    30.0 m  0.0356 rad s2  1.07 m s2

I


m r2

24.0 N  m
 0.750 kg   30.0 m 

2
 0.0356 rad s2

I  MR2  1.80 kg   0.320 m   0.184 kg  m2
2
 net   applied   resistive  I  , or F  r  f  R  I 
yielding
Page 8.22
Chapter 8
F 
8.39
(a)
F 
(b)
F 
I 
I  f  R
r
 0.184 kg  m2   4.50

rad s2   120 N   0.320 m 
4.50 
 0.184 kg  m2   4.50
10 2
m

rad s2   120 N   0.320 m 
2.80 
10 2
m
 872 N
 1.40 kN
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 

2
 1 rev   2 rad s
2 .00 s
Thus,   F  r  I  gives
I
F 

r
8.40
(a)
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
 m2 g  T2  m2 a m2
or
T2  m2  g  a 
[2]
Page 8.23
Chapter 8
  I 
For M :
 rT2  rT1  I 
T2  T1  I  r
or
[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
M

 m1  m2  2  a   m2  m1  g
or
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 .
8.41
The initial angular velocity of the wheel is zero, and the final angular velocity is
f 
v
50.0 m s

 40.0 rad s
r
1.25 m
Hence, the angular acceleration is
 
 f  i
t

40.0 rad s  0
 83.3 rad s2
0.480 s
The torque acting on the wheel is   fk  r , so   I  gives
fk 


110 kg  m 2 83.3 rad s2
I

r
1.25 m

 7.33  103 N
Thus, the coefficient of friction is
k 
fk
7.33  103 N

 0.524
n
1.40  10 4 N
Page 8.24
Chapter 8
8.42
(a)
The moment of inertia of the flywheel is
I 
1
1
2
MR2   500 kg   2 .00 m   1.00  103 kg  m 2
2
2
and the angular velocity is
rev   2 rad   1 min 

   5000
 524 rad s

min   1 rev   60 s 
Therefore, the stored kinetic energy is
1
1
I2 
1.00  103 kg  m 2
2
2

KEstored 
(b)
  524
rad s   1.37  108 J
2
A 10.0-hp motor supplies energy at the rate of
 746 W 
 7.46  103 J s
 1 hp 
P   10.0 hp  
The time the flywheel could supply energy at this rate is
t 
8.43
KEstored
P

1.37  108 J
 1.84  10 4 s  5.10 h
7.46  103 J s
The moment of inertia of the cylinder is
I 
1
1  w
1  800 N 
2
MR 2    R 2  
1.50 m   91.8 kg  m 2


2
2
2 g
2  9.80 m s 
The angular acceleration is given by
 

I

 50.0 N 1.50 m   0.817 rad s2
FR

I
91.8 kg  m 2
At t = 3.00 s, the angular velocity is
Page 8.25
Chapter 8
  i   t  0   0.817 rad s2  3.00 s   2 .45 rad s
and the kinetic energy is
KErot 
8.44
(a)
(b)
1
1
I2 
91.8 kg  m2
2
2

  2.45
rad s   276 J
2
Hoop:
I  MR2   4.80 kg   0.230 m   0.254 kg  m 2
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 R2  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:
Page 8.26
Chapter 8
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
Mv2 , 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  Irod  I3  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
2
and the rotational kinetic energy is
KEr 
(b)
1
1
I2 
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
2.00 kg   1.00 m   1.75 kg  m 2  1.92 kg  m 2

12
and
KEr 
8.46
1
1
I2 
1.92 kg  m 2
2
2

  2.50
rad s   6.00 J
2
Using conservation of mechanical energy,
 KE
trans
 KErot  PEg

f

 KEtrans  KErot  PEg
or
Page 8.27

i
Chapter 8
1
1
M vt2  I  2  0  0  0  M g  L sin  
2
2
Since I  2 5 MR2 for a solid sphere and vt  R when rolling without slipping, this becomes
1
1
M R2  2  M R 2 2  M g L sin  
2
5
and reduces to
 
8.47
(a)
10 gL sin 

7 R2

10 9.8 m s2
  6.0 m  sin 37
7  0.20 m 
2
 36 rad s
Assuming the disk rolls without slipping, the friction force between the disk and the ramp does no work. In
this case, the total mechanical energy of the disk is constant with the value
E  KEi  (PEg )i  0  Mgh  MgL sin  . When the disk gets to the bottom of the ramp, PEg  0 and
KE f  KEt  KEr  E  MgL sin  . Also, since the disk does not slip,   v R and
KEr 
1
1
I2 
2
2
1
2  v
 2 MR   R 
2
1
2

1
1
2
 2 M v   2 KEt
Then,
KEtotal  KEt 
1
KEt  E  MgL sin  or
2
31

M v2   M gL sin 

2  2
and
v
(b)
4 gL sin 

3

4 9.80 m s2
  4.50 m  sin 15.0
3
The angular speed of the disk at the bottom is
 
v
3.90 m s

 15.6 rad s
R
0.250 m
Page 8.28
 3.90 m s
Chapter 8
8.48
(a)
Assuming the solid sphere starts from rest, and taking y =0 at the level of the bottom of the incline, the total

mechanical energy will be split among three distinct forms of energy E  PEg

i
 mgh as the sphere rolls
down the incline. These are
rotational kinetic energy,
1
I2
2
translational kinetic energy,
1
mv2
2
and
gravitational potential energy, mgy
where y is the current height of the center of mass of the sphere above the level of the bottom of the incline.
(b)
The force of static friction, exerted on the sphere by the incline and directed up the incline, exerts a torque
about the center of mass giving the sphere an angular acceleration.
(c)
KEt 
I 
2
5
1
2
Mv2 and KEr 
1
2
I  2 where v  R (since the sphere rolls without slipping) and
MR2 for a solid sphere. Therefore,


2 MR2 5  2
KEr
I2 2
2 MR2 2
2




2
2
2
KEt  KEr
Mv 2  I 2
7
5 MR2 2  2 MR2 2
M  R   2 MR2 5  2
8.49
Using Wnet  KE f  KEi 
f 
8.50
1
2


I  2f  0 , we have
2F  s

I
2 Wnet

I
2  5.57 N   0.800 m 
4.00  10 4 kg  m 2
 149 rad s
The work done on the grindstone is Wnet  F  s  F   r     F  r       .
Thus, Wnet     
1
2
I  2f 
1
2
I i2 , or
Page 8.29
Chapter 8
 25.0 N  m 15.0 rev 
2 rad 
1

0.130 kg  m 2  2f  0

1 rev 
2


This yields
rad   1 rev 
 30.3 rev s
s   2 rad 

 f   190

8.51
(a)
KEtrans 
KErot 
(b)

(c)
8.52
1
1
2
mvt2   10.0 kg  10.0 m s   500 J
2
2
1 2
1 1
  v2 
I    mR 2   t2 
 R 
2
2 2
1
1
2
mvt2   10.0 kg  10.0 m s   250 J
4
4
KEtotal  KEtrans  KErot  750 J
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
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

Page 8.30
Mr 2 ,
Chapter 8
This gives
 
8.53
(a)

2mg yi  y f
m 
1
2
M


r2

2  3.00 kg  9.80 m s2

 3.00 kg+
1
2
  4.00 m 
 5.00 kg    0.600 m 2
 10.9 rad s
The arm consists of a uniform rod of 10.0 m length and the mass of the seats at the lower end is negligible.
The center of gravity of this system is then located at the
geometric center of the arm, located 5.00 m from the upper end.
From the sketch at the right, the height of the center of gravity above the zero level is
ycg  10.0 m   5.00 m  cos  .
(b)
When   45.0  , ycg  10.0 m   5.00 m  cos 45.0  6.46 m
and

PEg  mgycg   365 kg  9.80 m s2
(c)
  6.46 m  
2.31  104 J
In the vertical orientation,  = 0 and cos = 1, giving
ycg  10.0 m  5.00 m  5.00 m . Then,

PEg  mgycg   365 kg  9.80 m s2
(d)
  5.00 m  
1.79  104 J
Using conservation of mechanical energy as the arm starts
from rest in the 45° orientation and rotates about the upper
end to the vertical orientation gives
 
1
I end  2f  mg ycg
2
f
 
 0  mg ycg
i
or
f 
For a long, thin rod: Iend  mL2 3 . Equation [1] then becomes
Page 8.31
   
2mg  ycg  ycg
i

I end
f


[1]
Chapter 8
   
2 m g  ycg  ycg
i

m L2 3
f 

6 9.80 m s2

f



   y 
6 g  ycg

i
cg f


L2
  6.46 m  5.00 m 
 10.0 m 2
 0.927 rad s
Then, from   r  , the translational speed of the seats at the lower end of the rod is
  10.0 m   0.927 rad s   9.27 m s
8.54
8.55


(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)
The rotational speed of Earth is
E 
2
 35.0
rad s   2.72 kg  m 2 s
2 rad 
1d

 7.27  10 5 rad s

1 d  8.64  10 4 s 
2

Lspin  I sphere E   M E RE2   E
5

2
  5.98  10 24 kg 6.38  10 6 m
5


Page 8.32
2   7.27  10 5

rad s  7.08  10 33 J  s
Chapter 8
(b)
For Earth’s orbital motion,
 orbit 
2 rad
1y
1y

 3.156  107

 1.99  10 7 rad s
s 
and using data from Table 7.3, we find


2
Lorbit  I point orbit  ME Rorbit
orbit


 5.98  1024 kg 1.496  1011 m
8.56
(a)
2 1.99  107

rad s  2.67  10 40 J  s
Yes, the bullet has angular momentum about an axis through the
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  I bullet   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 
mB ri
1
Mdoor L2  mB r 2
3



 0.005 kg   0.900 m  1.00  103
ms

1
2
2
18.0 kg  1.00 m    0.005 kg   0.900 m 

3
Page 8.33
Chapter 8
yielding  f  0.749 rad s .
(d)
The kinetic energy of the door-bullet system immediately after impact is
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 
or
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  103 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
8.58
Using conservation of angular momentum, Laphelion  Lperihelion .




Thus, mra2 a  mrp2  p . Since  = 1/r at both aphelion and perihelion, this is equivalent to
 mra2  a


ra  mrp2  p rp , giving
 rp 
 0.59 A.U. 
a     p  
 54 km s   0.91 km s
 35 A.U. 
 ra 
8.59
The initial moment of inertia of the system is

2
Ii  mi ri2  4  M  1.0 m    M 4.0 m 2


Page 8.34

Chapter 8
The moment of inertia of the system after the spokes are shortened is

2
I f  m f rf2  4  M  0.50 m    M 1.0 m 2



From conservation of angular momentum, If f = Ii i, or
 Ii 
 i   4  2.0 rev s   8.0 rev s
 If 
f  
8.60

From conservation of angular momentum: I child  I m-g -r

f


 f  Ichild  I m -g -r i
i
where I m-g-r  275 kg  m2 is the constant moment of inertia of the merry-go-round.
Treating the child as a point object, Ichild  mr 2 where r is the distance the child is from the rotation axis. Conservation of angular momentum then gives
f
  25.0 kg  1.00 m 2  275 kg  m 2 
 mri2  I m-g-r 
  14.0 rev min 
  2
 i  
  25.0 kg   2.00 m 2  275 kg  m 2 
 mrf  I m-g-r 


or
 f  11.2
8.61
rev  2 rad 
min  1 rev 
 1 min 
 60.0 s   1.17 rad s
The moment of inertia of the cylinder before the putty arrives is
Ii 
1
1
2
M R2   10.0 kg  1.00 m   5.00 kg  m 2
2
2
After the putty sticks to the cylinder, the moment of inertia is
I f  Ii  mr 2  5.00 kg  m2   0.250 kg  0.900 m   5.20 kg  m2
2
Conservation of angular momentum gives If f = Ii i, or
Page 8.35
Chapter 8
 Ii 
 5.00 kg  m2 
 7.00 rad s   6.73 rad s
 i  
 5.20 kg  m2 
 If 
f  
8.62
The total moment of inertia of the system is


Itotal  I masses  I student  2 mr 2  3.0 kg  m2
2
Initially, r = 1.0 m, and Ii  2   3.0 kg  1.0 m    3.0 kg  m 2  9.0 kg  m 2 .


Afterward, r = 0.30 m, so
2
I f  2   3.0 kg  0.30 m    3.0 kg  m 2  3.5 kg  m 2


(a)
From conservation of angular momentum, If f = Ii i, or
 Ii 
 9.0 kg  m2 
 0.75 rad s   1.9 rad s
 i  
 3.5 kg  m2 
 If 
f  
(b)
8.63
KEi 
1
1
2
Ii i2 
9.0 kg  m 2  0.75 rad s   2.5 J
2
2
KE f 
1
1
2
I f  2f 
3.5 kg  m 2  1.9 rad s   6.3 J
2
2




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,
f
2
 Ii 
 mri2 
 0.400 m 
   i   2  i  
 2 .00 rad s   5.12 rad s
 0.250 m 
 If 
 mrf 
Page 8.36
Chapter 8
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
 0.250 m   5.12 rad s    0.400 m 

2
2
2
 2 .00
2
rad s  

This yields Wnet  5.99  102 J .
8.64


For one of the crew, Fc  m ac becomes n  m vt2 /r  mr i2 . We require n = mg, so the initial angular velocity must be i 
g/r . From conservation of angular momentum, I f  f  Ii i or  f  (Ii /I f )i . Thus,
the angular velocity of the station during the union meeting is
f
 I 
  i
 If 
 5.00  108 kg  m 2  150 65.0 kg 100 m 2  g
g
g

 
 1.12
2
r
r
 5.00  108 kg  m 2  50 65.0 kg 100 m   r


The centripetal acceleration experienced by the managers still on the rim is
ac  r  2f  r 12.3 m s2  1.12 
8.65
(a)
2
g
2
 1.12  9.80 m s2  12.3 m s2
r

From conservation of angular momentum, If f = Ii i, so
 Ii
 If
f  

 I1 
o
 i  
 I1  I 2 

2
(b)

 I

 I
1

1
1
  I
KE f  I f  2f   I1  I 2   1  o2   1   I1 o2    1  KEi
2
2
 I1  I2 
 I1  I2   2
  I1  I2 
or
Page 8.37
Chapter 8
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 Ii  3.7  102 kg  m2 . At the end, when r = 1.0,
the moment of inertia is I f  1.3  102 kg  m2 .
(a)
From conservation of angular momentum, If f = Ii i, or
 Ii 
 3.7  102 kg  m2 
 0.40 rad s   1.14  rad s  3.6 rad s
 i  
 1.3  102 kg  m2 
 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  10 2 kg  m 2  0.40 


2
s 
2
s 




2
or KE  5.4  102 J . The difference is the work done by the man as he walks inward.
8.67
(a)
The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that
of the woman so the total angular momentum maintains a constant value of Ltotal  Lwoman  Ltable  0 .
Since the final angular momentum is Ltotal  Iw w  It t  0 , we have
Page 8.38
Chapter 8
I 
 mw r 2 
I 
t    w  w   
 It 

t
 mw r 
 w 
 r     I   w
 t 
or
  60.0 kg  2 .00 m  
   1.50 m s   0.360 rad s
500 kg  m 2


t   
Hence, table  0.360 rad s counterclockwise .
(b)
Wnet  KE  KE f  0 
Wnet 
8.68
(a)
1
1
mw2  I t t2
2
2
1
1
2
2
60.0 kg 1.50 m s  
500 kg  m 2  0.360 rad s   99.9 J

2
2


In the sketch at the right, choose an axis perpendicular to
the page and passing through the indicated pivot. Then,
with  = 30.0, the lever arm of the force P is observed to be

5.00 cm
5.00 cm

 5.77 cm
cos 
cos 30.0
and  = 0 gives
 P  5.77 cm   150 N  30.0 cm   0
so
P 
(b)
150 N  30.0 cm 
5.77 cm
 780 N
Fy  0  n  P cos 30.0  0 , giving
n  P cos 30.0   780 N  cos30.0  675 N
Page 8.39
Chapter 8
Fx  0  f  F  P sin 30.0  0 , or
f  P sin 30.0  F   780 N sin 30.0  150 N  240 N
The resultant force exerted on the hammer at the pivot is
R 
f 2  n2 
 240 N 2   675 N 2
 716 N
at   tan1 (n/f )  tan1(675 N/240 N)  70.4 , or
R  716 N at 70.4 above the horizontal to the right
8.69
(a)
Since no horizontal force acts on the child-boat system, the center of gravity of this system will remain stationary, or
xcg 
mchild xchild  mboat xboat
 constant
mchild  mboat
The masses do not change, so this result becomes mchild xchild + mboatxboat = constant.
Thus, as the child walks to the right, the boat will move to the left .
(b)
Measuring distances from the stationary pier, with away from the pier being positive, the child is initially at
(xchild)i = 3.00 m and the center of gravity of the boat is at (xboat)i = 5.00. At the end, the child is at the right
end of the boat, so (xchild)f = (xboat)f + 2.00 m. Since the center of gravity of the system does not move,
we have  mchild xchild  mboat xboat  f 
 mchild xchild
 mboat xboat i , or
mchild  xchild  f  mboat   xchild  f  2.00 m   mchild  3.00 m   mboat  5.00 m 


and
 xchild  f

 xchild  f

mchild  3.00 m   mboat  5.00 m  2.00 m 
mchild  mboat
 40.0 kg   3.00 m    70.0 kg   5.00 m  2.00 m 
40.0 kg  70.0 kg
Page 8.40
 5.55 m
Chapter 8
(c)
When the child arrives at the right end of the boat, the greatest distance from the pier that he can reach is
xmax   xchild  f  1.00 m  5.55 m  1.00 m  6.55 m
This leaves him 0.45 m short of reaching the
turtle .
8.70
(a)
Consider the free-body diagram of the block given at
the right. If the +x-axis is directed down the incline,
Fx = max gives
mg sin 37.0  T  m at , or T  m  g sin 37.0  at 


T   12.0 kg   9.80 m s2 sin 37.0   2.00 m s 2 
 46.8 N
(b)
Now, consider the free-body diagram of the pulley.
Choose an axis perpendicular to the page and passing
through the center of the pulley,  = I  gives
 at 
T  r  I 
 r 
or
I 
(c)
8.71
 46.8 N   0.100 m 
T  r2

at
2.00 m s2
a 
2
 0.234 kg  m2
 2.00 m s2 
  i   t  0   t  t  
 2.00 s   40.0 rad s
 r 
 0.100 m 
If the ladder is on the verge of slipping, f   fs max  s n at both the
floor and the wall.
From Fx = 0, we find f1 – n2 = 0, or
n2 = sn1
[1]
Also, Fx = 0 gives n1 – w + sn2.= 0
Page 8.41
Chapter 8
Using Equation [1], this becomes


n 1 w  s s n1  0
or
n 1
w
w

 0.800 w
2
1  s
1.25
[2]
Thus, Equation [1] gives
n2  0.500  0.800 w   0.400 w
[3]
Choose an axis perpendicular to the page and passing through the lower end of the ladder. Then,  = 0 yields
L

 w  cos    n2  L sin    f2  L cos    0
2

Making the substitutions n2 = 0.400 w and f2  s n2  0.200 w , this becomes
L

 w  cos     0.400 w   L sin     0.200 w   L cos    0
2

and reduces to
 0.500  0.200 
sin   
 cos 

0.400
Hence, tan  = 0.750 and  = 36.9 .
8.72
Page 8.42
Chapter 8
We treat each astronaut as a point object, m, moving at speed v in a circle of radius r = d/2. Then the total angular
momentum is


L  I1  I 2  2  mr 2     2mr
 r 


(a)

Li  2mi ri  2  75.0 kg   5.00 m s   5.00 m 
Li  3.75  103 kg  m2 s
(b)
KEi 
1
1
1

m1 12i  m2 22i  2  mi2 


2
2
2
KEi   75.0 kg  5.00 m s   1.88  103 J= 1.88 kJ
2
8.73
(c)
Angular momentum is conserved: L f  Li  3.75  103 kg  m2 s .
(d)
f 
(e)
1

2
KE f  2  m 2f    75.0 kg  10.0 m s   7.50 kJ
2

(f)
Wnet  KE f  KEi  5.62 kJ
(a)

 d
Li  2  M      M  d
 2

(b)
1

KEi  2  M i2   M  2
2

(c)
L f  Li  M  d
Lf
 
2 mrf

3.75  103 kg  m2 s
 10.0 m s
2  75.0 kg  2.50 m 
Page 8.43
Chapter 8
8.74
Lf
M d
 2
2 M d 4 
(d)
f 
(e)
1

2
KE f  2  M  2f   M  2    4 M  2
2

(f)
Wnet  KE f  KEi  3M 2

2 Mrf


Choose an axis that is perpendicular to the page
and passing through the left end of the scaffold.
Then  = 0gives
  750 N   1.00 m    345 N  1.50 m 
  500 N   2 .00 m   1000 N   2 .50 m 
 TR  3.00 m   0
or
TR  1.59  103 N= 1.59 kN
Then,
Fy  0  TL   750  345  500  1000  N  1.59  103 N  1.01 kN
8.75
(a)
From conservation of angular momentum, L f  I f  f  Iii  Li , or

 Ii 
 i  
 If 

f  
2
5
2
5
2
2
 Ri 
MRi2 
 1.50  109 m 




 i

 i
 15.0  103 m   0.010 0 rev d 
MR2f 
 Rf 
giving
 f  1.00  108
(b)
 t  f


rev  2 rad  
1 d
 7.27  103 rad s



4
d  1 rev   8.64  10 s 


 R f  f  15.0  103 m 7.27  103 rad s  1.09  108 m s (which is about one-third the
Page 8.44
Chapter 8
speed of light).
8.76
(a)
Taking PEg = 0 at the level of the horizontal axis passing
through the center of the rod, the total energy of the rod
in the vertical position is
E  KE  PEg
 0  m1g   L   m2 g   L  
(b)
 m1  m2  gL
In the rotated position of Figure P8.76b, the rod is
in motion and the total energy is
E  KEr  PEg 

1
I total  2  m1gy1  m2 gy2
2
Figure P8.76
1
m1L2  m2 L2  2  m1g   L sin    m2 g   L sin  
2


or
E 
(c)
 m1  m2  L2 2
2
  m1  m2  gL sin 
In the absence of any nonconservative forces that do work on the rotating system, the total mechanical energy
of the system is constant. Thus, the results of parts (a) and (b) may be equated to yield an equation that can be
solved for the angular speed, , of the system as a function of angle .
(d)
In the vertical position, the net torque acting on the system is zero, net = 0. This is because the lines of action
of both external gravitational forces (m1g and m2g)pass through the pivot and hence have zero lever arms
about the rotation axis. In the rotated position, the net torque (taking clockwise as positive) is
 net    m1g  L cos    m2 g  L cos    m1  m2  gL cos 
Note that the net torque is not constant as the system rotates. Thus, the angular acceleration of the rotating
system, given by  = net/I, will vary as a function of . Since a net torque of varying magnitude acts on the
system, the angular momentum of the system will change at a nonuniform rate.
(e)
In the rotated position, the angular acceleration is
Page 8.45
Chapter 8
 
8.77
 net
I

 m1  m2  gL cos 
m1
L2
 m2
L2

 m1  m2  g cos 
 m1  m2  L
Let mp be the mass of the pulley, m1 be the mass of the sliding block, and m2 be the mass of the counterweight.
(a)
The moment of inertia of the pulley is I 
1
m p R2p and its angular velocity at any time is,  = /Rp where 
2
is the linear speed of the other objects. The friction force retarding the sliding block is
fk  k n  k  m1 g 
Choose PEg  0 at the level of the counterweight when the sliding object reaches the second photogate.
Then, from the work–energy theorem,

Wnc  KEtrans  KErot  PEg
 fk  s 

f

 KEtrans  KErot  PEg

i
2
1
11
 f 
m1  m2  2f   m p R 2p   2   0

  Rp 
2
22

1
11
  2 
m1  m2 i2   m p R 2p   i2   m2 gs

  Rp 
2
22
or
1
2
1  2
1

 m1  m2  2 m p   f  2
1  2

 m1  m2  2 m p  i  m2 gs  k  m1 g   s
This reduces to
f 
i2 
2  m2  k m1  gs
1
m1  m2  m p
2
and yields
f 


2
2  0.208 kg  9.80 m s2  0.700 m 
m

0.820

 1.63 m s

s 
1.45 kg
Page 8.46
Chapter 8
8.78
vf
1.63 m s
 54.2 rad s
0.030 0 m
(b)
f 
(a)
The frame and the center of each wheel moves forward at  = 3.35 m/s and each wheel also turns at angular
Rp

speed  = /R. The total kinetic energy of the bicycle is KE = KEt + KEr, or




1
 1

m frame  2 mwheel  2  2  I wheel  2 


2
2
 2 
1
1

m frame  2 mwheel  2 
mwheel R 2  2 
2
2
R 
KE 


This yields
KE 

(b)


1
m frame  3 mwheel v2
2
1
2
 8.44 kg  3  0.820 kg   3.35 m s   61.2 J
2
Since the block does not slip on the roller, its forward speed must
equal that of point A, the uppermost point on the rim of the roller.
That is,   v AE where v AE is the velocity of A relative to Earth.
Since the roller does not slip on the ground, the velocity of point O
(the roller center) must have the same magnitude as the tangential
speed of point B (the point on the roller rim in contact with the
ground). That is, vOE  R  O . Also, note that the velocity of point A relative
to the roller center has a magnitude equal to the tangential speed R, or vAO  R  O .
From the discussion of relative velocities in Chapter 3, we know that vAE  vAO  vOE . Since all of these
velocities are in the same direction, we may add their magnitudes getting vAE  vAO  vOE , or
  O  O  2O  2R .
The total kinetic energy is KE = KEt + KEr, or
Page 8.47
Chapter 8
1
1
mstone 2  2  mtree
2
 2
1
1

  mstone  mtree   2
2

4
KE 
This gives KE 
KE 
8.79
2
  
 1
2
 2    2  2 I tree  

 2 
1
 mtree R 2  2 
2
 4R 
1
3

m
 mtree   2 , or

2  stone
4
1
3

2
844 kg   82.0 kg   0.335 m s   50.8 J

2
4

We neglect the weight of the board and assume that
the woman’s feet are directly above the point of support
by the rightmost scale. Then, the free-body diagram
for the situation is as shown at the right.
From Fy  0 , we have Fg1  Fg2  w  0 , or w  380 N  320 N=700 N .
Choose an axis perpendicular to the page and passing through point P.
Then    0 gives w  x  Fg1  2 .00 m   0 or
Fg1  2.00 m 
x 
8.80
w

 380 N   2.00 m 
700 N
 1.09 m
Choose PEg = 0 at the level of the base of the ramp. Then, conservation of mechanical energy gives
 KE
trans
 KErot  PEg

f

 KEtrans  KErot  PEg

i
2
 
1
1
mi2 
mR 2  i   0
 R
2
2

0  0   mg  s  sin   

or
s 
i2
gsin 
R2 i2
 3.0 m   3.0 rad s   24 m

gsin 
9.80 m s2 sin 20
2


2

Page 8.48
Chapter 8
8.81
Choose an axis perpendicular to the page and passing
through the center of the cylinder. Then, applying  = I
to the cylinder gives
 2T   R
1

1
a 
1
  M R 2     M R 2   t  or T  M at
2

2
 R
4
[1]
Now Fy = may apply to the falling objects to obtain
 2m  g  2 T
(a)
  2m  at or at  g 
Mg  M 

T
4
 4 m 
which reduces to T 
(a)
M mg
M  4m
From Equation [2] above,
at  g 
8.82
[2]
Substituting Equation [2] into [1] yields
T 
(b)
T
m
1  M mg 
Mg
 g 



m  M  4m 
M  4m
4 mg
M  4m
A smooth (that is, frictionless) wall cannot exert a force parallel to its surface. Thus, the only force the vertical wall can exert on the upper end of the ladder is a horizontal normal force.
(b)
Consider the free-body diagram of the ladder given at the
right. If the rotation axis is perpendicular to the page and
passing through the lower end of the ladder, the lever
arm of the normal force n 2 that the wall exerts on the
upper end of the ladder is
d2 = L sin 
Page 8.49
Chapter 8
(c)
The lever arm of the force of gravity, m g , acting on the ladder is
d   L 2  cos  
(d)
 L cos  
2
Refer to the free-body diagram given in part (b) of this solution and make use of the fact that the ladder is in
both translational and rotational equilibrium.


Fy  0  n1  m g  mp g  0 or n1  m  mp g


When the ladder is on the verge of slipping, f1   f1 max  s n1  s m  m p g .


Then Fx  0  n2  f1 , or n2  s m  m p g .
Finally,   0  n2  L sin    m g  (L/2) cos    mp gx cos   0 where x is the maximum distance
the painter can go up the ladder before it will start to slip. Solving for x gives
L

n2  L sin    m g  cos  
m

 m 
2

x 
 s 
 1 L tan   
L
m p g cos 
 mp

 2m p 
and using the given numerical data, we find
 30 kg 
 30 kg

x   0.45  
 1  4.0 m  tan 53   
  4.0 m   2.5 m
 80 kg

 2  80 kg  
8.83
The large mass (m1 = 60.0 kg) moves in
a circular path of radius r1 = 0.140 m, while the
radius of the path for the small mass (m2 = 0.120 kg) is
r2   r1
 3.00 m  0.140 m  2.86 m
The system has maximum angular speed
when the rod is in the vertical position as
shown at the right.
We take PEg = 0 at the level of the horizontal
Page 8.50
Chapter 8
rotation axis and use conservation of energy to find:

KE f  PEg

f

 KEi  PEg

1
1

2
2
 2 I1 max  2 I 2 max    m2 gr2  m1gr1   0  0

i
Approximating the two objects as point masses, we have I1  m1r12 and I2  m2r22 . The energy conservation
equation then becomes
max 
1
2
2
  m1r1  m2r2  g
 m1r12  m2r22  max
2  m1r1  m2 r2  g
m1r12

m2 r22

and yields

2   60.0 kg   0.140 m    0.120 kg   2.86 m   9.80 m s 2
 60.0 kg   0.140 m 
2
  0.120 kg   2.86 m 

2
or max  8.56 rad s . The maximum linear speed of the small mass object is then
 2 max
8.84
(a)
 r2max   2.86 m   8.56 rad s   24.5 m s
Note that the cylinder has both translational and rotational
motion. The center of gravity accelerates downward while
the cylinder rotates around the center of gravity. Thus, we
apply both the translational and the rotational forms of
Newton’s second law to the cylinder:
Fy  ma y  T  mg  m  a 
or
T  m  g  a
  I  
[1]
 Tr  I   a r 
For a uniform, solid cylinder, I 
1
mr 2 so our last result becomes
2
 mr 2   a 
2T
Tr  
  or a 
 2   r 
m
[2]
Substituting Equation [2] into Equation [1] gives T = mg – 2T, and solving for T yields T = mg/3.
(b)
From Equation [2] above,
Page 8.51
Chapter 8
2T
2  mg 

 2g 3
m
m  3 
a 
(c)
Considering the translational motion of the center of gravity, 2y  02y  2ay y gives
y 
 2g 
0  2
 h  
 3 
4 gh 3
Using conservation of energy with PEg at the final level of the cylinder gives
 KE
t
 KEr  PEg

Since   y r and I 
yielding  y 
8.85

 KEt  KEr  PEg
f
1
2
mr 2 , this becomes

i
or
1
2
m2y 
1
2
I  2  0  0  0  mgh
2
3
1
11
  y 
m 2y   m r 2  
 mgh , or m 2y  mgh
  r 2 
2
22
4
4 gh 3 .
Considering the shoulder joint as the pivot, the second
condition of equilibrium gives
  0 
w
 70 cm    Fm sin 45   4.0 cm   0
2
or
Fm 
w  70 cm 
2  4.0 cm  sin 45 
 12.4 w
Recall that this is the total force exerted on the arm by a set of two muscles. If we approximate that the two muscles
of this pair exert equal magnitude forces, the force exerted by each muscle is
Page 8.52
Chapter 8
Feach
Fm
12.4 w

 6.2 w  6.2  750 N   4.6  10 3 N  4.6 kN
2
2

muscle
8.86
Observe that since the torque opposing the rotational motion of the gymnast is constant, the work done by nonconservative forces as the gymnast goes from position 1 to position 2 (an angular displacement of  /2 rad) will be the
same as that done while the gymnast goes from position 2 to position 3 (another angular displacement of  /2 rad).
Choose PEg = 0 at the level of the bar, and let the distance from the bar to the center of gravity of the outstretched

body be reg. Applying the work–energy theorem, Wnc  KE  PEg

f


 KE  PEg , to the rotation from posii
tion 1 to position 2 gives
 Wnc 12


1
2
 

I 22  0  0  mgrcg or  Wnc 12 
1
2
I 22  mgrcg
[1]
Now, apply the work–energy theorem to the rotation from position 2 to position 3 to obtain
 Wnc 23

 
  21 I 32  mg rcg  


1
2

I 22  0 or  Wnc 23 
1
2
I 32 
1
2
I 22  mgrcg
[2]
Since the frictional torque is constant and these two segments of the motion involve equal angular displacements,  Wnc 23   Wnc 12 . Thus, equating Equation [2] to Equation [1] gives
1
2
I 32 
1
2
I 22  mgrcg 
which yields 32  222 , or 3 
8.87
(a)
1
2
I 22  mgrcg
22 
2  4.0 rad s   5.7 rad s .
Free-body diagrams for each block and the
pulley are given at the right. Observe that
the angular acceleration of the pulley will be
clockwise in direction and has been given a
negative sign. Since    I  , the positive
sense for torques and angular acceleration
must be the same (counterclockwise).
For m1: Fy  may  T1  m1g  m1  a 
Page 8.53
Chapter 8
T1  m1  g  a  [1]
For m2:
Fx  max  T2  m2 a
[2]
For the pulley:   I   T2r  T1r  I  a r  or
 I 
T1  T2   2  a
r 
[3]
Substitute Equations [1] and [2] into Equation [3] and solve for a to obtain
a 
m1 g
 I   m1  m2
r2
or
a 
(b)
 4.00 kg   9.80 m s2 
 0.500 kg  m2   0.300 m 2  4.00 kg  3.00 kg

 3.12 m s2

Equation [1] above gives: T1   4.00 kg  9.80 m s2  3.12 m s2  26.7 N ,


and Equation [2] yields: T2   3.00 kg  3.12 m s2  9.37 N .
8.88
(a)
(b)
Fy  0

nF  120 N  mmonkey g  0


nF  120 N  10.0 kg  9.80 m s2  218 N
(c)
When x = 2L/3, we consider the bottom end of the ladder as our pivot and obtain
 bottom  0
end

 L

2L

  120 N  
cos 60.0     98.0 N  
cos 60.0    nW L sin 60.0   0
 2

 3


or
Page 8.54

Chapter 8
 60.0 N   196 3  N  cos 60.0 
nW  
 72.4 N
sin 60.0
Then,
Fx  0
(d)

T  nW  0 or T  nW  72.4 N
 0 yields
When the rope is ready to break, T  nW  80.0 N . Then  bottom
end
L

  120 N   cos 60.0     98.0 N  x cos 60.0    80.0 N   L sin 60.0    0
2

or
  80.0 N  sin 60.0    60.0 N  cos 60.0   L
x  
 0.802 L  0.802  3.00 m   2.41 m
 98.0 N  cos 60.0 
(e)
If the horizontal surface were rough and the rope removed, a horizontal static friction force directed toward
the wall would act on the bottom end of the ladder. Otherwise, the analysis would be much as what is done
above. The maximum distance the monkey could climb would correspond to the condition that the friction
force have its maximum value, s nF , so you would need to know the coefficient of static friction to solve
part (d).
Page 8.55