Convert the rpm values to angular velocities.

0


130 rev min



1 rev
1 min
60 sec

13.6 rad s
 

280 rev min



1 rev
1 min
60 sec

29.3 rad s
( a ) The angular acceleration is found from Eq. 10-3a.
 
  t
0

4.0 s

3.93 rad s
2 
3.9 rad s
2
( b ) To find the components of the acceleration, the instantaneous angular velocity is needed.
  
0
 t

13.6 rad s


3.93 rad s
2
 
2.0 s
 
21.5 rad s
The instantaneous radial acceleration is given by a
R a
R
 
2 r
 
2 r .


21.5 rad s
 
0.35 m


160 m s
2
The tangential acceleration is given by a tan
  r
 a tan
  r .

3.93rad s
2
 
0.35 m


1.4 m s
2
( a ) The direction of

1 is along the axle of the wheel, to the left. That is the
 i
ˆ direction.
The direction of
 is also along its axis of rotation, so it is straight up. That is the
2
 ˆ direction. That is also the angular velocity of the axis of the wheel.
( b ) At the instant shown in the textbook, we have the vector relationship as shown in the diagram.
  
1
2
 
2
2
 
44.0 rad s

35.0 rad s

2 
56.2 rad s

2


  tan

1

2

2
 tan

1
35.0
44.0

38.5


1
( c ) Angular acceleration is given by α
, α 
 d
ω
.
dt
Since

1
 ω
2
, and ω is a constant
2 d ω
1 .
ω is rotating counterclockwise about the z axis with the
1 dt z x

α
α
 d
ω dt angular velocity of
ω
1
 
1

 cos

2 t
ˆ i 

and so if the figure is at t = 0, then
2
, sin

2 t
ˆ j

.
 d

ω
1
 ω
2 dt

 d
ω
1 dt
 d

1

 cos

2 dt t i
ˆ  sin

2 t
ˆ j

  
1 2
 t

0

  
1 2
  
44.0 rad s

35.0 rad s
 ˆ j  
1540 rad s
2
 sin

2 t i
ˆ  cos

2 t
ˆ j

We are given that
 
8.5
t

15.0
t 2

1.6
t 4 .
( a )
( b )



 d d
 dt
 dt

8.5
 

30.0
 t
 t
6.4
2 , t
3 , where
where


is in rad/sec and
is in t is in sec.
2 rad sec and t is in sec.
( c )
  

8.5

30.0 3.0

6.4 3.0
3 
91rad s
   
 
2 
140 rad s
2
( d ) The average angular velocity is the angular displacement divided by the elapsed time.
 avg

 
 t

      
3.0 s

2.0 s
    
2   
4


    
2   
4
1.0 s

38 rad s
( e ) The average angular acceleration is the change in angular velocity divided by the elapsed time.
 avg

 
 t

      
3.0 s

2.0 s

8.5

 

 
3
8.5

 

 
3

92 rad s
2
1.0 s
For each torque, use Eq. 10-10c. Take counterclockwise torques to be positive.
( a ) Each force has a lever arm of 1.0 m.
   about

1.0 m

56 N sin 30
  
1.0 m
   
17m N
C
( b ) The force at C has a lever arm of 1.0 m, and the force at the top has a lever arm
P of 2.0 m.
   about

2.0 m

56 N sin 30
 

1.0 m
 
  
10 m N (2 sig fig)
The negative sign indicates a clockwise torque.

( a ) The torque exerted by the frictional force is
  rF fr
The force of friction is assumed to be tangential to the clay, and so
 
90 .
 total
 rF fr sin
 

1
2

0.12 m
   
 
0.090 m N
( b ) The time to stop is found from
   o t , with a final angular velocity of 0. The angular acceleration can be found from
 total

I

.
The net torque (and angular acceleration) is negative direction of rotation since the object is slowing. t

 
 o

 

I o


0



1.6 rev s
0.090 m N




0.11kg m
2



12 s
F fr
( a ) To calculate the moment of inertia about the y axis (vertical), use the following.
I
 
M R
2 ix
 m

0.50 m

2 
M

0.50 m

2  m

1.00 m

2 
M

1.00 m

2
  m

M
 
0.50 m

1.00 m

2  
5.3 kg
 
0.50 m
 
1.00 m

2 
6.6 kg m
2
( b ) To calculate the moment of inertia about the x -axis (horizontal), use the following.
I
 
M R
2 i iy
 
2 m

2 M

0.25 m

2 
0.66 kg m
2
( c ) Because of the larger I value, it is ten times harder to accelerate the array about the vertical axis .
( a ) The free body diagrams are shown. Note that only the forces producing torque are shown on the pulley. There would also be a gravity force on the pulley
(since it has mass) and a normal force from the pulley’s suspension, but they are not shown.
( b ) Write Newton’s second law for the two blocks, taking the positive x direction as shown in the free body diagrams.
F
TB
F
NB y y
F
NA x

A
F
TA

A
F
TA
F
TB

B

B x

m
A
:

F x

F
TA
 m g
F
TA
 m
A
 g sin

A
 sin a


A
 
 
8.0kg
 
9.80 m s
2
 sin 32
 
1.00 m s
2

  m
B
:

F x
 m g
F
TB
 m
B
 g sin sin


B
B


F
TB
 m a a




10 .0kg
 
9.80 m s
2
 sin 61
 
1.00 m s
2

49.55 N

75.71 N

76 N
( c ) The net torque on the pulley is caused by the two tensions. We take clockwise torques as positive.
  

F
TB

F
TB

R


75.71N

49.55 N

0.15 m


3.924m N

3.9m N
Use Newton’s second law to find the rotational inertia of the pulley. The tangential acceleration of the pulley’s rim is the same as the linear acceleration of the blocks, assuming that the string doesn’t slip.
I


 
I
 

I a
R
 
F
F
TB

F
TB

R
2

 a
TB

F
TB

R

75.71 N

49.55 N

0.15 m
1.00 m s
2

2

0.59 kg m 2
 
168.9
 
170

.
We assume that m
B
 m
A
, and so m will accelerate down,
B m will accelerate up, and the
A pulley will accelerate clockwise. Call the direction of acceleration the positive direction for each object. The masses will have the same acceleration since they are connected by a cord.
The rim of the pulley will have that same acceleration since the cord is making it rotate, and so
  pulley a R .
From the free-body diagrams for each object, we have the following.


F y A

F
TA
 m g
A
 m a
A

F
TA
 m g
A
 m a
A
F y B
 m g
B

F
TB
 m a
B

F
TB
 m g
B
 m a
B
 
I
R
  
F r

F r

I
 
I a
R
Substitute the expressions for the tensions into the torque equation, and solve for the acceleration.
F
TA
F
TA
 y m
A m
B
F
TB
F
TB
 y
m g
A
m g
B

a
 a
F R

F R

I

R
 m
B
 m
A

 m
A
 m
B

I R
2
 g
 m g
    m g
  
I a
R

If the moment of inertia is ignored, then from the torque equation we see that F
TB

F
TA
, and the acceleration will be a
I

0


 m
B
 m
A m
A
 m
B

 g .
We see that the acceleration with the moment of inertia included will be smaller than if the moment of inertia is ignored.
( a ) The kinetic energy of the system is the kinetic energy of the two masses, since the rod is treated as massless. Let A represent the heavier mass, and B the lighter mass.
K

1
2
I

1
2


A A
2

1
2
0.210 m
I

B B
2
 

1
2 m r
2

2
A A A
5.60 rad s
 

1
2 m r
2

2
A
7.00 kg
 

1
2 r
4.84 J
2

2
 m
A
 m
B

( b ) The net force on each object produces centripetal motion, and so can be expressed as mr
 2 .
F
A
 m r

2
A A A
 
4.00 kg

0.210 m

5.60 rad s

2 
26.3 N
F
B
 m r

2
B B B
 
3.00 kg

0.210 m

5.60 rad s

2 
19.8 N
These forces are exerted by the rod. Since they are unequal, there would be a net horizontal force on the rod (and hence the axle) due to the masses. This horizontal force would have to be counteracted by the mounting for the rod and axle in order for the rod not to move horizontally. There is also a gravity force on each mass, balanced of mass. by a vertical force from the rod, so that there is no net vertical force on either mass.
( c ) Take the 4.00 kg mass to be the origin of coordinates for determining the center x
CM
 m x
A A
 m x
B B m
A
 m
B


4.00 kg
  
3.00 kg

0.420 m

7.00 kg

0.180 m from mass A
So the distance from mass A to the axis of rotation is now 0.180 m, and the distance from mass B to the axis of rotation is now 0.24 m. Re-do the above calculations with these values.
K

1
2
I

A A
2

1
2
I

B B
2

1
2 m r
2

2
A A A

1
2 m r
2

2
B B A

1
2

2
 m r
2
A A
 m r
2
B B


1
2

5.60 rad s
 
4.00 kg

0.180 m

3.00 kg

0.240 m

2 
4.74 J
F
A
 m r

2
A A A
 
4.00 kg

0.180 m

5.60 rad s

2 
22.6 N
F
B
 m r

2
B B B
 
3.00 kg

0.240 m

5.60 rad s

2 
22.6 N
Note that the horizontal forces are now equal, and so there will be no horizontal force on the rod or axle.