10.29. A 1.50-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant
torque will bring it from rest to an angular speed of 1200 rpm in 2.5 s? (b) Through what angle
has it turned during that time? (c) Use W = z (2 – 1) to calculate the work done by the torque.
(d) What is the grinding wheel’s kinetic energy when it is rotating at 1200 rpm? Compare your
answer to the result in part (c).
Identify: Apply
wheel.

z
 I z
and constant angular acceleration equations to the motion of the
1 rev  2 rad .  rad/s  30 rev/min .
Set Up:
Execute:
(a)
 z  I z  I
z  0 z
t
.
s 
1 21.50 kg 0.100 m   1200 rev min  30 revradmin


2
z 
(b)
(c)
av t 
2.5 s
 0.377 N  m
 600 rev/min  2.5 s   25.0 rev  157 rad.
60 s/min
W    (0.377 N  m)(157 rad)  59.2 J
.
2
(d)
1
1

  rad/s  
K  I 2   (1/ 2)(1.5 kg)(0.100 m) 2   (1200 rev/min) 
   59.2 J
2
2
 30 rev/min  

.
the
same as in part (c).
Evaluate: The agreement between the results of parts (c) and (d) illustrates the work-energy
theorem