The rate of rotation of a solid disk with a radius of #8 m# and mass of #5 kg# constantly changes from #16 Hz# to #22 Hz#. If the change in rotational frequency occurs over #7 s#, what torque was applied to the disk?

1 Answer
Dec 15, 2016

Answer:

The torque is #=861.7 kgm^2s^(-1)#

Explanation:

The torque is #tau=Ialpha#

#alpha=(domega)/dt#

#alpha=6/7*2pi rads^(-1)=12/7pi rads^(-1)#

The moment of inertia of a solid disc #I=1/2mr^2#

#=1/2*5*8^2=160kgm^2#

So,

#tau=160*12/7pi kgm^2s^(-1)=861.7 kgm^2s^(-1)#