The rate of rotation of a solid disk with a radius of #4 m# and mass of #5 kg# constantly changes from #16 Hz# to #5 Hz#. If the change in rotational frequency occurs over #7 s#, what torque was applied to the disk?

1 Answer
Dec 10, 2016

Answer:

The torque is #=237kgm^2s^(-1)#

Explanation:

The torque is #tau=Ialpha#

#tau=I(domega)/dt#

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

#I=3*4^2/2=3*8=24 kg m^2#

#(domega)/dt=(16-5)/7*2pi=(22/7pi )rads^(-1)#

So, the torque is

#tau=24*22/7pi=237kgm^2s^(-1)#