The rate of rotation of a solid disk with a radius of #6 m# and mass of #5 kg# constantly changes from #11 Hz# to #15 Hz#. If the change in rotational frequency occurs over #3 s#, what torque was applied to the disk?

1 Answer
Dec 10, 2016

Answer:

The torque is #=754kgm^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=5*6^2/2=5*18=90 kg m^2#

#(domega)/dt=(15-11)/3*2pi=(8/3pi )rads^(-1)#

So, the torque is

#tau=90*8/3pi=240pi=754kgm^2s^(-1)#