The rate of rotation of a solid disk with a radius of #2 m# and mass of #5 kg# constantly changes from #27 Hz# to #18 Hz#. If the change in rotational frequency occurs over #3 s#, what torque was applied to the disk?

1 Answer
Jan 22, 2017

Answer:

The torque was #188.5Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

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

#=1/2*5*2^2= 10 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(27-18)/3*2pi#

#=((6pi) rads^(-2)#

So the torque is #tau=10*(6pi) Nm=60piNm=188.5Nm#