The rate of rotation of a solid disk with a radius of #6 m# and mass of #5 kg# constantly changes from #12 Hz# to #15 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

1 Answer
Feb 16, 2017

Answer:

The torque was #=141.4N#

Explanation:

The torque is the rate of change of angular momentum

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

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

#=1/2*5*6^2= 90 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15-12)/12*2pi#

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

So the torque is #tau=90*(pi/2) Nm=45piNm=141.4Nm#