The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #16 Hz# to #22 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

1 Answer
Jun 2, 2017

Answer:

The torque is #=15.7Nm#

Explanation:

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

For a solid disc, #I=(mr^2)/2#

#I=5*1^2/2=2.5kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(22-16)/12*2pi#

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

So the torque is #tau=2.5*(pi) Nm=5piNm=15.7Nm#