The rate of rotation of a solid disk with a radius of #6 m# and mass of #5 kg# constantly changes from #32 Hz# to #18 Hz#. If the change in rotational frequency occurs over #8 s#, what torque was applied to the disk?

1 Answer
Jun 29, 2017

Answer:

The torque was #=989.6Nm#

Explanation:

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

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

So, #I=5*(6)^2/2=90kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(32-18)/8*2pi#

#=(7/2pi) rads^(-2)#

So the torque is #tau=90*(7/2pi) Nm=315piNm=989.6Nm#