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 #24 Hz#. If the change in rotational frequency occurs over #3 s#, what torque was applied to the disk?

1 Answer
Jun 23, 2017

Answer:

The torque was #=2540.4Nm#

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#

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

The rate of change of angular velocity is

#(domega)/dt=(24-11)/3*2pi#

#=(26/3pi) rads^(-2)#

So the torque is #tau=90*(26/3pi) Nm=2450.4Nm#