The rate of rotation of a solid disk with a radius of #9 m# and mass of #5 kg# constantly changes from #24 Hz# to #15 Hz#. If the change in rotational frequency occurs over #5 s#, what torque was applied to the disk?

1 Answer
Apr 21, 2017

Answer:

The torque was #=2290.2Nm#

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia is #=I#

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

So, #I=(5*(9)^2)/2=202.5kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(24-15)/5*2pi#

#=(18/5pi) rads^(-2)#

So the torque is #tau=202.5*(18/5pi) Nm=2290.2Nm#