The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #5 Hz# to #17 Hz#. If the change in rotational frequency occurs over #6 s#, what torque was applied to the disk?

1 Answer
Jun 10, 2017

Answer:

The torque was #=31.42Nm#

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*(1)^2/2=5/2kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(17-5)/6*2pi#

#=(4pi) rads^(-2)#

So,

The torque is #tau=5/2*(4pi)=10pi=31.42Nm#