The rate of rotation of a solid disk with a radius of #2 m# and mass of #5 kg# constantly changes from #27 Hz# to #18 Hz#. If the change in rotational frequency occurs over #8 s#, what torque was applied to the disk?

1 Answer
May 21, 2017

Answer:

The torque was #=70.7Nm#

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*2^2/2=10kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(27-18)/8*2pi#

#=(9/4pi) rads^(-2)#

So the torque is #tau=10*(9/4pi) Nm=45/2piNm=70.7Nm#