The rate of rotation of a solid disk with a radius of #4 m# and mass of #2 kg# constantly changes from #5 Hz# to #1 Hz#. If the change in rotational frequency occurs over #3 s#, what torque was applied to the disk?

1 Answer
Jan 24, 2018

The torque is #=134.04Nm#

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

The mass of the disc is #m=2kg#

The radius of the disc is #r=4m#

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

So, #I=2*(4)^2/2=16kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(f_2-f_1)/t*2pi=(5-1)/3*2pi#

#=(8/3pi)rads^(-2)#

So,

The torque is #tau= 16*8/3pi=134.04Nm#