The rate of rotation of a solid disk with a radius of #8 m# and mass of #5 kg# constantly changes from #2 Hz# to #17 Hz#. If the change in rotational frequency occurs over #3 s#, what torque was applied to the disk?

1 Answer
May 13, 2017

Answer:

The torque was #=5026.55Nm#

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*(8)^2/2=160kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(17-2)/3*2pi#

#=(10pi) rads^(-2)#

So the torque is #tau=160*(10pi) Nm=(1600pi)Nm=5026.55Nm#