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 #12 Hz#. If the change in rotational frequency occurs over #5 s#, what torque was applied to the disk?

1 Answer
Jan 18, 2017

Answer:

The torque was #=2010.6Nm#

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of the disc is #I=1/2*mr^2#

#=1/2*5*8^2= 160 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(12-2)/5*2pi#

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

So the torque is #tau=160*(4pi) Nm=640piNm=2010.6Nm#