The rate of rotation of a solid disk with a radius of #5 m# and mass of #5 kg# constantly changes from #18 Hz# to #24 Hz#. If the change in rotational frequency occurs over #1 s#, what torque was applied to the disk?

1 Answer
Jun 8, 2017

Answer:

The torque was #=2356.2Nm#

Explanation:

The torque is the rate of change of angular momentum

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

Where the moment of inertiais #=I#

and #omega# is the angular velocity

The mass of the disc is #m=5kg#

The radius of the disc is #r=5m#

For the solid disc, #I=(1/2mr^2)#

So, #I=1/2*5*(5)^2=125/2kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(24-18)/1*2pi#

#=(12pi) rads^(-2)#

So the torque is #tau=125/2*(12pi) Nm=2356.2Nm#