Question

The rate of rotation of a solid disk with a radius of `8 m` and mass of `5 kg` constantly changes from `5 Hz` to `17 Hz`. If the change in rotational frequency occurs over `6 s`, what torque was applied to the disk?

Answer 1

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 a solid disc, rotating about the center is

`I=1/2*mr^2`

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

The rate of change of angular velocity is

`(domega)/dt=(17-5)/6*2pi`

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

So the torque is `tau=160*(4pi) Nm=2010.6Nm`