Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `32 Hz` to `18 Hz`. If the change in rotational frequency occurs over `8 s`, what torque was applied to the disk?

Answer 1

The torque was `=989.6Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

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

So, `I=5*(6)^2/2=90kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(32-18)/8*2pi`

`=(7/2pi) rads^(-2)`

So the torque is `tau=90*(7/2pi) Nm=315piNm=989.6Nm`