Question

The rate of rotation of a solid disk with a radius of `1 m` and mass of `5 kg` constantly changes from `16 Hz` to `22 Hz`. If the change in rotational frequency occurs over `12 s`, what torque was applied to the disk?

Answer 1

The torque is `=15.7Nm`

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`

`I=5*1^2/2=2.5kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(22-16)/12*2pi`

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

So the torque is `tau=2.5*(pi) Nm=5piNm=15.7Nm`