Question

The rate of rotation of a solid disk with a radius of `2 m` and mass of `5 kg` constantly changes from `27 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 `=70.7Nm`

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*2^2/2=10kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(27-18)/8*2pi`

`=(9/4pi) rads^(-2)`

So the torque is `tau=10*(9/4pi) Nm=45/2piNm=70.7Nm`