Question

The rate of rotation of a solid disk with a radius of `9 m` and mass of `5 kg` constantly changes from `24 Hz` to `15 Hz`. If the change in rotational frequency occurs over `5 s`, what torque was applied to the disk?

Answer 1

The torque was `=2290.2Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia is `=I`

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

So, `I=(5*(9)^2)/2=202.5kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(24-15)/5*2pi`

`=(18/5pi) rads^(-2)`

So the torque is `tau=202.5*(18/5pi) Nm=2290.2Nm`