Question

The rate of rotation of a solid disk with a radius of `4 m` and mass of `2 kg` constantly changes from `8 Hz` to `1 Hz`. If the change in rotational frequency occurs over `5 s`, what torque was applied to the disk?

Answer 1

The torque was `=140.7Nm`

Explanation:

The torque is the rate of change of angular momentum

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

For a solid disc, the moment of inertia is

`I=(mr^2)/2`

So, `I=2*(4)^2/2=16kgm^2`

The rate of change of angular velocity is

`(domega)/dt=((8-1))/5*2pi`

`=(14/5pi) rads^(-2)`

So the torque is `tau=16*(14/5pi) Nm=140.7Nm`