Question

The rate of rotation of a solid disk with a radius of `8 m` and mass of `5 kg` constantly changes from `2 Hz` to `17 Hz`. If the change in rotational frequency occurs over `3 s`, what torque was applied to the disk?

Answer 1

The torque was `=5026.55Nm`

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`

So, `I=5*(8)^2/2=160kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(17-2)/3*2pi`

`=(10pi) rads^(-2)`

So the torque is `tau=160*(10pi) Nm=(1600pi)Nm=5026.55Nm`