Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `11 Hz` to `24 Hz`. If the change in rotational frequency occurs over `3 s`, what torque was applied to the disk?

Answer 1

The torque was `=2540.4Nm`

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*(6)^2/2=90kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(24-11)/3*2pi`

`=(26/3pi) rads^(-2)`

So the torque is `tau=90*(26/3pi) Nm=2450.4Nm`