Question

The rate of rotation of a solid disk with a radius of `4 m` and mass of `5 kg` constantly changes from `21 Hz` to `12 Hz`. If the change in rotational frequency occurs over `8 s`, what torque was applied to the disk?

Answer 1

THe torque was `=282.7Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of a solid disc, is

`I=1/2*mr^2`

`=1/2*5*4^2= 40 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(21-12)/8*2pi`

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

So the torque is `tau=40*(9/4pi) Nm=(90pi)Nm=282.7Nm`