Question

The rate of rotation of a solid disk with a radius of `5 m` and mass of `5 kg` constantly changes from `11 Hz` to `24 Hz`. If the change in rotational frequency occurs over `1 s`, what torque was applied to the disk?

Answer 1

The torque was `=5105.1Nm`

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 the solid disc is `I=1/2*mr^2`

`=1/2*5*5^2= 125/2 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(24-11)/1*2pi`

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

So the torque is `tau=125/2*(26pi) Nm=125*13piNm=5105.1Nm`