Question

The rate of rotation of a solid disk with a radius of `2 m` and mass of `5 kg` constantly changes from `27 Hz` to `18 Hz`. If the change in rotational frequency occurs over `3 s`, what torque was applied to the disk?

Answer 1

The torque was `188.5Nm`

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*2^2= 10 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(27-18)/3*2pi`

`=((6pi) rads^(-2)`

So the torque is `tau=10*(6pi) Nm=60piNm=188.5Nm`