Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `12 Hz` to `15 Hz`. If the change in rotational frequency occurs over `12 s`, what torque was applied to the disk?

Answer 1

The torque was `=141.4N`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertiaof a solid disc is `I=1/2*mr^2`

`=1/2*5*6^2= 90 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(15-12)/12*2pi`

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

So the torque is `tau=90*(pi/2) Nm=45piNm=141.4Nm`