Question

The rate of rotation of a solid disk with a radius of `9 m` and mass of `5 kg` constantly changes from `32 Hz` to `18 Hz`. If the change in rotational frequency occurs over `6 s`, what torque was applied to the disk?

Answer 1

The torque was `=2968.8Nm`

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*9^2= 405/2 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(32-18)/6*2pi`

`=((14pi)/3) rads^(-2)`

So the torque is `tau=405/2*(14pi)/3 Nm=2968.8Nm`