Question

The rate of rotation of a solid disk with a radius of `5 m` and mass of `5 kg` constantly changes from `18 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 `=2356.2Nm`

Explanation:

The torque is the rate of change of angular momentum

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

Where the moment of inertiais `=I`

and `omega` is the angular velocity

The mass of the disc is `m=5kg`

The radius of the disc is `r=5m`

For the solid disc, `I=(1/2mr^2)`

So, `I=1/2*5*(5)^2=125/2kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(24-18)/1*2pi`

`=(12pi) rads^(-2)`

So the torque is `tau=125/2*(12pi) Nm=2356.2Nm`