Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `5 Hz` to `3 Hz`. If the change in rotational frequency occurs over `5 s`, what torque was applied to the disk?

Answer 1

The torque is `=226.2Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

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

The radius of the disc is `r=6m`

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

So, `I=5*(6)^2/2=90kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(f_2-f_1)/t*2pi=(5-3)/5*2pi`

`=(4/5pi)rads^(-2)`

So,

The torque is `tau= 90*4/5pi=226.2Nm`