Question

The rate of rotation of a solid disk with a radius of `1 m` and mass of `5 kg` constantly changes from `6 Hz` to `24 Hz`. If the change in rotational frequency occurs over `6 s`, what torque was applied to the disk?

Answer 1

The torque was `=47.1Nm`

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

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

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

The radius of the disc is `r=1m`

So, the moment of inertia is `I=5*(1)^2/2=2.5kgm^2`

The rate of change of the angular velocity is

`(domega)/dt=(24-6)/6*2pi=6pirads^-2`

So,

The torque is

`tau=2.5*6pi=47.1Nm`