Question

The rate of rotation of a solid disk with a radius of `4 m` and mass of `5 kg` constantly changes from `12 Hz` to `5 Hz`. If the change in rotational frequency occurs over `7 s`, what torque was applied to the disk?

Answer 1

The torque was `=251.3Nm`

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 is `r=4m`

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

So, `I=1/2*5*(4)^2=40kgm^2`

And the rate of change of angular velocity is

`(d omega)/dt=(Deltaomega)/t=(24pi-10pi)/7=2pirads^-2`

So,

The torque is

`tau=40*2pi=251.3Nm`