Question

The rate of rotation of a solid disk with a radius of `1 m` and mass of `5 kg` constantly changes from `16 Hz` to `27 Hz`. If the change in rotational frequency occurs over `12 s`, what torque was applied to the disk?

Answer 1

The torque was `=14.4Nm`

Explanation:

The torque is the rate of change of angular momentum

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

mass, `m=5kg`

radius, `r=1m`

The moment of inertia of a solid disc is

`I=1/2*mr^2`

`=1/2*5*1^2= 2.5 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=((27-16))/12*2pi`

`=(11/6pi) rads^(-2)`

So the torque is `tau=2.5*(11/6pi) Nm=14.4Nm`