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 `27 Hz`. If the change in rotational frequency occurs over `8 s`, what torque was applied to the disk?

Answer 1

THe torque is `=206.2Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of a solid disc is `I=1/2mr^2`

`=1/2*1*5^2= 25/2=12.5 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(27-6)/8*2pi`

`=((21pi)/4)pi rads^(-2)`

So the torque is `tau=12.5*(21pi)/4 Nm=65.625piNm=206.2Nm`