Question

The rate of rotation of a solid disk with a radius of `9 m` and mass of `5 kg` constantly changes from `21 Hz` to `12 Hz`. If the change in rotational frequency occurs over `5 s`, what torque was applied to the disk?

Answer 1

The torque was `=2290Nm`

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 disk is `m=5kg`

The radius is `r=9m`

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

So, `I=1/2*5*(9)^2=202.5kgm^2`

The angular velocity is

`omega=2xxpixxf` where the frequency is `=f`

And the rate of change of angular velocity is

`alpha=(d omega)/dt=(Deltaomega)/t=(42pi-24pi)/5=(18/5pi)rads^-2`

The torque is

`tau=202.5*18/5pi=2290Nm`