Question

The rate of rotation of a solid disk with a radius of `5 m` and mass of `5 kg` constantly changes from `12 Hz` to `10 Hz`. If the change in rotational frequency occurs over `9 s`, what torque was applied to the disk?

Answer 1

The torque was `=87.27Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

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

The radius of the disc is `r=5m`

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

So, `I=5*(5)^2/2=62.5kgm^2`

The initial angular velocity is

`omega_0=2pif_0=2*12*pi=24pirads^-1`

The final angular velocity is

`omega_1=2pif_1=2*10*pi=20pirads^-1`

The time is `t=9s`

The angular acceleration is

`alpha=(omega_1-omega_0)/t=(24pi-20pi)/9=4/9pirads^-2`

So,

The torque is `tau= 62.5*4/9pi=87.27Nm`