Question

The rate of rotation of a solid disk with a radius of `2 m` and mass of `5 kg` constantly changes from `12 Hz` to `16 Hz`. If the change in rotational frequency occurs over `12 s`, what torque was applied to the disk?

Answer 1

The torque was `=20.94Nm`

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=2m`

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

So, `I=5*(2)^2/2=10kgm^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*16*pi=32pirads^-1`

The time is `t=12s`

The angular acceleration is

`alpha=(omega_1-omega_0)/t=(32pi-24pi)/12=2/3pirads^-2`

So,

The torque is `tau= 10*2/3pi=20.94Nm`