Question

The rate of rotation of a solid disk with a radius of `3 m` and mass of `4 kg` constantly changes from `2 Hz` to `5 Hz` over `8 s`. What torque would was applied to the disk at `2 s`?

Answer 1

The torque is `=42.41Nm`

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=4kg`

The radius of the disc is `r=3m`

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

So, `I=4*(3)^2/2=18kgm^2`

The initial angular velocity is

`omega_0=2pif_0=2*2*pi=4pirads^-1`

The final angular velocity is

`omega_1=2pif_1=2*5*pi=10pirads^-1`

The time is `t=8s`

The angular acceleration is

`alpha=(omega_1-omega_0)/t=(10pi-4pi)/8=3/4pirads^-2`

So,

The torque is `tau= 18*3/4pi=42.41Nm`