Question

A solid disk with a radius of `8 m` and mass of `6 kg` is rotating on a frictionless surface. If `72 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `3 Hz`?

Answer 1

The torque is `=3.82Nm`

Explanation:

The mass of the disc is `m=6kg`

The radius of the disc is `r=8m`

The power `=P` and the torque `=tau` are related by the following equation

`P=tau omega`

where `omega=` the angular velocity

Here,

The power is `P=72W`

The frequency is `f=3Hz`

The angular velocity is `omega=2pif=2*pi*3=6pirads^-1`

Therefore,

the torque is `tau=P/omega=72/(6pi)=3.82Nm`