Question

A solid disk with a radius of `2 m` and mass of `3 kg` is rotating on a frictionless surface. If `36 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `8 Hz`?

Answer 1

The torque is `=0.72Nm`

Explanation:

Apply the equation

`P=tau omega`

The power is `P=36W`

The frequency is `f=8Hz`

The angular velocity is

`omega=2pif=2xxpixx8=(16pi)rads^-1`

Therefore,

The torque is

`tau=P/omega=36/(16pi)=0.72Nm`