Question

A solid disk with a radius of `6 m` and mass of `9 kg` is rotating on a frictionless surface. If `120 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 `=2.39Nm`

Explanation:

mass of disc `=9kg`

radius of disc `=6m`

The power is related to the torque by the equation

`P=tau *omega`

`P=120W`

Frequency of rotation is `f=8Hz`

`omega=f*2pi`

angular velocity, `omega=8*2pi rads^-1`

torque, `tau=P/omega`

`=120/(16pi)=2.39Nm`