Question

A solid disk with a radius of `8 m` and mass of `6 kg` is rotating on a frictionless surface. If `60 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 `=3.18Nm`

Explanation:

mass of disc `=2kg`

radius of disc `=2m`

The power is related to the torque by the equation

`P=tau *omega`

`P=60W`

Frequency of rotation is `f=3Hz`

`omega=f*2pi`

angular velocity, `omega=3*2pi rads^-1`

torque, `tau=P/omega`

`=60/(6pi)=3.18Nm`