Question

A solid disk with a radius of `9 m` and mass of `8 kg` is rotating on a frictionless surface. If `360 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `6 Hz`?

Answer 1

The torque is `=9.55Nm`

Explanation:

The mass of the disc is `m=8kg`

The radius of the disc is `r=9m`

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=360W`

The frequency is `f=6Hz`

The angular velocity is `omega=2pif=2*pi*6=(12pi)rads^-1`

Therefore,

the torque is `tau=P/omega=360/(12pi)=9.55Nm`