Question

A solid disk with a radius of `6 m` and mass of `9 kg` is rotating on a frictionless surface. If `960 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `32 Hz`?

Answer 1

The torque is `=4.78Nm`

Explanation:

The mass of the disc is `m=9kg`

The radius of the disc is `r=6m`

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=960W`

The frequency is `f=32Hz`

The angular velocity is `omega=2pif=2*pi*32=(64pi)rads^-1`

Therefore,

the torque is `tau=P/omega=960/(64pi)=4.78Nm`