Question

A solid disk with a radius of `2 m` and mass of `2 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 `9 Hz`?

Answer 1

The torque is `=16.98Nm`

Explanation:

The mass of the disc is `m=2kg`

The radius of the disc is `r=2m`

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=9Hz`

The angular velocity is `omega=2pif=2*pi*9=18pirads^-1`

Therefore,

the torque is `tau=P/omega=960/(18pi)=16.98Nm`