Question

A solid disk with a radius of `12 m` and mass of `5 kg` is rotating on a frictionless surface. If `450 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `7 Hz`?

Answer 1

The torque is `=10.23Nm`

Explanation:

The mass of the disc is `m=5kg`

The radius of the disc is `r=12m`

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=450W`

The frequency is `f=7Hz`

The angular velocity is `omega=2pif=2*pi*7=(14pi)rads^-1`

Therefore,

the torque is `tau=P/omega=450/(14pi)=10.23Nm`