Question

A solid disk with a radius of `3 m` and mass of `4 kg` is rotating on a frictionless surface. If `120 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `8 Hz`?

Answer 1

The torque applied is 2.38 N-m.

Explanation:

This goes very simple, Power is the product of the torque and angular velocity, similar to `P=Fv`.
`=>P=tauomega`
So here angular acceleration comes with `omega=2pinu`.
Calculating the angular velocity with given frequency we get,
`omega=50.24rads^-1`
Now the above power equation becomes
`=>tau=P/omega`
Substituting the given values we get,
`tau=(120W)/(50.24rads^-1)`.

`:. tau=2.38`Newtons-metre.