Question

A solid disk with a radius of `5 m` and mass of `2 kg` is rotating on a frictionless surface. If `72 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `4 Hz`?

Answer 1

`2.865\ \text{Nm}`

Explanation:

frequency of rotation of solid disk is `f=4\ Hz` hence its angular velocity `\omega` is given as follows

`\omega=2\pi f=2\pi\cdot 4=8\pi \ text{rad/s}`

Now, the power `P=72\ W` imparted to the solid disk by applying torque `T` to create final angular velocity `\omega=8\pi \text{rad/s}` is given as follows

`P=T\times \omega`

`T=P/\omega`

`=72/{8\pi}`

`=9/\pi`

`=2.865\ \text{Nm}`