Question

A solid disk with a radius of `8 m` and mass of `8 kg` is rotating on a frictionless surface. If `480 W` of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at `6 Hz`?

Answer 1

`=12.73Nm`

Explanation:

Given

• `m-> "Mass of solid disk"=8kg`

• `r-> "Radius of solid disk"=8m`

• `P-> "Power used to solid disk"=480W`

• `n-> "Frequency of rotaion of solid disk"=6Hz`

We are to find out

• `tau->"Torque applied to solid disk"`

Formula

`"Power"(P)="Torque"(tau)xx"Angular velocity"(w)`

`=>"Power"(P)="Torque"(tau)xx2pixx"Frequency of rotation"(n)`

`=>480="Torque"(tau)xx2pixx6`

`"Torque"(tau)=480/(2pixx6)=12.73Nm`