Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `9 m`. If the object's angular velocity changes from `2 Hz` to `6 Hz` in `8 s`, what torque was applied to the object?

Answer 1

The torque was `=763.4Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt`

The moment of inertia is `=I`

For the object, `I=(mr^2)`

So, `I=(3*(9)^2)=243kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(6-2)/8*2pi`

`=(pi) rads^(-2)`

So the torque is `tau=243*(pi) Nm=763.4Nm`