Question

An object with a mass of `8 kg` is traveling in a circular path of a radius of `12 m`. If the object's angular velocity changes from `2 Hz` to `12 Hz` in `6 s`, what torque was applied to the object?

Answer 1

The torque was `=12063.7Nm`

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=8*(12)^2=1152kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(12-2)/6*2pi`

`=(10/3pi) rads^(-2)`

So the torque is `tau=1152*(10/3pi) Nm=12063.7Nm`