Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `2 m`. If the object's angular velocity changes from `1 Hz` to `6 Hz` in `1 s`, what torque was applied to the object?

Answer 1

The torque was `=251.3Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

For the object, `I=mr^2`

`I=2*2^2=8kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(6-1)/1*2pi`

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

So the torque is `tau=8*(10pi) Nm=80piNm=251.3Nm`