Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `4 m`. If the object's angular velocity changes from `1 Hz` to `19 Hz` in `3 s`, what torque was applied to the object?

Answer 1

The torque is `=1206.4Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of the object is `I=mr^2`

`=2*4^2= 32 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(19-1)/3*2pi`

`=(12pi) rads^(-2)`

So the torque is `tau=32*12pi Nm=384piNm=1206.4Nm`