Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `12 m`. If the object's angular velocity changes from `5 Hz` to `4 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque is `=542.9Nm`

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`

`=3*12^2= 432 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(5-4)/5*2pi`

`=((2pi)/5) rads^(-2)`

So the torque is `tau=432*(2pi)/5 Nm=864/5piNm=542.9Nm`