Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `3 m`. If the object's angular velocity changes from `12 Hz` to `7 Hz` in `8 s`, what torque was applied to the object?

Answer 1

The torque is `=67.9 Nm`

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=mr^2`

`=3*3^2= 27 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(12-7)/8*2pi`

`=((4pi)/5) rads^(-2)`

So the torque is `tau=27*(4pi)/5 Nm=108/5piNm=67.9Nm`