Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `5 m`. If the object's angular velocity changes from `1 Hz` to `8 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque was `=329.9Nm`

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*5^2= 75 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(8-1)/5*2pi`

`=((7pi)/5) rads^(-2)`

So the torque is `tau=75*(7pi)/5 Nm=105piNm=329.9Nm`