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 `8 Hz` in `3 s`, what torque was applied to the object?

Answer 1

The torque was `=469.1Nm`

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`

So, `I=2*(4)^2=32kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(8-1)/3*2pi`

`=(14/3pi) rads^(-2)`

So the torque is `tau=32*(14/3pi) Nm=469.1Nm`