Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `2 m`. If the object's angular velocity changes from `2 Hz` to `3 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque `=15.1Nm`

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`

For the object, `I=(mr^2)`

So, `I=3*(2)^2s=12kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(3-2)/5*2pi`

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

So the torque is `tau=12*(2/5pi) Nm=15.1Nm`