Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `7 m`. If the object's angular velocity changes from `6 Hz` to `7 Hz` in `3 s`, what torque was applied to the object?

Answer 1

The torque was `=307.9Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt`

The moment of inertia is `=I`

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

So, `I=(3*(7)^2)=147kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(7-6)/3*2pi`

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

So the torque is `tau=147*(2/3pi) Nm=307.9Nm`