Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `6 m`. If the object's angular velocity changes from `6 Hz` to `11 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque was `=678.58Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertiai is `=I`

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

So, `I=3*(6)^2=108kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(11-6)/5*2pi`

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

So the torque is `tau=108*(2pi) Nm=678.58Nm`