Question

An object with a mass of `7 kg` is traveling in a circular path of a radius of `9 m`. If the object's angular velocity changes from `5 Hz` to `3 Hz` in `8 s`, what torque was applied to the object?

Answer 1

The torque was `=890.6Nm`

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 a the object, `I=mr^2`

So, `I=7*(9)^2=567kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(5-3)/8*2pi`

`=(1/2pi) rads^(-2)`

So the torque is `tau=567*(1/2pi) Nm=890.6Nm`