Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `15 m`. If the object's angular velocity changes from `25 Hz` to `34 Hz` in `9 s`, what torque was applied to the object?

Answer 1

The torque is `=4241.2Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of the mass is `I=mr^2`

`=3*15^2= 675 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(34-25)/9*2pi`

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

So the torque is `tau=675*(2pi) Nm=1350piNm=4241.2Nm`