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 `16 Hz` to `17 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque is `=184.7Nm`

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 object is `I=mr^2`

`=3*7^2= 147 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(1)/5*2pi`

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

So the torque is `tau=147*(2pi)/5 Nm=294/5piNm=184.7Nm`