Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `2 m`. If the object's angular velocity changes from `2 Hz` to `6 Hz` in `5 s`, what torque was applied to the object?

Answer 1

`60.31 N.m`

Explanation:

Torque (`tau`) is defined as change in rate of angular momentum i.e `(dL)/dt` or, `(d(Iomega))/dt` or, `I (domega)/dt` (as `L=I omega`, where, `I` is the moment of inertia)

So, `I = mr^2`

Given , `m =3 Kg` and `r=2m`

So, `I= 12 Kg.m^2`

Now `(domega)/dt = ((nu 1-nu 2)/t)*2 pi` i.e `(6-2)/5*2pi (rad)/s^2`

So, `tau= (12*8/5 pi) N.m` i.e `60.31 N.m`