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

Answer 1

The torque was `=60.3Nm`

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

The mass of the object is `m=3kg`

The radius is `r=2m`

For the object, `I=mr^2`

So, `I=3*(2)^2=12kgm^2`

And the rate of change of angular velocity is

`(d omega)/dt=(Deltaomega)/t=(10pi-2pi)/5=(8/5pi)rads^-2`

So,

The torque is

`tau=12*8/5pi=60.3Nm`