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

Answer 1

The torque was `=554.2Nm`

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=7m`

For the object, `I=mr^2`

So, `I=3*(7)^2=147kgm^2`

And the rate of change of angular velocity is

`(d omega)/dt=(Deltaomega)/t=(34pi-28pi)/5=6/5pirads^-2`

So,

The torque is

`tau=147*6/5pi=554.2Nm`