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 `6 Hz` to `8 Hz` in `8 s`, what torque was applied to the object?

Answer 1

The torque was `=230.9Nm`

Explanation:

The torque is the rate of change of angular momentum

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

Where the moment of inertia is `=I`

and `omega` is the angular velocity

The mass is `m=3kg`

The radius is `r=7m`

For the object, `I=(mr^2)`

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

The rate of change of angular velocity is

`(domega)/dt=(8-6)/8*2pi`

`=(1/2pi) rads^(-2)`

So the torque is `tau=147*(1/2pi) Nm=147/2piNm=230.9Nm`