Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `7 m`. If the object's angular velocity changes from `7 Hz` to `9 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque was `=246.3Nm`

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 is `m=2kg`

The radius is `r=7m`

For the object, `I=mr^2`

`I=2*7^2=98kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(9-7)/5*2pi`

`=(4/5pi) rads^(-2)`

So the torque is `tau=98*(4/5pi) Nm=246.3Nm`