Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `15 m`. If the object's angular velocity changes from `4 Hz` to `23 Hz` in `9 s`, what torque was applied to the object?

Answer 1

The torque is `=8953.5Nm`

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 of the path is `r=15m`

For the object, `I=mr^2`

So, `I=3*(15)^2=675kgm^2`

And the rate of change of angular velocity is

`(d omega)/dt=(Deltaomega)/t=(2pif_1-2pif_2)/t`

`=(46pi-8pi)/9=(38/9pi)rads^-2`

So,

The torque is

`tau=675*38/9pi=8953.5Nm`