Question

An object with a mass of `8 kg` is traveling in a circular path of a radius of `12 m`. If the object's angular velocity changes from `9 Hz` to `6 Hz` in `6 s`, what torque was applied to the object?

Answer 1

The torque is `=3619.1Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

For the object, `I=mr^2`

The mass is `m=8 kg`

The radius of the path is `r=12m`

So the moment of inertia is

,`I=8*(12)^2=1152kgm^2`

The angular acceleration is

`alpha=(2pif_1-2pif_0)/t=(18-12)pi/6=pirads^-2`

So,

The torque is

`tau=Ialpha=1152*pi=3619.1Nm`