Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `2 m`. If the object's angular velocity changes from `3 Hz` to `8 Hz` in `1 s`, what torque was applied to the object?

Answer 1

The torque is `=251.3Nm`

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=2 kg`

The radius of the path is `r=2m`

So the moment of inertia is

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

The angular acceleration is

`alpha=(2pif_1-2pif_0)/t=(16-6)pi/1=10pirads^-2`

So,

The torque is

`tau=Ialpha=8*10pi=251.3m`