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 `2 Hz` to `4 Hz` in `1 s`, what torque was applied to the object?

Answer 1

The torque is `=100.5Nm`

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=2kg`

The radius is `r=2m`

For the object, `I=mr^2`

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

And the rate of change of angular velocity is

`(d omega)/dt=(Deltaomega)/t=(8pi-4pi)/1=(4pi)rads^-2`

So,

The torque is

`tau=8*4pi=100.5Nm`