Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `5 m`. If the object's angular velocity changes from `7 Hz` to `8 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque is `=157.1Nm`

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=5m`

For the object, `I=mr^2`

So, `I=3*(5)^2=75kgm^2`

And the rate of change of angular velocity is

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

`=(16pi-14pi)/5=(2/5pi)rads^-2`

So,

The torque is

`tau=75*2/5pi=157.1Nm`