Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `7 m`. If the object's angular velocity changes from `6 Hz` to `9 Hz` in `8 s`, what torque was applied to the object?

Answer 1

The torque is `=21.21Nm`

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 moving in a circular motion, `I=(mr^2)/2`

The mass is `m= 3kg`

The radius is `r=7m`

So, `I=3*(7)^2/2=73.5kgm^2`

The rate of change of angular velocity is

`alpha=(domega)/dt=(9-6)/8*2pi`

`=(3/4pi) rads^(-2)`

So the torque is `tau=9*(3/4pi) Nm=21.21Nm`