Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `4 m`. If the object's angular velocity changes from `2 Hz` to `9 Hz` in `2 s`, what torque was applied to the object?

Answer 1

The torque is `=703.7Nm`

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 of the path is `r=4m`

For the object, the moment of inertia is `I=mr^2`

So, `I=2*(4)^2=32kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(f_2-f_1)/t*2pi=(9-2)/2*2pi`

`=(7pi)rads^(-2)`

So,

The torque is `tau= 32*7pi=703.7Nm`