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 `4 Hz` to `9 Hz` in `3 s`, what torque was applied to the object?

Answer 1

The torque applied to the object is 53.33 Newton-metre.

Explanation:

As we know torque `tau = Ialpha`
So angular acceleration `alpha`= (change in angular velocity `omega)//(`change in time t)
`=> alpha = (omega_2-omega_1)//dt`
`=>alpha=5//3 m/s^2`
And torque now becomes `tau= mr^2alpha[I=mr^2]`
By substituting all values we get,
`tau=2kgxx(4m)^2xx(5/3m/s^2)=53.33`Newton-metre.