Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `15 m`. If the object's angular velocity changes from `25 Hz` to `28 Hz` in `9 s`, what torque was applied to the object?

Answer 1

Angular acceleration `alpha=(omega_f-omega_i)/t= 2pi((n_f-n_i)/t)=2pi(28-25)/9=(2pi)/3rads^-2`
Moment of inertia

`I=mr^2=3xx15^2=675kgm^2`

Torque
`tau=Ixxalpha=675xx(2pi)/3=450pi J`