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

Answer 1

Torque: `\tau = \frac{dL}{dt}=\frac{d}{dt}(I\omega)= I\dot(\omega)=2\pi.mr^2\dot{f} = 16\pi \quad N.m`

Explanation:

`\omega = 2\pif; \qquad \dot{\omega} = 2\pi\dot{f}; \qquad I = mr^2`

Torque: `\tau = \frac{dL}{dt}=\frac{d}{dt}(I\omega)= I\dot(\omega)=2\pi.mr^2\dot{f}`

`m=2\quad kg; \qquad r = 4\quad m; \dot{f}=\frac{\Delta f}{\Delta t}=(5Hz-3Hz)/(3\quad s)=1s^{-2}`

`\tau = 2\pi.(2\quad kg)(4\quad m)^2(1s^{-2})=16\pi \quad N.m`