# 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?

Mar 9, 2016

Torque: $\setminus \tau = \setminus \frac{\mathrm{dL}}{\mathrm{dt}} = \setminus \frac{d}{\mathrm{dt}} \left(I \setminus \omega\right) = I \setminus \dot{\setminus \omega} = 2 \setminus \pi . m {r}^{2} \setminus \dot{f} = 16 \setminus \pi \setminus \quad N . m$

#### Explanation:

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

Torque: $\setminus \tau = \setminus \frac{\mathrm{dL}}{\mathrm{dt}} = \setminus \frac{d}{\mathrm{dt}} \left(I \setminus \omega\right) = I \setminus \dot{\setminus \omega} = 2 \setminus \pi . m {r}^{2} \setminus \dot{f}$

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

$\setminus \tau = 2 \setminus \pi . \left(2 \setminus \quad k g\right) {\left(4 \setminus \quad m\right)}^{2} \left(1 {s}^{- 2}\right) = 16 \setminus \pi \setminus \quad N . m$