# An object with a mass of  3 kg is traveling in a circular path of a radius of 2 m. If the object's angular velocity changes from  2 Hz to  6 Hz in  5 s, what torque was applied to the object?

Jan 27, 2018

$60.31 N . m$

#### Explanation:

Torque ($\tau$) is defined as change in rate of angular momentum i.e $\frac{\mathrm{dL}}{\mathrm{dt}}$ or, $\frac{d \left(I \omega\right)}{\mathrm{dt}}$ or, $I \frac{\mathrm{do} m e g a}{\mathrm{dt}}$ (as $L = I \omega$, where, $I$ is the moment of inertia)

So, $I = m {r}^{2}$

Given , $m = 3 K g$ and $r = 2 m$

So, $I = 12 K g . {m}^{2}$

Now $\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \left(\frac{\nu 1 - \nu 2}{t}\right) \cdot 2 \pi$ i.e $\frac{6 - 2}{5} \cdot 2 \pi \frac{r a d}{s} ^ 2$

So, $\tau = \left(12 \cdot \frac{8}{5} \pi\right) N . m$ i.e $60.31 N . m$