Question

A solid body rotates about a stationary axis with an angular deceleration $\beta \propto \sqrt{\omega }$ where $\omega$ is its angular velocity. If at the initial moment of time, its angular velocity was equal to ${\omega }_0$, then the mean angular velocity of the body averaged over the whole time of rotation till it comes to rest is,

• A. $\frac{{\omega }_0}{\sqrt{3}}$
• B. $\frac{{\omega }_0}{3}$
• C. $\frac{{\omega }_0}{2}$
• D. $\frac{{\omega }_0}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{{\omega }_0}{3}$

Given, $\beta$ is deceleration.
So, $\beta =\frac{-d\omega }{dt}\propto \sqrt{\omega }$
or $\frac{-d\omega }{dt}=K\sqrt{\omega }$
where $K$ is the proportionality constant.
i.e $-\underset{{\omega }_0}{\overset{\omega }{\int }}\frac{d\omega }{\sqrt{\omega }}=\underset{0}{\overset{t}{\int }}K.dt$
Integrating both sides and applying limits,
$\Rightarrow \sqrt{\omega }=\sqrt{{\omega }_0}-\frac{Kt}{2}$ ... (1)

When, $\omega =0$, total time of rotation,
$t=\frac{2\sqrt{{\omega }_0}}{K}$
As we know, Average angular velocity ${\omega }_{av}=\frac{\int \omega .dt}{\int dt}$
From (1),
$\omega ={\left(\sqrt{{\omega }_0}\right)}^2+{\left(\frac{Kt}{2}\right)}^2-2\sqrt{{\omega }_0}\frac{Kt}{2}$

Thus, ${\omega }_{av}=\frac{\underset{0}{\overset{\frac{2\sqrt{{\omega }_0}}{K}}{\int }}({\omega }_0+\frac{K^2{t}^2}{4}-Kt\sqrt{{\omega }_0})dt}{\frac{2\sqrt{{\omega }_0}}{K}}$
${\omega }_{av}=\frac{{[{\omega }_{0}t+\frac{K^2{t}^3}{3\times 4}-\frac{K}{2}\sqrt{{\omega }_0}{t}^2]}_0^{\frac{2\sqrt{{\omega }_0}}{K}}}{\frac{2\sqrt{{\omega }_0}}{K}}$
$\Rightarrow {\omega }_{av}=\frac{{\omega }_0}{3}$