Question

A solid body rotates with deceleration about a stationary axis with an angular deceleration $|\alpha |=k\sqrt{\omega }$; where 'k' is a constant and $\omega$ is the angular velocity of the body. If the initial angular velocity is ${\omega }_0$, then mean angular velocity of the body averaged over the whole time of rotation is

• A. ${\omega }_0$
• B. $\frac{{\omega }_0}{2}$
• C. $\frac{{\omega }_0}{3}$
• D. $\frac{{\omega }_0}{4}$

Answer 1

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

$-\frac{d\omega }{dt}=k{\omega }^{1/2}$
$\Rightarrow -{\int }_{{\omega }_0}^{\omega }\frac{d\omega }{{\omega }^{1/2}}=k{\int }_0^{t}dt$
$\Rightarrow -2{\left[\sqrt{\omega }\right]}_{{\omega }_0}^{\omega }=kt$
$\Rightarrow 2\sqrt{{\omega }_0}-2\sqrt{\omega }=kt$
$\Rightarrow 2\sqrt{\omega }=2\sqrt{{\omega }_0}-kt;\omega ={(\sqrt{{\omega }_0}-\frac{1}{2}kt)}^2$
The body comes to a stop $(\omega =0)$ in time
$t_0=\frac{2\sqrt{{\omega }_0}}{k}$
$\omega ={(\sqrt{{\omega }_0}-\frac{kt}{2})}^2$
$\Rightarrow {\omega }_{=}{\omega }_0+\frac{k^2{t}^2}{4}-kt\sqrt{{\omega }_0}$
Since $<{\omega }_{ar}>=\frac{{\int }_0^{t_0}\omega dt}{{\int }_0^{0}dt}=\frac{{\omega }_0{t}_0+\frac{k^2}{4}\frac{t_0^3}{3}-k\sqrt{{\omega }_0}\frac{t_0^2}{2}}{t_0}$
$\Rightarrow <{\omega }_{av}>={\omega }_0+\frac{k^2}{4}\frac{t_0^2}{3}-k\sqrt{{\omega }_0}.\frac{t_0}{2}$
$\Rightarrow <{\omega }_{ar}>={\omega }_0+\frac{k^2}{12}\frac{4{\omega }_0}{k^2}-\frac{k\sqrt{{\omega }_0}2\sqrt{{\omega }_0}}{2k}$
$\Rightarrow <{\omega }_{av}>={\omega }_0+\frac{{\omega }_0}{3}-{\omega }_0$
$\Rightarrow <{\omega }_{av}>=\frac{{\omega }_0}{3}.$