Two identical thin rods are moving on a smooth table, as shown. Both of them are rotating with angular speed $\omega$, in clockwise sense about their centres. Their centres have velocity V in opposite directions. The rods collide at their edge and stick together. Length of each rod is L. 
(a) For what value of $\frac{\mathrm{V}}{\mathrm{\omega L}}$ there will be no motion after collision ?
(b) If the ratio $\frac{\mathrm{V}}{\mathrm{\omega L}}$ is half the value found in (a)above, what fraction of kinetic energy is lost in the collision? 

 

. (a) Linear momentum of the system is zero. Hence, the centre of mass of the system (point P) is at rest after collision. Let angular speed after collision be $\omega$₀. Applying angular momentum conservation about point P gives:

${\mathrm{L}}_{\mathrm{BC}}={\mathrm{L}}_{\mathrm{AC}}$ 
$\frac{{\mathrm{ML}}^2}{12}\mathrm{\omega }\left(\mathrm{CLK}\right)+\mathrm{MV}\frac{\mathrm{L}}{2}\left(\mathrm{ACLK}\right)+\frac{{\mathrm{ML}}^2}{12}\mathrm{\omega }\left(\mathrm{CLK}\right)+\mathrm{MV}\frac{\mathrm{L}}{2}\left(\mathrm{ACLK}\right)=2\frac{{\mathrm{ML}}^2}{3}{\mathrm{\omega }}_0\left(\mathrm{CLK}\right)$ 
Where CLK stands for clockwise and ACLK stands for anticlockwise.

$\Rightarrow \frac{{\mathrm{ML}}^2}{6}\mathrm{\omega }-\mathrm{MVL}=\frac{2}{3}{\mathrm{ML}}^2{\mathrm{\omega }}_0\Rightarrow {\mathrm{\omega }}_0=\frac{\mathrm{\omega }}{4}-\frac{3\mathrm{V}}{2\mathrm{L}}$
For no motion ${\mathrm{\omega }}_0=0\Rightarrow \frac{\mathrm{V}}{\mathrm{\omega L}}=\frac{1}{6}$
$\text{ (b) If }\frac{\mathrm{V}}{\mathrm{\omega L}}=\frac{1}{12};{\mathrm{\omega }}_0=\frac{\mathrm{\omega }}{8}$

$$\begin{gathered} {\mathrm{KE}}_{\text{inital }}=\left[\frac{1}{2}\frac{{\mathrm{ML}}^2}{12}{\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{MV}}^2\right]\times 2=\left[\frac{{\mathrm{ML}}^2}{12}{\mathrm{\omega }}^2+\mathrm{M}{\left(\frac{\mathrm{\omega L}}{12}\right)}^2\right]=\frac{13}{144}{\mathrm{ML}}^2{\mathrm{\omega }}^2 \\ {\mathrm{KE}}_{\text{final }}=\frac{1}{2}\cdot \frac{2{\mathrm{ML}}^2}{3}{\mathrm{\omega }}_0^2=\frac{{\mathrm{ML}}^2}{3}{\left(\frac{\mathrm{\omega }}{8}\right)}^2=\frac{{\mathrm{ML}}^2{\mathrm{\omega }}^2}{192} \\ \mathrm{\Delta KE}=\left[\frac{13}{144}-\frac{1}{192}\right.⌋{\mathrm{ML}}^2{\mathrm{\omega }}^2=\frac{49}{576}{\mathrm{ML}}^2{\mathrm{\omega }}^2 \\ \therefore \frac{\mathrm{\Delta KE}}{{\mathrm{KE}}_{\text{inital }}}=\frac{49}{52} \end{gathered}$$