Two beads each of mass m are welded at the ends of two light rigid rods each of length l. If the pivots are smooth then find the ratio of translational and rotational kinetic energy of system.

We know translational kinetic energy ${\mathrm{K}}_{\mathrm{t}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{C}}^2$ and rotational kinetic energy ${\mathrm{K}}_{\mathrm{r}}=\frac{1}{2}{\mathrm{I}}_{\mathrm{C}}{\mathrm{\omega }}^2$

Distance of centre of mass from pivot end,

${\mathrm{r}}_{\mathrm{C}}=\frac{\mathrm{ml}+\mathrm{m}\left(2\mathrm{l}\right)}{\mathrm{m}+\mathrm{m}}=\frac{3}{2}\mathrm{l}$

Moment of inertia of system about centre of mass,

${\mathrm{I}}_{\mathrm{C}}=\mathrm{m}{\left(\frac{\mathrm{l}}{2}\right)}^2+\mathrm{m}{\left(\frac{\mathrm{l}}{2}\right)}^2=\frac{{\mathrm{ml}}^2}{2}$

Angular velocity of rod $\mathrm{\omega }=\frac{\mathrm{v}}{2\mathrm{l}}$

velocity of centre of mass, ${\mathrm{v}}_{\mathrm{cm}}=\mathrm{\omega }\left(\frac{3}{2}\mathrm{l}\right)=\left(\frac{\mathrm{v}}{2\mathrm{l}}\right)\left(\frac{3}{2}\mathrm{l}\right)=\frac{3}{4}\mathrm{v}$

Translational kinetic energy of system,

${\mathrm{K}}_{\text{transiational }}=\frac{1}{2}{\mathrm{M}}_{\text{total }}\cdot {\mathrm{v}}_{\mathrm{cm}}^2=\frac{1}{2}\left(2\mathrm{m}\right){\left(\frac{3}{4}\mathrm{v}\right)}^2=\frac{9}{16}{\mathrm{mv}}^2$

${\mathrm{K}}_{\text{roltaional }}=\frac{1}{2}{\mathrm{I}}_{\mathrm{CM}}{\mathrm{\omega }}^2=\frac{1}{2}\left(\frac{{\mathrm{ml}}^2}{2}\right){\left(\frac{\mathrm{v}}{2\mathrm{l}}\right)}^2=\frac{1}{16}{\mathrm{mv}}^2$

This gives $\frac{{\mathrm{K}}_{\mathrm{t}}}{{\mathrm{K}}_{\mathrm{r}}}=9$.