Question

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, find the ratio of translational and rotational kinetic energy of system.

Answer 1

We know translational energy $K_t=\frac{1}{2}m{v}_C^2$ and rotational kinetic energy $K_r=\frac{1}{2}{I}_C{\omega }^2$,
Distance of centre of mass from pivot end
$r_C=\frac{ml+m\left(2l\right)}{m+m}=\frac{3}{2}l$
Moment of inertia of system about centre of mass
and $I_C=m{\left(\frac{l}{2}\right)}^2+m{\left(\frac{l}{2}\right)}^2=\frac{m{l}^2}{2}$
Angular velocity of rod $\omega =\frac{v}{2l}$
Velocity of centre of mass
$v_{cm}=\omega \left(\frac{3}{2}l\right)=\left(\frac{v}{2l}\right)\left(\frac{3}{2}l\right)=\frac{3}{4}v$
Translational kinetic energy of system
$K_{translational}=\frac{1}{2}{M}_{total}\cdot v_{cm}^2=\frac{1}{2}\left(2m\right){\left(\frac{3}{4}v\right)}^2=\frac{9}{16}m{v}^2$
This gives $\frac{K_t}{K_r}=9$.