Question

A uniform rod $AB$ of mass $M$, length $L$ is placed against the vertical wall of a container. Now container is filled with a liquid which has half of the density of the density of rod. If end $B$ is given a gentle push, then calculate the velocity of centre of mass of rod when it becomes horizontal is $\sqrt{\frac{3gL}{n}}$.
Neglect any frictions, viscosity and kinetic energy of water molecules. Find $n$.

Answer 1

Using Archimedes principle, buoyant force on the rod is $={\rho }_{l}Vg=\frac{{\rho }_{b}Vg}{2}=\frac{Mg}{2}$
Apply work energy theorem,
$W_{F_b}=\Delta K+\Delta U$
Here
$W_{F_b}$ is work done by force of buoyancy,
$\Delta K$ is change in KE when rod becomes horizontal
$\Delta U$ is change in PE when rod becomes horizontal

$\frac{Mg}{2}\frac{L}{2}=\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2-0+(0-\frac{MgL}{2})$

$\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2=\frac{MgL}{2}-\frac{Mg}{2}\frac{L}{2}$

$\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2=\frac{MgL}{4}$
$\therefore \omega =\sqrt{\frac{3g}{2L}}$
So, $v_{cm}=\frac{L}{2}\omega =\sqrt{\frac{3gL}{8}}$