Question

A uniform vertical cylinder is released from rest when its lower end just touches the liquid surface of a deep lake. Calculate maximum displacement of cylinder (in meter).
Take $l=4\text{m}$ and $\frac{\sigma }{\rho }=\frac{1}{2}$

Answer 1

$Al\sigma g-Ax\rho g=Al\sigma a$


$a=g-\frac{\rho gx}{\sigma l}$
$\underset{0}{\overset{v}{\int }}vdv=\underset{0}{\overset{x}{\int }}gdx-\frac{\rho gx}{\sigma l}dx$
$\Rightarrow \frac{v^2}{2}=gx-\frac{\rho g}{\sigma l}\frac{x^2}{2}$
At maximum displacement, $v=0$
$\Rightarrow x=\frac{2\sigma l}{\rho }=2\times \frac{1}{2}\times 4=4\text{m}$