Question

A chain $AB$ of length $l$ is located in a smooth horizontal tube so that its fraction of length $h$ hangs freely and touches the surface of the table with its end $B$ (figure shown above). At a certain moment the end $A$ of the chain is set free. If the velocity with which this end of the chain slip out of the tube is given as $v=\sqrt{xgh\ln \left(\frac{l}{h}\right)}$. Find $x$

Answer 1

Let the length of the chain inside the smooth horizontal tube at an arbitrary instant is $x$.
From the equation,
$m\overrightarrow{w}=\overrightarrow{F}+\overrightarrow{u}\frac{dm}{dt}$
as $\overrightarrow{u}=0$, $\overrightarrow{F}\upuparrows \overrightarrow{w}$, for the chain inside the tube
$\lambda xw=T$ where $\lambda =\frac{m}{l}$ (1)
Similarly for the overhanging part,
$\overrightarrow{u}=0$
Thus $mw=F$
or $\lambda hw=\lambda hg-T$ (2)
From (1) and (2),
$\lambda (x+h)w=\lambda hg$ or $(x+h)v\frac{dv}{ds}=hg$
or, $(x+h)v\frac{dv}{(-dx)}=gh$,
[As the length of the chain inside the tube decreases with time, $ds=-dx$.]
or, $vdv=-gh\frac{dx}{x+h}$
Integrating, ${\int }_0^{v}vdv=-gh{\int }_{l-h}^0\frac{dx}{x+h}$
or, $\frac{v^2}{2}=gh\ln \left(\frac{l}{h}\right)$ or $v=\sqrt{2gh\ln \left(\frac{l}{h}\right)}$