A uniform chain of length L lies on a smooth horizontal table with one-third of its length hanging over the edge. The chain is now released and it starts sliding. If g is the acceleration due to the Earth's gravity, then the chain's velocity when it just slips off the table is:

If we choose the surface of the table as the zero potential energy level, the initial potential energy level, the initial potential energy of the chain is ${\mathrm{U}}_{\mathrm{i}}=-\frac{\mathrm{m}}{3}\mathrm{g}\frac{\mathrm{l}}{6}=-\frac{1}{18}\mathrm{mgl}$ and the final potential energy is ${\mathrm{U}}_{\mathrm{f}}=-\mathrm{mg}\frac{\mathrm{I}}{2}$. If v is the required velocity, then using the energy conservation principle, we have

$\begin{array}{l}{\mathrm{K}}_2+{\mathrm{U}}_2={\mathrm{K}}_1+{\mathrm{U}}_1\\ \Rightarrow \frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}\mathrm{mgl}=-\frac{1}{18}\mathrm{mgl}\\ \Rightarrow \mathrm{v}=\sqrt{\frac{8\mathrm{gl}}{9}}\end{array}$