A uniform chain of mass $m$ and length $l$ is placed on a smooth table so that one-third length hangs freely as shown in the figure. Now the chain is released, with what velocity chain slips off the table?

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Solution

Mass of hanging chain part is $\frac{M}{3}$ and its CM is $\frac{l}{6}$ below table.

Hence, initial potential energy is $U_{1} = - \frac{M}{3} g \frac{l}{6} = \frac{M g l}{18}$

When whole chain slips off, then hanging mass is $M$ and the CM of hanging part is $\frac{l}{2}$ below table.

Hence , final potential energy $U_{2} = - M g \frac{l}{2}$

Since total energy is conserved, hence-

$U_{2} + K_{2} = U_{1} + K_{1}$

$⟹ \frac{1}{2} M v^{2} - \frac{M g l}{2} = - \frac{M g l}{18} + 0$

$⟹ v^{2} = \frac{8 g l}{9}$

$⟹ v = \frac{2 \sqrt{2 g l}}{3}$

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A uniform chain of mass m and length l is placed on a smooth table so that one-third length hangs freely as shown in the figure. Now the chain is released, with what velocity chain slips off the table?

A uniform chain of mass $m$ and length $l$ is placed on a smooth table so that one-third length hangs freely as shown in the figure. Now the chain is released, with what velocity chain slips off the table?