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Question

A uniform chain of length ‘L’ is placed on a smooth table of height ‘h’ (h>L) with a length ‘l’ hanging from the edge of table. The chain begins to slide down the table. When the end of the chain is about to leave the edge of table its velocity is $\sqrt{\frac{K g (L^{2} - l^{2})}{2 L}}$ . Where K is ______

Moderate

Solution

$\begin{array}{l} W = \frac{1}{2} m (v^{2} - u^{2}) \\ \int m g d y = \frac{1}{2} λ \cdot L (v^{2} - 0^{2}) \\ λ g \int_{l}^{L} y d y = \frac{1}{2} λ L v^{2} \\ \frac{2 g}{L} {[\frac{y^{2}}{2}]}_{l}^{L} = v^{2} \Rightarrow v = \sqrt{\frac{2 g (L^{2} - l^{2})}{2 L}} \end{array}$

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