A uniform chain of length 2L is hanging in an | vedclass

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(C) Let $\lambda$ be the linear mass density of the chain. Total length is $2L$,so mass $M = 2L\lambda$.When end $B$ is displaced by $x$,the length on

Vedclass · Accepted Answer

$\sqrt{gL}$

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(C) Let $\lambda$ be the linear mass density of the chain. Total length is $2L$,so mass $M = 2L\lambda$.When end $B$ is displaced by $x$,the length on the right side becomes $L+x$ and on the left side becomes $L-x$.The net force $F_{net}$ acting on the chain is the difference in weights of the two sides:$F_{net} = \lambda(L+x)g - \lambda(L-x)g = 2\lambda xg$.The total mass of the chain is $M = 2L\lambda$. Thus,the acceleration $a$ is:$a = \frac{F_{net}}{M} = \frac{2\lambda xg}{2L\lambda} = \frac{xg}{L}$.Using the relation $v \frac{dv}{dx} = a$,we integrate:$\int_{0}^{v} v \, dv = \int_{0}^{L} \frac{g}{L} x \, dx$.$\frac{v^2}{2} = \frac{g}{L} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{g}{L} \cdot \frac{L^2}{2} = \frac{gL}{2}$.$v^2 = gL \Rightarrow v = \sqrt{gL}$.

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