A uniform flexible chain of mass m and length | vedclass

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(B) Let the chain have mass $m$ and length $2l$. Initially,it hangs symmetrically over the pin,so $l$ length is on each side. The center of mass of ea

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$\sqrt{gl}$

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(B) Let the chain have mass $m$ and length $2l$. Initially,it hangs symmetrically over the pin,so $l$ length is on each side. The center of mass of each half is at a distance $l/2$ below the pin.When the chain is displaced,let it slip by a distance $x$. The new lengths on the two sides are $(l+x)$ and $(l-x)$.Using the Conservation of Mechanical Energy $(COME)$: The decrease in Potential Energy $(P.E.)$ equals the increase in Kinetic Energy $(K.E.)$.Initially,the center of mass of each half is at $h_i = -l/2$ relative to the pin. Total $P.E._i = 2 	imes (m/2)g(-l/2) = -mgl/2$.Finally,when the chain leaves the pin,one end is at $2l$ and the other is at $0$. The center of mass is at $h_f = -l$. Total $P.E._f = mg(-l) = -mgl$.Change in $P.E. = P.E._i - P.E._f = (-mgl/2) - (-mgl) = mgl/2$.Equating to $K.E. = \frac{1}{2}mv^2$,we get $\frac{1}{2}mv^2 = mgl/2$,which simplifies to $v = \sqrt{gl}$.

$A$ uniform flexible chain of mass $m$ and length $2l$ hangs in equilibrium over a smooth horizontal pin of negligible diameter. One end of the chain is given a small vertical displacement so that the chain slips over the pin. The speed of the chain when it leaves the pin is

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