A smooth semicircular tube AB of radius R is | vedclass

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(C) Let the mass of the chain be $M$. The length of the chain is $L = \pi R$. The mass per unit length is $\lambda = M/(\pi R)$.Initially,the center o

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$\sqrt {2gR\left( {\frac{2}{\pi } + \frac{\pi }{2}} \right)} $

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(C) Let the mass of the chain be $M$. The length of the chain is $L = \pi R$. The mass per unit length is $\lambda = M/(\pi R)$.Initially,the center of mass of the semicircular chain is at a height $h_i = \frac{2R}{\pi}$ from the horizontal diameter.Initial potential energy $PE_i = Mgh_i = Mg \frac{2R}{\pi}$. Initial kinetic energy $KE_i = 0$.When the chain emerges from the tube,it hangs vertically. The center of mass of the chain of length $\pi R$ is at a distance $L/2 = \pi R / 2$ below the horizontal diameter.Final potential energy $PE_f = -Mg \frac{\pi R}{2}$. Final kinetic energy $KE_f = \frac{1}{2} Mv^2$.By the law of conservation of mechanical energy: $KE_i + PE_i = KE_f + PE_f$.$0 + Mg \frac{2R}{\pi} = \frac{1}{2} Mv^2 - Mg \frac{\pi R}{2}$.$MgR \left( \frac{2}{\pi} + \frac{\pi}{2} \right) = \frac{1}{2} Mv^2$.$v^2 = 2gR \left( \frac{2}{\pi} + \frac{\pi}{2} \right)$.$v = \sqrt{2gR \left( \frac{2}{\pi} + \frac{\pi}{2} \right)}$.

$A$ smooth semicircular tube $AB$ of radius $R$ is fixed in a vertical plane and contains a heavy flexible chain of length $\pi R$. Find the velocity $v$ with which it will emerge from the open end $B$ of the tube,when slightly displaced.

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