A chain of length L and mass m is placed upon | vedclass

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(C) Let the mass per unit length of the chain be $\lambda = \frac{m}{L}$.Initially,a length $b$ of the chain is on the incline. The center of mass of

Vedclass · Accepted Answer

$\sqrt {  \frac{{g\,\sin \,	heta }}{L}\left( {{L^2} - {b^2}} ight)}$

Vedclass · Answer

(C) Let the mass per unit length of the chain be $\lambda = \frac{m}{L}$.Initially,a length $b$ of the chain is on the incline. The center of mass of this part is at a vertical depth of $\frac{b}{2} \sin 	heta$ below the horizontal surface.The potential energy of the chain initially is $PE_i = -(\lambda b) g (\frac{b}{2} \sin 	heta) = -\frac{m g b^2}{2L} \sin 	heta$.Finally,the entire chain of length $L$ is on the incline. The center of mass of the entire chain is at a vertical depth of $\frac{L}{2} \sin 	heta$ below the horizontal surface.The potential energy of the chain finally is $PE_f = -(m) g (\frac{L}{2} \sin 	heta) = -\frac{m g L}{2} \sin 	heta$.By the law of conservation of mechanical energy,the loss in potential energy equals the gain in kinetic energy:$PE_i - PE_f = KE_f - KE_i$$-\frac{m g b^2}{2L} \sin 	heta - (-\frac{m g L}{2} \sin 	heta) = \frac{1}{2} m v^2 - 0$$\frac{m g}{2L} \sin 	heta (L^2 - b^2) = \frac{1}{2} m v^2$$v^2 = \frac{g \sin 	heta}{L} (L^2 - b^2)$$v = \sqrt{\frac{g \sin 	heta}{L} (L^2 - b^2)}$

$A$ chain of length $L$ and mass $m$ is placed upon a smooth surface as shown. The length of the chain on the incline is $b$. Calculate the velocity of the chain when its end reaches point $B$.

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