A heavy, flexible uniform chain of length π r and mass λ π r lies in a smooth semicircular tube AB of the radius r . Assuming a slight disturbance to start the chain in motion, find the velocity v with which it will emerge from the end B of the tube.

Step-by-step solution

Since friction absent, we can apply the law of conservation of energy. Centre of gravity of a semicircular arc is at a distance 2 π r from the centre. Initial potential energy = λ π r g 2 r π Final potential energy = λ π r g - π r 2 When the chain is completely slipped off the tube, all the links of the chain have the same velocity v . Kinetic energy of chain = 1 2 λ π r v 2 From COE, λ π rg 2 π r = λ π r g - π r 2 + 1 2 λ π r v 2 From which we find, v = 2 g r 2 π + π 2