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A heavy, flexible, uniform claim of length pi r and mass lambda pi r lies in a smooth semicircular tube AB of radius r. Assuming a slight disturbance to start the claim in motion, find the velocity w with which it will emerge from the end B of the tube. |
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Answer» Solution :Since friction is absent, we can apply the law of conservation of energy. Centre of gravity of a SEMICIRCULAR are is at a distance `((2r)/(pi))` from centre. Initial potential energy `= (lambda pi r) g((2r)/(pi))`, Final potential energy `= (lambda pi r) g((-pir)/(2))` when the CHAIN COMPLETELY slips off the TUBE, all the links of the chain have the same velocity v. Kinetic energy of chain `= (1)/(2) (lambda pi r) v^(2)` , From COE, `lambda pi r((2r)/(pi))g = (lambda pi r)g((-pi r)/(2))+ (1)/(2) (lambda pi r)v^(2)` From which we find `v = sqrt(2gr ((2)/(pi) + (pi)/(2)))` |
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