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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.

Answer» <html><body><p></p>Solution :Since friction is absent, we can apply the law of conservation of energy. <br/> Centre of gravity of a <a href="https://interviewquestions.tuteehub.com/tag/semicircular-1200436" style="font-weight:bold;" target="_blank" title="Click to know more about SEMICIRCULAR">SEMICIRCULAR</a> are is at a distance `((2r)/(pi))` from centre. <br/> Initial potential energy `= (lambda pi r) g((2r)/(pi))`, Final potential energy `= (lambda pi r) g((-pir)/(2))` <br/> when the <a href="https://interviewquestions.tuteehub.com/tag/chain-913601" style="font-weight:bold;" target="_blank" title="Click to know more about CHAIN">CHAIN</a> <a href="https://interviewquestions.tuteehub.com/tag/completely-409686" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETELY">COMPLETELY</a> slips off the <a href="https://interviewquestions.tuteehub.com/tag/tube-1428592" style="font-weight:bold;" target="_blank" title="Click to know more about TUBE">TUBE</a>, all the links of the chain have the same velocity v. <br/> Kinetic energy of chain `= (<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(2) (lambda pi r) v^(2)` , <br/> From COE, `lambda pi r((2r)/(pi))g = (lambda pi r)g((-pi r)/(2))+ (1)/(2) (lambda pi r)v^(2)` <br/> From which we find `v = sqrt(2gr ((2)/(pi) + (pi)/(2)))`</body></html>


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