1.

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»

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)))`


Discussion

No Comment Found