First slide
Law of conservation of angular momentum
Question

A hollow straight tube of length l and mass m can turn freely about its centre (fixed) on a smooth horizontal table. Another smooth uniform rod of same length and mass is fitted into the tube so that their centres coincide. The system is set in motion with an initial angular velocity ω0. The angular velocity of the rod at an instant when the rod slips out of the tube is:

 

Moderate
Solution

Initial angular momentum about O is

L1 = 2(ml212)ω0 = ml26ω0

Final angular momentum about O is

Lf =  = [ml212+(ml212+ml2)]ω = 76ml2ω

Angular momentum will be conserved

Li = Lf or ω = ω07

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