The rod of mass m and Length l

Question

A rod of mass m and length l fits into a hollow tube of same length and mass . The tube is rotated with an angular velocity w and the rod slips through the hollow surface.  The angular values the rod slips out of the tube

Comments

IS the rod hollow or solid?

Answer 1

Solution :Initial angular momentum about OO is.
Li=2(ml212)ω0=ml26ω0Li=2(ml212)ω0=ml26ω0
Final angular momentum about OO is
Lf=Iω=[ml212+(ml212+ml2)]ω=76ml2ωLf=Iω=[ml212+(ml212+ml2)]ω=76ml2ω
Angular momentum will be conserved
Li=LfLi=Lf or ω=ω07ω=ω07.