A bird of mass `m` is flying horizontally with speed `v_(0)` suddenly strikes a vertical rod of mass `3m` and length `L` hinged at its base at distance `L//3` from the top of the rod. After collision the bird drops to the ground (but soon recovers to fly happly away). What is the angular velocity of the rod
(a) just after it is hit by the bird and
(b) just as it reaches the ground?

Explanation

Applying the conservation on angular momentum about `O`
`mv_(0)(2L)/(3)=I_(0)omega=(3mL^(2))/(3)omega`
`omega=(2v_(0))/(3L)`
Applying the energy conservation between (`2`) and (`3`)
`3mg((L)/(2))+(1)/(2)I_(0)omega^(2)=(1)/(2)I_(0)omega'^(2)`
`(3mgL)/(2)+(1)/(2)I_(0)(omega'^(2)-omega^(2))=(1)/(2)(3mL^(2))/(3)(omega'^(2)-omega^(2))`
`omega'=sqrt(omega^(2)+(3g)/(L))=sqrt((4v_(0)^(2))/(9L^(2))+(3g)/(L))`