A rod of length `l` forming an angle `theta` with the horizontal strikes a frictionless floor at A with its centre of mass velocity `v_(0)` and no angular velocity. Assuming that the impact at A is perfectly elastic. Find the angular velocity of the rod immediately after the impact.

Explanation

Let `v=` lnear velocity of rod after impact (upwards)
`omega=` angular velocity of rod
and `J=` linear impulse at A during impact
then, `J=triangleP=P_(f)P_(i)`
`J=mv-(-mv_(0))`
`thereforeJ=m(v+v_(0))` ..(i)
Angular impulse `=triangleL`
`thereforeJ((l)/(2)costheta)=Iomega=(ml^(2))/(12)omega` ..(ii)

Collision is elastic `(e=1)`
`therefore` relative speed of approach `=` relative speed of separation at point of impact
`v_(0)=v+(l)/(2)omegacostheta` ..(iii)
Solving above equations we get
`omega=(6v_(0)costheta)/(l(1+3costheta^(2)theta))`