A uniform bar of mass $M$ and length $L$ collides with a horizontal surface. Before collision, velocity of centre of mass was $v_{0}$ and no angular velocity. Just after collision, velocity of centre of mass of bar becomes $v$ in upward direction as shown. Angular velocity $ω$ of the bar just after impact is:

Question

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Answer 1

Angular momentum will be conserved about point A just before and just after collision.

$m v_{0 .} \frac{L}{2} c o s θ = \frac{m L ^{2}}{3} . ω - m v . \frac{L}{2} c o s θ$

$ω = \frac{3 ( v _{0} + v ) c o s θ}{2 L} .$

Answer 2

Angular momentum will be conserved about point A just before and just after collision.

m v_{0 .} \frac{L}{2} c o s θ = \frac{m L ^{2}}{3} . ω - m v . \frac{L}{2} c o s θ

ω = \frac{3 ( v _{0} + v ) c o s θ}{2 L} .