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A uniform rod of length L, moving down with linear velocity $v_{0}$ and zero angular velocity, collides with a horizontal surface. Just after collision, the linear velocity of its centre of mass becomes v in the upward direction as shown in the figure. The angular velocity of the rod just after the impact is

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$\frac{6 (v - v_{0}) \cos θ}{L}$

$\frac{2 (v_{0} + v) \cos θ}{3 L}$

$\frac{3 (v - v_{0}) \cos θ}{2 L}$

$\frac{6 (v_{0} + v) \cos θ}{L}$

answer is D.

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Detailed Solution

Angular momentum equation about point of impact.

$m v_{0} \frac{L}{2} \cos θ = \frac{m L^{2}}{12} ω - m v \frac{L}{2} \cos θ \Rightarrow \frac{ω L}{6} = (v_{0} + v) \cos θ \Rightarrow ω = \frac{6 (v_{0} + v) \cos θ}{L}$

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