Consider a uniform rod of mass M= 4m and length L pivoted about its center. A mass m moving with velocity $ν$ making angle $θ = \frac{π}{4}$ to the rod’s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:

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$\frac{3}{7} \frac{ν}{L}$

$\frac{4}{7} \frac{ν}{L}$

$\frac{3}{7 \sqrt{2}} \frac{ν}{L}$

$\frac{3 \sqrt{2}}{7} \frac{ν}{L}$

answer is A.

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Detailed Solution

$\begin{array}{l} M o m e n t o f i n e r t i a o f r o d a n d m a s s s y s t e m = I = \frac{(4 m) L^{2}}{12} + m {(\frac{L}{2})}^{2} \\ A p p l y i n g C o n s e r v a t i o n o f A n g u l a r M o m e n t u m a b o u t O, \\ L_{o i} = L_{o f} \\ \Rightarrow m V r S i n θ = I ω \\ \Rightarrow \frac{m V}{\sqrt{2}} \times \frac{L}{2} = [\frac{4 m L^{2}}{12} + \frac{m L^{2}}{4}] \times ω \\ (∵ r = \frac{L}{2} a n d θ = 45 °) \\ \Rightarrow ω = \frac{6 V}{7 \sqrt{2} L} = \frac{3 \sqrt{2} V}{7 L} \end{array}$