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A thin uniform rod of length l and mass m suspended vertically is free to rotate about a smooth horizontal axis which passes through its end A. A point mass m moving horizontally with velocity $v_{0}$ hits the lower end B of the rod and sticks to it. After collision the rod just reaches the horizontal position. Which of the following statements is WRONG?

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The value of $v_{0}$ is $\sqrt{4 g l}$

The angular acceleration of the rod when it reaches the horizontal position is $\frac{9 g}{8 l}$

The magnitude of reaction force by the hinge on the rod in horizontal position is $\frac{5 mg}{8 l}$

The magnitude of acceleration of centre of mass of the system (rod + point mass) when the rod is in horizontal position is $\frac{27 g}{32} .$

answer is C.

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Detailed Solution

${mv}_{0} l = \frac{4}{3} {ml}^{2} ω \Rightarrow ω = \frac{3 v_{0}}{4 l}$

$\frac{1}{2} \frac{4}{3} m l^{2} {(\frac{3 v_{0}}{4 l})}^{2} = 3 mg \frac{l}{2} \Rightarrow v_{0}$

$= \sqrt{4 g l}$

$3 mg \frac{l}{2} = \frac{4}{3} {ml}^{2} α \Rightarrow α = \frac{9 g}{8 l}$

$a_{cm} = α r = \frac{9 g}{8 l} \times \frac{3 l}{4} = \frac{27 g}{32}$

$2 mg - F = 2 m \times \frac{27 g}{32} \Rightarrow F$ $= \frac{5 mg}{16}$

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