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A small bead of mass M is launched from the end of a horizontal rod at a speed u along the rod having same mass M. The rod is free to rotate in a vertical plane about a end. Simultaneously the rod is given a angular velocity $ω$ as shown in the figure. Find the angular acceleration of the rod.
[The rod is hinged at an end]

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$\frac{3}{4 L} (\frac{3 g}{2} + u ω)$

$\frac{9 g}{2 L} + u ω$

$\frac{9 g}{8 L}$

$\frac{3}{4 L} (\frac{3 g}{2} - 2 u ω)$

answer is D.

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Detailed Solution

$τ = \frac{d L}{d t}$

$m g \frac{1}{2} \cos θ + M g x \cos θ = \frac{d}{d t} [- (\frac{m l^{2}}{3} + M x^{2}) ω^{1}]$

$= - (\frac{m l^{2}}{3} + M x^{2}) \frac{d ω^{1}}{d t} = ω^{1} (0 + 2 M \times \frac{d x}{d t})$

$(M g \cos θ) (\frac{l}{2} + x) = - M (\frac{l^{2}}{2} + x^{2}) α - ω^{1} [2 M x (- V)]$

$(g \cos θ) (\frac{l}{2} + x) = - (\frac{l^{2}}{2} + x^{2}) α + 2 ω^{1} V$

At the initial moment,

$θ = 0^{0}, x = l, V = u, ω^{1} = ω$

$g (1) (\frac{l}{2} + x) = - (\frac{l^{2}}{3} + x^{2}) α + 2 l ω u$

$\frac{3}{2} g l = - \frac{4}{3} l^{2} α + 2 l ω u$

$\frac{4}{3} α l = 2 ω u - \frac{3 g}{2} \Rightarrow α = \frac{3}{2} \frac{ω u}{l} - \frac{g}{8 l}$

$α = - \frac{3}{4 l} [\frac{3 g}{2} - 2 ω u]$

-ve sign indicates that $α$ is outwards i.e., opposite to $ω$ .

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