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A thin rod of mass 0.9kg and length 1m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1kg moving in straight line with velocity 80m/s hits the rod at its bottom most point and sticks to its (see figure). If the angular speed (in rad/s) of the rod immediately after the collision will be 5X, then find the value of X.

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answer is 4.

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Detailed Solution

$\vec{L_{i}} = \vec{L_{f}}$

$mvL = Iω \Rightarrow mvL = (\frac{{ML}^{2}}{3} + {mL}^{2}) ω$

$\Rightarrow 0 .1 \times 80 \times 1 = (\frac{0 .9 \times 1^{2}}{3} + 0 .1 \times 1^{2})$
$\begin{array}{l} 8 = (\frac{3}{10} + \frac{1}{10}) ω \\ 8 = \frac{4}{10} ω \Rightarrow ω = 20 rad / s \end{array}$

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