A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and travelling with uniform speed ‘V’ collide with two ends of the rod simultaneously and inelastically as shown. The marbles get struck to the rod after the collision and continue to move with the rod. If m = M/6 and V = L mts/sec, then the time taken by the rod to rotate through $π$ /2 is

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$π$ s

1 s

$π$ /2 s

2 $π$ s

answer is D.

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Detailed Solution

From Law of Conservation of Angular Momentum :

L_i = L_f
$mv \frac{L}{2} + mv \frac{L}{2} = [\frac{{ML}^{2}}{12} + 2 m {(\frac{L}{2})}^{2}] ω$

$mvL = [\frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{2}] ω$

$\frac{M}{6} vL = \frac{{ML}^{2}}{6} ω [∴ m = \frac{M}{6}]$

$V = Lω [∴ V = Lm / \sec]$

$ω = 1 rad / s^{- 1} \Rightarrow t = \frac{θ}{ω} = \frac{π / 2}{1} = \frac{π}{2} s$

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