JEE Advanced · PhysicsMedium

$A thin uniform rod of mass M and length L is pivoted at one end and hangs vertically. A small particle of mass m moving horizontally with velocity v strikes the other end of the rod and sticks to it. The angular velocity of the system immediately after the collision is:$

A. $\frac{m v L}{\frac{M L ^{2}}{3} + m L ^{2}}$
B. $\frac{m v L}{\frac{M L ^{2}}{12} + m L ^{2}}$
C. $\frac{m v}{\frac{M L}{3} + m L}$
D. $\frac{m v L}{\frac{M L ^{2}}{3}}$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{m v L}{\frac{M L ^{2}}{3} + m L ^{2}}$

Solution

$Initial angular momentum L_{i} = m v L .$
$Final moment of inertia I_{f} = \frac{1}{3} M L^{2} + m L^{2} .$
$Conservation of angular momentum gives m v L = I_{f} ω .$
$Solving for ω yields ω = \frac{m v L}{\frac{1}{3} M L ^{2} + m L ^{2}} .$
$Hence the answer is (A).$

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