JEE Main 2019 — Physics solution

JEE Main2019Physics

A thin smooth rod of length L and mass M is rotating freely with angular speed $ω_{0}$ about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be :

A
$\frac{M ω _{0}}{M + 3 m}$
B
$\frac{M ω _{0}}{M + m}$
C
$\frac{M ω _{0}}{M + 6 m}$
D
$\frac{M ω _{0}}{M + 2 m}$

Solution

Initial angular momentum = Final Angular Momentum

$\frac{M L ^{2}}{12} ω_{0} = (\frac{M L ^{2}}{12} + 2 \frac{m L ^{2}}{4}) ω \Rightarrow ω = \frac{M ω _{0}}{M + 6 m}$

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