A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from… | JEE Main 2020 (5 Sep, Shift 2)

Q58JEE Main 2020 · 5 SEP · SHIFT 2Difficult

Angular Momentum & its Conservation

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

Numerical answer type. Work it out, then open the answer.

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Answer: 20

Angular momentum about the pivot is conserved: $mvL = I\omega$ .
$mvL = 0.1(80)(1) = 8\ \text{kg m}^2\text{/s}$
$I = \frac{1}{3}(0.9)(1)^2 + 0.1(1)^2 = 0.3 + 0.1 = 0.4\ \text{kg m}^2$
$\omega = \frac{8}{0.4} = 20\ \text{rad/s}$
Answer: 20 rad/s

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