Physics · Rotational Dynamics

JEE Advanced 2020 — Paper 2 — Question 7

A rod of mass $m$ and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_{M}$ is achieved for $x=x_{M}$ . Then

Option A:
$\omega=\frac{3 v x}{L^{2}+3 x^{2}}$
Correct
Option B:
$\omega=\frac{12 \mathrm{vx}}{\mathrm{L}^{2}+12 \mathrm{x}^{2}}$
Option C:
$\mathrm{x}_{\mathrm{M}}=\frac{\mathrm{L}}{\sqrt{3}}$
Correct
Option D:
$\omega_{\mathrm{M}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \sqrt{3}$
Correct

Answer: A, C, D

Step-by-step solution

Conserving angular momentum about pivot.

$\Rightarrow \operatorname{mvx}=\left(\frac{\mathrm{mL}^{2}}{3}+m x^{2}\right) \omega$

$\Rightarrow \omega=\frac{3 v x}{L^{2}+3 x^{2}}$

For $\omega$ to be maximum $\Rightarrow \frac{d \omega}{d x}=0$ , this gives $x=\frac{L}{\sqrt{3}}$ and $\omega_{\max }=\frac{\sqrt{3} v}{2 L}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Angular Momentum and its Conservation