Question

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

• A. $\omega=\frac{3 vX }{ L ^{2}+3 x ^{2}}$
• B. $\omega=\frac{12 vx }{ L ^{2}+12 x ^{2}}$
• C. $x_{M}=\frac{L}{\sqrt{3}}$
• D. $\omega_{ M }=\frac{ V }{2 L } \sqrt{3}$

Answer 1

The Correct Option is A, C, D

(A) $\omega=\frac{3 vX }{ L ^{2}+3 x ^{2}}$
(C) $x_{M}=\frac{L}{\sqrt{3}}$
(D) $\omega_{ M }=\frac{ V }{2 L } \sqrt{3}$</div>