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

Answer: (A), (C), (D)

Conserving angular momentum about pivot.
$\Rightarrow mvx =\left(\frac{ mL ^{2}}{3}+ mx ^{2}\right) \omega$
$\Rightarrow \omega=\frac{3 vx }{ L ^{2}+3 x ^{2}}$
For $\omega$ to be maximum
$\Rightarrow \frac{d \omega}{d x}=0$,
this gives $x=\frac{L}{\sqrt{3}}$
and $\omega_{\max }=\frac{\sqrt{3} v}{2 L}$