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 horiozontally 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. $\omega =\frac{3vx}{L^2+3x^2}$
• B. $\omega =\frac{12vx}{L^2+12x^2}$
• C. $x_m=\frac{L}{\sqrt{3}}$
• D. ${\omega }_m=\frac{v}{2L}\sqrt{3}$

Answer 1

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

${\omega }_m=\frac{v}{2L}\sqrt{3}$


On applying law of conservation of angular momentum before and after collision about the pivot point, we get
$mvx=(\frac{m{L}^2}{3}+m{x}^2)\omega$
$\Rightarrow \omega =\frac{mvx}{\frac{m{L}^2}{3}+m{x}^2}=\frac{3vx}{L^2+3x^2}$

For maximum $\omega ,\frac{d\omega }{dx}=0$
$\Rightarrow \frac{(L^2+3x^2)\left(3v\right)-\left(3vx\right)(0+6x)}{{(L^2+3x^2)}^2}=0$$\Rightarrow 3v{L}^2+9v{x}^2-18v{x}^2=0$$\Rightarrow 3v{L}^2=9v{x}^2$
$\Rightarrow x=\sqrt{\frac{L^2}{3}}=\frac{L}{\sqrt{3}}$

and, ${\omega }_m=\frac{3v\times \frac{L}{\sqrt{3}}}{L^2+3\times \frac{L^2}{3}}=\frac{\sqrt{3}vL}{2L^2}$
$\Rightarrow {\omega }_m=\frac{v}{2L}\sqrt{3}$

Thus, options $\left(A\right)$, $\left(C\right)$ and $\left(D\right)$ are correct.

Why this question? Concept Involved - Law of conservation of angular momentum and maxima and minima. Tip - For maxima or minima of $y\left(x\right)$, take $\frac{dy}{dx}=0$ If $\frac{d^{2}y}{d{x}^2}<0$, then maxima and if $\frac{d^{2}y}{d{x}^2}>0$, then minima. Importance in $JEE$ $-$ Asked in Advanced $2020$.