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. $\omega =\frac{3vx}{L^2+3x^2}$
• 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)

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

When bullet strikes the rod there is no external torque acting the system of rod & bullet $({\tau }_{ext}=0)$

By the angular momentum conservation about the suspension point,
$L_i=L_f$
$\Rightarrow mv{r}_{\perp }=I\omega$
$\because r_{\perp }=x,I=\frac{m{L}^2}{3}+m{x}^2$
$\Rightarrow mvx=(\frac{m{L}^2}{3}+m{x}^2)\omega$
$\therefore \omega =\frac{mvx}{\frac{m{L}^2}{3}+m{x}^2}=\frac{3vx}{L^2+3x^2}$
For ${}^{\prime }{\omega }^{\prime }$ to be maximum $\Rightarrow \frac{d\omega }{dx}=0$
$\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$
$\Rightarrow x=\sqrt{\frac{L^2}{3}}=\frac{L}{\sqrt{3}}$
$\Rightarrow x_M=\frac{L}{\sqrt{3}}$

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

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