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

The correct option is D

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

Step 1: Given data

Mass of the rod $=m$

Length of the rod $=L$

Mass of the bullet $=m$

Speed of the bullet $=v$

Angular speed $={\omega }_M$

Bullet strikes the rod horizontally at a distance $=x$

Moment of inertia of the system about pivoted end $=I$

Step 2: Formula

$\overrightarrow{v}$ strikes the rod here a torque is act on the system, the net external torque is zero

Therefore angular momentum is conserved

According to the law of conservation of angular momentum

Initial angular momentum $=$ Final angular momentum

$mvx=I\omega$

$I=I_{rod}+I_{bullet}$

Step 3: Find the final angular speed:

Using the parallel axis theorem

The parallel axis theorem is used for finding the moment of inertia of the area of a rigid body whose axis is parallel to the axis of the known moment body, and it is through the centre of gravity of the object.

$=\left(\frac{m{L}^2}{12}+m\times {\left(\frac{L}{2}\right)}^2\right)+m\times x^2$

$$\begin{gathered} =\frac{m{L}^2}{12}+\frac{m{L}^2}{4}+m{x}^2 \\ =\frac{m{L}^2}{3}+m{x}^2 \end{gathered}$$

Therefore,

$\begin{array}{rcl}mvx& =& \left(\frac{m{L}^3}{3}+m{x}^2\right)\omega \\ mvx& =& m\left(\frac{L^2}{3}+x^2\right)\omega \\ \omega & =& \frac{3mvx}{m\left(L^2+3x^2\right)}\\ \omega & =& \frac{3vx}{\left(L^2+3x^2\right)}\end{array}$

Option A is correct.

Step 3 : Find the maximum angular speed ${\omega }_M$ and maximum distance $x_M$

for maximum

$\frac{d\omega }{dt}=0$

$\begin{array}{rcl}\frac{\left(L^2+3x^2\right).3v-3vx\left(6x\right)}{L^2+3x^2}& =& 0\\ 3v{L}^2+9v{x}^2-18v{x}^2& =& 0\\ 3v{L}^2-9v{x}^2& =& 0\\ x^2& =& \frac{3v{L}^2}{9v}\\ x& =& \frac{L}{\sqrt{3}}\\ & & \end{array}$

$x_M=\frac{L}{\sqrt{3}}$

$\begin{array}{rcl}\omega & =& {\omega }_M\\ & =& \frac{3v\times {\displaystyle \frac{L}{\sqrt{3}}}}{L^2+3x{\displaystyle \frac{L^2}{3}}}\\ & =& \frac{3vL}{\sqrt{3}\times 2L^2}\\ {\omega }_M& =& \frac{\sqrt{3}v}{2L}\end{array}$

Options C and D are correct.

Hence options A, C, and D are the correct answers.