Question

Consider a uniform rod of mass $M=4m$ and length $L$ pivoted about its centre. A mass $m$ moving with a velocity $V$ making an angle $\theta =\frac{\pi }{4}$ to the rod's long axis collides with one end of the rod, and sticks to it. The angular speed of the rod-mass system just after the collision is:

• A. $\frac{3}{7\sqrt{2}}\frac{V}{L}$
• B. $\frac{3}{7}\frac{V}{L}$
• C. $\frac{3\sqrt{2}}{7}\frac{V}{L}$
• D. $\frac{4}{7}\frac{V}{L}$

Answer 1

The correct option is C $\frac{3\sqrt{2}}{7}\frac{V}{L}$



Angular momentum of the rod-mass system about point $O$ is

$\Rightarrow L=(mg\times 0)+\frac{mV}{\sqrt{2}}\times \frac{L}{2}=\frac{mVL}{2\sqrt{2}}$

Let, $I$ be the moment of inertia about $O$ of the rod-mass system, and, $\omega$ be the angular speed just after collision.

$I=\frac{4m{L}^2}{12}+\frac{m{L}^2}{4}=\frac{7}{12}m{L}^2$

$L=I\omega$

$\Rightarrow \frac{mVL}{2\sqrt{2}}=\frac{7}{12}m{L}^2\times \omega$

$\therefore \omega =\frac{6V}{7\sqrt{2}L}=\frac{3\sqrt{2}V}{7L}$

Hence, option $\left(C\right)$ is correct.