Question

A uniform rod of mass $m$ and length $L$ is released from the given position as shown in the figure. At its lowest position, a small particle of equal mass $m$ sticks with rod during the collision. The angular velocity of the combined system just after the collision is:

• A. $\sqrt{\frac{3g}{L}}$
• B. $\frac{1}{4}\sqrt{\frac{3g}{L}}$
• C. $\sqrt{\frac{g}{3L}}$
• D. $\frac{2}{3}\sqrt{\frac{3g}{L}}$

Answer 1

The correct option is B $\frac{1}{4}\sqrt{\frac{3g}{L}}$

Change in potential energy $=\frac{15w^2}{2}$

$\Rightarrow w^2=\frac{mg\frac{L}{2}}{\frac{I}{2}}$

$\Rightarrow w^2=\frac{mg\frac{L}{2}}{\frac{m{L}^2}{3}\times \frac{1}{2}}$

$\Rightarrow w=\sqrt{\frac{3g}{L}}$

angular momentum is conserb after collision inertia will be $I^{\prime }$

$\Rightarrow Iw=I^{\prime }{w}^{\prime }$

$\Rightarrow \frac{m{L}^2}{3}\sqrt{\frac{3g}{L}}=(\frac{{mL}^2}{3}+{mL}^2)w^{\prime }$

$\Rightarrow w^{\prime }=\frac{1}{4}\sqrt{\frac{3g}{L}}$

Hence, the answer is $\frac{1}{4}\sqrt{\frac{3g}{L}}.$