Question

A uniform rod of length $l$ & mass $M$ is free to rotate about frictionless pin through one end. The rod is released from rest in a horizontal position. Find momentum of system (rod) when rod is at its lowest position.

• A. $M\sqrt{5gl}$
• B. $M\sqrt{3gl}$
• C. $M\sqrt{2gl}$
• D. $\frac{M}{2}\sqrt{3gl}$

Answer 1

The correct option is D $\frac{M}{2}\sqrt{3gl}$

$\because$COM is at $\frac{l}{2}$.
By conservation of energy:
loss in PE = gain in KE$\Rightarrow Mg\frac{l}{2}=\frac{1}{2}Ml\frac{l}{3}{\omega }^2$
$\Rightarrow {\omega }^2=\frac{6gl}{2l^2}=\frac{3g}{l}$
$\Rightarrow \omega =\sqrt{\frac{3g}{l}}$
$v_{CM}=\frac{l}{2}\omega =\frac{l}{2}\sqrt{\frac{3g}{l}}=\sqrt{\frac{3gl}{4}},p=M{v}_{CM}=M\sqrt{\frac{3gl}{4}}$