Physics - Center of Mass Question with Solution | TestHub

When the rod is released from a horizontal position, its center of mass is at a height of $L/2$ from the pivot. When it becomes vertical, the center of mass is at a distance $L/2$ below the pivot. By conservation of mechanical energy, the decrease in potential energy equals the increase in rotational kinetic energy. $Mg(L/2)=\frac{1}{2}I{\omega }^2$. For a rod pivoted at one end, $I=\frac{1}{3}M{L}^2$. So, $Mg(L/2)=\frac{1}{2}(\frac{1}{3}M{L}^2){\omega }^2$. This simplifies to $gL=\frac{1}{3}{L}^2{\omega }^2$, so ${\omega }^2=\frac{3g}{L}$. The velocity of the center of mass is $v_{CM}=\omega r_{CM}=\omega (L/2)$. Therefore, $v_{CM}=\sqrt{\frac{3g}{L}}\cdot \frac{L}{2}=\sqrt{\frac{3g{L}^2}{4L}}=\sqrt{\frac{3gL}{4}}=\frac{\sqrt{3gL}}{2}$.