Question 84: Centre of Mass and Rotational Motion

Question

A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is $\_\_\_\_$ .
( $\mathrm{g}=$ gravitational acceleration)

Accepted Answer

The rod is pivoted at a distance $\mathrm{L} / 3$ from one end. Using the parallel axis theorem, the moment of inertia about this pivot point: $\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2$
Where, moment of inertia about the centre of mass $\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}$ and distance between the center of mass and the pivot $\mathrm{d}=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6}$.
$$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2=\frac{\mathrm{ML}^2}{12}+\mathrm{M}\left(\frac{\mathrm{~L}}{6}ight)^2 $$
$$ \mathrm{I}=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{36}=\frac{3 \mathrm{ML}^2+\mathrm{ML}^2}{36}=\frac{4 \mathrm{ML}^2}{36}=\frac{\mathrm{ML}^2}{9} $$
When the rod falls from a vertical to a horizontal position,
Initial height of COM (relative to the pivot) $\mathrm{h}_1=\frac{\mathrm{L}}{6}$ above the pivot.
Final height of COM, $\mathrm{h}_2=0$ (it is at the same horizontal level as the pivot when lying on the table).
Loss in Potential Energy:
$$ \Delta U=M g \Delta h=M g\left(\frac{L}{6}ight) $$
The loss in potential energy is converted entirely into rotational kinetic energy ( $\mathrm{K}_{\mathrm{rot}}$ ) at the moment of impact.
$$ \Delta U=K_{\mathrm{rot}} $$
$$\Rightarrow $$ $\operatorname{Mg}\left(\frac{L}{6}ight)=\frac{1}{2} I \omega^2$
$\Rightarrow \frac{\mathrm{MgL}}{6}=\frac{1}{2} 	imes\left(\frac{\mathrm{ML}^2}{9}ight) \omega^2$
$\Rightarrow \frac{\mathrm{g}}{6}=\frac{\mathrm{L} \omega^2}{18}$
$\begin{aligned} & \Rightarrow \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \ & \Rightarrow \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}\end{aligned}$

Answer

$\sqrt{\frac{3}{2} \frac{g}{L}}$

Answer

$\sqrt{\frac{3 g}{L}}$

Answer

$\frac{3}{\sqrt{2}} \sqrt{\frac{g}{L}}$

Answer

$\frac{1}{\sqrt{2}} \sqrt{\frac{g}{L}}$

Centre of Mass and Rotational Motion

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