Question 390: Centre of Mass and Rotational Motion

Question

Two identical thin rods of mass $M \mathrm{~kg}$ and length $L \mathrm{~m}$ are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point $P$ and perpendicular to the plane of the rods is $\frac{x}{12} \mathrm{ML}^2 \mathrm{~kg} \mathrm{~m}^2$. The value of $x$ is $\_\_\_\_$ .

Accepted Answer

The moment of inertia of a uniform rod of mass M and length L about an axis passing through its end and perpendicular to its length is :
$$ \mathrm{I}_{\mathrm{end}}=\frac{1}{3} \mathrm{ML}^2 $$
The vertical rod has one of its ends at point P . Thus, for the vertical rod :
$$ \mathrm{I}_1=\frac{1}{3} \mathrm{ML}^2 $$
The horizontal rod is connected to the bottom of the vertical rod. Its centre is at a distance L from point P . To find its moment of inertia about point P , we use the parallel axis theorem :
$$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 $$
Where :
$\mathrm{I}_{\mathrm{cm}}$ is the moment of inertia about its own centre of mass $\mathrm{I}_{\mathrm{cm}}=\frac{1}{12} \mathrm{ML}^2$.
d is the distance from point P to the center of the horizontal rod, $\mathrm{d}=\mathrm{L}$.
Substituting these values :
$$ \mathrm{I}_2=\frac{1}{12} \mathrm{ML}^2+\mathrm{M}(\mathrm{~L})^2 $$
$$\Rightarrow $$ $$ \mathrm{I}_2=\frac{1}{12} \mathrm{ML}^2+\mathrm{ML}^2=\frac{13}{12} \mathrm{ML}^2 $$
The total moment of inertia is the sum of $\mathrm{I}_1$ and $\mathrm{I}_2$ :
$\mathrm{I}_{	ext {total }}=\mathrm{I}_1+\mathrm{I}_2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{1}{3} \mathrm{ML}^2+\frac{13}{12} \mathrm{ML}^2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{4 \mathrm{ML}^2+13 \mathrm{ML}^2}{12}$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{17}{12} \mathrm{ML}^2$
$\Rightarrow \frac{\mathrm{x}}{12} \mathrm{ML}^2=\frac{17}{12} \mathrm{ML}^2 \Rightarrow \mathrm{x}=17$

Centre of Mass and Rotational Motion

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