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Step-by-Step Solution
Step 1: Identify the masses and the axis
Four equal masses (each of mass $m$) are placed at the corners of a square of side $l$. Label the corners as A, B, C, and D in a clockwise or counterclockwise manner. The axis for calculating the moment of inertia passes through corner A and is parallel to the diagonal DB.
Step 2: Assign coordinates for clarity (optional but helpful)
Choose a convenient coordinate system:
• Let A be at $(0, 0)$.
• Let B be at $(l, 0)$.
• Let C be at $(l, l)$.
• Let D be at $(0, l)$.
Step 3: Determine the direction of the axis
The diagonal DB goes from $(0, l)$ to $(l, 0)$, so its direction vector can be taken as $(l, -l)$. An axis parallel to this vector and passing through A (which is at $(0, 0)$) can be represented by the same direction vector $(l, -l)$.
Step 4: Calculate the perpendicular distances of each mass from the axis
Use the formula for the perpendicular distance of a point $(x, y)$ from a line passing through the origin with direction vector $(a, b)$:
$$
\text{Distance} = \frac{|\,\vec{r} \times \vec{v}\,|}{|\vec{v}|},
$$
where $\vec{r}$ is the position vector of the point, and $\vec{v} = (a, b)$ is the direction vector of the line.
Mass at A $(0, 0)$: This point is on the axis, so distance $= 0$.
Mass at B $(l, 0)$:
• $\vec{r}_B = (l, 0)$
• $\vec{v} = (l, -l)$ and $|\vec{v}| = \sqrt{l^2 + (-l)^2} = l\sqrt{2}.$
• Cross product magnitude: $|\vec{r}_B \times \vec{v}| = |\,l \times (-l) - 0 \times l\,| = l^2.$
• Perpendicular distance $= \dfrac{l^2}{l\sqrt{2}} = \dfrac{l}{\sqrt{2}}.$
Mass at C $(l, l)$:
• $\vec{r}_C = (l, l)$
• Cross product magnitude: $|\vec{r}_C \times \vec{v}| = |\,l \times (-l) - l \times l\,| = | -l^2 - l^2 | = 2l^2.$
• Perpendicular distance $= \dfrac{2l^2}{l\sqrt{2}} = \sqrt{2}\,l.$
Mass at D $(0, l)$:
• $\vec{r}_D = (0, l)$
• Cross product magnitude: $|\vec{r}_D \times \vec{v}| = |\,0 \times (-l) - l \times l\,| = l^2.$
• Perpendicular distance $= \dfrac{l^2}{l\sqrt{2}} = \dfrac{l}{\sqrt{2}}.$
Step 5: Calculate the moment of inertia
Moment of inertia of a single mass $m$ at a distance $r$ from the axis is $mr^2$. Sum the contributions of all four masses:
For A: $m \cdot 0^2 = 0$.
For B: $m \left(\dfrac{l}{\sqrt{2}}\right)^2 = m \left(\dfrac{l^2}{2}\right) = \dfrac{ml^2}{2}.$
For C: $m \bigl(\sqrt{2}\,l\bigr)^2 = m(2l^2) = 2ml^2.$
For D: $m \left(\dfrac{l}{\sqrt{2}}\right)^2 = m \left(\dfrac{l^2}{2}\right) = \dfrac{ml^2}{2}.$
Adding them up:
$$
I = 0 + \dfrac{ml^2}{2} + 2ml^2 + \dfrac{ml^2}{2} = 3ml^2.
$$
So, the moment of inertia of the system about the given axis is $3ml^2.$
Final Answer
$3ml^2$
