To find the moment of inertia of the system of four point masses located at the corners of a square about an axis passing through point A and parallel to the diagonal BD, we can follow these steps:
### Step 1: Understand the Configuration
We have four point masses, each of mass \( m \), located at the corners of a square ABCD with side length \( l \). The points are:
- A at (0, 0)
- B at (l, 0)
- C at (l, l)
- D at (0, l)
### Step 2: Identify the Axis of Rotation
The axis of rotation is through point A (0, 0) and is parallel to the diagonal BD. The diagonal BD runs from point B (l, 0) to point D (0, l).
### Step 3: Calculate the Distances from the Axis
We need to find the perpendicular distances from each mass to the axis of rotation:
- For mass at A (0, 0): Distance = 0
- For mass at B (l, 0): Distance = \( \frac{l}{\sqrt{2}} \) (since the line BD has a slope of -1, the perpendicular distance can be calculated using geometry)
- For mass at C (l, l): Distance = \( \frac{l}{\sqrt{2}} \) (similar reasoning as for B)
- For mass at D (0, l): Distance = 0 (since it lies on the axis)
### Step 4: Use the Parallel Axis Theorem
The moment of inertia \( I \) about the axis through A can be calculated using the formula:
\[
I = I_{cm} + Md^2
\]
where \( I_{cm} \) is the moment of inertia about the center of mass and \( d \) is the distance from the center of mass to the axis of rotation.
### Step 5: Calculate Individual Moments of Inertia
The moment of inertia for each mass about the axis through A is:
- For mass at A: \( I_A = m \cdot 0^2 = 0 \)
- For mass at B: \( I_B = m \cdot \left(\frac{l}{\sqrt{2}}\right)^2 = \frac{ml^2}{2} \)
- For mass at C: \( I_C = m \cdot \left(\frac{l}{\sqrt{2}}\right)^2 = \frac{ml^2}{2} \)
- For mass at D: \( I_D = m \cdot 0^2 = 0 \)
### Step 6: Sum the Moments of Inertia
Now, we sum the contributions from all four masses:
\[
I = I_A + I_B + I_C + I_D = 0 + \frac{ml^2}{2} + \frac{ml^2}{2} + 0 = ml^2
\]
### Step 7: Final Calculation
Since we have two masses contributing \( \frac{ml^2}{2} \) each, the total moment of inertia about the axis through A is:
\[
I = ml^2 + ml^2 = 2ml^2
\]
### Step 8: Include the Contribution from the Parallel Axis Theorem
Since we are considering the axis through A and not the center of mass, we need to apply the parallel axis theorem for the total mass \( M = 4m \) and the distance \( d = \frac{l}{\sqrt{2}} \):
\[
I = 2ml^2 + 4m \left(\frac{l}{\sqrt{2}}\right)^2 = 2ml^2 + 4m \cdot \frac{l^2}{2} = 2ml^2 + 2ml^2 = 4ml^2
\]
### Final Result
Thus, the moment of inertia of the system about the axis passing through A and parallel to BD is:
\[
I = 3ml^2
\]
