Animated Solution for Physics - Rotational Motion: Four point masses, each of value $m$, are placed at the corners of a square ABCD of side $l$. The moment of inertia of this system about an axis passing through A and parallel to BD is

Have you ever looked at a physics problem and realized that the math is trivial, but the geometry is the real puzzle? This classic problem from AIEEE 2006 is exactly that. We have four identical masses placed at the corners of a square, and we need to find the moment of inertia about a very specific axis. Let's dive into the elegant geometry of this setup!

Imagine a square $ABCD$ with side length $l$. At each corner, there is a point mass $m$. The problem asks for the moment of inertia about an axis passing through corner $A$ and parallel to the diagonal $BD$.

To make our lives infinitely easier, let's orient the square so that this axis of rotation is perfectly vertical. Since the diagonals of a square are always perpendicular to each other ($AC\perp BD$), and our axis is parallel to $BD$, the axis must be perpendicular to the diagonal $AC$.

For a system of discrete point masses, the moment of inertia is simply the sum of the products of each mass and the square of its perpendicular distance from the axis of rotation:

Our goal is to find these perpendicular distances ($r_A,r_B,r_C,r_D$) for all four masses.

Let's look at our vertically oriented axis passing through $A$.

1. Mass A: Since it lies exactly on the axis of rotation, its perpendicular distance is zero.

2. Masses B and D: The line segment connecting $B$ and $D$ is parallel to our axis. The perpendicular distance from the axis to this line is exactly half the length of the diagonal $AC$. The full diagonal of a square with side $l$ is $\sqrt{2}l$. Therefore, half the diagonal is $\frac{\sqrt{2}l}{2}=\frac{l}{\sqrt{2}}$.

3. Mass C: This mass is at the opposite end of the diagonal $AC$. Since the axis is perpendicular to $AC$ at point $A$, the perpendicular distance to $C$ is the full length of the diagonal.

Now, we just plug these distances into our master equation:

Let's square the terms carefully:

Adding them all up:

And there we have it! The total moment of inertia is $3m{l}^2$.

The Parallel Axis Theorem

Want to solve this like a pro? Use the Parallel Axis Theorem!

First, find the moment of inertia about the diagonal $BD$. Masses $B$ and $D$ lie on this axis, so they contribute nothing. Masses $A$ and $C$ are at a distance of $\frac{l}{\sqrt{2}}$.

Now, shift this axis to point $A$. The distance between the parallel axes is $d=\frac{l}{\sqrt{2}}$, and the total mass of the system is $M_{total}=4m$.

Both methods yield the exact same result. Physics is beautifully consistent!