Animated Solution for Physics - Rotational Motion: Four point masses, each of mass $m$ are fixed at the corners of a square of side $l$. The square is rotating with angular frequency $\omega$, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is

A Journey into Angular Momentum

Imagine a rigid square frame with four identical point masses, each of mass $m$, securely fixed at its corners. This entire system is spinning with an angular frequency $\omega$. But here is the catch: it is not spinning around its center. Instead, the axis of rotation passes directly through one of the corners and runs perfectly parallel to the square's diagonal.

Our mission is to determine the total angular momentum of this rotating system. To do this, we must rely on a fundamental principle of rotational dynamics.

The angular momentum $L$ of a rigid body rotating about a fixed axis is elegantly described by the equation:

Here, $I$ represents the moment of inertia of the system about that specific axis of rotation, and $\omega$ is the angular frequency. Since $\omega$ is already given, our primary challenge is to calculate the moment of inertia, $I$.

For a system of discrete point masses, the total moment of inertia is simply the algebraic sum of the individual moments of inertia of each mass. The formula is:

In this equation, $m_i$ is the mass of the $i$-th particle, and $r_i$ is its perpendicular distance from the axis of rotation. Let's break down these distances for our four masses.

Let's analyze the position of each mass relative to our specific axis of rotation.

1. The Mass on the Axis: Consider the mass located exactly at the corner where the axis passes through. Because it lies directly on the axis, its perpendicular distance is zero ($r_A=0$). Consequently, its contribution to the total moment of inertia is zero:

2. The Adjacent Masses: Next, look at the two masses adjacent to the first one. We need to find their perpendicular distances to the axis. The diagonal of a square of side $l$ has a length of $l\sqrt{2}$. Because our axis is parallel to the diagonal, the perpendicular distance from these adjacent corners to the axis is exactly half the length of the diagonal. Therefore, for both of these masses, the distance is:

3. The Farthest Mass: Finally, consider the fourth mass, which is located at the corner diagonally opposite to the first mass. The perpendicular distance from this farthest corner to our axis is exactly equal to the full length of the square's diagonal. Thus, its distance is:

Now that we have all the perpendicular distances, we can substitute them back into our moment of inertia formula:

Let's carefully square the terms inside the parentheses:

Adding these terms together, the two halves combine to make one whole, giving us:

We have successfully found the total moment of inertia of the system! The final step is to calculate the angular momentum by multiplying this result by the angular frequency $\omega$:

And there we have it! A beautiful and elegant result derived purely from the geometry of the square and the fundamental principles of rotational physics.