Animated Solution for Physics - Rotational Motion: Consider a uniform square plate of side $a$ and mass $m$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

A Square Plate and a Corner Axis

Imagine you are holding a uniform square plate of mass $m$ and side length $a$.

Instead of spinning it like a frisbee through its center, we want to find its moment of inertia when it rotates about an axis passing straight through one of its corners, completely perpendicular to the plate itself.

This is a classic rotational mechanics problem that tests your ability to shift axes intelligently.

Parallel Axis Theorem

To tackle this, we bring out a very powerful tool from our physics toolbox: the Parallel Axis Theorem.

This theorem is a lifesaver when dealing with off-center rotations. It states that the moment of inertia about any random axis $I$ is simply the moment of inertia about a parallel axis passing through the center of mass $I_{cm}$, plus the mass of the object multiplied by the square of the perpendicular distance $d$ between those two axes.

Our first mission is to find the moment of inertia about the center of mass, $I_{cm}$.

For a square plate, if we consider an axis perpendicular to its plane through the center, we can use the Perpendicular Axis Theorem. We add the moment of inertia of the two perpendicular axes lying in the plane of the plate.

This $\frac{m{a}^2}{6}$ is our starting anchor.

Next up, we need to figure out exactly how far our corner axis is from the center axis. This is pure geometry.

The diagonal of a square with side $a$ is given by the Pythagorean theorem as $a\sqrt{2}$. Since the center is exactly halfway along the diagonal, our distance $d$ is simply half of that.

Now comes the fun part: putting the pieces together. We take our Parallel Axis Theorem equation and substitute what we've found.

Let's crunch the numbers. Squaring the distance term gives us $\frac{a^2}{2}$.

To add these fractions, we find a common denominator of $6$, turning the second term into $\frac{3m{a}^2}{6}$. Adding them gives $\frac{4m{a}^2}{6}$, which simplifies perfectly to our final answer.

A Faster Path

Before we wrap up, here is a brilliant pro-tip for you! You could actually bypass the parallel axis theorem entirely by using the Perpendicular Axis Theorem right at the corner.

The moment of inertia about the two edges meeting at the corner are each $\frac{m{a}^2}{3}$. Since the axis we want is perpendicular to both these edges at their intersection, we just add them together!

It is always great to have multiple ways to solve a problem. Physics is beautiful when you look at it from different angles!