Animated Solution for Physics - Rotational Motion: For a uniform rectangular sheet, shown in the above figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O^{\prime }$ (corner point) is

## The Tale of Two Axes: Unraveling the Moment of Inertia of a Rectangular Sheet

Rotational inertia, or the moment of inertia, is a beautiful physical quantity. It tells us how much an object resists a change in its rotational motion. But here's the catch: it doesn't just depend on the mass of the object; it depends heavily on where that mass is distributed relative to the axis of rotation.

In this problem, we are presented with a classic scenario: a uniform rectangular sheet. We need to compare its resistance to rotation about two different axes. Let's dive into the mechanics of this setup.

Imagine a uniform rectangular sheet lying flat. We are given its dimensions: a width $B=60\text{ cm}$ and a length $L=80\text{ cm}$. We are interested in two specific axes of rotation, both perpendicular to the plane of the sheet.

The first axis passes straight through the geometric center, which is also the center of mass, denoted as $O$. The second axis passes through one of the corner points, denoted as $O^{\prime }$. Our goal is to find the ratio of the moment of inertia about $O$ ($I_O$) to the moment of inertia about $O^{\prime }$ ($I_{O^{\prime }}$).

To begin, we need the foundational formula for the moment of inertia of a rectangular sheet about an axis passing through its center of mass and perpendicular to its plane. This is a standard result derived using the perpendicular axis theorem ($I_z=I_x+I_y$):

Let's substitute our given dimensions into this master equation. We have $L=80\text{ cm}$ and $B=60\text{ cm}$:

We will leave it in this raw, unsimplified form because it will make our final ratio calculation much cleaner.

Now, we need to find the moment of inertia about the corner, $O^{\prime }$. Since the axis through $O^{\prime }$ is parallel to the axis through the center of mass $O$, we can invoke the powerful Parallel Axis Theorem:

Here, $d$ is the perpendicular distance between the two parallel axes. Looking at our rectangle, the distance from the center $O$ to the corner $O^{\prime }$ forms the hypotenuse of a right-angled triangle. The legs of this triangle are half the length ($\frac{80}{2}=40\text{ cm}$) and half the width ($\frac{60}{2}=30\text{ cm}$).

Using Pythagoras' theorem, we can find $d$:

Now, we substitute $d$ back into our parallel axis theorem equation:

To add these terms easily, let's express $M(2500)$ with a denominator of $12$:

We have successfully found expressions for both $I_O$ and $I_{O^{\prime }}$. The final step is to simply take their ratio:

Notice how beautifully the $\frac{M}{12}$ terms cancel out. This tells us that the ratio is completely independent of the mass of the sheet!

The moment of inertia about the corner is exactly four times greater than the moment of inertia about the center. This makes perfect physical sense: when rotating about the corner, the bulk of the sheet's mass is distributed much farther away from the axis of rotation, drastically increasing its rotational inertia.