Animated Solution for Physics - Rotational Motion: For the given uniform square lamina $ABCD$, whose centre is $O$

The moment of inertia of a uniform square lamina is a beautiful demonstration of symmetry in physics. Let's dive into this problem and uncover the elegant relationships between different axes of rotation!

Imagine a uniform square lamina $ABCD$ with side length $d$ and mass $M$. Let's draw two axes passing through its center $O$: one parallel to the sides $AD$ and $BC$ (let's call it $EF$), and another parallel to $AB$ and $CD$ (let's call it $GH$).

Because a square is perfectly symmetric, the mass distribution around the axis $EF$ is identical to the mass distribution around the axis $GH$. Therefore, their moments of inertia must be equal:

Now, let's apply the Perpendicular Axes Theorem. This theorem states that for a planar body, the moment of inertia about an axis perpendicular to the plane ($I_Z$) is the sum of the moments of inertia about two mutually perpendicular axes in the plane ($I_X$ and $I_Y$) that intersect at the same point.

Applying this to our axes $EF$ and $GH$:

What if we choose the diagonals $AC$ and $BD$ as our in-plane axes? They also pass through the center $O$ and, crucially, they intersect at $90^{\circ }$. By the exact same symmetry argument, $I_{AC}=I_{BD}$. Applying the theorem again:

Look at what we've found! Both $2I_{EF}$ and $2I_{AC}$ are equal to the same $I_Z$. This means:

This is a profound result: The moment of inertia of a uniform square lamina is the same about ANY axis passing through its center in its plane!

The problem asks us to find the relationship involving the edge $AD$. The axis $AD$ is parallel to our central axis $EF$. To find $I_{AD}$, we use the Parallel Axis Theorem:

We know the standard formula for the moment of inertia of a square plate about a central axis parallel to its edge is $\frac{M{d}^2}{12}$. Substituting this in:

Let's find a common denominator to add these fractions:

We have $I_{AD}=\frac{M{d}^2}{3}$ and $I_{EF}=\frac{M{d}^2}{12}$. Notice that $\frac{M{d}^2}{3}$ is exactly $4$ times $\frac{M{d}^2}{12}$. Therefore:

This perfectly matches our option (c). By leveraging symmetry and the fundamental theorems of moment of inertia, we've elegantly solved the problem without any complex integration!