Animated Solution for Physics - Rotational Motion: A symmetric lamina of mass $M$ consists of a square shape with a semi-circular section over of the edge of the square as shown in figure. The side of the square is $2a$. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is $1.6M{a}^2$. The moment of inertia of the lamina about the tangent $AB$ in the plane of the lamina is .........

Imagine you are an architect tasked with designing a perfectly balanced, beautiful structure. You are handed a square plate, and to make it aesthetically pleasing, you attach identical semi-circular petals to each of its four edges. What you have just created is the symmetric lamina described in our problem—a shape that is as mathematically elegant as it is visually striking.

But in the world of physics, beauty is often synonymous with symmetry, and symmetry is our most powerful tool. The problem asks us to find the moment of inertia of this complex-looking lamina about a tangent line $AB$. At first glance, calculating the moment of inertia for a composite shape involving a square and four semi-circles seems like a nightmare of integration. But fear not! We don't need to integrate anything. We just need to let the symmetry do the heavy lifting.

Let's break down the physical reality of our lamina. We place the center of the square exactly at the origin of our coordinate system, which we will call point $O$. The $x$-axis and $y$-axis run parallel to the sides of the square, cutting right through the middle of the semi-circular petals.

Because the shape is identical in all four directions, the mass distribution along the $x$-axis is completely indistinguishable from the mass distribution along the $y$-axis. If you were to close your eyes and someone rotated the lamina by $90^{\circ }$, you wouldn't be able to tell the difference when you opened them.

This physical indistinguishability translates directly into a profound mathematical truth: the moment of inertia about the $x$-axis must be exactly equal to the moment of inertia about the $y$-axis.

This simple realization is the golden key that unlocks the entire problem.

Perpendicular Axis Theorem

The problem generously provides us with the moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the lamina. In our coordinate system, this is the $z$-axis. We are given:

Now, how do we connect this perpendicular axis to the axes lying in the plane? Enter the Perpendicular Axis Theorem. This beautiful theorem states that for any planar body (a 2D lamina), the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in the plane and intersecting at the same point.

Mathematically, this is expressed as:

Since we have already established through symmetry that $I_x=I_y$, we can substitute this into our theorem:

Now, we simply plug in the given value for $I_z$:

Dividing both sides by 2, we find the moment of inertia about the $x$-axis:

Take a moment to appreciate what just happened. Without performing a single complex integral, we have determined the moment of inertia of this intricate shape about its central axis. This is the power of symmetry combined with fundamental theorems.

Parallel Axis Theorem

We are not done yet. The question doesn't ask for $I_x$; it asks for the moment of inertia about the tangent $AB$.

Let's locate this tangent $AB$. It lies in the plane of the lamina and touches the very top of the upper semi-circle. Notice that this tangent line is perfectly parallel to our $x$-axis. Whenever we need to find the moment of inertia about an axis that is parallel to an axis passing through the center of mass, we must summon the Parallel Axis Theorem.

The Parallel Axis Theorem states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the center of mass, plus the total mass of the body multiplied by the square of the perpendicular distance between the two axes.

We already know $I_x$. The only missing piece of the puzzle is $d$, the perpendicular distance from the $x$-axis to the tangent $AB$.

Let's trace the path from the center $O$ to the tangent $AB$. First, we move from the center of the square to its top edge. Since the total side length of the square is $2a$, this distance is simply $a$. Next, we move from the edge of the square to the top of the semi-circle. This distance is exactly the radius of the semi-circle. Since the diameter of the semi-circle is the side of the square ($2a$), its radius is also $a$.

Therefore, the total distance $d$ is the sum of these two segments:

We now have all the ingredients required to compute the final answer. Let's substitute our known values into the Parallel Axis Theorem equation.

This is where many students make a heartbreaking, silly mistake. When squaring the distance term, you must square the entire expression $(2a)$, not just the $a$.

Substituting this back into our equation:

Now, it's just simple addition.

And there we have it! The moment of inertia of the lamina about the tangent $AB$ is $4.8M{a}^2$.

By trusting in symmetry and methodically applying the Perpendicular and Parallel Axis Theorems, we transformed a seemingly impossible geometry problem into a straightforward algebraic calculation. Always remember, in physics, a complex setup is often just a disguise for a beautifully simple underlying principle.