Animated Solution for Physics - Rotational Motion: Seven identical circular planar discs, each of mass $M$ and radius $R$ are welded symmetrically as shown in the figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is

Imagine you are tasked with finding the moment of inertia of a complex, multi-component system. At first glance, a structure made of seven identical discs welded together might look intimidating. However, physics provides us with elegant tools to break down such problems into bite-sized, manageable pieces. Let's embark on this journey to find the moment of inertia of this beautiful symmetric arrangement about an axis passing through point $P$.

We have seven identical circular planar discs, each with mass $M$ and radius $R$. They are welded symmetrically: one in the center and six surrounding it, forming a perfect hexagonal pattern. Our goal is to calculate the moment of inertia of this entire arrangement about an axis normal to the plane and passing through a specific point $P$ on the outer edge of one of the outer discs.

To tackle this, we will rely heavily on the Parallel Axis Theorem, which 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 times the square of the distance between the two axes:

The System's Core Inertia

Before we can find the inertia about $P$, we must first determine the moment of inertia of the entire system about its center of mass, $O$. Due to the perfect symmetry, the center of mass of the whole system lies exactly at the center of the central disc.

Let's calculate the moment of inertia of the system about $O$ ($I_O$) by summing the contributions of all seven discs.

1. The Central Disc: The moment of inertia of the central disc about its own center $O$ is simply:

2. The Six Outer Discs: Consider one of the outer discs. Its center is located at a distance of $2R$ from $O$ (since the discs are tangent to each other, the distance between their centers is $R+R=2R$). Using the Parallel Axis Theorem for this single outer disc, its moment of inertia about $O$ is:

Since there are six identical outer discs, their total contribution to the inertia about $O$ is $6\times \frac{9}{2}M{R}^2=27M{R}^2$.

3. Total Inertia about O: Adding the central disc and the six outer discs together, we get the total moment of inertia of the system about $O$:

Shifting to Point P

Now that we have the moment of inertia of the entire system about its center of mass $O$, we can make the final leap to point $P$.

Point $P$ is located on the outer edge of one of the outer discs. The straight line from $O$ to $P$ passes through the center of that outer disc. Therefore, the total distance $d$ from $O$ to $P$ is the distance to the outer disc's center ($2R$) plus the radius of the outer disc ($R$):

We apply the Parallel Axis Theorem one last time, but now for the entire system. The total mass of the system is $7M$.

Substituting our known values:

To add these, find a common denominator:

And there we have it! By breaking the complex structure into individual discs and applying the Parallel Axis Theorem systematically, we arrived at the elegant final answer. This method is a testament to the power of symmetry and fundamental theorems in physics.