Animated Solution for Physics - System of Particles and Rotational Motion: The disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\left(\frac{xa}{3\pi },\frac{xa}{3\pi }\right)$, where $x$ is ............. (Round off to the nearest integer) ($a$ is an area as shown in the figure)

The Secret of the Quarter Disc

Imagine trying to balance an irregularly shaped object on the tip of your finger. The exact point where it balances perfectly is its Center of Mass (CM). In physics, especially in mechanics, knowing the center of mass of standard geometric shapes is like having a superpower. It simplifies complex rotational dynamics into elegant, solvable equations.

In this problem, we are dealing with a classic geometry: a uniform quarter disc. Let's dive into the mechanics of finding its center of mass and see how symmetry plays a beautiful role.

We are given a uniform quarter disc placed in the first quadrant of the $X-Y$ plane. The question mentions "($a$ is an area as shown in the figure)". However, if we look closely at the physical dimensions and the diagram, it is clear that $a$ represents the radius of the quarter disc, not the area. This is a common typographical error in exam papers, and recognizing the physical dimensions helps us stay on the right track.

Because the disc is uniform, its surface mass density $\sigma$ is constant. This means the mass is distributed evenly across its entire area.

Before we jump into formulas, let's look at the shape. A quarter disc in the first quadrant is perfectly symmetric about the line $y=x$. If you were to fold the $X-Y$ plane along this diagonal line, the two halves of the quarter disc would overlap perfectly.

What does this mean for the center of mass? It implies that the center of mass must lie exactly on this line of symmetry. Therefore, the $X$-coordinate and the $Y$-coordinate of the center of mass must be identical:

For a uniform quarter disc of radius $R$, the coordinates of the center of mass are a standard result derived using polar integration. By integrating the position vector over the area of the disc, we find:

These are incredibly important standard results that you should memorize for competitive exams like JEE and NEET. They save you from doing tedious double integrals during a high-pressure test.

In our specific problem, the radius of the quarter disc is given as $R=a$. Substituting this into our standard formula, we get the coordinates of the center of mass as:

The question states that the center of mass is located at the position $\left(\frac{xa}{3\pi },\frac{xa}{3\pi }\right)$.

Now, we simply compare our calculated coordinates with the given coordinates:

By comparing the numerators, it is crystal clear that:

And there we have it! A seemingly complex problem reduced to a simple comparison, all thanks to knowing our standard results.

Understanding the center of mass of a quarter disc opens the door to analyzing other shapes. For instance, what if we had a semi-circular disc placed symmetrically on the $X$-axis?

Due to symmetry along the $Y$-axis, the $X$-coordinate would be exactly zero ($X_{CM}=0$). However, the $Y$-coordinate would remain the same as that of the quarter disc:

Similarly, for a solid hemisphere, the center of mass lies at a distance of $\frac{3R}{8}$ from the base, and for a hollow hemisphere, it lies at $\frac{R}{2}$. Memorizing these standard values is a secret weapon that will significantly boost your speed and accuracy in physics.