Animated Solution for Physics - System of Particles: A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$, such that the circumferences of the discs coincide. The centre of mass of the new disc is $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

A Disc with a Hole

Imagine you are holding a perfectly uniform, solid circular disc. Balancing it on your finger is easy—you just place your finger exactly at the geometric center. But what happens if someone comes along and punches a large circular hole right near the edge? Suddenly, the symmetry is broken. The disc is now heavier on one side and lighter on the other. If you try to balance it at the original center, it will immediately tip over.

This is exactly the scenario we are facing in this classic physics problem. We start with a large circular disc of radius $2R$. Then, a smaller disc of radius $R$ is carved out from it, such that the edge of the smaller disc touches the edge of the larger one. Our mission is to find the new balancing point—the center of mass of this crescent-like remaining shape. The problem tells us that this new center of mass is located at a distance of $\alpha R$ from the original center, and we need to find the value of $\alpha$.

Negative Mass

If you were to solve this using brute-force calculus, you would have to set up a double integral over a very awkward, non-symmetric boundary. It would be a nightmare of trigonometry and limits. But physicists love elegance, and there is a brilliant shortcut for problems involving cavities: The Concept of Negative Mass.

Instead of looking at the remaining shape as a single complex object, we use the principle of superposition. We imagine the final shape as the combination of two simpler objects: 1. A complete, solid large disc (with positive mass). 2. A smaller disc (with negative mass) placed exactly where the hole is.

When you "add" the negative mass to the positive mass, they cancel out in that region, perfectly simulating the empty cavity! This allows us to work entirely with highly symmetric, easy-to-handle shapes.

To use our negative mass trick, we first need to figure out the masses of our two imaginary components. Since the original disc is uniform, its mass is directly proportional to its surface area.

Let's denote the mass of the complete large disc as $M_1=M$. The radius of this large disc is $2R$, so its area is $\pi (2R{)}^2=4\pi R^2$.

Now, let's look at the smaller disc that was removed. Its radius is $R$, so its area is $\pi R^2$. Notice that the area of the smaller disc is exactly one-fourth the area of the large disc ($\frac{\pi R^2}{4\pi R^2}=\frac{1}{4}$). Because mass is proportional to area, the mass of the removed disc must be one-fourth of the total mass. So, the mass of the removed disc is $M_2=\frac{M}{4}$.

Consequently, the mass of the actual remaining shape is simply the difference between the two:

Now that we have our masses, we need to set up a coordinate system. The smartest choice is to place the origin $(0,0)$ exactly at the center of the large complete disc.

Let's locate the centers of our components: - The center of the complete large disc is at the origin, so $x_1=0$. - The smaller disc is cut out such that its edge touches the large disc's edge. This means its center is shifted to the right by a distance $R$. So, $x_2=R$. - The problem states that the center of mass of the remaining shape is at a distance $\alpha R$ from the center. Since we removed mass from the right side, the remaining shape is heavier on the left. Therefore, the center of mass must shift to the left. Its coordinate will be $x_{\text{rem}}=-\alpha R$.

The fundamental definition of the center of mass tells us that the moment of mass of the complete system must equal the sum of the moments of its parts. In our superposition model, the complete disc is made of the remaining part plus the removed part.

Therefore, the moment of the remaining part plus the moment of the removed part must equal the moment of the complete disc:

Let's plug our values into this balancing equation. We know that $M_1{x}_1=M\cdot 0=0$. This makes perfect sense—the complete disc is balanced at the origin.

Substituting the rest of the values:

This equation is beautifully simple. We can immediately divide both sides by $\frac{M}{4}$ and by $R$, as neither is zero. This leaves us with:

Moving the $3\alpha$ to the other side:

And finally, solving for $\alpha$:

The center of mass of the remaining crescent shape is located exactly at a distance of $\frac{R}{3}$ to the left of the original center. The negative mass trick turned a terrifying calculus problem into a few lines of basic algebra. Whenever you see a cavity, remember this trick—it is one of the most powerful tools in your physics arsenal!