A circular hole of radius (a/2) is cut out of a circular disc of radius a as shown in figure. The ce

Q: Why do we subtract the mass in the formula?

It's a mathematical trick called the Negative Mass Concept. By treating the hole as an object with negative mass, we can use the standard center of mass formula without needing to integrate over the awkward remaining shape.

Q: Why is the mass of the hole $M/4$ and not $M/2$?

Because this is a 2D disc, mass scales with the area, not just the radius. Area is proportional to the square of the radius. Since the radius is halved, the area (and thus the mass) becomes $(1/2)^2 = 1/4$ of the original.

Q: How did we figure out the center of the hole is at $3a/2$?

Look at the diagram. The large disc is centered at $a$. The hole spans from the center ($a$) to the right edge ($2a$). The midpoint of the hole is exactly halfway between $a$ and $2a$, which is $a + a/2 = 1.5a$ or $3a/2$.

Q: Why did the center of mass shift to $5a/6$?

Originally, the center was at $a$. We removed mass from the right side (the hole). Naturally, the object becomes heavier on the left side, so the balance point (center of mass) shifts to the left, to $5a/6$ (which is about $0.833a$).

Q: Would this method work if the object was a solid sphere instead of a disc?

Absolutely! The only difference is that for a 3D sphere, mass scales with volume ($r^3$). So a hole with half the radius would have a mass of $(1/2)^3 = 1/8$ of the original sphere.

Question

A circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $a$ as shown in figure. The centroid of the remaining circular portion with respect to point $O$ will be

Options:

(A) $\frac{1}{6}a$

(B) $\frac{10}{11}a$

(C) $\frac{5}{6}a$

(D) $\frac{2}{3}a$

Sigma Percentile · Accepted Answer

Have you ever looked at a Swiss cheese and wondered where its center of mass lies? When a portion of a uniform body is removed, finding the new balance point can seem like a daunting integration problem. But fear not! Physics offers a beautifully elegant shortcut: the **Negative Mass Concept**.

### Analyzing the Setup
Imagine you have a complete, uniform circular disc of radius $a$. Its center of mass is perfectly in the middle, at a distance $a$ from the origin $O$. Now, we carve out a smaller circular hole of radius $\frac{a}{2}$. This hole is tangent to the center of the large disc and its right edge.

Instead of dealing with the awkward crescent-like shape that remains, we treat the system as a superposition of two complete shapes:
1. A solid, complete large disc of positive mass $M$.
2. A smaller solid disc of **negative mass** $-m$ placed exactly where the hole is.

### The Mass Ratio
Since the original disc is uniform, its mass is directly proportional to its area. The area of a circle is $\pi r^2$. 
The large disc has a radius $a$, so its mass $M \propto \pi a^2$.
The hole has a radius $\frac{a}{2}$, so its mass $m \propto \pi \left(\frac{a}{2}ight)^2 = \frac{\pi a^2}{4}$.
This means the mass of the hole is exactly one-fourth of the large disc: $m = \frac{M}{4}$.

### Locating the Centers
Next, we need the coordinates of their individual centers of mass. 
The large disc is centered at $x_1 = a$.
The hole is located halfway between the center of the large disc and its right edge. So, its center is at $x_2 = a + \frac{a}{2} = \frac{3a}{2}$.

### The Master Equation
Now, we bring in our center of mass formula, modified for the negative mass:
$$x_{CM} = \frac{M x_1 - m x_2}{M - m}$$

Let's substitute our values carefully:
$$x_{CM} = \frac{M(a) - \left(\frac{M}{4}ight)\left(\frac{3a}{2}ight)}{M - \frac{M}{4}}$$

### Final Calculation
Time for some satisfying algebra. In the numerator, we have $Ma - \frac{3}{8}Ma$, which simplifies to $\frac{5}{8}Ma$. 
In the denominator, the total remaining mass is $M - \frac{M}{4} = \frac{3}{4}M$.

Dividing the two:
$$x_{CM} = \frac{\frac{5}{8}Ma}{\frac{3}{4}M} = \frac{5}{8} 	imes \frac{4}{3} a = \frac{5}{6} a$$

And there we have it! The new center of mass is at $\frac{5}{6}a$. Notice how it has shifted slightly to the left (from $a$ to $0.833a$). This makes perfect physical sense—since we removed mass from the right side, the balance point must shift to the heavier left side.

Next time you see a cavity problem, whether it's a disc, a sphere, or a cylinder, remember the negative mass trick. It turns a calculus nightmare into a simple algebra puzzle!

Animated Solution for Physics - System of Particles: A circular hole of radius $(\frac{a}{2})$ is cut out of a circular disc of radius $a$ as shown in figure. The centroid of the remaining circular portion with respect to point $O$ will be

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Visualized Solution

$S y s t e m S e t u p$

$N e g a t i v e M a ss C o n ce pt$

$M a ss C a l c u l a t i o n$

$C oor d ina t eso f C e n t er s$

$S u b s t i t u t i o n$

$S im pl i f i c a t i o n$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Centre of Mass

Analyzing the Setup Imagine you have a complete, uniform circular disc of radius $a$

The Mass Ratio Since the original disc is uniform, its mass is directly proportional to its area

Locating the Centers Next, we need the coordinates of their individual centers of mass

The Master Equation

Final Calculation Time for some satisfying algebra

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