A square shaped hole of side l = a/2 is curved out at a distance d = a/2 from the centre O of a unif

Q: Why do we use the concept of negative mass? Is it a real thing?

Negative mass isn't a real physical entity. It's a brilliant mathematical trick! Instead of integrating over a complex shape with a hole, we pretend the object is whole (positive mass) and add a 'negative mass' exactly where the hole is. The math works out perfectly and saves us from doing complicated calculus.

Q: Why does the center of mass shift to the left (negative x-axis)?

Think of a seesaw. If you take away weight from the right side, the right side goes up, and the left side goes down. To balance it again, you have to move the fulcrum (the center of mass) towards the heavier side, which is the left side in this case.

Q: Does the value of mass per unit area ($\sigma$) matter?

No, it doesn't! As long as the disc is uniform, $\sigma$ is constant everywhere. When you plug it into the center of mass formula, it appears in every term in both the numerator and denominator, so it completely cancels out.

Q: How do we approximate $8\pi - 2$ without a calculator?

You should memorize the value of $\pi$ to at least two decimal places: $\pi \approx 3.14$. So, $8 \times 3.14 = 25.12$. Subtracting 2 gives $23.12$. Since the question asks for the nearest integer, the answer is simply 23.

Q: What if the hole was not centered on the x-axis?

If the hole was off-axis, you would have to calculate both the x-coordinate ($X_{CM}$) and the y-coordinate ($Y_{CM}$) of the new center of mass separately, using the exact same negative mass formula for both dimensions.

Question

A square shaped hole of side $l = \frac{a}{2}$ is curved out at a distance $d = \frac{a}{2}$ from the centre $O$ of a uniform circular disc of radius $a$. If the distance of the centre of mass of the remaining portion from $O$ is $-a/X$, value of $X$ (to the nearest integer) is ……… .

Sigma Percentile · Accepted Answer

### The Magic of Negative Mass

Imagine you have a perfectly uniform circular pizza, and someone takes a square bite out of it. The pizza is no longer perfectly balanced at its center. The point where it balances—its center of mass—has shifted. But how do we calculate exactly where this new balance point is?

This is where the brilliant concept of **negative mass** comes into play. Instead of trying to calculate the center of mass of the weird, bitten shape directly, we imagine the system as a complete, untouched pizza, plus a "negative" square pizza placed exactly where the bite was taken.

### Setting Up the Problem

Let's define our system mathematically. We have a uniform circular disc of radius $a$. Its center is at the origin, so its center of mass is at $x_1 = 0$. Let's call its mass $m_1$. If the mass per unit area is $\sigma$, then:

$m_1 = \sigma \pi a^2$

Now, we remove a square of side $l = \frac{a}{2}$. The center of this square is at a distance $d = \frac{a}{2}$ from the origin. So, the center of mass of this "negative" piece is at $x_2 = \frac{a}{2}$. Its mass, $m_2$, is:

$m_2 = \sigma l^2 = \sigma \left(\frac{a}{2}\right)^2 = \sigma \frac{a^2}{4}$

### The Master Equation

The formula for the center of mass of a two-particle system is $X_{CM} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$. However, since we are *removing* mass, we treat $m_2$ as negative. The formula beautifully adapts to:

$X_{CM} = \frac{m_1x_1 - m_2x_2}{m_1 - m_2}$

Let's substitute our values into this equation:

$X_{CM} = \frac{(\sigma \pi a^2)(0) - (\sigma \frac{a^2}{4})(\frac{a}{2})}{\sigma \pi a^2 - \sigma \frac{a^2}{4}}$

Notice how $\sigma$ and $a^2$ are present in every single term? This is a common pattern in these problems. They cancel out completely, leaving us with a purely geometric ratio:

$X_{CM} = \frac{0 - \frac{a}{8}}{\pi - \frac{1}{4}}$

### Final Calculation

Let's simplify the denominator by taking a common factor:

$X_{CM} = \frac{-\frac{a}{8}}{\frac{4\pi - 1}{4}} = \frac{-a}{2(4\pi - 1)}$

The negative sign perfectly aligns with our intuition: if you remove mass from the right side, the balance point must shift to the left to compensate.

The problem states that this distance is $-\frac{a}{X}$. By comparing our result with this given form, we can easily find $X$:

$X = 2(4\pi - 1) = 8\pi - 2$

Now, we just need to calculate the numerical value. Using $\pi \approx 3.14$:

$X \approx 8(3.14) - 2 = 25.12 - 2 = 23.12$

Rounding to the nearest integer, we get our final answer: **23**.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

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

$C a l c u l a t in g M a sses$

$I d e n t i f y in g C oor d ina t es$

$A ppl y in g t h e F or m u l a$

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

$F ina l C a l c u l a t i o n$

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

The Sigma Insight: Centre of Mass

The Magic of Negative Mass

Setting Up the Problem

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Negative Mass Concept

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Masses

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Coordinates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward