Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. Th

Q: Why can we treat the shapes as point masses?

Because the ink has uniform thickness, the mass is distributed uniformly. For any uniform symmetric object, its center of mass lies at its geometric center. By the principle of superposition, a system of extended objects can be treated as a system of point masses located at their respective centers of mass.

Q: Why did we ignore the x-coordinates?

The problem only asks for the y-coordinate of the center of mass. However, if you look closely, the system is perfectly symmetric about the y-axis. For every mass at a positive x, there is an equal mass at a negative x. Therefore, the x-coordinate of the center of mass is exactly zero.

Q: What if the ink thickness wasn't uniform?

If the thickness varied, the mass wouldn't be directly proportional to the length or circumference. We would need to know the exact mass distribution function (linear mass density) and perform integration to find the center of mass for each component first.

Q: Why is the mass of the outer circle $6m$ while the others are $m$?

The problem explicitly states these mass values. Physically, this could mean the outer circle has a much larger circumference (roughly 6 times the length of the other components) assuming the ink thickness is the same everywhere.

Q: Does the radius of the inner circles matter?

No, the radius of the inner circles doesn't affect the calculation, as long as we know their mass and the location of their centers. The center of mass of a uniform circle is always at its center, regardless of its radius.

Question

Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6 m$. The coordinates of the centres of the different parts are : outer circle $(0, 0)$, left inner circle $(-a, a)$, right inner circle $(a, a)$, vertical line $(0, 0)$ and horizontal line $(0, - a)$. The $y$ - coordinate of the centre of mass of the ink in this drawing is

Options:

(A) $\frac{a}{10}$

(B) $\frac{a}{8}$

(C) $\frac{a}{12}$

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

Sigma Percentile · Accepted Answer

Finding the center of mass of a complex, extended object might seem like a daunting task that requires heavy calculus and integration. However, physics is often about finding elegant shortcuts. When an object is composed of several simpler, uniform geometric shapes, we can use the powerful principle of superposition.

In this problem, we are presented with a drawing resembling a face, made entirely of ink with a uniform line thickness. Because the thickness is uniform, the mass of any part of the drawing is directly proportional to the amount of ink used. Instead of integrating over the entire complex shape, we can break it down into five distinct, simple components: the outer circle, the two inner circles (the "eyes"), the vertical line (the "nose"), and the horizontal line (the "mouth").

### Analyzing the Setup

The beauty of uniform, symmetric shapes is that their center of mass lies exactly at their geometric center. This allows us to treat each of the five components as a single point mass located at its respective center. Let's list out our "arsenal" of point masses and their $y$-coordinates:

1. **The Outer Circle:** We are given that its mass is $6m$, and its center is at the origin $(0, 0)$. So, $y_1 = 0$.
2. **The Left Inner Circle:** Its mass is $m$, and its center is at $(-a, a)$. So, $y_2 = a$.
3. **The Right Inner Circle:** Its mass is $m$, and its center is at $(a, a)$. So, $y_3 = a$.
4. **The Vertical Line:** Its mass is $m$, and its center is at the origin $(0, 0)$. So, $y_4 = 0$.
5. **The Horizontal Line:** Its mass is $m$, and its center is at $(0, -a)$. So, $y_5 = -a$.

### The Master Equation

For a system of discrete point masses, the $y$-coordinate of the center of mass is given by the weighted average of their individual $y$-coordinates:

$$y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4 + m_5 y_5}{m_1 + m_2 + m_3 + m_4 + m_5}$$

This equation perfectly mirrors our physical setup. The numerator represents the total "moment of mass" about the $x$-axis, and the denominator is simply the total mass of the entire drawing.

### Final Calculation

Now, we carefully substitute our values into the master equation:

$$y_{CM} = \frac{(6m)(0) + (m)(a) + (m)(a) + (m)(0) + (m)(-a)}{6m + m + m + m + m}$$

Let's simplify the numerator. The outer circle and the vertical line contribute nothing since their centers lie exactly on the $x$-axis ($y=0$). The two inner circles contribute $ma + ma = 2ma$. The horizontal line contributes $-ma$. Adding these up, the numerator becomes $2ma - ma = ma$.

Next, we calculate the total mass in the denominator. Adding the mass of the outer circle ($6m$) to the four other components ($m$ each) gives us a total mass of $10m$.

Plugging these back into our fraction:

$$y_{CM} = \frac{ma}{10m}$$

The mass $m$ beautifully cancels out, leaving us with our final answer:

$$y_{CM} = \frac{a}{10}$$

**Physical Intuition:** Does this answer make sense? The center of mass is at a positive $y$-coordinate, meaning it is slightly above the $x$-axis. If we look at the drawing, the two "eyes" (mass $2m$ total) are at $y = a$, while the "mouth" (mass $m$) is at $y = -a$. There is more mass in the upper half of the drawing than in the lower half, so it perfectly aligns with our intuition that the center of mass should be pulled slightly upwards!

Select Answer:

Visualized Solution

Visualizing the System

The Center of Mass Formula

Identifying Masses and Coordinates

Setting up the Equation

Simplifying the Numerator

Calculating Total Mass

Final Calculation

The Sigma Insight: Centre of Mass

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the System

The Center of Mass Formula

Identifying Masses and Coordinates

Setting up the Equation

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