Three particles each of mass `2kg` are at the corners of an equilateral triangle of side `sqrt 3 m`. If one of the particles is removed, the shift in the centre of mass is

Options

A. `0.2m`
B. `0.5 m`
C. `0.4 m`
D. `0.3m`

Explanation

To solve the problem of finding the shift in the center of mass when one particle is removed from an equilateral triangle, we can follow these steps:

### Step 1: Understand the System
We have three particles, each of mass \(2 \, \text{kg}\), located at the corners of an equilateral triangle with a side length of \(\sqrt{3} \, \text{m}\). Let's label the corners as points A, B, and C.

### Step 2: Determine the Coordinates of the Particles
Assuming point A is at the origin \((0, 0)\), we can find the coordinates of points B and C:
- Point A: \( (0, 0) \)
- Point B: \( (\sqrt{3}, 0) \)
- Point C: To find the coordinates of C, we note that it is at a \(60^\circ\) angle from the x-axis, so:
- \( x_C = \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} \)
- \( y_C = \frac{1}{2} \cdot \sqrt{3} = \frac{\sqrt{3}}{2} \)
Thus, the coordinates are:
- Point C: \( \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) \)

### Step 3: Calculate the Original Center of Mass
The center of mass (CM) for three particles is given by:
\[ X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \]
\[ Y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \]
Substituting the values:
- \(m_1 = m_2 = m_3 = 2 \, \text{kg}\)
- \(x_1 = 0, y_1 = 0\)
- \(x_2 = \sqrt{3}, y_2 = 0\)
- \(x_3 = \frac{3}{2}, y_3 = \frac{\sqrt{3}}{2}\)

Calculating \(X_{CM}\):
\[ X_{CM} = \frac{2 \cdot 0 + 2 \cdot \sqrt{3} + 2 \cdot \frac{3}{2}}{2 + 2 + 2} = \frac{2\sqrt{3} + 3}{6} = \frac{\sqrt{3} + \frac{3}{2}}{3} \]
Calculating \(Y_{CM}\):
\[ Y_{CM} = \frac{2 \cdot 0 + 2 \cdot 0 + 2 \cdot \frac{\sqrt{3}}{2}}{6} = \frac{\sqrt{3}}{6} \]

### Step 4: Remove One Particle and Calculate New Center of Mass
Let's remove particle C. Now we have particles A and B only:
- \(m_1 = m_2 = 2 \, \text{kg}\)
- \(x_1 = 0, y_1 = 0\)
- \(x_2 = \sqrt{3}, y_2 = 0\)

Calculating new \(X_{CM}\):
\[ X'_{CM} = \frac{2 \cdot 0 + 2 \cdot \sqrt{3}}{2 + 2} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]
Calculating new \(Y_{CM}\):
\[ Y'_{CM} = \frac{2 \cdot 0 + 2 \cdot 0}{4} = 0 \]

### Step 5: Calculate the Shift in Center of Mass
Now we find the shift in the center of mass:
- Original \(Y_{CM} = \frac{\sqrt{3}}{6}\)
- New \(Y'_{CM} = 0\)

The shift in the Y-coordinate:
\[ \Delta Y_{CM} = Y'_{CM} - Y_{CM} = 0 - \frac{\sqrt{3}}{6} = -\frac{\sqrt{3}}{6} \]

### Final Answer
The shift in the center of mass when one particle is removed is \(-\frac{\sqrt{3}}{6} \, \text{m}\) in the downward direction.

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