Three identical spheres of mass M each are placed at the corners of an equilateral triangle of side 2 m. Taking one of the corners as the origin, the position vector of the centre of mass is

Options

A. `sqrt3(hati-hatj)`
B. `(hati)/(sqrt3)+hatj`
C. `hati+hatj//3`
D. `hati+hatj//sqrt3`

Explanation

To find the position vector of the center of mass of three identical spheres placed at the corners of an equilateral triangle, we can follow these steps:

### Step 1: Define the positions of the spheres
Let's denote the three spheres located at the corners of an equilateral triangle with side length 2 m. We will place one corner at the origin (0, 0) and define the positions of the other two corners based on this.

1. **Sphere 1 (at origin)**:
- Position vector \( \mathbf{r_1} = (0, 0) \)

2. **Sphere 2** (2 m along the x-axis):
- Position vector \( \mathbf{r_2} = (2, 0) \)

3. **Sphere 3** (at the third corner):
- To find the coordinates of the third sphere, we can use the properties of an equilateral triangle. The height \( h \) of the triangle can be calculated using:
\[ h = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \]
- The coordinates of Sphere 3 will be:
- Position vector \( \mathbf{r_3} = (1, \sqrt{3}) \) (1 m along the x-axis and \( \sqrt{3} \) m along the y-axis).

### Step 2: Calculate the center of mass
The formula for the center of mass \( \mathbf{R_{cm}} \) of a system of particles is given by:
\[ \mathbf{R_{cm}} = \frac{m_1 \mathbf{r_1} + m_2 \mathbf{r_2} + m_3 \mathbf{r_3}}{m_1 + m_2 + m_3} \]
Since all spheres have the same mass \( M \), we can simplify the equation:
\[ \mathbf{R_{cm}} = \frac{M \mathbf{r_1} + M \mathbf{r_2} + M \mathbf{r_3}}{M + M + M} = \frac{\mathbf{r_1} + \mathbf{r_2} + \mathbf{r_3}}{3} \]

### Step 3: Substitute the position vectors
Substituting the position vectors into the formula:
\[ \mathbf{R_{cm}} = \frac{(0, 0) + (2, 0) + (1, \sqrt{3})}{3} \]
Calculating the sum:
\[ \mathbf{R_{cm}} = \frac{(0 + 2 + 1, 0 + 0 + \sqrt{3})}{3} = \frac{(3, \sqrt{3})}{3} = (1, \frac{\sqrt{3}}{3}) \]

### Step 4: Write the final position vector
Thus, the position vector of the center of mass is:
\[ \mathbf{R_{cm}} = (1, \frac{\sqrt{3}}{3}) \]

### Final Answer
The position vector of the center of mass is \( (1, \frac{\sqrt{3}}{3}) \).

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