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

Answer

To find the position vector of the center of mass of three identical spheres placed at the corners of an equilateral triangle with a side length of 2 m, we can follow these steps:

### Step 1: Define the Coordinates of the Masses
We place the triangle in the Cartesian coordinate system. Let’s assign the following coordinates to the three masses located at the corners of the triangle:
- Mass \( M_1 \) at point A (0, 0)
- Mass \( M_2 \) at point B (2, 0)
- Mass \( M_3 \) at point C (1, \( \sqrt{3} \)) (This is derived from the height of the equilateral triangle, which is \( \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \))

### Step 2: Use the Center of Mass Formula
The position vector of the center of mass \( \vec{R}_{cm} \) for a system of particles is given by:

\[ \vec{R}_{cm} = \frac{M_1 \vec{r}_1 + M_2 \vec{r}_2 + M_3 \vec{r}_3}{M_1 + M_2 + M_3} \]

Since all the masses are identical (mass \( M \)), we can simplify this to:

\[ \vec{R}_{cm} = \frac{M(\vec{r}_1 + \vec{r}_2 + \vec{r}_3)}{3M} = \frac{\vec{r}_1 + \vec{r}_2 + \vec{r}_3}{3} \]

### Step 3: Substitute the Coordinates
Now we substitute the coordinates of the three masses into the formula:

\[ \vec{R}_{cm} = \frac{(0, 0) + (2, 0) + (1, \sqrt{3})}{3} \]

### Step 4: Calculate the Sum of the Coordinates
Now, we calculate the sum of the coordinates:

\[ \vec{R}_{cm} = \frac{(0 + 2 + 1, 0 + 0 + \sqrt{3})}{3} = \frac{(3, \sqrt{3})}{3} = (1, \frac{\sqrt{3}}{3}) \]

### Step 5: Write the Position Vector
Thus, the position vector of the center of mass is:

\[ \vec{R}_{cm} = 1 \hat{i} + \frac{\sqrt{3}}{3} \hat{j} \]

### Final Answer
The position vector of the center of mass is:

\[ \vec{R}_{cm} = \hat{i} + \frac{\sqrt{3}}{3} \hat{j} \]