Four particles, each of mass 1kg, are placed at the corners of a square OABC of side 1m. 'O' is at the origin of the coordinate system. OA and OC are aligned along positive x-axis and positive y-axis respectively. The position vector of the centre of mass is (in 'm')

Options

A. `hat(i)+hat(j)`
B. `(1)/(2)(hat(i)+hat(j))`
C. `(hat(i)-hat(j))`
D. `(1)/(2)(hat(i)-hat(j))`

Answer

To find the position vector of the center of mass of the four particles placed at the corners of a square, we can follow these steps:

### Step 1: Identify the coordinates of the particles
The square OABC has its corners at the following coordinates:
- O (0, 0)
- A (1, 0)
- B (1, 1)
- C (0, 1)

### Step 2: Assign masses to the particles
Each particle has a mass of 1 kg. Therefore, the total mass \( M \) of the system is:
\[ M = 1 \, \text{kg} + 1 \, \text{kg} + 1 \, \text{kg} + 1 \, \text{kg} = 4 \, \text{kg} \]

### Step 3: Calculate the x-coordinate of the center of mass
The x-coordinate of the center of mass \( x_{cm} \) is given by the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4}{M} \]
Substituting the values:
\[ x_{cm} = \frac{1 \cdot 0 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 0}{4} = \frac{0 + 1 + 1 + 0}{4} = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Calculate the y-coordinate of the center of mass
The y-coordinate of the center of mass \( y_{cm} \) is given by the formula:
\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{M} \]
Substituting the values:
\[ y_{cm} = \frac{1 \cdot 0 + 1 \cdot 0 + 1 \cdot 1 + 1 \cdot 1}{4} = \frac{0 + 0 + 1 + 1}{4} = \frac{2}{4} = \frac{1}{2} \]

### Step 5: Write the position vector of the center of mass
The position vector of the center of mass can be expressed in vector form as:
\[ \vec{r}_{cm} = x_{cm} \hat{i} + y_{cm} \hat{j} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \]

### Final Answer
The position vector of the center of mass is:
\[ \vec{r}_{cm} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \, \text{m} \]

---