Three particle each of 1 kg mass are placed at corners of a right angled triangle AOB, 'O' being the origin of coordinate system (OA and OB along the x-direction and +ve y-direction). If OA = OB = 1m, the position vector of centre of mass (in metres) is

Options

A. `(hat(i)+hat(j))/(3)`
B. `(hat(i)-hat(j))/(3)`
C. `(2(hat(i)+hat(j)))/(3)`
D. `hat(i)-hat(j)`

Answer

To find the position vector of the center of mass of three particles each of mass 1 kg placed at the corners of a right-angled triangle AOB, we can follow these steps:

### Step 1: Define the Coordinates of the Points
- The origin O is at (0, 0).
- Point A is at (1, 0) since it lies 1 meter along the x-axis.
- Point B is at (0, 1) since it lies 1 meter along the y-axis.

### Step 2: Identify the Masses
- Each particle has a mass of 1 kg.
- Therefore, we have:
- Mass at O (m1) = 1 kg at coordinates (0, 0)
- Mass at A (m2) = 1 kg at coordinates (1, 0)
- Mass at B (m3) = 1 kg at coordinates (0, 1)

### Step 3: Calculate the Total Mass
- Total mass (M) = m1 + m2 + m3 = 1 kg + 1 kg + 1 kg = 3 kg.

### Step 4: Calculate the Coordinates of the Center of Mass
The coordinates of the center of mass (CM) can be calculated using the formula:

\[ \text{CM} = \left( \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}, \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} \right) \]

Substituting the values:
- For the x-coordinate:
\[ x_{CM} = \frac{(1 \cdot 0) + (1 \cdot 1) + (1 \cdot 0)}{3} = \frac{0 + 1 + 0}{3} = \frac{1}{3} \]

- For the y-coordinate:
\[ y_{CM} = \frac{(1 \cdot 0) + (1 \cdot 0) + (1 \cdot 1)}{3} = \frac{0 + 0 + 1}{3} = \frac{1}{3} \]

### Step 5: Write the Position Vector of the Center of Mass
The position vector of the center of mass is given by:
\[ \vec{r}_{CM} = x_{CM} \hat{i} + y_{CM} \hat{j} = \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} \]

### Final Answer
Thus, the position vector of the center of mass is:
\[ \vec{r}_{CM} = \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} \text{ (in metres)} \]

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