A body of mass 4 kg is attached to another body of mass 2 kg with a massless rod. If the 4 kg mass is at (2i + 5j)m and 2 kg mass is at (4i + 2j)m, the centre of mass of the system is at (in m)

Options

A. `8i+12j`
B. `(1)/(2)(8i+12j)`
C. `(1)/(3)(8i+12j)`
D. `(1)/(4)(8i+12j)`

Answer

To find the center of mass of the system consisting of two bodies, we can use the formula for the center of mass \( \vec{R}_{cm} \):

\[ \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \]

where:
- \( m_1 \) and \( m_2 \) are the masses of the two bodies,
- \( \vec{r}_1 \) and \( \vec{r}_2 \) are the position vectors of the two bodies.

### Step 1: Identify the masses and their position vectors
Given:
- Mass \( m_1 = 4 \, \text{kg} \) at position \( \vec{r}_1 = 2\hat{i} + 5\hat{j} \, \text{m} \)
- Mass \( m_2 = 2 \, \text{kg} \) at position \( \vec{r}_2 = 4\hat{i} + 2\hat{j} \, \text{m} \)

### Step 2: Substitute the values into the center of mass formula
Substituting the values into the center of mass formula:

\[ \vec{R}_{cm} = \frac{4(2\hat{i} + 5\hat{j}) + 2(4\hat{i} + 2\hat{j})}{4 + 2} \]

### Step 3: Calculate the numerator
Calculating the terms in the numerator:

\[ 4(2\hat{i} + 5\hat{j}) = 8\hat{i} + 20\hat{j} \]
\[ 2(4\hat{i} + 2\hat{j}) = 8\hat{i} + 4\hat{j} \]

Adding these results together:

\[ (8\hat{i} + 20\hat{j}) + (8\hat{i} + 4\hat{j}) = 16\hat{i} + 24\hat{j} \]

### Step 4: Calculate the total mass
The total mass is:

\[ m_1 + m_2 = 4 + 2 = 6 \, \text{kg} \]

### Step 5: Divide the result by the total mass
Now, substituting back into the equation:

\[ \vec{R}_{cm} = \frac{16\hat{i} + 24\hat{j}}{6} \]

### Step 6: Simplify the result
This simplifies to:

\[ \vec{R}_{cm} = \frac{16}{6}\hat{i} + \frac{24}{6}\hat{j} = \frac{8}{3}\hat{i} + 4\hat{j} \]

### Final Result
Thus, the center of mass of the system is at:

\[ \vec{R}_{cm} = \left(\frac{8}{3}\hat{i} + 4\hat{j}\right) \, \text{m} \]

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