To find the position of the center of mass of the two uniform rods A and B placed end to end, we will follow these steps:
### Step 1: Identify the lengths and linear densities of the rods
- Length of rod A, \( L_A = 5 \, \text{m} \)
- Length of rod B, \( L_B = 3 \, \text{m} \)
- Linear density of rod A, \( \lambda_A = 3 \, \text{kg/m} \)
- Linear density of rod B, \( \lambda_B = 2 \, \text{kg/m} \)
### Step 2: Calculate the masses of the rods
- Mass of rod A, \( m_A = L_A \times \lambda_A = 5 \, \text{m} \times 3 \, \text{kg/m} = 15 \, \text{kg} \)
- Mass of rod B, \( m_B = L_B \times \lambda_B = 3 \, \text{m} \times 2 \, \text{kg/m} = 6 \, \text{kg} \)
### Step 3: Determine the positions of the center of mass of each rod
- The center of mass of rod A is located at \( \frac{L_A}{2} = \frac{5}{2} = 2.5 \, \text{m} \) from the left end (point A).
- The center of mass of rod B is located at \( L_A + \frac{L_B}{2} = 5 + \frac{3}{2} = 5 + 1.5 = 6.5 \, \text{m} \) from the left end (point A).
### Step 4: Set up the equation for the center of mass of the system
The center of mass \( x_{cm} \) of the system can be calculated using the formula:
\[
x_{cm} = \frac{m_A x_A + m_B x_B}{m_A + m_B}
\]
Where:
- \( x_A = 2.5 \, \text{m} \) (position of center of mass of rod A)
- \( x_B = 6.5 \, \text{m} \) (position of center of mass of rod B)
### Step 5: Substitute the values into the equation
\[
x_{cm} = \frac{(15 \, \text{kg} \times 2.5 \, \text{m}) + (6 \, \text{kg} \times 6.5 \, \text{m})}{15 \, \text{kg} + 6 \, \text{kg}} = \frac{(37.5 + 39)}{21} = \frac{76.5}{21} \approx 3.643 \, \text{m}
\]
### Step 6: Calculate the position of the center of mass from the interface
The interface is at \( 5 \, \text{m} \) (the end of rod A). Therefore, the distance from the interface to the center of mass is:
\[
\text{Distance from interface} = 6.5 - 3.643 = 2.857 \, \text{m}
\]
### Step 7: Final adjustment to find the distance from the interface
Since the center of mass is located towards rod B, we need to find the distance from the interface:
\[
\text{Distance from interface} = 5 - 3.643 = 1.357 \, \text{m}
\]
### Conclusion
The position of the center of mass from the interface is approximately \( 1.36 \, \text{m} \).
