Two point masses m and 3m are placed at distance r. The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the joining the point masses is

Options

A. `(3)/(5)mr^(2)`
B. `(3)/(4)mr^(2)`
C. `(3)/(2)mr^(2)`
D. `(6)/(7)mr^(2)`

Answer

To find the moment of inertia of a system of two point masses \( m \) and \( 3m \) placed at a distance \( r \) about an axis passing through the center of mass (COM) and perpendicular to the line joining the two masses, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the masses and their positions**:
- Let the two point masses be \( m \) and \( 3m \).
- They are separated by a distance \( r \).

2. **Calculate the position of the center of mass (COM)**:
- The formula for the center of mass \( x_{cm} \) for two point masses is given by:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
- Let \( m_1 = m \) (located at \( x_1 = 0 \)) and \( m_2 = 3m \) (located at \( x_2 = r \)):
\[ x_{cm} = \frac{m \cdot 0 + 3m \cdot r}{m + 3m} = \frac{3mr}{4m} = \frac{3r}{4} \]

3. **Determine the distances from the COM to each mass**:
- The distance from the COM to mass \( m \) (located at \( 0 \)):
\[ d_1 = x_{cm} - 0 = \frac{3r}{4} \]
- The distance from the COM to mass \( 3m \) (located at \( r \)):
\[ d_2 = r - x_{cm} = r - \frac{3r}{4} = \frac{r}{4} \]

4. **Calculate the moment of inertia \( I_{cm} \)**:
- The moment of inertia about the COM is given by:
\[ I_{cm} = m d_1^2 + 3m d_2^2 \]
- Substituting the distances:
\[ I_{cm} = m \left(\frac{3r}{4}\right)^2 + 3m \left(\frac{r}{4}\right)^2 \]
- Calculating each term:
\[ I_{cm} = m \cdot \frac{9r^2}{16} + 3m \cdot \frac{r^2}{16} \]
\[ = \frac{9mr^2}{16} + \frac{3mr^2}{16} = \frac{12mr^2}{16} = \frac{3mr^2}{4} \]

5. **Final Result**:
- The moment of inertia of the system about the axis passing through the center of mass is:
\[ I_{cm} = \frac{3mr^2}{4} \]