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 consisting of two point masses \( m \) and \( 3m \) placed at a distance \( r \) from each other about an axis passing through the center of mass (CM) of the system and perpendicular to the line joining the point masses, we can follow these steps:

### Step 1: Determine the position of the center of mass (CM)
The center of mass \( x_{cm} \) for two point masses can be calculated using the formula:

\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

Here, let \( m_1 = m \) (mass at position \( 0 \)) and \( m_2 = 3m \) (mass at position \( r \)). Therefore, the positions are:

- \( x_1 = 0 \)
- \( x_2 = r \)

Substituting the values:

\[ x_{cm} = \frac{m \cdot 0 + 3m \cdot r}{m + 3m} = \frac{3mr}{4m} = \frac{3r}{4} \]

### Step 2: Calculate distances from the center of mass
Now, we need to find the distances \( d_1 \) and \( d_2 \) from the center of mass to each mass:

- For mass \( m \):
\[ d_1 = x_{cm} - 0 = \frac{3r}{4} \]

- For mass \( 3m \):
\[ d_2 = r - x_{cm} = r - \frac{3r}{4} = \frac{r}{4} \]

### Step 3: Calculate the moment of inertia (I)
The moment of inertia \( I \) about the center of mass is given by:

\[ I = m \cdot d_1^2 + 3m \cdot d_2^2 \]

Substituting \( d_1 \) and \( d_2 \):

\[ I = m \left(\frac{3r}{4}\right)^2 + 3m \left(\frac{r}{4}\right)^2 \]

Calculating each term:

\[ I = m \cdot \frac{9r^2}{16} + 3m \cdot \frac{r^2}{16} \]

\[ I = \frac{9mr^2}{16} + \frac{3mr^2}{16} = \frac{(9 + 3)mr^2}{16} = \frac{12mr^2}{16} = \frac{3mr^2}{4} \]

### Final Answer
Thus, the moment of inertia of the system about the axis passing through the center of mass is:

\[ \boxed{\frac{3mr^2}{4}} \]

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