Two balls of masses `m` and `2m` are attached to the ends of a light rod of length `L`. The rod rotates with an angular speed `omega` about an axis passing through the center of mass of system and perpendicular to the plane. Find the angular momentum of the system about the axis of rotation.

Options

A. `(2)/(3)momegaL^(2)`
B. `(1)/(3)omega^(2)Lm`
C. `(2)/(3)omega^(2)Lm`
D. `(1)/(3)Lm`

Answer

To find the angular momentum of the system consisting of two balls of masses `m` and `2m` attached to a light rod of length `L`, we will follow these steps:

### Step 1: Determine the Center of Mass (COM)
We need to find the position of the center of mass of the system. The formula for the center of mass \( x_{cm} \) is given by:

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

Where:
- \( m_1 = m \) (mass of the first ball)
- \( x_1 = 0 \) (position of the first ball)
- \( m_2 = 2m \) (mass of the second ball)
- \( x_2 = L \) (position of the second ball)

Substituting the values:

\[ x_{cm} = \frac{m \cdot 0 + 2m \cdot L}{m + 2m} = \frac{2mL}{3m} = \frac{2L}{3} \]

### Step 2: Calculate the Distance from the COM to Each Mass
Now, we need to find the distances from the center of mass to each mass:
- Distance from \( m \) to \( x_{cm} \): \( r_1 = x_{cm} - x_1 = \frac{2L}{3} - 0 = \frac{2L}{3} \)
- Distance from \( 2m \) to \( x_{cm} \): \( r_2 = x_2 - x_{cm} = L - \frac{2L}{3} = \frac{L}{3} \)

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

\[ I = m_1 r_1^2 + m_2 r_2^2 \]

Substituting the values:

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

Calculating each term:

\[ I = m \cdot \frac{4L^2}{9} + 2m \cdot \frac{L^2}{9} = \frac{4mL^2}{9} + \frac{2mL^2}{9} = \frac{6mL^2}{9} = \frac{2mL^2}{3} \]

### Step 4: Calculate Angular Momentum (L)
The angular momentum \( L \) of the system is given by:

\[ L = I \omega \]

Substituting the expression for \( I \):

\[ L = \left(\frac{2mL^2}{3}\right) \omega \]

### Final Result
Thus, the angular momentum of the system about the axis of rotation is:

\[ L = \frac{2mL^2 \omega}{3} \]