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To solve this problem, we need to use the conservation of angular momentum. Here's the detailed breakdown:

### Given Information:
1. Two masses, each \(2 \, \text{kg}\), are attached to a \(1 \, \text{m}\) rigid rod of negligible mass.
2. The rod is pivoted at its center, with the masses located \(0.5 \, \text{m}\) on each side from the pivot.
3. A \(0.4 \, \text{kg}\) piece of putty is dropped on the right mass at a speed of \(3.3 \, \text{m/s}\) and sticks to it.

### Required:
The angular velocity (\(\omega\)) of the system just after the putty sticks.

### Solution:

1. **Calculate the initial angular momentum** of the putty relative to the pivot point:
   \[
   L_{\text{initial}} = m_{\text{putty}} \cdot v \cdot r
   \]
   where:
   - \( m_{\text{putty}} = 0.4 \, \text{kg} \)
   - \( v = 3.3 \, \text{m/s} \)
   - \( r = 0.5 \, \text{m} \) (distance from the pivot to the mass where the putty sticks)

   Substituting these values:
   \[
   L_{\text{initial}} = 0.4 \cdot 3.3 \cdot 0.5 = 0.66 \, \text{kg} \cdot \text{m}^2/\text{s}
   \]

2. **Calculate the moment of inertia (I)** of the system after the putty sticks:
   - Each \(2 \, \text{kg}\) mass is \(0.5 \, \text{m}\) from the pivot.
   - The putty, now stuck to the right mass, also contributes to the moment of inertia at \(0.5 \, \text{m}\) from the pivot.

   So,
   \[
   I = (2 \cdot 2 \cdot 0.5^2) + (0.4 \cdot 0.5^2)
   \]
   Breaking this down:
   \[
   I = (2 \cdot 2 \cdot 0.25) + (0.4 \cdot 0.25)
   \]
   \[
   I = (2 \cdot 0.5) + 0.1 = 1.0 + 0.1 = 1.1 \, \text{kg} \cdot \text{m}^2
   \]

3. **Use Conservation of Angular Momentum**:
   Since there is no external torque, the initial angular momentum of the putty must equal the final angular momentum of the entire system.

   \[
   L_{\text{initial}} = I \cdot \omega
   \]
   Solving for \(\omega\):
   \[
   \omega = \frac{L_{\text{initial}}}{I} = \frac{0.66}{1.1} = 0.6 \, \text{rad/s}
   \]

### Answer:
The angular velocity of the masses just after the putty hits and sticks to the right mass is:
\[
\boxed{0.6 \, \text{rad/s}}
\]

This matches option (B) in the problem.

---

Would you like further details on any part of the solution? Here are some follow-up questions you may find helpful:

1. How does the conservation of angular momentum apply to rotating systems?
2. Why is the moment of inertia calculated based on distances from the pivot point?
3. How would the result change if the rod itself had a non-negligible mass?
4. What would happen if the putty didn’t stick and instead bounced off?
5. How do we handle angular momentum when forces are applied at different angles?

**Tip:** When using conservation laws, ensure you account for all masses and their respective distances from the pivot to correctly calculate the moment of inertia.

Two masses, each 2 kg, are attached to a uniform 1-meter long rigid rod of negligible mass. The rod is free to rotate about an axis through the center of the rod perpendicular to the plane of the paper. When the rod is in the horizontal position, 0.4 kg of putty is dropped on the mass to the right with a speed of 3.3 m/s, and sticks to it. What is the angular velocity of the masses just after the putty hits the mass to the right?

Solve for the angular velocity of a rotating rod with attached masses when 0.4 kg of putty impacts the mass on the right at 3.3 m/s. This physics problem covers the concepts of angular momentum conservation and moment of inertia calculations, suitable for advanced high school or college-level students.

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