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To solve this problem, we need to calculate the angular acceleration \( \alpha \) of the system when the rod makes a 45.3° angle with the horizontal. Here's a step-by-step approach:

### Given Information:
- Mass of the rod, \( M_r = 3.9 \, \text{kg} \)
- Length of the rod, \( L = 2.3 \, \text{m} \)
- Mass of particle \( m_1 = 4.5 \, \text{kg} \)
- Mass of particle \( m_2 = 2.6 \, \text{kg} \)
- Angle with the horizontal, \( \theta = 45.3^\circ \)

### Step 1: Moment of Inertia of the System
The moment of inertia \( I \) about the center (pivot point) for a system consisting of a uniform rod and two point masses at each end is calculated by:

1. **Moment of Inertia of the Rod**:
   Since the rod rotates about its center, its moment of inertia \( I_{\text{rod}} \) is:
   \[
   I_{\text{rod}} = \frac{1}{12} M_r L^2
   \]

2. **Moment of Inertia of Point Masses**:
   The point masses \( m_1 \) and \( m_2 \) are located at a distance of \( \frac{L}{2} \) from the pivot. Therefore, their moments of inertia are:
   \[
   I_{m_1} = m_1 \left(\frac{L}{2}\right)^2
   \]
   \[
   I_{m_2} = m_2 \left(\frac{L}{2}\right)^2
   \]

3. **Total Moment of Inertia**:
   Adding these together gives the total moment of inertia of the system:
   \[
   I = I_{\text{rod}} + I_{m_1} + I_{m_2}
   \]

### Step 2: Torque Due to Gravity
The gravitational torque acts on each mass when the rod is at an angle \( \theta \) with the horizontal.

1. **Torque on Rod**:
   The gravitational force on the rod acts at its center of mass, which is at the pivot, so it contributes no torque about the pivot.

2. **Torque on Point Masses**:
   Each mass contributes a torque due to the gravitational force \( mg \), where \( g = 9.8 \, \text{m/s}^2 \). The torque for each mass is:
   \[
   \tau_{m_1} = m_1 g \cdot \frac{L}{2} \cdot \sin(\theta)
   \]
   \[
   \tau_{m_2} = m_2 g \cdot \frac{L}{2} \cdot \sin(\theta)
   \]

3. **Total Torque**:
   The total torque \( \tau \) about the pivot is the sum of the torques from \( m_1 \) and \( m_2 \):
   \[
   \tau = \tau_{m_1} + \tau_{m_2}
   \]
   \[
   \tau = \left(m_1 + m_2\right) g \cdot \frac{L}{2} \cdot \sin(\theta)
   \]

### Step 3: Calculate Angular Acceleration
Using Newton's second law for rotation, \( \tau = I \alpha \), we can solve for the angular acceleration \( \alpha \):
\[
\alpha = \frac{\tau}{I}
\]
Substitute the expressions for \( \tau \) and \( I \) to find \( \alpha \).

Let’s calculate it using the given values:

1. **Moment of Inertia of the Rod**:
   \[
   I_{\text{rod}} = \frac{1}{12} \times 3.9 \times (2.3)^2 = 1.43525 \, \text{kg} \cdot \text{m}^2
   \]

2. **Moment of Inertia of Point Masses**:
   \[
   I_{m_1} = 4.5 \times \left(\frac{2.3}{2}\right)^2 = 5.9475 \, \text{kg} \cdot \text{m}^2
   \]
   \[
   I_{m_2} = 2.6 \times \

A uniform rigid rod with mass Mr = 3.9 kg, length L = 2.3 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 4.5 kg and m2 = 2.6 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 45.3° with the horizontal?

Learn how to calculate the angular acceleration of a rotating system consisting of a uniform rod with attached masses. This problem includes concepts such as moment of inertia, torque, and Newton's second law for rotation, with detailed steps for determining the angular acceleration when the rod makes an angle of 45.3° with the horizontal.

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