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 \