Let's solve this problem step by step:

Given Data:

1. Mass of the bar (M): $3.80 \, \text{kg}$
2. Length of the bar (L): $80.0 \, \text{cm} = 0.80 \, \text{m}$
3. Mass of each ball (m): $2.50 \, \text{kg}$
4. The axle is frictionless and located at the center of the bar.

The problem has three parts:

Part (a): Finding angular acceleration ($\alpha$) after one ball falls off

Step 1: Calculating torque ($\tau$)

The bar is horizontal, so torque arises only from the gravitational force on the remaining ball. After one ball falls off, the torque due to the remaining ball is:

$\tau = \text{force} \times \text{distance} = m g \cdot \frac{L}{2}$

where $g = 9.8 \, \text{m/s}^2$. Substituting:

$\tau = 2.50 \cdot 9.8 \cdot 0.40 = 9.8 \, \text{N·m}$

Step 2: Moment of inertia ($I$)

The total moment of inertia after one ball falls off consists of:

1. Moment of inertia of the bar: For a uniform bar pivoted at its center: $I_\text{bar} = \frac{1}{12} M L^2$

2. Moment of inertia of the remaining ball: For a point mass at distance $L/2$ from the center: $I_\text{ball} = m \left(\frac{L}{2}\right)^2$

Total moment of inertia: $I = I_\text{bar} + I_\text{ball} = \frac{1}{12} M L^2 + m \left(\frac{L}{2}\right)^2$

Substituting values: $I = \frac{1}{12} (3.80) (0.80)^2 + (2.50) \left(0.40\right)^2$ $I = 0.2027 + 0.4000 = 0.6027 \, \text{kg·m}^2$

Step 3: Angular acceleration ($\alpha$)

The angular acceleration is related to torque and moment of inertia: $\alpha = \frac{\tau}{I}$

Substitute $\tau = 9.8 \, \text{N·m}$ and $I = 0.6027 \, \text{kg·m}^2$: $\alpha = \frac{9.8}{0.6027} \approx 16.26 \, \text{rad/s}^2$

Part (b): Does angular acceleration remain constant?

As the bar begins to swing downward, the torque decreases because the perpendicular component of the gravitational force reduces. Specifically, torque is proportional to $\sin \theta$, where $\theta$ is the angle the bar makes with the vertical.

Thus, the angular acceleration decreases as the bar swings downward.

Part (c): Angular velocity ($\omega$) when the bar swings to a vertical position

At the vertical position, all potential energy of the remaining ball is converted to rotational kinetic energy. The potential energy of the ball relative to the pivot is: $U = m g \frac{L}{2}$

Rotational kinetic energy is: $K = \frac{1}{2} I \omega^2$

From energy conservation ($U = K$): $m g \frac{L}{2} = \frac{1}{2} I \omega^2$

Solve for $\omega$: $\omega = \sqrt{\frac{2 m g \frac{L}{2}}{I}}$

Substitute values: $\omega = \sqrt{\frac{2 (2.50) (9.8) (0.40)}{0.6027}}$ $\omega = \sqrt{\frac{19.6}{0.6027}} \approx \sqrt{32.52} \approx 5.70 \, \text{rad/s}$

Final Answers:

(a) Angular acceleration: $\alpha \approx 16.26 \, \text{rad/s}^2$
(b) Angular acceleration decreases as the bar swings.
(c) Angular velocity at vertical position: $\omega \approx 5.70 \, \text{rad/s}$

Would you like detailed clarifications on any of these steps?

Here are 5 related questions for deeper understanding:

1. How does the moment of inertia affect the angular acceleration?
2. What happens if the bar's mass was different? How would $\alpha$ change?
3. Why does torque depend on the angle $\theta$?
4. How is energy conservation applied to rotational motion?
5. How would the motion change if the axle had friction?

Tip: Always check if forces or torques change with the angle when analyzing rotational motion!