A tall, cylindrical chimney starts falling over as shown in Fig. Treat the chimney as a thin rigid rod of length L = 55.0 m. At the instant it makes as angle of `theta=60.0^(@)`m with the vertical, what is its angular speed `omega_(f)`?

Explanation

To find the angular speed \( \omega_f \) of a falling cylindrical chimney modeled as a thin rigid rod at an angle of \( \theta = 60^\circ \) with the vertical, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the initial and final states
- **Initial State (Point A)**: The chimney is vertical, and its center of mass is at a height \( \frac{L}{2} \) from the ground.
- **Final State (Point B)**: The chimney is at an angle of \( 60^\circ \) with the vertical. The center of mass is now at a height of \( \frac{L}{2} \cos(60^\circ) \).

### Step 2: Write the conservation of energy equation
The total mechanical energy at the initial state must equal the total mechanical energy at the final state:
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