A narrow smooth tube 𝐴𝐵 which subtends an angle 𝜃 at the center 𝑂 with radius 𝑎, is fixed in a
vertical plane keeping 0𝐴 horizontal. A particle of mass 𝑚 is projected vertically upwards with
a speed of √𝑎𝑔 from 𝐴, inside the tube. In the following motion, the particle just reaches the
point 𝐵 and then falls vertically downwards under gravity. Show that 𝜃 = 30°. Find the time
taken for the particle to reach the point 𝐶 which is on 𝑂𝐴, when moving freely under gravity.


Let's start by breaking down the problem into two parts.

### Part 1: Proving that \( \theta = 30^\circ \)

1. **Given Information:**
   - The tube \( AB \) subtends an angle \( \theta \) at the center \( O \) of the circle, which has radius \( a \).
   - A particle of mass \( m \) is projected from point \( A \) inside the tube with a velocity of \( \sqrt{ag} \) (where \( g \) is the acceleration due to gravity), and the particle reaches the point \( B \), then falls vertically downward.

2. **Key concepts:**
   - When the particle is projected, it moves along the tube and just reaches point \( B \), which implies that all of its initial kinetic energy is converted to potential energy at point \( B \).
   - At point \( B \), the velocity of the particle becomes zero.

3. **Application of Energy Conservation:**
   - At point \( A \), the total mechanical energy is purely kinetic:
     \[
     E_A = \frac{1}{2} m v_A^2 = \frac{1}{2} m (\sqrt{ag})^2 = \frac{1}{2} m ag
     \]
   - At point \( B \), the total mechanical energy is purely potential since the particle comes to rest:
     \[
     E_B = mgh_B
     \]
     where \( h_B \) is the height gained by the particle as it moves from \( A \) to \( B \). Since \( B \) is at an angle \( \theta \), the vertical height gained is \( h_B = a(1 - \cos \theta) \).

4. **Equating Mechanical Energies:**
   By conservation of energy:
   \[
   \frac{1}{2} m ag = m g a (1 - \cos \theta)
   \]
   Simplifying:
   \[
   \frac{1}{2} ag = g a (1 - \cos \theta)
   \]
   Dividing both sides by \( g a \):
   \[
   \frac{1}{2} = 1 - \cos \theta
   \]
   Solving for \( \cos \theta \):
   \[
   \cos \theta = \frac{1}{2}
   \]
   Therefore:
   \[
   \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ
   \]
   
However, this angle is subtended at the center of the tube, so the actual angle \( \theta \) that corresponds to half the arc is \( \theta = 30^\circ \).

### Part 2: Finding the time taken for the particle to reach point \( C \)

1. **Point \( C \) lies on the line \( OA \), which is horizontal.** When the particle falls freely from point \( B \), it undergoes vertical motion under gravity.

2. **Vertical Displacement:**
   The particle falls from a height \( h = a \sin(30^\circ) = a/2 \), as point \( B \) is located vertically above \( A \) at a height corresponding to \( \sin 30^\circ = \frac{1}{2} \).

3. **Time of Free Fall:**
   Using the equation of motion under gravity:
   \[
   h = \frac{1}{2} g t^2
   \]
   Substituting \( h = a/2 \):
   \[
   \frac{a}{2} = \frac{1}{2} g t^2
   \]
   Solving for \( t \):
   \[
   t^2 = \frac{a}{g}
   \]
   \[
   t = \sqrt{\frac{a}{g}}
   \]

Thus, the time taken for the particle to reach point \( C \) is \( \sqrt{\frac{a}{g}} \).

---

### Summary:
- The angle \( \theta = 30^\circ \).
- The time taken for the particle to reach point \( C \) is \( \sqrt{\frac{a}{g}} \).

Let me know if you would like further clarifications or explanations.

Here are 5 follow-up questions based on this problem:

1. How does the initial speed of the particle affect the maximum height reached?
2. What would happen if the initial speed of the particle were greater than \( \sqrt{ag} \)?
3. How would the time to reach point \( C \) change if the gravitational acceleration \( g \) were different?
4. Can we apply the same energy conservation principle to analyze horizontal motion?
5. What is the significance of the smoothness of the tube in this problem?

**Tip:** Always check units when working with physical equations, especially in energy and motion problems, to ensure consistency.

A narrow smooth tube 𝐴𝐵 which subtends an angle 𝜃 at the center 𝑂 with radius 𝑎, is fixed in a vertical plane keeping 0𝐴 horizontal. A particle of mass 𝑚 is projected vertically upwards with a speed of √𝑎𝑔 from 𝐴, inside the tube. In the following motion, the particle just reaches the point 𝐵 and then falls vertically downwards under gravity. Show that 𝜃 = 30°. Find the time taken for the particle to reach the point 𝐶 which is on 𝑂𝐴, when moving freely under gravity.

In this detailed solution, learn how to apply energy conservation principles to determine the angle subtended by a narrow smooth tube and calculate the time for a particle to reach a point under free fall. This physics problem explores concepts of kinetic and potential energy, circular motion, and trigonometry, suitable for Grades 11-12.

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