A uniform rod of length 2l is placed with one end in contact with the horizontal table and is then inclined at an angle `alpha` to the horizontal and allowed to fall. When it becomes horizontal, its angular velocity will be

Options

A. `omega = sqrt(((3g sin alpha)/(2l)))`
B. `omega = sqrt(((2l)/(3g sin alpha)))`
C. `omega = sqrt(((g sin alpha)/(l)))`
D. `omega = sqrt(((l)/(g sin alpha)))`

Answer

To solve the problem, we will use the principles of conservation of energy and the moment of inertia of the rod. Here are the steps to find the angular velocity of the rod when it becomes horizontal:

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- The rod is initially inclined at an angle \( \alpha \) to the horizontal.
- The length of the rod is \( 2L \).
- The mass of the rod is \( m \).
- When the rod becomes horizontal, we need to find its angular velocity \( \omega \).

2. **Calculate the Initial Potential Energy:**
- The center of mass of the rod is at a distance \( L \) from the hinge (the point in contact with the table).
- The height of the center of mass when the rod is inclined at angle \( \alpha \) is given by:
\[ h = L \sin \alpha \]
- Therefore, the initial potential energy \( PE_i \) is:
\[ PE_i = mgh = mg(L \sin \alpha) \]

3. **Calculate the Final Potential Energy:**
- When the rod becomes horizontal, the height of the center of mass is 0, so the final potential energy \( PE_f \) is:
\[ PE_f = 0 \]

4. **Calculate the Initial and Final Kinetic Energy:**
- Initially, the rod is at rest, so the initial kinetic energy \( KE_i \) is:
\[ KE_i = 0 \]
- When the rod is horizontal, it has rotational kinetic energy given by:
\[ KE_f = \frac{1}{2} I \omega^2 \]
- The moment of inertia \( I \) of the rod about the hinge point is:
\[ I = \frac{1}{3} m (2L)^2 = \frac{4}{3} m L^2 \]

5. **Apply Conservation of Energy:**
- According to the conservation of mechanical energy:
\[ PE_i + KE_i = PE_f + KE_f \]
- Substituting the values we calculated:
\[ mg(L \sin \alpha) + 0 = 0 + \frac{1}{2} \left(\frac{4}{3} m L^2\right) \omega^2 \]

6. **Simplify the Equation:**
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g L \sin \alpha = \frac{2}{3} L^2 \omega^2 \]
- Rearranging gives:
\[ \omega^2 = \frac{3g \sin \alpha}{2L} \]

7. **Find the Angular Velocity:**
- Taking the square root of both sides:
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]

### Final Answer:
The angular velocity \( \omega \) of the rod when it becomes horizontal is:
\[ \omega = \sqrt{\frac{3g \sin \alpha}{2L}} \]