In the above problem, if the rod is released from horizontal position, the angular velocity of the rod as it passes the vertical position is (l=length fo rod)

Options

A. `sqrt((12g)/(5l))`
B. `sqrt((2g)/(3l))`
C. `sqrt((3g)/(l))`
D. `sqrt((3g)/(7l))`

Answer

To solve the problem of finding the angular velocity of a rod as it passes the vertical position after being released from a horizontal position, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
- We have a uniform rod of length \( l \) that is hinged at one end and initially held in a horizontal position.

2. **Determine the Initial and Final Positions**:
- The initial position of the rod is horizontal, and the final position we are interested in is vertical.

3. **Calculate the Change in Potential Energy**:
- The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from the hinge.
- When the rod is horizontal, its height relative to the lowest point (vertical position) is \( \frac{l}{2} \).
- The loss in gravitational potential energy (PE) as the rod moves from horizontal to vertical is given by:
\[ \Delta PE = mgh = mg \cdot \frac{l}{2} \]

4. **Relate Potential Energy to Kinetic Energy**:
- As the rod falls, the loss in potential energy is converted into kinetic energy (KE).
- The kinetic energy of a rotating body is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
- For a rod rotating about one end, the moment of inertia \( I \) is:
\[ I = \frac{1}{3} ml^2 \]

5. **Set Up the Energy Conservation Equation**:
- Equating the loss in potential energy to the gain in kinetic energy:
\[ mg \cdot \frac{l}{2} = \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 \]

6. **Simplify the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \cdot \frac{l}{2} = \frac{1}{6} l^2 \omega^2 \]
- Multiply both sides by 6:
\[ 3gl = l^2 \omega^2 \]

7. **Solve for Angular Velocity \( \omega \)**:
- Rearranging gives:
\[ \omega^2 = \frac{3g}{l} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{3g}{l}} \]

8. **Final Result**:
- The angular velocity of the rod as it passes the vertical position is:
\[ \omega = \sqrt{\frac{3g}{l}} \]

### Conclusion:
The correct answer is option C: \( \sqrt{\frac{3g}{l}} \).