A rod of length L is hinged from one end. It is brought to the horizontal position and released. The angular velocity of the rod when it is in the vertical position is `sqrt(Kg//L)`. Then find the value of K.

Answer

To solve the problem, we will use the work-energy theorem, which relates the work done by forces to the change in kinetic energy of the system.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A rod of length \( L \) is hinged at one end and is initially held in a horizontal position. When released, it swings down to a vertical position.

2. **Identify Forces and Work Done**:
- The only force doing work on the rod as it swings down is gravity. The gravitational force acting on the center of mass of the rod is \( mg \), where \( m \) is the mass of the rod and \( g \) is the acceleration due to gravity.

3. **Calculate the Work Done by Gravity**:
- The center of mass of the rod, which is located at a distance \( \frac{L}{2} \) from the hinge, falls a vertical distance of \( \frac{L}{2} \) when the rod moves from horizontal to vertical.
- The work done by gravity \( W \) is given by:
\[ W = mg \cdot h = mg \cdot \frac{L}{2} \]

4. **Apply the Work-Energy Theorem**:
- According to the work-energy theorem, the work done by gravity is equal to the change in kinetic energy of the rod:
\[ W = KE_{final} - KE_{initial} \]
- Since the rod starts from rest, \( KE_{initial} = 0 \). Therefore:
\[ mg \cdot \frac{L}{2} = KE_{final} \]

5. **Calculate the Final Kinetic Energy**:
- The final kinetic energy when the rod is vertical can be expressed as:
\[ KE_{final} = \frac{1}{2} I \omega^2 \]
- Here, \( I \) is the moment of inertia of the rod about the hinge, which is given by \( I = \frac{1}{3} m L^2 \).
- Thus, we have:
\[ KE_{final} = \frac{1}{2} \cdot \frac{1}{3} m L^2 \cdot \omega^2 \]

6. **Equate the Work Done to the Kinetic Energy**:
- Setting the work done equal to the kinetic energy gives:
\[ mg \cdot \frac{L}{2} = \frac{1}{2} \cdot \frac{1}{3} m L^2 \cdot \omega^2 \]
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \cdot \frac{L}{2} = \frac{1}{6} L^2 \cdot \omega^2 \]

7. **Solve for \( \omega^2 \)**:
- Rearranging the equation gives:
\[ \omega^2 = \frac{6g}{L} \]

8. **Relate to Given Angular Velocity**:
- The problem states that the angular velocity \( \omega \) is given by:
\[ \omega = \sqrt{\frac{Kg}{L}} \]
- Setting the two expressions for \( \omega^2 \) equal to each other:
\[ \frac{6g}{L} = \frac{Kg}{L} \]

9. **Find the Value of \( K \)**:
- Cancel \( g \) and \( L \) (assuming they are not zero):
\[ K = 6 \]

### Final Answer:
The value of \( K \) is \( 6 \).