A uniform rod is kept vertically on a horizontally smooth surface at a point O. IF it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain

Options

A. at O
B. at a distance less than `l/2` from O
C. at a distance `l/2` from O
D. at a distance larger than `l/2` from O

Answer

To solve the problem, we need to analyze the motion of the uniform rod as it falls after being slightly rotated from the vertical position.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A uniform rod of length \( L \) is placed vertically on a smooth horizontal surface at point \( O \). The center of mass of the rod is located at a distance \( \frac{L}{2} \) from point \( O \).

2. **Initial Position of the Center of Mass**:
- When the rod is vertical, the center of mass is at \( \frac{L}{2} \) above point \( O \).

3. **Rotation and Release**:
- When the rod is slightly rotated and released, it will begin to fall due to gravity. The lower end of the rod (point \( O \)) will remain in contact with the ground.

4. **Analyzing the Motion**:
- As the rod falls, the center of mass will follow a path that is influenced by the pivot point at \( O \). Since there are no horizontal forces acting on the rod (the surface is smooth), the horizontal position of the center of mass will not change.

5. **Final Position of the Center of Mass**:
- After the rod has fallen and comes to rest on the horizontal surface, the center of mass will still be located at a distance \( \frac{L}{2} \) from point \( O \) because the horizontal position of the center of mass does not change during the fall.

6. **Conclusion**:
- Therefore, the lower end of the rod will remain at point \( O \), and the center of mass will be at a distance \( \frac{L}{2} \) from point \( O \) when the rod is lying flat on the surface.

### Final Answer:
The lower end of the rod will remain at a distance \( \frac{L}{2} \) from point \( O \).

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