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 of a uniform rod kept vertically on a smooth horizontal surface and released after a slight rotation, we can analyze the motion of the rod step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform rod of length \( L \) that is placed vertically at a point \( O \) on a smooth horizontal surface. The rod is initially at rest.

**Hint**: Visualize the rod standing straight up at point \( O \) on the ground.

2. **Initial Conditions**:
- When the rod is slightly rotated and released, it begins to fall due to gravity. The center of mass of the rod is located at its midpoint, which is at a distance \( L/2 \) from point \( O \).

**Hint**: Remember that the center of mass of a uniform rod is at its midpoint.

3. **Forces Acting on the Rod**:
- The only force acting on the rod is its weight \( mg \) acting downwards through the center of mass. There are no horizontal forces acting on the rod since the surface is smooth.

**Hint**: Identify the forces acting on the rod and their directions.

4. **Motion of the Center of Mass**:
- Since there are no horizontal forces, the center of mass will move vertically downward. The horizontal position of the center of mass remains unchanged during the fall.

**Hint**: Consider how the absence of horizontal forces affects the trajectory of the center of mass.

5. **Position of the Lower End**:
- As the rod falls, the lower end (point \( O \)) will remain fixed on the surface. The rod will rotate about this point. The center of mass will fall straight down while the lower end stays at point \( O \).

**Hint**: Think about the pivot point and how the rod rotates around it.

6. **Final Position of the Rod**:
- When the rod has completely fallen, the lower end will remain at point \( O \), and the center of mass will be directly above point \( O \) at a distance \( L/2 \) vertically downward.

**Hint**: Visualize the final position of the rod after it has fallen.

7. **Conclusion**:
- The lower end of the rod remains at point \( O \), and the distance from point \( O \) to the center of mass is \( L/2 \).

**Hint**: Summarize the key points: the lower end stays at \( O \), and the center of mass is \( L/2 \) below it.

### Final Answer:
The lower end of the rod will remain at point \( O \), and the center of mass will be at a distance \( L/2 \) below point \( O \) when the rod falls down.

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