One end of a uniform rod of mas m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity `omega`. The force exerted by the clamp on the rod has a horizontal component

Options

A. `momega^2l`
B. `zero`
C. `mg`
D. `1/2momega^2l`

Answer

To solve the problem of finding the horizontal component of the force exerted by the clamp on a rotating uniform rod, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a uniform rod of mass \( m \) and length \( l \) that is clamped at one end. The rod rotates about the clamped end with a uniform angular velocity \( \omega \).

2. **Consider an Element of the Rod**:
- Take a small element of the rod at a distance \( x \) from the clamped end, with a small length \( dx \).

3. **Identify Forces Acting on the Element**:
- The tension in the rod at distance \( x \) can be denoted as \( T \). At the next segment of the rod, the tension will be \( T + dT \).

4. **Centripetal Force Requirement**:
- For the small element \( dx \) to rotate in a circular path, it requires a centripetal force. This centripetal force is provided by the difference in tension:
\[ F_C = T - (T + dT) = -dT \]

5. **Express the Centripetal Force**:
- The mass of the small element \( dm \) can be expressed as:
\[ dm = \frac{m}{l} dx \]
- The centripetal force required for this mass rotating at radius \( x \) is given by:
\[ F_C = dm \cdot \omega^2 \cdot x = \frac{m}{l} dx \cdot \omega^2 \cdot x \]

6. **Set Up the Equation**:
- Equating the expressions for centripetal force:
\[ -dT = \frac{m}{l} \cdot \omega^2 \cdot x \cdot dx \]
- Rearranging gives:
\[ dT = -\frac{m \omega^2}{l} x \, dx \]

7. **Integrate to Find Tension**:
- Integrate \( dT \) from \( T \) (at \( x \)) to \( 0 \) (at \( L \)):
\[ \int_{T}^{0} dT = -\int_{x}^{L} \frac{m \omega^2}{l} x \, dx \]
- This results in:
\[ T = \frac{m \omega^2}{2l} (L^2 - x^2) \]

8. **Find the Tension at the Clamped End**:
- To find the tension at the clamped end (where \( x = 0 \)):
\[ T(0) = \frac{m \omega^2}{2l} L^2 \]

9. **Conclusion**:
- The horizontal component of the force exerted by the clamp on the rod is equal to the tension at the clamped end:
\[ F_{horizontal} = T(0) = \frac{m \omega^2 L}{2} \]

### Final Answer:
The horizontal component of the force exerted by the clamp on the rod is:
\[ F_{horizontal} = \frac{m \omega^2 L}{2} \]