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 unifrom angular velocity `omega`. Theforce 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 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 \( \omega \). The goal is to determine the horizontal component of the force exerted by the clamp on the rod.

**Hint:** Visualize the rod and its rotation. Remember that each point along the rod is moving in a circular path about the clamped end.

### Step 2: Identify Forces Acting on the Rod
The rod experiences a gravitational force acting downward (its weight) and a horizontal force due to the rotation. The clamp must exert a force to counteract the centrifugal effect caused by the rotation of the rod.

**Hint:** Consider the forces acting on a small segment of the rod to analyze the centrifugal effect.

### Step 3: Calculate the Centrifugal Force on a Small Element
Consider a small element of the rod at a distance \( R \) from the clamped end with a small length \( dR \). The mass of this small element \( dM \) can be expressed as:
\[ dM = \frac{m}{l} dR \]
The centrifugal force \( dF \) acting on this small element due to its rotation is given by:
\[ dF = dM \cdot R \cdot \omega^2 = \left(\frac{m}{l} dR\right) \cdot R \cdot \omega^2 \]

**Hint:** Use the formula for centrifugal force and remember that the force depends on the distance from the axis of rotation.

### Step 4: Integrate to Find Total Horizontal Force
To find the total horizontal force \( F \) exerted by the clamp on the rod, integrate \( dF \) from \( R = 0 \) to \( R = l \):
\[ F = \int_0^l dF = \int_0^l \left(\frac{m}{l} R \omega^2 dR\right) \]
This simplifies to:
\[ F = \frac{m \omega^2}{l} \int_0^l R dR \]
The integral \( \int_0^l R dR \) evaluates to \( \frac{l^2}{2} \):
\[ F = \frac{m \omega^2}{l} \cdot \frac{l^2}{2} = \frac{m l \omega^2}{2} \]

**Hint:** Make sure to perform the integration correctly and apply the limits properly.

### Step 5: Conclusion
The horizontal component of the force exerted by the clamp on the rod is:
\[ F = \frac{m l \omega^2}{2} \]

This means the correct option is \( \frac{m l \omega^2}{2} \).

**Hint:** Review the final expression to ensure it matches the physical context of the problem.