A uniform rod of mass m. length L, area of cross- secticn A is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity \(omega\) in a horizontal plane If Y is the Young's modulus of the material of rod, the increase in its length due to rotation of rod is

Options

A. `(momega^(2)L^(2))/(AY)`
B. `(momega^(2)L^(2))/(2AY)`
C. `(momega^(2)L^(2))/(3AY)`
D. `(2momega^(2)L^(2))/(AY)`

Explanation

To find the increase in length of a uniform rod of mass \( m \), length \( L \), and area of cross-section \( A \) when it is rotated about one of its ends with a constant angular velocity \( \omega \), we can follow these steps:

### Step 1: Define the problem and parameters
We have a uniform rod rotating about one end. We denote:
- Length of the rod: \( L \)
- Mass of the rod: \( m \)
- Cross-sectional area: \( A \)
- Angular velocity: \( \omega \)
- Young's modulus of the material: \( Y \)

### Step 2: Consider a small element of the rod
Let’s consider a small element of the rod at a distance \( x \) from the axis of rotation with a small length \( dx \). The mass of this small element \( dm \) can be expressed as:
\[ dm = \frac{m}{L} dx \]
where \( \frac{m}{L} \) is the mass per unit length (linear density) of the rod.

### Step 3: Calculate the centripetal force acting on the small element
The small element experiences a centripetal force due to its rotation, which can be given by:
\[ dF = dm \cdot \omega^2 \cdot x = \left(\frac{m}{L} dx\right) \cdot \omega^2 \cdot x \]
Thus, we have:
\[ dF = \frac{m \omega^2}{L} x \, dx \]

### Step 4: Relate the centripetal force to the tension in the rod
The tension in the rod at a distance \( x \) from the axis of rotation provides the necessary centripetal force. The tension \( T \) in the rod can be expressed using Young's modulus:
\[ Y = \frac{T/A}{\frac{dL}{dx}} \]
Rearranging gives:
\[ dL = \frac{A}{Y} dF \]

### Step 5: Substitute for \( dF \) and integrate
Substituting \( dF \) into the expression for \( dL \):
\[ dL = \frac{A}{Y} \cdot \frac{m \omega^2}{L} x \, dx \]
Now, we integrate this expression from \( x = 0 \) to \( x = L \) to find the total increase in length \( \Delta L \):
\[ \Delta L = \int_0^L \frac{A m \omega^2}{Y L} x \, dx \]

### Step 6: Perform the integration
The integral can be computed as follows:
\[ \Delta L = \frac{A m \omega^2}{Y L} \int_0^L x \, dx = \frac{A m \omega^2}{Y L} \left[\frac{x^2}{2}\right]_0^L = \frac{A m \omega^2}{Y L} \cdot \frac{L^2}{2} = \frac{A m \omega^2 L}{2Y} \]

### Final Result
Thus, the increase in length of the rod due to rotation is:
\[ \Delta L = \frac{A m \omega^2 L}{2Y} \]