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 o 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 solve the problem of finding the increase in length of a uniform rod rotating about an axis, we can follow these steps:

### Step 1: Understand the Problem
We have a uniform rod of mass \( m \), length \( L \), and cross-sectional area \( A \) that is rotating about one of its ends with a constant angular velocity \( \omega \). We need to find the increase in its length due to this rotation.

### Step 2: Consider a Small Element of the Rod
Take a small element of the rod at a distance \( x \) from the axis of rotation with a length \( dx \). The mass of this small element can be expressed as:
\[ dm = \frac{m}{L} dx \]
where \( \frac{m}{L} \) is the mass per unit length (denoted as \( \mu \)).

### Step 3: Calculate the Centripetal Force
The centripetal force acting on this small element due to its rotation is given by:
\[ dT = dm \cdot \omega^2 \cdot x = \left(\frac{m}{L} dx\right) \cdot \omega^2 \cdot x \]

### Step 4: Relate Tension to the Length of the Rod
The tension \( T \) at a distance \( x \) must balance the centripetal force of all the elements from \( x \) to \( L \). Therefore, we can integrate the tension from \( x \) to \( L \):
\[ T = \int_{x}^{L} \left(\frac{m}{L} \cdot \omega^2 \cdot x\right) dx \]

### Step 5: Perform the Integration
The integral can be simplified:
\[ T = \frac{m \omega^2}{L} \int_{x}^{L} x \, dx = \frac{m \omega^2}{L} \left[\frac{x^2}{2}\right]_{x}^{L} = \frac{m \omega^2}{L} \left(\frac{L^2}{2} - \frac{x^2}{2}\right) \]
This gives:
\[ T = \frac{m \omega^2}{2L} (L^2 - x^2) \]

### Step 6: Calculate the Increase in Length
Using Young's modulus \( Y \), we know that:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{T}{A}}{\frac{\Delta L}{L}} \]
Rearranging gives us:
\[ \Delta L = \frac{T \cdot L}{A \cdot Y} \]

### Step 7: Substitute for T
Substituting \( T \) into the equation for \( \Delta L \):
\[ \Delta L = \frac{\left(\frac{m \omega^2}{2L} (L^2 - x^2)\right) \cdot L}{A \cdot Y} \]

### Step 8: Integrate to Find Total Increase in Length
To find the total increase in length, we need to integrate \( \Delta L \) over the length of the rod from \( 0 \) to \( L \):
\[ \Delta L = \int_{0}^{L} \frac{m \omega^2}{2A Y} \left(L^2 - x^2\right) dx \]
Calculating this integral gives:
\[ \Delta L = \frac{m \omega^2}{2A Y} \left[L^2 x - \frac{x^3}{3}\right]_{0}^{L} = \frac{m \omega^2}{2A Y} \left[L^3 - \frac{L^3}{3}\right] = \frac{m \omega^2 L^3}{6A Y} \]

### Final Result
Thus, the increase in length of the rod due to rotation is:
\[ \Delta L = \frac{m \omega^2 L^3}{6A Y} \]