A heavy metallic rod of non-uniform mass distribution, is hanged vertically from a rigid support. The linear mass density varies as `lambda_((x))=k xx x` where x is the distance measured along the length of rod, from its lower end. Find extension in rod due to its own weight.

Answer

To find the extension in a heavy metallic rod of non-uniform mass distribution due to its own weight, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Linear Mass Density**:
The linear mass density of the rod is given by:
\[ \lambda(x) = kx \]
where \( k \) is a constant and \( x \) is the distance from the lower end of the rod.

2. **Determine the Mass of an Element**:
Consider an infinitesimal element of the rod of length \( dx \) at a distance \( x \) from the lower end. The mass \( dm \) of this element can be expressed as:
\[ dm = \lambda(x) \cdot dx = kx \cdot dx \]

3. **Calculate the Weight of the Element**:
The weight \( dW \) of this infinitesimal mass is given by:
\[ dW = dm \cdot g = kx \cdot dx \cdot g \]
where \( g \) is the acceleration due to gravity.

4. **Determine the Force Acting on the Element**:
The force \( F \) acting on the element due to the weight of the rod above it can be calculated by integrating the weights of all the elements above it. The total weight acting on the element at distance \( x \) is:
\[ F = \int_x^L dW = \int_x^L kx' \cdot g \, dx' = kg \int_x^L x' \, dx' \]
Evaluating the integral:
\[ F = kg \left[ \frac{x'^2}{2} \right]_x^L = kg \left( \frac{L^2}{2} - \frac{x^2}{2} \right) = kg \frac{L^2 - x^2}{2} \]

5. **Calculate the Stress**:
The stress \( \sigma \) in the rod at position \( x \) is given by:
\[ \sigma = \frac{F}{A} = \frac{kg \frac{L^2 - x^2}{2}}{A} \]
where \( A \) is the cross-sectional area of the rod.

6. **Relate Stress to Strain**:
Using Young’s modulus \( Y \):
\[ Y = \frac{\sigma}{\epsilon} \implies \epsilon = \frac{\sigma}{Y} \]
The strain \( \epsilon \) can be expressed in terms of the change in length \( \delta L \):
\[ \epsilon = \frac{d\delta L}{dx} \]

7. **Express the Change in Length**:
Therefore, we can write:
\[ d\delta L = \frac{kg (L^2 - x^2)}{2A Y} \, dx \]

8. **Integrate to Find Total Extension**:
To find the total extension \( \delta L \), integrate from \( x = 0 \) to \( x = L \):
\[ \delta L = \int_0^L d\delta L = \int_0^L \frac{kg (L^2 - x^2)}{2A Y} \, dx \]
Evaluating the integral:
\[ \delta L = \frac{kg}{2A Y} \left[ L^2x - \frac{x^3}{3} \right]_0^L = \frac{kg}{2A Y} \left( L^3 - \frac{L^3}{3} \right) = \frac{kg L^3}{6A Y} \]

9. **Final Result**:
The total extension in the rod due to its own weight is:
\[ \delta L = \frac{kg L^3}{6A Y} \]