A uniform material rod of length L is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is w. Find increase in length of the rod. It is given that density and Young’s modulus of the rod are r and Y respectively.

Answer

To solve the problem of finding the increase in length of a uniform rod of length \( L \) when it is rotated about a vertical axis through one of its ends with an angular speed \( \omega \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Rod
When the rod is rotated, each small element of the rod experiences a centrifugal force due to the rotation. The centrifugal force acting on an infinitesimal element \( dx \) at a distance \( x \) from the axis of rotation is given by:
\[ dF = \rho A \, dx \cdot \omega^2 x \]
where \( \rho \) is the density of the rod, \( A \) is the cross-sectional area, and \( dx \) is the infinitesimal length of the element.

### Step 2: Calculate the Tension in the Rod
The tension \( T \) at a distance \( x \) from the end of the rod can be expressed as the integral of the centrifugal force from \( x \) to \( L \):
\[ T(x) = \int_x^L dF = \int_x^L \rho A \omega^2 x \, dx \]
This integral will give us the total tension at point \( x \).

### Step 3: Evaluate the Integral
The integral can be evaluated as follows:
\[ T(x) = \rho A \omega^2 \int_x^L x \, dx = \rho A \omega^2 \left[ \frac{x^2}{2} \right]_x^L = \rho A \omega^2 \left( \frac{L^2}{2} - \frac{x^2}{2} \right) \]
Thus, we have:
\[ T(x) = \frac{\rho A \omega^2}{2} (L^2 - x^2) \]

### Step 4: Calculate Stress and Strain
The stress \( \sigma \) at point \( x \) is given by:
\[ \sigma = \frac{T(x)}{A} = \frac{\rho \omega^2}{2} (L^2 - x^2) \]
The strain \( \epsilon \) is related to stress by Young's modulus \( Y \):
\[ \sigma = Y \epsilon \Rightarrow \epsilon = \frac{\sigma}{Y} = \frac{\rho \omega^2}{2Y} (L^2 - x^2) \]

### Step 5: Relate Strain to Change in Length
The change in length \( dL \) of an infinitesimal segment \( dx \) can be expressed as:
\[ dL = \epsilon \, dx = \frac{\rho \omega^2}{2Y} (L^2 - x^2) \, dx \]

### Step 6: Integrate to Find Total Increase in Length
To find the total increase in length \( \Delta L \), we integrate \( dL \) from \( x = 0 \) to \( x = L \):
\[ \Delta L = \int_0^L dL = \int_0^L \frac{\rho \omega^2}{2Y} (L^2 - x^2) \, dx \]
Evaluating this integral:
\[ \Delta L = \frac{\rho \omega^2}{2Y} \left[ L^2 x - \frac{x^3}{3} \right]_0^L = \frac{\rho \omega^2}{2Y} \left( L^3 - \frac{L^3}{3} \right) = \frac{\rho \omega^2 L^3}{2Y} \cdot \frac{2}{3} = \frac{\rho \omega^2 L^3}{3Y} \]

### Final Result
Thus, the increase in length of the rod is:
\[ \Delta L = \frac{\rho \omega^2 L^3}{3Y} \]