A uniform rod of length L is rotated in a horizontal plane about a...

Given, length of uniform $\operatorname{rod}=L$ angular speed $=\omega$ and density of rod $=\rho$ When rod is rotated in a horizontal plane, centrifugal force is responsible for increasing the length of rod, which is also equal to stress. If $d l$ be the change in length of small element $d x$ due to rotation, by application of force $d F$.
Young's modulus, $$\begin{aligned}& Y=\frac{\text { stress }}{\text { strain }}=\frac{d F / A}{d l / x}\\ \Rightarrow \quad & d l=\frac{d F \cdot x}{A Y}\\ \Rightarrow \quad & d F=\frac{A Y d l}{x} \end{aligned}$$Where, $d F=$ centrifugal force due to mass of the element of length, $d x$. $$\begin{aligned}d F & =d m \cdotx \omega^2 \\ d F \text{ and } =\frac{m}{L} \cdotx d x \omega^2 \\ & \quad \ldots \text { (ii) }\\ & {\left[\because \text { Change in mass }d m=\frac{m}{L}d x\right] } \end{aligned}$$From Eqs. (i) and (ii), we get $$\begin{aligned}\frac{A Y d l}{x} \text{ and } =\frac{m}{L} \cdotx d x \omega^2 \\ \Rightarrow \quad d l \text{ and } =\frac{m x^2 \omega^2}{L A Y} \cdotd x \end{aligned}$$⇒ Total change in length, $$\begin{aligned}l \text{ and } =\intd l=\int_0^L \frac{m x^2 \omega^2}{L A Y}d x \\ \text{ and } =\frac{m \omega^2 L^2}{3 A Y}=\frac{m}{A L} \cdot \frac{\omega^2 L^3}{3 Y}=\frac{\rho \omega^2 L^3}{3 Y} \end{aligned}$$