A steel uniform rod of length $2L$ cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young’s modulus for steel, find the extension in the length of the rod.

Let us consider an element of width $d r$ at a distance $r$ from the given axis of rotation as shown in the diagram.

Suppose tension in the rod at $r$ and $d r$ distance are $\mathrm{T}(r)$ and $\mathrm{T}(r+d r)$

Centripetal force acting on small length element

$\mathrm{F} =d m r \omega^{2} \quad\left[\because \mathrm{F}_{\mathrm{C}}=m r \omega^{2}\right]$

$\therefore \mathrm{F} =\mu r \omega^{2} d r$

${\left[\because \mu=\frac{d m}{d r} \Rightarrow d m=\mu d r\right]}$

$\therefore \mathrm{T}(r)-\mathrm{T}(r+d r)=\mu r \omega^{2} d r$

${[\because \mathrm{T}(r)-\mathrm{T}(r+d r)=\text { resultant force } \mathrm{F}]}$

$\therefore-d \mathrm{~T}=\mu r \omega^{2} d r$

Centripetal force and tension force are opposite to each other hence negative sign present.

Integrating on both side,

$\quad \int_{\mathrm{T}=\mathrm{T}}^{\mathrm{T}=0} d \mathrm{~T}=\mu \omega^{2} \int_{r}^{l} r d r$

$\quad-[\mathrm{T}]_{\mathrm{T}}^{0}=\mu \omega^{2}\left[\frac{r^{2}}{2}\right]_{r}^{l}$

$\therefore-[0-\mathrm{T}]=\frac{\mu \omega^{2}}{2}\left[l^{2}-r^{2}\right]$

$\therefore \mathrm{T}=\frac{\mu \omega^{2}}{2}\left(l^{2}-r^{2}\right)$

Now if $\Delta r$ is the extension in the element of length $d r$