$A$ 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 its center with angular velocity $\omega$. If $Y$ is the Young's modulus for steel,find the total extension in the length of the rod.

(D) Let the rod have length $2L$,mass $M$,and cross-sectional area $A$. The linear mass density is $\mu = \frac{M}{2L}$.
Consider an element of length $dr$ at a distance $r$ from the center of rotation.
The mass of this element is $dm = \mu dr$.
The centripetal force required for this element is $dF = (dm) r \omega^2 = \mu \omega^2 r dr$.
This force is provided by the difference in tension $T(r)$ across the element: $dT = -dF = -\mu \omega^2 r dr$.
Integrating from $r$ to $L$ (where tension is zero at the end $r=L$):
$\int_{T(r)}^{0} dT = -\int_{r}^{L} \mu \omega^2 r dr \Rightarrow -T(r) = -\mu \omega^2 \left[ \frac{r^2}{2} \right]_r^L \Rightarrow T(r) = \frac{\mu \omega^2}{2} (L^2 - r^2)$.
The extension $d(\Delta L)$ in the element $dr$ is given by $d(\Delta L) = \frac{T(r) dr}{AY}$.
The total extension $\Delta L$ for one half of the rod (from $0$ to $L$) is:
$\Delta L = \int_0^L \frac{\mu \omega^2}{2AY} (L^2 - r^2) dr = \frac{\mu \omega^2}{2AY} \left[ L^2 r - \frac{r^3}{3} \right]_0^L = \frac{\mu \omega^2}{2AY} \left( L^3 - \frac{L^3}{3} \right) = \frac{\mu \omega^2 L^3}{3AY}$.
Substituting $\mu = \frac{M}{2L}$,the extension for one half is $\frac{M \omega^2 L^2}{6AY}$.
Since the rod has two halves,the total extension is $2 \times \frac{M \omega^2 L^2}{6AY} = \frac{M \omega^2 L^2}{3AY}$.