A thin uniform rod of length $\mathcal{l}$ and mass $m$ rotates uniformly with an angular velocity $\omega$ in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance $x$ from the rotation axis.

Choose a small element of width $dx$ at a distance $x$ from one end of the rod. The mass of the element $dm=\frac{m}{\mathcal{l}}dx$. Let $T$ be the tension in the rod at a distance $x$. By Newton’s second law for the motion of element of mass $dm$, we have

$$\begin{gathered} T-(T+dT)=\left(dm\right) {\omega }^{2}x \\ or -dT=\left(\frac{m}{\mathcal{l}}dx\right) {\omega }^{2}x \end{gathered}$$

Integrating above equation, we get

$$\begin{gathered} -{\int }_T^{0}dT=\frac{m{\omega }^2}{\mathcal{l}}{\int }_x^{l}x dx \\ -{\left|T\right|}_T^0=\frac{m{\omega }^2}{2\mathcal{l}}{\left|x^2\right|}_x^l \\ -(0-T)=\frac{m{\omega }^2}{2\mathcal{l}}\left({\mathcal{l}}^2-x^2\right) \\ T=\frac{m{\omega }^2}{2\mathcal{l}}\left({\mathcal{l}}^2-x^2\right) \end{gathered}$$

At $x=0, T=T_{max}=\frac{m{\omega }^2\mathcal{l}}{2}$

At $x=\frac{\mathcal{l}}{2}, T=\frac{3m{\omega }^2\mathcal{l}}{8}$