Why does a spinning rod create transverse waves?

Consider a coordinate system with the $x$-axis parallel to the initial position of the rod and let $y(x)$ describe the shape of the spinning rod.

The potential energy of a small piece of the spinning rod is the sum of the elastic energy, $\frac{\kappa y''^2}{2} dx$ (here we assume that $y' \approx 0$) and the energy due to centripetal force $\frac{-\rho \omega^2 y^2}{2}dx$, where $\rho$ is the linear density and $\kappa$ characterises the stiffness of the rod (Young's modulus multiplied by the cross-section area of the rod). Thus $$U = \int_0^L \frac{\kappa y''^2}{2} - \frac{\rho \omega^2 y^2}{2}dx$$ where $L$ is the length of the rod.

In a stable position we must have $\delta U=0$

Also we have $y(0) = 0$

The extremum can be found by calculus of variations - the solution will be $$y(x) = Ash(ax) + Bsin(ax)$$ where $a = (\frac{\omega ^ 2 \rho}{\kappa})^\frac{1}{4}$ and $A, B$ are solutions of a linear system $$\begin{cases} Ash(aL) - Bsin(aL) = 0 \\ Ach(aL) - Bcos(aL) = 0 \end{cases}$$ If $B$ was zero, we would indeed have bent all the way to the tip without the node in a hyperbolic shape. But it is due to this additional $Bsin(ax)$ term that we have a node. Indeed, if your pen was longer, I would say you might have been able to observe multiple nodes, the position of the nodes, $x_n$, given by the equation $$y(x_n) = Ash(ax_n) + Bsin(ax_n) = 0$$