Why does a spinning rod create transverse waves? I attached a ballpoint pen refill to a DC motor and made it spin very fast. Instead of just turning along its axis, the refill started to wobble around to make a transverse wave. You can see that there is a node located about 10 cm away from the base of the refill. (You can click on the images to see video)

I know that a spinning object tends to spread out its mass away from the axis of rotation. But then, I would expect the tube to bend away from the axis all the way from base to tip. Why would it incline towards the axis instead, near the node and form a wave?

Note that these transverse waves only arise if there is some kind of perturbation. 
I confirmed this by dipping the spinning refill in a glass of water (which would dampen any vibrations) and found that it spins stably along its axis.
 A: 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
$$y(x_n) = Ash(ax_n) + Bsin(ax_n) = 0$$ 
A: My guess is that at high rotation speeds any small defect of excentricity of your refill will generate a rotating force due to rotation unbalance. This force may in turn excite a fundamental flexural mode of your refill with a clamped and a free edge condition. The mode shape (in particular, the position of the nodes) should then depend on the rotation speed. Maybe you could test this hypothesis.
