# 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.

• if you repeat the experiment with a different refill (say an empty one), does the node appear at the same point? Mar 27, 2020 at 8:45
• @SuperfastJellyfish I'll check... Mar 27, 2020 at 8:47
• @Superfast Jellyfish No, I tried with a different refill and the node appears 1 cm farther away. imgur.com/siBVlUw Mar 27, 2020 at 8:58
• I wonder if it is the same effect as this: physics.stackexchange.com/questions/22268/… with your refill acting as a stiff kind of rope. Mar 27, 2020 at 10:30
• @KrishnanandJ: Were you able to see this with your own eyes? If not, this might be due to the rolling shutter effect. Mar 28, 2020 at 7:15

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 $$y(x_n) = Ash(ax_n) + Bsin(ax_n) = 0$$

• You have a mistake in the compuationt - only $sinh(x), cosh(x)$ linear combinations are solutions. The wave-like shape seems to correspond to the $cosh(x)$ part, given that the boundary condition cannot be enforced perfectly.
– Void
Mar 28, 2020 at 10:03
• The equation I got was $\kappa y'''' - \rho \omega ^ 2 y = 0$. $sin(ax)$ and $cos(ax)$ satisfy it. The equation you will get with Lagrangian similar to the one I have, you will get a 4-th order ODE, so a general solution will be a linear combination of 4 functions. Mar 28, 2020 at 10:13

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.