Stability of vertical magnetic chains with only rolling motion

Johannes Schönke

Reply to the Comments of Ingo Rehberg

It is agreed that in an actual experiment the initial motion out of the vertical equilibrium state is likely to be dominated by a rolling motion. Presentations of the corresponding stability analyses for chains of two and three balls can be found in the supplementary information below.

The critical value of alpha = 3 for N = 2 balls is confirmed. For N = 3 balls, the critical value is alpha = 18.55 being only about 6% below the sliding case, for which the corresponding critical value is 19.68. This indicates that for larger values of N the stability thresholds for the sliding and rolling cases are likely to be very close to one another. Therefore, the idealized analysis in the original paper still provides a reasonable approximation and the results can be compared to experimental measurements for chains of sufficient length.

Supplementary information

Case 1: stability of a heavy magnetic ball rolling on a fixed magnetic ball

We use equation (2.4) of the paper for the dimensionless potential energy \[ E=\frac{\alpha}{6}\sum_{i< j}\frac{(\mathbf{n}_i\cdot\mathbf{n}_j)x_{ij}^2-3(\mathbf{n}_i\cdot\mathbf{x}_{ij})(\mathbf{n}_j\cdot\mathbf{x}_{ij})}{x_{ij}^5} +\mathbf{e}_2\cdot\sum_i\mathbf{w}_i\,. \] With the angle \(t\) defined as in Figure 1, and

\[\mathbf{w}_1=(0,0)\,,\] \[\mathbf{n}_1=\mathbf{e}_2=(0,1)\,,\] \[\mathbf{w}_2=\mathbf{x}_{12}=(\sin t,\cos t)\,,\] \[\mathbf{n}_2=(\sin(2t),\cos(2t))\,,\] \[ x_{12}=1\,,\]

we then obtain \[ E=-\frac{\alpha}{12}\Big[\cos(2t)+3\Big]+\cos t\,, \] which leads to the critical \(\alpha\) for stability \[\frac{d^2 E}{d t^2}=\frac{\alpha}{3}\cos(2t)-\cos t\,,\] \[\frac{d^2 E}{d t^2}\bigg|_{t=0}=0\quad\Leftrightarrow\quad\alpha_{\sf c}=3\,.\]

Figure 1: sketch of the system indicating the angle \(t\) in red.
Figure 2: plot of the energy \(E(t)\) for different values of \(\alpha\).

Case 2: stability of two rolling heavy magnetic balls on a fixed magnetic ball

With the same energy as used above and with the angles \(t\) and \(s\) defined as in Figure 3 we have \[ E=-\frac{\alpha}{12}\big[M_{12}+M_{23}+M_{13}\big]+G_2+G_3 \] with \[ M_{12}=\cos(2t)+3\,, \] \[ M_{23}=\cos(2s)+3\,, \] \[ M_{13}=\frac{3\cos\big(\frac{t+3s}{2}\big)\cos\big(\frac{3t+s}{2}\big)-\cos(2t+2s)}{4\cos^3\big(\frac{t+s}{2}\big)}\,, \] \[ G_2=\cos t\,, \] \[ G_3=\cos t+\cos(2t+s)\,. \]
The determinant of the Hessian matrix \(\mathbf{H}=\begin{pmatrix}E_{tt}&E_{ts}\\E_{st}&E_{ss}\end{pmatrix}\) evaluated at \(t=s=0\) is \[ \det \mathbf{H}\big|_{t=s=0}=2-65\alpha/24+323\alpha^2/2304\,, \] the two roots of this polynomial are given by \[ \alpha_{1,2}=(65\pm\sqrt{3579})\,48/323\,, \] the larger root being \(\alpha_1\approx 18.55\). After the vertical solution (\(t=s=0\)) destabilizes below \(\alpha_1\), there are two symmetric (left/right) bend solutions which then destabilize around \(\alpha\approx 17.58\) and the tower crashes down.

Figure 3: sketch of the system indicating the angle \(t\) in red and \(s\) in yellow.