Rigid body motion

(Under construction. Last modified: .)

This animation illustrates the so called "Tennis Racket Theorem" (also known as the "Dzhanibekov Effect"). There are many videos, articles and discussions online (eg this and this). But it is still quite mysterious, both theoretically and intuitively.

One considers the motion a free rigid body, whose center of mass is fixed at the origin, consisting of 6 masses: $m_i$ at $\pm \mathbf{x}_i$ (lab coordinates), $i=1,2,3,$ with $\mathbf{x}_i\cdot\mathbf{x}_j=\delta_{ij}$. The angular velocity is $\boldsymbol{\omega}=\sum_i \mathbf{x}_i \times \dot{\mathbf{x}}_i$, so that $\dot{\mathbf{x}}_i=\boldsymbol{\omega}\times\mathbf{x}_i$. The moments of intertia are $I_1=m_2+m_3,\ I_2=m_1+m_3,\ I_3=m_1+m_2.$ If $m_1\gt m_2\gt m_3$, then $I_1\lt I_2\lt I_3$. Thus rotation with $\boldsymbol{\omega}=\mathbf{x}_i=const,$ is admissable, but for $i=2$ it is unstable. In the animation one takes initially $\mathbf{x}_1=(1, 0 ,0)$, $\mathbf{x}_2=(0,1,0),$ $\mathbf{x}_3=(0,0,1)$, and $\boldsymbol{\omega}$ in the $xy$ plane, near $(0,1,0)$, forming angle $\psi$ with $(0,1,0)$.

This is equivalent to the motion of a solid uniform box of dimensions $(a,b,c)=(\sqrt{m_1},\sqrt{m_2},\sqrt{m_3})$, which is what the animation shows.

The plots on the right show the $y$-coordinate of $\mathbf{x}_2$ (red) and the $x$-coordinate of $\mathbf{x}_1$, as a function of time.

Animations were done with Mathematica, following a slight generalization of David Brown's video. We do not use Euler's eqns, which are more suitable for body coordinates. It's quicker to write directly 2nd order ODEs for the $\mathbf{x}_i$'s: \begin{aligned} \ddot{\mathbf{x}}_1 &=-\lVert \dot{\mathbf{x}}_1\rVert^2\,\mathbf{x}_1 -\frac{2m_2}{m_1+m_2}(\dot{\mathbf{x}}_1\cdot\dot{\mathbf{x}}_2)\mathbf{x}_2 -\frac{2m_3}{m_1+m_3}(\dot{\mathbf{x}}_1\cdot\dot{\mathbf{x}}_3)\mathbf{x}_3\\[4pt] \ddot{\mathbf{x}}_2 &=-\lVert \dot{\mathbf{x}}_2\rVert^2\,\mathbf{x}_2 -\frac{2m_3}{m_2+m_3}(\dot{\mathbf{x}}_2\cdot\dot{\mathbf{x}}_3)\mathbf{x}_3 -\frac{2m_1}{m_2+m_1}(\dot{\mathbf{x}}_2\cdot\dot{\mathbf{x}}_1)\mathbf{x}_1\\[4pt] \ddot{\mathbf{x}}_3 &=-\lVert \dot{\mathbf{x}}_3\rVert^2\,\mathbf{x}_3 -\frac{2m_1}{m_3+m_1}(\dot{\mathbf{x}}_3\cdot\dot{\mathbf{x}}_1)\mathbf{x}_1 -\frac{2m_2}{m_3+m_2}(\dot{\mathbf{x}}_3\cdot\dot{\mathbf{x}}_2)\mathbf{x}_2 \end{aligned}

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