Archives for February 21st, 2017 | Open System

We will start with a simple road. Like that in Nebraska in the last post,

except that we take a nice blue sky with some puffy clouds:

Instead of T-handle (as in Taming the T-handle ) we take the nice, asymmetric, but as much symmetric as possible, spinning top, with principal moments of inertia $I_1=1,I_2=2,I_3=3.$
The x-axis, with the smallest moment of inertia, is along the two bronze spheres. The y-axis, with the middle moment of inertia, is along blue-red line. The z-axis, whose moment of inertia is the sum of the other two, is vertical.

The vertical z-axis is the natural axis to try to spin the thing. Imagine our top is floating in space, in zero gravity. We take the z-axis between our fingers, and spin the device. If our hand is not shaking too much, our top will nicely spin about the z-axis. This is the most stable axis for spinning.

The corresponding solution of Euler’s equation is $\vec{\omega}=(0,0,\omega_3),$ where $\omega_3$ is a constant. The solution of the attitude matrix equation is $Q(t)=\exp(W(t\vec{\omega})).$
The square of angular momentum vector is $I_3^2\omega_3^2,$ the doubled kinetic energy is $I_3\omega_3^2,$ the parameter $d$ that we were using in the previous notes is $1/I_3.$ The parameter $m$ is just zero, $m=0.$ In what follows for simplicity we will take $\omega_3=1.$

We will start with describing the spin history in $\mathbf{R}^3$ using stereographic projection described in Chiromancy in the rotation group.

In fact, we already did it in Dzhanibekov effect – Part 1, and Dzhanibekov effect – Part 2, but that was at the very beginning of this series, and we did not know yet what we were doing! So, now we do it again, in a more “legal” way.

Rotation about an axis $\mathbf{k}$ by an angle $\theta$ can be described by (see Eq. (4) in Putting a spin on mistakes) an $\mathrm{SU}(2)$ matrix:

(1) $\begin{equation*}U(t)=\exp(i\frac{t}{2}\vec{k}\cdot \vec{s})=\cos\frac{t}{2}I+i\sin\frac{t}{2}(\vec{k}\cdot\vec{s}).\end{equation*}$

In fact in Putting a spin on mistakes I have made a mistake that I corrected only at this moment, when checking it again: I have forgotten the imaginary $i$ in the formula!

We will be rotating about the z-axis, so we take $\mathbf{k}=(0,0,1).$ Then $\vec{k}\cdot\vec{s}=s_3.$ Therefore

(2) $\begin{equation*}U(t)=\begin{bmatrix}\cos t/2+i\sin t/2&0\\0&\cos t/2 -i\sin t/2\end{bmatrix}.\end{equation*}$

Comparing with Eq. (1) in Chiromancy in the rotation group, we get

(3) $\begin{equation*} \begin{align} W(t)&=\cos t/2,\\ X(t)&=0,\\ Y(t)&=0,\\ Z(t)&=\sin t/2. \end{align} \end{equation*}$

The stereographic projection (Eq. (4) in Chiromancy in the rotation group ) is

(4) $\begin{equation*} \begin{align} x(t)&=0,\\ y(t)&=0,\\ z(t)&=\frac{\sin t/2}{1-\cos t/2}. \end{align} \end{equation*}$

We can use trigonometric formulas to simplify:

(5) $\begin{equation*}z(t)=\cot t/4.\end{equation*}$

Here is the plot of $\cot t/4$ from $0$ to $2\pi$ :

At $t=0$ we have $z=\infty.$ That is OK, because at $t=0$ the frame of the body coincides with the laboratory frame. The rotation $Q(t)$ is the identity, its stereographic image is at infinity.
At $t=2\pi$ we get $x=y=z=0.$ That is the point representing the matrix $U=-I.$ It also describes the identity rotation in $\mathbf{R}^3.$ We get a straight path, the positive part of the z-axis. It is like the road in Nebraska at the top.