Towards the road less traveled with spin

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.

In the next posts we will learn how to travel more dangerous, less traveled roads.

8 thoughts on “Towards the road less traveled with spin

    1. “Keep on asking, and you will receive what you ask for. Keep on seeking, and you will find. Keep on knocking, and the door will be opened to you.”

      Later in the evening, in the new note.

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