Spinning Top

Simple example from The Theoretical Minimum

I recently re-read The Theoretical Minimum - Classical mechanics, an excellent book. The authors give an example of the use of the Hamiltonian formulation of mechanics to illustrate how much easier it makes things (compared to the Newtonian formulation) and that inspired me to have a go myself.

The example they give in the book is of a negligible mass charged sphere spinning in a magnetic field pointing in the z direction. In this case the Hamiltonian $H = T + V$ is proportional to $L_z$ where $\mathbf{L}$ is the angular momentum $L_x \hat{i} + L_y \hat{j} + L_z \hat{k}$. Let's say $H=\omega^2L_z$. The book explains that for any function $G(q,p)$ of the generalized coordinates $q_i$ and the conjugate momenta $p_i$ that
$$
\dot{G} = \{G,H\}
$$
i.e the time derivative of $G$ is the poisson bracket of $G$ and the Hamiltonian. And indeed this does make the solution simple because the components of angular momentum satisfy
$$
\begin{align}
\{L_x,L_z\} &= -L_y\\
\{L_y,L_z\} &= L_x\\
\{L_z,L_z\} &= 0\\
\end{align}
$$
and noticing that the LHS of the above equations are $\dot{L_i}/\omega^2$ we see that $\mathbf{L}$ stays the same magnitude but precesses around the z-axis.

Something more complicated: A spinning top

That was easy, I thought, let's see if I understand this correctly!  The example I chose was a spinning top and I fully expected it to turn out as trivial as the example Leonard Susskind and George Habrovsky provided.  I got there in the end but it was a lot more work and I found it was sooo easy to make a mistake that led to pages and pages of calculations thrown in the bin!

Anyway, here goes.  Our goal is going to be to explain why spinning tops precess, and why they don't fall down, and we're going to do this using Hamiltonian mechanics.

We're going to use spherical coordinates with the basis vectors $\hat{r}, \hat{\theta}, \hat{\phi}$ and the following identities which you can convince yourself of using an orange and a marker pen

$$
\begin{matrix}
\frac{\partial \hat{r}}{\partial r} = 0 &
\frac{\partial \hat{\theta}}{\partial r}= 0 &
\frac{\partial \hat{\phi}}{\partial r} = 0 \\
\frac{\partial \hat{r}}{\partial \theta} = \hat{\theta} &
\frac{\partial \hat{\theta}}{\partial \theta} = -\hat{r} &
\frac{\partial \hat{\phi}}{\partial \theta} = 0 \\
\frac{\partial \hat{r}}{\partial \phi} = sin\theta \hat{\phi}&
\frac{\partial \hat{\theta}}{\partial \phi} = cos\theta \hat{\phi} &
\frac{\partial \hat{\phi}}{\partial \phi} = -sin\theta \hat{r} -cos\theta \hat{\theta}\\
\end{matrix}
$$

Our medium term goal is to work out the kinetic energy $T$, but to do that we're going to need to know a bit about our spinning top.  To make things simple we're assuming that all its mass is around the outside of a circle of radius $l$ and that the centre of that circle is a distance of $r$ from the pivot.  We'll choose the pivot to be the origin, meaning that the centre of the circle is at location $r\hat{r}$ and a small mass $dm$ is at location

$$
\overrightarrow{u} = r\hat{r} + l\ cos\alpha\ \hat\theta + l\ sin\alpha\ \hat \phi
$$
for some $\alpha$ between $0$ and $2\pi$. 

Differentiating this w.r.t. time using the identities above we get
$$
\begin{align}
\overrightarrow{v} &= \dot{r}\hat{r} + r\dot\theta\hat\theta + r\ sin\theta\ \dot\phi\hat\phi\ + \\
&-l\dot\alpha\ sin\alpha\ \hat\theta + l\ cos\alpha\ (-\dot\theta\hat{r} + \dot\phi\ cos\theta \hat\phi)\ + \\
&l\dot\alpha\ cos\alpha \hat\phi + l\ sin\alpha\ (-\dot\phi\ sin\theta\ \hat{r} -\dot\phi\ cos\theta \hat\theta)
\end{align}
$$

Gathering together the coefficients of the unit vectors, and noting that $\dot{r}$ is zero, this becomes

$$
\hat{r}(-l\dot\theta\ cos\alpha - l\dot\phi\ sin\theta\ sin\alpha)\ +\\

\hat{\theta}(r\dot\theta -l\dot\alpha\ sin\alpha - l\dot\phi\ cos\theta\ sin\alpha)\ +\\

\hat{\phi}(r\dot\phi\ sin\theta + l\dot\phi\ cos\theta\ cos\alpha + l\dot\alpha\ cos\alpha)
$$

We're now going to calculate $\frac{1}{2\pi}\int{v^2d\alpha}$ because it's something we'll need later on.  Note that $v^2$ is the sum of the squares of the components, and that we can ignore any terms like $cos\alpha$, or $sin\alpha$, or even $cos\alpha\ sin\alpha$, and that $\frac{1}{2\pi}\int{sin^2\alpha\ d\alpha}$ is just $\frac{1}{2}$

$$
\begin{align}
\frac{1}{2\pi}\int{v^2d\alpha} = &\frac{l^2\dot{\theta}^2}{2} + \frac{l^2\dot{\phi}^2sin^2\theta}{2}\ +\\
&r^2\dot{\theta}^2 + \frac{(l\dot\alpha + l\dot\phi\ cos\theta)^2}{2}\ + \\
&r^2\dot{\phi}^2sin^2\theta + \frac{(l\dot\alpha + l\dot\phi\ cos\theta)^2}{2}
\end{align}
$$

This allows us to work out the kinetic energy $T$ since

$$
\begin{align}
T &= \frac{1}{2}\int{dm\ v^2}\\
&= \frac{1}{2}\int{\rho dl\ v^2}\\
&= \frac{1}{2}\int{\rho l\ d\alpha\ v^2}\\
&= \frac{1}{2}\rho l \int{v^2 d\alpha}\\
&= \frac{m}{2}\frac{1}{2\pi} \int{v^2 d\alpha}\\
\end{align}
$$
where $\rho$ is the mass per unit length and $m$ is the total mass.  Putting this together gives

$$

T = \frac{m(l^2+2r^2)}{4}\dot{\theta}^2 + \frac{m\ sin^2\theta(l^2+2r^2)}{4}\dot{\phi}^2 + \frac{ml^2(\dot\alpha+\dot\phi cos\theta)^2}{2}
$$ 

It's worth pausing for a moment and asking "does this look right?"  (I say that because I made many mistakes before getting to this point.) Well, if $\theta = 0$ and $\dot\theta = 0$ then $\dot\alpha$ and $\dot\phi$ should be interchangeable, and - sure enough - they contribute equally to the final term, which is the only non-zero term.  This term also has the form $ml^2\omega^2/2$ which is what we'd expect.  Conversly, if $\dot\alpha$ and $\dot\phi$ are both zero then only the first term contributes, which also has the form $mk^2\omega^2/2$ as expected.  In the case where $l = 0$ this $k$ is just $r$.  For non-zero $l$, we have $k^2 = r^2 + \frac{1}{2}l^2$ i.e. $l$ is contributing half of what we'd expect from Pythagoras.  However, only two points on the spinning top are as far as $\sqrt{r^2+l^2}$ from the axis of rotation so it seems reasonable.  That all gives us a bit of confidence, and we can move on....

The potential energy $V$ has a much simpler form - it's just proportional to the height of the centre of mass

$$
V = mgr\ cos\theta
$$ 

The Lagrangian $L$ is $T-V$ and the Hamiltonian $H$ is $T+V$, but the latter is more useful if written in terms of the conjugate momenta instead of $\dot\theta, \dot\phi, \dot\alpha$, which are

$$
\begin{align}
p_\theta &\equiv \frac{\partial L}{\partial \dot\theta} = \frac{m}{2}(l^2+2r^2)\dot\theta \\
p_\phi &\equiv \frac{\partial L}{\partial \dot\phi} = \frac{m}{2}sin^2\theta(l^2+2r^2)\dot\phi + ml^2(\dot\alpha+\dot\phi cos\theta)cos\theta \\
p_\alpha &\equiv \frac{\partial L}{\partial \dot\alpha} = ml^2(\dot\alpha+\dot\phi cos\theta)\\
\end{align}
$$

We need to do a bit of reorganization to get $H$ in the format we want, but after a bit of work we get
$$
H = \frac{p_\theta^2}{A} + \frac{(p_\phi - cos\theta\ p_\alpha)^2}{sin^2\theta\ A} + \frac{p_\alpha^2}{2B} + mgr\ cos\theta
$$

where $A=m(l^2+2r^2)$ and $B=ml^2$.  This is where it starts to get interesting because we can apply the general Hamiltonian equations of motion

$$
\dot{q_i} = \frac{\partial H}{\partial p_i} \\
\dot{p_i} = -\frac{\partial H}{\partial q_i}
$$

Now since $\alpha$ and $\phi$ don't appear in $H$ we can safely say that $p_\alpha$ and $p_\phi$ are conserved.  Let's see if we can explain precession!  Well, rewriting $p_\phi$ we get

$$
p_\phi = \frac{m}{2}sin^2\theta(l^2+2r^2)\dot\phi + p_\alpha cos\theta
$$

This tells us that if $\dot\phi$ starts off zero, then as $\theta$ increases $\dot\phi$ must increase because $cos\theta$ decreases and everything else in the equation above is constant.  This is precession explained in an equation!

What about stability?  Why don't spinning tops fall over?  Looking again at our formula for $H$, which is always conserved, we see that $p_\theta$ can only increase if the sum of the other terms decreases.  As $\theta$ increases the potential term $mgr\ cos\theta$ does decrease, but what happens to the second term $\frac{(p_\phi - cos\theta\ p_\alpha)^2}{sin^2\theta\ A}$?

Consider the usual case where $\theta$ starts off at (or near) zero.  From their definitions $p_\alpha$ must be equal (or nearly equal) to $p_\phi$.  This means that the second term is actually a constant times $(1-cos\theta)^2/sin^2\theta$.  For small $\theta$ we can approximate this by $\theta^2/4$ and $cos\theta$ by $1-\theta^2/2$, and so the top is stable provided:

$$
\frac{p_\alpha^2}{A}\frac{\theta^2}{4} > mgr\frac{\theta^2}{2}
$$

which, with a bit of rearranging, gives

$$
\omega^2 > \frac{gr(2 + 4\frac{r^2}{l^2})}{l^2}
$$

where $\omega$ is the spin rate, or more pedantically $\dot\alpha + \dot\phi cos\theta$.  It's worth factoring out a "shape constant" $k = \frac{r}{l}$ to get

$$
\omega > \sqrt{\frac{gk(2 + 4k^2)}{l}}
$$

In other words, to get stability you have to make it short and squat, and it helps to make the whole thing bigger, or take it somewhere with a lower gravitational field.  Yay!