[NOTES/CM-09005] Heavy Symmetrical Top with One Point Fixed

Setting up Lagrangian
The problem of solving for motion of a symmetrical rigid body moving under the influence of gravity with one point fixed is best approached using the Lagrangian formulation. We shall see that the solution can be reduced to quadratures.

Choosing the principal axes as the body axes, the Lagrangian can be written as
\begin{equation}
{\cal L} = \frac{1}{2}( I_1 \omega_1^2 +I_2\omega_2^2 + I_3\omega_3^2) - mgh
\cos\theta .\Label{EQ72}
\end{equation}
where $h$ is the distance of the centre of mass from the fixed point. We will use Euler angles as generalized coordinates. The angular velocity \(\pmb{\omega}= \omega_1, \omega_2, \omega_3\) in terms of generalized coordinates is
\begin{eqnarray} \omega_1&=&\dot{\theta} \cos \psi +\dot{\phi} \sin \theta \sin \psi \\ \omega_2&=& \dot{\phi} \sin \theta \cos \psi -\dot{\theta} \sin \psi \\ \omega_3&=&\dot{ \phi} \cos \theta +\dot{\psi} \end{eqnarray}
Taking $I_1=I_2$, substituting for angular velocity velocity $\vec{\omega}$ in terms of the Euler angles and their time derivatives, the Lagrangian takes the form
\begin{equation} {\cal L} =\frac{1}{2} I_1 \big(\dot{\theta}^2 +\dot{\phi}^2\sin^2\theta\big) + \frac{1}{2}I_3\big(\dot{\psi} + \dot{\phi}\cos\theta\big)^2 - mgh\cos\theta . \end{equation}

Find cyclic coordinates and conserved quantities
The angles $\psi, \phi$ do not appear in the Lagrangian:
\begin{equation} \frac{\partial {\cal L}}{\partial \phi}=0, \qquad \frac{\partial {\cal L}}{\partial \psi}=0. \end{equation}
Hence corresponding generalised momenta are constants of motion and we have
\begin{equation} \pp[{\cal L}]{\dot{\phi}} = \text{const},\, p_\phi, \qquad \pp[{\cal L}]{\dot{\psi}} = \text{const},\, p_\psi. \end{equation}
Differentiating the Lagrangian, we get
\begin{eqnarray} (I_1 \sin^2\theta + I_3\cos^2\theta)\dot{\phi} + I_3\dot{\psi} \cos\theta = p_\phi \Label{EQ76}\\ I_3(\dot{\psi}+\dot{\phi}\cos\theta) = p_\psi \Label{EQ77}. \end{eqnarray}

Solve for velocities in terms of canonical momenta

Solving \eqref{EQ76}- \ref{EQ77} for $\dot{\phi}$ and $\dot{\psi}$ we get
\begin{eqnarray} \dot{\phi}&=& \frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\Label{EQ79}\\ \dot{\psi}&=&\frac{p_\psi}{I_3} - \cos\theta\left( \frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\right).\Label{EQ80} \end{eqnarray}

Obtain Hamiltonian as function of coordinates and momenta
The Hamiltonian, or the total energy, is conserved. Therefore,
\begin{eqnarray} H &=& p_\phi\dot{\phi} +p_\theta \dot{\theta} + p_\psi\dot{\psi} -L\\ &=&\frac{1}{2} I_1 \big(\dot{\theta}^2 +\dot{\phi}^2\sin^2\theta\big) + \frac{1}{2}I_3\big(\dot{\psi} + \dot{\phi}\cos\theta\big)^2 + mgh\cos\theta =E.\Label{EQ78} \end{eqnarray}
The expression for total energy can now be expressed in terms of $\theta, \dot{\theta}$ and constants of motion $p_\phi,p_\psi$. Thus we get
\begin{equation}
E =
\frac{1}{2}I_1\dot{\theta}^2+\frac{(p_\phi-p_\psi\cos\theta)^2}{2I_1\sin^2\theta
} + m g h\cos\theta + \frac{p_\phi^2}{2I_3},\Label{EQ81}
\end{equation}
We introduce \(E\Prime\) by
\begin{equation}
E^\prime = E - \frac{p_\phi^2}{2I_3} \Label{EQ82},
\end{equation}
\(E\Prime\) is a constant of motion and we have
\begin{equation}
E^\prime = \frac{1}{2}I_1\dot{\theta}^2+
\frac{(p_\phi-p_\psi\cos\theta)^2}{2I_1\sin^2\theta} +m gh\cos\theta,\Label{EQ83}.
\end{equation}
The energy is a function of \(\dot{\theta},\theta\) and constants of motion.

Solve for $\dot{\theta}$ and integrate

The equation \eqref{EQ83} is similar to that for a particle moving in effective potential \(V_{eff}\) one dimension.
Solving \eqref{EQ81} for $\dot{\theta}$ gives

\begin{equation}
\Big(\dd[\theta]{t}\Big)=\sqrt{ \frac{2}{I_1}\big(
E^\prime-V_{\text{eff}}(\theta)\big)},\Label{EQ84}
\end{equation}
where
\begin{equation}
V_{\text{eff}}(\theta) =
\frac{(p_\phi-p_\psi\cos\theta)^2}{2I_1\sin^2\theta} + \mu gh
\cos\theta\Label{EQ85}.
\end{equation}
%
Integrating \eqref{EQ84} gives $\theta$ as an implicit function of time:
\begin{equation}
t =\sqrt{\frac{I_1}{2}} \int \frac{d\theta}{\sqrt{(E^\prime -
V\text{eff}(\theta))}}. \Label{EQ86}
\end{equation}
The integral in \eqref{EQ86} is an elliptic integral. Knowing $\theta$ as a function of time, the solution for $\phi,\psi$ can be obtained by integrating \eqref{EQ79} and \eqref{EQ80}
\begin{eqnarray}\Label{EQ79A}
\phi&=&\int dt \left(\frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\right)\\
\psi&=&\int dt \left\{\frac{p_\psi}{I_3} - \cos\theta\left(
\frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\right)\right\}.\Label{EQ80A}
\end{eqnarray}
This reduces the problem of getting time dependence of the Euler angles to quadratures.

Solution for $\phi$ and $\psi$ as functions of $\theta$

Alternatively, the angle \(\phi\) and \(\psi\) can be obtained as functions of \(\theta\) as follows. Use expressions for generalized velocities \(\dot{\phi}, \dot{\psi}\) to obtain \(\dd[\phi]{\theta}\) and \(\dd[\psi]{\theta}\) in terms of \(\theta\) and \(\dot{\theta}\)
\begin{equation}
\dd[\phi]{\theta} = \Big(\dd[\phi]{t} \Big)\Big/\Big( \dd[\theta]{t}\Big)
= \Big(\frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\Big)
\Big/ \Big(\dd[\theta]{t}\Big). \Label{EQ87}
\end{equation}
Similarly,
\begin{equation}
\dd[\psi]{\theta} = \Big(\dd[\psi]{t} \Big)\Big/\Big( \dd[\theta]{t}\Big)
= \left(\frac{p_\psi}{I_3} - \cos\theta\Big(
\frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\Big)\right)\left/
\Big(\dd[\theta]{t}\Big)\right. .\Label{EQ88}
\end{equation}
Note that \eqref{EQ84} gives \(\dot{\theta}\) as a function of \(\theta\) and constants of motion. Therefore Eqs. \eqref{EQ87} and \eqref{EQ88} can now be integrated to express $\psi,\phi$ in terms of $\theta$.

For a discussion of the resulting motion for different values of the constants of motion we refer the reader to the books by Landau Lifshitz and by Goldstein.

Qualitative discussion of solution
The exact solution of equation motion requires higher transcendental functions. A lot of information about qualitative behaviour of the top can be obtained by graphical considerations.
It should be noted that \eqref{EQ83} is very similar to equation \begin{equation} E = \frac{1}{2} m \dot{x}^2 + V(x) \end{equation} for motion in one dimension in potential \(V(x)\). Thus the methods available for discussion of motion in one dimension can be used about possible motions of the top from from the plot of \(V_\text{eff}(\theta)\) as function of \(\theta\).