Lagrange Equations for Top with One Fixed Point

We can analyze the motion of a spinning top using the Lagrange equations for the Euler angles. Let us assume that the top has its lowest point (tip) fixed on a surface. We will use the fixed point as the origin. The rotation about the origin will be described by the Euler angles so that all the kinetic energy is contained in the rotation.

\bgroup\color{black}$\displaystyle T=T_{rot} $\egroup

For a symmetric top, we can immediately write things in terms of the rotation about principal axes of inertia.

\bgroup\color{black}$\displaystyle T={1\over 2}\sum\limits_i I^{(i)}\omega_i^2 $\egroup

(Remember we have three principal moments of inertia but they don't make up a vector.) We have already written \bgroup\color{black}$ \omega$\egroup in terms of the Euler angles.

\bgroup\color{black}$ \displaystyle \vec{\omega}=(\dot{\phi}\sin\theta\sin\psi+\...

Recalling that \bgroup\color{black}$ \theta$\egroup is the angle between the Inertial \bgroup\color{black}$ z$\egroup axis and the \bgroup\color{black}$ z$\egroup axis in the rotating frame, the potential energy can be written,

\bgroup\color{black}$\displaystyle U=mgh\cos\theta $\egroup

where \bgroup\color{black}$ h$\egroup is the height of the center of mass of the top above the fixed tip.
Note that we are assuming that the symmetry axis of the top is the \bgroup\color{black}$ z$\egroup axis so that \bgroup\color{black}$ I^{(2)}=I^{(1)}\equiv I^{(12)}$\egroup. We can now write the kinetic energy in terms of the Euler angles.

  $\displaystyle T= {1\over 2} I^{(12)}\left(\omega_1^2+\omega_2^2\right)+{1\over 2}I^{(3)}\omega_3^2$    
  $\displaystyle \omega_1^2= \left(\dot{\phi}\sin\theta\sin\psi+\dot{\theta}\cos\psi\right)^2$    
  $\displaystyle \omega_1^2= \dot{\phi}^2\sin^2\theta\sin^2\psi+2\dot{\phi}\dot{\theta}\sin\theta\sin\psi\cos\psi+\dot{\theta}^2\cos^2\psi$    
  $\displaystyle \omega_2^2= \left(\dot{\phi}\sin\theta\cos\psi-\dot{\theta}\sin\psi\right)^2$    
  $\displaystyle \omega_2^2 = \dot{\phi}^2\sin^2\theta\cos^2\psi-2\dot{\phi}\dot{\theta}\sin\theta\sin\psi\cos\psi+\dot{\theta}^2\sin^2\psi$    
  $\displaystyle \omega_1^2 + \omega_2^2= \dot{\phi}^2\sin^2\theta +\dot{\theta}^2$    
  $\displaystyle \omega_3^2= \left(\dot{\phi}\cos\theta+\dot{\psi} \right)^2$    
  $\displaystyle T= {1\over 2} I^{(12)}\left(\dot{\phi}^2\sin^2\theta +\dot{\theta}^2\right)+{1\over 2}I^{(3)}\left(\dot{\phi}\cos\theta+\dot{\psi} \right)^2$    
  $\displaystyle U= mgh\cos\theta$    
  $\displaystyle L= {1\over 2} I^{(12)}\left(\dot{\phi}^2\sin^2\theta +\dot{\theta...
...1\over 2}I^{(3)}\left(\dot{\phi}\cos\theta+\dot{\psi} \right)^2 - mgh\cos\theta$    

\bgroup\color{black}$\displaystyle {d\ \over dt}\left({\partial L\over\partial\dot{q}_i}\right)-{\partial L\over\partial q_i}=0 $\egroup

Since the Lagrangian does not depend on \bgroup\color{black}$ \phi$\egroup or \bgroup\color{black}$ \psi$\egroup (cyclic), so the momenta are conserved.

\bgroup\color{black}$\displaystyle p_\phi={\partial L\over\partial \dot{\phi}}=\...
...^{(3)}\cos^2\theta\right)\dot{\phi}+I^{(3)}\cos\theta\dot{\psi}= const. $\egroup

This is the angular momentum about the \bgroup\color{black}$ \hat{z}_I$\egroup axis.

\bgroup\color{black}$\displaystyle p_\psi={\partial L\over\partial \dot{\psi}}=I^{(3)}\left(\dot{\psi}+\dot{\phi}\cos\theta\right)=const. $\egroup

This is the angular momentum about the \bgroup\color{black}$ \hat{z}_B$\egroup axis. This is reasonable since one can see that the torque is along the line of nodes. The actual values of \bgroup\color{black}$ p_\phi$\egroup and \bgroup\color{black}$ p_\psi$\egroup are set by initial conditions in the problem.

So \bgroup\color{black}$ p_\phi$\egroup and \bgroup\color{black}$ p_\psi$\egroup are constants of the motion and we can solve the equations for \bgroup\color{black}$ \dot{\phi}$\egroup and \bgroup\color{black}$ \dot{\psi}$\egroup.

$\displaystyle \dot{\phi}$ $\displaystyle ={p_\phi-p_\psi\cos\theta\over I^{(12)}\sin^2\theta}$    
$\displaystyle \dot{\psi}$ $\displaystyle ={p_\psi\over I^{(3)}}-{\left(p_\phi-p_\psi\cos\theta\right)\cos\theta\over I^{(12)}\sin^2\theta}$    

There is a third Lagrange equation but it will be easier to understand the motion of the top by using the total energy equation, along with the two conserved momenta.

$\displaystyle E$ $\displaystyle = {1\over 2} I^{(12)}\left(\dot{\phi}^2\sin^2\theta +\dot{\theta}...
...1\over 2}I^{(3)}\left(\dot{\phi}\cos\theta+\dot{\psi} \right)^2 + mgh\cos\theta$    
$\displaystyle p_\psi$ $\displaystyle = I^{(3)}\left(\dot{\psi}+\dot{\phi}\cos\theta\right)$    
$\displaystyle E$ $\displaystyle = {1\over 2} I^{(12)}\left(\dot{\phi}^2\sin^2\theta +\dot{\theta}^2\right) +{1\over 2}{p_\psi^2\over I^{(3)}} + mgh\cos\theta$    
$\displaystyle E-{1\over 2}{p_\psi^2\over I^{(3)}}$ $\displaystyle = {1\over 2} I^{(12)}\left(\dot{\phi}^2\sin^2\theta +\dot{\theta}^2\right)+ mgh\cos\theta$    
$\displaystyle E-{1\over 2}{p_\psi^2\over I^{(3)}}$ $\displaystyle = {1\over 2} I^{(12)}\dot{\theta}^2 +{\left(p_\phi-p_\psi\cos\theta\right)^2\over 2I^{(12)}\sin^2\theta}+ mgh\cos\theta$    
$\displaystyle E-{1\over 2}{p_\psi^2\over I^{(3)}}$ $\displaystyle = {1\over 2} I^{(12)}\dot{\theta}^2 +V(\theta)=const.$    


This is very much like a central force problem with the mass oscillating back and forth in the potential. \bgroup\color{black}$ \dot{\theta}$\egroup goes to zero at the limits. The motion will be limited between the some angles \bgroup\color{black}$ \theta_1$\egroup and \bgroup\color{black}$ \theta_2$\egroup at which \bgroup\color{black}$ E-{1\over 2}{p_\psi^2\over I^{(3)}}= V(\theta)$\egroup. This oscillation of \bgroup\color{black}$ \theta$\egroup as the angular momentum precesses is called nutation.


Jim Branson 2012-10-21