Euler angles for a symmetric top

With the convention in which $\theta$ is inclination, $\varphi$ is precession, and $\psi$ is body-axis spin, an axisymmetric body's kinetic energy is

For a fixed point a distance $a$ from the centre of mass and inclination measured from the upward vertical,The spin and precession angles are cyclic.

This is the Lagrangian of a Lagrange top, one of the Lagrange, Euler, and Kovalevskaya tops.

A Lagrange top is an axisymmetric rigid body with one point fixed on its symmetry axis, moving in a uniform gravitational field with its center of mass on that axis.

For a Lagrange top whose inclination $\theta$ is measured from the upward vertical, steady precession at angular speed $\Omega$ obeysThus steady precession at a prescribed $0<\theta <\pi /2$ exists precisely when

After fixing axial and vertical angular momenta, the heavy symmetric top has energy $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A fixed angle strictly between zero and $\pi$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$. Thus $P(u)=2A(1-u^2)(E-V_{\mathrm{eff}}(u))$ has a repeated interior root. A vertical axis belongs to the excluded endpoints: the upright stationary top has $P(u)=2Amga(1-u{)}^2(1+u)$, whose only repeated root is one. An assertion of an interior common root therefore needs the nonvertical hypothesis.

With transverse and axial moments $I,J$, the Hamiltonian is

Fixing the cyclic momenta $p_{\psi }=\lambda$ and $p_{\varphi }=\mu$ leaves the canonical pair $(\theta ,p_{\theta })$ and a one-degree-of-freedom Hamiltonian with effective potential

At the upright spinning state, $p_{\varphi }=p_{\psi }=\lambda$. The reduced potential is locally minimized at zero inclination whenso sufficiently large axial angular momentum stabilizes the gravitationally inverted top.