Reduce the heavy symmetric top to one angle

Take an ideal rigid body with a fixed, frictionless pivot, a symmetry axis e₃ pointing from that pivot toward the center of mass, and equal transverse moments I about the pivot. Let the axial moment about the same pivot be I₃. The center of mass lies distance ℓ along e₃. Use an upward inertial vertical and define θ as the angle from that vertical to e₃. Then gravitational potential is Mgℓ cos θ; its largest value occurs with the center of mass directly above the pivot.

Adopt active body-to-space Euler rotations R = Rz(φ)Rx(θ)Rz(ψ). The kinetic energy separates into transverse motion with rates θ̇ and φ̇ sin θ, and axial angular velocity ω₃ = ψ̇ + φ̇ cos θ. Although ψ̇ describes one coordinate rate, it is not by itself the physical axial spin component. This distinction matters whenever the axis is precessing. The chosen convention fixes the signs and allows all subsequent formulas to be checked against the same geometry.

Neither φ nor ψ appears explicitly in the Lagrangian. Their conjugate momenta are therefore conserved even while θ varies. The momentum pψ equals the axial component of angular momentum, I₃ω₃, so the axial spin stays constant in this ideal model. The momentum pφ equals the component of angular momentum along the inertial vertical. Gravity exerts no torque component around that vertical, while its torque also has no component about the body symmetry axis. These are two different projections, not the full angular-momentum vector.

Solve the pφ equation for the precession rate away from sin θ = 0. Substitute that rate into the energy to obtain a kinetic term Iθ̇²/2 plus an effective potential. The additional constant pψ²/(2I₃) can be subtracted from the total energy without changing θ dynamics. If a denominator appears singular near a vertical orientation, investigate the conserved-momentum numerator and the coordinate chart together. A vertical configuration is not automatically forbidden just because this nonvertical Euler-angle expression is inconvenient.

Allowed inclinations satisfy reduced energy greater than or equal to the effective potential. A simple turning point is an inclination where equality holds and θ̇ vanishes before changing sign. Motion between two such points is nutation. Constant-inclination precession requires a stationary point of the effective potential together with the matching reduced energy. A local minimum supports small oscillations of inclination; a local maximum does not. These are statements at fixed conserved pφ and pψ, so comparing unrelated momenta would change the effective problem.

Inspect the quadratic-function lab near a stable reduced equilibrium. Identify x with a small dimensionless displacement from that equilibrium and positive a with a scaled curvature. Predict that doubling a doubles the tangent slope at the same x and makes the local potential steeper. Check the readouts, then return to the actual Ueff to assess the top. The displayed parabola is only a local expansion: it contains no angular-momentum barriers, no pivot geometry, and no calculated precession or nutation trajectory.