Solution

Measure

θ

from the upward vertical, let

e_{3}

point from the support along the symmetry axis, and use the support as the reference point for all moments of inertia. Choose

e_{2} = (z - cos θ e_{3}) / sin θ

and

e_{1} = e_{2} \times e_{3}

. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal Euler angles rotations add as vectors:

ω = \dot{θ} e_{1} + \dot{ϕ} z + \overset{χ}{˙} e_{3} .

(1)

Resolving the vertical direction gives

ω = (\dot{θ}, \dot{ϕ} sin θ, \dot{ϕ} cos θ + \overset{χ}{˙}), J = (I_{1} \dot{θ}, I_{1} \dot{ϕ} sin θ, I_{3} [\dot{ϕ} cos θ + \overset{χ}{˙}]) .

(2)

A convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle

χ

. For one conventional choice they are

\dot{θ} cos χ + \dot{ϕ} sin θ sin χ

and

- \dot{θ} sin χ + \dot{ϕ} sin θ cos χ

. Since

I_{1} = I_{2}

, that rotation changes neither kinetic energy nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.^[ref]

For a frictionless fixed support, the support does no work and gravity is conservative, so the total energy is constant. The heavy symmetric top Lagrangian is

L = \frac{I _{1}}{2} (\dot{θ}^{2} + \dot{ϕ}^{2} sin^{2} θ) + \frac{I _{3}}{2} (\overset{χ}{˙} + \dot{ϕ} cos θ)^{2} - m g h cos θ .

(3)

The two cyclic coordinates

χ, ϕ

give

p_{χ} = J_{3} = I_{3} (\overset{χ}{˙} + \dot{ϕ} cos θ), p_{ϕ} = J_{z} = I_{1} \dot{ϕ} sin^{2} θ + J_{3} cos θ .

(4)

Equivalently, gravitational torque has no component along the vertical or symmetry axis. Eliminate

\dot{ϕ} = (J_{z} - J_{3} cos θ) / (I_{1} sin^{2} θ)

and the axial spin from

E = T + V

E = \frac{I _{1}}{2} \dot{θ}^{2} + U_{eff} (θ), U_{eff} = \frac{( J _{z} - J _{3} cos θ ) ^{2}}{2 I _{1} sin ^{2} θ} + m g h cos θ + \frac{J _{3}^{2}}{2 I _{3}} .

(5)

This is the heavy symmetric top reduction at fixed

J_{3}, J_{z}

. A constant inclination must satisfy the reduced equation of motion

I_{1} \ddot{θ} = - U_{eff}^{'}

, not merely have zero instantaneous velocity. Differentiation at fixed conserved momenta gives, for

0 < θ < π

and

Ω = \dot{ϕ}

U_{eff}^{'} = sin θ [J_{3} Ω - I_{1} Ω^{2} cos θ - m g h] .

(6)

Therefore uniform precession of a heavy symmetric top requires

I_{1} cos θ Ω^{2} - J_{3} Ω + m g h = 0.

(7)

For a prescribed upward inclination

0 < θ < π /2

, real precession rates exist only if the quadratic discriminant is nonnegative:

Ω_{\pm} = \frac{J _{3} \pm J _{3}^{2} - 4 I _{1} m g h c o s θ}{2 I _{1} c o s θ}, ∣ J_{3} ∣_{m i n} (θ) = 2 I_{1} m g h cos θ .

(8)

For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is

2 I_{1} m g h

; strict inequality gives gyroscopic stabilization of an inverted symmetric top. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As