Consider the special situation where θ is constant.
That means that the angle of the symmetry axis of the top to the vertical is constant.
If θ is constant, θdot = θddot = 0.
Since both L_{3} and L_{z} are constant, the expression for L_{z} leads to the conclusion that φdot is constant; call this constant φdot = Ω.
The expression for L_{3} leads to the conclusion that ψdot is constant; use the quantity ω_{3} = ψdot + φdot cosθ.
The θ equation becomes, dividing out a common factor sinθ

λ_{1}Ω² cosθ - λ_{3}ω_{3}Ω + MgR = 0

This is a quadratic equation for Ω with roots

Ω = (2λ_{1}cosθ)^{-1}[λ_{3}ω_{3} ± √(λ_{3}²ω_{3}² - 4λ_{1}MgR cosθ)]

Note two things. (i) Since Ω must be real, the factor in the square root must be positive,

λ_{3}²ω_{3}² - 4λ_{1}MgR cosθ ≥ 0

or

ω_{3}² ≥ (4λ_{1}/λ_{3}²)MgR cosθ

Therefore, the top must spin sufficiently fast for steady precession to occur.

(ii) Two values exist for Ω, one large and one small.

A common situation is one where the top spins very rapidly, such that we can take λ_{3}²ω_{3}² >> 4λ_{1}MgR cosθ and expand the square root to obtain approximations for the two values of Ω

Ω ≈ (2λ_{1}cosθ)^{-1}[λ_{3}ω_{3} ± (λ_{3}ω_{3} - 2λ_{1}MgR cosθ/λ_{3}ω_{3})].

This yields the approximations

Ω ≈ λ_{3}ω_{3}/λ_{1}cosθ - MgR/λ_{3}ω_{3}

for the large solution, and

Ω ≈ MgR/λ_{3}ω_{3}

for the small solution.

In the general case, we must allow θ to vary. Recall that θ is the angle between the axis of symmetry and the z-axis, vertical in this case. As the axis of symmetry precesses in φ a variation of θ produces the nodding motion of the axis. This is called nutation, latin for nodding.

It turns out that for this situation, the energy can be expressed as a function of the angle θ alone. This simplifies the analysis of the problem. Begin with E = T + U, Start with the kinetic energy for a symmetric top as given in Equation~(10.105),

T = ½λ1(φdot² sin²θ + θdot²) + ½λ3(ψdot + φdot cosθ)²,

and potential energy U = MgR cosθ.
Use the relations for the constants of the motion
L_{3} = λ_{3}(ψdot + φdot cosθ)

and

L_{z} = &lambda_{1}φdot sin²θ + L_{3}cosθ

to eliminate ψdot and φdot in the expression for the total energy

E = T + U

= ½λ_{1}θdot² + [(L_{z} - L_{3}cosθ)²]/[2&lambda_{1}sin²θ] + L_{3}²/[2λ_{3}] + U = ½&lambda_{1}θdot² + U_{eff}(θ)

where

U_{eff} = (L_{z} - L_{3}cosθ)²/[2λ_{1}sin²θ] + L_{3}²/[2λ_{3}] + MgR cosθ.

We are treating an idealized problem of a frictionless top where the energy is constant, or at least working in an approximation where the friction is small and the energy is approximately constant. Therefore, we can write

θdot² = 2[E - U_{eff}(θ)]/&lambda_{1}

In principal, this can be integrated to find θ as a function of t. It is a little easier if we use the clever substitution u = cosθ which yields udot = -θdot sinθ, or θdot = -udot/sinθ = -udot/√(1 - u²). Upon substituting this into the expression for θdot² we find

udot² = 2(1 - u²)(E - MgRu)/&lambda_{1} - (L_{z} - L_{3}u)²/&lambda_{1}²

or
udot² = f(u).

Turning points of the motion occur for udot=0, or therefore at the roots of the equation f(u) = 0. The equation in udot² can be integrated to yield

t = ∫ du/√(f(u)).

© 2008 Robert Harr