PHY5210 W08

Special Case: Steady Precession

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 L3 and Lz are constant, the expression for Lz leads to the conclusion that φdot is constant; call this constant φdot = Ω. The expression for L3 leads to the conclusion that ψdot is constant; use the quantity ω3 = ψdot + φdot cosθ. The θ equation becomes, dividing out a common factor sinθ

λ1Ω² cosθ - λ3ω3Ω + MgR = 0

This is a quadratic equation for Ω with roots

Ω = (2λ1cosθ)-13ω3 ± √(λ3²ω3² - 4λ1MgR cosθ)]

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

λ3²ω3² - 4λ1MgR cosθ ≥ 0


ω3² ≥ (4λ13²)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 λ3²ω3² >> 4λ1MgR cosθ and expand the square root to obtain approximations for the two values of Ω

Ω ≈ (2λ1cosθ)-13ω3 ± (λ3ω3 - 2λ1MgR cosθ/λ3ω3)].

This yields the approximations

Ω ≈ λ3ω31cosθ - MgR/λ3ω3

for the large solution, and

Ω ≈ MgR/λ3ω3

for the small solution.

General Case: Nutation

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

L3 = λ3(ψdot + φdot cosθ)


Lz = &lambda1φdot sin²θ + L3cosθ

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

E = T + U
= ½λ1θdot² + [(Lz - L3cosθ)²]/[2&lambda1sin²θ] + L3²/[2λ3] + U = ½&lambda1θdot² + Ueff(θ)


Ueff = (Lz - L3cosθ)²/[2λ1sin²θ] + L3²/[2λ3] + 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 - Ueff(θ)]/&lambda1

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)/&lambda1 - (Lz - L3u)²/&lambda1²
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