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
This is a quadratic equation for Ω with roots
Note two things. (i) Since Ω must be real, the factor in the square root must be positive,
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
This yields the approximations
for the large solution, and
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),
to eliminate ψdot and φdot in the expression for the total energy
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
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
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