We derived a set of equations for treating the motion of a rigid body in a rotating frame where the principal axes of the body are fixed. These are the Euler equations:

or in component form

λ_{1}ω_{1}dot - (λ_{2} - λ_{3})ω_{2}ω_{3} = Γ_{1}

λ_{2}ω_{2}dot - (λ_{3} - λ_{1})ω_{3}ω_{1} = Γ_{2}

λ_{3}ω_{3}dot - (λ_{1} - λ_{2})ω_{1}ω_{2} = Γ_{3}

This is a set of coupled, nonlinear differential equations. We solved these for certain special cases only. Now that we've learned the Lagrangian formulation of mechanics, we can learn how to treat them in the Lagrangian formalism. Since the Lagrangian is simply kinetic minus potential energy in an inertial system, we need a way to express the kinetic energy of a rotating body in an inertial system. This is normally accomplished with by defining 3 angular coordinates that specify the orientation of the body relative to fixed spatial coordinates. The 3 angular coordinates are called Euler angles.

Since the principal axes represent a rotating coordinate system, they are non-inertial and a Lagrangian cannot be expressed in terms of them. A fixed, inertial coordinate system is needed, but then we need a way to relate the principal axes to it. The Euler angles relate the principal axes of a rotating object to a fixed coordinate system.

The transformation relates (x,y,z) to (**e**_{1}, **e**_{2}, **e**_{3}).
First, rotate about the z axis by $phi; until the x axis lies in the plane z-e_{3} plane.
Call this set of axes (x', y', z') and note that z' = z.
Second, rotate about the y' axis by $theta; until the z' = z axis is aligned with the e_{3} axis.
Call this set of axes (**e**_{1}', **e**_{2}', **e**_{3}') and note that **e**_{3}' = **e**_{3} and **e**_{2}' = y'.
Finally, rotate about the **e**_{3}' = **e**_{3} axis by ψ until **e**_{1}' and **e**_{2}' are aligned with **e**_{1} and **e**_{2}.

Now we can express the angular velocity of the body in terms of the time derivatives of the Euler angles

For an axially symmetric object, withe the symmetry axis in the 3-direction, **e**_{1}' and **e**_{2}' serve as principal axes equally as well as **e**_{1} and **e**_{2}.
Therefore, we can express ω in terms of body axes by transforming **z**hat = **e**_{3}cosθ - **e**_{1}'sinθ yielding

L_{3} = λ_{3}ω_{3} = λ_{3}(ψdot + φdot cosθ)

L_{z} = **L**⋅**z**hat = λ_{1}φdot sin²θ + λ_{3}(ψdot + φdot cosθ) cosθ

L_{z} = λ_{1}φdot sin²θ + L_{3} cosθ

T = ½ **ω**^{T} **I** **ω** = ½λ_{1}(φdot² sin²θ + θdot²) + ½λ_{3}(ψdot + φdot cosθ)²

U = MgR cosθ

L = T - U = ½λ_{1}(φdot² sin²θ + θdot²) + ½λ_{3}(ψdot + φdot cosθ)² - MgR cosθ

∂L/∂θ = (d/dt)(∂L/∂θdot)

λ_{1}φdot² sinθcosθ - λ_{3}(ψdot + φdot cosθ)φdot sinθ + MgR sinθ = (d/dt)(λ_{1}θdot) = λ_{1}θddot

∂L/∂φ = 0 = (d/dt)(∂L/∂φdot)

λ_{1}φdot sin²θ + λ_{3}(ψdot + φdot cosθ)cosθ = p_{φ} = L_{z} = const.

∂L/∂ψ = 0 = (d/dt)(∂L/∂ψdot)

λ_{3}(ψdot + φdot cosθ) = p_{ψ} = L_{3} = const.

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)).