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
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 (e1, e2, e3). First, rotate about the z axis by $phi; until the x axis lies in the plane z-e3 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 e3 axis. Call this set of axes (e1', e2', e3') and note that e3' = e3 and e2' = y'. Finally, rotate about the e3' = e3 axis by ψ until e1' and e2' are aligned with e1 and e2.
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, e1' and e2' serve as principal axes equally as well as e1 and e2. Therefore, we can express ω in terms of body axes by transforming zhat = e3cosθ - e1'sinθ yielding
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