PHY6200 W07

Rotational Motion of Rigid Bodies


ω = -φdot sinθ e1' + θdot e2' +(ψdot + φdot cosθ)e3.
L = -λ1φdot sinθ e1' + λ1θdot e2' + λ3(ψdot + φdot cosθ)e3.
L3 = λ3ω3 = λ3(ψdot + φdot cosθ)
Lz = Lzhat = λ1φdot sin²θ + λ3(ψdot + φdot cosθ) cosθ
Lz = λ1φdot sin²θ + L3 cosθ
T = ½ ωT I ω = ½λ1(φdot² sin²θ + θdot²) + ½λ3(ψdot + φdot cosθ)²

Motion of a Spinning Top

U = MgR cosθ
L = T - U = ½λ1(φdot² sin²θ + θdot²) + ½λ3(ψdot + φdot cosθ)² - MgR cosθ
L/∂θ = (d/dt)(∂L/∂θdot)
λ1φdot² sinθcosθ - λ3(ψdot + φdot cosθ)φdot sinθ + MgR sinθ = (d/dt)(λ1θdot) = λ1θddot
L/∂φ = 0 = (d/dt)(∂L/∂φdot)
λ1φdot sin²θ + λ3(ψdot + φdot cosθ)cosθ = pφ = Lz = const.
L/∂ψ = 0 = (d/dt)(∂L/∂ψdot)
λ3(ψdot + φdot cosθ) = pψ = L3 = const.

Special Case: Steady Precession

ω3² ≥ (4λ13²)MgR cosθ
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)).

Since θ lies between 0 and π, and u = cosθ, we see that u must lie in the range -1 to +1. In fact, for a top spinning on a table, θ is restricted to the range from 0 to π/2, meaning that u lies in the range from 0 to 1. The derivative udot is real, and udot² is positive. Therefore, the turning points of the motion occur when θdot is zero, corresponding to udot = 0, or udot² = f(u) = 0. Clearly the turning points of the motion correspond to the solutions for f(u) = 0, under the restriction that u lies in the range from 0 to 1.

The function f(u) is a cubic polynomial in u with either 1, 2, or 3 real roots.

f(u) = 2(1 - u²)(E - MgRu)/&lambda1 - (Lz - L3u)²/&lambda1²

[Draw examples of the three cases.] Either one or two of the roots will lie in the allowed range, and the allowed motion occurs in the region where f(u) is positive. The case of one allowed root corresponds to a fixed value of θ, that is steady precession. Nutation corresponds to the case of two allowed roots, with θ varying between the two limiting values. There are two types of nutation, one where the top steadily precesses forward while nodding up and down, and one where the top precesses both forward and backward while nodding up and down. The first case corresponds to φdot being always positive (or always negative), and in the second case, φdot can be both positive and negative so it sometimes passes through zero. We can use the expression for Lz to write

φdot = (Lz - L3cosθ)/λ1sin²θ

from which we see that if Lz is larger than L3, then φdot can never vanish, meaning that the top always precesses forward. If Lz is less than L3, then φdot can vanish, and the top can precess backward during some parts of the motion.

Chapter 11: Coupled Oscillators and Normal Modes

In chapter 5 we studied systems with a single oscillator, subject to damping and driving forces, learning about the behavior of resonance. In this chapter we will consider systems of coupled oscillators and examine the behavior possible in such systems.

Two Masses and Three Springs

Our model system for this study consists of two masses, m1 and m2, each connected to a fixed support by springs of spring constants k1 and k3, and coupled to each other by a third spring of spring constant k2. There is some equilibrium position where the net force on each mass is zero. Measure the locations of each mass from their corresponding equilibrium position, call them x1 and x2. Then the kinetic energy of the system is

T = ½m1x1dot² + ½m2x2dot²

The potential energy of the system comes from stretching or compressing the springs. The change in length of k1 is x1 and the change in length of k3 is x2. The change in length of k2 is (x1 - x2); if both masses are displaced by an equal amount in the same direction, then k2 doesn't change length. The potential energy is

U = ½k1x1² + ½k2(x1 - x2)² + ½k3x2²

The Lagrangian for the system is

L = T - U = ½m1x1dot² + ½m2x2dot² - ½k1x1² - ½k2(x1 - x2)² - ½k3x2²

Lagrange's equations yield the following equations of motion

mx1ddot = -k1x1 - k2(x1 - x2) = -(k1 + k2)x1 + k2x2 mx2ddot = k2(x1 - x2) -k3x2 = k2x1 - (k2 + k3)x2

This can be written more compactly (and usefully) by introducing matrix notation. Let x be the column vector (x1, x2), then xdot is the column vector (x1dot, x2dot), and xddot is the column vector (x1ddot, x2ddot). The equations of motion can be written in the form

Mxddot = -Kx

where M is the diagonal matrix [(m1, 0), (0, m2)] and K is the matrix [((k1 + k2), -k2), (-k2, (k2 + k3))].

Three Identical Springs and Two Equal Masses

Two Weakly Coupled Oscillators: One Spring Different from the other Two

Lagrangian Approach: The Double Pendulum

The General Case

Three Coupled Pendula

Normal Coordinates*

© 2007 Robert Harr