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Rigid-body dynamics and stability — Examples

1 · Prerequisites

2 · Summary

All numerical formulas on this page use the SI convention of Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top and A steady precession of the heavy symmetric top checked on explicit data: the pivot balance, the two conserved momenta and the reduced energy: $t = T / (1 s)$ is dimensionless for physical elapsed time $T$ ; inertia entries are numerical values in $k g m^{2}$ , angular velocities in $s^{- 1}$ , angular momenta in $k g m^{2} s^{- 1}$ , and energies in joules. The angle paths $φ (t) = t$ and $ψ (t) = 2 t$ therefore have physical rates $1 s^{- 1}$ and $2 s^{- 1}$ , and every trigonometric argument is dimensionless. The escape bound means $T \leq (1 s) 4 \log 100$ ; no measured rates or times are reported.

These examples and counterexamples exercise the definitions and theorems of cm11-rigid-body-dynamics-and-stability on explicit model data. The first example (Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top) declares two inertia matrices, the symmetric $d i a g (1, 1, 2)$ and the asymmetric $d i a g (1, 2, 3)$ , and checks principal-axis spins and their invariants, the explicit non-uniform symmetric-top motion $ω (t) = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 1)$ with conserved angular momentum and energy, and the linearized frequencies of the asymmetric model at its three principal-axis spins. The counterexample (A torque-free motion of a symmetric rigid-material model with constant angular momentum and energy whose body angular velocity is not constant: the invariants do not force a uniform rotation) uses that same explicit motion to refute the claim that constant angular momentum and constant rotational kinetic energy force a constant body angular velocity: the invariants are $C = \frac{17}{4}$ and $E = \frac{9}{8}$ while $ω (\frac{π}{2}) \neq ω (0)$ .

The escape example (Explicit escape from the intermediate-axis spin of the model with inertia diag(1,2,3): the constants, the perturbed invariant level, and the escape-time bound) instantiates the instability of the intermediate-axis spin of the model with $d i a g (1, 2, 3)$ : the constants are $a = \sqrt{3}$ , $c_{0} = \frac{1}{4}$ , $δ_{0} = \frac{1}{4}$ , $ε_{0} = \frac{1}{2}$ ; the perturbed datum with $q = 1 / 400$ lies on the same invariant level at distance below $10^{- 2}$ from the spin, and the motion escapes the ball of radius $\frac{1}{2}$ within $4 \log 100 \approx 18.42$ time units. The last example (A steady precession of the heavy symmetric top checked on explicit data: the pivot balance, the two conserved momenta and the reduced energy) verifies a steady precession of the heavy-top branch on declared parameters $A_{c} = 2$ , $C_{c} = 1$ , $m = d = g = 1$ (so $A = 3$ ) and angles $φ (t) = t$ , $θ = π / 3$ , $ψ (t) = 2 t$ : the pivot balance holds identically in time, the conserved momenta are $a = \frac{7}{2}$ , $b = \frac{5}{2}$ , the reduced energy is $\frac{19}{4}$ , and the heavy-top system holds with constant nutation angle $π / 3$ .

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top

Physical scope

Worked checks inside the torque-free branch of the rigid-material model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope on the declared time window $I_{0} = R$ for both models and two declared inertia matrices: a symmetric model with body inertia matrix diag(1,1,2) and an asymmetric model with diag(1,2,3). The checks verify principal-axis spins and their invariants, an explicit non-uniform symmetric-top motion with conserved angular momentum and energy, and the linearized frequencies at the principal-axis spins of the asymmetric model. No measured value, no concrete body and no experiment is asserted; the numerical data are declared model parameters under the explicit numerical SI convention dimensionless t = T / (one second) in the Example.

Example

Numerical SI convention. Let $T$ denote physical elapsed time and fix $T_{*} = 1 s$ , $M_{*} = 1 k g$ and $D_{*} = 1 m$ . Below $t : = T / T_{*}$ is dimensionless and all quantities are numerical SI values: inertia is divided by $M_{*} D_{*}^{2}$ , angular velocity and its perturbations by $T_{*}^{- 1}$ , angular momentum by $M_{*} D_{*}^{2} / T_{*}$ , and energy by $M_{*} D_{*}^{2} / T_{*}^{2}$ . Primes differentiate with respect to $t$ . Restore physical angular velocity by $ω_{p h y s} (T) = ω (T / T_{*}) / T_{*}$ and angular momentum by $l_{p h y s} (T) = (M_{*} D_{*}^{2} / T_{*}) l (T / T_{*})$ . The displayed numerical rates $1$ , $2$ , $1 / \sqrt{3}$ mean physical rates $1 / T_{*}$ , $2 / T_{*}$ , $1 / (\sqrt{3} T_{*})$ ; $λ_{j}$ is restored by $λ_{j, p h y s} = λ_{j} / T_{*}^{2}$ . Thus $\cos t$ , $\sin t$ , $\cos (t / \sqrt{3})$ and the hyperbolic functions have dimensionless arguments; $t = π / 2$ means $T = π T_{*} / 2$ . For the reconstructed motion use $x_{p h y s} (T, ρ_{p h y s}) = D_{*} x (T / T_{*}, ρ_{p h y s} / D_{*})$ ; the chain rule restores $R_{p h y s}^{T} d R_{p h y s} / d T = [ω_{p h y s}]_{\times}$ and the dimensioned torque-free equation. This convention uses Dimensions and SI unit conventions of the non-relativistic classical mechanics framework and reports no measurement.

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), inherited for finite mass moments and inertial integrals through the rigid-material and torque-balance suppliers, and for the double-cover surjectivity used in the quaternion constructions below. Consider two declared models of the rigid-material branch Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope, each with body inertia matrix diagonal in the declared positively oriented orthonormal basis $(e_{1}, e_{2}, e_{3})$ and no other inertia data, each with time window $I_{0} = R$ (equivalently all physical times $T \in R$ ), each used in the torque-free branch of Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (iv):

Model S (symmetric): $I = d i a g (1, 1, 2)$ in kilogram metres squared in the declared units (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); the principal moments are $I_{1} = I_{2} = 1$ , $I_{3} = 2$ .
Model A (asymmetric): $I = d i a g (1, 2, 3)$ ; the principal moments are $I_{1} = 1 < I_{2} = 2 < I_{3} = 3$ .

Verify the following.

Principal-axis spins. For each model, each $j$ and each $Ω \in R$ , the constant path $ω \equiv Ω e_{j}$ is the body angular velocity of a torque-free motion; along it $l = Ω I_{j} e_{j}$ is constant, $C = Ω^{2} I_{j}^{2}$ and $E = \frac{1}{2} Ω^{2} I_{j}$ . In Model S with $j = 3$ and $Ω = 2$ this gives $l = 4 e_{3}$ , $C = 16$ , $E = 4$ ; in Model A with $j = 1$ and $Ω = 1$ it gives $l = e_{1}$ , $C = 1$ , $E = \frac{1}{2}$ .
A non-uniform symmetric-top motion. For Model S the path $ω (t) = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 1) (t \in R)$ satisfies the torque-free Euler equations $l^{'} = l \times (I^{- 1} l)$ and is the body angular velocity of a torque-free motion of Model S; along it $l (t) = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 2), C = \frac{17}{4}, E = \frac{9}{8}, ∥ ω (t) ∥_{2}^{2} = \frac{5}{4},$ and $ω$ is not constant: $ω (\frac{π}{2}) - ω (0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ .
Linearization of Model A at its principal-axis spins. At the spin $Ω e_{1}$ with $Ω = 1$ the linearized system of Linearization of the torque-free spin of the rigid-material model at a principal-axis spin: the oscillatory/exponential dichotomy and linear stability has $λ_{1} = - \frac{1}{3}$ , $κ_{1} = 1 / \sqrt{3}$ , and the pair $(δ_{2}, δ_{3}) = (\cos (t / \sqrt{3}), - \frac{1}{\sqrt{3}} \sin (t / \sqrt{3}))$ solves the transverse equations; at the intermediate spin $e_{2}$ with $Ω = 1$ the linearized system has $λ_{2} = \frac{1}{3} > 0$ and the solution with $δ_{1} (0) = 1$ , $δ_{3} (0) = 0$ has $δ_{1} (t) = \cosh (t / \sqrt{3})$ , which is unbounded on $[0, \infty)$ ; at the extreme spin $e_{3}$ with $Ω = 1$ one has $κ_{3} = 1$ .

Facts & Assumptions

Given: The two models S and A of the Example with time window $I_{0} = R$ and their declared diagonal inertia matrices, the torque-free branch, and the data $j$ , $Ω$ of each claim.

[L1]

Model data and nondegeneracy: for each model the body inertia matrix $I = \int (∥ ρ ∥_{2}^{2} I_{3} - ρ ρ^{T}) d ν$ is diagonal with the displayed positive entries in the declared basis, and the model is nondegenerate (Inertia matrix, spatial inertia tensor, inertia tensor about a point and moment of inertia about an axis of a rigid-material model clauses (a) and (d)-(e), Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values clauses (a) and (d), The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L2]

Principal-axis spins: for a nondegenerate torque-free model, every axis $j$ , rate $Ω$ and initial orientation there is a torque-free motion with constant body angular velocity $Ω e_{j}$ , and along the motions $l = Ω I_{j} e_{j}$ , $C = Ω^{2} I_{j}^{2}$ , $E = \frac{1}{2} Ω^{2} I_{j}$ (Principal-axis uniform rotations of the torque-free rigid-material model: existence, a fixed spatial direction, the invariants, and the converse classification clauses 1-3).

[L3]

Torque-free equations and invariants: $L_{c} = R l$ , and $l = I ω$ satisfies $l^{'} = l \times (I^{- 1} l)$ , and $⟨ l, l ⟩$ and $\frac{1}{2} ⟨ l, I^{- 1} l ⟩$ are constant along solutions; the latter is $E = \frac{1}{2} ω^{T} I ω$ (Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clauses 2-3, Euler's equations with torque for the rigid-material model, the rotational power identity, the torque-free reduction and frame covariance clause 4).

[L4]

Realisability: for every continuous $ω : R \to R^{3}$ there is a $C^{2}$ rotation path with body angular velocity $ω$ if $ω$ is $C^{1}$ ; and a $C^{2}$ rotation path with any prescribed initial orientation exists by the surjectivity of the quaternion cover (The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval clauses 1-2, The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts and its second remark).

[L5]

Linearization: the derivative of the torque-free spin field at $Ω e_{j}$ and the dichotomy with $λ_{j} = Ω^{2} (I_{a} - I_{j}) (I_{j} - I_{b}) / (I_{a} I_{b})$ and $κ_{j} = \sqrt{∣ λ_{j} ∣}$ , with the explicit $\cos, \sin$ and $\cosh, \sinh$ solutions (Linearization of the torque-free spin of the rigid-material model at a principal-axis spin: the oscillatory/exponential dichotomy and linear stability clauses 1-2).

[L6]

Calculus: $(\sin)^{'} = \cos$ , $(\cos)^{'} = - \sin$ , $(\sinh)^{'} = \cosh$ , $(\cosh)^{'} = \sinh$ , the sum, product and chain rules, $\sin^{2} + \cos^{2} = 1$ , and $\cosh t \geq 1$ with $\cosh t \to + \infty$ ; cross products are bilinear and alternating (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The addition formulas for sine and cosine, The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The cross product is bilinear, alternating, and orthogonal to both factors, The cross product in $R^{3}$ ).

[L7]

Quarter-turn values: $π = 2 γ > 0$ for the smallest positive cosine zero $γ$ , and $\sin (π / 2) = 1$ , $\cos (π / 2) = 0$ (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two, Quarter-turn values and shifts by pi/2 and pi); square roots denote unique nonnegative roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct; instantiate the general statements on the declared diagonal data and check the displayed formulas by differentiation and expansion

1.1givenL1L2algebra

By [L2] applied to each model, each $j$ and each $Ω$ there is a torque-free motion with constant body angular velocity $Ω e_{j}$ , and $l = Ω I_{j} e_{j}$ , $C = Ω^{2} I_{j}^{2}$ , $E = \frac{1}{2} Ω^{2} I_{j}$ along it; for Model S with $I_{3} = 2$ , $j = 3$ , $Ω = 2$ this gives $l = 4 e_{3}$ , $C = (2 \cdot 2)^{2} = 16$ , $E = \frac{1}{2} \cdot 4 \cdot 2 = 4$ , and for Model A with $I_{1} = 1$ , $j = 1$ , $Ω = 1$ it gives $l = e_{1}$ , $C = 1$ , $E = \frac{1}{2}$ , which is claim 1.

1.2givenL1L3L6algebra

For the path $ω$ of claim 2 in Model S, $l = I ω = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 2)$ by [L1], so $l^{'} = (- \frac{1}{2} \sin t, \frac{1}{2} \cos t, 0)$ by [L6], while $l \times (I^{- 1} l) = l \times (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 1)$ has components $(\frac{1}{2} \sin t \cdot 1 - 2 \cdot \frac{1}{2} \sin t, - (\frac{1}{2} \cos t \cdot 1 - 2 \cdot \frac{1}{2} \cos t), \frac{1}{2} \cos t \cdot \frac{1}{2} \sin t - \frac{1}{2} \sin t \cdot \frac{1}{2} \cos t) = (- \frac{1}{2} \sin t, \frac{1}{2} \cos t, 0)$ by the coordinate formula for the cross product in [L6]; hence $l^{'} = l \times (I^{- 1} l)$ , so $l$ solves the torque-free equations and $ω$ satisfies the Euler equations by [L3].

1.3givenL5L6algebra

For claim 3, [L5] gives $λ_{j} = Ω^{2} (I_{a} - I_{j}) (I_{j} - I_{b}) / (I_{a} I_{b})$ ; for Model A and $j = 1$ , with $a = 2$ , $b = 3$ : $λ_{1} = \frac{(2 - 1) (1 - 3)}{2 \cdot 3} = - \frac{1}{3}$ and $κ_{1} = 1 / \sqrt{3}$ ; for $j = 2$ , with the cyclic choice $a = 3$ , $b = 1$ : $λ_{2} = \frac{(3 - 2) (2 - 1)}{3 \cdot 1} = \frac{1}{3}$ and $κ_{2} = 1 / \sqrt{3}$ ; for $j = 3$ , with $a = 1$ , $b = 2$ : $λ_{3} = \frac{(1 - 3) (3 - 2)}{1 \cdot 2} = - 1$ and $κ_{3} = 1$ . In the first case the linearized field of [L5] satisfies $A e_{2} = - \frac{1}{3} e_{3}$ and $A e_{3} = e_{2}$ , and the pair $δ_{2} (t) = \cos (t / \sqrt{3})$ , $δ_{3} (t) = - \frac{1}{\sqrt{3}} \sin (t / \sqrt{3})$ satisfies $δ_{2}^{'} = δ_{3}$ and $δ_{3}^{'} = - \frac{1}{3} δ_{2}$ by [L6], so it solves the transverse linearized equations.

2.1step 1.2givenL1L6L7algebra

The quantities of claim 2: $∥ ω (t) ∥_{2}^{2} = \frac{1}{4} \cos^{2} t + \frac{1}{4} \sin^{2} t + 1 = \frac{1}{4} + 1 = \frac{5}{4}$ and $C = ⟨ l, l ⟩ = \frac{1}{4} \cos^{2} t + \frac{1}{4} \sin^{2} t + 4 = \frac{17}{4}$ by the Pythagorean identity in [L6]; and $E = \frac{1}{2} ω^{T} I ω = \frac{1}{2} (\frac{1}{4} \cos^{2} t + \frac{1}{4} \sin^{2} t + 2 \cdot 1) = \frac{1}{2} (\frac{1}{4} + 2) = \frac{9}{8}$ . The difference $ω (\frac{π}{2}) - ω (0) = (- \frac{1}{2}, \frac{1}{2}, 0) \neq 0$ has norm $\frac{\sqrt{2}}{2}$ , so $ω$ is not constant.

3.1step 1.2step 2.1givenL3L4L6

Since $ω$ is $C^{\infty}$ , hence $C^{1}$ , [L4] supplies a $C^{1}$ rotation path $R : R \to S O (3)$ with body angular velocity $ω$ and any prescribed initial orientation; then $R^{'} = R [ω]_{\times}$ is $C^{1}$ , so $R$ is of class $C^{2}$ , and $x : I_{0} \times R^{3} \to R^{3}$ , $x (t, ρ) : = R (t) ρ$ (that is, $c \equiv 0$ ), is a motion of Model S on its declared window $I_{0} = R$ by Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope clause (ii) with body angular velocity $ω$ and $l$ as in step 1.2. Its balance holds: $L_{c} = R l$ by [L3], so $(R l)^{'} = R^{'} l + R l^{'} = R (ω \times l + l^{'}) = 0$ because $l^{'} = l \times (I^{- 1} l) = l \times ω = - ω \times l$ from step 1.2 and the alternation of the cross product in [L6]; hence the motion is torque-free and realizes claim 2.

4.1step 1.3givenL5L6algebra∎

In the intermediate case $j = 2$ , the same formulas give $A e_{1} = - \frac{1}{3} e_{3}$ and $A e_{3} = - e_{1}$ , so the transverse system is $δ_{1}^{'} = - δ_{3}$ , $δ_{3}^{'} = - \frac{1}{3} δ_{1}$ ; with $δ_{1} (0) = 1$ , $δ_{3} (0) = 0$ the $\cosh, \sinh$ solution of [L5] is $δ_{1} (t) = \cosh (t / \sqrt{3})$ , $δ_{3} (t) = - \frac{1}{\sqrt{3}} \sinh (t / \sqrt{3})$ , and $δ_{1}$ is unbounded on $[0, \infty)$ by [L6]; hence the linearized system at the intermediate spin of Model A has an unbounded solution, which is claim 3.

Remarks

The two models are declared, not measured. The matrices $d i a g (1, 1, 2)$ and $d i a g (1, 2, 3)$ are model parameters in the declared units; no body is asserted to have these principal moments and no measurement is reported.
What the checks exhibit. Claim 2 exhibits a torque-free motion whose body angular velocity is not constant while both invariants and the spatial angular momentum are constant; claim 3 exhibits the oscillatory and the exponentially growing linearized behaviour on explicit data. The nonlinear consequences of the latter are proved in Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat and instantiated on the companion page.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

A torque-free motion of a symmetric rigid-material model with constant angular momentum and energy whose body angular velocity is not constant: the invariants do not force a uniform rotation

Physical scope

A refutation inside the torque-free branch of the rigid-material model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope: on the declared symmetric model with body inertia matrix diag(1,1,2) there is a torque-free motion whose angular momentum about the centre of mass and rotational kinetic energy are both constant while its body angular velocity varies with time. The item refutes only the displayed false statement; it asserts no measurement, no concrete body and no failure of the torque-free equations, of which the exhibited motion is an exact solution.

Statement refuted

Let a model of the rigid-material branch Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope be used in the torque-free branch of Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (iv), and let $x (t, ρ) = c (t) + R (t) ρ$ be a motion of the resulting model on a time window containing $[0, \infty)$ . Suppose that along the motion the angular momentum about the centre of mass $L_{c}$ and the rotational kinetic energy $\frac{1}{2} ω^{T} I ω$ are constant in time. Then the body angular velocity $ω$ of the motion is constant in time.

Facts & Assumptions

Use the numerical SI convention of Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top: $t = T / T_{*}$ with $T_{*} = 1 s$ , inertia entries in $M_{*} D_{*}^{2}$ with $M_{*} = 1 k g$ and $D_{*} = 1 m$ , $ω_{p h y s} (T) = ω (T / T_{*}) / T_{*}$ and energy in $M_{*} D_{*}^{2} / T_{*}^{2}$ . Thus $t = π / 2$ means $T = π T_{*} / 2$ ; constancy or nonconstancy is preserved by these fixed nonzero scales. Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried from the supplier for its integral and quaternion constructions. No measured data enter this counterexample (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

Given: The model S of Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top, namely the rigid-material model with body inertia matrix $d i a g (1, 1, 2)$ in the declared basis, used torque-free, and the path $ω (t) = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 1)$ .

[F1]

Model data: $ν$ is the centred reference mass distribution of model S with body inertia matrix $I = d i a g (1, 1, 2)$ in the declared orthonormal basis, and the model is nondegenerate (Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope clauses (i)-(ii), Inertia matrix, spatial inertia tensor, inertia tensor about a point and moment of inertia about an axis of a rigid-material model clause (a), Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top).

[F2]

The path $ω$ is the body angular velocity of a torque-free motion of model S; along that motion $L_{c}$ is constant with $C = ∥ L_{c} ∥_{2}^{2} = \frac{17}{4}$ by Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clause 1, and the rotational kinetic energy is the constant $E = \frac{9}{8}$ ; and $ω$ satisfies the torque-free Euler equations $l^{'} = l \times (I^{- 1} l)$ with $l = I ω = (\frac{1}{2} \cos t, \frac{1}{2} \sin t, 2)$ (Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top claim 2, Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clauses 2-3).

[F3]

Values of the path: $ω (0) = (\frac{1}{2}, 0, 1)$ and $ω (\frac{π}{2}) = (0, \frac{1}{2}, 1)$ , using $\cos 0 = \sin \frac{π}{2} = 1$ and $\sin 0 = \cos \frac{π}{2} = 0$ with the sine and cosine of Sine and cosine defined by their real power series, the definition of $π$ in Pi as twice the smallest positive zero of cosine, and the exact quarter-turn values of Quarter-turn values and shifts by pi/2 and pi together with the zero values of The derivatives of sine and cosine are cosine and minus sine; the Euclidean norm of the difference is computed with The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ .

Counterexample

technique · direct; instance the hypotheses of the refuted statement on the explicit model and motion of [Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top](/physics/item/ex-cm11-free-top-checks) and exhibit two times at which the body angular velocity differs

1.1givenF1F2

The motion of [F2] is a torque-free motion of model S on $R$ , its angular momentum about the centre of mass is constant (with $∥ L_{c} ∥_{2}^{2} = \frac{17}{4}$ ), and its rotational kinetic energy is constantly $\frac{9}{8}$ ; hence all hypotheses of the refuted statement are satisfied.

2.1step 1.1givenF3algebra

By [F3], $ω (\frac{π}{2}) - ω (0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ , whose Euclidean norm is $\frac{\sqrt{2}}{2} \neq 0$ ; hence $ω (\frac{π}{2}) \neq ω (0)$ , so the body angular velocity of the motion is not constant in time.

3.1step 1.1step 2.1given∎

The hypotheses of the refuted statement hold along the motion by step 1.1, while its asserted conclusion fails by step 2.1; therefore the statement is false.

Remarks

What the counterexample shows and does not show. It shows that conservation of the angular momentum about the centre of mass and of the rotational kinetic energy does not force a constant body angular velocity, so no proof of uniform rotation may rely on those invariants alone. It does not show that the torque-free equations are inconsistent: the exhibited path is an exact solution of them, and the two invariants are precisely the reason the motion is bounded in $l$ and $ω$ .
Where uniform rotation does occur. Constant body angular velocity characterises the principal-axis spins, as proved in Principal-axis uniform rotations of the torque-free rigid-material model: existence, a fixed spatial direction, the invariants, and the converse classification clause 4; the present motion is not one of them because $ω (t)$ is not parallel to a fixed principal direction.

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Explicit escape from the intermediate-axis spin of the model with inertia diag(1,2,3): the constants, the perturbed invariant level, and the escape-time bound

Physical scope

A worked instance of the intermediate-axis instability of Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat on the declared model with body inertia matrix diag(1,2,3) and spin datum $l_{0} = e_{2}$ : the instability constants, a perturbed datum on the same invariant level with the checked proximity bound, and the resulting escape time and escape radius. The numerical data are declared model parameters; no measurement, no concrete body and no experiment is asserted.

Example

Use the numerical SI convention of Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top: $t = T / (1 s)$ is dimensionless, inertia entries are in $1 k g m^{2}$ , and $l, L, q, δ_{0}, ε_{0}$ and all distances in angular-momentum space are numerical values in $1 k g m^{2} s^{- 1}$ . Restore the growth rate by $c_{0, p h y s} = c_{0} / (1 s)$ and the escape time by $T = t (1 s)$ ; $δ_{0} / q$ is dimensionless. The datum below is model data, not an observation.

Adopt the model A of Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top: the rigid-material model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope with body inertia matrix $I = d i a g (1, 2, 3)$ in the declared orthonormal basis, used in the torque-free branch of Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (iv). Its principal moments are $I_{1} = 1 < I_{2} = 2 < I_{3} = 3$ (Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values clauses (a) and (d)). Take the intermediate-axis spin datum $l_{0} : = L e_{2}$ with $L = 1$ , and let $q : = \frac{1}{400}$ and $l^{q} : = (q, \sqrt{1 - 4 q^{2}}, - \sqrt{3} q),$ so that $q = 1 / 400$ and $\sqrt{1 - 4 q^{2}} = \sqrt{39999 / 40000}$ . Verify:

The instability constants of Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat clause 3.

$a = \sqrt{3}, c_{0} = \frac{1}{4}, δ_{0} = \frac{1}{4}, ε_{0} = \frac{1}{2} .$

The perturbed datum lies on the invariant level of the spin. $C (l^{q}) = C (l_{0}) = 1$ and $E (l^{q}) = E (l_{0}) = \frac{1}{4}$ , where $C (z) = ⟨ z, z ⟩$ and $E (z) = \frac{1}{2} ⟨ z, I^{- 1} z ⟩$ are the invariants of Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clause 2.
The perturbed datum is close to the spin. $∥ l^{q} - l_{0} ∥_{2} < \frac{1}{100}$ and $q < \min {δ_{0}, \frac{1}{100}}$ .
The escape. There is a torque-free motion of model A whose $l$ -path has $l (0) = l^{q}$ and satisfies $∥ l (0) - e_{2} ∥_{2} < \frac{1}{100}, ∥ l (t) - e_{2} ∥_{2} \geq \frac{1}{2}$ at some time $0 \leq t \leq 4 \log 100 \approx 18.42$ ; in particular the intermediate-axis spin $e_{2}$ of model A is not Lyapunov stable, with an explicit effective escape at the displayed constants.

Facts & Assumptions

Given: The model A with $I = d i a g (1, 2, 3)$ and principal moments $I_{1} = 1 < I_{2} = 2 < I_{3} = 3$ , the spin datum $l_{0} = e_{2}$ , the number $q = 1 / 400$ and the datum $l^{q}$ of the statement.

[L1]

Model data: $I = d i a g (1, 2, 3)$ is symmetric positive definite in the declared basis, so model A is nondegenerate with the displayed principal moments (Inertia matrix, spatial inertia tensor, inertia tensor about a point and moment of inertia about an axis of a rigid-material model clause (a), Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values clauses (a) and (d), Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope, Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top).

[L2]

Instability statement: for a nondegenerate torque-free model with $I_{1} < I_{2} < I_{3}$ and $L \neq 0$ , the spin $L e_{2}$ is not Lyapunov stable, and with $a$ , $c_{0}$ , $δ_{0}$ , $ε_{0}$ as in Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat clause 3, for every $δ > 0$ and every $q \in (0, \min {δ_{0}, δ})$ whose prescribed datum additionally satisfies $∥ l^{q} - L e_{2} ∥_{2} < δ$ , there is a torque-free motion with $∥ l (0) - L e_{2} ∥_{2} < δ$ and a time $t \leq c_{0}^{- 1} \log (δ_{0} / q)$ with $∥ l (t) - L e_{2} ∥_{2} \geq ε_{0}$ ; the motion's datum is the $l$ -path with $l (0) = l^{q}$ of the statement (Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat clause 3).

[L3]

Invariants and global existence: $C$ and $E$ are constant along every solution of $l^{'} = l \times (I^{- 1} l)$ , which has a unique global maximal solution through every datum (Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clauses 2 and 4).

[L4]

Arithmetic: $\sqrt{3} \cdot \sqrt{3} = 3$ and the nonnegative square root is unique, $\sqrt{4} = 2$ , and for $0 \leq s \leq 1$ the elementary ordered-field inequalities used below hold; squares of nonnegative reals are ordered as their bases are (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

Proof

technique · direct; compute the instability constants from the principal moments, check the invariants and the proximity of the perturbed datum by explicit arithmetic, and instantiate the escape statement

1.1givenL1L4algebra

With $I_{1} = 1$ , $I_{2} = 2$ , $I_{3} = 3$ and $L = 1$ : $1 / I_{1} - 1 / I_{2} = \frac{1}{2}$ , $1 / I_{2} - 1 / I_{3} = \frac{1}{6}$ , so $a^{2} = \frac{1 / 2}{1 / 6} = 3$ and $a = \sqrt{3} > 0$ by [L4]; then $c_{0} = a (1 / I_{2} - 1 / I_{3}) \frac{\sqrt{3}}{2} ∣ L ∣ = \sqrt{3} \cdot \frac{1}{6} \cdot \frac{\sqrt{3}}{2} = \frac{3}{12} = \frac{1}{4}$ by [L4]. Further $1 + a^{2} = 1 + 3 = 4$ , so $δ_{0} = \frac{∣ L ∣}{2 \sqrt{1 + a^{2}}} = \frac{1}{2 \cdot 2} = \frac{1}{4}$ by [L4] and $ε_{0} = \frac{∣ L ∣}{2} = \frac{1}{2}$ , which is claim 1.

2.1step 1.1givenL1L3L4algebra

Writing $s = \sqrt{1 - 4 q^{2}}$ , one has $s^{2} = 1 - 4 q^{2}$ ; hence $C (l^{q}) = q^{2} + s^{2} + 3 q^{2} = 4 q^{2} + 1 - 4 q^{2} = 1$ and $E (l^{q}) = \frac{1}{2} (q^{2} + \frac{s^{2}}{2} + \frac{3 q^{2}}{3}) = \frac{1}{2} (q^{2} + \frac{1 - 4 q^{2}}{2} + q^{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ by [L4]. For $l_{0} = e_{2}$ : $C (l_{0}) = 1$ and $E (l_{0}) = \frac{1}{2} ⟨ e_{2}, I^{- 1} e_{2} ⟩ = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ . This is claim 2, and by [L3] the perturbed datum and the spin datum lie on the same invariant level.

2.2step 1.1givenL4algebra

For the proximity bound: $∥ l^{q} - l_{0} ∥_{2}^{2} = q^{2} + (s - 1)^{2} + 3 q^{2} = 4 q^{2} + (1 - s)^{2}$ , and $1 - s = \frac{1 - s^{2}}{1 + s} = \frac{4 q^{2}}{1 + s} \leq 4 q^{2}$ because $s \geq 0$ ; with $q = \frac{1}{400}$ , $4 q^{2} = \frac{4}{160000} = \frac{1}{40000}$ and $(1 - s)^{2} \leq (4 q^{2})^{2} = \frac{16}{400^{4}} = \frac{1}{16 \cdot 10^{8}} < \frac{1}{10^{9}}$ , so $∥ l^{q} - l_{0} ∥_{2}^{2} \leq \frac{1}{40000} + \frac{1}{10^{9}} < \frac{1}{10000}$ , whence $∥ l^{q} - l_{0} ∥_{2} < \frac{1}{100}$ by [L4] applied to the nonnegative squares. Moreover $q = \frac{1}{400} < \frac{1}{100} = \min {δ_{0}, \frac{1}{100}}$ by step 1.1. This is claim 3.

3.1step 1.1step 2.2givenL2L3algebra∎

Apply clause 3 of Nonlinear stability of the torque-free spin of the rigid-material model about an extreme principal axis, and instability about the intermediate axis, with an explicit escape time and the orientation caveat with $δ = \frac{1}{100}$ and $q = \frac{1}{400}$ , which is admissible by step 2.2 and [L2]: there is a torque-free motion of model A with $∥ l (0) - e_{2} ∥_{2} < \frac{1}{100}$ and a time $0 \leq t \leq c_{0}^{- 1} \log (δ_{0} / q)$ with $∥ l (t) - e_{2} ∥_{2} \geq ε_{0}$ . With the constants of step 1.1, $c_{0}^{- 1} \log (δ_{0} / q) = 4 \log (\frac{1 / 4}{1 / 400}) = 4 \log 100$ and $ε_{0} = \frac{1}{2}$ , so the escape-time bound and escape radius of claim 4 follow; by [L3] the motion's $l$ -path is defined for all times, so the escape time is finite as displayed.

Remarks

The escape is effective. The initial perturbation has $∥ l (0) - e_{2} ∥_{2} < 10^{- 2}$ while the escape radius is $\frac{1}{2}$ at a time at most $4 \log 100 \approx 18.421$ in the declared units; this is a quantitative instance of the statement that the intermediate-axis spin is not Lyapunov stable, not merely a linearized indication.
What is declared. The inertia matrix, the spin datum and the perturbation parameter $q$ are declared model data in the framework's units; no measurement is reported and no concrete body is asserted to have inertia $d i a g (1, 2, 3)$ .

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A steady precession of the heavy symmetric top checked on explicit data: the pivot balance, the two conserved momenta and the reduced energy

Physical scope

A worked steady precession of the heavy symmetric top branch of Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot on declared parameters $A_{c} = 2$ , $C_{c} = 1$ , $m = 1$ , $d = 1$ , $g = 1$ (so the pivot transverse moment is $A = 3$ ) and the explicit angles $φ (t) = t$ , $θ = π / 3$ , $ψ (t) = 2 t$ : the pivot balance is verified identically in time, the two conserved momenta are $a = 7 / 2$ and $b = 5 / 2$ , the reduced energy is $19 / 4$ , and the heavy-top system holds, so the reduction conclusions apply. All formulas use numerical SI values with dimensionless dimensionless t = T / (one second) as defined in the Example; no measurement, no concrete top and no experiment is asserted.

Example

Numerical SI convention. Let $T$ denote physical elapsed time and fix $T_{*} = 1 s$ , $M_{*} = 1 k g$ and $D_{*} = 1 m$ . All formulas below use numerical SI values: $t : = T / T_{*}$ is dimensionless, angles are in radians, positions and $d$ are divided by $D_{*}$ , masses by $M_{*}$ , inertia entries by $M_{*} D_{*}^{2}$ , and $g$ by $D_{*} / T_{*}^{2}$ . Primes denote differentiation with respect to $t$ . Restore physical angular velocity as $ω_{p h y s} (T) = ω (T / T_{*}) / T_{*}$ , angular momentum as $J_{p h y s} (T) = (M_{*} D_{*}^{2} / T_{*}) J (T / T_{*})$ , energy as $(M_{*} D_{*}^{2} / T_{*}^{2}) E$ , and torque as $(M_{*} D_{*}^{2} / T_{*}^{2}) τ (T / T_{*})$ . Thus the displayed rates $φ^{'} = 1$ and $ψ^{'} = 2$ mean physical rates $1 s^{- 1}$ and $2 s^{- 1}$ , and every trigonometric argument is dimensionless. The physical placement is $x_{p h y s} (T, ρ_{p h y s}) = D_{*} x (T / T_{*}, ρ_{p h y s} / D_{*})$ ; its derivatives restore the pivot balance and heavy-top equations with the same SI data. These are unit conventions of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, not measured inputs.

Adopt the heavy symmetric top branch Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot with declared parameters $A_{c} = 2, C_{c} = 1, m = 1, d = 1, g = 1$ in the framework's units (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework), so that the body inertia matrix is $I = d i a g (2, 2, 1)$ in the declared basis and the pivot transverse moment is $A = A_{c} + m d^{2} = 2 + 1 = 3, C = C_{c} = 1.$ Let $θ_{0} : = π / 3$ and consider the angles $φ (t) = t, θ (t) = θ_{0}, ψ (t) = 2 t (t \in R),$ the orientation path $R (t) : = Θ (φ (t), θ (t), ψ (t))$ of The Euler-angle parametrization and the heavy symmetric top system, the centre of mass $c (t) : = d R (t) e_{3}$ , and the motion $x (t, ρ) : = c (t) + R (t) ρ$ . Verify:

It is a motion of the branch satisfying the pivot balance. The pivot constraint $c (t) + R (t) (- d e_{3}) = 0$ holds, and with $ω (t) = (- \frac{\sqrt{3}}{2} \cos (2 t), \frac{\sqrt{3}}{2} \sin (2 t), \frac{5}{2})$ and $γ (t) = R (t)^{T} e_{3}$ , $\tilde{I} ω^{'} + ω \times \tilde{I} ω = - m g d (e_{3} \times γ) (t \in R),$ where $\tilde{I} = d i a g (3, 3, 1)$ ; hence the motion satisfies the adopted balance of Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot clause (v) and the conclusions of Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation clauses 1-2 hold along it.
The conserved momenta and the reduced energy. With $φ^{'} = 1$ , $θ^{'} = 0$ , $ψ^{'} = 2$ , $a = A φ^{'} \sin^{2} θ_{0} + b \cos θ_{0} = \frac{7}{2}, b = C (ψ^{'} + φ^{'} \cos θ_{0}) = \frac{5}{2}, E = U (θ_{0}) = \frac{19}{4} .$
The heavy-top system holds, and $θ$ is constant. The three equations of The Euler-angle parametrization and the heavy symmetric top system hold for the triple, and $A θ^{''} = - U^{'} (θ_{0}) = 0$ , so the motion is a steady precession: $φ^{'} = 1$ and $ψ^{'} = 2$ are constant and the nutation angle is constantly $π / 3$ .

Facts & Assumptions

Given: The branch parameters $A_{c} = 2$ , $C_{c} = 1$ , $m = 1$ , $d = 1$ , $g = 1$ , the derived constants $A = 3$ , $C = 1$ , the angle data $θ_{0} = π / 3$ , $φ (t) = t$ , $ψ (t) = 2 t$ , and the path $R = Θ (φ, θ_{0}, ψ)$ with $c = d R e_{3}$ .

[L1]

Branch kinematics and dynamics: the motions of the branch satisfy the pivot constraint $c = d R e_{3}$ and the balance $J_{p}^{'} = τ_{p}$ with $J_{p} = R \tilde{I} ω$ , $τ_{p} = - m g d (R e_{3}) \times e_{3}$ , so the reduction conclusions of Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation clauses 1-4 hold on any interval carrying an Euler-angle description (Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot clauses (ii)-(v), Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation clauses 1-4).

[L2]

Euler-angle kinematics: for $R = Θ (φ, θ, ψ)$ one has $ω = (- φ^{'} \sin θ \cos ψ + θ^{'} \sin ψ, φ^{'} \sin θ \sin ψ + θ^{'} \cos ψ, ψ^{'} + φ^{'} \cos θ)$ and $γ = R^{T} e_{3} = (- \sin θ \cos ψ, \sin θ \sin ψ, \cos θ)$ (Angular velocity in Euler angles: $ω_{3} = ψ^{'} + φ^{'} \cos θ$ and $ω_{1}^{2} + ω_{2}^{2} = θ^{' 2} + φ^{' 2} \sin^{2} θ$ , Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation clause 2, The coordinate rotation matrices $R_{z} (a)$ and $R_{y} (a)$ : values in $S O (3)$ , composition law, and derivatives).

[L3]

The heavy-top system and its reduction: the three displayed equations of The Euler-angle parametrization and the heavy symmetric top system, and the conclusions $A θ^{''} = - U^{'} (θ)$ , $φ^{'} = (a - b \cos θ) / (A \sin^{2} θ)$ , $ψ^{'} = b / C - (a - b \cos θ) \cos θ / (A \sin^{2} θ)$ of Reduction of the heavy symmetric top system: conserved momenta, one-dimensional $θ$ equation and conserved reduced energy clauses 2-4.

[L4]

Calculus and algebra: $(\sin)^{'} = \cos$ , $(\cos)^{'} = - \sin$ , the sum and product rules, and $\sin^{2} + \cos^{2} = 1$ ; the cross product is bilinear with the coordinate formula and is orthogonal to its factors (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, Parity and the Pythagorean identity for sine and cosine, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The cross product in $R^{3}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L5]

The constant $π = 2 γ$ , where $γ$ is the smallest positive zero of cosine; $\cos π = - 1$ , $\cos 0 = 1$ , $\cos (π / 2) = 0$ , $\sin 0 = 0$ ; cosine is strictly decreasing on $[0, π]$ and sine strictly increasing on $[0, π / 2]$ ; sine and cosine obey their addition formulas; and every nonnegative real has a unique nonnegative square root (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two, Quarter-turn values and shifts by pi/2 and pi, Signs, monotonicity intervals, and ranges of sine and cosine, The addition formulas for sine and cosine, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct; compute the body angular velocity, the vertical direction in body coordinates and the two sides of the pivot balance on the declared data, then read off the conserved momenta and the reduced energy

1.1L4L5algebra

Put $θ = π / 3$ , $c = \cos θ$ and $s = \sin θ$ . Since $0 < θ < π / 2$ , [L5] gives $0 < c < 1$ and $s > 0$ . The addition formulas [L5] and Pythagorean identity [L4] give $\cos π = \cos (3 θ) = 4 c^{3} - 3 c$ ; as $\cos π = - 1$ , $(c + 1) (2 c - 1)^{2} = 0$ . Because $c > 0$ , $c = 1 / 2$ . Then $s^{2} = 1 - c^{2} = 3 / 4$ and $s > 0$ , so the uniqueness of the nonnegative square root [L5] gives $s = \sqrt{3} / 2$ .

2.1givenstep 1.1L2L4algebra

By [L2] with $φ^{'} = 1$ , $θ^{'} = 0$ , $ψ^{'} = 2$ , $θ_{0} = π / 3$ and $ψ (t) = 2 t$ : $ω (t) = (- \sin θ_{0} \cos (2 t), \sin θ_{0} \sin (2 t), 2 + \cos θ_{0}) = (- \frac{\sqrt{3}}{2} \cos (2 t), \frac{\sqrt{3}}{2} \sin (2 t), \frac{5}{2})$ and $γ (t) = (- \frac{\sqrt{3}}{2} \cos (2 t), \frac{\sqrt{3}}{2} \sin (2 t), \frac{1}{2})$ by step 1.1 and [L2].

2.2step 1.1givenL1L4algebra

With $φ^{'} = 1$ , $θ^{'} = 0$ , $ψ^{'} = 2$ , $\sin^{2} θ_{0} = \frac{3}{4}$ and $\cos θ_{0} = \frac{1}{2}$ by [L4]: $b = C (ψ^{'} + φ^{'} \cos θ_{0}) = 1 \cdot (2 + \frac{1}{2}) = \frac{5}{2}$ and $a = A φ^{'} \sin^{2} θ_{0} + b \cos θ_{0} = 3 \cdot \frac{3}{4} + \frac{5}{2} \cdot \frac{1}{2} = \frac{9}{4} + \frac{5}{4} = \frac{7}{2}$ .

3.1step 2.1givenL1L4algebra

Compute both sides of the balance with $\tilde{I} = d i a g (3, 3, 1)$ : by [L4], $ω^{'} = (\sqrt{3} \sin (2 t), \sqrt{3} \cos (2 t), 0)$ , so $\tilde{I} ω^{'} = (3 \sqrt{3} \sin (2 t), 3 \sqrt{3} \cos (2 t), 0)$ , and with $\tilde{I} ω = (- \frac{3 \sqrt{3}}{2} \cos (2 t), \frac{3 \sqrt{3}}{2} \sin (2 t), \frac{5}{2})$ , the coordinate formula for the cross product gives $(ω \times \tilde{I} ω)_{1} = \frac{5 \sqrt{3}}{4} \sin (2 t) - \frac{15 \sqrt{3}}{4} \sin (2 t) = - \frac{5 \sqrt{3}}{2} \sin (2 t)$ , $(ω \times \tilde{I} ω)_{2} = - \frac{15 \sqrt{3}}{4} \cos (2 t) + \frac{5 \sqrt{3}}{4} \cos (2 t) = - \frac{5 \sqrt{3}}{2} \cos (2 t)$ and $(ω \times \tilde{I} ω)_{3} = 0$ . Hence the left side is $(\frac{\sqrt{3}}{2} \sin (2 t), \frac{\sqrt{3}}{2} \cos (2 t), 0)$ . On the right, $e_{3} \times γ = (- γ_{2}, γ_{1}, 0) = (- \frac{\sqrt{3}}{2} \sin (2 t), - \frac{\sqrt{3}}{2} \cos (2 t), 0)$ and $m g d = 1$ , so the right side is the same vector; the balance holds for every $t$ .

3.2step 2.2givenL3L4algebra

The reduced energy: $E = \frac{1}{2} A θ^{' 2} + \frac{(a - b \cos θ_{0})^{2}}{2 A \sin^{2} θ_{0}} + \frac{b^{2}}{2 C} + m g d \cos θ_{0} = \frac{(9 / 4)^{2}}{2 \cdot 3 \cdot (3 / 4)} + \frac{25 / 4}{2} + \frac{1}{2} = \frac{9}{8} + \frac{25}{8} + \frac{4}{8} = \frac{38}{8} = \frac{19}{4}$ , where $a - b \cos θ_{0} = \frac{7}{2} - \frac{5}{4} = \frac{9}{4}$ by step 2.2.

4.1step 2.1step 3.1givenL1

The constraint $c (t) + R (t) (- d e_{3}) = d R (t) e_{3} - d R (t) e_{3} = 0$ holds by the definition of $c$ ; hence with step 3.1 the data constitute a motion of the branch satisfying the adopted pivot balance, and clauses 1-2 of Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation apply.

5.1step 4.1step 2.2step 3.2givenL3L4algebra∎

The three equations of the heavy-top system of [L3] hold for the triple: the first and third are the statements that $a$ and $b$ of step 2.2 are constant, and the second is $0 = A φ^{' 2} \sin θ_{0} \cos θ_{0} - C (ψ^{'} + φ^{'} \cos θ_{0}) φ^{'} \sin θ_{0} + m g d \sin θ_{0} = 3 \cdot \frac{\sqrt{3}}{4} - \frac{5 \sqrt{3}}{4} + \frac{2 \sqrt{3}}{4}$ with $φ^{'} = 1$ , which vanishes by [L4]. Hence $(φ, θ, ψ)$ is a heavy-top trajectory with parameters $(A, C, m, d, g) = (3, 1, 1, 1, 1)$ , and $A θ^{''} = - U^{'} (θ_{0}) = 0$ ; the conservation of $a$ and $b$ and the formulas $φ^{'} = (a - b \cos θ_{0}) / (A \sin^{2} θ_{0}) = \frac{9 / 4}{9 / 4} = 1$ and $ψ^{'} = b / C - (a - b \cos θ_{0}) \cos θ_{0} / (A \sin^{2} θ_{0}) = \frac{5}{2} - \frac{9 / 4 \cdot 1 / 2}{9 / 4} = \frac{5}{2} - \frac{1}{2} = 2$ of [L3] are consistent with the data, so the motion is a steady precession with constant nutation angle $π / 3$ .

Remarks

A steady precession, not a sleeping top. The nutation angle stays at $π / 3$ while the precession and spin rates are constant; the sleeping-top configuration $θ = 0$ is a pole of the chart and is not covered by Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $θ$ equation, so this example makes no claim about it.
What is declared. The parameters, angles and rates are declared model data in the framework's units; the verification is an exact computation inside the branch and asserts no measured mass, length, duration or field strength, and no claim that a concrete top realises this data.

Rigid-body dynamics and stability — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Two explicit free-rotation models checked: principal-axis spins, a non-uniform symmetric-top motion with its invariants, and the linearized frequencies of an asymmetric top

Example

Facts & Assumptions

Proof

Remarks

A torque-free motion of a symmetric rigid-material model with constant angular momentum and energy whose body angular velocity is not constant: the invariants do not force a uniform rotation

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Explicit escape from the intermediate-axis spin of the model with inertia diag(1,2,3): the constants, the perturbed invariant level, and the escape-time bound

Example

Facts & Assumptions

Proof

Remarks

A steady precession of the heavy symmetric top checked on explicit data: the pivot balance, the two conserved momenta and the reduced energy

Example

Facts & Assumptions

Verification

Remarks

Sources