Rigid body stability and Poinsot s theorem

Transcription

1 Rigid body stability Rigid body stability and Ravi N Banavar banavar@iitb.ac.in 1 1 Systems and Control Engineering, IIT Bombay, India November 7, 2013

2 Outline 1 Momentum sphere-energy ellipsoid 2 The inertial ellipsoid

3 Outline 1 Momentum sphere-energy ellipsoid 2 The inertial ellipsoid

4 Energy-momentum picture Figure : Flow lines of the Euler equations - intersection of the momentum sphere and the kinetic energy ellipsoid (Courtesy: Introduction to Mechanics and Symmetry, J. E. Marsden and T. Ratiu)

5 Stability analysis Euler equations Π = Π Ω Equilibria in the angular momentum phase-space for distinct moments of inertia (I 1 > I 2 > I 3) Π e = {(Π 1, 0, 0), (0, Π 2, 0), (0, 0, Π 3)} Hamiltonian H(Π) = 1 < Π, Ω >. Kinetic energy alone does not help 2 in stability analysis. Casimir function C φ ( ) : R 3 R - similar to a constant of motion Amended Hamiltonian H C = 1 2 < Π, Ω > +Φ( 1 2 Π 2 )

6 Variations For Lyapunov stability, the first variation should be zero and the second variation should be sign definite. The first variation is δh C(Π e) = 0 δ 2 H C(Π e) > 0 (or < 0) δh C(Π e) = DH C(Π e) δπ = Ω e δπ + DΦ( 1 2 Πe 2 )Π e δπ Take Π e = (1, 0, 0). Then for stationarity DΦ( 1 2 Πe 2 ) = DΦ( 1 2 ) = 1 I 1

7 Stability The second variation is δ 2 H C(Π e) = D 2 H C(Π e) (δπ, δπ) = δω δπ + [D 2 Φ( 1 2 Πe 2 )Π e δπ]π e δπ +DΦ( 1 2 Πe 2 )δπ δπ The definiteness of the second variation is dependant on the definiteness of the matrix D 2 Φ( 1 2 ) ( 1 I 2 1 I 1 ) ( 1 I 3 1 I 1 ) For D 2 Φ( 1 ) > 0, I1 > I2, I1 > I3, the matrix is positive definite, and 2 for D 2 Φ( 1 ) < 0, I1 < I2, I1 < I3, the matrix is negative definite. 2

8 Conclusions A rigid body is Lyapunov stable while spinning about its largest or smallest principle inertia axis. The energy-casimir test is inconclusive for the intermediate axis since the earlier matrix is now sign-indefinite. Resort to spectral-stability test.

9 Spectral stability Linearized Euler equations at Π e δπ 1 δπ 2 δπ 3 = (I 3 I 1 ) I 1 I 3 (I 3 I 2 ) I 2 I δπ 1 δπ 2 δπ 3 Define a 1 = (I 3 I 1) I 1I 3, a 2 = (I 3 I 2) I 2I 3 Then eigen values of the matrix are Hence unstable. 0, ± a 1a 2

10 Outline 1 Momentum sphere-energy ellipsoid 2 The inertial ellipsoid

11 The inertia ellipsoid The kinetic energy defines an inertia ellipsoid (ɛ) in the body angular velocity space given by ɛ = {Ω R 3 :< Ω, I b Ω >= 2K} In the inertial space (u = RΩ), at any time instant t and an orientation R(t), this defines an ellipsoid R(ɛ) = {u R 3 :< u, RI b R T u >= 2K} We examine the properties of this ellipsoid in space.

12 ectures 5, 6 and 7 Theorem The moment of inertia ellipsoid in space rolls without slipping on the invariable plane. Figure : The inertia ellipsoid (Courtesy: Introduction to Mechanics and Symmetry, J. E. Marsden and T. Ratiu)

13 Proof of The spatial angular momentum π is constant. The gradient at any point u on the ellipsoid is given by R(ɛ) = 2RI b R T u A plane that is perpendicular to π and tangential to the inertia ellipsoid, satisfies at the point of tangency u t, the condition RI b R T u t = aπ for some a R u t = aω Since u t satisfies the equation of the kinetic energy ellipsoid 2K =< u t, RI b R T u t >=< aω, RI b R T aω > = a 2 < ω, RI b R T ω >= a 2 < Ω, I b Ω >= a 2 (2K) a = ±1

14 Proof of The point of tangency of the plane is ±ω. What is the equation of the tangential plane perpendicular to π? < u, π >= C. What is C? Since ω belongs to the plane, it satisfies the equation of the plane and hence < ω, π >= 2K = C. Since the point of tangency is ω, which is also the instantaneous axis of rotation, the ellipsoid rolls on the invariable plane.

15 Analytical solution: I 1 = I 2 Define Euler s equations are a 1 = (I 2 I 3) I 2I 3, a 2 = (I 3 I 1) I 3I 1, a 3 = (I 1 I 2) I 1I 2 Π 1 = a 1Π 2Π 3, Π 2 = a 2Π 3Π 1, Π 3 = a 3Π 1Π 2, I 1 = I 2 > I 3 a 3 = 0 Π 3 = constant, Π1 = a 1Π 2Π 3 = λπ 2, Π2 = a 2Π 3Π 1 = λπ 1

16 Analytical solution: I 1 = I 2 Π 3 = constant ( ) ( ) ( Π1 0 λ = Π 2 λ 0 Recall < π, ω >= 2K is the invariable plane. Π 1 Π 2 ) Ω = constant ω = constant π and ω make a constant angle. Say η is a unit vector along the axis of symmetry of the rigid body. Then, in the body-coordinates, η = (0, 0, 1). < π, η >= Π 3 = constant. π and η make a constant angle. π, ω and η are coplanar since < (π ω), η >=< R T (π ω), (0, 0, 1) >=< (Π Ω), (0, 0, 1) >= 0

17 The heavy top motions Figure : Heavy top motions (Courtesy: Introduction to Mechanics and Symmetry, J. E. Marsden and T. Ratiu)