Lectures D25-D26 : 3D Rigid Body Dynamics 12 November 2004 Outline Dynamics 16.07 Dynamics D25-D26 1 ?Review of Equations of Motion ?Rotational Motion ?Equations of Motion in Rotating Coordinates ?Euler Equations ?Example: Stability of Torque Free Motion ?Gyroscopic Motion - Euler Angles - Steady Precession ?Steady Precession with M = 0 Equations of Motion Dynamics 16.07 Dynamics D25-D26 2 Conservation of Linear Momentum ˙L = F, L = mvG Conservation of Angular Momentum ˙HG = MG, HG = IGω or ˙HO = MO, HO = IOω EquationsofMotioninRotatingCoordinates Dynamics 16.07 Dynamics D25-D26 3 Angular Momentum HG = IGω (or HO = IOω) Time variation - Non-rotating axes XYZ (I changes) ˙H = ˙Iω+I˙ω ... ˙I big problem! - Rotating axes xyz (I constant) ˙H = ( ˙H)xyz + ?×H = I(˙ω)xyz + ?×H EquationsofMotioninRotatingCoordinates Dynamics 16.07 Dynamics D25-D26 4 ( ˙H)xyz + ?×H = M or, ˙Hx?Hy?z +Hz?y = Mx ˙Hy?Hz?x +Hx?z = My ˙Hz?Hx?y +Hy?x = Mz xyz axis can be any right-handed set of axis, but . . . willchoosexyz(?) to simplifyanalysis(e.g.I constant) Example:ParallelPlaneMotion Dynamics 16.07 Dynamics D25-D26 5 ωx = ωy = 0, ωz negationslash= 0 Hx = ?Ixzωz, Hy = ?Iyzωz, Hz = Izωz Body fixed axis ? = ω (and z≡Z) ?Ixz ˙ωz +Iyzω2z = Mx (1) ?Iyz ˙ωz?Ixzω2z = My (2) Iz ˙ωz = Mz (3) Solve (3) for ωz, and then, (1) and (2) for Mx and My. Euler’s Equations Dynamics 16.07 Dynamics D25-D26 6 If xyz are principal axes of inertia ?Hx = Ixωx, Hy = Iyωy,Hz = Izωz ?? = ω Ix˙ωx?(Iy?Iz)ωyωz = Mx Iy ˙ωy?(Iz?Ix)ωzωx = My Iz ˙ωz?(Ix?Iy)ωxωy = Mz Euler’s Equations Dynamics 16.07 Dynamics D25-D26 7 ?Body fixed principal axes ?Right-handed coordinate frame ?Origin at: – Center of mass G (possibly accelerated) – Fixed point O ?Non-linear equations . . . hard to solve ?Solution gives angular velocity components . . . in unknown directions (need to integrate ω to determine orientation). Example:StabilityofTorqueFreeMotion Dynamics 16.07 Dynamics D25-D26 8 Body spinning about principal axis of inertia, ωz = ω, ωx = ωy = 0 Consider small perturbation ωx, ωy, lessmuchω After initial perturbation M = 0 Ix˙ωx?(Iy?Iz)ωyωz = 0 (1) Iy ˙ωy?(Iz?Ix)ωzωx = 0 (2) Iz ˙ωz?(Ix?Iy)ωxωybracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright small = 0 (3) Example:StabilityofTorqueFreeMotion Dynamics 16.07 Dynamics D25-D26 9 From (3) → ωz ≈ω≡ constant Differentiate (1) and substitute value of ˙ωy from (2), → . . . Ix¨ωx?(Iy?Iz)(Iz?Ix)I y ω2ωx = 0 or, ¨ωx?Aωx = 0, A = ?(Iz?Iy)(Iz?Ix)I xIy ω2 Solutions, ωx = Ae √At +Be? √At Example:StabilityofTorqueFreeMotion Dynamics 16.07 Dynamics D25-D26 10 ?A> 0→√A> 0→ Growth → Unstable Ix <Iz <Iy, or Iy <Iz <Ix ?A<0→√A=iα→Oscillatory→Stable Iz >Ix,Iy, or Iz <Ix,Iy GyroscopicMotion Dynamics 16.07 Dynamics D25-D26 11 ?Bodies symmetric w.r.t.(spin) axis Iz = I, Ix = Iy = I0 ?Origin at fixed point O (or at G) GyroscopicMotion Dynamics 16.07 Dynamics D25-D26 12 ?XYZ fixed axes ?xprimeyprimez body axes — angular velocity ω ?xyz “working” axes — angular velocity ? GyroscopicMotion Euler Angles Dynamics 16.07 Dynamics D25-D26 13 φ : Precession θ : Nutation ψ : Spin – position of xyz requires φ and θ – position of xprimeyprimez requires φ, θ and ψ Relation between (φ,θ,ψ) and ω,(and ?) ωx = ˙θ ?x = ˙θ ωy = ˙φsinθ ?y = ˙φsinθ ωz = ˙φcosθ+ ˙ψ ?z = ˙φcosθ GyroscopicMotion Euler Angles Dynamics 16.07 Dynamics D25-D26 14 Angular Momentum Hx = Ixωx = I0 ˙θ Hy = Iyωy = I0 ˙φsinθ Hz = Izωz = I( ˙φcosθ+ ˙ψ) Equation of Motion, ˙Hx?Hy?z +Hz?y = Mx ˙Hy?Hz?x +Hx?z = My ˙Hz?Hx?y +Hy?x = Mz GyroscopicMotion Euler Angles Dynamics 16.07 Dynamics D25-D26 15 become, I0(¨θ? ˙φ2sinθcosθ)+I ˙φsinθ( ˙φcosθ+ ˙ψ) = Mx I0(¨φsinθ+2˙φ˙θcosθ)?I˙θ( ˙φcosθ+ ˙ψ) = My I(¨ψ+ ¨φcosθ? ˙φ˙θsinθ) = Mz . . . not easy to solve!! GyroscopicMotion SteadyPrecession Dynamics 16.07 Dynamics D25-D26 16 ˙φ = constant, ¨φ = 0 θ = constant, ˙θ = ¨θ = 0 ˙ψ = constant, ¨ψ = 0 GyroscopicMotion SteadyPrecession Dynamics 16.07 Dynamics D25-D26 17 ˙φsinθ[I( ˙φcosθ+ ˙ψ)?I0 ˙φcosθ] = Mx 0 = My 0 = Mz Also, note that ( ˙H)xyz = 0→ H does not change in xyz axes External Moment Mx = mgzGsinθ GyroscopicMotion SteadyPrecession Dynamics 16.07 Dynamics D25-D26 18 Then, I ˙φ ˙ψ?(I0 ?I) ˙φ2 cosθ = mgzG If ˙ψgreatermuch ˙φ, or, θ≈ pi2, ˙φ = mgzG I ˙ψ ˙φ: precession velocity, ˙ψ: spin velocity SteadyPrecessionwithM=0 Dynamics 16.07 Dynamics D25-D26 19 MG = 0 ? HG = constant HGx=I0ωx =0 HGy=I0ωy =HGsinθ HGz=Iωz =HGcosθ tanθ=HGyH Gz =I0Iωyω z =I0I tanβ SteadyPrecessionwithM=0 DirectPrecession Dynamics 16.07 Dynamics D25-D26 20 From x-component of angular momentum equation, ˙φ = I ˙ψ (I0 ?I)cosθ If I0 >I, then β<θ (tanθ = (I0/I)tanβ), → ˙φ same sign as ˙ψ SteadyPrecessionwithM=0 RetrogradePrecession Dynamics 16.07 Dynamics D25-D26 21 ˙φ = I ˙ψ (I0 ?I)cosθ If I0 <I, then β>θ (tanθ = (I0/I)tanβ), → ˙φ and ˙ψ have opposite signs