Lectures d25-D26 3D Rigid body dynamics 12 November 2004
Lectures D25-D26 : 3D Rigid Body Dynamics 12 November 2004
Outline o Review of Equations of motion ● Rotational| Motion Equations of Motion in Rotating Coordinates ● Euler Equations o EXample: Stability of Torque Free Motion ● Gyroscopic Motion Euler Angles Steady Precession e Steady precession with M=0 AERO Dynamics 16.07 Dynamics D25-D26 1
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 Conservation of linear momentum L=F L= mVG Conservation of Angular Momentum HG= MG, HG =IGw or O=M O Ho= low AERO Dynamics 16.07 Dynamics D25-D26 2
Equations of Motion Dynamics 16.07 Dynamics D25-D26 2 Conservation of Linear Momentum L = F ˙ , L = mvG Conservation of Angular Momentum H˙ G = MG, HG = IGω or H˙ O = MO, HO = IOω
Equations of Motion in Rotating Coordinates Angular Momentum (or Ho =low) Time variation Non-rotating axes xYZ(I changes H=I心+D…. I big problen! Rotating axes yz (I constant) H=(H)agz+g×H Cyz +Ω×H AERO Dynamics 16.07 Dynamics D25-D26 3
Equations of Motion in Rotating Coordinates Dynamics 16.07 Dynamics D25-D26 3 Angular Momentum HG = IGω (or HO = IOω) Time variation - Non-rotating axes XY Z (I changes) H =˙ I˙ω + Iω˙ . . . I˙ big problem! - Rotating axes xyz (I constant) H˙ = (H) ˙ xyz + Ω × H = I( ˙ω)xyz + Ω × H
Equations of Motion in Rotating Coordinates (H)eyz+×H=M x -h,nz+h, nn=m H,-H2 Q+Hn2=My H2-H2nytHyn2x 之 yz axis can be any right-handed set of axis but will choose yz(@ to simplify analysis(e.g. I constant) AERO Dynamics 16.07 Dynamics D25-D26 4
Equations of Motion in Rotating Coordinates Dynamics 16.07 Dynamics D25-D26 4 (H) ˙ xyz + Ω × H = M or, H˙ x − HyΩz + HzΩy = Mx H˙ y − HzΩx + HxΩz = My H˙ z − HxΩy + HyΩx = Mz xyz axis can be any right-handed set of axis, but . . . will choose xyz (Ω) to simplify analysis (e.g. I constant)
Example: Parallel Plane Motion 0. ≠0 awag y 02z 之 Body fixed axis =w(and z= Z +I2 y Iy2Wx-Iazw2= My (3) Solve (3)for wx, and then, (1)and(2)for Mr and My AERO Dynamics 16.07 Dynamics D25-D26 5
Example: Parallel Plane Motion Dynamics 16.07 Dynamics D25-D26 5 ωx = ωy = 0, ωz 6= 0 Hx = −Ixzωz, Hy = −Iyzωz, Hz = Izωz Body fixed axis Ω = ω (and z ≡ Z) − Ixzω˙ z + Iyzω 2 z = Mx (1) −Iyzω˙ z − Ixzω 2 z = My (2) Izω˙ z = Mz (3) Solve (3) for ωz, and then, (1) and (2) for Mx and My
Euler's Equations If yz are principal axes of inertia h,=Lwh 之 之0z I2wx-(Iy-I2wyWx=M2 Iywy-(Ix-Irwxwr= M. Iz心2-(L a一1)0:t=xvz AERO Dynamics 16.07 Dynamics D25-D26 6
Euler’s Equations Dynamics 16.07 Dynamics D25-D26 6 If xyz are principal axes of inertia • Hx = Ixωx, Hy = Iyωy, Hz = Izωz • Ω = ω Ixω˙ x − (Iy − Iz)ωyωz = Mx Iyω˙ y − (Iz − Ix)ωzωx = My Izω˙ z − (Ix − Iy)ωxωy = Mz
Euler's Equations o body fixed principal axes o Right-handed coordinate frame Origin at Center of mass G(possibly accelerated Fixed point O o non-linear equations. . hard to solve Solution gives angular velocity components unknown directions(need to integrate w to determine orientation) AERO Dynamics 16.07 Dynamics D25-D26 7
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: Stability of Torque Free Motion Body spinning about principal axis of inertia 0 Consider small perturbation ,y9 After initial perturbation M=0 w 之 0 0 y 0 small AERO Dynamics 16.07 Dynamics D25-D26 8
Example: Stability of Torque Free Motion Dynamics 16.07 Dynamics D25-D26 8 Body spinning about principal axis of inertia, ωz = ω, ωx = ωy = 0 Consider small perturbation ωx, ωy, ω After initial perturbation M = 0 Ixω˙ x − (Iy − Iz)ωyωz = 0 (1) Iyω˙ y − (Iz − Ix)ωzωx = 0 (2) Izω˙ z − (Ix − Iy) ωxωy | {z } small = 0 (3)
Example: Stability of Torque Free Motion From(3)→d2≈≡ constant Differentiate(1)and substitute value of iy from(2) (Iy -I2)(Iz 0 , n-An=0,A=(1-b2- Solutions, W= AeVAt +Be-VAr AERO Dynamics 16.07 Dynamics D25-D26 9
Example: Stability of Torque Free Motion Dynamics 16.07 Dynamics D25-D26 9 From (3) → ωz ≈ ω ≡ constant Differentiate (1) and substitute value of ω˙ y from (2), → . . . Ixω¨x − (Iy − Iz)(Iz − Ix) Iy ω 2ωx = 0 or, ω¨x − Aωx = 0, A = − (Iz − Iy)(Iz − Ix) IxIy ω 2 Solutions, ωx = Ae √ At + Be− √ At