Angular Momentum(D) mn Angular momentum Z System in total ∑ r1×mr tion relative to Inertial frame mI If we assume that Collection of point (a) Origin of rotating Frame in Body cm masses mi at ri (b) Fixed Position Vectors ri in Body frame (Rigid Body) Angular momentum Decomposition The ote that p measured in the total ∑m×R+∑m2x inertial frame ANGULAR MOMENTUM 口H BODY OF TOTAL MASS WRT BODY ANGULAR INERTIAL ORIGIN MOMENTUM ABOUT CENTER OFMASS
Angular Momentum (I) Angular Momentum (I) Angular Momentum total 1 n i i i i H r mr = = × ∑ m1 mn mi X Y Z Collection of point masses mi at ri ri r1 rn rn ri r1 . . . System in motion relative to Inertial Frame If we assume that (a) Origin of Rotating Frame in Body CM (b) Fixed Position Vectors ri in Body Frame (Rigid Body) Then : BODY total 1 1 ANGULAR MOMENTUM OF TOTAL MASS W.R.T BODY ANGULAR INERTIAL ORIGIN MOMENTUM ABOUT CENTER OFMASS n n i i i i i i H H mRR m ρ ρ = = = ×+ × ∑ ∑ Note that i is measured in the inertial frame Angular Momentum Decomposition
Angular Momentum(l) For a rigid body -L BODY +Xp.=0×p we can write RELATIVE MOTION IN BODY And we are able to write. H=1 O RIGID BODY, CM COORDINATES h and o are resolved in BODY FraMe The vector of angular momentum in the body frame is the product of the 3x3 Inertia matrix and the 3xl vector of angular velocities. Inertia Matrix Real Symmetric; 3x3 Tensor; coordinate dependent Properties: y ∑m(吃+)12=21=∑ mi pi2 pi 21122123 12=∑m(听+)1=1=∑mP l31132/33 ∑ m1(0i+ )23=12=∑mP2P3
Angular Momentum (II) Angular Momentum (II) For a RIGID BODY we can write: ,BODY RELATIVE MOTION IN BODY ii i i ρ = +× =× ρ ωρ ωρ And we are able to write: H I = ω “The vector of angular momentum in the body frame is the product of the 3x3 Inertia matrix and the 3x1 vector of angular velocities. ” RIIGID BODY, CM COORDINATES H and are resolved in BODY FRAME Inertia Matrix Properties: 11 12 13 21 22 23 31 32 33 III II I I III = Real Symmetric ; 3x3 Tensor ; coordinate dependent ( ) ( ) ( ) 2 2 11 2 3 12 21 2 1 1 1 2 2 22 1 3 13 31 1 3 1 1 2 2 33 1 2 23 32 2 3 1 1 n n i i i ii i i i n n i i i ii i i i n n i i i ii i i i I m II m I m II m I m II m ρ ρ ρρ ρ ρ ρρ ρ ρ ρρ = = = = = = = + = =− = + = =− = + = =− ∑ ∑ ∑ ∑ ∑ ∑
Kinetic energy and euler equations Kinetic E Energy total ∑m2+,∑mP l=」 E-TRANS E-ROT For a RIGId bODY. CM coordinates with o resolved in body axis frame EROT =0.H=@l0 2 2 H=7-0×0 Sum of external and internal torques In a BODY-FIXED, PRINCIPAL AXES CM FRAME: Euler equations H1=101=71+(2-13)O203 No general solution exists Particular solutions exist for H2=1202=12+(l3-1)2O1 simple torques. Computer simulation usually required. H3=l303=13+(1-12)0O2
Kinetic Energy and Kinetic Energy and Euler Equations Equations 2 2 total 1 1 E-TRANS E-ROT 1 1 2 2 n n i ii i i E mR m ρ = = = + ∑ ∑ Kinetic Energy For a RIGID BODY, CM Coordinates with resolved in body axis frame ROT 1 1 2 2 T E HI = ⋅= ω ωω HT I =−× ω ω Sum of external and internal torques In a BODY-FIXED, PRINCIPAL AXES CM FRAME: 1 1 1 1 22 33 2 3 2 2 2 2 33 11 3 1 3 3 3 3 11 22 1 2 ( ) ( ) ( ) HI TI I HI TI I HI TII ω ωω ω ωω ω ωω = =+ − = =+ − = =+ − Euler Equations Equations No general solution exists. Particular solutions exist for simple torques. Computer simulation usually required.
Smb Torque Free Solutions of Eulers Eq TORQUE-FREE An important special case is the torque-free motion of a(nearly CASE symmetric body spinning primarily about its symmetry axis By these assumptions: @%,0,<<@=92 lxx=ly The components of angular velocity And the euler equations become then becomes O(t)=@ro CoS @,t a,(t)=@vo coS ant XX The o. is defined as the"natural 0.=K.K.9 K or“ nutation” frequency of the body: Z H . Qa H K pace 0 /Cone H and o never align v: nutation unless spun about Ⅰ.<Ⅰ.=1 angle a principal axIs
Torque Free Solutions of Torque Free Solutions of Euler’s Eq. TORQUE-FREE CASE: An important special case is the torque-free motion of a (nearly) symmetric body spinning primarily about its symmetry axis By these assumptions: , ωω ω xy z << = Ω xx yy I I ≅ And the Euler equations become: 0 x y zz yy x y xx K zz xx y y yy K z I I I I I I ω ω ω ω ω − =− Ω − = Ω = The components of angular velocity then become: ( ) cos ( ) cos x xo n y yo n t t t t ω ωω ω ωω = = The n is defined as the “natural ” or “nutation ” frequency of the body: 2 2 ωn xy = Ω K K Body Cone Space Cone Z H zxy III < = Z H Body Cone Space Cone zxy III > = : nutation nutation angle H and never align unless spun about a principal axis !