MMS I, Lecture 31 Content MM3 Short repetition of mm2 Rigid body Dynamics –Angular Momentum –Inertial Matrixes minus Dyadic(p. 334(down) – 336(top) –Principal Axes –Euler’s Rotational Equation of Motion minus Dyadic –Kinetic Rotational Energy

MMS I, Lecture 32 Rotation Representations RepresentationPar.CharacteristicsApplications Direction Cosine matrix 9 Nonsingular Intuitive Six redundant parameters Analytical studies Euler Angles3 Minimal set Clear physical representation Trigonometric functions in rotation matrix Singular Analytical studies Quaternions4 Easy orthogonality Not singular No clear physical representation One redundant parameter Widely used in simulation Preferred for global rotation

MMS I, Lecture 33 EULER ANGLES (BODY FIXED 3-2-1) Direction cosine: A U

MMS I, Lecture 34 EULER EIGENAXIS ROTATION Direction cosine:

MMS I, Lecture 35 QUATERNIONS Definition:Rules: Direction cosine: q 13 - q 13 0 q 4 q 4 1 =

MMS I, Lecture 36 QUATERNIONS-CONT’D Multiplication rule: i 2 = j 2 = k 2 = -1; ij = k = -ji ; q = q”O q’ = (q” 1 i + q” 2 j + q” 3 k + q” 4 )(q’ 1 i + q’ 2 j + q’ 3 k + q’ 4 )

MMS I, Lecture 37 KINEMATICS Direction Cosine: Euler Angles (3-2-1): Quaternions:

MMS I, Lecture 38 Kinematics and Dynamics Rigid body θ(t) = ω(t) · θ(t) · ·

MMS I, Lecture 39 Angular momentum for rigid body NIF n3n3 n2n2 n1n1 R0R0 R COM o dm rcrc r  Definitions: Center of mass COM v  dm = 0 External torque around O: M o = r x R dm ; r = r c +  v · · R = R o + r => R = R o + r · · · Absolute angular momentum around O: H o = r x R dm => v h o = r x r dm => v Relative angular momentum around O: Rotational motion in NIF (Newtonian Initial reference Frame) · · · · H o = ( ( r x R ) + ( r x R ) )dm v · · · h o = ((r x r ) + (r x r)) dm v · · · · ·

MMS I, Lecture 310 Angular momentum for rigid body cont. v · NIF n3n3 n2n2 n1n1 R0R0 R COM o dm rcrc r  · · · · H o = ( ( r x R ) + ( r x R ) )dm v · = M o + mr c x R o h o = (r x r ) dm v · · · M o = r x(R o + r )dm = h o + r xR o dm v · · · · = mr c x R o + h o · · · · · · · If R o fixed and/or r c = 0: M o = H o = h o If M o = 0 Angular momentum constant and unchanged Conservation of angular momentum · ·

MMS I, Lecture 311 Inertia Matrix NIF n3n3 n2n2 n1n1 RcRc R COM dm  BODY ω BN b1b1 b2b2 b3b3  N = = + ω BN x  N d  dt N d  dt B H H =  x R dm =  x (R c +  )dm vvv · ·· · = -  x (  x ω )dm v = - S(  ) S(  )ω dm v H = Jω J = - S(  ) S(  )dm Angular momentum about COM

MMS I, Lecture 312 MOMENTS OF INERTIA -S(ρ)S(ρ) = - 0 -ρ 3 ρ 2 ρ 3 0 -ρ 1 -ρ 2 ρ ρ 3 ρ 2 ρ 3 0 -ρ 1 -ρ 2 ρ 1 0 ρ ρ ρ 1 ρ 2 - ρ 1 ρ 3 - ρ 2 ρ 1 ρ ρ ρ 2 ρ 3 - ρ 3 ρ 1 - ρ 3 ρ 2 ρ ρ 2 2 = J 11 = (ρ ρ 3 2 ) dm J 22 = (ρ ρ 3 2 ) dm J 33 = (ρ ρ 2 2 ) dm J 12 = J 21 = - ρ 1 ρ 2 dm J 13 = J 31 = - ρ 1 ρ 3 dm J 23 = J 32 = - ρ 2 ρ 3 dm Moments of inertia Products of inertia vvvvvv

MMS I, Lecture 313 Principal Axes com {A} {B} H B = J B ω B H A = J A ω A C BA A->B H B = C BA H A = C BA J A ω A = C BA J A C BA ω B => -1 J B = C BA J A C BA ; ω B = C BA ω A T If H B should be a diagonal matrix: 1. find eigen values (elements of diagonal) 2. eigen vectors (row elements of DCM) 1.| λI – J| = 0 i i = 1,2,3 2.(λ i I - J) e i = 0 i = 1,2,3 C BA = J B J B J B33 e 11 e 12 e 13 e 21 e 22 e 23 e 31 e 32 e 33 J B = C BA J A C BA = = λ λ λ 3 T JBJB

MMS I, Lecture 314 Dynamic equations of rotation H = J ω BA J = - ∫ S(ρ) S(ρ) dm M = H = J ω BA + S(ω BA ) J ω BA Transportation theorem Rotating about Principal axis: J 11 ω 1 M 1 0 -ω 3 ω 2 J 11 ω 1 J 22 ω 2 = M 2 - ω 3 0 -ω 1 J 22 ω 2 J 33 ω 3 M 3 -ω 2 ω 1 0 J 33 ω 3 ω 1 = J M 1 - J (J 22 –J 33 ) ω 2 ω 3 ω 2 = J M 2 - J (J 11 –J 33 ) ω 1 ω 3 ω 3 = J M 3 - J (J 11 –J 22 ) ω 1 ω 2

MMS I, Lecture 315 = - ω S(  ) S(  )ω dm = ω H = ω T H = ω J ω T = (ω (  x(ω x  )dm = - ω (  x(  x ω )dm Using the trigonometric identity: (a x b) c = a (b x c) Kinetic Energy (scalar) T =   dm = (ω x  ) (ω x  )dm Rotational Kinetic Energy: (use transportation theorem and rigid body) vv v 1212 v v T Diagonal J: T = ( J 11 ω 1 + J 22 ω 2 + J 33 ω 3 )

MMS I, Lecture 316 Explorer 1, 1958 Unstable spin A realy good explanation you will find on:

MMS I, Lecture 317 Energy and momentum ellipsoids - 1 H T H = ( J 11 ω 1 ) 2 + (J 22 ω 2 ) 2 + ( J 33 ω 3 ) 2 = H H H 3 2 T = ( J 11 ω 1 + J 22 ω 2 + J 33 ω 3 ) J 11 < J 22 < J

MMS I, Lecture 318 Energy and momentum ellipsoids - 2

MMS I, Lecture 319 Energy and momentum ellipsoids - 3 Tmax