Supplementary Lecture Notes on Kinematics – Descriptions of Angular Orientations1. Euler’s Axis of RotationLet’s begin with Euler’s Theorem:Two arbitrary oriented dextral basis vectors, r and r∗, with common origin O can be made tocoincide with one another by rotating one of them through a certain angle about an axis whichpasses through O.OCAA*BDEθnrr*nx(nxr)nxrFigure for Euler’s Axis of RotationLet us construct now the rotational tensor (or operator) R in terms of the axis of rotation n and itsrotational angle φ. As shown in the figure above, let us rotate vector r around n by φ in a rigidmotion into r∗. Thus, we obtain for the new rotated vector r∗asr∗= OC + CB +BA∗= (r · n)n −cosφn × (n ×r) + sin φn × r={nn + cos φ(I − nn) + sin φn × I}·r= R · r(1)from which we identify the rotational tensor R to beR(n,φ)= nn + cos φ(I − nn) + sin φn × I (2)or its matrix equivalent asRT(n,φ)= nnT+ cos φ(I − nnT) + sin φ ˜n (3)1R(n,φ)= nnT+ cos φ(I − nnT) − sin φ ˜n (4)Note that there are nine direction cosines in R for which we must impose six constraints due toorthogonality.3i=1RjiRki= δjk, j, k = 1, 2, 3.(5)This leaves three coordinates to remain independent. Before we specialize R to various rotationalrepresentations, let us show that it is in fact a rotational operator. To see this, let us apply it to avector, r:If r is parallel to n ( r = αn) , then we haveR(n,φ)· r = αn = r (6)Hence, R(n,φ)leaves unchanged all vectors that are parallel to n.If r is perpendicular to n (r · n = 0), we haveR(n,φ)· r = cos φr +sin φn × r (7)Hence rhas been rotatedin its plane throughthe angle φ. In general, if ris anyvector, its componentparallel to n is preserved during the rotational operation, R(n,φ), and its component perpendicularto n will be rotated about n through the angle φ.2. Euler AnglesLet us consider a specific orthogonal basis, ijk and three successive rotations in the followingmanner. First, rotate the k-axis through angle ψ. The corresponding rotatinal operator from (3.39)isR(k,ψ)= kk + cosψ(I − kk) + sinψI ×k (8)in whichI = ii +jj + kk (9)Second, denote the new rotated i-axis as i. Now, rotate the i-axis through angle θ such thatR(i,θ) = ii+ cosθ(I − ii) + sin θI × i(10)Third, denote the new rotated k-axis as kand rotate the k-axis through angle φ so that we haveR(k,φ)= kk+ cos φ(I − kk) + sin φI ×k(11)Finally, if we express the compounded rotation to be effectively about the n-axis through an angleβ,wehaveR(n,β) = R(k,φ)· R(i,θ)· R(k,ψ) (12)2How do we use (12) to describe the rotation of a vector r? The answer is simplyr= R(k,φ)· R(i,θ)· R(k,φ)· r (13)Note, however, that the new rotated vector, r, has its three components in e =ijkTand not inb =ijkT!To obtain the matrix expression for (13), we noter ={re}Te ={rb}Tb (14)Pre-dot producting by b, we obtain{rb}=b · eT{re}=Rbe{re} (15)Sinceb = R(n,β)· e (16)we haveRbe= b · eT= (R · e) · eTor in component formRbeij= ej· R(n,β)· ei(17)It can be shown that (12) is identical with the familiar successive projectionRbe= R3R2R1(18)in whichR1=cosψ sinψ 0−sin ψ cosψ 0001R2=10 00 cosθ sin θ0 −sin θ cos θ(19)R3=cosφ sinφ 0−sin φ cosφ 00013. Euler ParametersLet us recall the dyadic operator (2) asR(n,β) = nn + cos β(I − nn) + sin βn ×I (20)and introduceq0= cosβ2, q = nsinβ2, {q}=q1q2q3T(21)3so thatq20+ q ·q = q20+q21+q22+q23= cos2β2+ sin2β2= 1 (22)The four parameters, (q0, q1, q2, q3), are called the Euler parameters. Substituting (21) and bymaking use of the constraint condition (22), we obtainR(q0, q) = (2q20− 1)I +2qq +2q0(q × I)(23)from which one obtains the corresponding matrix expressionR = (2q20− 1)I + 2{q}{q}T− 2q0˜q (24)which, when expanded, can be expressed asR =2(q20+q21) − 12(q1q2+q0q3) 2(q1q3−q0q2)2(q1q2−q0q3) 2(q20+q22) − 12(q2q3+q0q1)2(q1q3+q0q2) 2(q2q3−q0q1) 2(q20+q23) − 1(25)4. Angular VelocityLet us assume that b is attached to the moving particle, P. The time rate of b with respect to theinertially fixed basis vector, e, can be obtained from (16)Edbdt=dRdte + REdedt=dRdte =˙RRTb (26)We define a skew-symmetric matrix˜ω =−˙RRT=0 −ω3ω2ω30 −ω1−ω2ω10(27)The three components in the above equation are called the angular velocity components in theb-basis with respect to the e-basis, which can be written asE˙b = ω × b (28)where the angular velocity vector, ω, can be written asω =bω1b1+bω2b2+bω3b3(29)To show that˙RRTcan in fact be written as in (26), we noteb · bT= I (30)Time differentiation of the above expression yieldsE˙b · bT+ b ·E˙bT= 0 (31)4Let us assume for the momentE˙b = Cb (32)Substitution of (32) into (31) leads toCb · bT+ b ·bTCT= 0 ⇒ C + CT= 0 (33)Hence C is a skew-symmetric matrix and time differentiation of the body-fixed basis vector b yieldsEdb1dt= ω × b1Edb2dt= ω × b2(34)Edb3dt= ω × b35. Angular Velocity in Terms of Euler AnglesThe Euler angles have been derived in Section 2. The derivation of the angular velicity vector ω interms of the Euler angles can be carried out in several different ways. In this course we will employboth the rotation tensor (2) and the basic definition of the angular velocity components (27). First,we notice that the angular velocity due to the rotation, ψ, around the k-axis iseωψ=˙ψk (35)Second, the angular velocity due to the rotation, θ, around the i-axis after the rotation of ψ aroundthe k-axis iseωθ= R(k,ψ)·˙θi =˙θi(36)Third, the angular velocity due to the rotation, φ, around the k-axis after the two preceding twosuccessive rotation iseωφ= R(k,ψ)· R(i,θ)·˙φk =˙φk(37)Finally, the compounded angular velocity vector is obtained by summing the above three compo-nents:ω =eωψ+eωθ+eωφ=˙ψk +˙θ(cos ψi + sinψj) +˙φ(cosθk + sin θ · sin ψi −sin θ cosψ j)(38)or in matrix formseωieωjeωk=0 cosψ sinθ sin ψ0 cosψ −sinθ cos ψ1 0 cosθ˙ψ˙θ˙φ(39)Notice that the above resulting angular velocity vector ω is expressed in the e-system. The body-fixed components of ω can be obtained via the tensor transformation (4) to give5bωibωjbωk=sinθ sin φ cosψ 0cosθ sin φ −sin ψ 0cosθ 01˙ψ˙θ˙φ(40)6. Angular Velocity in Terms of