# CMchap2 - Chapter 2 Kinematics A“ The trajectory of a...

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Unformatted text preview: Chapter 2 Kinematics A“ The trajectory of a point particle is described by a curve r05) in Euclidean , 3 ' ' N g _ 44"” space R (see Flg. 2.1). 1/ (if, A > \ Fig. 2.1 / is“ / g The velocity and the acceleration a(t) are given respectively by 5/ f “a: 5? fig ,5" £17. " 2 ll. " t E —— , 2.1 M > d, < ) do (127' t E —-— : —-—— . 2.2 C“ ) dt alt? ( ) With respect to a space—ﬁxed frame {5,}, we can write M) = we 6.- , (2.3) '00?) = 7‘05) = 1131151, (2.4) a(t) = Ht) 2 ii 5,- , (2.5) M where the single dot and double dot above a quantity mean the ﬁrst and second ,5 Q‘f derivatives with respect to time, respectively, of that quantity. f \‘(N‘ To describe rigid body motions, in particular, rotations, it is often more T‘s, if? \ convenient to use body-ﬁxed frames (see Fig. 2.2). From now on we will if“ "V j always denote body—ﬁxed frames by {61-} and space-ﬁxed frames by {6,}. ’ f Fig. 2.2 35:: 5 659 With respect to a body-ﬁxed frame, the position vector r(t) of a point in the ’1‘ ‘ 1'1 rigid body is given by r(t) : (Iii e, , (2.6) 11 12 CHAPTER 2. KINEJVIATICS where the components mi remain; constant in time as the rigid body moves. We then have 1- dei v(t) = a: E. (2.7) The geometry of the rotational motion is completely described by a matrix of differentials (differential one-forms) Lag related to the moving axes 6) by .7 dei = wi ej . (2.8) In R3, ) is a 3 X 3 matrix of 1-forms, called a connection matrix of one- forms in the language of differential geometry. The “cl” in the above equation is more properly called a covariant derivative. The velocity of a point in the rigid body is then given by 'U(t) = ej , (2.9) where (ng is a matrix of derivatives (with respect to time) deﬁned by 90\$ 5 —t . (2.10) In R3 (and only in a space of dimension 3), we can deﬁne a quantity wk with 3 i“ (of E eijkwk. (2.11) {1117:}: v(t) : rial-7km,“ ej : Z (Elgijkwk ej = Z ejkiwkmi ej . (2.12) i j By the deﬁnition of the cross product (1.37) this is recognized to be the familiar equation for 'u in terms of the angular velocity “vector” w with components wk: 1) = w x 7' . (2.13) w is not really a vector (hence the quotes above) because it is deﬁned in terms of (pg which in turn is deﬁned in terms of Log, and the connection matrim 0f one—forms a)? is not a tensorz'al quantity, in the sense that it does not satisfy the transformation properties of a tensor (under a change of coordinate frames) as given by (1.30). This is an important point7 which can be demonstrated easily as follows. Suppose the frame {61-} is related to another {6;} by (see Table 1.1) I 6. = g? ej , (2.14) where is an element in a matrix group of transformations. We will rewrite the above equation in matrix notation as e’ 2 ye , (2.15) 13 where e and e’ are understood to be column matrices of basis vectors and g is a square matrix of appropriate dimension. The connection matrix of one-forms is given with respect to the frames 6 and e’ by the following matrix equations. de 2 we, de' 2 w’e' = w’ge. (2.16) But de’ 2 d(ge) : (dg)e + gwe. (2.17) Thus w'g = (19 + gw, (2.18) which implies the following important transformation rule for a connection ma— trix under a change of frames (a so—called gauge transformation). w’ : (alg)g“1 + gang—1 . (2.19) The presence of the first term on the right hand side makes to non—tensorial [compare (1.31)]. Note that since this term involves dg, it gives the effect of the dependence of the transformation matrix 9 on the particular location in space. Suppose {31-} is an orthonormal frame. Then ei -ej = 62-]- . (2.20) On differentiating, we obtain dei -ej + ei - dej : 0. (2.21) It follows from (2.8) that Mfek'€j+€i’w;el=o, (2.22) or of 6M 4mg. 6“ = 0. (2.23) Thus if {3)} is orthonormal, the connection matrix to is skew—symmetric: wj +wi- = 0 . (2.24) is then also skew—symmetric, and by (2.11) we can display it as a matrix in terms of the “components” of the angular velocity cut as follows: (903): —w3 0 ml . (2.25) 14 CHAPTER 2. KINEMATICS Iii“: A Suppose g is the space-dependent transformation matrix relating the body— ﬁxed frame {61'} to the space—ﬁxed frame {6]} at each point in space (Fig. 2.3). 91 Then, in matrix notation, e = 96, and thus 6 : 9—16. '3 g . XJ/ NM Fig. 2.3 x “5‘5 g}? T‘i‘}: Since d6 = O is a space-ﬁxed frame), we have s a? - 99-23 de:dg-5:(dg.g—1)e. (2.26) Comparison with (2.8) shows that (w?) = d9 - 9‘1 . (2.27) We will now Use the above equation to calculate a)? in terms of the spherical ‘ coordinates (736,925). The frame ﬁeld ({ei} as a function of spatial position) 5 s E appropriate to this system of coordinates is shown in Fi .2.4. We have 2 3 e1 : er : (sin6 cos qi) 61 + (sine sin qi) 62 + cos 0 63 , (2.28) 62 2 eg 2 (cos 6 cos 43) 61 + (cos 05in ¢) 62 — sin6 63 , (2.29) 63 2 e4, 2 —sind)61 +cos¢62. (2.30) Verify the above three equations. Then the matrix 9 relating e to 6 is given by sin 6’ cos gb sin 0 sin qt cos 6 9 : cos<9cos¢> cosBsinqﬁ —sin9 - (231) — sin 91) cos ()5 0 It is easily veriﬁed that g as given above is an orthogonal matrix: 9‘1 : gT. Exercise 2.2 Verify the above assertion. It follows from (2.27) that 0 d6 sinﬁdqﬁ 2 —d9 0 cos6d¢ ~ (232) —sin6d¢ —cos6dqb 0 15 Eq. (2.10) then gives d9 . qu :6 g 31nd 3g 9: = ___ _ . 2.33 (991) dtdqs 0 (M) cos0 dt ( l - 51116 It? — cos9 E 0 On comparison with (2.25), we have the following expressions for the compo- nents of the angular velocity: wlzcosdgt, w2:—sin0d, wazd. (2.34) Verify (2.32) by calculating the RHS of (2.27). Use the matrix for 9 given by (2.31). We are now in a position to give general expressions for the velocity 'u and the acceleration a in terms of spherical coordinates. For clearer presentation of the equations we will use boldface for the unit vectors in {61'}. Starting with r = 1" er , (2.35) we have v=f=rer+réT=re1+ré1. (2.36) Dividing (2.8) by dt we obtain __ deal to? w? 2 ~— 62 + —- 83 . 81“? dt dt Using the explicit expressions for Lug given by (2.32) we have (2.39) Eq. (2.36) then gives v :reT +7969 +r<i>sin6e¢ Note that the sum of last two terms on the RHS of the above equation is exactly equal to w x 7‘: er 89 €¢ wxr: écosg -gﬁsing :rdeg-l-Tdsindqu. (2.40) 7' 0 .0 16 CHAPTER 2. KINEMATICS This describes tangential motion (velocity tangent to the surface of a sphere), While the term involving 7'“ describes radial motion. Differentiation of (2.39) with respect to time yields a2?)=Fer+réT+(rd+rd)eg~i—r9ég . . . .. . 2.41 +(rgzﬁsinﬁ+r6¢cosr9+r¢>sin6)e¢+r¢sin0é¢. ( ) Similarly to (2.38), we can use (2.8) and (2.32) to obtain égz—der+dcosﬁe¢, é¢:—qlsin0er—dcosﬁeg . (2.42) Finally, the acceleration a is given by a : er —— r92 — rng sin2 0) + 89 (27*6'J + rd —— “52 sin 6' cos 0) (2.43) + 8.) (2rd sin0 + 2rdd3 cos 6 + rd; sin 0) . Verify the above equation for the acceleration. In (2.43) the term —er (7‘92 + mtg sin2 0) is called the centripetal acceler- ation (it is always directed along the radial direction and towards the origin of the coordinate system); while the sum of the terms proportional to 7A is called the CoriolIis acceleration. The sum of the rest of the terms, excluding the one proportional to 7", is called the tangential acceleration. ...
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CMchap2 - Chapter 2 Kinematics A“ The trajectory of a...

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