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Siacci's Resolution of the Acceleration Vector for a Space Curve Meccanica (2011) 46: 471–476 DOI 10.1007/s11012-010-9296-x BRIEF NOTE Siacci’s resolution of the acceleration vector for a space curve James Casey Received: 17 January 2009 / Accepted: 16 March 2010 / Published online: 9 June 2010 © The Author(s) 2010. This article is published with open access at Springerlink.com Abstract For motion of a material point along a space For spatial motions of a particle, Siacci succeeded curve, a kinematical decomposition, discovered by in showing that a similar resolution can be obtained [8]. Siacci, expresses the acceleration vector as the sum In this case, the two components are in the instanta- of two special oblique components in the osculating neous osculating plane to the path of the particle. One plane to the curve. A new proof of Siacci’s theorem is is tangent to the path, while the other is directed to- presented. wards the point where the perpendicular from a fixed origin meets the osculating plane. Although Siacci’s Keywords Siacci · Space curve · Kinematics · formulas are quite remarkable, his statement of the Classical mechanics · Central forces theorem is not altogether precise and his proof is cum- bersome.2 The purpose of this note is to present a proof of 1 Introduction Siacci’s theorem in space that is based on the Serret- Frenet formulae. Siacci’s theorem in the plane is in- For planar motions of a particle, a resolution of the cluded as a corollary. acceleration vector, due to Siacci [1], is well known and is particularly useful for problems in which angu- lar momentum is constant [2–6]. This resolution com- 2 Preliminaries prises a radial component and a tangential component, which, in general, are not orthogonal to each other.1 Consider a particle P of mass m(> 0) moving in a 3-dimensional Euclidean space E3 under the influence of arbitrary forces. Choose an arbitrary fixed origin O 3 1Whittaker [2, p. 21] gives a geometrical proof of Siacci’s theo- in E , and let x be the position vector of P at time t. rem in the plane. Grossman [3] has a more modern proof (but the Let C be the oriented curve traced out by P, and let s accompanying diagram is misdrawn). Francesco Siacci (1839– be the arc length of C corresponding to time t. 1907) held professorships in ballistics at the military academy { } in Genoa, and in mechanics at the Universities of Turin and Let et , en, eb be a right-handed orthonormal basis Naples [7]. He played an active role in the movement for unifi- consisting of the unit tangent vector cation of Italy, and in his later years was elected a Senator for dx life. A street in Rome is named in his honor. e = x = , (1) t ds J. Casey () Department of Mechanical Engineering, University of California, Berkeley, CA 94720, USA 2Whittaker [2, p. 24, Problem 12] repeats Siacci’s statement of e-mail: [email protected] the theorem. 472 Meccanica (2011) 46: 471–476 the unit normal vector en, and the unit binormal vec- The foregoing resolution of x is indicated in Fig. 1. tor eb. Recall the Serret-Frenet formulae The vector beb is the position vector of the perpendic- ular OB from O to the osculating plane, and r is the = =− + =− et κen, en κet τeb, eb τen, (2) position vector of P relative to B. BZ is the perpen- dicular from B to the tangent line SPT , so that the where κ and τ are the curvature and torsion functions. position vector of Z relative to B is −pe . The posi- For κ = 0, ρ = 1/κ is the radius of curvature of C. n tion vector of P relative to Z is qe . The velocity v and acceleration a of P at time t are t given by Substituting (8)in(1) and invoking (2)1,2,3, we ob- tain = ˙ = =˙ = ˙ =˙ + 2 v x vet ,vs, a v vet κv en, (3) et = q et + qκen − p en − p(−κet + τeb) + b eb where the superposed dot denotes d/dt. Along C, v + b(−τe ), (12) may be expressed as a function of s, and the accelera- n tion may then be written as which yields dv 2 a = v et + κv en, (4) q = 1 − κp, p = κq − τb, b = τp. (13) ds which always lies in the osculating plane Π to C at P. It follows from (11) and (13)1,2,3 that The angular velocity vector of the Serret-Frenet basis is rr = q − τpb = q − bb . (14) ω =˙s(τet + κeb) (5) With the aid of (8), the angular momentum vector in (7) takes the form and lies in the rectifying plane to C. Clearly, O H = m(ben + peb)v. (15) e˙t = ω × et , e˙n = ω × en, e˙b = ω × eb. (6) Let The angular momentum of P about O is O h = pv, w = bv. (16) H = x × mv = x × mvet (7) The two components, h and w, of angular momentum and lies in the plane perpendicular to et (i.e., the nor- mal plane to C). per unit mass may be interpreted as twice the rates at which the orthogonal projections of x sweep out area on the osculating plane Π and on the rectifying plane, 3 3 Siacci’s theorem respectively. Siacci’s idea is to resolve the acceleration vector Let the position vector of P be resolved on the Serret- a in (4) along the tangential direction SPT and the Frenet basis: radial direction BP in the osculating plane. To do this, we need to be able to express en in terms of r and et , = − + x qet pen beb, (8) whichinviewof(10), is possible if and only if p = 0. We may ensure that p never vanishes by making where the physical assumption that the binormal component of angular momentum is nonzero. Then, h = pv = 0. q = x · et , −p = x · en,b= x · eb. (9) Hence, in view of (11), r = 0 and we may define a unit Also, define a vector r in the osculating plane Π by vector er by r =−pe + qe . (10) 1 n t e = r. (17) r r Note that r2 = r · r = p2 + q2. (11) 3See [9] for a discussion of the areal velocity vector in space. Meccanica (2011) 46: 471–476 473 Fig. 1 A particle P travels along a space curve C. SPT is the to Π. BZ is perpendicular to SPT ,andr is the position vec- tangent line to C at P and Π is the osculating plane at P . OB is tor of P relative to B the perpendicular, from an arbitrary fixed origin O in space, Equation (10) then yields the desired resolution The tangential Siacci component may be put into a number of different forms. First note that q satisfies 1 the equations en = (−rer + qet ). (18) p κq = p + τb, q = rr + τpb = rr + bb , (21) Substituting (18)in(4), we obtain the fundamental form of Siacci’s resolution of the acceleration vector: which follow from (13)2 and (14). With the aid of (21)1, (16)1 and (13)3, we find that 2 2 κv r dv κv q 2 a =− er + v + et = Sr er + St et . 1 2 κv q p ds p St = (v ) + 2 p (19) 1 v2p τv2b = (v2) + + p p In general, the unit vectors er and et are not orthogonal 2 to each other and the two components of a in (19)are 2 2 = (h ) + τbh not equal to the orthogonal projections of a onto er 2p2 p3 and e . We call S and S , respectively, the radial and t r t 2 1 h tangential Siacci components of the acceleration. = (h2) + (b2) . (22) p2 p2 Invoking (16)1 with p = 0, we may express the ra- 2 dial Siacci component as Likewise, employing (21)2, we obtain 2 2 κrh 1 2 2 (r ) S =− . (20) St = (v ) + κv + τb . (23) r p3 2 2p 474 Meccanica (2011) 46: 471–476 If instead, we use the second expression for q in (21)2, The expression (19) now holds in the plane of the we may deduce that motion. Equations (25) and (26) follow immediately from (22) and (23). 2 1 2 κv 2 2 St = (v ) + (r + b ) 2 2p It may be noted that in the planar case, (14) and 2 (13)2 yield 1 κv = (v2) + (x · x) , (24) 2 2p rr = q, p = κrr . (28) where (10) and (8) have also been utilized. Summarizing the above derivation, we state Corollary 2 Suppose that the fixed plane in Corol- lary 1 passes through O, and that the angular momen- Siacci’s Theorem Let a particle P of mass m(> 0) tum of P never vanishes. Then, Siacci’s resolution (19) travel along a curve C in space, and suppose that the holds, with St being given by (25) and (26). binormal component of its angular momentum vector never vanishes. Then, the acceleration vector of P may Proof This is a special case of Corollary 1 with = be expressed as the sum of two oblique components, b 0. the Siacci components, as in (19). One Siacci compo- nent lies along the tangent to C, while the other one is directed from P towards the point where the perpen- 4 Illustrative examples dicular from an arbitrary fixed origin meets the oscu- lating plane. The radial component may also be writ- Example 1: Motion along a helix ten as in (20), and alternative expressions for the tan- gential component are given by (22), (23), and (24).
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