The classical spin-rotation coupling and the kinematic origin of inertia

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1 The classcal spn-rotaton couplng and the knematc orgn of nerta Loua Hassan Elzen Basher * December 15, 2018 Abstract Ths paper s prepared to show that a rgd body whch accelerates curvlnearly from ts center of mass relatve to a fxed pont must smultaneously accelerate angularly relatve to ts center of mass. Formulae whch couplng of the angular momentum and knetc energy due to nduced spn moton n the rgd body to the angular momentum and knetc energy due to rotatonal moton of the same spnnng rgd body have been derved. The paper also brngng to lght the nature of the forces whch cause nducton of spn moton n that rgd body and a formula whch couplng of ths hghlghted forces to the force whch causes rotaton of the rgd body has been also derved. Keywords : Rgd body and gyroscope moton; orgn of nerta; Spn-rotaton couplng; Mach s prncple; mass fluctuatons. PACS No. : D ; dc ; df ; Cc ; Cv. *Khartoum, Sudan. Postcode: Emal: louaelzen@gmal.com 1

2 Nomenclature m = mass of the rgd body m = mass element n the rgd body r CM = vector poston of the center of mass of the rgd body relatve to the axs of rotaton O r = vector poston of mass element m relatve to the axs of rotaton O of the rgd body, also poston vector of the pont n the space-fxed reference system S ρ = vector poston of mass element m relatve to the center of mass of the rgd body p = lnear momentum of the rgd body ω = rotatonal angular velocty of the rgd body relatve to the axs of rotaton O Ω = spn angular velocty relatve to the center of mass of the rgd body I O = moment of nerta of the rgd body relatve to the axs of rotaton O I CM = moment of nerta of the rgd body relatve to ts center of mass L = total angular momentum of the rgd body L R = rotatonal angular momentum of the rgd body relatve to the axs of rotaton O L S = coupled spn angular momentum of the rgd body relatve to ts center of mass T = total knetc energy of the rgd body T R = rotatonal knetc energy of the rgd body relatve to the axs of rotaton O T S = coupled spn knetc energy of the rgd body relatve to ts center of mass a = rectlnear acceleraton of the rgd body F = external rectlnear force acts on the center of mass of the rgd body ω = angular acceleraton of the rgd body relatve to the axs of rotaton O τ = total torque over the rgd body τ R = rotatonal torque of the rgd body relatve to the axs of rotaton O ( or the actve torque) τ S = coupled spn torque of the rgd body relatve to ts center of mass (or the nertal torque) r Q = poston vector of the reference pont Q n the space-fxed reference system S r = poston vector of the pont n the body-fxed reference system S u Q = velocty of the orgn of S -frame relatve to the orgn of S -frame u = velocty of the pont relatve to the orgn of S -frame u = velocty of the pont relatve to the orgn of S -frame a Q = rectlnear acceleraton of the orgn of S -frame relatve to the orgn of S -frame a = rectlnear acceleraton of the pont relatve to the orgn of S -frame a = rectlnear acceleraton of the pont relatve to the orgn of S -frame τ Euler = nertal torque occurs due to Euler force τ Corols = nertal torque occurs due to Corols force 2

3 1 Introducton One can defne the problem by the followng statement: dvson of the total angular momentum nto ts orbtal and spn parts s especally useful because t s often true (at least to a good approxmaton) that the two parts are separately conserved. [see 1, p. 370]. The statement brefs the common understandng wthn scentfc communty about spn-rotaton relaton for a rgd body n rotatonal moton. Nevertheless, we are gong to prove that the negaton of ths statement s what s true. We begn wth the dstncton between rotatonal (also crcular or orbtal) and spn moton of a rgd body. Hence, we defne rotatonal moton as the angular moton of center of mass of a rgd body relatve to a fxed pont whereas the dstance between the center of mass of the rgd body and the axs of rotaton remans fxed. The spn moton s the angular moton of a rgd body relatve to ts center of mass. Another thng s that; the analyss s gong to be on 3-dmensonal Eucldean space and wth a planar rgd body undergoes planar moton. The followng three subsectons (2.1), (2.2) and (2.3) can be consdered as the observaton of couplng phenomena and whch has been obtaned from mathematcal analyss of rotatonal moton of a rgd body. The last subsecton (Subsecton (2.4)) gves theoretcal explanaton to ths phenomena. 2 Analyss 2.1 Couplng of spn and rotatonal angular momentums Referrng to Fgure1, the rgd body A of mass m s free to rotate relatve to ts center of mass CM as t s also smultaneously free to rotate relatve to the fxed pont O (nertal frame). Thus, t s pvoted at these two ponts. It s known that the total angular momentum L of such rgd body n rotatonal moton s gven by[see 1, p. 369]: L = r CM p + ρ ρ m (1) 3

4 The frst term s the angular momentum (relatve to O) of the moton of the center of mass. The second s the angular momentum of the moton relatve to the center of mass. Thus, we can re-express Equaton (1) to say[see 1, p. 369] L = L moton of CM + L moton relatve to CM (2) Snce the mass s constraned to a crcle then the tangental velocty of the mass of the rgd body s ω r CM and ts lnear momentum s p = m(ω r CM ). Therefore, the total angular momentum equaton (Equaton (1)) becomes (assumng the moton s planar, thus both axses of rotaton O and CM are parallel): L = r CM m(ω r CM ) + ρ m (Ω ρ ) (3) Takng the frst term n the RHS and usng the poston vector equaton r = r CM + ρ, (4) one fnds (see appendx A, I) L = L R + L S (5) Takng the dot product of Equaton (5) wth tself, we get L 2 = L 2 R + L 2 S + 2 L R L S, (6) (snce L R and L S commute), and therefore L R L S = ½ ( ) L 2 L 2 R L 2 S. (7) Snce the total angular momentum (Equaton (5)) s conserved, t s mples that the rotatonal and coupled spn angular momentum are mutually exchange and that n order to conserve the total angular momentum, that s L = L R + L S (8) Equaton (7) and (8) negate the statement of uncouplng of rotatonal and spn angular momentums wth whch we had began the argument snce L R L S 0. Another thng we can notce s that f we fully do the dot product of Equaton (6) and then rearrange t, we obtan the parallel axs theorem. (see appendx A, II) 4

5 2.2 Couplng of spn and rotatonal knetc energes It s known that the knetc energy T of the rgd body A n ts rotatonal moton relatve to the axs of rotaton O s gven by[see 2, p. 206]: T = ½m (ω r CM ω r CM ) + ½I CM (Ω Ω) (9) where ω r CM s the tangental velocty of the center of mass of the rgd body relatve to the axs of rotaton O and Ω s the spn velocty relatve to the center of mass of the rgd body. Takng the frst term n the RHS of Equaton (9), one fnds (see appendx A, III) T = T R + T S (10) If we fully do the dot products of Equaton (10) and then rearrange t, we agan wll obtan the parallel axs theorem. (see appendx A, IV) 2.3 Couplng of the actng forces At ths secton we wll explore the couplng between the force causes the rotaton of the rgd body A and the force causes ts spn. Referrng to Fgure1, f an external force F acts on the center of mass of the rgd body, and snce the mass m s constraned to a crcle, then the tangental acceleraton of the rgd body s ω r CM, and snce F = ma, the total torque τ s gven by: τ = r CM F = r CM m( ω r CM ) (11) Substtutng Equaton (4) nto (11), one fnds (see appendx A, V) τ = τ R + τ S (12) Takng the dot product of Equaton (12) wth tself, we obtan τ 2 = τ 2 R + τ 2 S + 2 τ R τ S, (13) (snce τ R and τ S commute), and therefore τ R τ S = ½ ( ) τ 2 τ 2 R τ 2 S. (14) If we fully do the dot products of Equaton (13) and then rearrange t, we agan wll obtan the parallel axs theorem. (see appendx A, VI) 5

6 2.4 The nature of the forces whch cause the spn torque To fnd out the nature of the force behnds the spn torque τ S we are gong to take the knetc approach to fnd the same term that assgned to t and whch appears n Equaton (12), that s, I CM ( ω). Referrng to Fgure2, we have a space-fxed coordnate system S whch s a coordnate system wth the orgn fxed n space at pont O, and wth space-fxed drectons for the axes. We have also a body-fxed coordnate system S wth an arbtrary pont Q (reference pont) on the rgd body s selected as the coordnate orgn. Therefore, the quanttes n the reference systems S and S are related as follows[see 3, p ]: r = r Q + r (15) Takng the frst change n poston vector (Equaton (15)) wth respect to tme, yelds (see appendx B, VII) Rectlnear velocty[see 4, p. 17]: u Q = u u, and (16) Azmuthal velocty: ω r Q = ω r ω r. (17) If the rgd body A s rotatonally moves, then by lettng the coordnate orgn (pont Q) of the S -frame to be the center of mass of the rgd body then that yelds dentcally Equaton (5), the angular momentums couplng formula whch derved earler n Subsecton (2.1). (see appendx B, VIII) Takng the second change n poston vector (Equaton (15)) wth respect to tme, we obtan (see appendx B, IX) Azmuthal acceleraton: ω r Q = ω r + ( ω r ) (18) Corols acceleraton: 2ω u Q = 2ω u + ( 2ω u ) (19) 6

7 Centrpetal acceleraton: ω (ω r Q ) = ω (ω r ) + ( ω (ω r )) (20) Rectlnear acceleraton: a Q = a + ( a ) (21) The combnaton of these acceleratons gves the total acceleraton (acceleratons relatve to S, S -frames smultaneously): a Q + ω (ω r Q ) + ω r Q + 2ω u Q = Acceleratons relatve to S -frame (nertal frame) { }} { a + ω (ω r ) + ω r + 2ω u + Acceleratons relatve to S -frame (actve frame) { }} { ( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω u ) (22) From equatons (18) to (21) and by multplyng by mass m, rememberng that m = m, we obtan the forces actng over the rgd body, that s Azmuthal force: m ( ) ( ω r Q = m ( ω r ) + m ω r ) (23) Corols force: m ( ) 2ω u Q = m (2ω u ) + m ( 2ω u ) (24) Centrpetal force: m ( ω (ω r Q ) ) ( ( = m (ω (ω r )) + m ω ω r )) (25) Rectlnear force: ma Q = m a + m ( a ) (26) 7

8 The combnaton of these forces gves the total force: Total force (forces relatve to S, S -frames smultaneously) { }} { m[a Q + ω (ω r Q ) + ω r Q + 2ω u Q ] = Actve force (forces relatve to S -frame (nertal frame)) { }} { m [a + ω (ω r ) + ω r + 2ω u ] + Inertal force (forces relatve to S -frame (actve frame)) { }} { m [( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω u )] (27) and more generally, F total = F actve + F nertal (28) We generally found that every sngle force of the above forces, (equatons (23) to (26)), s a synthess of an actve force and an nertal force and smlarly the total force (Equaton (27)). If the coordnate orgn (pont Q) of the S -frame has been chosen to be a center of mass of a rgd body, say the rgd body A, that s, r = ρ where ρ s the poston vector of the pont relatve to the center of mass, and wth the help of Equaton (37) and ts frst and second dervatve wth respect to tme ( m ρ = m u = 0 and m ρ = m a = 0 snce ρ s constant). Then equatons (23) to (27) wll gve Azmuthal force: Corols force: m ( ω r CM ) = m ( ω r ) + 0 (29) m (2ω u CM ) = m (2ω u ) + 0 (30) 8

9 Centrpetal force: m (ω (ω r CM )) = m (ω (ω r )) + 0 (31) Rectlnear force: ma CM = m a + 0 (32) The combnaton of these forces gves the total force: m[a CM + ω (ω r CM ) + ω r CM + 2ω u CM ] = m [a + ω (ω r ) + ω r + 2ω u ] + 0 (33) The nertal forces have been neutralzed due to the balanced dstrbuton of the mass elements relatve to the center of mass of the rgd body, or n another word the net nertal forces s equal to zero, nevertheless the nertal acceleratons have not vanshed The contrbuton of Euler force to the spn torque If the rgd body A s rotatonally accelerates from ts center of mass then by cross multplyng Equaton (29) by r CM, we wll obtan the torque due to the tangental force. Hence, one fnds (see appendx C, X) τ = τ R + τ Euler (34) Ths force causes spn of the mass element m relatve to the center of mass of the rgd body (orgn of S -frame) n a drecton counter to the drecton of rotaton of the rgd body relatve to the axs of rotaton O (orgn of S -frame). Equaton (34) s dentcally Equaton (12) whch has been derved earler n Subsecton (2.3). Ths mples that τ Euler = τ S, that s, the counter rotaton of the rgd body relatve to ts center of mass occurs due to Euler nertal force. 9

10 2.4.2 The contrbuton of Corols force to the spn torque If the rgd body A s radal translatng whle unformly rotate from ts center of mass, then by cross multplyng Equaton (30) by r CM, we wll obtan the torque due to the Corols force. Therefore, Equaton (34) wll update to (see appendx C, XI): τ = τ R + τ S { }} { (τ Euler + τ Corols ) (35) Ths force also causes spn of the mass element m relatve to the center of mass of the rgd body n a drecton counter to the drecton of rotaton of the rgd body relatve to the axs of rotaton O. We also wll fnd the centrpetal and rectlnear forces do not contrbute to the spn of the rgd body (see appendx C, XII and XIII). Therefore, we can wrte the equaton of moton whch descrbes all forces couplng spn to rotaton: τ { }} { r CM m( ω r CM ) + r CM m(2ω u CM ) that s. { }} { = r m ( ω r ) + r m (2ω u ) τ R τ S { }} { + ρ m ( ω r ) + ρ m ( 2ω u ), (36) 3 Dscusson The results of the analyss suggest that dstant stars and celestal bodes have no sgnfcant effect on local nertal frames see Mach s hypothess [5 8, Berkeley, Mach, Scama, and others], whereas nertal forces have been completely derved from Gallean-Newtonan set of concepts [cf. 9, Prncpa] and have been shown that t have a knematc rather than a dynamcal orgn [cf. 10, p. 1476]. 10

11 If the rest frame determnes the nertal frames, t follow that nerta s not an ntrnsc property of matter, but arses as a result of the nteracton of matter wth the rest frame [cf. 8, p. 35] nertal forces are exerted by absolute space, not by matter and the whole of the nertal feld must not be due to sources [cf. 10, p. 1476]. Therefore, Newton s laws of moton can be accurate wthout need of reference to the physcal propertes of the unverse, such as the amount of matter t contans, whch does not mply that matter has nerta only n the presence of other matter [cf. 8, p. 35]. In opposte to the rectlnear moton where nerta presents as resstance of the mass to moton, the objects when move curvlnearly, ther nertas present as spn of ther masses. The fndng that nertal forces cause the masses of objects to oscllate when set nto acceleraton by external force (to spn or rotate the opposte phenomenon as the law of conservaton of momentum mples when move curvlnearly or to vbrate when move rectlnearly) s the counterpart of the Mach effect [cf. 11] (also referred to as Woodward effect) whch predcted fluctuatons n the masses of thngs that change ther nternal energes as they are accelerated by external forces [see 12, p. 4], whereas n Gallean-Newtonan verson as the fndngs says the fluctuatons show themselves as oscllatory mechancal motons of masses (spn, rotaton and vbraton) and not as a change n the magntude of the accelerated mass. Another thng s that the fluctuatons that occur n Gallean- Newtonan frame are measurable at the macroscopc mechancal systems whch n opposte to the Mach effect where measurable effect needs to be drven at a hgh frequency. The fndngs pont also toward absorpton of nternal energy by masses of thngs that are accelerated by external forces[see eq. 10]. Ths energy obtanable at any pont of the space[cf. 13, p. 58]. One of the mportant nsghts that we have ganed here s that when we utlze (dfferentate, substtute or both) the poston transformaton equaton whch couples 1 the poston vectors of a pont relatve to space-fxed and movng frames of 1 The magntude of the poston transformaton squared equaton (4) or (15) shows couplng term,.e., r 2 Q = r2 + r r r. (Compare ths equaton wth equatons (6) and (13).) Smlar equaton can be obtan from veloctes, acceleratons and forces transformatons equatons (16) to (21) and (23) to (26) and that by 11

12 reference[see eqs. 4, 15], we obtan the velocty and acceleratons transformaton equatons[see eqs ] whch couple those latter quanttes n space-fxed and movng frames, whereas from these latter transformatons we obtan transformaton equatons of forces, torques, angular momentum and rotatonal knetc energy[see eqs , 34, 35, 5, 10]. Whereas, the parallel axs theorem provdes the transformaton equaton of moment of nerta between the space-fxed and movng frames. These transformatons are what we perceve as nertal forces, momentums and energes, that s, t explan occurrence of nertal forces and snce the space of the fxed and movng frames may coupled 2 therefore any change n poston vector relatve to any one of these frames wll be faced by opposte change n poston vector n the other one (appears as resstance to moton). These transformatons could have been notced earler f Gallean transformaton was parameterzed wth constant acceleraton nstead of constant velocty, that s, r = r ½at 2, the nstantaneous poston vector, and ts frst dfferentaton ṙ = ṙ at s the nstantaneous nertal frame transformaton equaton, and ts second dfferentaton s r = r a whch can be rearrange to a = r + ( r ), where r s an nertal acceleraton. Snce a transformaton for almost every knematc and dynamc quantty has been obtaned, a sutable substtuton of t n classcal equatons (e.g., Newton s equatons of moton, etc.) wll help understandng the role of the space and nertal forces n physcs. The analyss cover only n detal the orbtal moton whereas the case of concdng of center of mass and center of rotaton (pure spn) s not covered n ths analyss and t need a specal mathematcal treatment to derve ts nertal forces wthout causng cancellaton of actve forces by nertal forces (both are equal and opposte). Fnally, a crude observaton of the reported phenomenon can be obtan easly by rotatng a metallc sold dsk pvoted at ts center or by rotatng a vessel of water contanng ce cubes and t can be exercse wth hands. dottng each one wth tself. 2 Snce two frames can couple to form a thrd frame, whch n ts turn can couple to a fourth one to form a ffth, etc., then, the logcal consequence s the formaton of a master contaner one-frame,.e., the statement of nfntely many nertal frames and absolute space becomes equvalent n the presence of couplng. 12

13 4 Concluson A rgd body that angularly moves n a curvlnear path, wll spn under the nfluence of nertal forces, exclusvely, Euler and Corols forces. The nertal force supples a curvlnearly movng rgd body wth an addtonal rotatonal knetc energy and angular momentum whch are ndependent from the rotatonal knetc energy and angular momentum that have been suppled by the actve force. The rotatonal (orbtal) angular momentum of a rgd body whch undergoes rotatonal acceleraton s mutually exchange wth ts spn angular momentum and that happens n order to conserve the total angular momentum of the rgd body. The angular moton of the center of mass of a rgd body relatve to a fxed pont s equvalent to superposton of angular motons of ts mass elements relatve to the that fxed pont and relatve to the center of mass. Ths s the spn-rotaton couplng theorem whch has been summarzed from the precedng analyss. The parallel axs theorem couplng of rotatonal (orbtal) dynamcs of a crcularly accelerated rgd body to ts spn dynamcs. It maps moment of nerta of a rgd body to a moment of nerta of pont mass. Any mechancal force (rectlnear, azmuthal, centrpetal and Corols) when acts over a rgd body, t de-synthess nto actve and nertal forces. 13

14 Appendces Appendx A I. Dervaton of the couplng formula of spn and rotatonal angular momentums Takng the frst term n the RHS of Equaton (3) and substtute the poston vector equaton r = r CM + ρ (Ths substtuton s the man devce whch brngs us to another level of analyss of these formulae and the results follow drectly from t), and snce m = m then we have r CM m(ω r CM ) = (r ρ ) m (ω (r ρ )), = r m (ω r ) ρ m (ω r ) r m (ω ρ ) + ρ m (ω ρ ), = r m (ω r ) ρ m (ω (r CM + ρ )) (r CM + ρ ) m (ω ρ ) + ρ m (ω ρ ), = r m (ω r ) m ρ (ω r CM ) ρ m (ω ρ ) r CM ω m ρ ρ m (ω ρ ) + ρ m (ω ρ ), snce ρ s the vector poston of mass element m relatve to the center of mass therefore from the defnton of the center of mass, we have[see 4, p. 98] m ρ = 0, (37) 14

15 whch mples that r CM m(ω r CM ) = r m (ω r ) ρ m (ω ρ ), (38) usng the dentty A (B C) = (A C) B (A B) C, (39) and usng the facts that ρ and ω are mutually orthogonal and so are r and ω. Therefore, one fnds r CM m(ω r CM ) = m r 2 ω m ρ 2 ω = I Oω + I CM ( ω), (40) where I O = m r 2, (41) s the moment of nerta of the rgd body relatve to the axs of rotaton O, whch s a perpendcular dstance r CM from the center of mass, and I CM = m ρ 2, (42) s the moment of nerta of the rgd body relatve to ts center of mass[see 14, p. 246]. Therefore, Equaton (3), the total angular momentum becomes L = I O ω + I CM ( ω) + I CM Ω, (43) The term I CM ( ω) s an addtonal angular momentum term relatve to the center of mass of the rgd body (spn angular momentum) and occurs due to the rgd body rotatonal moton relatve to the axs of rotaton O. The term I CM Ω can be consder as the ntal spn angular momentum that the rgd body acqured before t start ts rotatonal moton and snce Ω s arbtrary, so that t can be zero and have not to be a mandatory term of Equaton (43). Thus, one can wrtes L = I O ω + I CM ( ω), and by wrtng I O ω = L R and I CM ( ω) = L S, we obtan Equaton (5). 15

16 II. Dervaton of the parallel axs theorem from the couplng formula of spn and rotatonal angular momentums Takng the dot product of Equaton (5) wth tself, we get L 2 = L 2 R + L 2 S + 2 L R L S, usng Identty (39) to smplfy the term L = r CM (mω r CM ), and snce the moton s planar then r CM and ω are mutually orthogonal, so that we have ((r CM r CM ) mω) 2 = (I O ω) 2 + (I CM ( ω)) (I O ω) (I CM ( ω)), m 2 r 4 CM ω 2 = I 2 Oω 2 + I 2 CMω 2 2I O I CM ω 2, ( mr 2 CM) 2 ω 2 = (I O I CM ) 2 ω 2, (44) dvdng nto ω 2 and then takng the square root and rearrange, we get I O = mr 2 CM+I CM, whch s the parallel axs theorem[see 14, p. 249]. III. Dervaton of the couplng formula of spn and rotatonal knetc energes Takng the frst term n the RHS of Equaton (9) and usng the dentty (A B C D) = (A C) (B D) (A D) (B C), (45) and the fact that r CM and ω are mutually orthogonal and accompany wth sutable substtutons of Equaton (4), one obtans 16

17 ½m (ω r CM ω r CM ) = ½m (ω ω) (r CM r CM ) = ½m (ω ω) ( r ρ r ρ ), = ½m (ω ω) ((r r ) 2 ( r ρ ( ) + ρ ρ ) ), ( = ½ m (r r ) (ω ω) m r ρ ) (ω ω) ( + ½ m ρ ρ ) (ω ω), ( = ½ m (r r ) (ω ω) m rcm + ρ ρ ) (ω ω) ( + ½ m ρ ρ ) (ω ω), = ½ m (r r ) (ω ω) r CM m ρ (ω ω) ( m ρ ρ ) ( (ω ω) + ½ m ρ ρ ) (ω ω), (46) usng equatons (37), (41) and (42), we get ½m (ω r CM ω r CM ) = ½I O (ω ω) + ½I CM ( ω ω), (47) substtutng Equaton (47) back nto Equaton (9), the knetc energy formula becomes T = ½I O (ω ω) + ½I CM ( ω ω) + ½I CM (Ω Ω), (48) The term ½I CM (Ω Ω) can be consder as the ntal spn knetc energy whch s the rgd body have before t start ts rotatonal moton and snce Ω s arbtrary, so that t can be zero and have not to be a mandatory term of Equaton (48). Therefore, we can wrte T = ½I O (ω ω) + ½I CM ( ω ω), and by wrtng ½I O (ω ω) = T R and ½I CM ( ω ω) = T S, we obtan Equaton (10). 17

18 IV. Dervaton of the parallel axs theorem from the couplng formula of spn and rotatonal knetc energes Takng Equaton (10) and wrte t explctly usng Identty (45) to smplfy the term T (see Equaton (46)), so we have T = T R + T S, ½m (ω ω) (r CM r CM ) = ½I O (ω ω) + ½I CM ( ω ω), ( mr ) ( 2 CM ½ω 2 = I ) ( O ½ω 2 I ) CM ½ω 2, (49) dvdng nto ½ω 2 and rearrange, we obtan I O = mr 2 + I CM CM, whch s the parallel axs theorem. V. Dervaton of the couplng formula of spn and rotatonal actng forces Substtutng Equaton (4) nto (11), we have r CM m( ω r CM ) = (r ρ ) m( ω (r ρ )), = r m ( ω r ) ρ m ( ω r ) r m ( ω ρ ) + ρ m ( ω ρ ), = r m ( ω r ) ρ m ( ω (r CM + ρ )) (r CM + ρ ) m ( ω ρ ) + ρ m ( ω ρ ), = r m ( ω r ) m ρ ( ω r CM ) ρ m ( ω ρ ) r CM ω m ρ ρ m ( ω ρ ) + ρ m ( ω ρ ), usng Equaton (37), we obtan r CM m( ω r CM ) = r m ( ω r ) + ρ m ( ω ρ ), (50) 18

19 usng Identty (39) accompany wth the facts that ρ and ω are mutually orthogonal and so are r and ω and then smplfy usng equatons (41) and (42). Therefore we can wrte r CM m( ω r CM ) = I O ω + I CM ( ω). Puttng I O ω = τ R and I CM ( ω) = τ S, we obtan Equaton (12). VI. Dervaton of the parallel axs theorem from couplng formula of spn and rotatonal actng forces Takng the dot product of Equaton (12) wth tself, we get τ 2 = τ 2 R + τ 2 S + 2 τ R τ S, usng Identty (39) to smplfy the term τ = r CM m( ω r CM ), and snce the moton s planar therefore r CM and ω are mutually orthogonal, so that we have ((r CM r CM ) m ω) 2 = (I O ω) 2 + (I CM ( ω)) (I O ω) (I CM ( ω)), m 2 r 4 CM ω 2 = I 2 O ω 2 + I 2 CM ω 2 2I O I CM ω 2, ( mr 2 CM) 2 ω 2 = (I O I CM ) 2 ω 2, (51) dvdng nto ω 2 and then takng the square root and rearrange, we get I O = mr 2 CM+I CM, whch s the parallel axs theorem. Appendx B VII. Calculaton of the frst change n poston vector r = r Q + r Takng the frst change n poston vector (Equaton (15)) wth respect to tme, yelds[see 3, p. 97] ṙ = ṙ Q + ṙ, u + ω r = (u Q + ω r Q ) + (u + ω r ), The angular veloctes ω, ω and ω are due to the rotaton of the vectors poston r Q, r and r respectvely, and snce the moton s happenng to a rgd body therefore 19

20 we have ω = ω = ω. Hence, we can wrte u + ω r = (u Q + ω r Q ) + (u + ω r ), (52) and wth the help of Equaton (15), Equaton (52) gves u Q u + u = ω (r Q r + r ) = 0, (53) therefore we have u Q = u u, whch s Equaton (16) and ω r Q = ω r ω r, whch s Equaton (17). VIII. Retrevng of the couplng formula of spn and rotatonal angular momentums In Equaton (17), let r Q = r CM and r = ρ, then by cross multplyng by r CM and multply by m usng the fact that m = m, one obtans r CM m(ω r CM ) = r CM ω m r r CM ω m ρ, usng equatons (37) and (15), gves r CM m(ω r CM ) = (r ρ ) m (ω r ) r CM (0), = r m (ω r ) ρ m (ω (r CM + ρ )), = r m (ω r ) m ρ (ω r CM ) ρ m (ω ρ ), whch reduces by Equaton (37) to Equaton (38) and then reduces by Identty (39) to r CM m(ω r CM ) = I O ω + I CM ( ω), whch s Equaton (40) or (5). 20

21 IX. Calculaton of the second change n poston vector r = r Q + r Takng the second change n poston vector (Equaton (15)) wth respect to tme, yelds r = r Q + r, u + ω r + ω ṙ = ( u Q + ω r Q + ω ṙ Q ) + ( u + ω r + ω ṙ ), substtutng ṙ, ṙ Q and ṙ from Equaton (52), gves[see 3; 15, p. 97 ;p. 250] u + ω r + ω u + ω (ω r ) = ( u Q + ω r Q + ω u Q + ω (ω r Q )) + ( u + ω r + ω u + ω (ω r )), substtutng the values of u, u Q, u, gves a + ω r + 2ω u + ω (ω r ) = (a Q + ω r Q + 2ω u Q + ω (ω r Q )) + (a + ω r + 2ω u + ω (ω r )), (54) Regroupng by the types of acceleratons, we obtan 0 = [a Q (a a )] + [ ω r Q ( ω r ω r )] + [2ω u Q (2ω u 2ω u )] + [ω (ω r Q ) {ω (ω r ) ω (ω r )}], (55) wth the ad of Equaton (15) we fnd that the grouped terms n the second and fourth square brackets n Equaton (55) are equal to zeros, that s ω (r Q r + r ) = 0, and (56) ω (ω (r Q r + r )) = 0, (57) and Equaton (16) mples that the grouped terms n the thrd square brackets n Equaton (55) are equal to zero, that s 2ω (u Q u + u ) = 0, (58) 21

22 and thus equatons (56), (57) and (58) mply that the grouped terms n the frst square brackets n Equaton (55) are also equal to zero, that s a Q a + a = 0. (59) Therefore, equatons (56) to (59) can be rewrtten as followng: ω r Q = ω r + ( ω r ) (60) 2ω u Q = 2ω u + ( 2ω u ) (61) ω (ω r Q ) = ω (ω r ) + ( ω (ω r )) (62) a Q = a + ( a ) (63) The sum of equatons (60) to (63) s equals to Equaton (54): a Q + ω (ω r Q ) + ω r Q + 2ω u Q = a + ω (ω r ) + ω r + 2ω u + ( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω u ), whch s Equaton (22). Appendx C X. Calculaton of the contrbuton of Euler force to the spn torque Cross multply Equaton (29) by r CM, and wth the ad of Equaton (15) after lettng r Q = r CM and r = ρ then one fnds r CM m( ω r CM ) = (r ρ ) m ( ω r ) + r CM (0), = r m ( ω r ) + ρ m ( ω r ), = r m ( ω r ) + ρ F Euler,, = r m ( ω r ) + τ Euler. (64) 22

23 where τ Euler s an nertal torque occurs due to Euler force[see 15; 16, p. 251; p. 469] F Euler, = m ( ω r ) and whch acts on the mass element m. Then, wth the help of equatons (15) and (37), we obtan (see steps to Equaton (50)) { }} { τ Euler = ρ m ( ω r ) = ρ m ( ω ρ ), (65) τ S whch leads to Equaton (34). XI. Calculaton of the contrbuton of Corols force to the spn torque Cross multplyng Equaton (30) by r CM, and wth the help of equatons (15), (37) and (16) we wll obtan the torque due to Corols force, that s r CM m(2ω u CM ) = (r ρ ) m (2ω u ) + r CM (0), = r m (2ω u ) + ρ m ( 2ω u ), = r m (2ω u ) + ρ F Corols,, = r m (2ω u ) + τ Corols. (66) where τ Corols s an nertal torque occurs due to Corols force[see 15; 17, p. 251; p. 233] F Corols, = m ( 2ω u ) and whch acts on the mass element m. Then, wth the help of equatons (16) and (37), one fnds τ Corols = ρ m ( 2ω u ) = ρ m ( 2ω u ), (67) whch leads to Equaton (35). 23

24 XII. The null contrbuton of centrpetal force to the spn torque Cross multplyng Equaton (31) by r CM and accompany wth the help of equatons (15) and (37), we wll obtan the torque due to centrpetal force, that s r CM m(ω (ω r CM )) = (r ρ ) m (ω (ω r )) + r CM (0), = r m (ω (ω r )) + ρ m ( ω (ω r )), = r m (ω (ω r )) + ρ F Centrfugal,, = r m (ω (ω r )) + τ Centrfugal. (68) where τ Centrfugal s an nertal torque occurs due to centrfugal force[see 15, p. 251] F Centrfugal, = m ( ω (ω r )) and whch acts on the mass element m. Then, wth the help of equatons (15) and (37), one obtans τ Centrfugal = ρ m ( ω (ω r )) = ρ m ( ω (ω ρ )) (69) Usng Identty (39) and the fact that the moton s planar, Equaton (68) gves r CM m( ω 2 r CM ) = r m ( ω 2 r ) ρ m ( ω 2 ρ ), (70) snce the cross product of a vector wth tself s zero then all terms of Equaton (70) are zeros, that s, the centrpetal force does not contrbute to the counter spn of the rgd body. 24

25 XIII. The null contrbuton of rectlnear force to the spn torque Cross multplyng Equaton (32) by r CM, we wll obtan the torque due to the rectlnear force, that s r CM ma CM = (r ρ ) m a + r CM (0), = r m a + ρ m ( a ), = r m a + ρ F Rectlnear,, = r m a + τ Rectlnear, (71) where τ Rectlnear s an nertal torque occurs due to rectlnear force F Rectlnear, = m ( a ) and whch acts on the mass element m. Then, wth the help of equatons (59) and (37), one fnds τ Rectlnear = ρ m ( a ) = ρ m ( a ), (72) therefore Equaton (71) becomes r CM ma CM = r m a + ρ m ( a ), (73) snce the vectors r CM and a CM are parallel and so are r, a and ρ, a then all terms of Equaton (73) are zeros due to the cross product of parallel vectors. Therefore, the rectlnear force also does not contrbute to the counter spn of the rgd body.. We now can summarze all nertal forces F whch relatve to the space-fxed coordnate system S and act over mass element m : F = m [( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω u )], (74) and all nertal forces F whch relatve to the body-fxed coordnate system S wth orgn at center of mass: F = m [( a ) + ( ω (ω ρ )) + ( ω ρ ) + ( 2ω u )], = m [ 0 + ( ω (ω ρ )) + ( ω ρ ) + 0 ], (75) 25

26 where u = ρ = 0 and a = ρ = 0 snce ρ s constant. Hence and despte the fact that both rectlnear and Corols nertal forces whch are relatve to the body-fxed coordnate system S are predcted by an observer on S -frame to be equal to zero, ths observer stll observes acton of smlar forces over the rgd body on hs or her frame, specally the Corols effect where the rgd body spns whle the S -frame transpose-rotates relatve to an orgn fxed n space (see appendx C, XI). Therefore, that gves a clue that nertal forces whch are relatve to the bodyfxed coordnate system and that whch are relatve to the space-fxed coordnate system are not one n the same and n spte of equatons (65),(67),(69) and (72) are hnted to the opposte of that meanng 3. A smlar pattern can be recognzed n Equaton (38) where the negatve lnear momentum n the second term of RHS can be attrbuted to moton that relatve to the space-fxed coordnate system, that s, ρ m ( ω ρ ) = ρ m ( ω r ); the addtonal acqured momentum s due to the moton relatve to the space. 3 Here, the cause (space) of the effect (nertal forces) have been solated. The effect occurs due to the nteracton of the matter and space, or superposton of space-fxed and body-fxed frames of reference. 26

27 Fgure 1: Rotaton of the rgd body A relatve to the fxed pont O. 27

28 Fgure 2: Body-fxed S and space-fxed coordnate systems S. 28

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