The classical spinrotation coupling


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1 LOUAI H. ELZEIN 2018 All Rghts Reserved The classcal spnrotaton couplng Loua Hassan Elzen Basher Khartoum, Sudan. Postal code:11123 Abstract Ths paper s prepared to show that a rgd body whch accelerates crcularly 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 the nduced spn moton of the rgd body to the angular momentum and knetc energy due to the rotatonal moton of the same spnnng rgd body have been derved. The paper also brngng to lght the nature of the force whch causes the nducton of spn moton n a rgd body and a formula whch couplng of ths hghlghted force to the force whch causes the rotaton of the rgd body has been also derved.
2 LOUAI H. ELZEIN 1 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. [1] John Taylor. The statement brefs the common understandng wthn scentfc communty about spnrotaton relaton for a rgd body n crcular moton. Hence, here we are gong to prove that the negaton of ths statement s what s true. We begnnng wth the dstncton between rotatonal and spnnng moton therefore we defne rotaton wth the moton of the rgd body crcularly around a fxed pont or a center of rotaton whch s dfferent from ts center of mass whle the spnnng s the rotaton of the rgd body around ts center of mass. Another thng s that the analyss s gong to be wth planar rgd body n planar moton over Eucldean space. The followng three subsectons, 2.1, 2.2, 2.3, can consdered to be as the observaton of the couplng phenomena and whch has been obtaned from mathematcal analyss of the crcular moton of the rgd body. The last subsecton, 2.4, gves the theoretcal explanaton to ths phenomena. 2 Analyss 2.1 Couplng of the spnnng and rotatonal angular momentum Referrng to Fgure 1, a rgd body A of a mass m s free to rotate relatve to ts center of mass CM as t s also can be, smultaneously, free to rotate relatve to the pont O (nertal frame). Thus, t s pvoted at these two ponts. It s known that the total angular momentum L of the rgd body A n ts crcular moton s gven by[1]: (1) L = r CM p + ρ ρ m where r CM s the vector poston of the center of mass of the rgd body relatve to the axs of rotaton O, p s the lnear momentum and ρ s the vector poston of the element mass m relatve to the center of mass of the rgd body. 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 reexpress Equaton (1)
3 2 THE CLASSICAL SPINROTATION COUPLING to say[1] (2) L = L moton of CM + L moton relatve to CM Snce the mass s constraned to a crcle, the tangental velocty of the mass of the rgd body A s ω r CM where ω s the rotatonal angular velocty of the rgd body relatve to the axs of rotaton O whch s perpendcular to the plane of the moton and snce p = mω r CM, the total angular momentum equaton (Equaton (1)) becomes (assumng the moton s planar, hence both axses of rotaton, O and CM, are parallel): (3) L = r CM (mω r CM ) + ρ (Ω ρ )m where Ω s the rgd body spnnng angular velocty (arbtrary) relatve to ts center of mass. Takng the frst term n the RHS and usng the poston vector equaton 1 (4) r = r CM + ρ one fnds r CM (mω r CM ) = (r ρ ) (mω (r ρ )), = r (mω r ) ρ (mω r ) r (mω ρ ) + ρ (mω ρ ), = r (mω r ) ρ (mω (r CM + ρ )) (r CM + ρ ) (mω ρ ) + ρ (mω ρ ), = r (mω r ) ρ (mω r CM ) ρ (mω ρ ) r CM (mω ρ ) ρ (mω ρ ) + ρ (mω ρ ), = r (mω r ) ρ (mω r CM ) ρ (mω ρ ) r CM (mω ρ ), snce m = m, where m s a mass element, hence we can wrte r CM (mω r CM ) = r (m ω r ) m ρ (ω r CM ) ( ρ (m ω ρ ) r CM ω ) m ρ, 1 Ths substtuton s the man tool whch brngs us to another level of analyss of these formulae and the results follow drectly from t.
4 LOUAI H. ELZEIN 3 because ρ s the vector poston of a mass element m relatve to the center of mass, then from the defnton of the center of mass we have[2] (5) m ρ = 0, whch mples that (6) r CM (mω r CM ) = usng the dentty (7) r (m ω r ) A (B C) = (A C) B (A B) C ρ (m ω ρ ), and usng the facts that ρ and ω are mutually orthogonal and so are r and ω. Therefore, one wll fnd r CM (mω r CM ) = m r 2 ω m ρ 2 ω, (8) where (9) = I O ω I CM ω, I O = m r 2, s the moment of nerta relatve to the axs of rotaton O, whch s a perpendcular dstance r CM from the centre of mass, and (10) I CM = m ρ 2, s the moment of nerta of the rgd body relatve to ts center of mass[3]. Thus, Equaton (3), the total angular momentum, becomes L = I O ω I CM ω + ρ (Ω ρ )m, (11) = I O ω I CM ω + I CM Ω. The term I CM ω s an addtonal angular momentum term relatve to the center of mass of the rgd body (spnnng) and occurs due to the rgd body rotatonal or crcular moton relatve to the axs of rotaton O. The term I CM Ω can be consder as the ntal spnnng angular momentum that the rgd body acqured before t start ts crcular moton and snce Ω s arbtrary, so that t can be zero and have not to be a mandatory term of Equaton (11). Hence, we can wrte (12) L = I O ω I CM ω, = L R + L S.
5 4 THE CLASSICAL SPINROTATION COUPLING where L = r CM (mω r CM ) s the total angular momentum, L R = I O ω s the rotatonal angular momentum and L S = I CM ( ω) s the spnnng angular momentum. Snce the total angular momentum (Equaton (12)) s conserved, that mples the rotatonal and spnnng angular momentum are mutually exchangeable n order to conserve t, that s (13) L = L R + L S Ths property (Equaton (13)) negate the above statement whch we have begn wth. Another thng we can notce s that f we smplfy the term r CM (mω r CM ) n Equaton (8) usng the dentty (7) and rearrange t, that yelds I O ω = [(r CM r CM ) mω (r CM mω) r CM ] + I CM ω, snce moton s planar then r CM and ω are mutually orthogonal, so that I O ω = mr 2 CM ω + I CM ω, dotted wth ω and then dvde by ω 2, we have (14) I O = mr 2 CM + I CM. whch s nothng other than the parallel axs theorem[3]. 2.2 Couplng of the spnnng and rotatonal knetc energy 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[4]: (15) T = ½m (ω r CM ω r CM ) + ½I CM (Ω Ω) 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 an arbtrary spnnng velocty relatve to the center of mass of the rgd body. Takng the frst term n the RHS and usng the dentty (A B C D) = (A C) (B D) (A D) (B C), one obtans (16) ½m (ω r CM ω r CM ) = ½m (ω ω) (r CM r CM ),
6 LOUAI H. ELZEIN 5 substtute Equaton (4), usng the fact that ω and r CM are mutually orthogonal, we have 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 (r CM + ρ ρ ) (ω ω) + ½ m (ρ ρ ) (ω ω), = ½ m (r r ) (ω ω) m (ρ ρ ) (ω ω) ( r CM ) m ρ (ω ω) + ½ m (ρ ρ ) (ω ω), usng Equaton (5), (9) and (10), we obtan (17) ½m (ω ω) (r CM r CM ) = ½I O (ω ω) ½I CM (ω ω). An mportant thng has to be notce n Equaton (17) when we rearrange t as followng: ½I O (ω ω) = ½m (ω ω) (r CM r CM ) + ½I CM (ω ω), dong the dot product and then dvde nto ½ω 2, we wll obtan I O = mr 2 CM + I CM, whch agan s the parallel axs theorem. Snce there s no negatve knetc energy therefore one can understand that the negatve sgn s assgn to the drecton nstead of the magntude. Substtutng Equaton (17) back nto Equaton (15), the knetc energy formula becomes (18) T = ½I O (ω ω) + ½I CM ( ω ω) + ½I CM (Ω Ω). The term ½I CM (Ω Ω) can be consder as the ntal spnnng knetc energy whch s the rgd body have before t start ts crcular moton and snce Ω s arbtrary, so that t can be zero and have not to be a mandatory term of Equaton (18). Hence we can wrte (19) T = ½I O (ω ω) + ½I CM ( ω ω), = T R + T S.
7 6 THE CLASSICAL SPINROTATION COUPLING where T = ½m (ω r CM ω r CM ) s the total knetc energy, T R = ½I O (ω ω) s the rotatonal knetc energy and T S = ½I CM ( ω ω) s the spnnng knetc energy. 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 Fgure 1 f an external force F acts on the center of mass of the rgd body A such that t causes t to angularly accelerate wth angular acceleraton ω = (dω/dt) relatve to the axs of rotaton O where ω s the angular velocty of rotaton of the center of mass of the rgd body relatve to the axs of rotaton O and snce the mass m s constraned to a crcle, the tangental acceleraton of the mass of the rgd body A s ω r CM, and snce F = ma, the total torque equaton s gven by: (20) τ = r CM F, = r CM (m ω r CM ). where τ s the total torque and r CM s the vector poston of the center of mass of the rgd body relatve to the axs of rotaton O. Substtutng Equaton (4), we fnd r CM (m ω r CM ) = (r ρ ) (m ω (r ρ )), = r (m ω r ) r (m ω ρ ) ρ (m ω r ) + ρ (m ω ρ ), = r (m ω r ) (r CM + ρ ) (m ω ρ ) ρ (m ω (r CM + ρ )) + = ( r (m ω r ) r CM ω ρ (m ω ρ ), m ρ ) ρ (m ω ρ ) ρ (m ω ρ ) + m ρ ( ω r CM ) ρ (m ω ρ ),
8 LOUAI H. ELZEIN 7 usng Equaton (5), one fnds (21) r CM (m ω r CM ) = r (m ω r ) ρ (m ω ρ ), usng the dentty (7) accompany wth the facts that ρ and ω are mutually orthogonal and so are r and ω. Then smplfy usng Equaton (9) and (10). Therefore, one wll have r CM (m ω r CM ) = m r 2 ω m ρ 2 ω, (22) We can wrte t as: (23) = I O ω I CM ω. τ = τ R + τ S where τ = r CM (m ω r CM ) s the total torque. τ R = I O ω s the rotaton torque and τ S = I CM ω s the spn torque. If we rearrange Equaton (22) as followng: I O ω = r CM (m ω r CM ) + I CM ω, and usng the dentty (7) wth the fact that r CM and ω are mutually orthogonal, we obtan I O ω = mr 2 CM ω + I CM ω, dotted wth ω and dvde nto ω 2, we agan obtan I O = mr 2 CM + I CM, the parallel axs theorem. 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 (22), that s, I CM ω. Referrng to Fgure 2, we have a spacefxed coordnate system, S, whch s a coordnate system wth the orgn fxed n space at pont O, and wth spacefxed drectons for the axes. We have also a bodyfxed 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[5]: (24) r = r CM + r. where r s the poston vector of pont n the spacefxed reference system S. r s the poston vector of pont n the bodyfxed reference system
9 8 THE CLASSICAL SPINROTATION COUPLING S. r CM s the poston vector of the reference pont Q n the spacefxed reference system S. Frst change n poston vector: Takng the frst change n poston vector (Equaton (24)) wth respect to tme, yelds ṙ = ṙ CM + ṙ, v + ω 1 r = (v CM + ω 2 r CM ) + (v + ω 3 r ), where v s the velocty of pont relatve to the orgn of Sframe, v CM s the velocty of the orgn of S frame relatve to the orgn of Sframe, and v s the velocty of pont relatve to the orgn of S frame. The angular veloctes ω 1, ω 2 and ω 3 are due to the rotaton of the vectors poston r, r CM and r, respectvely. Snce the moton s happenng to a rgd body, therefore we have ω 1 = ω 2 = ω 3 = ω. Hence, we can wrte (25) (26) (27) (28) v + ω r = (v CM + ω r CM ) + (v + ω r ). Wth the help of Equaton (24), Equaton (25) yelds Therefore, we have v CM v + v = ω (r CM r + r ) = 0. v CM = v v, and ω r CM = ω r ω r. Second change n poston vector: Takng the second change n poston vector (Equaton (24)) wth respect to tme, yelds substtutng ṙ, ṙ CM r = r CM + r, v + ω r + ω ṙ = ( v CM + ω r CM + ω ṙ CM ) + ( v + ω r + ω ṙ ), and ṙ from Equaton (25) and (24), that gves[6] v + ω r + ω v + ω (ω r ) substtutng the values of v, v CM, v, that gves a + ω r + 2ω v + ω (ω r ) (29) = ( v CM + ω r CM + ω v CM + ω (ω r CM )) + ( v + ω r + ω v + ω (ω r )), = (a CM + ω r CM + 2ω v CM + ω (ω r CM )) + (a + ω r + 2ω v + ω (ω r )).
10 LOUAI H. ELZEIN 9 where a s the acceleraton of pont relatve to the orgn of Sframe, a CM s the acceleraton of the orgn of S frame relatve to the orgn of S frame, and a s the acceleraton of pont relatve to the orgn of S frame. Regroupng by acceleraton type, we obtan (30) 0 = [a CM (a a )] + [ ω r CM ( ω r ω r )] + [2ω v CM (2ω v 2ω v )] + [ω (ω r CM ) { ω (ω r ) ω (ω r ) } ]. Wth the ad of Equaton (24), we fnd that, the grouped terms n the second and fourth square brackets n Equaton (30) are equal to zeros, that s (31) (32) ω (r CM r + r ) = 0, and ω [ω (r CM r + r )] = 0. Equaton (27) mples that, the grouped terms n the thrd square brackets n Equaton (30) are equal to zero, that s (33) 2ω [v CM v + v ] = 0. Equatons (31), (32) and (33) mply that, the grouped terms n the frst square brackets n Equaton (30) are also equal to zero, that s (34) a CM a + a = 0. Thus, Equatons (31), (32), (33) and (34) can be rewrtten as followng: (35) (36) (37) (38) ω r CM = ω r ω r. 2ω v CM = 2ω v 2ω v. ω (ω r CM ) = ω (ω r ) ω (ω r ). a CM = a a. The sum of Equatons (35) to (38) equals to Equaton (29), that s a CM + ω (ω r CM ) + ω r CM + 2ω v CM (39) = a + ω (ω r ) + ω r + 2ω v + ( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω v ). The forces that actng over the rgd body: From Equatons (35) to (39), and by multplyng by mass m and rememberng that m = m, we obtan
11 10 THE CLASSICAL SPINROTATION COUPLING Tangental force: (40) m ω r CM = m ω r m ω r. Corols force: (41) 2ω mv CM = 2ω m v 2ω m v. Centrpetal force: ( ( ) (42) ω m(ω r CM ) = ω ω m r ) ω ω m r. Rectlnear force: (43) ma CM = m a m a. The combnaton of these forces gves the total force: T otalforce {}}{ m[a CM + ω (ω r CM ) + ω r CM + 2ω v CM ] = Actve force {}}{ m [a + ω (ω r ) + ω r + 2ω v ] (44) + Inertal force { }}{ m [( a ) + ( ω (ω r )) + ( ω r ) + ( 2ω v )]. We fnd that, generally, every sngle force of the above forces (Equatons (40) to (43)), s a synthess of an actve force and an nertal force. Smlarly, the total force (Equaton (44)). 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 (5) and ts frst and second dervatve wth respect to tme, then, Equatons (40) to (44) wll gve
12 LOUAI H. ELZEIN 11 Tangental force: (45) m ω r CM = m ω r. Corols force: (46) 2ω mv CM = 2ω m v. Centrpetal force: (47) ω m(ω r CM ) = ω ( ω m r ). Rectlnear force: (48) ma CM = m a. The combnaton of these forces gves the total force: m[a CM + ω (ω r CM ) + ω r CM + 2ω v CM ] (49) = m [a + ω (ω r ) + ω r + 2ω v ] The contrbuton of the Euler force to the spn torque If the rgd body A s crcularly acceleratng, then by cross multplyng Equaton (45) wth r CM, we wll obtan the torque due to the tangental force, that s r CM (m ω r CM ) = (r ρ ) (m ω r ), = = r (m ω r ) r (m ω r ) ρ (m ω (r CM + ρ )), m ρ ( ω r CM ) ρ (m ω ρ ), usng Equaton (5), we obtan (50) r CM (m ω r CM ) = r (m ω r ) ρ (m ω ρ ).
13 12 THE CLASSICAL SPINROTATION COUPLING Substtutng Equaton (24) nto the second term of RHS of Equaton (50) and then wth the help of Equaton (5), we fnd ρ [ (m ω ρ )] = = ρ [ m ω (r r CM )], ρ [ (m ω r )], = ρ F, = τ Euler. where τ Euler s an nertal torque whch occurs due to the Euler force[7], F = m ω r = m ω ρ, whch acts on the mass element m. Therefore, we fnd (51) τ Euler = ρ (m ω ρ ) = I CM ω. Ths force causes the spnnng 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. Hence, Equaton (50) becomes (52) r CM (m ω r CM ) = I O ω I CM ω, = τ R + τ Euler. Equaton (52) s dentcally Equaton (22) whch derved earler n subsecton (2.3), whch mples that τ Euler = τ S, that s, the counter rotaton of the rgd body relatve to ts center of mass occurs due to the Euler nertal force 2. Retrevng the angular momentum formula: If the rgd body A s crcularly movng, then by cross multplyng Equaton (28) wth r CM and then multply by m, usng the fact that m = m, then we wll obtan ( r CM (mω r CM ) = r CM ω ) ( m r r CM ω ) m ρ, 2 In ths case, Euler force acts lke a relatve feld of force.
14 LOUAI H. ELZEIN 13 usng Equaton (5), gves r CM (mω r CM ) = (r ρ ) (m ω r ), = = r (m ω r ) r (m ω r ) ρ (m ω (r CM + ρ )), m ρ (ω r CM ) ρ (m ω ρ ), whch reduces by Equaton (5) to (53) r CM (mω r CM ) = r (m ω r ) ρ (m ω ρ ). Equaton (53), dentcally, s Equaton (6), the angular momentums couplng formula whch derved earler n subsecton (2.1) The contrbuton of the Corols force to the spn torque If the rgd body A s radal translate whle unformly rotatng, then by cross multplyng Equaton (46) wth r CM, we wll obtan the torque due to the Corols force, that s ( r CM (2ω mv CM ) = r CM 2ω ) m v, = (r ρ ) (2ω m v ), (54) = r (2ω m v ) ρ (2ω m v ). Thus, the Corols nertal force contrbutes to the spnnng of the rgd body wth the nertal torque (55) τ Cor = ρ (2ω m v ) where τ Cor s an nertal torque whch occurs due to the Corols force, F Cor, = 2ω m v, whch acts on the mass element m. Ths force, also, causes the spnnng 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.
15 14 THE CLASSICAL SPINROTATION COUPLING The contrbuton of the centrpetal force to the spn torque If the rgd body A s unformly rotatng, then by cross multplyng Equaton (47) wth r CM, we wll obtan the torque due to the centrpetal force, that s [ ( r CM [ω (mω r CM )] = r CM ω ω )] m r, = = (r ρ ) [ω (m ω r )], r [ω (m ω r )] (56) ρ [ω (m ω r )], usng the dentty (7), rememberng that the moton s planar. Therefore, one fnds r CM [ mω 2 r CM ] = r [ m ω 2 r ] ρ [ m ω 2 r ]. Snce the cross product of a vector wth tself s zero, so we have Whch mples that r CM [ mω 2 r CM ] = 0, and r [ m ω 2 r ] = 0. ρ [ m ω 2 r ] = 0. Thus, the centrpetal force does not contrbute to the spnnng of the rgd body The contrbuton of the rectlnear force to the spn torque If the rgd body A s rectlnearly acceleratng, then by cross multplyng Equaton (48) wth r CM, we wll obtan the torque due to the rectlnear force, that s r CM ma CM = r CM m a, = (r ρ ) m a, (57) = r m a ρ m a.
16 LOUAI H. ELZEIN 15 Usng the facts that r CM and a CM are parallel to each other and so are r and a. Therefore, one fnds Whch mples that r CM ma CM = 0, and r m a = 0. ρ m a = 0. Hence, the rectlnear force does not contrbute to the spnnng of the rgd body. Now, we can wrte the equaton of moton whch descrbes all the forces that causng rotaton couplng to the spnnng: τ {}}{ r CM (m ω r CM ) + r CM (2ω mv CM ) (58) { }}{ = r (m ω r ) + r (2ω m v ) τ R τ S { [ }}{ ρ (m ω ρ ) + ] ρ (2ω m v ). where τ s the total torque, τ R s the rotaton torque or the actve torque, and τ S s the spn torque or the nertal torque. 3 Concluson A rgd body that angularly moves n a curvlnear path, wll spn under the nfluence of nertal force, exclusvely, the Euler and Corols forces. In opposte to the translatonal moton where nerta presents as resstance of the mass to moton, the mass when moves curvlnearly t spns under the nfluence of ts nerta. The nertal force supples a curvlnearly movng rgd body wth an addtonal knetc energy and angular momentum and these are ndependent from the knetc energy and angular momentum whch have been suppled by the actve force.
17 16 THE CLASSICAL SPINROTATION COUPLING The orbtal angular momentum of a rgd body whch undergoes crcular acceleraton s mutually exchangeable 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 the collectve angular moton of ts mass elements relatve to the that fxed pont and relatve to the center of mass. Ths s the spnrotaton couplng theorem whch has been summarzed from the precedng analyss. The parallel axs theorem couplng of the orbtal dynamcs of a crcularly accelerated rgd body to ts spn dynamcs. It reduces a moment of nerta of a rgd body to a moment of nerta of a pont mass. Every known mechancal force (rectlnear, tangental, centrpetal and Corols), when acts over a rgd body, t desynthess nto an actve force and an nertal force.
18 LOUAI H. ELZEIN 17 Fgure 1. Rotaton of the rgd body A relatve to the fxed pont O.
19 18 THE CLASSICAL SPINROTATION COUPLING Fgure 2. Bodyfxed S and spacefxed coordnate systems S.
20 LOUAI H. ELZEIN 19 References [1] John R. Taylor, Classcal Mechancs, Unversty Scence Books, n 2005, p. 368 to 370. [2] L. D. Landau and E. M. Lfshtz, Mechancs (Vol 1. 2nd ed.), Pergamon Press, n 1969, p. 98. [3] D. Kleppner and R. Kolenkow, An Introducton to Mechancs, Cambrdge Unversty Press, n 2014, p. 246 to 249. [4] Tom W. B. Kbble and Frank H. Berkshre, Classcal Mechancs (5th ed.), Imperal College Press, n 2004, p [5] Walter Benenson; John Harrs, Horst Stocker, Holger Lutz, Handbook of Physcs, SprngerVerlag New York. Inc., n 2002, p. 96 to 97. [6] Jerrold E. Marsden; Tudor S. Ratu, Introducton to Mechancs and Symmetry, Sprnger, n 1999, p. 250 to 251. [7] Davd Morn, Introducton to classcal mechancs: wth problems and solutons, Cambrdge Unversty Press, n 2008, p [8] Ferdnand P. Beer and E. Russell Johnston, Jr., Vector Mechancs for Engneers: Dynamcs (3th ed.), McGrawHll, Inc., n 1977, p. 616 to 617.
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