Spin-rotation coupling of the angularly accelerated rigid body

Transcription

1 Spn-rotaton couplng of the angularly accelerated rgd body Loua Hassan Elzen Basher Khartoum, Sudan. Postal code:11123 E-mal: November 1, 2017 All Rghts Reserved.

2 Abstract Ths paper s prepared to show that a rgd body whch undergoes an angular acceleraton 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 ths spn moton 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 ths spn moton and the formula whch couplng of ths hghlghted force to the force whch causes the rotaton of the rgd body has been also derved. Keywords: Classcal mechancs Rotatonal dynamcs PACS: dc I. 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 spn-rotaton 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. II. Analyss II.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. 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[2]: L = r O p + ρ ρ m (1 where r O 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, 2

3 we can re-express equaton(1 to say[2] L = L moton of CM + L moton relatve to CM (2 Snce the mass s constraned to a crcle, the tangental velocty of the mass of the rgd body A s ωˆk r O where ωˆk s the rotatonal angular velocty of the rgd body relatve to the axs of rotaton O, ˆk s a unt vector perpendcular to the plane of the moton and snce p = mωˆk r O, the total angular momentum equaton (equaton(1 becomes (assumng the moton s planar hence both axses of rotaton, O and CM, are parallel: L = r O (mωˆk r O + ρ (Ωˆk ρ m (3 where Ωˆk s the rgd body spnnng angular velocty (arbtrary relatve to ts center of mass. Takng the frst term n the RHS and usng the fact that r O = r ρ, one fnds r O (mωˆk r O = (r ρ (mωˆk (r ρ, = (r ρ (mωˆk r mωˆk ρ, = (r ρ (mωˆk r (r ρ (mωˆk ρ, = r (mωˆk r ρ (mωˆk r r (mωˆk ρ + ρ (mωˆk ρ, snce m = m, therefore r O (mωˆk r O = r (m ωˆk r ρ (m ωˆk r r (m ωˆk ρ + ρ (m ωˆk ρ, farther mathematcal smplfcaton can be done by dong the cross product usng the Cartesan coordnate system and convertng the summaton to ntegraton over the whole body[3][4]. Therefore, one wll fnd r O (mωˆk r O = I O ωˆk + I CM ωˆk ρ (m ωˆk r r (m ωˆk ρ, (4 where I O s the moment of nerta relatve to the axs of rotaton O whch s a perpendcular dstance r O from the centre of mass and I CM s the moment of nerta of the rgd body relatve to ts center of mass. Contnue the mathematcal smplfcaton usng the dentty A (B C = (A C B (A B C, we fnds r O (mωˆk r O = I O ωˆk + I CM ωˆk [ (ρ r m ωˆk [ ( ] (r ρ m ωˆk r m ωˆk ρ ( ] ρ m ωˆk r, 3

4 snce ρ and ˆk are mutually orthogonal and so are r and ˆk then we have r O (mωˆk r O = I O ωˆk + I CM ωˆk = I O ωˆk + I CM ωˆk 2 (ρ r m ωˆk (r ρ m ωˆk, (r ρ m ωˆk, = I O ωˆk + I CM ωˆk 2 = I O ωˆk + I CM ωˆk 2 = I O ωˆk + I CM ωˆk 2 ((r O + ρ ρ m ωˆk, (r O ρ m ωˆk 2 m ρ r O cos γ ωˆk 2 (ρ ρ m ωˆk, m ρ 2 ωˆk, where γ s the angle between r O and ρ, mass, and m ρ 2 = I CM. Hence we have m ρ = 0 (property of the center of r O (mωˆk r O = I O ωˆk I CM ωˆk. (5 Thus the total angular momentum (equaton(3 becomes L = I O ωˆk I CM ωˆk + ρ (Ωˆk ρ m, = I O ωˆk I CM ωˆk + I CM Ωˆk. (6 where I CM ωˆk 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 Ωˆk can be consder as the ntal spnnng angular momentum that the rgd body acqured before t start ts crcular moton and snce Ωˆk s arbtrary, so that t can be zero and therefore have not to be a mandatory term of equaton(6. Hence, we can rewrte equaton(6 as: L = I O ωˆk I CM ωˆk, = L r + L s. (7 where L = r O (mωˆk r O s the total angular momentum, L r = I O ωˆk s the rotatonal angular momentum and L s = I CM ωˆk s the spnnng angular momentum. Snce the total angular momentum (equaton(7 s conserved then that mples the rotatonal and spnnng angular momentum are mutually exchangeable n order to conserve t, that s L = L r + L s (8 Ths property (equaton(8 negate the above statement whch we have begn wth. Another thng we can notce s that f we smplfy the term r O (mωˆk r O n equaton(5 usng the dentty A (B C = (A C B (A B C and rearrange t, that yelds [ ( ] I O ωˆk = (r O r O mωˆk r O mωˆk r O + I CM ωˆk, 4

5 snce moton s planar then r O and ˆk are mutually orthogonal, so that I O ωˆk = mr 2 O ωˆk + I CM ωˆk, dvdng by ω and dotted wth ˆk, we have I O = mr 2 O + I CM. (9 whch s nothng other than the parallel axs theorem. Thus the theorem dsplay the couplng of spn and rotaton moment of nerta whch leads to the couplng of spnnng and rotatonal moton of a rgd body when moves crcularly. II.2. Couplng of the effected 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 αˆk = (dω/dtˆk 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 αˆk r O and snce F = ma, the total torque equaton s gven by: τ = r O F, = r O (mαˆk r O. (10 where τ s the total torque and r O s the vector poston of the center of mass of the rgd body relatve to the axs of rotaton O. From fgure(1 we have r O = r ρ. Hence one can wrte τ = (r ρ (mαˆk (r ρ, = r (mαˆk (r ρ ρ (mαˆk (r ρ, = r (mαˆk r r (mαˆk ρ ρ (mαˆk r + ρ (mαˆk ρ, snce m = m, therefore we have ( ( τ = r m αˆk r r m αˆk ρ ( ρ m αˆk r ( + ρ m αˆk ρ, ( ( = r m αˆk r (r O + ρ m αˆk ρ ( ( ρ m αˆk (r O + ρ + ρ m αˆk ρ, 5

6 τ = r (m αˆk r r O ( αˆk m ρ ρ (m αˆk ρ ρ (m αˆk ρ + m ρ (αˆk r O ρ (m αˆk ρ, from the propertes of the center of mass we have m ρ = 0. Therefore, one fnds r O (mαˆk r O = r (m αˆk r ρ (m αˆk ρ. (11 Equaton(11 can be mathematcally smplfed by dong the cross product usng the Cartesan coordnate system and convertng the summaton to ntegraton over the whole body. Therefore one wll fnd r O (mαˆk r O = I O αˆk I CM αˆk (12 where I O s the moment of nerta relatve to the axs of rotaton O whch s a perpendcular dstance r O from the centre of mass and I CM s the moment of nerta of the rgd body relatve to ts center of mass. We can wrte t as: τ = τ r + τ s (13 where τ = r O (mαˆk r O s the total torque. τ r = I O αˆk s the rotaton torque and τ s = I CM αˆk s the spn torque. Thus the total torque τ s a synthess of two torques; the rotaton torque τ r and the spn torque τ s. There s another thng that has to be notce n equaton(12 when we rearrange t as followng: I O αˆk = r O (mαˆk r O + I CM αˆk, and usng the dentty A (B C = (A C B (A B C to obtan [ ( ] I O αˆk = (r O r O mαˆk r O mαˆk r O + I CM αˆk, snce moton s planar then r O and ˆk are mutually orthogonal, so that I O αˆk = mr 2 αˆk + I O CM αˆk, dvdng by α and dotted wth ˆk, we get I O = mr 2 + I O CM. (14 Agan we obtaned the parallel axs theorem. 6

7 How one can understand the negatve sgn of the spn torque τ s Referrng to equatons(12 and (13 we fnd τ s = I CM αˆk and the crucal queston s; Does the negatve sgn assgn to the magntude α or to the drecton ˆk? or n another words; Does the spn of the rgd body s de-acceleraton n the same drecton of ts rotaton or acceleraton n opposte drecton of ts rotaton?. To answer ths queston we are gong to take another approach to fnd the same term that assgned to the spn τ s. Referrng to fgure(1 and snce the rgd body A s angularly accelerates relatve to the axs of rotaton O therefore an nertal force F wll occur at any element mass m that composes ts total mass m. Ths nertal force s known as Euler force[5], F Euler, such that: F = F Euler = m dω dt ˆk r, = m αˆk r. (15 where αˆk = (dω/dtˆk s the same angular acceleraton of rotaton of the center of mass of the rgd body relatve to the axs of rotaton O and r s the vector poston of the pont where the acceleraton s measured relatve to the axs of the rotaton O. We know that the Euler force s an nertal force whch mples that t acts n a drecton that opposte to the drecton of the force appled (the actve force. Therefore, the negatve sgn that occurs n equaton(15 s assgned to the drecton ˆk. Thus, we can rewrte t: F = m α( ˆk r (16 The torque that acts on any element mass m due to the Euler force F s gven wth: τ = ρ F (17 where τ s the torque of any element mass m relatve to the center of mass of the rgd body A whch occurs due to Euler force F and ρ s the vector poston of the same element mass m relatve to the center of mass of the rgd body. Hence, the total nertal torque, τ nr, that acts on the rgd body due to the Euler force F Euler s gven by: τ nr = = ρ F, ρ (m α( ˆk r, from fgure(1 we have r = r O + ρ. Hence one can wrte τ nr = ρ (m α( ˆk r O + m α( ˆk ρ, = m ρ (α( ˆk r O + ρ (m α( ˆk ρ, 7

8 from the propertes of the center of mass we know that obtan m ρ = 0. Therefore, we τ nr = ρ (m α( ˆk ρ. (18 Equaton(18 can be mathematcally smplfed by dong the cross product usng the Cartesan coordnate system and convertng the summaton to ntegraton over the whole body. Therefore, one wll fnd τ nr = I CM α( ˆk. (19 where α( ˆk = (dω/dt( ˆk s the angular acceleraton of the rgd body A relatve to ts center of mass, where t s equal n magntude to the rgd body angular acceleraton relatve to the axs of rotaton O. Therefore, the nertal torque τ nr acts over the rgd body A where τ nr 0 such that t causes t to rotate relatve to ts center of mass 1 n a drecton counter to the drecton of rotaton of ts center of mass relatve to the axs of rotaton O. Therefore, we retreve the spn torque term of equaton(12, that s; the term that assocated wth the total nertal torque τ nr (equaton(19 s dentcal wth the one that assocated wth the spn torque τ s occurs as a second term n equaton(12. Hence, one can rewrte equaton(12, the total torque τ, and equaton(7, the total angular momentum, n a more precse manner: where I CM α( ˆk = τ s and r O (mαˆk r O = I O αˆk + I CM α( ˆk, (20 L = I O ωˆk + I CM ω( ˆk, (21 where I CM ω( ˆk = L s. Thus, the negatve sgn ponts to the drecton of spnnng of the rgd body. II.3. 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[6][7]: ( T = ½m (ωˆk r O ωˆk r O + ½I CM Ωˆk Ωˆk, (22 where ωˆk r O s the tangental velocty of the center of mass of the rgd body relatve to the axs of rotaton O and Ωˆk s an arbtrary spnnng velocty relatve to the center of mass of the rgd body. When one uses the dentty (A B C D = (A C (B D (A D (B C, he wll get [( ( ] ( T = ½m ωˆk ωˆk (r O r O (ωˆk r O r O ωˆk + ½I CM Ωˆk Ωˆk, 1 The observaton of ths phenomenon can be obtan easly by rotatng a metallc sold dsk pvoted at ts center or by rotatng a vessel contanng ce cubes floatng on water and can be exercse usng one s hands. 8

9 snce ˆk and r O are mutually orthogonal, so that ( ( T = ½m ωˆk ωˆk (r O r O + ½I CM Ωˆk Ωˆk, (23 ( ( = ½m ωˆk ωˆk (r ρ r ρ + ½I CM Ωˆk Ωˆk, ( = ½m ωˆk ωˆk [(r r 2 (r ρ + (ρ ρ ] + ½I CM (Ωˆk Ωˆk, = ½ ( m (r r ωˆk ωˆk ( m (r ρ ωˆk ωˆk + ½ ( ( m (ρ ρ ωˆk ωˆk + ½I CM Ωˆk Ωˆk, = ½I O (ωˆk ωˆk ( ( m (r ρ ωˆk ωˆk + ½I CM ωˆk ωˆk + ½I CM (Ωˆk Ωˆk, = ½I O (ωˆk ωˆk + ½I CM (Ωˆk Ωˆk, ( ( m (r O + ρ ρ ωˆk ωˆk + ½I CM ωˆk ωˆk = ½I O (ωˆk ωˆk ( m (ρ ρ ωˆk ωˆk ( + ½I CM (ωˆk ωˆk + ½I CM Ωˆk Ωˆk, ( m (r O ρ ωˆk ωˆk ( = ½I O (ωˆk ωˆk ½I CM ωˆk ωˆk + ½I CM (Ωˆk Ωˆk, m ρ r O cos γ ( ωˆk ωˆk where γ s the angle between r O and ρ. Agan m ρ = 0. Therefore, we obtan ( ( T = ½I O (ωˆk ωˆk ½I CM ωˆk ωˆk + ½I CM Ωˆk Ωˆk. (24 When subtractng equaton(23 from (24, we agan wll obtan the parallel axs theorem and f we subtract equaton(22 from (24 we obtan ( ( ½m (ωˆk r O ωˆk r O = ½I O ωˆk ωˆk ½I CM ωˆk ωˆk. (25 Snce there s no negatve knetc energy therefore one can understand that the negatve sgn s assgn to drecton. Thus, the correct knetc energy formula: ( ( T = ½I O (ωˆk ωˆk + ½I CM ω( ˆk ω( ˆk + ½I CM Ωˆk Ωˆk. (26 The term ½I CM (Ωˆk Ωˆk can be consder as the ntal spnnng knetc energy that the rgd body have before t start ts crcular moton and snce Ωˆk s arbtrary, 9

10 so that t can be zero and therefore have not to be a mandatory term of equaton(26. Hence we can wrte equaton(26 as: ( T = ½I O (ωˆk ωˆk + ½I CM ω( ˆk ω( ˆk, where T = ½m (ωˆk r O ωˆk r O = T r + T s. (27 s the total knetc energy, T r = ½I O (ωˆk ωˆk s the rotatonal knetc energy and T s = ½I CM (ω( ˆk ω( ˆk knetc energy. III. Concluson s the spnnng A rgd body that angularly accelerates n a crcular path wll spn under the nfluence of nertal force, specally the Euler force. 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. In the case of the angular acceleraton of a rgd body n a crcular path, both the actve torque and the nertal (fcttous torque that are causng the rgd body to rotate and spn respectvely are real torques and occur due to the acton of real forces and these forces are de-synthess from one real orgn, the total force whch n return causes a total torque. In another words, the total torque de-synthess to actve torque (causes the rotaton of the rgd body and nertal torque (causes the spnnng of the rgd body and the three (total, actve and nertal are real torques. The nertal force supples an angularly acceleratng rgd body wth an addtonal knetc energy and angular momentum and these are ndependent from the knetc energy and angular momentum whch are be suppled by the actve force. The parallel axs theorem couplng of the orbtal moton of an angularly accelerated rgd body to ts spn moton. The orbtal angular momentum of a rgd body whch undergoes angular acceleraton s mutually exchangeable wth ts spn angular momentum and that happens n order to conserve the total angular momentum of the rgd body. 10

11 Fgure 1: Rotaton of the rgd body A relatve to the fxed pont O. 11

12 References [1] John R. Taylor, Classcal Mechancs, Unversty Scence Books, n 2005, p [2] John R. Taylor, Classcal Mechancs, Unversty Scence Books, n 2005, p [3] Danel Flesch, A Student s Gude to Vectors and Tensors, Cambrdge Unversty Press 2012, p. 163 to 164. [4] G. E. Hay, Vector and Tensor Analyss, Dover Publcatons, Inc., n 1953, p. 80 to 82. [5] Davd Morn, Introducton to classcal mechancs: wth problems and solutons, Cambrdge Unversty Press 2008, p [6] Paul, Burton, Knematcs and Dynamcs of Planar Machnery, Prentce Hall, n [7] Ucker, John J.; Pennock, Gordon R.; Shgley, Joseph E., Theory of Machnes and Mechansms (4th ed., Oxford Unversty Press, n