(Solution Manual) Classical Dynamics of Particles and Systems

Thornton,Marion, 5

Physics

Solution Manual

5th Edition

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Solution Manual for Classical Dynamics of Particles and Systems, The downloadable file consists of 497 pages, and the total size of the file is 3.38 MB. Purchase now only for highly discounted price.

Author(s): Thornton, Marion

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CHAPTER 0 Contents Preface v Problems Solved in Student Solutions Manual vii 1 Matrices, Vectors, and Vector Calculus 1 2 Newtonian Mechanics—Single Particle 29 3 Oscillations 79 4 Nonlinear Oscillations and Chaos 127 5 Gravitation 149 6 Some Methods in The Calculus of Variations 165 7 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics 181 8 Central-Force Motion 233 9 Dynamics of a System of Particles 277 10 Motion in a Noninertial Reference Frame 333 11 Dynamics of Rigid Bodies 353 12 Coupled Oscillations 397 13 Continuous Systems; Waves 435 14 Special Theory of Relativity 461 iii iv CONTENTS CHAPTER 0 Preface This Instructor’s Manual contains the solution s to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems , Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is in tended for use only by instructors using Classical Dynamics as a textbook, and it is not available to stud ents in any form. A Student Solutions Manual containing solutions to about 25% of the end- of-chapter problems is available for sale to students. The problem numbers of those solutions in the Student Solutions Manual are listed on the next page. As a result of surveys received from users, I continue to add more worked out examples in the text and add additional prob lems. There are now 509 problems, a significant number over the 4th edition. The instructor will find a large array of prob lems ranging in difficulty from the simple “plug and chug” to the type worthy of the Ph.D . qualifying examinations in classical mechanics. A few of the problems are quite challenging. Many of them require numerical methods. Having this solutions manual should pr ovide a greater appreciation of what the authors intended to accomplish by the statement of the problem in those cases where the problem statement is not completely clear. Please inform me when either the problem statement or solutions can be improved. Specific help is encouraged. The instructor will also be able to pick and choose different levels of difficulty when assignin g homework problems. And since students may occasionally need hints to work some problems, this manual will allow the instructor to take a quick peek to see how the students can be helped. It is absolutely forbidden for the st udents to have access to this manual. Please do not give students solutions from this manual. Postin g these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in the decrease of the usefulness of the text. The author would like to acknowledge the a ssistance of Tran ngoc Khanh (5th edition), Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of previous versions, went over user comments, and worked out solutions for new problems. Without their help, this manual would not be possible. The author would appreciate receiving reports of suggested improvements and suspecte d errors. Comments can be sent by email to stt@virginia.edu, the more detailed the better. Stephen T. Thornton Charlottesville, Virginia v vi PREFACE CHAPTER 1 Matrices, Vectors, and Vector Calculus 1-1. x2 = x2′ x1′ 45˚ x1 x3′x345˚ Axes and lie in the plane. 1x′3x′13xx The transformation equations are: 11 3 cos 45 cos 45 xx x= °− ° ′ 22xx=′ 33 1 cos 45 cos 45 xx x= °+ ° ′ 1111 22xx=−′3x 22xx=′ 3111 22xx=−′3x So the transformation matrix is: 110 22 01 0 110 22−     1 2 CHAPTER 1 1-2. a) x1AB CD αβγ OE x2x3 From this diagram, we have cosOE OA α= cosOE OB β= (1) cosOE OD γ= Taking the square of each equation in (1) and adding, we find 2222cos cos cos OA OB OD αβγ  ++ = + +2 2 2OE  (2) But 22OA OB OC +=2 (3) and 22OC OD OE +=2 (4) Therefore, 22 2OA OB OD OE ++ =2 (5) Thus, 222cos cos cos 1 αβγ+ += (6) b) x3 AA ′ x1x2 OED CBθ C′B′E′D′ First, we have the following trigonometric relation: 222c os OE OE OE OE EE θ2 ′ ′+− =′ (7) MATRICES, VECTORS, AND VECTOR CALCULUS 3 But, 22 2 2 22 2cos cos cos cos cos cosEE OB OB OA OA OD OD OE OE OE OE OE OEβ βα γγ     ′′ ′ ′=−+− + −           ′′=− +− ′′   ′+− ′α  (8) or, 22 2222 2 2 2 22cos cos cos cos cos cos 2 cos cos cos cos cos cos 2 cos cos cos cos cos cosEE OE OE OE OE OE OE OE OEα βγ α β αα ββ γ γγ α αβ βγ′′  =+ + + + + ′′′  ′−+ + ′′′  ′=+ − + + ′′ ′ γ ′     (9) Comparing (9) with (7), we find cos cos cos cos cos cos cosθ αα ββ γ γ =++ ′ ′′ (10) 1-3. x1e3′x2x3 O e1e2e3A e2′ e1′e2 e1e3 Denote the original axes by , , , and the corresponding unit vectors by e,, . Denote the new axes by , , and the corresponding unit vectors by 1x2x3x1 2e3e 1x′2x′3x′1′e, 2′e, e. The effect of the rotation is ee , , e . Therefore, the transformati on matrix is written as: 3′ 13 ′→21 ′ ee→3→2′e ( ) ( ) ( ) ()()() ()()()11 12 13 21 22 23 31 32 33cos , cos , cos , 010 cos , cos , cos , 0 0 1 100 cos , cos , cos , ′′′ λ   ′′′ ==    ′′′   ee ee ee ee ee ee ee ee ee 1-4. a) Let C = AB where A, B, and C are matrices. Then, ij ik k j kCA= B∑ (1) ()t ji jkk i k i jkij kkCC A B B A == = ∑ ∑ 4 CHAPTER 1 Identifying ()t kiikB=B and ()t jkkjAA= , () ()()tt it j ik k j kCB A =∑ (2) or, ()t tCA BB A==tt (3) b) To show that ()1 11ABB A− −−= , () ( )11 11ABB A I B A A B−− −−== (4) That is, ()11 1 1ABB A A I A A A I−− − −= == (5) ( )()11 1 1BA A B BI B BB I−− − −= == (6) 1-5. Take λ to be a two-dimensional matrix: 11 12 11 22 12 21 21 22λλλ λλ λλλλ== − (1) Then, ( )( ) ( ) ( )( ) ( )( )()2 22 22 22 22 22 22 11 22 11 22 12 21 12 21 11 21 12 22 11 21 12 22 22 2 22 2 2 2 2 2 22 11 12 21 11 12 11 21 11 22 12 21 12 22 2 22 22 11 12 22 21 11 21 12 222 2λ λλ λλλλ λλ λλ λλ λλ λλ λ λ λ λ λ λ λλ λλλλ λλ λλ λλ λ λλ λ=− ++ + − + =+ ++ − + + =+ +− + (2) But since λ is an orthogonal transformation matrix, ijkj ik jλλδ= ∑ . Thus, 22 22 11 12 21 22 11 21 12 221 0λλλλ λλ λλ+ =+= += (3) Therefore, (2) becomes 21 λ= (4) 1-6. The lengths of line segments in the jx and jx′ systems are 2 j jLx=∑ ; 2 i iL= x ′ ′∑ (1) MATRICES, VECTORS, AND VECTOR CALCULUS 5 If , then LL=′ 22 j i jixx= ′ ∑ ∑ (2) The transformation is ii jj jx λ =′ x∑ (3) Then, (4) 2 ,ji kk ji k ki ki kixx xxλλ λλ =   =∑∑ ∑ ∑ ∑∑AA A AA Aix But this can be true only if ik i k iλλδ= ∑ AA (5) which is the desired result. 1-7. x1(1,0,1)x3 x2 (1,0,0) (1,1,0)(0,1,0)(1,1,1)(0,0,1)(0,1,1) (0,0,0) There are 4 diagonals: 1D, from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D; 1 2D, from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = ; 2D 3D, from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = ; and 3D 4D, from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D. 4 The magnitudes of the diagonal vectors are 1234 3 ====DDDD The angle between any two of these diagonal vectors is, for example, ( )( )12 121,1,1 1,1,1 1cos33θ⋅− ⋅= ==DD DD 6 CHAPTER 1 so that 11cos 70.53θ−= =°  Similarly, 13 23 34 14 24 13 14 23 24 341 3⋅⋅⋅ ⋅⋅=====DD DD DD DD DD DD DD DD DD DD± 1-8. Let θ be the angle between A and r. Then, 2A⋅= Ar can be written as 2cosArA θ= or, cos rA θ= (1) This implies 2QPOπ= (2) Therefore, the end point of r must be on a plane perpendicular to A and passing through P. 1-9. 2 =+ −Ai jk 23=− + +Bi j k a) 32 −= − −AB ij k () ( )1222 231 ( 2)  −= + − + −AB 14 −=AB b) component of B along AB Aθ The length of the component of B along A is B cos θ. cos AB θ ⋅= AB 261 3 6cos or2 66 Aθ⋅− + −== =ABB The direction is, of course, along A. A unit vector in the A direction is ()12 6+− ij k MATRICES, VECTORS, AND VECTOR CALCULUS 7 So the component of B along A is ()122+−ij k c) 33cos 61 4 27 ABθ⋅== =AB; 13cos 27θ−= 71θ ° d) 21 1 1 1 212 131 21 2 323 1−−×= − = − +−−−ij k AB i j k 57 ×= + +AB ij k e) 32 −= − −AB ij k 5 +=− + AB i j () () 31 15 0−×+= −− −ij k AB AB 2 () () 10 2 14 −× += + +AB AB i j k 1-10. 2s i n c o sbt b t ω ω =+ri j a) 222c o s s i n 2s in c osbt bt bt btωω ωω 2ω ωωω ω== − == − − = −vr i j av i j r  1222 2 22 2 1222speed 4 cos sin 4c o s s i nbt b bt tωω ωω ωωωt   == +   =+v 122speed 3 cos 1 btωω  =+  b) At 2 tπω = , sin 1 tω=, cos 0 tω= So, at this time, bω=−vj , 22bω =−ai So, 90θ ° 8 CHAPTER 1 1-11. a) Since ( )ijk jk i jkAB ε ×= ∑ AB , we have () ( ) () (), 12 3 3 2 21 3 3 1 31 2 2 1 123 123 123 123 123 123 123 123 123()ijk j k i ij kABC CA B A B CA B A B CA B A B CCC AAA AAA AAA CCC BB B BBB BBB CCCε ×⋅ = =− −− +− == −== ⋅∑∑ AB C ABC × (1) We can also write (123 123 123 123 123 123()CCC BBB BBB CCC AAA AAA×⋅ = − = = ⋅×AB C BCA ) (2) We notice from this result that an even nu mber of permutations leaves the determinant unchanged. b) Consider vectors A and B in the plane defined by e, . Since the figure defined by A, B, C is a parallelepiped, area of the base, but 1 32e 3 ×= ×ABe ⋅= eC altitude of the parallelepiped. Then, ( )( )3 area of the base = altitude area of the base = volume of the parallelepiped⋅×=⋅ × ×CAB C e 1-12. O A BC ha bc a – c c – b b – a The distance h from the origin O to the plane defined by A, B, C is MATRICES, VECTORS, AND VECTOR CALCULUS 9 ( )( ) () () ()h⋅−×−=−×− ⋅× − × + ×=×−×+× ⋅×=×+× + ×aba cb ba cb abcacab bcacab ab c abbcca (1) The area of the triangle ABC is: () () () ( ) () ()111 222× − =− × − = − × − ba cb ac ba cb ac A=− (2) 1-13. Using the Eq. (1.82) in the text, we have ( )( ) ( )2A φ ×= × × = ⋅ − ⋅ = − A AX X A A A A X A X AB from which ( ) 2A×+=BA AXφ 1-14. a) 12 12 1 0 1 21 03 1 0 12 1 29 2 01 113 533−−      =− =       AB −  Expand by the first row. 29 19 1 212 133 5 3 53− −=+ +AB 104=−AB b) 12 1 21 9 7 03 1 43 1 39 20 1 10 5 2−          ==             AC 97 13 9 52   =    AC 10 CHAPTER 1 c) ()12 1 8 5 03 1 2 3 20 1 9 4−      == − −       ABC A BC 55 35 25 14−−   =−    ABC d) ?tt− = AB B A 12 1 12 9 ( f ro m p art ) 533 201 10 2 1 15 11 1 2 3 0 22 3 023 1 1 1 1 93tt− =−       =− = − −      −   AB a BA    034 30 6 46 0tt−−    −=    −  AB B A 1-15. If A is an orthogonal matrix, then 2 21 1001 00 1 0 0 00 0 1 00 0 0 10 0 1 0 0 02 0 010 002 0 0 1t aa a a aa a a a a=      −=      −  0 1         =       AA 1 2a= MATRICES, VECTORS, AND VECTOR CALCULUS 11 1-16. x3 P r θ x2 x1a r θa r cos θ constant ⋅=ra cos constant ra θ= It is given that a is constant, so we know that cos constant r θ= But cos r θ is the magnitude of the component of r along a. The set of vectors that satisfy all have the same component along a; however, the component perpendicular to a is arbitrary. constant ⋅=ra Th is us the surface represented by constant a plane perpendicular to .⋅=ra a 1-17. a AθbB cC Consider the triangle a, b, c which is formed by the vectors A, B, C. Since () (2 222) A B=− = −⋅ − = −⋅ +CAB CA B A B AB (1) or, 2 222c os AB A B θ =+ −C (2) which is the cosine law of plane trigonometry. 1-18. Consider the triangle a, b, c which is formed by the vectors A, B, C. Aα CB γβ b ca 12 CHAPTER 1 =− CAB (1) so that ( ) ×=− × CB AB B (2) but the left-hand side and the right-hand side of (2) are written as: 3 sinBC α ×= CB e (3) and ( )3 sinAB γ − × =× − × =× =AB BABBBAB e (4) where e is the unit vector perpendicular to the triangle abc. Therefore, 3 sin sin BC AB α γ = (5) or, sin sinCA γ α= Similarly, sin sin sinCAB γ αβ== (6) which is the sine law of plane trigonometry. 1-19. x2 a α x1a2 b2 a1 b1b β a) We begin by noting that ( )2 222c o s ab a b αβ −= + − −ab (1) We can also write that ( )( ) () ( ) ( ) ( ) () ()22 2 11 22 22 22 2 22 2 22cos cos sin sin sin cos sin cos 2 cos cos sin sin 2c o s c o s s i n s i nab ab ab ab ab a b ab a bαβ αβ α αβ β α β α αβ αβ−=− +− =− +− =+ ++ − + =+− +ab β (2) MATRICES, VECTORS, AND VECTOR CALCULUS 13 Thus, comparing (1) and (2), we conclude that ()cos cos cos sin sinαβα β α−= + β (3) b) Using (3), we can find ( ) sinαβ− : () () ( ) ( ) ()2 22 2 2 22 2 2 22 2 2 2sin 1 cos 1 cos cos sin sin 2cos sin cos sin 1 cos 1 sin sin 1 cos 2cos sin cos sin sin cos 2sin sin cos cos cos sin sin cos cos sinαβ αβ αβ αβ α α β β α βα β α α β αβ α β α β αβ αβ αβ−= − − =− − − =− − − − − =− + =−β (4) so that ()sin sin cos cos sinαβα β α−= − β j (5) 1-20. a) Consider the following two cases: When i≠ 0ijδ= b u t 0ijkε≠. When i=j 0ijδ≠ b u t 0ijkε=. Therefore, 0ijk ij ijεδ = ∑ (1) b) We proceed in the following way: When j = k, 0ijk ijjε ε== . Terms such as 11 11 0jεε=A . Then, 12 12 13 13 21 21 31 31 32 32 23 23 ijk jk iiiiii jkεεε εε εε εε εε εε ε=+++++ ∑ A AAAAAA = Now, suppose i , then, 1 ==A 123 123 132 132 112 jkεε εε=+= + ∑ 14 CHAPTER 1 for , . For 2 i==A213 213 231 231 112 jkεε εε=+= + ∑ = 3 i==A , 312 312 321 321 2 jkεε εε= += ∑ . But i = 1, gives . Likewise for i = 2, 2=A 0 jk=∑ 1=A ; i = 1, 3 =A ; i = 3, 1 =A ; i = 2, A ; i = 3, . Therefore, 3= 2=A ,2ijk jki jkεεδ= ∑ AA (2) c) () () () () () () ( ) ( )123 123 312 312 321 321 132 132 213 213 231 231 1 1 1 111 11 11 1 1ijk ijk ijkεε ε εε εε εε εε εε ε=+++++ = ⋅ + ⋅ +− ⋅− +− ⋅− +− ⋅− + ⋅∑ or, 6ijk ijk ijkεε = ∑ (3) 1-21. ( )ijk jk i jkAB ε ×= ∑ AB (1) ( )ijk jki ij kABC ε ×⋅ = ∑∑ ABC (2) By an even permutation, we find ijki jk ijkABC ε =∑ ABC (3) 1-22. To evaluate ijkm k kεε∑ A we consider the following cases: a) : 0 for all , ,ijk mk iik mk kkij i m εε εε == =∑∑ AA A b) :1 for 0 for ijk mk ijk imk kkij jmεε εε == = = =≠∑∑ A A and , m kij≠ ij ic) :0 for 1 for and ,ijk mk ijk ik kkim j jkεε εε == = ≠ =− = ≠∑∑ AA A A d) :0 for 1 for and ,ijk mk ijk jmk kkjm mi ki jεε εε == = ≠ =− = ≠∑∑ A A MATRICES, VECTORS, AND VECTOR CALCULUS 15 e) :0 for 1 for and ,ijk mk ijk jk kkjm i ikεε εε == = ≠ == ≠∑∑ AA A A ij jk mf) : 0 for all , ,ijk mk ijk k kkmi εε εε == = ∑∑ AA A A g) : This implies that i = k or i = j or m = k. or i≠A Then, for all 0ijk mk kεε = ∑ A , , ,ij mA h) for all or : 0ijk mk kjm εε ≠= ∑ A A , , ,ij mA Now, consider i jmi m j δδδ δ−AA and examine it under the same conditions. If this quantity behaves in the same way as the sum ab ove, we have verified the equation ijkm k i jmi m j kεεδ δ δ δ=− ∑ AA A a) : 0 for all , ,ii m i m i ij i m δδδ δ =− =AA A b) : 1 if , , 0 if ii jm im ji ij jmmijm δδδ δ =− = = ≠ =≠A c) : 1 if , , 0 if ij i i i j im j ij jδδδ δ =− =− = =≠AA AA A≠ mi d) : 1 if , 0 if im i m ji imδδδ δ =− = − = =≠AA A AAA ≠ e) : 1 if , 0 if im m i m m jm i m iδδδ δ =− = = =≠AA AA A≠ all,,j f) :0 for ij i l j miδδδ δ =− =AA A AA g) , : 0 for all , , ,ij m i m j im i j m δδδ δ ≠− =AA AA h) ,: 0 f o r a l l , , ,ij m i m i jm i j m δδδ δ ≠− =AA AA Therefore, ijkm k i jmi m j kεεδ δ δ δ=− ∑ AA A (1) Using this result we can prove that ( )( )( ) ××=⋅ −⋅AB C A C BA B C 16 CHAPTER 1 First ( )ijk jk i jkBCε ×= ∑ BC . Then, ( ) [ ] ( ) ( ) ()()mn m mn m njk j k n mn mn jk mn njk m j k mn jkn m j k jkmn jkmn lmn jkn m j k jkm n jl km k jm m j k jkm mm m m m m m m mm m mABC A B C AB C AB C AB C AB C ABC A B C B A C C A B BCεε ε εε εε εε δδ δ δ×× = × = == = =−  =−= −   =⋅ −⋅∑∑ ∑ ∑∑ ∑∑ ∑ ∑∑ ∑ ∑AB C AC ABAA A AA A AA AA AA  Therefore, () ( ) ( ) ××= ⋅ −⋅AB C A C BA B C (2) 1-23. Write ( )jmm j mAB ε ×= ∑ ABAA A ( )krs r s k rsCDε ×= ∑ CD Then, MATRICES, VECTORS, AND VECTOR CALCULUS 17 () ( ) [] ( ) ( )ijk j m m krs r s i jk m rs ijk j m krs m r s jk mrs j m ijk rsk m r s jm r s k j m ir js is jr m r s jm r s jm m i j m i j jm jm j m i jm jm jAB CD AB CD AB CD AB CD AB CD AB DC DAB Cεε ε εε ε εε ε εδ δδ δ ε εε  ××× =     = = =− =−  =−∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑AB CDAA A AA A AA A AA A AA A A AA A A () ()j mi m iiCAB D CD  =−∑ ABD ABCA A Therefore, [() () ]() () ××× = −AB CD A B D C A B C D 1-24. Expanding the triple vector product, we have ( ) ()( ) × ×= ⋅− ⋅ eA eA e ee A e (1) But, ()⋅= Aee A (2) Thus, () ( )= ⋅+ × × A e A eeA e (3) e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e. 18 CHAPTER 1 1-25. er eφ eθθ φ The unit vectors in spherical coordinates are ex pressed in terms of rectangular coordinates by ( ) () ()cos cos , cos sin , sin sin , cos , 0 sin cos , sin sin , cosrθ φθ φθ φ θ φφ θφθφ θ =−  =−   =e e e (1) Thus, ( ) cos sin sin cos , cos cos sin sin , cosθ φ θφ θθφ φθφ θθφ θθ =− − − −e     cosr φ θ φθ+e  =− (2) e Similarly, ( ) cos , sin , 0φ φφ φφ=− −e   cos sinr θ φ θφ θ −e =− (3) e sinr φ θ φ θθ =+ee   e (4) Now, let any position vector be x. Then, rr=xe (5) ( ) sin sinrr rrrr r rr rφθ φθφθ θ φθ θ=+= + + =+ +xe e e e e ee e       r (6) ( ) ( ) ( )( ) ( )22 2 2sin cos sin sin 2 sin 2 cos sin sin 2s incosrr rrr r r r r r r rrr r r r rrrφφ θ θ φ θφθ θ φθ φθ φθ θ θ θ φθ θ φθ φθ φ θ θ θθφ θ θ=+ + + + + + + + =+ + + −− ++ −xe e e ee e                  r e e e    (7) or, MATRICES, VECTORS, AND VECTOR CALCULUS 19 () ()22 2 2 2 221sin sin cos 1sinsinrdrr r r rrd t drrd tθ φθφ θ θφ θθ φθθ  == − − + −    + xa e e e      (8) 1-26. When a particle moves along the curve ( ) 1c o s rk θ =+ (1) we have 2sin cos sinrk rkθθ θ θθ θ =−    =− +        (2) Now, the velocity vector in pola r coordinates is [see Eq. (1.97)] rrrθθ =+ve e   (3) so that ( )2 22 22 22 2 2 2 2 22sin 1 2 cos cos 22 c o svr r kk kθ θ θθ θθ== + =+ + + =+ v    θθ ) (4) and is, by hypothesis, constant. Therefore, 2v (2 221 c o sv kθθ=+ (5) Using (1), we find 2v krθ= (6) Differentiating (5) and using the expression for r, we obtain  ()22 2 2 2sin sin 4 41 c o svv r kθθθ θ== + (7) The acceleration vector is [see Eq. (1.98)] ( )( )22r rr r rθ θθ θ =− + +ae      e (8) so that 20 CHAPTER 1 ( )() ()() () ()2 22 22 22 2 2 2cos sin 1 cos sincos 1 cos21 c o s 1c o s2c o s 121 c o s 31c o s2rrr kk k k kθ θθ θ θ θ θ θθθ θθθ θθθθ θθ⋅= − =− + − +  =− + + + +   −=− + + +  +ae       θ =− (9) or, 23 4rv k⋅= −ae (10) In a similar way, we find 2sin 3 41 c o sv kθθ θ⋅= −+ae (11) From (10) and (11), we have () ()2 2 r θ =⋅+ ⋅aa e a e (12) or, 232 41 c osv k θ=+a (13) 1-27. Since ( )()() ××=⋅ −⋅ rv r r r vr v r we have ()[] () ( )[] ( ) ()()( ) () () ( )222dd dt dt rv×× = ⋅ − ⋅ =⋅ + ⋅ −⋅ − ⋅ −⋅ =+ ⋅ − + ⋅rv r r r vr v r rra rvv rvv vvr rar ar v v r r a (1) Thus, ()[] () ( )2rdt×× = + ⋅ − ⋅ +rv r ar v v r r a v2 d (2)