Marion Solutions Manual

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 [email protected], the more detailed the better. Stephen T. Thornton Charlottesville, Virginia v edeewerwrwrgbfdgdyyy5y hfhfhfh fghfh fhfh 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) MATRICES, VECTORS, AND VECTOR CALCULUS 21 1-28. () () ln lni i ix∂=∂∑ grad r r e (1) where 2 i ix =∑ r (2) Therefore, () 2 21lni ii i ix x x x∂=∂ =∑rr r (3) so that ()21lnii ix=∑ grad r e r (4) or, ()2lnr=rgrad r (5) 1-29. Let describe the surface S and 29r=121 xyz++= describe the surface S. The angle θ between and at the point (2,–2,1) is the angle betw een the normals to these surfaces at the point. The normal to is 2 1S2S 1S () ( ) ( ) ()22 2 1 123 2 ,2,1 12399 222 442xyzSr x y z xyz====− =+ + =+ + =−+grad grad grad eee eee2− (1) In , the normal is: 2S () ( ) ()2 2 12 3 2 , 2 , 12 31 2 2x 1 y zSx yz z= =− ==+ +− =+ + =++grad grad ee e ee e (2) Therefore, 22 CHAPTER 1 () () () () () ( )12 12 123 1 23cos 442 2 66SS SSθ⋅= −+ ⋅ + +=grad grad grad grad eee e ee (3) or, 4cos 66θ= (4) from which 16cos 74.29θ−= =° (5) 1-30. ()()3 1ii ii ii ii ii iixx xxφψ ψφ ixφψφ ψφφψ=ψ  ∂ ∂∂==+  ∂∂ ∂   ∂∂=+∂∂∑∑ ∑∑grad e e ee Thus, () φψφ ψ ψ φ=+ grad grad grad 1-31. a) ()123 2 1 12 2 12 2 222n n n ii j ij ii n ii j ij n ii j ij n ii irrxxx nxx xn x xn r= − − −   ∂ ∂  == ∂∂      =   =  =∑∑∑ ∑∑ ∑∑ ∑grad e e e e e (1) Therefore, ()2 n nrn r−= grad r (2) MATRICES, VECTORS, AND VECTOR CALCULUS 23 b) ()() () () ()33 11 12 2 12 2 ii ii ii ij ij i ii j ij i i ifr fr rfrxr frxxr frxxr fx rd r== −∂∂ x∂==∂ ∂∂  ∂ ∂= ∂∂  ∂= ∂  ∂=∑∑ ∑∑ ∑∑ ∑grad e e e e e (3) Therefore, ()() frfrrr∂=∂rgrad (4) c) () () () ( )()122 2 22 22 12 2 12 2 1 2 21 22 222 2ln ln ln 122 2 123j ij ii ij j i i j j ij ij i i ii j j ij i j i j irrxxx xx x x xxx xxx x xx xrr− − − − − ∂ ∂ == ∂∂   ⋅  ∂= ∂       ∂ = ∂   ∂=− +  ∂   =− +∑∑∑ ∑ ∑ ∑ ∑∑ ∑∑ ∑ ∑ ∑∇ 2 42 223 1r rr r  =− + = (5) or, ()2 21ln rr= ∇ (6) 24 CHAPTER 1 1-32. Note that the integrand is a perfect differential: () ( 22ddab a bdt dt⋅+ ⋅= ⋅ + ⋅rr rr rr rr   ) (1) Clearly, ()2222 c on ab d tar b r ⋅+ ⋅ = + +∫rr rr    st. (2) 1-33. Since 2dr r dtrrr2r r− == −rr rr r   (1) we have 2rddt dtrr d t r −=    ∫∫rr r  (2) from which 2rdtrr r− =+∫rr rC (3) where C is the integration constant (a vector ). 1-34. First, we note that ( )d dt× =×+× AA AAAA    (1) But the first term on the right-hand side vanishes. Thus, ( ) ( )ddt dtdt×= ×∫∫AA AA  (2) so that ( )dt × =×+ ∫AA AAC  (3) where C is a constant vector. MATRICES, VECTORS, AND VECTOR CALCULUS 25 1-35. x zy We compute the volume of the intersection of the two cylinders by dividing the intersection volume into two parts. Part of the common volume is that of one of the cylinders, for example, the one along the y axis, between y = –a and y = a: ()2 122 Va a3a π π == (1) The rest of the common volume is formed by 8 equal parts from the other cylinder (the one along the x-axis). One of these parts extends from x = 0 to x = a, y = 0 to 2ya x=−2, z = a to 22z ax=− . The complementary volume is then 22 22 200 22 22 0 33 21 0 338 8 8s in32 1623aa x a x a a aVd xd yd z d x ax axa xa xaxa aa π−− −=   =− − −  =− − =−∫∫ ∫ ∫ (2) Then, from (1) and (2): 3 1216 3aVV V=+= (3) 26 CHAPTER 1 1-36. dz xy c2 = x2 + y2 The form of the integral suggests the use of the divergence theorem. (1) SVd⋅=∇ ⋅∫∫Aa A dv Since ∇⋅ , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is 1=A (2) 2 Vdv c d π=∫ 1-37. z y xR To do the integral directly, note that A , on the surface, and that . 3 rR=e 5r dd a=ae 33 244 SSdRd a R R R π π ⋅= = × =∫∫Aa (1) To use the divergence theorem, we need to calculate ∇⋅A. This is best done in spherical coordinates, where A . Using Appendix F, we see that 3 rr=e ( )2 215rrrr2r∂∇⋅ = =∂AA (2) Therefore, ()222 00 0sin 5 4R Vdv d d r r dr Rππ5θθφ π ∇⋅ = = ∫∫A∫∫ (3) Alternatively, one may simply set dv in this case. 24r drπ= MATRICES, VECTORS, AND VECTOR CALCULUS 27 1-38. xz y C x2 + y2 = 1z = 1 – x2 – y2 By Stoke’s theorem, we have () Sd Cd ∇×⋅= ⋅∫∫Aa A s (1) The curve C that encloses our surface S is the unit circle that lies in the xy plane. Since the element of area on the surface da is chosen to be outward from the origin, the curve is directed counterclockwise, as required by the right-hand rule. Now change to polar coordinates, so that we have ddθθ=se and sin cos θ θ =+Aik on the curve. Since sin θ θ ⋅=− ei and 0 θ⋅=ek , we have ( )22 0sin Cdπdθθπ ⋅=− = − ∫∫As (2) 1-39. a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then (1 , 2 , 0 )=−AB ; (1 , 0 , 3 )=−AC ; and (6,3,2) ×=AB AC . Any vector perpendicular to pl ane (ABC) must be parallel to ×AB AC , so the unit vector perpendicular to plane (ABC) is (6 7,3 7,2 7)=n b) Let’s denote D = (1,1,1) and H = ( x,y,z) be the point on plane (ABC) closest to H. Then (1 ,1 ,1 xyz=− − −DH ) is parallel to n given in a); this means 16213x y−==− and 16312x z−==− Further, (1 , ,xy=− )z AH is perpendicular to n so one has 6( 1) 3 2 0 xy z −++= . Solving these 3 equations one finds H ( , , ) (19 49,34 49,39 49) xyz== and 5 7DH = 1-40. a) At the top of the hill, z is maximum; 02 and 6zyxx∂== − −∂18 02 8zxyy28∂= =−+∂ 28 CHAPTER 1 so x = –2 ; y = 3, and the hill’s height is max[ z]= 72 m. Actually, this is the max value of z, because the given equation of z implies that, for each given value of x (or y), z describes an upside down parabola in term of y ( or x) variable. b) At point A: x = y = 1, z = 13. At this point, two of the tangent vectors to the surface of the hill are 1 (1,1)(1, 0, ) (1, 0, 8)z x∂=∂t= and −2 (1,1)(0,1, ) (0,1,22)z y∂==∂t Evidently tt is perpendicular to the hill surface, and the angle 12 (8, 22,1) ×= − θ between this and O z axis is 22 2(0,0,1) (8, 22,1) 1s23.43 82 21⋅− ++coθ= so θ = 87.55 degrees. = c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is steepest. Evidently, dy = (tan α)dx and d 2 2 6 8 18 28 22(tan 1)z xdy ydx xdx ydy dx dy dx α =+ −−−+= − then 22cos 1 22(tan 1) 22 2 cos ( 45)dx dy dx dz dxα α α+ −=− +tanβ== The hill is steepest when tanβ is minimum, and this happens when α = –45 degrees with respect to W-E axis. (note that α = 135 does not give a physical answer). 1-41. 2( 1 ) aa ⋅=− AB then if only a = 1 or a = 0. 0 ⋅=AB CHAPTER 2 Newtonian Mechanics — Single Particle 2-1. The basic equation is ii Fmx= (1) a) ()()() ,iiFx t f x gt m x == ii: Not integrable (2) b) ()()() ,iiFx t f x gt m x == ii () ()i iidxmf xgdt= t ()()i iigt dxdtfx m= : Integrable (3) c) ( )()() ,ii i i iFx x f x gx m x ==i: Not integrable (4) 2-2. Using spherical coordinates, we can writ e the force applied to the particle as rrF FFθθφ φ = ++ Fe e e (1) But since the particle is constrained to move on the surface of a sphere, there must exist a reaction force that acts on the particle. Therefore, th e total force acting on the particle is rrF−e total F Fmθθ φφ= += Fe e r (2) The position vector of the particle is rR=re (3) where R is the radius of the sphere and is const ant. The acceleration of the particle is rR == ar e  (4) 29 30 CHAPTER 2 We must now express in terms of , rereθe, and φe. Because the unit vectors in rectangular coordinates, e, , e, do not change with time, it is convenient to make the calculation in terms of these quantities. Using Fig. F-3, Appendix F, we see that 1 2e3 123 12 3 12sin cos sin sin cos cos cos cos sin sin sin cosr θ φθ φθ φθ θ φθ φ φφ =++ θ =+− =− + ee e e e e ee eee (5) Then ( )( ) 12 sin sin cos cos cos sin sin cos sin sinr φθ3 φ θφ θθφ θθφ φθφ θ φθ θ=− + + + − =+ee e e ee    θ (6) Similarly, cosr θφ θ φ =− +eee    θ (7) sin cosr φθ φ θφ =− −ee e   θ (8) And, further, ( )( )( )22 2 2sin sin cos 2 cos sinrr θφ φ θθ θφ θ θ θ φ θφ θ =− + + − + +ee e e        (9) which is the only second time derivative needed. The total force acting on the particle is total r mm R= = Fr   e (10) and the components are ( ) ( )2sin cos 2c o s s i nFm R Fm Rθ φθφθ θ θφθ φθ=− =+   (11) NEWTONIAN MECHANICS—SINGLE PARTICLE 31 2-3. y xv0P βα ℓ The equation of motion is m=Fa (1) The gravitational force is the on ly applied force; therefore, 0x yFm x F my mg==    == −  (2) Integrating these equations and using the initial conditions, ( ) ()0 00c os 0s inxt v yt vα α==   ==   (3) We find () ()0 0cos sinxt v yt v g tα α=   =−   (4) So the equations for x and y are () ()0 2 0cos 1sin2xt vt yt vt g tα α=    =−  (5) Suppose it takes a time t to reach the point P. Then, 0 00 2 00 0cos cos 1sin sin2vt vt g tβα βα=    =− A A (6) Eliminating A between these equations, 0 0 002s i n 2 1cos tan 02v vtggααβgt − += (7) from which 32 CHAPTER 2 ()0 02sin cos tanvtgα αβ =− (8) 2-4. One of the balls’ height can be described by 2 00 2 yy v tg t=+ − . The amount of time it takes to rise and fall to its initial height is therefore given by 02vg . If the time it takes to cycle the ball through the juggler’s hands is 0.9 s τ= , then there must be 3 balls in the air during that time τ. A single ball must stay in the air for at least 3τ, so the condition is 023vg τ≥ , or . 1 013.2 m sv−≥⋅ 2-5. mgN point of maximum accelerationflightpath planeer a) From the force diagram we have ( )2 r mm v R −=Ng e . The acceleration that the pilot feels is ( )2 r mm vR =+ Ng e , which has a maximum magnitude at the bottom of the maneuver. b) If the acceleration felt by the pilot must be less than 9 g, then we have ( )12 233 3 0 ms 12.5 km88 9.8msvRg− −⋅⋅ ≥=⋅⋅ (1) A circle smaller than this w ill result in pilot blackout. 2-6. Let the origin of our coordinate system be at th e tail end of the cattle (or the closest cow/bull). a) The bales are moving initially at the speed of the plane when dropped. Describe one of these bales by the parametric equations 00 xx v t= + (1) NEWTONIAN MECHANICS—SINGLE PARTICLE 33 2 01 2yy g t=− (2) where , and we need to solve for . From (2), the time the bale hits the ground is 080 m y=0x 02yg τ= . If we want the bale to land at ()30 m xτ=− , then ()0xx v0 τ τ = − . Substituting and the other values, this gives -1s044.4 m v=⋅0 210 x m − . The rancher should drop the bales 210 m behind the cattle. b) She could drop the bale earlier by any amount of time and not hit the cattle. If she were late by the amount of time it takes the bale (or the plane) to travel by 30 m in the x-direction, then she will strike cattle. This time is given by ( )0 30 m 0.68 s v . 2-7. Air resistance is always anti-parallel to the velocity. The vector expression is: 2 11 22wcA v cA vvρ=− =−vWwρv a (1) Including gravity and setting , we obtain the parametric equations net m=F 2xb x x y=− +   2 (2) 22yb y x yg=−+    − (3) where 2wAm bc ρ= . Solving with a computer using the given values and , we find that if the rancher drops the bale 210 m be hind the cattle (the answer from the previous problem), then it takes  4.44 s to land  62.5 m behind the cattle. This means that the bale should be dropped at  178 m behind the cattle to land 30 m behind. This solution is what is plotted in the figure. The time error she is allo wed to make is the same as in the previous problem since it only depends on how fast the plane is moving. -31.3 kg mρ=⋅ –200 –180 –160 –140 –120 –100 –80 –60 –40020406080 With air resistance No air resistancex (m)y (m) 34 CHAPTER 2 2-8. P Q xy v0 αh From problem 2-3 the equations for the coordinates are 0cos xv t α = (1) 2 01sin2yv t g t α =− (2) In order to calculate the time when a projective reaches the ground, we let y = 0 in (2): 2 01sin 02vt g t α− = (3) 02sinvtgα = (4) Substituting (4) into (1) we find the re lation between the range and the angle as 2 0sin 2vxgα = (5) The range is maximum when 22πα=, i.e., 4πα=. For this value of α the coordinates become 0 2 02 1 2 2vxt vxt g t=    =− (6) Eliminating t between these equations yields 22 2 000vvxx ygg− += (7) We can find the x-coordinate of the projectile when it is at the height h by putting y = h in (7): 22 2 000vv hxxgg− += (8) This equation has two solutions: 22 2 00 10 22 2 00 20422 422vvxvgg vvxvgg=− − gh gh   =+ − (9) NEWTONIAN MECHANICS—SINGLE PARTICLE 35 where corresponds to the point P and to Q in the diagram. Therefore, 1x2x 2 0 21 0 4vdx x v g hg=−= − (10) 2-9. a) Zero resisting force ( ): 0rF= The equation of motion for the vertical motion is: dvFma m mgdt== = − (1) Integration of (1) yields 0 vg t v =−+ (2) where v is the initial velocity of the projectile and t = 0 is the initial time. 0 The time required for the projectile to reach its maximum height is obtained from (2). Since corresponds to the point of zero velocity, mt mt () 0 0mvt v g t == −m, (3) we obtain 0 mvtg= (4) b) Resisting force proportional to the velocity ( )rF kmv=− : The equation of motion for this case is: dvFmm gkdt== − − mv (5) where – kmv is a downward force for mtt<′ and is an upward force for mtt>′. Integrating, we obtain ()0 kt gk v gvt ekk− +=− + (6) For t , v(t) = 0, then from (6), mt= ( ) 0 1mktgvek= − (7) which can be rewritten as 0ln 1mkvktg =+     (8) Since, for small z (z  1) the expansion 36 CHAPTER 2 ()211ln 1233zzz z += − + (9) is valid, (8) can be expressed approximately as 2 00 0 1123 2mvk v k vtgg g    =− + −     … (10) which gives the correct result, as in (4) for the limit k → 0. 2-10. The differential equation we are asked to solve is Equation (2.22), which is . Using the given values, the plots are shown in the figure. Of course, the reader will not be able to distinguish between the results shown here and the analytical results. The reader will have to take the word of the author that the graphs were obtained using numerical methods on a computer. The results obtained were at most within 10xk=−  x 8− of the analytical solution. 0 5 10 15 20 25 30510v vs t t (s)v (m/s) 0 2 04 06 08 00510 100v vs x x (m)v (m/s)0 5 10 15 20 25 30050100x vs t t (s)x (m) 2-11. The equation of motion is 2 2 2dxm kmv mgdt=− + (1) This equation can be solved exactly in th e same way as in problem 2-12 and we find NEWTONIAN MECHANICS—SINGLE PARTICLE 37 2 0 21log2gk vxkg kv  −=  −  (2) where the origin is taken to be the point at which v0v= so that the initial condition is ( )00 xv v== . Thus, the distance from the point 0 vv= to the point v1v= is ()2 0 01 2 11log2gk vsv vkg kv  −→=  −  (3) 2-12. The equation of motion for the upward motion is 2 2 2dxmm kvdt=− − mg (1) Using the relation 2 2d x dv dv dx dvvdt dt dx dt dx== = (2) we can rewrite (1) as 2vd vdxkv g=−+ (3) Integrating (3), we find ( )2 1log2kv g x Ck+ =− + (4) where the constant C can be computed by using the initial condition that when x = 0: 0 vv= ( )2 01log2Ck vkg = + (5) Therefore, 2 0 21log2kv gxkk v g+=+ (6) Now, the equation of downward motion is 2 2 2dxmm kvdt=− + mg (7) This can be rewritten as 2vd vdxkv g=−+ (8) Integrating (8) and using th e initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find 38 CHAPTER 2 21log2gxkg k v=− (9) At the highest point the velocity of the particle must be zero. So we find the highest point by substituting v = 0 in (6): 2 0 1log2hkv gxkg+= (10) Then, substituting (10) into (9), 2 0 211log log22kv g g kg k g k+=−v (11) Solving for v, 2 0 2 0gvkvgvk= + (12) We can find the terminal velocity by putting x → ∞ in(9). This gives tgvk= (13) Therefore, 0 22 0t tvvv vv= + (14) 2-13. The equation of motion of the particle is ( )32 dvmm kv adt=− + v (1) Integrating, ( )22dvkd t vv a=− +∫ ∫ (2) and using Eq. (E.3), Appendix E, we find 2 22 21ln2vkt Caa v=−+ + (3) Therefore, we have 2 22At vCeav−=′+ (4) NEWTONIAN MECHANICS—SINGLE PARTICLE 39 where 22A ak≡ 0 vv= and where C′ is a new constant. We can evaluate C′ by using the initial condition, at t = 0: 2 0 2 0vCav=′2+ (5) Substituting (5) into (4) and rearranging, we have 122 1At AtaCe d xvCe d t− − ′=−′ = (6) Now, in order to integrate (6), we introduce At−≡ue so that du = –Au dt. Then, 12 122 211At AtaCe a Cu d uxd tCe A Cu u aC d u A Cu u− − ′′ ==  −−′′   ′ −+′∫∫ =−∫ (7) Using Eq. (E.8c), Appendix E, we find ()1sin 1 2axC uCA−=− +′ ′′ (8) Again, the constant C″ can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1: (1sin 1 2aC )CA−=− −′′′ (9) Therefore, we have ( ) ()11sin 2 1 sin 2 1At ae C xCA−− −=− + − − + ′′ Using (4) and (5), we can write 22 22 11 0 22 22 01sin sin2va va a k va va−−x   −+ −+−  =    ++     (10) From (6) we see that v → 0 as t → ∞. Therefore, 22 1 22lim sin sin (1)2 tva va1 π−− →∞−+= = +  (11) Also, for very large initial velocities, () 022 11 0 22 0lim sin sin 12 vva vaπ−− →∞−+= −= − +  (12) Therefore, using (11) and (12) in (10), we have 40 CHAPTER 2 ()2xtkaπ→∞ = (13) and the particle can never move a distance greater than 2kaπ for any initial velocity. 2-14. y xdαβ a) The equations for the projectile are 0 2 0cos 1sin2xv t yv t g tα α= =− Solving the first for t and substituting into the second gives 2 22 01tan2c osgxyxvαα=− Using x = d cos β and y = d sin β gives 22 22 0 2 22 0cossin cos tan2c o s cos0c ostan2c o sgdddv gddvβββ αα βsin β αβα=− =− +  Since the root d = 0 is not of interest, we have ( ) ()22 0 2 2 0 22c o s t a n s i n c o s cos 2c o s s i nc o s c o ss i n cosvdg v gβα β α β α αβ αβ β−= −= ()2 0 22c o ss i n cosvdgα αβ β−= (1) NEWTONIAN MECHANICS—SINGLE PARTICLE 41 b) Maximize d with respect to α () () ()(2 0 220s insin coscos cos2cosv dddg) α αβ α αβ αβαβ== − − + − −  ( ) cos 2 0 αβ− = 22παβ−= 42πβα=+ c) Substitute (2) into (1) 2 0 max 22cos sincos 4 2 4 2v gdπβπ β β   =+  −       Using the identity () ()11n sin 2 cos sin22siA BA B A −= + − B we have 22 00 max 22sin sin1s i n 2 2 cos 2 1 sinvv ggdπββ β β−  −=⋅ =  −  ()2 0 max1s i nvdg β=+ 2-15. mg θmg sin θ The equation of motion along the plane is 2sindvmm g k mdtθ =− v (1) Rewriting this equation in the form 21 sindvdtgkvkθ= − (2) 42 CHAPTER 2 We know that the velocity of the particle continues to increase with time (i.e., 0 dv dt >), so that ()2sin gk v θ>. Therefore, we must use Eq. (E.5a), Appe ndix E, to perform the integration. We find 1 11tanh sin sinvtCkgg kkθθ−=+   (3) The initial condition v(t = 0) = 0 implies C = 0. Therefore, ( ) sin tanh sing dxvg kkdθθ = tt= (4) We can integrate this equation to obtain the displacement x as a function of time: ( ) sin tanh singxg kkθθ = ∫tdt Using Eq. (E.17a), Appendix E, we obtain ( ) ln cosh sin sin singk t g k gkθ θ θ=xC +′ (5) The initial condition x(t = 0) = 0 implies C′ = 0. Therefore, the relation between d and t is ( )1ln cosh sin dg kktθ = (6) From this equation, we can easily find ()1cosh sindke t gk θ− = (7) 2-16. The only force which is applied to the articl e is the component of the gravitational force along the slope: mg sin α. So the acceleration is g sin α. Therefore the velocity and displacement along the slope for upward motion are described by: ( ) 0 sin vv g t α =− (1) (2 01sin2xv t g t )α =− (2) where the initial conditions ( )0 0v ==vt and ( )0 xt 0 == have been used. At the highest position the velocity becomes ze ro, so the time required to reach the highest position is, from (1), 0 0sinvtg α= (3) At that time, the displacement is NEWTONIAN MECHANICS—SINGLE PARTICLE 43 2 0 01 2s i nvxg α= (4) For downward motion, the velocity and the displacement are described by ( ) sin vg t α = (5) (2 1sin2xg α = )t (6) where we take a new origin for x and t at the highest position so that the initial conditions are v(t = 0) = 0 and x(t = 0) = 0. We find the time required to move from the highest position to the starting position by substituting (4) into (6): 0 sinvtg α=′ (7) Adding (3) and (7), we find 02 sinvtg α= (8) for the total time required to return to the initial position. 2-17. v0 Fence35˚ 0.7 m60 m The setup for this problem is as follows: 0cos xv t θ = (1) 2 001sin2yy v t g t θ =+ − (2) where and . The ball crosses the fence at a time 35θ=D 007 m y=. ( ) 0cos Rvτ θ = , where R = 60 m. It must be at least h = 2 m high, so we also need 2 00 sinv gτθ τ − 2 hy−= . Solving for , we obtain 0v ()2 2 0 0 2 cos sin cosgRv Rh y θ θθ=  −−  (3) which gives v . 1 025 4 m s−.⋅ 44 CHAPTER 2 2-18. a) The differential equation here is the same as that used in Prob lem 2-7. It must be solved for many different values of v in order to find the minimum required to have the ball go over the fence. This can be a computer-intensive and time -consuming task, although if done correctly is easily tractable by a person al computer. This minimum is 0 0v135.2 m s−⋅  3m, and the trajectory is shown in Figure (a). (We take the density of air as 13 k gρ−=. ⋅ .) 01 02 03 04 05 06 0051015 With air resistance No air resistance fence height fence rangex (m)y (m) b) The process here is the same as for part (a), but now we have v fixed at the result just obtained, and the elevation angle θ must be varied to give th e ball a maximum height at the fence. The angle that does this is  0.71 rad = 40.7°, and the ball now clears the fence by 1.1 m. This trajectory is shown in Figure (b). 0 01 02 03 04 05 06 005101520 Flight Path fence height fence rangex (m)y (m) NEWTONIAN MECHANICS—SINGLE PARTICLE 45 2-19. The projectile’s motion is described by ( ) ()0 2 0cos 1sin2xv t yv t g tα α =    =− (1) where v is the initial velocity. The distance from the point of projection is 0 2rx y=+2 (2) Since r must always increase with time, we must have : 0r> 0xx yyrr+= > (3) Using (1), we have ()23 2 2 013sin22yy g t g v t v t α += − +0 xx (4) Let us now find the value of t which yields 0 xx yy+ =  (i.e., 0 r= ): 2 0 0 sin 39s i n 822v vtggαα =± − (5) For small values of α, the second term in (5) is imaginary. That is, r = 0 is never attained and the value of t resulting from the condition 0 r= is unphysical. Only for values of α greater than the value for which the radicand is zero does t become a physical time at which does in fact vanish. Therefore, the maximum value of α that insures for all values of t is obtained from r 0r> 2 max 9s i n 8 0 α −= (6) or, max22sin3α = (7) so that max 70.5 α ≅ ° (8) 2-20. If there were no retardation, the range of the projectile would be given by Eq. (2.54): 2 0 0 sin 2vRgθ = (1) The angle of elevation is therefore obtained from 46 CHAPTER 2 () ( ) ()0 2 0 2 2sin 2 1000 m 9.8 m/sec 140 m/sec 0.50Rg vθ= × = = (2) so that 015 θ=° (3) Now, the real range R′, in the linear approximation, is given by Eq. (2.55): 2 00413 sin 2 4 sin13kVRRg vk v ggθ θ=−′    =−     (4) Since we expect the real angle θ to be not too different from the angle 0θ calculated above, we can solve (4) for θ by substituting 0θ for θ in the correction term in the parentheses. Thus, 2 00 0sin 2 4s in13gR kvvgθ θ′= −   (5) Next, we need the value of k. From Fig. 2-3(c) we find the value of km by measuring the slope of the curve in the vicinity of v = 140 m/sec. We find ( )( ) 110 N 500 m/s 0.22 kg/s km≅ ≅ . The curve is that appropriate for a projectile of mass 1 kg, so the value of k is 10.022 seck− (6) Substituting the values of the variou s quantities into (5) we find 17.1 θ= °. Since this angle is somewhat greater than 0θ, we should iterate our solution by using this new value for 0θ in (5). We then find 17.4 θ=° . Further iteration does not substantially change the value, and so we conclude that 17.4θ= ° If there were no retardation, a projectile fired at an angle of 17.4° with an initial velocity of 140 m/sec would have a range of ( )2 2140 m/sec sin 34.8 9.8 m/sec 1140 mR°=  NEWTONIAN MECHANICS—SINGLE PARTICLE 47 2-21. x3 αv0 x2 x1 Assume a coordinate system in which the projectile moves in the 2xx3− plane. Then, 20 2 30cos 1sin2xv t xv t g tα α=   =−  (1) or, ()22 33 2 02 01cos sin2xx vt vt g t αα=+ =+ − re e e3e (2) The linear momentum of the projectile is ( ) ( ) 02 0 cos sin mm v v g t αα3  == + − e e  pr (3) and the angular momentum is ( )( ) ( ) ( )2 02 0 3 0 2 0 cos sin cos sinvt vt g t m v v g t αα α α 3  =× = + − × + −  Lrp e e e e (4) Using the property of the unit vectors that 3 i j ijkε ×= eee , we find ( )2 01cos2mg v t α =L1e (5) This gives ( ) 0cos mg v t α =−L 1e (6) Now, the force acting on the projectile is 3 mg=−Fe (7) so that the torque is () () ()2 02 0 3 011cos sin2 cosvt vt g t m g mg v tαα α =× = + − −     =−NrF e e e e3 which is the same result as in (6). 48 CHAPTER 2 2-22. xyz eB E Our force equation is ( ) q= +× FE v B (1) a) Note that when E = 0, the force is always perpendicular to the velocity. This is a centripetal acceleration and may be analyzed by elemen tary means. In this case we have also so that ⊥vB vB ×=vB . 2 centripetalmvma qvBr== (2) Solving this for r cmv vrqB ω== (3) withcqB m ω≡/ . b) Here we don’t make any assumptions about the relative orientations of v and B, i.e. the velocity may have a component in the z direction upon entering the field region. Let , with xyz=++rij k =vr and . Let us calculate first the v × B term. =ar ( 00xyz B y x B×= = −ij k vB i j    ) (4) The Lorentz equation (1) becomes ( ) y z m qBy q E Bx qE== + − +Fr i j   k (5) Rewriting this as component equations: cqBxymyω ==   (6) y cqE E qByx xmm Bω=− + =− −  y  (7) zqEzm= (8) NEWTONIAN MECHANICS—SINGLE PARTICLE 49 The z-component equation of motion (8) is easily integrable, with the constants of integration given by the initial conditions in the problem statement. ()2 002zqEztzz t tm=+ +  (9) c) We are asked to find expressions for and , which we will call and x yxvyv, respectively. Differentiate (6) once with respect to time, and substitute (7) for yv 2 y xc y c xE vv vBωω== − −  (10) or 22 y xc xcE vvBωω+= (11) This is an inhomogeneous differential equation that has both a homogeneous solution (the solution for the above equation with the right si de set to zero) and a particular solution. The most general solution is the sum of both, which in this case is () ()12cos siny xc cE vC t C tBωω =++ (12) where C and C are constants of integration. This result may be substituted into (7) to get 1 2 yv  () () 12cos sin y cc cCt C v ct ω ωω ω =− − (13) () () 12sin cosyc cvC t C t ωω =− + + K (14) where K is yet another constant of integration. It is found upon substitution into (6), however, that we must have K = 0. To compute the time averages, note that both sine and cosine have an average of zero over one of their periods 2c T πω ≡ . 0yE xyB= ,=  (15) d) We get the parametric equations by simply integrating the velocity equations. () ()12sin cosy cc ccE CCtBωωωω=−+x tD+ xt (16) () ()12cos sincc y ccCCyt t ωωωω=+ D+ (17) where, indeed, D and x yD are constants of integration. We may now evaluate all the C’s and D’s using our initial conditions ()0c xA ω =− , ()0y xE B = , ()00y =, ()0y = A. This gives us , and gives the correct answer 1 xy CDD== 0=2C=A () ()cosy c cE Axt t tBωω−=+ (18) 50 CHAPTER 2 ( () s i nc cAyt t )ωω= (19) These cases are shown in the figure as (i) y AEB> , (ii) y AEB< , and (iii) y AEB= . (i) (ii) (iii) 2-23. () () atFtm atk te−== (1) with the initial conditions x(t) = v(t) = 0. We integrate to get the velocity. Showing this explicitly, ()() (0) 0vt tt vkat d t t e d tmα−= ∫∫ (2) Integrating this by parts and using our initial conditions, we obtain 211 1()t kvt t emα αα α  −       =− +  (3) By similarly integrating v(t), and using the integral (2) we can obtain x( t). ()322211 2t kxt t emα ααα α−   =− + + +       (4) To make our graphs, substi tute the given values of m = 1 kg, 11 N sk−= ⋅, and . 10.5 sα−= ()2txt t e−= (5) () ( )242 2tvt t e−=− + (6) () ( )216 4 4 4ttt t α−=− + + + e (7) NEWTONIAN MECHANICS—SINGLE PARTICLE 51 0 5 10 15 20050100 tx(t) 0 5 10 15 20024 tv(t) 0 5 10 15 2000.51 ta(t) 2-24. mg mgFfFfN BN′d= length of incline s= distance skier travels along level ground θ xmg sin θmg cos θ yy x While on the plane: FN∑ s o Nm cos 0y m g m y θ =− = =  c o sg θ = sinx f Fm∑ ; g F θ =− cosfF Nm g µ µθ = = sin cos mg mg mx θµθ− = So the acceleration down the plane is: ( ) 1 sin cos constant ag θµ θ =−= 52 CHAPTER 2 While on level ground: ; Nm=′ g gfFm µ=− So becomes xFm=∑x mg mxµ−=  The acceleration while on level ground is 2 constant ag µ=−= For motion with constant acceleration, we can get the velocity and position by simple integration: xa= 0 vxa tv==+ (1) 2 001 2xx v t a t−= + (2) Solving (1) for t and substituting into (2) gives: 0 vvta−= ( ) ( )2 00 0 01 2vvv vvxxaa−−−= + ⋅ or ( )22 00 2ax x v v− =− Using this equation with the initial and final points being the top and bottom of the incline respectively, we get: 2 = speed at bottom of incline 2 1 B ad V=BV Using the same equation for motion along the ground: 2 22B as V=− (3) Thus ad 1 as=−2 ( ) 1 sin cos ag θµ =− θ 2ag µ=− So ( ) sin cosgd gs θµθ µ−= Solving for µ gives sin cosd dsθµθ=+ Substituting θ = 17°, d = 100 m, s = 70 m gives 0.18µ= Substituting this value into (3): NEWTONIAN MECHANICS—SINGLE PARTICLE 53 22B gs Vµ− =− 2BVg sµ = 15.6 m/secBV= 2-25. a) At A, the forces on the ball are: N mg The track counters the gravitational force and provides centripetal acceleration 2Nm gm v R−= Get v by conservation of energy: 0top top topET U m gh = += + 2 102AA AET U m v = += + 2top AEEv g=→ = h So 2 Nm gm g h R= + 21hNm gR=+  b) At B the forces are: mgN 45˚ 2 2cos 45 2Nm v Rm g mv R mg= +° =+ (1) Get v by conservation of energy. From a), totalEm gh = . At B, 2 1 2v m gh =+ ′ Em 54 CHAPTER 2 RR h′45˚ RRcos 45 2 ˚=R2 2RR=+ h′ or 11 2hR=−′  So becomes: total B BET U =+ 2 1112 2mgh mgR mv=− +  Solving for 2v 2 121 2gh gR v − −=   Substituting into (1): 232 2hNm gR  =+ −      c) From b) 222B h R R =− +vg ()12 22 vg h R R   =− +  d) This is a projectile motion problem 45˚ 45˚ B A Put the origin at A. The equations: 00 x xx v t= + 2 001 2y yy v t g t=+ − become 22Bv Rx=+ t (2) 21 2 2Bvyh t g t=+ −′ (3) Solve (3) for t when y = 0 (ball lands). NEWTONIAN MECHANICS—SINGLE PARTICLE 55 222B gt v t h 0 − −=′ 222 8 2BBvv gtg±+ h′= We discard the negative root since it give s a negative time. Substituting into (2): 222 8 2 22BB Bvv g h v Rxg  ±+ ′  =+    Using the previous expressions for and h′ yields Bv ()12 22 2321 22h h R R xR =−+ + − +   e) , with , so Ux has the shape of the track. () ()Ux m g yx = (0)yh = ( ) 2-26. All of the kinetic energy of the block goes into compressing the spring, so that 2222 mv kx = , or 23 m xv m k= . , where x is the maximum compression and the given values have been substituted. When th ere is a rough floor, it exerts a force kmgµ in a direction that opposes the block’s velocity. It therefore does an amount of work kmgdµ in slowing the block down after traveling across the floor a distance d. After 2 m of floor, the block has energy 22k mv mgd µ− , which now goes into compressing the sp ring and still overcoming the friction on the floor, which is 22k kx mgx µ+ . Use of the quadratic formula gives 2 22 mg mg mgd mv kk k kµµ µ =− + + − x (1) Upon substitution of the given values, the result is  1.12 m. 2-27. 0.6 m To lift a small mass dm of rope onto the table, an amount of work ()( ) 0 dW dm g z z= − must be done on it, where 006 mz=. is the height of the table. The total amount of work that needs to be done is the integration over all th e small segments of rope, giving 02 0 00() ( )2z gzWd zgzzµµ =− = ∫ (1) When we substitute ( )( ) 04 k g 4 m mLµ== . , we obtain 0 18 J W . . 56 CHAPTER 2 2-28. m Mv vv4 v3 before collisionafter collision The problem, as stated, is completely one-dimensional. We may therefore use the elementary result obtained from the use of our conservation theorems: energy (since the collision is elastic) and momentum. We can factor the momentum conservation equation 11 22 13 24mv mv mv mv + =+ (1) out of the energy conservation equation 222 11 22 13 241111 2222mv mv mv mv +=+2 4 (2) and get 13 2vvvv+ =+ (3) This is the “conservation” of relative velocities that motivates the definition of the coefficient of restitution. In this problem, we initially have the superball of mass M coming up from the ground with velocity 2g =v , while the marble of mass m is falling at the same velocity. Conservation of momentum gives h ()34 Mvmv M v m v+ −= + (4) and our result for elastic collisions in one dimension gives 3() vv v v4 + =− + (5) solving for and and setting them equal to 3v4v 2itemgh , we obtain 23 1marblehα α−h = +  (6) 213 1superballhα α−h = +  (7) where mMα≡ . Note that if 13α< , the superball will bounce on the floor a second time after the collision. NEWTONIAN MECHANICS—SINGLE PARTICLE 57 2-29. mg cos θ mg sin θmgFf N y xθ 1tan 0.08 4.6 cos 0 cos sin cosy xf fFN m g my Nm g Fm g F mx FN m gθ θ θ θ µ µθ−= =° =− == = =− = ==∑ ∑  so ()sin cos sin cosmx mg mg xgθµθ θµ θ= − =−  Integrate with respect to time ( ) 0 sin cos xg t x θµ θ = −  + (1) Integrate again: (2 001sin cos2xx x t g t ) θµθ =+ + −  (2) Now we calculate the time required for the driver to stop for a given (initial speed) by solving Eq. (1) for t with . 0x 0x= ()10sin cosxtgθµθ−=− −′ Substituting this time into Eq. (2) gives us th e distance traveled before coming to a stop. () () () () ()2 00 221100 2 2101sin cos2 1sin cos sin cos2 cos sin2xx x t g t xxxggg xxgθµ θ θµθ θ µθ µθ θ− − −−= + −′′ ′ − − + − ∆= −  ∆= 58 CHAPTER 2 We have 4.6 θ=° , 0.45µ= , . 29.8 m/secg= For , . 025 mph 11.2 m/sec x== 17.4 meters x∆= If the driver had been going at 25 mph, he could only have skidded 17.4 meters. Therefore, he was speeding How fast was he going? gives . 30 meters x∆≥032.9 mph x≥ 2-30. 1 Tt t2 =+ (1) where T = total time = 4.021 sec. = the time required for the balloon to reach the ground. 1t = the additional time required for the sound of the splash to reach the first student. 2t We can get t from the equation 1 2 001 2yy y t g t=+ −  ; 00 0 yy= = When , y = –h; so ( h = height of building) 1tt= 2 11 2h−gt −= or 12h g=t distance sound travels speed of soundh vt== Substituting into (1): 2hh gv+ T= or 20hhTvg+ −= This is a quadratic equation in the variable h. Using the quadratic formula, we get: 22 4 2112 2T gg v gT v V g v−±+ h −± == +     Substituting 331 m/secV= 29.8 m/secg= 4.021 secT= and taking the positive root because it is the physically acceptable one, we get: NEWTONIAN MECHANICS—SINGLE PARTICLE 59 128.426 mh= h = 71 meters 2-31. For , example 2.10 proceeds as is until the equations following Eq. (2.78). Proceeding from there we have 00 x≠ 00 Bxα= ≠ 0 Az α= so ()00 0 cos sinzxxx t t α ααα−= + ( )00 yy y t−=  ()00 0 cos sinxzzzt t α ααα−= − + Note that () ( )22 2200 00 22z xxx zzα α−+ −= + Thus the projection of the motion onto the x–z plane is a circle of radius ( )1222 001xzα+ . ( )1222 00 0So the motion is unchanged except for a change in the the helix. The new radius is .mxzqB+ radius of 2-32. The forces on the hanging mass are T mg The equation of motion is (calling downward positive) mg T ma−= or ( ) Tm ga= − (1) The forces on the other mass are 60 CHAPTER 2 y x FfN T θ2mg2mg cos θ 2mg sin θ The y equation of motion gives 2c os Nm g m y0 θ −= = or 2c os Nm g θ = The x equation of motion gives ( ) 2c osfk kFNm g µ µθ == 2s in 2 c osk Tm g m g m a θ µθ −− = (2) Substituting from (1) into (2) 2s in 2 c os 2k mg mg mg ma θ µθ −− = When 0 θθ=, a = 0. So 00 2s i n 2 c o s 0k gg g θ µθ −− = ( )00 122 001sin cos2 sin 1 sink kθµ θ θµ θ=+ =+ − Isolating the square root, squaring both sides and rearranging gives ( )22 2 0011s ins in4kkµθθµ+− − 0= Using the quadratic formula gives ( )2 0 213 4sin 21kk kµ µθ µ±+= + 2-33. The differential equation to solve is 22 12W tcA v vyg gmvρ     = −= −     (1) where 2t gcwA vm ρ = is the terminal velocity. The initial conditions are , and . The computer integrations for parts (a), (b), and (c) are shown in the figure. 0100 m y= 00 v= NEWTONIAN MECHANICS—SINGLE PARTICLE 61 05050100 10 t (s) 05–10–50 10 t (s) 05–50 10 t (s)0246050100 t (s)y (m) 0246–200 t (s)v (m/s) 0246–50 t (s)a (m/s2)0 5 10 15050100 t (s) 0 5 10 15–50 t (s) 0 5 10 15–50 t (s) d) Taking as the density of air, the terminal velocities are 32.2, 8.0, and 11.0 (all ) for the baseball, ping-pong ball, and raindr op, respectively. Both the ping-pong ball and the raindrop essentially reach their terminal velo cities by the time they hit the ground. If we rewrite the mass as average density times volume, then we find that 313 k g mρ−=. ⋅ -1ms⋅ tm aterial vR ρ∝ . The differences in terminal velocities of the three objects can be explained in terms of their densities and sizes. e) Our differential equation shows that the effect of air resistance is an acceleration that is inversely proportional to the square of the te rminal velocity. Since the baseball has a higher terminal velocity than the ping-pong ball, the magnitude of its deceleration is smaller for a given speed. If a person throws the two objects wi th the same initial velocity, the baseball goes farther because it has less drag. f) We have shown in part (d) that the terminal ve locity of a raindrop of radius 0.004 m will be larger than for one with radius 0.002 m (-190 m s. ⋅) by a factor of 2. 2-34. FR y mg Take the y-axis to be positive downwards. The initial conditions are 0 yy== at t = 0. 62 CHAPTER 2 a) RF vα= The equation of motion is dvmy m mg vdtα == − md vdtmg v α=− Integrating gives: ()lnmmg v t C αα−− =+ Evaluate C using the condition v = 0 at t = 0: ()lnmmg Cα− = So () () ln lnmmmg v mg t ααα−− + = or ln ln 1mg v tv mm gα αα   −= −     mg−=   Take the exponential of both sides and solve for v: 1rm vemgα α−=− 1tm vemgα α−=− (1tm mgveα α−=− ) (1) ()1tm mgdy e dtα α−=− Integrate again: tm mg myC t eα αα− += +  y = 0 at t = 0, so: 22 mg mCm gααα== tm mg mmyt eα αα α−  =− + +   (2) Solve (3) for t and substitute into (4): 1tm vemgα α−−= (3) NEWTONIAN MECHANICS—SINGLE PARTICLE 63 ln 1mvtmgα α=− −   ln 1 1 ln 1mg mg mm v m v vm vymg mg g mgαα αα α α α α    =− − −+− =− − −           α  (4) ln 1mg mvyvmgα αα   =− + −          b) 2 RFv β= The equation of motion becomes: 2 dvmm gdtvβ =− 2md vdtmg v β=− Integrate and apply the initial condition v = 0 at t = 0: 2dvdtmg mvβ β= −∫ ∫ From integral tables 1 221tanhdx x ax a a−=−∫; so 1 1tanhvtCaa mβ−= + where mgaβ≡ 1 1t a n h000 Cα−==+ so: 1 1tanhvtaa mβ−= Solving for v: tanhatvamβ= (5) tanhdy atadt mβ= From integral tables tanh ln cosh ud u u = ∫ So ln coshmayCmtβ β+= 64 CHAPTER 2 Apply the conditions at y = 0 and t = 0 ()ln cosh 0 ln 1 0mmCββ= == So ln coshmaymtβ β= (6) Solving (5) for t: 1tanhmvta αβ−= Substituting into (6): 1ln cosh tanhmvya β−  =   Use the identity: 11 21tanh cosh 1u u−−= −, where 1u<. (In our case 1u< as it should be because 2vv am gβ= ; and the condition that 1u< just says that gravity is stronger than the retarding force, which it must be.) So ( )1212 21ln cosh cosh ln 1 1mm vm gβββ β−− ==  − yv mg − ( )2ln 12myv ββ=− − mg NEWTONIAN MECHANICS—SINGLE PARTICLE 65 2-35. 0 5 10 15 20 25 30 3502468101214 x (km)y (km) 0 0.02 0.04 0.06 0.080102030 k (1/s)Range (km) We are asked to solve Equations (2.41) and (2.42), for the values k = 0, 0.005, 0.01, 0.02, 0.04, and 0.08 (all in ), with initial speed v1s− 1 0600 m s−= ⋅ and angle of elevation 60 θ=° . The first figure is produced by numerical solution of the differe ntial equations, and agrees closely with Figure 2-8. Figure 2-9 can be most closely reproduc ed by finding the range for our values of k, and plotting them vs. k. A smooth curve could be drawn, or more ranges could be calculated with more values of k to fill in the plot, but we chose here to just connect the points with straight lines. 66 CHAPTER 2 2-36. θ hy x R Put the origin at the initial point. The equations for the x and y motion are then ( ) 0cos xv t θ = ()2 01sin2yv t g t θ =− Call τ the time when the projectile lands on the valley floor. The y equation then gives ()2 01sin2hv g θττ −= − Using the quadratic formula, we may find τ 22 0 0sin 2 sin vg v ggθ θτ+=+h (We take the positive since 0 τ>.) Substituting τ into the x equation gives the range R as a function of θ. 2 2 0cos sin sinvRgθθ θ =+2x + (1) where we have defined 2 0 2 xg hv ≡2. To maximize R for a given h and , we set 0v 0 dR d θ=. The equation we obtain is 2 22 2 2 22sin cosos sin sin sin 0 sinx xθθθθ θθ θ−− + + +c (2) = Although it can give () xx θ= , the above equation canno t be solved to give ( ) x θθ= in terms of the elementary functions. The optimum θ for a given x is plotted in the figure, along with its respective range in units of 2 0vg . Note that x = 0, which among other things corresponds to h = 0, gives the familiar result θ = 45° and 2 0 Rvg= . NEWTONIAN MECHANICS—SINGLE PARTICLE 67 012345678901020304050 10 xθ 01234567890510 10 xR/(v02/g)45˚ 1 2-37. vxα= 2dv dx x α=− Since dv dv dx dvvdt dx dt dx== then 2dv dvFm m v mdt dx x xα α  == = −   ()23Fxm x α =− 2-38. ()nvx a x−= a) ( )( )1 nn dv dv dx dvFm m m v m a x n a xdt dx dt dx−−== = = −− ()()21 2 nFx m n a x−+=− b) ()n dxvx a xdt−== nxd x a d t = Integrate: 1 1nxat Cn+ =++ C = 0 using given initial conditions 68 CHAPTER 2 ( )11nxn+=+ at ()[]()111nxn a t+=+ c) Substitute x(t) into F(x): () ( ) {}()( ) 2111 21nnFt m n a n a t− ++ =− + () ( ) []() ( ) 21 1 21nnFt m n a n a t−+ +=− + 2-39. a) vFeβα=− v dvedt mβα=− ved v dmβtα−=− ∫ ∫ 1vetmβCα β−− =− + 0 vv= at t = 0, so 01veCβ β−− = ()01v veemβ βtα β− −−− = − Solving for v gives ()01lnv tvt emβ αβ β−  =− +   b) Solve for t when v = 0 01v temβ αβ−+ = 01v mteβ αβ−  =−  c) From a) we have 01lnv tdx e dtmβ αβ β−  =− +   NEWTONIAN MECHANICS—SINGLE PARTICLE 69 Using () () ln lnax bax b dx ax b xa++= + − ∫ we obtain 00ln1vv tteemmtmββ αβ αβ βα β−− xC   ++    += − −        Evaluating C using x = 0 at t = 0 gives 0 0 v vmCeβ αβ−= So 00 0 2lnvv mv tm t te emmββ αβ αβ αβ β αβ−−0vβ−xe =− + − + +      Substituting the time required to stop from b) gives the distance required to stop 0 011v mxe vβ αββ β−   =− +        2-40. y anat(x(t),y(t)) x Write the velocity as v(t) = v(t)T(t). It follows that ()tndd v dtv adt dt dt= =+= +vTaT T aN (1) where N is the unit vector in the direction of ddTt. That N is normal to T follows from ( ) 0dd t=⋅ TT . Note also is positive definite. na a) We have 2 254 c o s y vA x t α α =+ = −  . Computing from the above equation, 22s in 54 c o stA t dvadt tα α α== − (2) We can get from knowing a in addition to . Using nata2 2 2y aA x α =+ =  , we get 22 22c o s 1 54 c o snttaa a A tαα α−=− = − (3) b) Graphing versus t shows that it has maxima at na tnα π= , where 2 naA α= . 70 CHAPTER 2 2-41. a) As measured on the train: 0iT=; 2 1 2f=Tm v 2 1 2Tm v ∆= b) As measured on the ground: 2 1 2iTm= u; ()2 1 2fTm v u =+ 2 1 2Tm v m v ∆= + u c) The woman does an amount of work equal to the kinetic energy gain of the ball as measured in her frame. 2 1 2Wm v = d) The train does work in order to keep moving at a constant speed u. (If the train did no work, its speed after the woman threw the ball would be slightly less than u, and the speed of the ball relative to the ground would not be u + v.) The term mvu is the work that must be supplied by the train. Wm v u= 2-42. Rθθ θb R From the figure, we have () ( 2 ) c o s s i nhR b R θ θθ =+ + θ, and the potential is Um( ) ( ) gh θ θ = . Now compute: sin cos2dU bmg Rdθ θθθ =− +   (1) 2 2cos sin2dU bmg R Rdθ θθθ  =− −     (2) NEWTONIAN MECHANICS—SINGLE PARTICLE 71 The equilibrium point (where 0 ddU θ=) that we wish to look at is clearly θ = 0. At that point, we have ( )222 dU d m g R b θ=− , which is stable for 2>Rb and unstable for Rb . We can use the results of Problem 2-46 to obtain stability for the case 2</ 2 Rb= , where we will find that the first non-trivial result is in fourth order and is negative. We therefore have an equilibrium at θ = 0 which is stable for 2 Rb> and unstable for 2 Rb≤ . 2-43. 32Fk x k xα =− + ()4 2 211 24xUx F d x k x kα=− = −∫ To sketch U(x), we note that for small x, U(x) behaves like the parabola 21 2kx. For large x, the behavior is determined by 4 21 4xkα− U(x) E0 E1 E2 E3 = 0 E4x1x2x3x4x5x ()2 1 2Em v U x =+ For E , the motion is unbounded; the particle may be anywhere. 0E= For E (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may remain at rest where it is, but if perturbed s lightly, it will move away from the equilibrium. 1E= What is the value of ? We find the x values by setting 1E 0dU dx=. 320kx kx α =− x = 0, ± α are the equilibrium points ()22 1111 244UE kkk2α αα ±== − = α For E , the particle is either bounded and oscillates between 2E=2x− and ; or the particle comes in from ±∞ to ± and returns to ±∞. 2x 3x 72 CHAPTER 2 For E , the particle is either at the stable equilibrium point x = 0, or beyond . 30=4 xx=± For E, the particle comes in from ±∞ to 4 5x± and returns. 2-44. m1T m1gm2T T m2gθ From the figure, the forces acting on the masses give the equations of motion 111mm g x T = −  (1) 222 2c o s mm g T x θ = −  (2) where is related to by the relation 2x1x ()2 1 2 24bxx−d = − (3) and ( )1 cos 2 db xθ =−. At equilibrium, 12 0 xx= =   and Tm1g = . This gives as the equilibrium values for the coordinates 1 102 124 4mdxb mm=− −2 (4) 2 202 124mdx mm= −2 (5) We recognize that our expression is identical to Equation (2.105), and has the same requirement that 10x 21 2 mm < for the equilibrium to exist. Wh en the system is in motion, the descriptive equations are obta ined from the force laws: ( )21 1 1 2() (4mbxmgxx− 2)gx −=   − 0 (6) To examine stability, let us expand the coordina tes about their equilibrium values and look at their behavior for small displacements. Let 11 1xxξ≡− and 222xx0 ξ≡ − . In the calculations, take terms in 1ξ and 2ξ, and their time derivatives, only up to first order. Equation (3) then becomes 21 2 (mm1) ξ ξ − . When written in terms of these new coordinates, the equation of motion becomes ( ) ()3222 12 1 1 12 1 24 4gm m mm m m dξ ξ− =−+ (7) NEWTONIAN MECHANICS—SINGLE PARTICLE 73 which is the equation for simple harmonic motion . The equilibrium is therefore stable, when it exists. 2-45. and 2-46. Expand the potential about the equilibrium point () 1 01 !i i i induUx xid x∞ =+= ∑ (1) The leading term in the force is then (1 ) (1 ) 01()n n ndU d UFx xdx n dx+ +=− =− ! (2) The force is restoring for a stable point, so we need ( )0 Fx>< 0 and ( )0 Fx 0 <>. This is never true when n is even (e.g., Uk ), and is only true for odd when 3x= n(1 ) (nndU d x+ 1 ) 00+  <. 2-47. We are given ( ) 0 () U ax xa=+ Ux for . Equilibrium points are defined by 0x> 0 dU dx =, with stability determined by 2dU2d x at those points. Here we have 0 21 dU aUdx x a =− +   (1) which vanishes at x = a. Now evaluate 2 0 3 220 aU dU a dx = >   (2) indicating that the equilibrium point is stable. 0 0.5 1 1.5 20510152025 x/aU(x)/U0 74 CHAPTER 2 2-48. In the equilibrium, the gravitational force an d the eccentric force acting on each star must be equal 32 22 22 /2 2Gm mv mG d dvdd d v mGππτ =⇒ = ⇒ = = 2-49. The distances from stars to the center of mass of the system are respectively 2 1 12dmrmm=+ and 1 2 12dmrmm=+ At equilibrium, like in previous problem, we have 32 22 12 1 1 2 1 1 2 11 2 1 1222 () ()m m v Gm r dvdr d m m v Gm mππτ =⇒ = ⇒ = =+ +Gm The result will be the same if we consider th e equilibrium of forces acting on 2nd star. 2-50. a) 00 0 22 2 0 2 0 22 2() 11 1tmv mv mv dFFd F t v tdt vv vmcc c =⇒ = = ⇒ =   −− − ∫22 2t Ft c+ 22 2 2 00 2 0() ()tcF txt vtd t m mFc = + −  ∫⇒= b) tv c) From a) we find 0 2 21vmt vFc= − Now if 010F m= , then NEWTONIAN MECHANICS—SINGLE PARTICLE 75 when 2=vc , we have 0.55 year 10 3ct== when v = 99% c, we have 996.67 years 10 199c==t 2-51. a) 2 0 2 0()mv dv dv bmb v d t v tdt v m btv m=− ⇒ =− ⇒ =+ ∫∫ Now let v(t) = v0/1000 , one finds 0999138.7 hoursmtvb== . tv b) 0 0() l ntbtv m mx t vdtbm+ ==  ∫ We use the value of t found in question a) to find the corresponding distance ( ) ln(1000) 6.9 kmmxtb== 2-52. a) 2 0 224() 1Ux dU xFxdx a a=− =− −  b) xU When F = 0, there is equilibrium; further when U has a local minimum (i.e. 0 dF dx <) it is stable, and when U has a local maximum (i.e. 0 dF dx >) it is unstable. 76 CHAPTER 2 So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0 is a stable equilibrium position. c) Around the origin, 00 2244Ux U kFk xamω ≈− ≡− ⇒ = =ma d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential, 2 0 min max 0 min2 2U mvUU vm== ⇒ = e) From energy conservation, we have 2 2 22 00 min 222122Ux U mv dx xvad t m+= ⇒ = = − mv a We note that, in the ideal case, because the initial velocity is th e escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression, 0 2 2 2 000 0 2 28exp 1 ln ( )28 81 exp 1xUatma md x m a a xtUU ax x Uta ma −    +== ⇒ =−   − +     ∫tx  tx 2-53. F is a conservative force when there ex ists a non-singular potential function U(x) satisfying F(x) = – grad (U(x)). So if F is conservative, its components satisfy the following relations y xF F yx∂ ∂=∂ ∂ and so on. a) In this case all relations above are satisfied, so F is indeed a conservative force. 2 1(,)2xUb xayz bx c U ayzx cx f y zx∂= + + ⇒ =− − − +∂F=− (1) where is a function of only y and z 1(,)fy z 2(,)yUaxz bz U ayzx byz f x zy∂= + ⇒ =− − +∂F=− (2) NEWTONIAN MECHANICS—SINGLE PARTICLE 77 where is a function of only x and z 2(,)fx z 3(,)Uaxy by c U ayzx byz f y zzFz∂= + + ⇒ =− − +∂=− (3) then from (1), (2), (3) we find that 2 2 2bxU axyz byz cx C=− − − − + where C is a arbitrary constant. b) Using the same method we find that F in this case is a conserva tive force, and its potential is exp( ) ln Uz x y z C=−− −+ c) Using the same method we find that F in this case is a conserva tive force, and its potential is ( using the result of problem 1-31b): ln Ua r=− 2-54. a) Terminal velocity means final steady velocity (here we assume that the potato reaches this velocity before the impact with the Earth) when the total force acting on the potato is zero. mg = kmv and consequently 1000 m/sk vg= = . b) 00 0()x vF dx dv vdvgk v d t d xdt m v g kv g kv− + ⇒ == − ⇒ = − ⇒++ ∫∫dv== 0 max 2 0ln 679.7 mgg vxkk g k v=+ =+ where v is the initial velocity of the potato. 0 2-55. Let’s denote and 0xv0yv the initial horizontal and vertical velocity of the pumpkin. Evidently, 0x 0y vv= in this problem. 0xx f xx xx f xxvv dv dv dxmkv d t xdt v kv kmF− == − ⇒ − = − = ⇒= (1) where the suffix f always denote the final value. From the second equality of (1), we have 0fkt x xf x xdvdt v v ekv−−= ⇒ = (2) Combining (1) and (2) we have (01fkt x fvxk−=− )e (3) 78 CHAPTER 2 Do the same thing with the y-component, and we have 0 2 00l nyy yf y yf yy f yy ydv dv g kv v v dy gm F mg mkv dt ydt v g kv k g kv k+ − == − − ⇒ − = − = ⇒ = = +++ (4) and ( )0fkt y yf f ydv dt g kv g kv egk v−⇒ + = ++−= (5) From (4) and (5) with a littl e manipulation, we obtain 01fkt f ygkt egk v−−=+ (6) (3) and (6) are 2 equations with 2 unknowns, ft and k. We can eliminate ft, and obtain an equation of single variable k. () ()( )00 01fyxkt g kv gvx fvxek−+=− Putting and 142 mfx=0 00 38.2 m/s 2xyv===vv we can numerically solve for k and obtain K= 0.00246 1s−. CHAPTER 3 Oscillations 3-1. a) 421 0 2gram cm 10 dyne/cm 11 1 0 1 0 sec cmsec22 1 0 gram 2 g ram 2k mππ π π−⋅ ⋅= = ν== or, 01.6 Hz ν≅ (1) 0 012sec10πτν== or, 00.63 sec τ≅ (2) b) 24 2 1110 3 dyne-cm22Ek A == × × so that 44.5 10 ergE=× (3) c) The maximum velocity is attained when the to tal energy of the oscillator is equal to the kinetic energy. Therefore, 24 max 4 max14.5 10 erg2 24 . 51 0v100mv =× ××= 79 80 CHAPTER 3 or, max 30 cm/sec v = (4) 3-2. a) The statement that at a certain time 1tt= the maximum amplitude has decreased to one- half the initial value means that 1 01 2t enxA e Aβ−==0 (1) or, 11 2teβ−= (2) so that 11ln 2 0.69 ttβ== (3) Since , 110 sect= 26.9 10 secβ1 − −=× (4) b) According to Eq. (3.38), the angular frequency is 2 102ω ωβ=− (5) where, from Problem 3-1, 1 010 sec ω−= . Therefore, () ( ) ()2 2 2 1 2 6110 6.9 10 110 1 6.9 10 sec2ω− − −=− × ≅− × (6) so that 5 110(1 2 40 10 ) sec2νπ1 − −=− . × (7) which can be written as ( )10 1 ν νδ= − (8) where 52.40 10δ−=× (9) That is, 1ν is only slightly different from 0ν. OSCILLATIONS 81 c) The decrement of the motion is defined to be e1βτ where 111 τ ν= . Then, 11.0445 eβτ 3-3. The initial kinetic energy (equal to the total energy) of the oscillator is 2 01 2mv, where m = 100 g and v . 01 cm/sec= a) Maximum displacement is achieved when the to tal energy is equal to the potential energy. Therefore, 22 0011 22mv kx = 2 00 410 11 c10 10mxvk== × = m or, 01cm10x= (1) b) The maximum potential energy is 24 max 01110 1022Uk x2−== × × or, max 50 ergs U = (2) 3-4. a) Time average: The position and velocity for a simple harmonic oscillator are given by 0 sin xA t ω = (1) 0cos xA0t ω ω = (2) where 0 km ω= The time average of the kinetic energy is 2 11 2t tTm xτ τ+ =∫dt (3) where 02πτω= is the period of oscillation. 82 CHAPTER 3 By inserting (2) into (3), we obtain 22 2 01cos2t tTm A tτ ωωτ+ = ∫ 0dt (4) or, 22 0 4mATω= (5) In the same way, the time average of the potential energy is 2 22 0 211 2 1sin2 4t t t tUk xdt kA t dt kAτ ττ ωτ+ += = =∫ ∫ (6) and since 2 0km ω= , (6) reduces to 22 0 4mAUω= (7) From (5) and (7) we see that TU= (8) The result stated in (8) is reasonable to expe ct from the conservation of the total energy. ETU=+ (9) This equality is valid instantaneously, as well as in the average. On the other hand, when T and U are expressed by (1) and (2), we notice that they are described by ex actly the same function, displaced by a time 2τ: 22 2 0 0 2 2 0 0cos2 sin2mATt mA tUtωω ωω=    = (10) Therefore, the time averages of T and U must be equal. Then, by taking time average of (9), we find 2ETU= = (11) b) Space average: The space averages of the kine tic and potential energies are OSCILLATIONS 83 2 011 2A Tm xA=∫dx (12) and 2 2 0 0011 22AAmUk xdxxAAω==∫2dx∫ (13) (13) is readily integrated to give 22 0 6mAUω= (14) To integrate (12), we notice that from (1) and (2) we can write ( ) ( )22 2 2 2 2 2 00 0 22 2 0cos 1 sin xA t A Ax0t ω ωω ω ω== − =− (15) Then, substituting (15) into (12), we find 2 22 0 0 2 3 3 02 23AmTA xA m AAAω ωdx   =−  =− ∫ (16) or, 22 026mATω= (17) From the comparison of (14) and (17), we see that 2TU= (18) To see that this result is reasonable, we plot T = T(x) and U = U(x): 2 22 0 2 22 0112 1 2xTm AA Um xω ω  =−      = (19) U = U(x) T = T(x) AEnergy –AOxmA2 02ω Em A ==const.1 22 02ω And the area between T(x) and the x-axis is just twice that between U(x) and the x-axis. 84 CHAPTER 3 3-5. Differentiating the equation of motion for a simple harmonic oscillator, 0 sin xA t ω = (1) we obtain 00cos xA t t ω ω ∆=∆ (2) But from (1) 0 sinxtAω= (3) Therefore, ()2 0 cos 1 txω=− A (4) and substitution into (2) yields 22 0xt A x ω∆∆= − (5) Then, the fraction of a complete period that a si mple harmonic oscillator spends within a small interval ∆x at position x is given by 22 22 0 2tx x A xA τ ωτ π∆∆ ∆== −− x (6) –A1 –A2 –A3 A3 A2 A1∆t⁄τ x This result implies that the harmonic os cillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for tτ∆ at x = ±A. This occurs because at these points x = 0, and (2) is not valid. 3-6. x1x2k xm1 m2 Suppose the coordinates of m and are and x and the length of the spring at equilibrium is . Then the equations of motion for m and are 1 2 m1x2 1 A2m ( )11 1 2mx k x x =−− + A (1) ( )22 2 1mx k x x =−− + A (2) OSCILLATIONS 85 From (2), we have (12 2 21xm x k x kk=+ −  A) (3) Substituting this expression into (1), we find ()2 12 2 1 2 2 20dmmx m m k xdt + + = (4) from which 12 2 12mmxmm2kx+=− (5) Therefore, oscillates with the frequency 2x 12 12mmkmmω+= (6) We obtain the same result for . If we notice that the reduced mass of the system is defined as 1x 1211 1 mmµ=+ (7) we can rewrite (6) as kωµ= (8) k µ This means the system oscillates in the same way as a system consisting of a single mass µ. Inserting the given values, we obtain µ  66.7 g and 12.74 rad sω−⋅  . 3-7. A hb hs Let A be the cross-sectional area of the floating body, its height, the height of its submerged part; and let ρ and bhsh 0ρ denote the mass densities of the body and the fluid, respectively. The volume of displaced fluid is therefore VsA h= . The mass of the body is b M Ahρ= . 86 CHAPTER 3 There are two forces acting on the body: that due to gravity ( Mg), and that due to the fluid, pushing the body up (00 s gV gh Aρ ρ− −= ). The equilibrium situation occurs when the total force vanishes: 0 00 bMg gV gAh gh Asρ ρρ= − =− (1) which gives the relation between and : shbh 0sbhhρ ρ= (2) For a small displacement about the equilibrium position ( ), (1) becomes sshh→+ x ( ) 0 bb s MxA h xg Ah g h x ρρρ==−+  A (3) Upon substitution of (1) into (3), we have 0 bAh x gxAρ ρ=− (4) or, 00 bxg xhρ ρ+ =  (5) Thus, the motion is oscillator y, with an angular frequency 2 0 bsgg Aghh Vρωρ== = (6) where use has been made of (2), and in the la st step we have multiplied and divided by A. The period of the oscillations is, therefore, 22V gAπτπω== (7) Substituting the given values, 01 8 sτ . . 3-8. y O m s2a2a xℓ The force responsible for the motion of the pend ulum bob is the component of the gravitational force on m that acts perpendicular to the straight portion of the suspension string. This component is seen, from the figure (a) below, to be cos Fma mv mg α = == − (1) OSCILLATIONS 87 where α is the angle between the vertical and the tangent to the cycloidal path at the position of m. The cosine of α is expressed in terms of the differ entials shown in the figure (b) as cosdy dsα= (2) where 2ds dx dy=+2 (3) ααm mgFdxdy Sds (a) (b) The differentials, dx and dy, can be computed from the defining equations for x(φ) and y(φ) above: ( ) 1c o s sindx a d dy a dφφ φφ =−  =−  (4) Therefore, () ()222 222 2 2 22 21c o s s i n 2 1c o s i n2ds dx dy ad a ad2d 4sφ φφ φφ φφ=+ =− + =− = (5) so that 2s i n2ds a dφφ = (6) Thus, sin 2s i n2 cos cos2dy a d dsadφφ φφ φα−= =− = (7) The velocity of the pendulum bob is 88 CHAPTER 3 2s i n2 4c os2ds dvadt dt dadtφφ φ==  =−   (8) from which 2 24c os2dvadtφ  =−   (9) Letting cos2zφ≡ be the new variable, and substituting (7) and (9) into (1), we have 4 maz mgz− = (10) or, 04gzza+ =  (11) which is the standard equation for simple harmonic motion, 2 00 zzω+ =  (12) If we identify 0gω=A (13) where we have used the fact that . 4a=A Thus, the motion is exactly isochronous, independ ent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673). 3-9. The equation of motion for 0 0tt≤≤ is ( ) ( ) 0 mx k x x F kx F kx=− − + =− + +0 (1) while for , the equation is 0tt≥ ( )0 mx k x x kx kx= −−= − +0 (2) It is convenient to define 0 xxξ=− which transforms (1) and (2) into mkξξ F =−+; 0 0tt≤≤ (3) mkξ ξ=−; (4) 0tt≥ OSCILLATIONS 89 The homogeneous solutions for both (3) and (4) are of familiar form ()it ittA e B eω ωξ−=+ , where km ω= . A particular solution for (3) is Fk ξ= . Then the general solutions for (3) and (4) are it it FAeB ekω ωξ− −=+ + ; 0 0tt≤≤ (5) it itCe Deω ωξ− +=+ ; t (6) 0t≥ To determine the constants, we use the initial conditions: ( )0 0 xt x== and x(t = 0) = 0. Thus, ( ) ( ) 00ttξξ−− 0 === = (7) The conditions give two equations for A and B: ()0 0FABk iA Bω=++  =−  (8) Then 2FABk== − and, from (5), we have () 0 1c o sFxx tkξ ω−=− = − ; 0 0tt≤≤ (9) Since for any physical motion, x and must be continuous, the values of x ( )0ttξ−= and ( )0ttξ−= are the initial conditions for ()tξ+ which are needed to determine C and D: () () ()00 0000 001c o s sinit it it itFtt t C e D ek Ftt t i C e D ekωω ωωξω ξω ω ω− + − +== − = +     == = −  (10) The equations in (10) can be rewritten as: ()00 000 01c o s sinit it it itFCe De tk iFCe De tkωω ωωω ω− −+= −  − −= (11) Then, by adding and subtracting one from the other, we obtain () ()00 0012 12it it it itFCe ek FDe ekωω ωω− −=−    =− (12) 90 CHAPTER 3 Substitution of (12) into (6) yields () ( ) () () ()00 00 0112 2 cos cosit it it it it t it t it itFee e ek Fee e ek Ftt tkωω ω ω ωω ωωξ ωω−− + −− − −  =− + −    =− + −  =− − (13) Thus, () 00 cos cos ; Ftt tttkωω  −= −− ≥ 0 xx (14) 3-10. The amplitude of a damped oscillator is expressed by () ( )1 costxt A e tβω δ−= + (1) Since the amplitude decreases to 1 after n periods, we have e 121 nT nπββω= = (2) Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped oscillations), 22 102ω ωβ=− , we have 2 22 2 1 01 1 221124nnωωω ωππ  =+ = +     (3) Therefore, 12 1 22 0114nω ωπ− =+   (4) so that 1 22 2118nω ωπ≅− 3-11. The total energy of a damped oscillator is () () ()2 11 22Et m xt k xt =+ 2 (1) where () ( )1 costxt A e tβω δ−= − (2) () ( ) ( )11 1 cos sintA e t tβxt β ωδ ω ωδ−  =− − − −  (3) OSCILLATIONS 91 22 10ω ωβ=− , 0k mω= Substituting (2) and (3) into (1), we have () ( ) () () () ()2 22 2 2 2 11 1 11 1cos sin 2s in c ost AEt e m k t m t mt tβ 2β ωδ ω ωδ βωω δω δ− =+ − +  +−− − (4) Rewriting (4), we find the expression for E(t): () () ()2 22 2 2 2 10 1 cos2 sin 22t mAe t tββω δβω βω δω− =− + −0 −+ Et (5) Taking the derivative of (5), we find the expression for dE dt: ( ) () ()2 22 3 01 222 2 01 024 c os2 4s in2 2t dE mAet tββω β ω δ βωβ ω δ β ω− =− −  −− − −2 dt (6) The above formulas for E and dE reproduce the curves shown in Figure 3-7 of the text. To find the average rate of energy loss for a lightly damped oscillator, let us take dt 0 βω . This means that the oscillator has time to complete some number of periods before its amplitude decreases considerably, i.e. the term 2teβ− does not change much in the time it takes to complete one period. The cosine and sine terms will average to nearly zero compared to the constant term in dE dt , and we obtain in this limit 222 0t dEmA edtββω−− (7) 3-12. mgmg sin θθ ℓ The equation of motion is sin mm g θ θ −=A (1) singθ θ =− A (2) If θ is sufficiently small, we can approximate sin θθ≅, and (2) becomes 92 CHAPTER 3 gθ θ =− A (3) which has the oscillatory solution ()0cos t0t θ θω= (4) where 0 g ω= A and where 0θ is the amplitude. If there is the retarding force 2mg θA, the equation of motion becomes sin 2 mm g m g θ θ −= + A θA (5) or setting sin θθ≅ and rewriting, we have 2 002θω θ ω θ 0 + +=   (6) Comparing this equation with the standard equation for damped motion [Eq. (3.35)], 2 0 2xxx βω 0 + +=   (7) we identify 0ω β=. This is just the case of critical damping , so the solution for θ(t) is [see Eq. (3.43)] ()( )0ttA B t eωθ−=+ (8) For the initial conditions ()0 0θ θ= and θ(0) = 0, we find () ()0 001tttωθθω−=+ e 3-13. For the case of critical damping, 0 βω= . Therefore, the equation of motion becomes 22xx x ββ 0 + +=   (1) If we assume a solution of the form ()()txt yteβ−= (2) we have 22tt ttxy e y e xy e y e y eββ βββ ββ−− −− − =− tβ =− +     (3) Substituting (3) into (1), we find ye (4) 2222 2tt tt t tye ye ye ye yeββ ββ β βββ β β β−− −− − −−++− + 20= 0or, y= (5) Therefore, ()yt A B t=+ (6) OSCILLATIONS 93 and () ( )txt A B teβ−=+ (7) which is just Eq. (3.43). 3-14. For the case of overdamped oscillations, x(t) and ()xt are expressed by ()2 12t txt e A e Aeβ ω − −2tω  =+  (1) () ( )( )22 2 12 1 2 2 2tt te A e Ae A e A eωω ωβω− −−+ + + − 2tωω− txtβ (2) where 2 22 0 ω βω=− . Hyperbolic functions are defined as cosh2yyeey−+= , sinh2yyeey−−= (3) or, cosh sinh cosh sinhyyey ey− =+ y y  =−  (4) Using (4) to rewrite (1) and (2), we have ()( )() ()12 2 12 2 cosh sinh cosh sinh t t A A t A A tββ ω ω xt   =− + + −  (5) and ()() () () () ()12 1 2 2 22 2 2 2cosh sinh cosh sinh cosh sinhxt t t A A t t A Atββ ω β ω ω βω ω ω =− − +  −+ − t (6) 3-15. We are asked to simply plot the fo llowing equations from Example 3.2: () ( )1 costA e tβxt ω δ−=− (1) ( ) ( )11 1 () c o s s i ntA e t tβvt β ωδ ω ωδ−  =− − + −  (2) with the values A = 1 cm ,1 01 rad s ω−=⋅ , 101 sβ−=. , and δ = π rad. The position goes through x = 0 a total of 15 times before dropping to 0. 01 of its initial amplitude. An exploded (or zoomed) view of figure (b), shown here as figure (B), is the best for determining this number, as is easily shown. 94 CHAPTER 3 –1 –0.5 0 0.5 1–1–0.500.51 x (cm)v (cm/s)0 5 10 15 20 25 30 35 40 45 50–101 0.5 –0.5x(t) (cm) v(t) (cm/s) t (s)(b) (c) 0 5 10 15 20 25 30 35 40 45 50 550 t (s)x (cm)0.01 –0.01(B) 3-16. If the damping resistance b is negative, the equation of motion is 2 0 2xxx βω 0 − +=   (1) where 2bm 0 β≡− > because b < 0. The general solution is just Eq. (3.40) with β changed to – β: () ( ) ( )22 22 10 2 exp expte A t A tββω βω=− + −0−xt (2) From this equation, we see that the motion is not bounded, irrespective of the relative values of 2β and 2 0ω. The three cases distinguished in Section 3.5 now become: a) If 2 02ω β> , the motion consists of an oscillatory solution of frequency 22 10ω ωβ=− , multiplied by an ever-increasing exponential: OSCILLATIONS 95 ()1 12t it itxt e A e Aeβ ω −1ω  =+  (3) b) If 2 02ω β= , the solution is ()( )txt A B teβ=+ (4) which again is ever-increasing. c) If 2 02ω β< , the solution is: ()2 12t txt e A e Aeβ ω −2tω  =+  (5) where 22 20ω βω β= −≤ (6) This solution also increases continuously with time. The tree cases describe motions in which the part icle is either always moving away from its initial position, as in cases b) or c), or it is oscillating around its initial position, but with an amplitude that grows with the time, as in a). Because b < 0, the medium in which the particle mo ves continually gives energy to the particle and the motion grows without bound. 3-17. For a damped, driven oscillator, the equation of motion is 2 0 2cxxx A ost β ω =+=  ω (1) and the average kinetic energy is expressed as ( )22 222 2 04 4mATω 2ω ωω= −+ β (2) Let the frequency n octaves above 0ω be labeled 1ω and let the frequency n octaves below 0ω be labeled 2ω; that is 10 202 2n nω ω ω ω−= = (3) The average kinetic energy for each case is ( )122 2 0 222 2 2 2 00 02 4 2( 4)2n nnmA ωω 2T ω ωω= −+ β (4) ( )222 2 0 222 2 2 2 00 02 4 2( 4)2n nnmA ωω 2T ω ωω− −−= −+ β (5) Multiplying the numerator and denominator of (5) by , we have 42n 96 CHAPTER 3 ( )222 2 0 222 2 2 2 00 02 4 2( 4)2n nnmA ωω 2T ω ωω= −+ β Hence, we find 1TTω= 2ω (6) and the proposition is proven. 3-18. Since we are near resonance and there is only light damping, we have 0 R ω ωω , where ω is the driving frequency. This gives 02 Qω β  . To obtain the total energy, we use the solution to the driven oscillato r, neglecting the transients: () ( ) cos xt D t ω δ = − (1) We then have ()2 22 2 2 2 2 2 011 1sin ( ) cos22 2 2mDEm x k x t t m D ωω δ ω ω δ ω  =+ = − + − 2 0 (2) The energy lost over one period is () ()2 022Tmx x d t m Dβ πω β ⋅= ∫ (3) where 2T πω = . Since 0 ωω , we have 0 energy lost over one period 4 2EQ ω πβπ (4) which proves the assertion. 3-19. The amplitude of a damped oscillator is [Eq. (3.59)] ( )222 2 0 4AD 2ω ωω= −+ β (1) At the resonance frequency, 2 0 R2ωωω β== − , D becomes 22 0 2RAD βωβ= − (2) Let us find the frequency, ω = ω′, at which the amplitude is 1 2RD: ( )22 222 20011 22 2 4RAAD βω β ω ωω β2== − −+′′ (3) Solving this equation for ω′, we find OSCILLATIONS 97 122 22 2 00 2 022 1βωω β β ωω =− ± −′    (4) For a lightly damped oscillator, β is small and the terms in 2β can be neglected. Therefore, 22 020 ω ωβ≅±′ ω (5) or, 0 01βωωω ≅±′    (6) which gives ( )( ) 00 2 ω ωβ ωβ ∆= + − − = β (7) We also can approximate Rω for a lightly damped oscillator: 22 02R 0 ω ωβ ω=− ≅ (8) Therefore, Q for a lightly damped oscillator becomes 00 2Qω ω β ω≅≅∆ (9) 3-20. From Eq. (3.66), ( )(222 2 2 0sin 4Ax )tωω δ ωω ω β−= −+ − (1) Therfore, the absolute value of the velocity amplitude v is given by ( )0222 2 0 4Av 2ω ω ωω= −+ β (2) The value of ω for v a maximum, which is labeled 0 vω, is obtained from 00 vv ωωω=∂=∂ (3) and the value is 0 vω ω= . Since the Q of the oscillator is equal to 6, we ca n use Eqs. (3.63) and (3.64) to express β in terms of 0ω: 2 2 0 146ωβ= (4) We need to find two frequencies, 1ω and 2ω, for which 0m ax2 vv= , where ( ) max 0 0vv ωω == . We find 98 CHAPTER 3 ( )max 222 2 022 24v AA 2ω βω ωω== −+ β (5) Substituting for β in terms of 0ω from (4), and by squaring an d rearranging terms in (5), we obtain ( ) ( )222 2 2 01 ,2 1 ,2 02073ωω ω ω− −= (6) from which 22 0 1,2 1,2 0 1,2 021 73 6ω ωω ω ω −= ± ≅ ± ω (7) Solving for 1ω, 2ω we obtain 0 1,2 012ωω ω≅±± (8) It is sufficient for our purposes to consider 1ω, 2ω positive : then 0 1012ωω ω ≅ + ; 0 2120ωω ω ≅−+ (9) so that 0 126ωωω ω∆= − = (10) A graph of vs. ω for Q = 6 is shown. 0v vA max=2β A 22β ω0 121 60ω ω0 12 ω0v0 ∆ω 3-21. We want to plot Equation (3.43), and its derivative: ()( )txt A B teβ−=+ (1) () ( ) [ ]tvt B A B t eββ−=− + (2) where A and B can be found in te rms of the initial conditions 0 Ax= (3) 0 Bv x0β = + (4) OSCILLATIONS 99 The initial conditions used to produce figure (a) were ( )( )00 24 xv,=−, , (1 4),, , ( , , and , where we take all x to be in cm, all v in cm(4 1),− 1 4),− (1 4 )−, − ( 4 0)−,1s−⋅, and . Figure (b) is a magnified view of figure (a). The dashed line is the path that all paths go to asymptotically as t → ∞. This can be found by taking the limits. 11 s−β= lim ( )t tvt B t eββ− →∞=− (5) lim ( )t txt B t eβ− →∞= (6) so that in this limit, v = –βx, as required. –4 –2 0 2 4–4–3–2–101234 x (cm)v (cm/s) –0.5 –0.25 0 0.25 0.5–0.4–0.200.20.4 x (cm)v (cm/s)(a) (b) 3-22. For overdamped motion, the position is given by Equation (3.44) ()1 12txt A e Ae2t β β −=+− (1) 100 CHAPTER 3 The time derivative of the above eq uation is, of course, the velocity: ()1 11 22tvt A e A e2t β βββ−=− −− 2 (2) a) At t = 0: 01xA A= + (3) 01 1 2 vAA2 β β =−− (4) The initial conditions and v can now be used to solve for the integration constants 0x0 1A and 2A. b) When 10 A=, we have vx02 0β=− and () ()2 vt xt β=− quite easily. For 10 A≠, however, we have ()1teβ 11A1x vt β β−=− →− as t → ∞ since 12β β< . 3-23. Firstly, we note that all the δ = π solutions are just the negative of the δ = 0 solutions. The 2 δπ= solutions don’t make it all the way up to the initial “amplitude,” A, due to the retarding force. Higher β means more damping, as one might expect. When damping is high, less oscillation is observable. In particular, 209 β=. would be much better for a kitchen door than a smaller β, e.g. the door closing ( δ = 0), or the closed door being bumped by someone who then changes his/her mind and does not go through the door ( 2 δπ= ). OSCILLATIONS 101 –1–0.500.51β2 = 0.1, δ = π/2β2 = 0.9, δ = 0 β2 = 0.9, δ = π/2 0 5 10 15β2 = 0.9, δ = πβ2 = 0.5, δ = 0 β2 = 0.5, δ = π/2 0 5 10 15–10.500.51β2 = 0.1, δ = π 0 5 10 15β2 = 0.5, δ = π–0.500.51β2 = 0.1, δ = 0 –1 3-24. As requested, we use Equations (3.40), (3. 57), and (3.60) with the given values to evaluate the complementary and particular solutions to the driven oscillator. The amplitude of the complementary function is constant as we vary ω, but the amplitude of the particular solution becomes larger as ω goes through the resonance near 109 6 r a d s−. ⋅, and decreases as ω is increased further. The plot closest to resonance here has 111 ωω=., which shows the least distortion due to transients. These figures are shown in figure (a). In figure (b), the 16 ωω= plot from figure (a) is reproduced along with a new plot with 220 m spA−= ⋅. 102 CHAPTER 3 –101ω/ω1 = 1/3 01 0 2 0 3 0 –1–0.500.5 01 02 03 0ω/ω1 = 6 t (s)–1–0.500.5ω/ω1 = 3 01 0 2 0 3 0 t (s)–202ω/ω1 = 1.1 01 0 2 0 t (s)30–101ω/ω1 = 1/9 01 0 2 0 3 0 xc xp xLegend: (a) 0 5 10 15 20 25 30–1–0.500.5 0 5 10 15 20 25 30–1–0.500.5 (b)Ap = 1 Ap = 20 3-25. This problem is nearly identical to the prev ious problem, with the exception that now Equation (3.43) is used instead of (3.40) as the complementary solution. The distortion due to the transient increases as ω increases, mostly because the complementary solution has a fixed amplitude whereas the amplitude due to the particular solution only decreases as ω increases. The latter fact is because there is no resonance in this case. OSCILLATIONS 103 05 1 0–101ω/ω1 = 1/9 05 1 0–1–0.500.51ω/ω1 = 1/3 05 1 0–1–0.50ω/ω1 = 1.1 05 1 0–1–0.50ω/ω1 = 3 05 1 0–1–0.50ω/ω1 = 6 05 1 0–1–0.50ω/ω1 = 6, Ap = 6 xc xp x Legend: 3-26. The equations of motion of this system are ( ) ()11 1 1 1 2 22 22 1 2 1cos mx k x b x x F t mx bx b x xω =− − − +  =− − −       (1) The electrical analog of this system can be constructed if we substitute in (1) the following equivalent quantities: mL ; 11→1kC→ ; ; x → q 1bR→1 mL ; 22→0 F ε→ ; 22bR→ Then the equations of the equivalent electrical circuit are given by () ()11 1 1 2 1 0 22 22 1 2 11cos 0Lq R q q q tC Lq Rq R q qεω+− + =  ++ − =       (2) Using the mathematical device of writing exp( iωt) instead of cos ωt in (2), with the understanding that in the results only the real pa rt is to be considered, and differentiating with respect to time, we have 104 CHAPTER 3 ( ) () ( )1 11 1 1 2 0 22 2 2 1 2 1 0it ILI R I I i eC LI R I R I Iωωε+− + =  ++ − =        (3) Then, the equivalent electrical ci rcuit is as shown in the figure: ε0 cos ωtL1 L2R1R2I1(t) I2(t) 1 2C The impedance of the system Z is 11Zi L i ZCωω1 = −+ (4) where is given by 1Z 11211 1 ZRR i L2ω=++ (5) Then, ( ) ()22 12 2 1 2 2 1 1 2 22 12 2RRR R L i L R Z RR Lωω ω  ++ + = ++ (6) and substituting (6) into (4), we obtain () ()() ()2 22 22 12 2 1 2 12 1 1 2 2 2 22 12 21RRR R L i RL L R R LC Z RR Lωω ω ωω ω   ++ + + − + +     = ++ (7) 3-27. From Eq. (3.89), () ( 0 11cos sin2nn na a n t b n t ) Ft ω ω∞ ==+ + ∑ (1) We write () ( 0 11cos2n nFt a c n t )n ω φ∞ ==+ − ∑ (2) which can also be written using trigonometric relations as ()0 11cos cos sin sin2nn nt a c nt ntn F ω φω∞ ==+ + φ     ∑ (3) Comparing (3) with (2), we notice that if there exists a set of coefficients c such that n OSCILLATIONS 105 cos sinnn n nn nca cbφ φ=    =  (4) then (2) is equivalent to (1). In fact, from (4), 222 tannnn n n ncab b aφ =+    =  (5) with and b as given by Eqs. (3.91). nan 3-28. Since F(t) is an odd function, F(–t) = – F(t), according to Eq. (3.91) all the coefficients vanish identically, and the b are given by na n () ()0 0 0 0sin sin sin 11cos cos 2cos 0 cos 4 for odd 0 for evennbF t ntdt nt d t nt d t nt ntnn nn nn nπ ω π ω π ω π ω π ω π ωωωπ ωωωπ ωωωπω ω ππ π− − −= ′′ ′ =− + ′′ ′′     =+ − ′′   =−    ∫ ∫∫   = (1) Thus, ()() ()21 24 210, 1, 2, 0n nbnn bπ+=+= = … (2) Then, we have ()44 4sin sin 3 sin 535t t tωωωππ π= ++ … Ft (3) + 106 CHAPTER 3 F(t) t –11 –π⁄ω π⁄ω Terms 1 + 2 t –0.8490.849 –π⁄ω π⁄ω Terms 1 + 2 + 3 t –1.0991.099 –π⁄ω π⁄ω Terms 1 + 2 + 3 + 4 t –0.9180.918 –π⁄ω π⁄ω 3-29. In order to Fourier analyze a fu nction of arbitrary period, say 2P τ ω = instead of 2πω, proportional change of scale is necessary. Analytically, such a change of scale can be represented by the substitution txPπ= or Pxtπ= (1) for when t = 0, then x = 0, and when 2P tτ ω == , then 2x πω = . Thus, when the substitution tP x π = is made in a function F(t) of period 2Pω′, we obtain the function ()PxF fxπ= (2) and this, as a function of x , has a period of 2πω. Now, f(x) can, of course, be expanded according to the standard formula, Eq. (3.91): () ( 0 11cos sin2nn na a nx b nx ) fx ω ω∞ ==+ + ∑ (3) where OSCILLATIONS 107 () ()2 0 2 0cos sinn naf x nxd bf x nxdπ ω π ωωωπ ωωπ= x x′ ′′    = ′ ′′∫ ∫ (4) If, in the above expressions, we make the inverse substitutions txPπ= and dx dtPπ= (5) the expansion becomes ()0 1cos sin2nn nt t nt nt a PFt a bPP P Pπ π ωπ ωπ π∞ =fF      =⋅ = = + +            ∑ (6) and the coefficients in (4) become () ()2 0 2 0cos sinP n P nntaF tPP ntbF t dPPω ωωπ ω ωπ ω ′ = dt t′ ′   ′  = ′ ′  ∫ ∫ (7) For the case corresponding to this problem, the period of F(t) is 4π ω, so that P = 2π. Then, substituting into (7) and replacing the integral limits 0 and τ by the limits 2τ− and 2τ+, we obtain () ()2 2 2 2cos22 sin22n nntaF t ntbF tπ ω π ω π ω π ωω ω π ω ω π− − ′ = dt dt′ ′     ′ = ′ ′  ∫ ∫ (8) and substituting into (6), the expansion for F(t) is ()0 1cos sin22nn nnt nt aa bωω∞ = 2Ft   =+ +         ∑ (9) Substituting F(t) into (8) yields 2 0 2 0sin cos22 sin sin22n nntat ntbtπ ω π ωω ωωπ ω ωωπ ′ = dt dt′ ′   ′  = ′ ′  ∫ ∫ (10) Evaluation of the integrals gives 108 CHAPTER 3 () ( )2 01 21;0 f or 22 0 e ve 02 4odd 4n nbb n n aa a n n nπ== ≠ n  == ≥=  −  −   (11) and the resulting Fourier expansion is ()357 14 4 4 4sin cos cos cos cos23 2 5 2 21 2 45 2tt tFt tωω ω ωωππ π π=+ − − − …t+ (12) 3-30. The output of a full-wave rectifier is a periodic function F(t) of the form ()sin ; 0 sin ; 0tt Ft ttπωω πωω− −< ≤ =  << (1) The coefficients in the Fourier representation are given by () ()0 0 0 0sin cos sin cos sin sin sin sinn nat ntdt t n bt ntdt t nπ ω π ω π ω π ωωωω ω ωπ ωωω ω ωπ− − =− + tdt tdt′ ′′ ′ ′′        =− + ′ ′′ ′ ′′   ∫∫ ∫∫ (2) Performing the integrations, we obtain ( )()24;i f e v e n o r 0 1 0; i f o d d 0f or all n nn na n bnπ   −=       =  (3) The expansion for F(t) is ()24 4cos 2 cos 431 5t tωππ π=− − … Ft (4) ω The exact function and the sum of the first three terms of (4) are shown below. OSCILLATIONS 109 Sum of first three terms F(t)sin ωt ωt–π⁄2 π⁄21 –π π.5 3-31. We can rewrite the forcing function so that it consists of two forcing functions for t > τ: ()() ()()00 0t Ftat tm atat tτ τ ττττ  < =   −−> << (1) During the interval 0 < t < τ, the differential equation which describes the motion is 2 0 2atxxx βωτ+ +=   (2) The particular solution is , and substituting this into (2), we find pxC t=+ D 22 00 2atCC tD βω ωτ+ += (3) from which 2 0 2 020 0CD aCβω ωτ +=   −=  (4) Therefore, we have 4 002,aDC2a β ωτω=− =τ (5) which gives 2 002 paaxt4β ωτω=−τ (6) Thus, the general solution for 0 < t < τ is ()11 2 002cos sint aae A t B t tβ 4xtβωωωτω−=+ + τ− (7) and then, 110 CHAPTER 3 ()11 1 1 1 2 0cos sin sin costt axt e A t B t e A t B tβββω ω ω ω ωωτ−−=− + + − + +      (8) The initial conditions, x(0) = 0, ()0x 0= , implies 4 0 2 22 10 02 21aA aBβ ωτ β ωωτ ω=    =−    (9) Therefore, the response function is ()2 11 22 2 2 00 10 022cos 1 sint t aee t t tβ β 2xtβ ββωωωτω ω ω ω− − =+ −   +− (10) For the forcing function ( ) at τ τ−− in (1), we have a response similar to (10). Thus, we add these two equations to obtain the total response function: () () ( ) ()2 11 22 2 00 10 1122cos cos 1 sin sint tt taext e t e t te tβ ββ βββωω τω ω ω ωω τ τ− − ωτ = −− +  −      ×+ − − +  (11) When τ → 0, we can approximate eβτ as 1 + βτ, and also 11 sinωτω τ≅ , co1s 1ωτ≅. Then, () () () () ()2 11 1 1 22 2 0 00 10 11 1 1 3 1 11 22 2 00 1 1022cos 1 cos sin 1 sin 1 sin cos 2 21c os s int t ttaext e t t t tt t aet e tβ β τ ββββωβ τ ω ωτ ωωτ ω ω ω ωβ τ ω ωτ ω τ βω ββωωωω ω ωω− − → −−    →− + + +  −       ×− + − +   =− − − +    2 (12) If we use 22 10ω ωβ=− , the coefficient of 1 sinteβtω− becomes 1βω. Therefore, ()1 2 0 011c os s intt ae t e t−− → 1 xt  →− −   ββ τβωωωω (13) This is just the response for a step function. 3-32. a) Response to a Step Function : From Eq. (3.100) ()0Ht is defined as OSCILLATIONS 111 ()0 0 100, ,tt Ht at t<  =  >  (1) With initial conditions ( ) 00 xt = and ( ) 00 xt = , the general solution to Eq. (3.102) (equation of motion of a damped linear oscillator) is given by Eq. (3.105): ()()()() () ()0 0 10 10 0 2 01 01 cos sin for 0ftt tt aext e t t t t t t xt t tβ β βωωωω−− −−  =− − − − >        =<  or (2) where 22 10ω ωβ=− . For the case of overdamping, 2 02ω β< , and consequently 2 1i2 0 ω βω=− is a pure imaginary number. Hence, ( ) 1 cos ttω −0 and ( ) 10ttω − sin are no longer oscillatory functions; instead, they are transformed into hyperbolic functions. Thus, if we write 2 202ω βω=− (where 2ω is real), ( ) ( ) ( ) () () ()10 20 20 10 20 20cos cos cosh sin sin sinhtt i tt tt tt i tt i ttωω ω ωω ω −= −= −  −= −= − (3) The response is given by [see Eq. (3.105)] ()()()() () ()0 0 20 10 0 2 02 01 cosh sinh for 0ftt tt aext e t t t t t t xt t tβ β βωωωω−− −−  =− − − − >        =<  or (4) For simplicity, we choose t , and the solution becomes 00= ()() 2 2 0201c osh s inht t H ee t tβ β βωωω− − 2ω xt =− −    (5) This response is shown in (a) below for the case 05 β ω = . b) Response to an Impulse Function (in the limit τ → 0): From Eq. (3.101) the impulse function ( ) 01,It t is defined as ()0 01 0 1 10 , 0tt It t a t t t tt<   = << >  (6) For tt12 0 τ −= → in such a way that aτ is constant = b, the response function is given by Eq. (3.110): 112 CHAPTER 3 ()()()0 10 1sin for tt be t t t tβωω−−=−0> xt (7) Again taking the “spike” to be at t = 0 for simplicity, we have () ()1 1sin for 0t bxt e t tβωω−= > (8) For 2 12ii2 0 ω ωβ== − ω (overdamped oscillator), the solution is ()2 2sinh ; 0t bxt e t tβωω−= > (9) This response is shown in (b) below for the case 05 β ω = . 012345678900.51 10 ω0txH 002() ω 012345678900.51 10 ω0t()xbω2(a) (b) 3-33. a) In order to find the maximum amplitude of th e response function shown in Fig. 3-22, we look for such that 1t ()xt given by Eq. (3.105) is maximum; that is, ()( ) 10 ttxt t =∂ =∂ (1) From Eq. (3.106) we have ()( ) ()2 1 2 010sintxt Hetβ β 1t ω ωωω−∂ =+ ∂  (2) OSCILLATIONS 113 For 0 0.2β ω = , 22 10 0.980 ω ωβ ω=− = . Evidently, 11tπω= makes (2) vanish. (This is the absolute maximum, as can be seen from Fig. 3-22.) Then, substituting into Eq. (3.105), the maximum amplitude is given by () ()1 1 2 max 01axt xt eβπ ω ω−   == +       (3) or, ()1 2 01.53axtω≅ (4) b) In the same way we find the maximum amplitud e of the response function shown in Fig. 3-24 by using x(t) given in Eq. (3.110); then, ()( )()() ()0 1110 10 1cos sintt ttttxt be t t t ttβ βωωω−− ==∂  =− − −  ∂   (5) If (5) is to vanish, is given by 1t ()11 1 10 1111tan tan 4.9ω 11.37ttω βω ω−−−= = =  (6) Substituting (6) into Eq. (3.110), we obtain (for 0 0.2β ω = ) () () (11.37 1 max 0sin 1.370.98bxt eβ ω ω−× == ) xt (7) or, ()1 00.76axtτ ω≅ (8) 3-34. The response function of an undamped ( β = 0) linear oscillator for an impulse function I(0,τ), with 02πτω= , can be obtained from Eqs. (3.105) and (3.108) if we make the following substitutions: 10 01 00; 20;ttβω ω πτω==   == =   (1) (For convenience we have assumed that th e impulse forcing function is applied at t = 0.) Hence, after substituting we have 114 CHAPTER 3 () () () ()0 2 00 00 2 0000 21c o s 0 2cos 2 cos 0xt t at t axt wt wt txtπωωω ππτωω=<    =− < <      =− − = >=  (2) This response function is shown below. Since the oscillator is undamped, and since the impulse lasts exactly one period of the oscillator, the oscill ator is returned to its equilibrium condition at the termination of the impulse. t02a ω 2 0π ω0π ω2 02a ω 3-35. The equation for a driven linear oscillator is ()0 2 2xx w x f β++=  t where f(t) is the sinusoid shown in the diagram. f(t) I II III ta Region I: x = 0 (1) Region II: xx2 0 2 x a sint β ω ++= ω 0 (2) Region III: xx2 0 2 x βω+ +=  (3) The solution of (2) is ( ) 11 sin cost P xe A tB t xβωω−=+ + (4) in which ( ) sinPxD a t ω δ = − (5) where ( )22 2 01 4D 2ω ωβ ω= −+ (6) OSCILLATIONS 115 1 2 02tan2βωδω ω−=− (7) Thus, ( ) ( )11 sin cos sintA tB t D a tβxe ω ωω−=+ + δ− (8) The initial condition x(0) = 0 gives sin Ba D δ = (9) and ()00x = 1 cos 0 BA D a β ωω δ −+ + = or, () 1sin cosaDA βδ ωδω=− (10) The solution of (3) is ()1 sin costxt e A t B tβ 1 ω ω−=+ ′′    (11) We require that ()xt and ()xt for regions II and III match at  tπω= . The condition that II IIIxxπ π ω ω =    gives ( ) ( ) ( ) sin cos sin sin cos B D a eA Bβπ ω βπ ωeA φ φπ δ φ−−++ − = + ′′ φ where 1ωφ πω= or, sincot cotsinaDA B e ABβπωδφφφ+= + + ′′ (12) The condition that II IIIxxπ π ω ω =    gives ( ) ( ) ( ) () ()11 11sin cos cos cos sin sin cos cos sineA B a D eA B eA B eA Bβπ ω βπ ω βπ ω βπ ωβ φφ ω π δ ω φ ω βφ φ ω φ ω φ−− −−−+ + −+ − =− + + − ′′ ′ ′φ or, ( ) ( ) () ( )11 11cos sin sin cos sin cos sin cos cosAB AB eβπ ωωφ β φ ωφ β φ aD β φω φ ω φβ φ ω δ−− + ′′ =− + − + − or, 11 11 1sin cos sin cos cos cos sin cos sin cos sinDaAB A B eβπ ω ω φ βφ ω φ βφ ω δ ω φβ φ ω φβ φ ω φβ φ   ++−= − − ′′   −−    −  (13) 116 CHAPTER 3 Substituting into (13) from (12), we have ( ) ( ) () () () ()11 1 11 1 1cos sin cos sin cos sin cos sin sin cos sin cos sin cos sin sin cos cos sin sin sin cos sinB aDeβπ ωωφ β φφ ωφ β φ φ ωφ β φ φ ωφ β φφ ωφ β φ φ δω δ ω φβ φ φ φ ω φβ φ−+ +′− −+ +  =+ +  −−  (14) from which ()1 1sin sin cos sin cos sinaD D eβπ ωBa δ δω φ β φ ω δ φω  =+ − +′  (15) Using (12), we can find A′: sincot cotsinaD eB Bβπ ωδAA φ φφ=+ + −′′ (16) Substituting for A, B, and B′ from (10), (9), and (15), we have ( ) 11 1cossin cos sin 1 cosaDAa D e eβπ ω βπ ω ωδ ββδ φφωω ω =+ + −+ ′  φ (17) Thus, we obtained all constants giving us the response functions explicitly. 3-36. With the initial conditions, ()0xt x0= and ()0xt x0=  , the solution for a step function for given by Eq. (3.103) yields 0tt> 00 10 2 2 01 1;xx aaAx A2 01β β ω ωωω ω=− = + − (1) Therefore, the response to ()0Ht for the initial conditions above can be expressed as ()()() () ()()()()0 0000 01 0 1 0 11 10 10 2 01cos sin 1 cos sin for tt tt ttxxx t e x tt tt aet t e t tβ βββωωωω βωωωω−− −− −− =− + + −    +− − − − >  0tt (2) The response to an impulse function ( ) ()01 1,It t Ht = , for the above initial conditions will then be given by (2) for tt and by a superposition of solutions for 0 t <<1 ()0Ht and for ()1Ht 1t> ( taken individually for . We must be careful, howeve r, because the solution for t must be equal that given by (2) for . This can be insured by using as a solution for 1tt> 1tt= )1Ht Eq. (3.103) with initial conditions ()00= x , ()00x = , and using t instead of t in the expression. 1 0 The solution for t is then 1t> OSCILLATIONS 117 ()()() () ()0 00 01 0 1 0 1 11cos sintt xxt e x tt tt x tβ βωωωω−−  =− + + −   x (3) + where ()( ) () ( ) () ()0 111 0 1 0 2 0 10 10 11cos cos sin sin for ttaext e t t t t ett t te ttβ βτ βτβτ ωτ ωω ββωτβ ωωωω−−=− −− − +  +− −− − 1> (4) We now allow a → ∞ as τ → 0 in such a way that aτ = b = constant; expanding (3) for this particular case, we obtain ()()() ()0 00 01 0 1 0 111cos sintt xx bxt e x t t t t t tβ βωωωωω−−  =− + + + −    0> (5) which is analogous to Eq. (3.119) but for initial conditions given above. 3-37. Any function ()Ft m can be expanded in terms of step functions, as shown in the figure below where the curve is the sum of the vari ous (positive and negative) step functions. In general, we have () ()2 0 2n n n nFtxxxm Htβω∞ =−∞ ∞ =−∞++= =∑ ∑  (1) where ()() 0nn n nat tt n Ht tt nτ τ>=  =  <=  (2) Then, since (1) is a linear equation, the solution to a superposition of functions of the form given by (2) is the superposition of the so lutions for each of those functions. According to Eq. (3.105), the solution for ()nHt for is ntt> ()()()() ( 1 2 011c os s inn ntt tt n nna ee tt ttβ β βωωω−− −− =− − − −  ) 1 n ω xt (3) then, for ()()n nFtHtm∞ =−∞=∑ (4) the solution is 118 CHAPTER 3 () ()()()() () () () () ()11 2 0111c os s inn ntt tt nn n nn nn nnext H t e t t t t m H tG t F tG tβ β βωωωω−− ∞ −− =−∞ ∞∞ =−∞ =−∞ =−− −   ==∑ ∑∑n− (5) where ()()()() () 11 2 0111c os s in ; 0n ntt tt nn n neet t t tmGt ttβ β βωωωω−− −−  −− − −    =    < ntt≥ (6) or, comparing with (3) ()() , 0nn n nxtm a tt Gt tt≥  = n <  (7) Therefore, the Green’s function is the response to the unit step. Hn(t)Ft m() tn+1tn t 3-38. The solution for x(t) according to Green’s method is () ( ) ( ) ()()0 101, sin sint tt ttxt Ft Gtt d t Fet e t tmγ βωωω−∞ − −− ′ ′= ′′ ′ dt − = ′ ′∫ ∫′ (1) Using the trigonometric identity, () () ()11 1 11sin sin cos cos2tt t t t t ωω ω ω ω ω ωω1t     −= + − − − + ′′ ′ ′      (2) we have OSCILLATIONS 119 ()()()()()0 11 11 1 00cos cos2tt t tt Fext d t e t t d t e t tmβ βγ βγωω ω ωω ωω− −− ′′  =+ − − − ′′ ′   ∫∫ +′ (3) Making the change of variable, ( )11 z tt ωωω =+ − ′ , for the first integral and ( )11 yt t ωωω =− + ′ for the second integral, we find ()() ()() ()11 11 1 110 11 1cos cos2tt zy tt t ttFe eed z e z d y e ymβγ ω βγ ω βγ βγ ωω β ωω ωω ωω ωω ωωωω ω ω ω−− − −−− +− + − =− +−∫∫1−xt (4) After evaluating the integrals and rearranging terms, we obtain () () () () () () []() () []()0 22 22 11 2 22 1 2 22 1 11sin2c os sin2c ost tFxtm tet tetγ βω βγ ωω βγ ωω ωγβ ω β γ ω ωω ωβγ ω βγ ω ωω− −=  −+ + −+ −   ×− + − + −     +− + − + −    (5) 3-39. ()sin 0 02tt Ft tωπ ω πωπ<<  =  <<  ω From Equations 3.89, 3.90, and 3.91, we have () () 0 11cos sin2nn nFt a a n t b n t ω ω∞ ==+ + ∑ () 02cosnaF t n tτωτ= ′′∫dt′ () 02sinnbF t n tτωτ= ′′∫dt′ 0sin cosnat nπωtdtωωωπ= ′′ ∫′ 002sin atπωdtωωπ π== ′′ ∫ 10sin cos 0 at tπωdtωωωπ== ′′ ′ ∫ 120 CHAPTER 3 ()() ()() ()0 0cos 1 cos 12s in cos21 21nnt ntan t n t d tnnπω πω ωω ωωωωππ ω ω  −+ ′≥= = − − ′′ ′  −+  ∫ Upon evaluating and simplifying, the result is ( )22 even 1 0 onn na nπ −=   dd n = 0,1,2, … 00 by inspectionb= 2 101sin2btπωdtωωπ== ′′ ∫ ()() ()() ()0 0sin 1 sin 12s in sin21 21nnt ntbn t nt d tnnπω πω ωω ωωωωππ ω −+ ′′≥= = − = ′′ ′ −+ ∫0ω So ()( )2 2 2,4,6,11 2sin cos2 1 nFt t n t nω ωπ π∞ ==+ + −∑ … or, letting n → 2n ()( )2 1,2,11 2sin cos 22 14 nt n t nFt ω ωπ π∞ ==+ + −∑ … The following plot shows how well the first four terms in the series approximate the function. Sum of first four terms t π ω2π ω1.0 0.5 3-40. The equation describing the car’s motion is ()2 2sindymk yadttω =− − where y is the vertical displacement of the car from its equilibrium position on a flat road, a is the amplitude of sine-curve road, and k = elastic coefficient = 100 9.898000 N/m0.01dm g dy× ×== OSCILLATIONS 121 02174vπωλ== rad/s with and λ being the car’s speed and wavelength of sine-curve road. The solution of the motion equation can be cast in the form 0v () ()2 0 0 22 0cos sinayt B t tωω βωωω=+ +− with 0 9.9 rad/sk mω== We see that the oscillation with angular frequency ω has amplitude 2 0 22 00.16 mmaAω ωω== −− The minus sign just implies that the spring is compressed. 3-41. a) The general solution of the given differential equation is (see Equation (3.37)) () () ( ) ( )22 22 10 2 exp exp exp xt t A t A t ββ ω β=− − + −− 0ω and () () ( ) ( ) () ( ) ( )22 22 10 2 22 22 22 01 0 2 0() e x p e x p e x p exp exp expvt x t t A t A t tA t A tβ β βω βω ββω βω βω0 == − − − + −− ′   +− − − − − −  at t = 0, , vt 0 ()xt x=0 () v=⇒ 00 1022 01 2vxβAx β ω+ =+−  and 00 1022 01 2vxAxβ βω+ =−−  (1) b) i) Underdamped, 0 2ωβ= In this case, instead of using above parameterizat ion, it is more convenient to work with the following parameterization () ( )22 0 () e x p c o sxt A t t β ωβ δ =− − − (2) () ( ) ( )22 22 22 00 0 () e x p c o s s i nvt A t t t ββ ωβδ ωβ ωβδ=− − − − + − − − (3) Using initial conditions of x(t) and v(t), we find 2 2 00 022 00xv x2A β ωβ ωβ+ + −= − and 0 0 22 0tan ( )v xβ δ ω β+ = − 122 CHAPTER 3 In the case 0 2ωβ= , and using (6) below we have 0 0021 130 33 3v x=+ =− ⇒ =− δδωtan ° 2 2 00 0 0 00 2 00 022 33xvvA xxxωωω=+ + = so finally 00 021 3() e x p c o s 3 022 3x t t =− +  ωω xt (4) ° ii) Critically damped, 0 βω= , using the same parameterization as in i) we have from (2) and (3): () ( ) 0 () e x p e x pxt A t x t0 β ω = −= − (5) and 00 0 0 00 () () e x p (vt x t x t v x ω ω == − −⇒ = − ′ ω (6) iii) Overdamped, 0 βω= , returning to the original parameterization (1) we have (always using relation (6)), () () ( ) ( ) ()()()()()(22 22 10 2 0 00 0exp exp exp 31 31exp 3 2 exp 3 2 23 23xt t A t A t xxtββ ω β ω )0t ω ω=− − + −− +−+ − =− + (7) Below we show sketches for equations (4), (5), (7) tx Underdamped Critically damped Overdamped 3-42. a) ( )2 00 sin 0 mx x F t ωω+ − ′′ = (1) The most general solution is 00 ( ) sin cos sinxt a t b t A t ω ωω =++ where the last term is a particular solution. OSCILLATIONS 123 To find A we put this particular solution (the last term) into (1) and find ()0 22 0FA mω ω= − At t = 0, x = 0, so we find b = 0, and then we have xt00 0 () s i n s i n () c o s c o s a t A t vt a t A t ω ωω ω ω =+ ⇒ = + ω At t = 0, 00avAω ω−=⇒ = ⇒ () ()()0 00 00 01() s i n s i nFt tmxt ω ωω ωωω ω ω ω=−+− b) In the limit 0 ω ω→ one can see that 3 00()6Ftxtmω→ The sketch of this function is shown below. tx 3-43. a) Potential energy is the elastic energy: 2 1() ( )2Ur kr a =− , where m is moving in a central force field. Then the effective potential is (see for example, Chapter 2 and Equation (8.14)): 22 2 221() () ( )22 2effllU r k r amr mr=+ = − + Ur where lm2v rm r ω == is the angular momentum of m and is a conserved quantity in this problem. The solid line below is U ; at low values of r, the dashed line represents ()effr 2 1( )2kr a=−()Ur , and the solid line is dominated by 2 22l mr. At large values of r, 2( )k r a−1() ()2effUrU r ≅= . 124 CHAPTER 3 Potential energy rUr kr a() ( )=−1 22l mr2 22 Ureff() b) In equilibrium circular motion of radius , we have 0r ()()0 2 00 0 0 0kr aa m rmrωω kr−−= ⇒ = c) For given (and fixed) angular momentum l, V(r) is minimal at r0, because 0() 0rr=Vr = ′ , so we make a Taylor expansion of V(r) about ; 0r 22 2 00 0 00 0 0 03( ) ( ) 1( ) () ( ) () ( ) () . . .22mr r K r rVr Vr r r V r r r V rω −−=+ − + − + ≈ = ′′ ′2 2 where 2 0 3Km ω = , so the frequency of oscillation is 0 0 03( )3kr a K mmωωr−== = 3-44. This oscillation must be underdamped oscilla tion (otherwise no period is present). From Equation (3.40) we have () ( ) 1 () e x p c o sxt A t t β ωδ = −− so the initial amplitude (at t = 0) is A. Now at 184tTπ ω== 18(4 ) exp cos (8 )xT Aπβ πδω= −− The amplitude now is 18expAπβω− , so we have OSCILLATIONS 125 18exp 1A A eπβω−= and because 2 02 1 β ωω=− , we finally find 1 2 08 64 1ω π ω π= + 3-45. Energy of a simple pendulum is 2 2mglθ where θ is the amplitude. For a slightly damped oscillation ( ) exp( ) ttθ θβ≈ − . Initial energy of pendulum is 2 2mglθ. Energy of pendulum after one period, 2l gπ=T , is 22() e x p (2 )22mgl mglTTθ θβ =− So energy lost in one period is () ()221e x p2 222mgl mglTT mg2lT θ βθ β θ −− ≈ = β So energy lost after 7 days is 22(7 days)(7 days) mgl T mglTθβ θβ = This energy must be compensated by potential energy of the mass M as it falls h meters: 21 2(7 days) 0.01 s(7 days)Mhh mglmlθβ βθMg−=⇒ = = Knowing β we can easily find the coefficient Q (see Equation (3.64)) 2 22 02217822 2Rg lβωβ ω ββ β−−== = =Q 126 CHAPTER3 CHAPTER 4 Nonlinear Oscillations and Chaos 4-1. dd ℓ0ℓ0 (a) (b) (c)ℓ = ℓ0 + d ℓ = ℓ0 + dm ms x θ The unextended length of each spring is , as shown in (a). In order to attach the mass m, each spring must be stretched a distance d, as indicated in (b). When the mass is moved a distance x, as in (c), the force acting on the mass (neglecting gravity) is 0A ( )0 2s i Fk s nθ =− − A (1) where 2s2x = +A (2) and 2sinx xθ= +A2 (3) Then, () ()22 22 022 22 122 22222 21 21 1kx kxFx x x d xx dd xkx kx x− =− + − =− + − −++    −−− =− − +   +    AA A A AA AA AA A  =− (4) 127 128 CHAPTER 4 Expanding the radical in powers of 22xA and retaining only the first two terms, we have () ()2 2 2 2 3 3121 12 121 12 2dxFx k x ddkx kd kdxx −≅− − −    x   − =− − − +     −=− −A AA A AA A A AA (5) The potential is given by (6) () ()Ux Fxd x =−∫ so that ()( )2 34kd kdUx x x−=+A AA4 (7) 4-2. Using the general procedure explained in Sect ion 4.3, the phase diagram is constructed as follows: ·xx xE1U(x) E6E5E4E3E2 E6E5E4 E3E2E1 NONLINEAR OSCILLATIONS AND CHAOS 129 4-3. The potential () ( )33xλ=− Ux has the form shown in (a) below. The corresponding phase diagram is given in (b): xx ·xU(x) E5E4 E2 E1E2E3 E1(a) (b) 4-4. Differentiation of Rayleigh ’s equation above yields ( )22 0 3 xab x x x ω 0 −−+     = (1) The substitution, 03byy xa=  (2) implies that 0 0 033 3yaxby yaxby yaxby=    =    =     (3) When these expressions are su bstituted in (1), we find 2 2 0 2 00 03033 3yy y aa b a aaby b b y y byω−− +     0y= (4) Multiplying by 03bya and rearranging, we arrive at van der Pol’s equation: ( )22 2 00 2 00ayy y y yyω − −+ =    (5) 130 CHAPTER 4 4-5. a) A graph of the functions ()2 1 1 fx x x =+ + and ()2 tan fx x = in the region 0x 2π ≤≤ shows that there is an intersection (i.e., a solution) for 1x 38π≅ . tan x Ox π⁄2x2 + x +1 The procedure is to use this approximate soluti on as a starting poi nt and to substitute 138 x π= into ()1fx and then solve for ()1 11 tanxf−x  = . If the result is within some specified amount, say 10 , of 4−38π, then this is our solution. If the result is not within this amount of the starting value, then use the result as a new starting po int and repeat the calculation. This procedure leads to the following values: 1x ()2 11 1 1 1 fx x x =++ ()1 11 tan fx−   Difference 1.1781 3.5660 1.2974 0.11930 1.2974 3.9806 1.3247 0.02728 1.3247 4.0794 1.3304 0.00573 1.3304 4.1004 1.3316 0.00118 1.3316 4.1047 1.3318 0.00024 1.3318 4.1056 1.3319 0.00005 Thus, the solution is x = 1.3319. Parts b) and c) are solved in exac tly the same way with the results: b) x = 1.9151 c) x = 0.9271 4-6. For the plane pendulum, the potential energy is 1c o s um g θ =−     A (1) If the total energy is larger than , all values of θ are allowed, and the pendulum revolves continuously in a circular path. The potential energy as a function of θ is shown in (a) below. 2mgA U 2mgℓ mgℓ –π –π⁄2 π⁄2 Oθ π (a) Since T = E – U(θ), we can write NONLINEAR OSCILLATIONS AND CHAOS 131 (22 2 111c o s22v m E m g ) Tm θ θ == = −− AA (2) and, therefore, the phase path s are constructed by plotting ()12 221c o s Em gmθθ   =− −  AA (3) versus θ. The phase diagram is shown in (b) below. E = 2mgℓ E = 3mgℓ E = mgℓ θ·θ –ππ−π 2π 2Em g=3 2ℓ Em g=5 2ℓ (b) 4-7. Let us start with the equation of motion for the simple pendulum: 2 0sin θ ω=− θ (1) where 2g≡A ω . Put this in terms of the horizontal component by setting sin yx≡=A θ. Solving for θ and taking time derivatives, we obtain 2 3222 (1 ) 1yy y y yθ=+− − (2) Since we are keeping terms to third order, we need to get a better handle on the term. Help comes from the conservation of energy: 2y 22 01cos cos2mm g m g −= − AA Aθ θθ (3) where 0θ is the maximum angle the pendulum makes, and serves as a convenient parameter that describes the total energy. When written in terms of , the above equation becomes (with the obvious definition for ) y 0y ( )2 22 0 221 11yyyω=− − −−2 0y (4) Substituting (4) into (2), and the result into (1) gives 132 CHAPTER 4 ( )222 031 21 0 yy y yω+− − − 0= (5) Using the binomial expansion of the square root s and keeping terms up to third order, we can obtain for the x equation of motion 2 2 0 0 23312g xxx x30 + +− = AAω (6) 4-8. For x > 0, the equation of motion is 0 mx F=− (1) If the initial conditions are ()0xA =, ()0x 0= , the solution is ()2 0 2Fxt A tm=− (2) For the phase path we need () xx x= , so we calculate ()02(Fxx A xm) =±  − (3) Thus, the phase path is a parabola with a vertex on the x-axis at x = A and symmetrical about both axes as shown below. t = 0·x Ax t=1 4τ Because of the symmetry, the period τ is equal to 4 times the time required to move from x = A to x = 0 (see diagram). Therefore, from (2) we have 024mA Fτ= (4) 4-9. The proposed force derives from a potential of the form () ()2 21 2 1 2kx x a Ux xx a x x a δδ <  =    + −>    which is plotted in (a) below. NONLINEAR OSCILLATIONS AND CHAOS 133 U(x) xa –a OE3 E2 E1E7E6E5E4 (a) For small deviations from the equilibrium position ( x = 0), the motion is just that of a harmonic oscillator. For energies , the particle cannot reach regions with x < –a, but it can reach regions of x > a if . For the possibility exists that the particle can be trapped near x = a. 6 EE< 4EE>4 EE E<<2 A phase diagram for the system is shown in (b) below. (b)·x xE1E3E4E2 E7… 4-10. The system of equations that we need to solve are 0.05 sin 0.7 cosxy yy x t ω  =  −− +     (1) The values of ω that give chaotic orbits are 0.6 and 0. 7. Although we may appear to have chaos for other values, construction of a Poincaré plot that samples at the forcing frequency show that they all settle on a one period per drive cycle orbit. This occurs faster for some values of ω than others. In particular, when ω = 0.8 the plot looks chaotic until it locks on to the point . The phase plot for ω = 0.3 shown in the figure was produced by numerical integration of the system of equations (1) with 1 00 points per drive cycle. The box encloses the point on the trajectory of the system at the star t of a drive cycle. In addition, we also show Poincaré plot for the case ω = 0.6 in figure, integrated over 80 00 drive cycles with 100 points per cycle. ( 2.50150,0.236439)− 134 CHAPTER 4 –1.5 –1 -0.5 0 0.5 1 1.5–1-0.500.51 –4–3–3 –2 –1 0 1 2 3 4–2–10123 4-11. The three-cycle does indeed occur where in dicated in the problem, and does turn chaotic near the 80th iteration. This value is approximate, however, and depends on the precision at which the calculations are performed. The behavior returns to a three-cycle near the 200th iteration, and stays that way until approxim ately the 270th iteration, although some may see it continue past the 300th. 100 200 300 400 50000.20.40.60.81x iteration NONLINEAR OSCILLATIONS AND CHAOS 135 4-12. 0 0.2 0.4 0.6 0.8 100.1 0.050.20.25 0.15x1 = 0.4 x1 = 0.75 0 0.2 0.4 0.6 0.8 100.10.2 0.050.25 0.15 These plots are created in the manner described in the text. They are created with the logistic equation ( ) 10.9 1nnxx+=⋅ −nx (1) The first plot has the seed value as asked for in the text. Only one additional seed has been done here ( ) as it is assumed that the reader could easily produce more of these plots after this small amount of practice. 10.4x= 10.75x= 136 CHAPTER 4 4-13. 30 32 34 36 38 40 42 44 46 48 5000.20.40.60.81 x1 = 0.7 x1 = 0.700000001iteration 30 32 34 36 38 40 42 44 46 48 5001 x1 = 0.7 x1 = 0.7000000001iteration0.20.40.60.8 The plots are created by iteration on the initia l values of (i) 0.7, (ii) 0.700000001, and (iii) 0.7000000001, using the equation ( )2 12.5 1nnxx+=⋅ −nx (1) A subset of the iterates from (i) and (ii) are plotted together, and clearly diverge by n = 39. The plot of (i) and (iii) clearly diverge by n = 43. 4-14. 20 22 24 26 28 30 32 3400.20.40.60.81 x1 = 0.9 x1 = 0.9000001 fractional difference iteration NONLINEAR OSCILLATIONS AND CHAOS 137 The given function with the given initial values are plotted in the figure. Here we use the notation and , with 10.9x=10.9000001y= () 1nnxf x+= and () 1nyf y+=n where the function is () ( )22.5 1 fx x x =⋅− (1) The fractional difference is defined as xy x− , and clearly exceeds 30% when n = 30. 4-15. A good way to start finding the bifurcations of the function f(α,x) = α sin πx is to plot its bifurcation diagram. 0 0.2 0.4 0.6 0.8 100.20.40.60.81 αx One can expand regions of the diagram to give a rough estimate of the location of a bifurcation. Its accuracy is limited by the fact that the map does not converge very rapidly near the bifurcation point, or more precisely, the Lyapunov exponent approaches zero. One may continue undaunted, however, with the help of a graphical fractal generating software application, to estimate quite a few of the period doublings nα.Using Fractint for Windows, and Equation (4.47) to compute the Feigenbaum constant, we can obtain the following: n α δ 1 0.71978 2 0.83324 4.475 3 0.85859 4.611 4 0.86409 4.699 5 0.86526 4.680 6 0.86551 4.630 7 0.865564 4.463 8 0.8655761 One can see that although we should obtain a better value of δ as n increases, numerical precision and human error quickly degrade the qu ality of the calculation. This is a perfectly acceptable answer to this question. One may compute the nα to higher accuracy by other means, all of which are a great deal more complicated. See, for example, Exploring Mathematics with Mathematica , which exploits the vanishing Lyapunov exponent. Using thei r algorithm, one obtains the following: 138 CHAPTER 4 n α δ 1 0.719962 2 0.833266 4.47089 3 0.858609 4.62871 4 0.864084 4.66198 5 0.865259 4.65633 6 0.865511 5.13450 7 0.865560 Note that these are shown here only as reference, and the student may not necessarily be expected to perform to this degree of sophistica tion. The above values are only good to about , but this time only limited by machine pr ecision. Another alternative in computing the Feigenbaum constant, which is not requested in the problem, is to use the so-called ”supercycles,” or super-stable points , which are defined by 610− nR ()12 11,22n n fR−= The values obey the same scaling as the bifurcation points, and are much easier to compute since these points converge faster than for other α (the Lyapunov exponent goes to – ∞). See, for example, Deterministic Chaos: An Introduction by Heinz Georg Schuster or Chaos and Fractals: New Frontiers of Science by Peitgen, Jürgens and Saupe. As a result, the estimates for δ obtained in this way are more accurate than those obta ined by calculating the bifurcation points. nR 4-16. The function y = f(x) intersects the line y = x at 0 xx=, i.e. is defined as the point where 0x () 0xf x=0. Now expand f(x) in a Taylor series, so that near we have 0x ()()( ) ( ) 00 0 0 fxf x x x x x x ββ+− = +−  (1) where 0xdf dxβ≡ (2) Now define 0 nnxx ε≡− . If we have very close to , then 1x0x1ε should be very small, and we may use the Taylor expansion. The equation of iteration () 1 x+nxf=n n becomes 1nε βε+ (3) If the approximation (1) remains valid from the initial value, we have 11n nε βε+ . a) The values xx0 nn ε −= form the geometric sequence . 2 11 1, , , εβ εβ ε … b) Clearly, when 1 β< we have stability since lim 0nnε →∞= Similarly we have a divergent sequence when 1β>, although it will not really be exponentially divergent since the approximation (1) becomes in valid after some number of iterations, and normally the range of allowable is restricted to some subset of the real numbers. nx NONLINEAR OSCILLATIONS AND CHAOS 139 4-17. 02468 1 0 1 2 1 4 1 6 1 800.20.40.6 20 α = 0.4 α = 0.7iteration The first plot (with α = 0.4) converges rather rapidly to zero, but the second (with α = 0.7) does appear to be chaotic. 4-18. 0 0.2 0.4 0.6 0.8 100.20.40.60.81 αx The tent map always converges to zero for α < 0.5. Near α = 0.5 it takes longer to converge, and that is the artifact seen in the figure. There exists a “hole” in the region 0.5 < α < 0.7 (0.7 is approximate), where the iterations are chaotic bu t oscillate between an upper and lower range of values. For α > 0.7, there is only a single range of ch aos, which becomes larger until it fills the range (0,1) at α = 1. 4-19. From the definition in Equation (4.52) the Lyapunov exponent is given by 1 01lim ln in ni xdf nd xλ− →∞== ∑ (1) The tent map is defined as () ()2f or 0 21 f o r 1 2 1xx fx xxα α<<  = 12 − <<  (2) This gives 2 df dx α= , so we have 140 CHAPTER 4 () (1lim ln 2 ln 2 nn n) λ α →∞α−= = (3) As indicated in the discussion below Equation (4.52), chaos occurs when λ is positive: 12α> for the tent map. 4-20. –1.5 –1 –0.5 0 0.5 1 1.5–0.4–0.200.20.4 xy 4-21. –1.5 –1 –0.5 0 0.5 1 1.5–0.4–0.200.20.4 xy The shape of this plot (the attrac tor) is nearly identical to that obtained in the previous problem. In Problem 4-20, however, we can clearly see the first few iterations (0,0), (1,0), (–0.4,0.3), NONLINEAR OSCILLATIONS AND CHAOS 141 whereas the next iteration (1.076,–0.12) is almost on the attractor. In this problem the initial value is taken to be on the attractor already, so we do not see any transient points. 4-22. The following. system of differential equations were integrated numerically (1) 30.1 cosxy yy x B  =  −− +     t using different values of B in the range [9.8,13.4], and with a variety of initial conditions. The integration range is over a large number of driv e cycles, throwing away the first several before starting to store the data in order to reduce the effects of the transient response. For the case B = 9.8, we have a one period per three drive cycle orbit. The phase space plot (line) and Poincaré section (boxes) for this case are overlaid and shown in figure (a). All integrations are done here with 100 points per driv e cycle. One can experiment with B and determine that the system becomes chaotic somewhere between 9.8 and 9.9. The section for B = 10.0, created by integrating over 8000 drive cycles, is shown in figure (b). If one further experiments with different values of B, and one is also lucky enough to have the right initial conditions, (0,0) is one that works, then a transition will be found for B in the range (11.6,11.7). As an example of the different results one can get depending on the initial conditions, we show two plots in figure (c). One is a phase plot, overlaid with its section, for B = 12.0 and the initial condition (0,0). Examination of the time evolution reveals th at it has one period per cycle. The second plot is a Poincaré section for the same B but with the initial condition (10,0), clearly showing chaotic motion. Note that the section looks quite similar to the one for B = 10.0. Another transition is in the range (13.3,13.4), where the orbits become re gular again, with one period per drive cycle, regardless of initial conditions. The phase plot for B = 13.4 looks similar to the one with B = 12.0 and initial condition (0,0). To summarize, we may enumerate the above transition points by B, , and . Circumventing the actual task of computing where these transition points are, we do know that , 11 , and 13.31 2B3B 1 9.8 9.9 B<<2 .6 11.7B<<313.4 B< < . We can then describe the behavior of the system by region. • : one period per three drive cycles 1 BB< • : chaotic 1BB B<<2 3 • : mixed chaotic/one period per drive cycle (depending on initial conditions) 2BB B<< • : one period per drive cycle 3BB< We should remind ourselves, though, that the above list only applies for B in the range we have examined here. We do not know the behavior when B < 9.8 and B > 13.4, without going beyond the scope of this problem. 142 CHAPTER 4 –3 –2 –1 0 1 2 3–505 xy(a) 2.4 2.6 2.8 3 3.2 3.4 3.6–50510 xy(b) –3 –2 –1 0 1 2 3–505 xy(c) 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8–10–50510 xy(d) NONLINEAR OSCILLATIONS AND CHAOS 143 4-23. The Chirikov map is defined by 1 sinnnpp K+ nq = − (1) 1nn nqq p1 + + = − (2) The results one should get from doing this proble m should be some subset of the results shown in figures (a), (b), and (c) (for K = 0.8, 3.2, and 6.4, respectively). These were actually generated using some not-so-random initial points so that a reasonably complete picture could be made. What look to be phase paths in the figures are actually just different points that come from iterating on a single initial condition. For example, in figure (a), an ellipse about the origin (just pick one) comes from iterating on any one of the points on it. Above the ellipses is chaotic orbit, then a five ellipse orbit (all five come from a single initial condition), etc. The case for K = 3.2 is similar except that there is an orbit outside of which the system is always undergoing chaotic motion. Finally, for K = 6.4 the entire space is filled with chaotic orbits, with the exception of two small lobes. Inside of these lobes are regular or bits (the ones in the left are separate from the ones in the right). –1 –0.5 0 0.5 1–1–0.500.51 q⁄πp⁄π(a) –1 –0.5 0 0.5 1–1–0.500.51 q⁄πp⁄π(b) 144 CHAPTER 4 –1 –0.5 0 0.5 1–1–0.500.51 q⁄πp⁄π(c) 4-24. a) The Van de Pol equation is ()2 22 2 0 2dx d xxa xdt dtωµ+=− Now look for solution in the form 0 () c o s ()xt b t ut ω = + (1) we have 00sindx dubtdt dtωω=− + and 22 2 00 22cosdx dubtdt dtωω=− + Putting these into the Van de Pol equation, we obtain {} { }2 22 2 2 00 0 0 0 2() ()( ) cos ( ) 2 ( ) cos sindut d utut b t u t b ut t a b tdt dtωµ ω ω ω ω+= − + + − − + From this one can see that u(t) is of order µ (i.e. ~ ( ) uO µ), which is assumed to be small here. Keeping only terms up to order µ, the above equation reads {}2 32 2 00 0 0 0 2 22 2 00()( ) sin cos sin sin sin 344dutut b t t ab tdt bbba t tωµ ω ω ω ω ω µω ω ω+= −− + =− − −  0 0 0t (where we have used the identity 2 00 0 4 sin cos sin sin 3 tt tω ωω=+ ω ) This equation has 2 frequencies (0ω and 30ω), and is complicated. However, if then the term 2b=a 0 sin tω disappears and the above equation becomes NONLINEAR OSCILLATIONS AND CHAOS 145 23 00 2()() s i n34dut but tdt0 ω µω ω += We let , and the solution for this equation is 2b=a 33 00 00() s i n3 s i n332 4bat tµµut ω ωωω=− =− So, finally putting this form of u(t) into (1), we obtain one of the exact solutions of Van de Pol equation: 3 00 0() 2 c o s s i n34aut a t tµω ωω=− b) See phase diagram below. Since 0.05 µ= is very small, then actually the second term in the expression of u(t) is negligible, and the phase diagram is very close to a circle of radius b = 2a = 2. x –2–112.x –2 –1 1 2 4-25. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because 0.07 µ= is a very small value, the limit cycle is very close to a circle of radius b = 2a = 2. a) In this case, see figure a), the phase diagram starts at the point ( x = 1, x′ = 0) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution). 146 CHAPTER 4 x.x –2 –1 1 2 –2–112 –2 –1 1 2 3 –2–112b) In this case, see figure b), the phase diagram starts at the point ( x = 3, x′ = 0) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution). x.x 4-26. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because 0.5 µ= is not a small value, the limit cycle is NOT close to a circle (see problem 4-24 above). a) In this case, see figure a), the phase diagram starts at the point ( x = 1, x′ = 0) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution). NONLINEAR OSCILLATIONS AND CHAOS 147 x.x –2 –1 1 2 –2–112 –2 –1 1 2 3 –2–112b) In this case (see figure below), the phase diagram starts at the point ( x = 3, x′ = 0) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-24 for exact expression of stable solution). x.x 148 CHAPTER4 CHAPTER 5 Gravitation 5-1. a) Two identical masses : The lines of force (dashed lines) and the equipo tential surfaces (solid lines) are as follows: b) Two masses, +M and –M : In this case the lines of force do not continue ou tward to infinity, as in a), but originate on the “negative” mass and terminate on the positive mass. This situation is similar to that for two electrical charges, + q and – q; the difference is that the elec trical lines of force run from + q to – q. 149 150 CHAPTER 5 5-2. Inside the sphere the gravit ational potential satisfies ()24Gr φ πρ ∇= (1) Since ρ(r) is spherically symmetric, φ is also spherically symmetric. Thus, ()2 214 rGrr rφπρ∂∂=∂∂r (2) The field vector is independent of the radial distance. This fact implies rφ∂ ∂ = constant ≡ C (3) Therefore, (2) becomes 24CGrπρ = (4) or, 2C Grρπ= (5) 5-3. In order to remove a particle from the surfac e of the Earth and transport it infinitely far away, the initial kinetic energy must equal the work required to move the particle from e rR= to r = ∞ against the attractive gravitational force: 2 0 21 2 ee RMmGd r mr∞= ∫v (1) where eM and are the mass and the radius of the Earth, respectively, and is the initial velocity of the particle at . eR0v e rR= Solving (1), we have the expression for : 0v 02e eGMvR= (2) Substituting G , , , we have 11 3 26.67 10 m /kg s−=× ⋅245.98 10 kgeM=×66.38 10 meR=× 011.2 km/sec v≅ (3) 5-4. The potential energy corresponding to the force is 2 2 32dx mkUF dx m kxx=− = =−∫∫ 2 (1) The central force is conservative and so the tota l energy is constant and equal to the potential energy at the initial position, x = d: GRAVITATION 151 22 2 211 1constant22 2kx mxd== − = − 2km Em (2) Rewriting this equation in integrable form, 00 22 2 2211dddx d xdxk dxkxd=− =− − −∫∫dt (3) where the choice of the negative sign for the radical insures that x decreases as t increases. Using Eq. (E.9), Appendix E, we find 0 22 ddtd xk=− (4) or 2dtk= 5-5. The equation of motion is 2Mmmx Gx=− (1) Using conservation of energy, we find 211 2xG M E G Mxx1 ∞−= =−  (2) 112dxGMdt x x∞ =− −    (3) where is some fixed large distance. Therefore, the time for the particle to travel from x∞ x∞ to x is ()1 2 112xx xxxx dx xx GM GMxx∞∞∞ ∞ ∞=− =−− −∫∫td x Making the change of variable, , and using Eq. (E.7), Appendix E, we obtain 2xy→ ()1sin2x xx xx x xGM xtx ∞− ∞ ∞∞ ∞ =− −     (4) If we set x = 0 and 2 xx∞= in (4), we can obtain the time for the particle to travel the total distance and the first half of the distance. 152 CHAPTER 5 32 0 01 2xxTd t GM ∞∞==∫ (5) 2 32 121122x xxTd t GMπ∞ ∞∞  == +    ∫ (6) Hence, 12 012T Tπ π+ = Evaluating the expression, 12 00.818T T= (7) or 12 09 11T T≅ (8) 5-6. α θ φz xyrsP r2drd(cos θ)dφ Since the problem has symmetry around the z-axis, the force at the point P has only a z-component. The contribution to the fo rce from a small volume element is ()2 2cos coszdg G r dr d dsρθφ =− α (1) where ρ is the density. Using coscoszr sθα−= and integrating over the entire sphere, we have () ( )12 2 3222 01 0coscos 2c o sa zzrrdr d d rz r zπθρθ φ θ+ −−=− +−∫∫ ∫gG (2) Now, we can obtain the integral of cos θ as follows: GRAVITATION 153 ( )() ( )()1 3222 1 11222 1coscos 2c o s 2c o s c o szrId rz r z rz r z dzθθ θ θ θ+ − + −−= +− ∂=− + −∂∫ ∫ Using Eq. (E.5), Appendix E, we find ( )1 1222 1 212c o s 22Ir z rzzr z zz zθ+ −  ∂=− − + −  ∂  ∂=− =∂ (3) Therefore, substituting (3) into (2) and performing the integral with respect to r and φ, we have 3 2 3 2223 41 3zagGz Gazρ π πρ=−⋅ =− (4) But 3 4 3aπρ is equal to the mass of the sphere. Thus, 21 zgG Mz=− (5) Thus, as we expect, the force is the same as that due to a point mass M located at the center of the sphere. 5-7. dx PRsx The contribution to the potential at P from a small line element is dGsdxρΦ=−A (1) where Mρ=AA is the linear mass density. Integrating ov er the whole rod, we find the potential 2 22 21 MG xR−Φ=− +∫A AAdx (2) 154 CHAPTER 5 Using Eq. (E.6), Appendix E, we have 2 2 2 22 22224ln ln 24RMG MGx x R R−  ++     − + + =−Φ=  −+ +  A AAA AA AA 22 224ln 4GM R R  ++Φ=−   +−    AA A AA (3) 5-8. zz y xra rdrdθdzz0 θα rz z2 02+−() Since the system is symmetric about the z-axis, the x and y components of the force vanish and we need to consider only the z-component of the force. The contribution to the force from a small element of volume at the point ( r,θ,z) for a unit mass at (0,0,0z) is () () ()2 2 0 0 322 2 0coszrdrd dzdg G rz z zzr d r dd zG rz zθρ α θρ=− +− −=− +− (1) where ρ is the density of the cylinder and where we have used ( ) ()0 2 2 0coszz rz zα−= +−. We can find the net gravitational force by integrating (1) over the entire volume of the cylinder. We find ()2 0 322 200 00a zzzrdr d d z rz zπ ρθ−=−gG   +− ∫∫ ∫A Changing the variable to , we have 0 xz z=− 0 03222 02z a z zxdxgG r dr rxπρ− =  + ∫∫A (2) Using the standard integral, GRAVITATION 155 ( )32221 xdx ax ax2−= ± ±∫ (3) we obtain ()22 2 20 0 02a zrrdr rz rzπρ gG    =− −  + +− ∫A (4) Next, using Eq. (E.9), Appendix E, we obtain ()2 22 0 2z a z a z πρ2 0 gG =− + − − + +   AA (5) Now, let us find the force by first computing the potential. The contribution from a small element of volume is ()2 2 0rdrd dzdG rz zθρ Φ=− +− (6) Integrating over the entire volume, we have ()2 2 20000ardG d z d d r rz zπ ρθ Φ=− +−∫∫∫A (7) Using Eq. (E.9), Appendix E, again, we find () (2 2 00 02 dza zz z πρ )z dG Φ=− + − − −    ∫A (8) Now, we use Eqs. (E.11) and (E .8a), Appendix E, and obtain ( )() () ()2 22 0 2 2 00 0 2 22 22 2 0 00 0 02l n222 1ln 2 222 2z aGa z z z z aaz z za zπρ −2 a Φ=− − + − + − − + − +      ++ − − + + − +  AAA AAA Thus, the force is 156 CHAPTER 5 ()() ()() () ()0 2 2 2 2 2 0 0 2 022 22000 0 0 22 2 2 0 22 0 022 22 00 01 11222 2 1 11 22 2zz za z agG a zzaz z z a z za z aaz az z zaπρ− −−+ − ∂Φ =− =− + − + +∂+− −− + −+   −+ −+ − − ++− + + A A AA AA A A (9) or, ()2 22 0 2z a z a z πρ2 0 gG =− + − − + +   AA (10 and we obtain the same result as in (5). In this case, it is clear that it is considerab ly easier to compute the force directly. (See the remarks in Section 5.4.) 5-9. θPr Ra The contribution to the potential at the point P from a small line element dA is dGrρΦ=− ∫AA (1) where ρA is the linear mass density which is expressed as 2M aρπ=A . Using 222c o s rR a a R θ =+ − and dA = adθ, we can write (1) as 2 22 02 2c o sGM d Ra a Rπθ π θΦ=− +−∫ (2) This is the general expre ssion for the potential. If R is much greater than a, we can expand the integrand in (2) using the binomial expansion: 122 222 222 221112 c o s 2c o s 1312 c os 2 c os28aa RRR Ra a R aa aa RR RR Rθ θ θθ−=− − +−    =+ − + − +     …1 (3) GRAVITATION 157 If we neglect terms of order 3a R and higher in (3), the potential becomes 2 22 2 22 0 22 2231c os c os22 2 3222GM a a adRR R GM a a RRπ θ θθπ ππ ππ Φ=− + − +    − +∫ =− (4) or, ()2 2114GM aRRR Φ≅ − +    (5) We notice that the first term in (5) is the potential when mass M is concentrated in the center of the ring. Of course this is a very rough a pproximation and the first correction term is 2 34GMa R− . 5-10. P R xr a R sin θR cos θ θφ Using the relations ()2 2sin 2 sin cos xR a a R θ θ =+ − φ (1) 22 2 2 2cos 2 sin cosR R a a R rx θ θ =+ = + − φ (2) 2M aρπ=A (the linear mass density), (3) the potential is expressed by 2 2 0 22 12s i n c o sd GM dGrR aa RR−− =Φ=  −−  ∫∫AAπρ φ π θφ (4) If we expand the integrand and neglect terms of order ()3aR and higher, we have 1222 22 22131 2 sin cos 1 sin cos sin cos22aa a a a RR R R R2 2θ φθ φ−−− ≅+ − + θ φ (5) Then, (4) becomes 158 CHAPTER 5 22 2 221322 s in22 2GM a a RRRπ πππθ − − +Φ≅    Thus, ()2 2 21311 sin22GM aRRRθ   − − − Φ≅        (6) 5-11. P a r z dmθ The potential at P due to a small mass element dm inside the body is 222c o sdm dmdG Gr za z a θΦ=− =− +− (1) Integrating (1) over the entire volume and dividing the result by the surface area of the sphere, we can find the average field on the surface of the sphere due to dm: 2 222 02s in 1 4 2c o saveaddma za z aππθ θ πdG θ Φ= −  +−  ∫ (2) Making the variable change cos θ = x, we have ( )1 22 12 2aveGd xdd m za zax+ −Φ= − +−∫ (3) Using Eq. (E.5), Appendix E, we find ( ) ( ) () ()22 22 11222 2aveGdd m z a za z aza za za za Gdmza Gdmza zΦ= − − + − + + + −−++=− =− (4) This is the same potential as at the center of the sphere. Since the average value of the potential is equal to the value at the center of the sphere at any arbitrary element dm, we have the same relation even if we integrate over the entire body. GRAVITATION 159 5-12. O Pdm Rrr' Let P be a point on the spherical surface. The potential dΦ due to a small amount of mass dm inside the surface at P is GdmdrΦ=− (1) The average value over th e entire surface due to dm is the integral of (1) over dΩ divided by 4 π. Writing this out with the help of the figure, we have 2 0 22s i n 4 2c osaved Gdm Rr R rπdπ θθ π θΦ= − +− ′ ′∫ (2) Making the obvious change of variable and performing the integration, we obtain 1 2 1 2 4 2aveGdm du Gdm R Rr Ru r π−Φ= − = − +− ′ ′∫d (3) We can now integrate over all of the mass and get ave Gm R Φ =− . This is a mathematical statement equivalent to the problem’s assertion. 5-13. R1 R2R0 ρ2ρ1 0R= position of particle. For , we calculate the force by assuming that all mass for which is at r = 0, and neglect mass for which . The force is in the radially inward direction ( −). 10RRR<<2 0 rR< e0 rR> r The magnitude of the force is 2 0GMmFR= where M = mass for which 0 rR< 160 CHAPTER 5 ( )33 11 0 1 244 33MR R R3πρπ =+ − ρ So ( )333 11 20 21 2 04 3rGmRRRRπρρρ =− + −Fe ()3 12 1 20 2 04 3rRGm RRρρπρ  −=− +    Fe 5-14. Think of assembling the sphere a shell at a time ( r = 0 to r = R). For a shell of radius r, the incremental energy is dU = dm φ where φ is the potential due to the mass already assembled, and dm is the mass of the shell. So 2 22 3333444MMr dr r drRRρπ ππ== =rdrdm Gm rφ=− where 3 3rmMR= So 22 33 0 2 4 6 03 3R r RUd u Mrd r G M r RR GMrd rR==   =−      =−∫ ∫ ∫ 23 5GMUR=− 5-15. When the mass is at a distance r from the center of the Earth, the force is in the inward radial direction and has magnitude rF: m r 3 24 3rGmFrrπρ=  where ρ is the mass density of the Earth. The equation of motion is GRAVITATION 161 3 24 3rGmFm r rrπρ == −   or where 20 rrω+=24 3Gπρω= This is the equation for simple harmonic motion. The period is 23TGπ π ω ρ== Substituting in values gives a period of about 84 minutes. 5-16. z y xM h rθ rh22+ For points external to the sphere, we may consider the sphere to be a point mass of mass M. Put the sheet in the x-y plane. Consider force on M due to the sheet. By symmetry, 0xyF F= = ( )22 0coszz rGMdmFd F rhθ∞ === +∫∫ With 2s dm rdr ρπ = and 22sh rhθ= +co we have ( ) ( )3222 0 1222 02 12 2zs r zs zsrdrFG Mh rh FG Mh rh FG Mπρ πρ πρ∞ = ∞= +     =−  +  =∫ Th e sphere attracts the sheet in the -direction with a force of magnitude 2sz GMπρ 162 CHAPTER 5 5-17. y x Earthmoon (not to scale)water Start with the hint given to us. The expression for and xgyg are given by 332me xGM x GM x DRg=− ; 3m yGM y GM ygDR=− −3e (1) where the first terms come from Equations (5.54) and the second terms come from the standard assumption of an Earth of uniform density. The or igin of the coordinate system is at the center of the Earth. Evaluating the integrals: max2 max 3302 2xme xGM GM xgd xDR=−∫; max2 max 330 2yme yy GM GMgd yDR=− −∫  (2) To connect this result with Example 5.5, let us write (1) in the following way ( )2 22 max max max max 3322me y GMxxDR+= − 2yGM (3) The right-hand side can be factored as () () ()()max max max max 3222eGM GMxy xy R h gRR+− =3eh= 2R (4) If we make the approximation on the left-hand side of (3) that , we get exactly Equation (5.55). Turning to the ex act solution of (3), we obtain 22 max maxxy 33 3 3 33 3 32 2 2em e em eMM M M RD R DhR MM M M RD R D+− − = ++ −m m (5) Upon substitution of the pr oper values, the answer is  0.54 m, the same as for Example 5.5. Inclusion of the centrifugal term in does not change this answer significantly. xg GRAVITATION 163 5-18. From Equation (5.55), we have wi th the appropriate substitutions 2 3 3 moon 2 sun 33 2 3 2m me s s s esGM r gD hM GM r hM gRR D== (1) Substitution of the known values gives 322 11 moon 30 8 sun73 5 0 1 0 k g 1 495 10 m221 993 10 kg 3 84 10 mh h .× .×= .× . × . (2) 5-19. ωearthωmoon Because the moon’s orbit about the Earth is in the same sense as the Earth’s rotation, the difference of their frequencies will be half the ob served frequency at which we see high tides. Thus tides earth moon11 1 2TTT=− (1) which gives T  12 hours, 27 minutes. tides 5-20. The differential potential created by a thin loop of thickness dr at the point (0,0, z) is ()()2 22 22 222 2222() () ()Gr drM G M dr G Mdz z dz z R zRR R zr zrπ π−− −Φ= = ⇒ Φ=Φ= + − ++∫ Then one can find the gravity acceleration, 22 2222 ˆˆ()dG M zRk kdz R zR zgzΦ +−=− =−  + where is the unit vector in the z-direction. ˆk 164 CHAPTER 5 5-21. (We assume the convention that D > 0 means m is not sitting on the rod.) The differential force dF acting on point mass m from the element of thickness dx of the rod, which is situated at a distance x from m, is ( ) 22()LD DGMLm d x GMm dx GMmdF F dFxL x+ =⇒ = = =+ ∫∫DLD And that is the total gravitational force acting on m by the rod. CHAPTER 6 Some Methods in the Calculus o f Variations 6-1. If we use the varied function ( ) ( ) ,s in1 yx x xα απ =+− (1) Then ( 1c os 1dyxdxαπ π ) =−− (2) Thus, the total length of the path is () ()2 1 0 11222 2 01 2 cos 1 cos 1dySd xdx xx απ π α π π=+   − + −∫ ∫2 =− (3) dx  Setting ( ) 1xu π−≡ , the expression for S becomes 12 22 2 01121 c o s c o s2π απ α ππ=− +∫Su (4) u du  The integral cannot be performed directly since it is, in fact, an elliptic integral . Because α is a small quantity, we can expand the integrand and obtain 2 22 2 22 2 021 1 1 11 cos cos cos cos22 82Su u u uπ απ α π απ α ππ  =− − − − +    ∫…du (5) If we keep the terms up to co and perform the integration, we find 2su 222216S πα =+ (6) which gives 165 166 CHAPTER 6 22 8Sπαα∂=∂ (7) Therefore 00 aS α=∂=∂ (8) and S is a minimum when α = 0. 6-2. The element of length on a plane is 2dS dx dy=+2 (1) from which the total length is ()( ) 22 2 11 1, 2 22 ,1xy x xy xdySd x dydx=+ = + ∫∫dx (2) If S is to be minimum, f is identified as 2 1dyfdx=+  (3) Then, the Euler equation becomes 21ddydx dy 0 + =′′ (4) where dyydx=′ . (4) becomes 20 1y d dx y′ = +′ (5) or, 21y y′ +′ = constant ≡ C (6) from which we have 2 21CyC=′− = constant ≡ a (7) Then, ya xb= + (8) This is the equation of a straight line. SOME METHODS IN THE CALCULUS OF VARIATIONS 167 6-3. The element of distance in three-dimensional space is 22dS dx dy dz=+ +2 (1) Suppose x, y, z depends on the parameter t and that the end points are expressed by ()()() ( ) 11 11 11 ,, xt yt zt , ()()() ( ) 22 22 22 ,, xt yt zt . Then the total distance is 2 12 22 t tdy dx dzSdt dt dt  =+ +    ∫dt (2) The function f is identified as 22 2f xyz= ++ (3) Since 0fff xyz∂∂∂===∂∂∂, the Euler equations become 0 0 0fd dt x fd dt y fd dt z ∂=∂ ∂= ∂  ∂=∂    (4) from which we have 122 2 222 2 322 2constant constant constantxC xyz yC xyz zC xyz  =≡  ++    =≡  ++    =≡ ++      (5) From the combination of these equations, we have 12 23y x CC y z CC=   =    (6) If we integrate (6) from t to the arbitrary t, we have 1 168 CHAPTER 6 1 1 12 1 1 23yy xx CC yy zz CC− − =  − −= (7) On the other hand, the integration of (6) from to gives 1t2t 21 21 12 21 21 23yy xx CC yy zz CC− − =  − −= (8) from which we find the constants , C, and C. Substituting these constants into (7), we find 1C2 3 1 1 21 21 21yy xx zz xx yy zz− −−==−−−1 (9) This is the equation expressing a straight line in three-dimensional space passing through the two points ( ) 11 1,,xyz , ( ) 222,,xyz . 6-4. xyz 1 2dS φρ The element of distance along the surface is 22dS dx dy dz=+ +2 (1) In cylindrical coordinates ( x,y,z) are related to ( ρ,φ,z) by cos sinx y zzρ φ ρ φ=    = =  (2) from which sin cosdx d dy d dz dzρ φφ ρφ φ=−    = =  (3) SOME METHODS IN THE CALCULUS OF VARIATIONS 169 Substituting (3) into (1) and integrat ing along the entire path, we find 2 12 22 2 2 2 1Sd dz zφ φd ρφρ =+ = + ∫∫φ (4) where dzzdφ= . If S is to be minimum, 22f z ρ≡ + must satisfy the Euler equation: 0ff zz φ∂ ∂∂− =∂∂ ∂  (5) Since 0f z∂=∂, the Euler equation becomes 220z z φρ∂=∂ +  (6) from which 22z z ρ+ = constant ≡ C (7) or, 2 21CzCρ =− (8) Since ρ is constant, (8) means constantdz dφ= and for any point along the path, z and φ change at the same rate. The curve described by this condition is a helix . 6-5. ds (x1,y1)(x2,y2) y xz The area of a strip of a surface of revolution is 222 dA ds dx dy ππ=× =× +2 (1) Thus, the total area is 170 CHAPTER 6 2 1221x xA xyπ=+∫dx (2) where dyydx= . In order to make A a minimum, 21 fxy≡+  must satisfy equation (6.39). Now 2 21 1fyx f xy y y∂=+∂ ∂=∂ +    Substituting into eq uation (6.39) gives () ( )2 22 22 1222 211 11 11 1xy ddyx ydx dx yy y x yd yd x y y−x     += +− =    ++   +− + =+      Multiplying by 21y+ and rearranging gives ( )21dy dx x yy−= +  Integration gives 2 21ln ln ln21yxay−+=+  where ln a is a constant of integration. Rearranging gives ( )2 221 1y xa= − Integrating gives 1coshxybaa−=+ or coshybxaa−= which is the equation of a catenary. SOME METHODS IN THE CALCULUS OF VARIATIONS 171 6-6. (x1,y1) (x2,y2)2a0 θ = 0θ = π xy If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum point of the cycloid at (2 a,0) the parametric equations are ( ) ()1c o s sinxa yaθ θ θ =+  =+ (1) Since the particle starts from rest at the point ( ) 11,xy , the velocity at any elevation x is [cf. Eq. 6.19] ( )1 2 vg x x =− (2) Then, the time required to reach the point ( ) 22,xy is [cf. Eq. 6.20] ()2 1122 11 2x xytgx x +′=−∫dx (3) Using (1) and the derivatives obtained therefrom, (3) can be written as 1 212 1 01c o s cos cosatgθ θθdθθθ= += − ∫  (4) Now, using the trigonometric identity, 2cos 2 cos 21 θ θ += , we have 1 122 1 0 22 1 0cos2 cos cos22 cos2 sin sin22datg da gθ θθθ θ θ θθ θ θ= − = −∫ ∫ (5) Making the change of variable, sin 2z θ = , the expression for t becomes 1sin2 22 1 02 sin2ad ztgzθ θ= −∫ (6) The integral is now in standard form: 172 CHAPTER 6 1 22sindx x a ax−= =∫ (7) Evaluating, we find atgπ= (8) Thus, the time of transit from ( ) 11,xy to the minimum point does not depend on the position of the starting point. 6-7. vc n1 1= vc n2 2=n1 n2θ1 θ1 θ2(n2 > n1) db xa The time to travel the path shown is (cf. Example 6.2) 21y dstvv+′==∫∫dx (1) Although we have v = v(y), we only have 0 dv dy ≠ when y = 0. The Euler equation tells us 20 1y d dxvy′ =  +′ (2) Now use vc n= and y′ = –tan θ to obtain n sin θ = const. (3) This proves the assertion. Alternatively, Ferm at’s principle can be proven by the method introduced in the solution of Problem 6-8. 6-8. To find the extremum of the followi ng integral (cf. Equation 6.1) (), J fy x d x=∫ we know that we must have from Euler’s equation 0f y∂=∂ This implies that we also have SOME METHODS IN THE CALCULUS OF VARIATIONS 173 0f Jdxyy∂ ∂= =∂∂∫ giving us a modified form of Euler’s equation. Th is may be extended to several variables and to include the imposition of auxiliary conditions sim ilar to the derivation in Sections 6.5 and 6.6. The result is () 0j j j iig Jxyyλ∂ ∂+ =∂∂∑ when there are constraint equations of the form ( ),0jigyx = a) The volume of a parallelepiped with sides of lengths , b, is given by 1a1 1c 111 Va b c= (1) We wish to maximize such a volume under the condit ion that the parallelepiped is circumscribed by a sphere of radius R; that is, (2) 222 111 4 abc R++=2 We consider , , c as variables and V is the function that we want to maximize; (2) is the constraint condition: 1a1b1 { } 111,, 0 gabc = (3) Then, the equations for the solution are 11 11 110 0 0g V aa g V bb g V ccλ λ λ ∂ ∂+= ∂∂ ∂ ∂ +=∂∂ ∂ ∂+= ∂∂  (4) from which we obtain 11 1 11 1 11 120 20 20bc a ac b ab cλ λ λ+=    += +=  (5) Together with (2), these equations yield 1112 3abc R=== (6) Thus, the inscribed parallelepiped is a cube with side 2 3R. 174 CHAPTER 6 b) In the same way, if the parallelepiped is now circumscribed by an ellipsoid with semiaxes a, b, c, the constraint condition is given by 222 111 2221444abc abz= == (7) where , , c are the lengths of the sides of the pa rallelepiped. Combining (7) with (1) and (4) gives 1a1b1 22 11 22abc abc2 1 2= = (8) Then, 11122,, 33aa bb cc===2 3 (9) 6-9. The average value of the square of the gradient of ( ) 123,,xxxφ within a certain volume V is expressed as ()2 123 2 22 12 1231 1Id xdxdxV Vx x xφ φφφ=∇    ∂∂∂=+ +    ∂∂∂   ∫∫∫ ∫∫∫ 3 dxdxdx v (1) In order to make I a minimum, 2 22 12fxxx3φ φφ    ∂∂∂=++    ∂∂∂    must satisfy the Euler equation: 3 10 i i iff x xφ φ =∂∂ ∂− =∂∂ ∂∂∂∑ (2) If we substitute f into (2), we have 3 10 i iixxφ =∂∂=∂∂∑ (3) which is just Laplace’sequation: 20φ∇ = (4) Therefore, φ must satisfy Laplace’s equation in order that I have a minimum value. SOME METHODS IN THE CALCULUS OF VARIATIONS 175 6-10. This problem lends itself to the method of solution suggested in the solution of Problem 6-8. The volume of a right cylinder is given by (1) 2VRπ= H The total surface area A of the cylinder is given by ( )2 bases side 22 2 AAA RR H R R ππ π=+ = + H =+ (2) We wish A to be a minimum. (1) is the constraint condition, and the other equations are 0 0g A RR g A HHλ λ∂ ∂ +=∂∂ ∂ ∂ +=∂∂  (3) where . 20 gV R H π =− = The solution of these equations is 1 2RH= (4) 6-11. y dsRθ }1 2a The constraint condition can be found from the relation ds = Rdθ (see the diagram), where ds is the differential arc length of the path: ( )1222ds dx dy Rd θ =+ = (1) which, using , yields 2ya x= 2214 ax d x R d θ += (2) If we want the equation of constraint in other than a differential form, (2) can be integrated to yield ( )2 141l n2 424xa x a x a xaθ+= + + + +221 AR (3) where A is a constant obtained from the initial cond itions. The radius of curvature of a parabola, , is given at any point ( x,y) by 2ya x=012≥r . The condition for the disk to roll with one and only one point of contact with the parabola is a 0 Rr<; that is, 176 CHAPTER 6 1 2Ra< (4) 6-12. The path length is given by 221 sd s y z d== + + ′′ ∫∫x (1) and our equation of constraint is ( )22 22,, 0gxyz x y z ρ = ++−= (2) The Euler equations with undetermined multipliers (6.69) tell us that 222 1yd g dydx dy yzλ λ′ == ++′′ (3) with a similar equation for z. Eliminating the factor λ, we obtain 22 220 11y dd z yd x zd x yz yz ′ ′ 11 − =   ++ ++′′ ′′  (4) This simplifies to ( ) ( ) ( )( )22 2211 zy y z y y y z z yz y z z y y z z  ++ − + − ++ − + =′′ ′ ′ ′ ′ ′′ ′ ′′ ′′ ′ ′ ′ ′ ′′ ′ ′′ 0 (5) ( ) ( ) 0 yy zz z y yz yy zz y z − −+ = ′′ ′ ′ ′ ′′ ′′ ′ ′ ′ ′′zy++ (6) and using the derivative of (2), ( ) ( ) zxz y y xy z−= − ′′′ ′ ′′ (7) This looks to be in the simplest form we can make it, but is it a plane? Take the equation of a plane passing through the origin: AxB yz+ = (8) and make it a differential equation by taking derivatives (giving A + By′ = z′ and By″ = z″) and eliminating the constants. The substitution yields (7) exactly. This confirms that the path must be the intersection of the sphere with a pl ane passing through the origin, as required. 6-13. For the reason of convenience, without lost of generality, suppose that the closed curve passes through fixed points A(-a,0) and B(a,0) (which have been chosen to be on axis Ox). We denote the part of the closed curve above and below the Ox axis as and respectively. (note that and ) 1()yx2()yx 10y>20 y< The enclosed area is () 12 1 2 1 2 12(, ) ( ) ( ) ( ) ( ) (, )aa a a aa a aJyy yx d x yx d x yx yxd x f yyd x −− − −=−= − =∫∫∫ ∫ SOME METHODS IN THE CALCULUS OF VARIATIONS 177 The total length of closed curve is () ( ) () () () { } ()22 2 2 12 1 2 1 1 12,1 1 1 1 ,aa a a aa a aKy y y d x y d x y y d x gy y d x −− − −=+ ++ = + + + = ′′ ′ ′ ′ ′ ′′ ∫∫∫ ∫ Then the generalized versions of Eq. (6.78) (see textbook) for this case are 1 2 1111101 1()ffg g y dd d yd x y yd x y d x yλλ ∂∂∂∂ ′ −+− = ⇒ −  ∂∂∂∂ ′′  +′   0= (1) 2 2 2222201 1( )ffg g y dd d yd x y yd x y d x yλλ ∂∂∂∂ ′ −+− = ⇒ −  ∂∂∂∂ ′′  +′   0= (2) Analogously to Eq. (6.85); from (1) we obtain ( )( )2 2 2 11 2 xA y A λ − +− = (3) from (2) we obtain ( )( )2 2 2 12 2 xB y B λ − +− = (4) where constants A’s, B’s can be dete rmined from 4 initial conditions ( ) 1,0 xa y=± = and ( ) 2,0 xa y=±= We note that and , so actually (3) and (4) altogeth er describe a circular path of radius 10y<20 y> λ. And this is the sought configuration that renders maximum enclosed area for a given path length. 6-14. It is more convenient to work with cylindrical coordinates ( r,φ,z) in this problem. The constraint here is z = 1 – r , then dz = –dr ( )22 2 22 2 22 s dr r d dz dr r d2d φ β =+ += + where we have introduced a new angular coordinate 2φβ= In this form of , we clearly see that the space is 2-dime nsional Euclidean flat, so the shortest line connecting two given points is a straight line given by: 2ds ()00 0 0 coscos2rrrφφ ββ==− −  this line passes through the endpoints ( r = 1, 2πφ=±) , then we can determine unambiguously the shortest path equation cos22() cos2rπ φφ= and z = 1 – r 178 CHAPTER 6 Accordingly, the shortest connecting length is 2 2 2 222 2sin 22drrdπ πldπφφ−=+ =∫ 6-15. 2 1 2 0[]dyIy y d xdx    =−     ∫ a) Treating I[y] as a mechanical action, we find the corresponding Euler-Lagrange equation 2 2()dyyxdx=− Combining with the boundary conditions ( x = 0, y = 0) and ( x = 1, y = 1), we can determine unambiguously the functional form of () ( s i n ) ( s i n 1 )yx x = . b) The corresponding minimum value of the integral is 2 11 2 2 001[ ] cos 2 cot (1) 0.642sin 1dyy d x d x xdx=− = = = ∫∫Iy c) If x = y then I[y] = ( 2 3 ) = 0.667. 6-16. a) S is arc length 22 2 222 9114dy dy dzS d xd yd z d x d x x L ddx dx dx  =+ + =+ +=+ + =      ∫∫ ∫x∫ Treating S and L like a mechanical action and Lagrangian respectively, we find the canonical momentum associated with coordinate y 2914dy L dxpdy dyxdx dxδ δ==  ++    Because L does not depend on y explicitly, then E-L equation implies that p is constant (i.e. 0 dp dx =), then the above equation becomes 32 229199 41111 4 4xdy pp yd x x A xdx p p+= + = + −− ∫B+ =⇒ where A and B are constants. Using boundary conditions ( x = 0, y = 0) and ( x = 1, y = 1) one can determine the arc equation unambiguously SOME METHODS IN THE CALCULUS OF VARIATIONS 179 32 3289() 1 113 8 4x yx    =+ −   −    and 32zx= b) z x y 00.250.50.751 00.250.50.75100.250.50.751 00.250.50.1 0 6-17. a) Equation of a ellipse 2 2 221y x ab+ = which implies 2abxy≤ because 2 2 222xy y x ab a b≤+ so the maximal area of the rectangle, whose corners lie on that ellipse, is Max[ A] = Max[4 xy] = 2 i ab. This happens when 2ax= and 2by= b) The area of the ellipse is 0A abπ= ; so the fraction of rectangle area to ellipse area is then 0[] 2Max A A π= 6-18. One can see that the surface xy = z is “locally” symmetric with respect to the line xy=− = − z where x > 0, y < 0, z < 0. This line is a parabola. Th is implies that if the particle starts from point (1,-1,-1) (which belongs to the symmetry line) under gravity ideally will move downward along this line. Its velocity at altitude z (z < –1) can be found from the conservation of energy. () 2( 1 )vz gz=−+ 180 CHAPTER6 CHAPTER 7 Hamilton’s Principle — Lagrangian and Hamiltonian D ynamics 7-1. Four coordinates are necessary to comp letely describe the disk. These are the x and y coordinates, the angle θ that measures the rolling, and the angle φ that describes the spinning (see figure). φ θ xy Since the disk may only roll in one directio n, we must have the following conditions: cos sindx dy R d φ φθ + = (1) tandy dxφ = (2) These equations are not integrable, and because we cannot obtain an equation relating the coordinates, the constraints are nonholonomic. This means that al though the constraints relate the infinitesimal displacements, they do not dictate the relations between the coordinates themselves, e.g. the values of x and y (position) in no way determine θ or φ (pitch and yaw), and vice versa. 7-2. Start with the Lagrangian ( )( )22 0 cos sin cos2mat m g Lv θ θθ θ=+ + + +AA A θ  (1) () ()2 22 00 2c os2mva t va t m g cos θ θθ  + + + +AA A θ =+ (2) 181 182 CHAPTER 7 Now let us just compute ()2 0 cosdL dmv a t mdt dtθ θθ∂  =+ +    ∂ AA (3) ( )2 0 cos sinma m v at m θ θθ =− − +  AA θA (4) ()0 sin sinLmv a t m g θ θθ∂+ −∂AA θ =− (5) According to Lagrange’s equations, (4) is equal to (5). This gives Equation (7.36) sin cos 0g aθθθ= +AA=  (6) To get Equation (7.41), start with Equation (7.40) cos sine gae θ θη η−=− (7) A and use Equation (7.38) tanea gθ=− (8) to obtain, either through a trigonometric identi ty or a figure such as the one shown here, g aθega22+ 22sg ga+coeθ= 22sinea gaθ= + (9) Inserting this into (7), we obtain 22agη η+=− A (10) as desired. We know intuitively that the period of the pe ndulum cannot depend on whether the train is accelerating to the left or to the right, which implies that the sign of a cannot affect the frequency. From a Newtonian point of view, the pe ndulum will be in equilibrium when it is in line with the effective acceleration. Since the acce leration is sideways and gravity is down, and the period can only depend on the magnitude of the effective acceleration, the correct form is clearly 22ag+ . HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 183 7-3. θ ρ φθR If we take angles θ and φ as our generalized coordinates, the kinetic energy and the potential energy of the system are ()22 1 22Tm R I1ρθ  =− + φ (1) ( )cos UR R m ρθ =− −g (2) where m is the mass of the sphere and where U = 0 at the lowest position of the sphere. I is the moment of inertia of sphere with respect to any diameter. Since ()225Im ρ = , the Lagrangian becomes () ()222 2 11cos25U m R m R R m g ρθ ρ φ ρ θ  =−= − + − − −  LT (3)  When the sphere is at its lowest position, the points A and B coincide. The condition A0 = B0 gives the equation of constraint: ()( ) ,fRθφ ρθ ρ φ 0 = −− = (4) Therefore, we have two Lagrange’s equa tions with one undetermined multiplier: 0 0f Ld L dt f Ld L dtλθθ θ λφφ φ∂ ∂∂  −+ = ∂∂ ∂   ∂ ∂∂ −+ =  ∂∂ ∂    (5) After substituting (3) and fRθ ρ ∂∂ = − and fφ ρ ∂∂= − into (5), we find ( ) ( ) ( )2sin 0 Rm g m R R ρθ ρ θλ ρ − − + −=  −− (6) 2 205mρφ λ ρ − −= (7) From (7) we find λ: 2 5m λ ρφ =−  (8) or, if we use (4), we have ()2 5mR λ ρθ =− −  (9) Substituting (9) into (6), we find th e equation of motion with respect to θ : 184 CHAPTER 7 2sin θ ω=− θ ) (10) where ω is the frequency of small oscillations, defined by (5 7g Rωρ=− (11) 7-4. y m xr θ If we choose ( r,θ ) as the generalized coordinates, the kinetic energy of the particle is ( ) ( )22 2 2 2 11 22Tm x y m r r θ =+ =+    (1) Since the force is related to the potential by Ufr∂=−∂ (2) we find AUrα α= (3) where we let U(r = 0) = 0. Therefore, the Lagrangian becomes ( )22 2 1 2ALm r r rαθα=+ −   (4) Lagrange’s equation for the coordinate r leads to 210 mr mr Arαθ−− +=  (5) Lagrange’s equation for the coordinate θ leads to ( )20dmrdtθ= (6) Since is identified as the angular momentum , (6) implies that angular momentum is conserved. Now, if we use A, we can write (5) as 2mrθ=A 2 1 30 mr Armrα−− +=A (7) Multiplying (7) by , we have r 2 1 30rmrr Ar rmrα−− +A  = (8) HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 185 which is equivalent to 2 2 21022dd d Amr rdt dt mr dtα α   + +       A = (9) Therefore, () 0dTUdt+ = (10) and the total energy is conserved. 7-5. φxy rm Let us choose the coordinate system so that the x-y plane lies on the vertical plane in a gravitational field and let the gravitational potential be zero along the x axis. Then the kinetic energy and the potential energy are expressed in terms of the generalized coordinates ( r,φ) as (22 2 1 2Tm r r )φ =+   (1) sinAUr m grαφα=+ (2) from which the Lagrangian is ()22 2 1sin2AU m r r r m g rαLT φ φα=−= + − −   (3) Therefore, Lagrange’s equation for the coordinate r is 21sin 0 mr mr Ar mgαφ−φ − ++ =  (4) Lagrange’s equation for the coordinate φ is ()2cos 0dmr mgrdtφφ+ =  (5) Since 2mrφ is the angular momentum along the z axis, (5) shows that the angular momentum is not conserved. The reason, of course, is that the particle is subject to a torque due to the gravitational force. 186 CHAPTER 7 7-6. y M Sm α αφ ξx Let us choose ξ,S as our generalized coordinates. The x,y coordinates of the center of the hoop are expressed by ()cos sin cos sinxS r yr Sξα α α α=+ +    =+ −  A (1) Therefore, the kinetic energy of the hoop is ( ) ( )( )22 2 hoop 2 2211 22 11cos sin22Tm x y I mS S I=+ + =+ + − +   φ  ξ αα φ (2) Using and 2Im r= Sr φ= , (2) becomes 22 hoop122 cos2Tm S S ξ ξα   =+ +  (3) In order to find the total kinetic energy, we need to add the kinetic energy of the translational motion of the plane along the x-axis which is 2 plane1 2TM ξ = (4) Therefore, the total kinetic energy becomes ()2 21cos2Tm S mM m S ξ ξ =++ +  α  (5) The potential energy is ( ) cos sin Um g ym g r S α α == + −     A (6) Hence, the Lagrangian is () ( )2 21cos cos sin2S mM m S m g r S lm ξ ξα α =++ + − + − α      A (7) from which the Lagrange equations for ξ and S are easily found to be 2c os s in mS m mg ξα α 0 + −  = (8) () cos 0 mM m S ξα + +  = (9) HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 187 or, if we rewrite these equations in the form of uncoupled equations by substituting for ξ and , we have S ()2 2cos2s i sin cos 2c osmSgmM mg mM mαα ααξn 0 α −− = +    =− +−   (10) Now, we can rewrite (9) as () cos 0dmM m Sdtξα + +  = (11) where we can interpret ( ) mM ξ + as the x component of the linear momentum of the total system and cos mS α  as the x component of the linear momentum of the hoop with respect to the plane. Therefore, (11) means that the x component of the total linear momentum is a constant of motion. This is the expected result because no external force is applied along the x-axis. 7-7. x1y1y2 x2my m xφ1 φ2 If we take ( )12,φφ as our generalized coordinates, the x,y coordinates of the two masses are 11 11cos sinx yφ φ=    = A A (1) 21 21cos cos sin sinx y2 2φ φ φφ=+    =+ AA AA (2) Using (1) and (2), we find the kinetic energy of the system to be ( ) ( ) () ()22 22 11 22 2 222 112 1 2 1 2 1 2 22 2 12 1 2 1 222 2s in s inc os c os2 22 cos2mmTx y x y m mφ φφ φ φ φ φ φ φ φφ φ φ φ φ=+ ++  = +++ +  + −    A    A =+ (3) 188 CHAPTER 7 The potential energy is Um12 1 2c o s c o s gx m gx m g2 φ φ =− − =− +     A (4) Therefore, the Lagrangian is ()22 2 12 1 2 1 2 11cos 2 cos cos2m g2 Lm φ φφ φ φφ φ φ=+ + − + +        AA (5) from which () () () ()2 12 1 2 1 1 22 12 1 2 1 2 12 1 2 2 2 22 21 1 2 2sin 2 sin 2c os sin sin cosLmm Lmm Lmm g Lmmg φφφ φ φφ φφ φ φφ φφφ φ φφ φφ φ φφ∂ =− −∂ ∂=+ − ∂  ∂ =− − −∂    ∂=+ − ∂ AA AA AA AA (6) The Lagrange equations for 1φ and 2φ are () ()2 12 12 2 12 1 cos sin 2 sin 0gφφ φ φφ φ +− +− +   A2φφ (7) = () ()2 21 12 1 12 2 cos sin sin 0gφφ φφ φ φφ φ+− −− +   A=  (8) 7-8. v1v2 θ2 θ1U1 U2y x Let us choose the x,y coordinates so that the two regions are divided by the y axis: ()1 20 0Ux Ux Ux<  =  >  If we consider the potential energy as a function of x as above, the Lagrangian of the particle is ( ) ()22 1 2Lm x y U x=+ −  (1) Therefore, Lagrange’s equa tions for the coordinates x and y are HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 189 ()0dU xmxdx+ =  (2) 0 my= (3) Using the relation x xx x dp dP dP P dd xmx mxdt dt dx dt m dx== = =  (4) (2) becomes ()0xx dU x Pd P md x d x+ = (5) Integrating (5) from any point in the region 1 to any point in the region 2, we find ()22 110xx dU x Pd Pdx dxmd x d x+ = ∫∫ (6) 2122 21 022xxPPUUmm− +−= (7) or, equivalently, 22 11 211 22mx U mx U += +2 (8) Now, from (3) we have 0dmydt= and is constant. Therefore, my 1my my2 = (9) From (9) we have 2 111 22my my =2 2 (10) Adding (8) and (10), we have 2 11 211 22mv U mv U += +2 2 2 (11) From (9) we also have 11 2sin sinmv mv θ θ = (12) Substituting (11) i nto (12), we find 12 1 21 2 21 1sin1sinvU U vTθ θ  −== +    (13) 190 CHAPTER 7 This problem is the mechanical analog of the refr action of light upon passing from a medium of a certain optical density into a medium with a different optical density. 7-9. Oxy M mφ αξ Using the generalized coordinates given in the fi gure, the Cartesian coordinates for the disk are (ξ cos α, –ξ sin α), and for the bob they are ( A sin φ + ξ cos α, –A cos φ – ξ sin α). The kinetic energy is given by ( )22 22 disk bob bob bob11 1 22 2T M I m x y ξθ=+= + + +   TT (1) Substituting the coordinate s for the bob, we obtain () (22 22 11 1cos22 2m I m mξθ φφ ξ φ =++ + + +  AA )aTM (2) The potential energy is given by ( )disk bob disk bob sin cos U U Mgy mgy M m g mgU ξ α = + = − + − A φ =+ (3) Now let us use the relation ξ = Rθ to reduce the degrees of freedom to two, and in addition substitute 22 IM R= for the disk. The Lagrangian becomes () ()2 22 31 1cos sin cos42 2LTU M m m m a Mm g m g ξ φφ ξ φ ξ α=−= + + + ++ + +  AA A φ (4) The resulting equations of motion for our two generalized coordinates are () () ()2 3sin cos sin 02Mm Mm g m ξα φ φα φ φ  +− + + + − + =  A α (5) ()1cos sin 0gφξ φ α φ+ ++ = AA (6) 7-10. MMMMyx x –y S HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 191 Let the length of the string be A so that ( ) Sx y−−= A (1) Then, xy=− (2) a) The Lagrangian of the system is 22 2 11 22Mx My Mgy My Mgy=+− = −   L (3) Therefore, Lagrange’s equation for y is 2dL LMy Mgdt y y0∂ ∂− =+ =∂∂ (4) from which 2gy=− (5) Then, the general solution for y becomes ()2 14gyt t C t C2 =−+ + (6) If we assign the initial conditions ( )0 yt 0 == and ( )0 yt 0 ==  , we find ()2 4gyt t=− (7) b) If the string has a mass m, we must consider its kinetic energy and potential energy. These are 2 string1 2T = my (8) 2 string22ym g mUy g =− =−AAy (9) Adding (8) and (9) to (3), the total Lagrangian becomes 22 1 22mgLM y M g y m y y=−+ + A2 (10) Therefore, Lagrange’s equation for y now becomes ()2mgMm y yM g 0 + −+ = A (11) In order to solve (11), we arrange this equation into the form ()2mg MMm y ym += −  A A (12) 192 CHAPTER 7 Since 22 2dM dydt m dt−=A 2y, (12) is equivalent to ()2 22mg dM Mydt m M m m −= − +  A Ay A (13) which is solved to give t MyA e Bmteγ γ−−= +A (14) where ()2mg Mmγ=+ A (15) If we assign the initial condition ( )00 yt==; ( )0 yt 0 ==  , we have 2MABm=+ =−A Then, () ()1c o s hMyt tmγ =−A (16) 7-11. x x′m yy′ φer′ The x,y coordinates of the particle are ( ) ()cos cos sin sinxR tR t yR tR tωφ ω ωφ ω =+ +  =+ + (1) Then, ( )( ) () ()sin sin cos cosxR t R t yR tR tω ω φω φω ω ω φω φω =− − + +   =+ + +     (2) Since there is no external force, the potential energy is constant and can be set equal to zero. The Lagrangian becomes HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 193 ( ) () ()22 2 22 2 21 2 2c2Lm x y mRR R os ω φω ω φω φ=+  + + + =+ (3) from which ()2sinLmRωφω φφ∂=− +∂ (4) (2cosdL dmRdt dt) φωω φφ∂  =+ + ∂ (5) Therefore, Lagrange’s equation for φ becomes 2sin 0 φω φ+ =  (6) which is also the equation of motion for a simple pendulum. To make the result appear reasonable, note that we may write the acceleration felt by the particle in the rotating frame as ( )2 r Rω=+ ′′ ai e (7) where the primed unit vectors are as indicate d in the figure. The part proportional to r′e does not affect the motion since it has no contributio n to the torque, and the part proportional to i′ is constant and does not contribute to the torque in the same way a constant gravitational field provides a torque to the simple pendulum. 7-12. rm θ Put the origin at the bottom of the plane ( )22 2 1sin2U m r r m g r LT θ θ =−= + −   ; t θαθ α= = ()2 22 1sin2Lm r rm gr t α α =+ −  Lagrange’s equation for r gives 2sin mr m r mg t α α =− or 2sin rr g tα α −= −  (1) 194 CHAPTER 7 The general solution is of the form p hr rr=+ where is the general solution of the homogeneous equation and hr 20 rrα−=pr is a particular solution of Eq. (1). So tt hrA e B eα α−=+ For pr, try a solution of the form rC sinp tα = . Then 2sinprC t α α =− . Substituting into (1) gives 22sin sin sin Ct C t g t α αα α −− = − α 22gCα= So ()2sin2tt gr t Ae Be tαααα−=+ + We can determine A and B from the initial conditions: ()0 0rr = (2) ()00r = (3) (2) implies 0rA=+ B (3) implies 22gAB0α=− + Solving for A and B gives: 00 2211 22 22ggAr Brα α =− =+      ()00 2211sin22 22 2tt gg gr e r e tαα 2rt ααα α−  =− ++ +   or () () 0 2cosh sin sinh2gr t t t rt α ααα=+ − 7-13. a) θa b m HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 195 21sin2 cos cos sinxa t b yb xa tb ybθ θ θ θ θθ=− =− =− =   ( ) ( )22 22 2 21 2 12c os c os2Lm x y m gy mat a t b b m g b θ θθ=+ − =− + +  θ dL L dt θ θ∂ ∂=∂ ∂ gives 2cos sin sin mat b mb mat b mgbdtdθ θθ θ −+ = −  θ  This gives the equation of motion sin cos 0g a bbθθθ+ −=  b) To find the period for small os cillations, we must expand sin θ and cos θ about the equilibrium point 0θ. We find 0θ by setting 0 θ= . For equilibrium, 00 sin cosga θ θ = or 0 tana gθ= ga θ0ag22+ Using the first two terms in a Taylor series expansion for sin θ and cos θ gives ()() () ( ) 000 ff fθθθ θθ θ= +−′ θ ( ) 00 sin sin cos0 θ θθ θ θ+−  ( ) 00 cos cos sin0 θ θθ θ θ−−  0 tana gθ= implies 022sina agθ= +, 022cosg agθ= + 196 CHAPTER 7 Thus () 0221sin ag g agθ θθ+− + () 0221cos ga a agθ θθ−+ + Substituting into the eq uation of motion gives () () 0022 22g aag g ga a ba g ba g0θ θθ θθ+ − − − + ++=+ This reduces to 22 22 0ga ga bbθ θθ+++= The solution to this inhomoge neous differential equation is 0 cos sinA B θθω θ ω θ = ++ where ( )1422 12ga bω+ = Thus ( )12 14222 2 bT gaπ π ω== + 7-14. θa b m 2sin 1cos2 cos sinxb ya t b xb ya tbθ θ θθ θ θ= =− = =+   ( ) ( )22 2 2 2 2 112s in22x y m b a ta bt Tm θ θθ =+ = + +   HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 197 21cos2g ym g a t b Um θ == −   ( )22 2 2 2 112s in c os22U m b a t a b t m g b a t θθ θ θ=−= + + + −  LT Lagrange’s equation for θ gives 2sin cos sindmb mabt mabt mgbdtθ θθ θ += − θ  ba2sin cos cos sinb a bt a bt g b θ θθ θθ θ ++ = −   θ sin 0ag bθθ++ =  For small oscillations, sin θ  θ 0ag bθθ++ =  . Comparing with gives 20 θω θ+= 22bTagππω==+ 7-15. k mθ b = unextended length of spring A = variable length of spring ( )22 2 1 2Tm θ =+ AA () ()22 11cos22b m gykb m g Uk θ =− + =− − AA A ( ) ()2 22 2 11cos22U m b m g LT θ θ =−= + − − + AA A A Taking Lagrange’s equations for A and θ gives ()2:cdmm kb m gdtos θ θ  =− − + AA AA 198 CHAPTER 7 2:sdmm gdtin θ θθ  =−AA This reduces to ()2cos 0 2sin 0kbgm gθθ θθ θ−+ − − = ++ = AA A    AAA 7-16. θ mbx = a sin ωt For mass m: sin sin cos cos cos sinxa tb yb xa tb ybω θ θ ω ωθ θθθ= + =− =+ =   Substitute into ( )22 1 2Tm x y =+  Um g y= and the result is ( )22 2 2 2 1cos 2 cos cos cos2LTU m a t a b t b m g b ω ωω θ ω θ θ =−= + + +  θ Lagrange’s equation for θ gives ( cod)2s cos cos sin sin mab t mb mabw t mgbdtω ωθ θ θ ωθ += − − θ 22sin cos cos sin cos sin sin ab t ab t b ab t gbω ωθ ω θ ωθ θ ω θ ωθ −− + =− −    θ or 2sin sin cos 0g atbbθθ ω ω θ +− = HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 199 7-17. θ θA BCy h mgq Using q and θ (= ω t since θ (0) = 0), the x,y coordinates of the particle are expressed as () ()cos sin cos sin sin cos sin cosxh q h tq t t yh q h tq t tθ θω ω θ θω ω=+ = +   =−= −  (1) from which sin cos sin cos sin cosxh t q t q t yh tq tq tω ωωω ω ωωωω ω=− + +    =+ −   (2) Therefore, the kinetic energy of the particle is ( ) ( )22 22 22 21 2 1 2Tm x y mh q q m h q ω ω=+ =+ + −  ω (3) The potential energy is ( ) sin cos Um g ym g h tq t ω ω == − (4) Then, the Lagrangian for the particle is 22 22 2 111sin cos222mh mq mq mgh t mgq t mh q L ω ωω ω + − + −  ω t=+ (5) Lagrange’s equation for the coordinate is 2cos qq gω ω −= (6) The complementary solution and the partic ular solution for (6) are written as () ( ) ()2cos cos2c Pqt A it gqt tω δ ωω=+    =−  (7) so that the general solution is () ()2cos cos2gqt A i t t ω δω=+ − ω (8) Using the initial conditions, we have 200 CHAPTER 7 () ()20c os 02 0s in 0gqA qi Aδω ωδ=− =  =− =  (9) Therefore, δ = 0, 22gAω= (10) and () (2cos cos2gqt i t t ) ω ωω=− (11) or, () ()2cosh cos2gqt t t ω ωω=− (12) q(t) tg 22ω In order to compute the Hamiltonian, we first find the canonical momentum of q. This is obtained by Lp mq m hqω∂==−∂ (13) Therefore, the Hamiltonian becomes 22 2 22 2 111sin cos222Hp q L mq m hq m h m q mq mgh t mgq t m qh ω ωω ω ω=− =− − − − + − +   ω so that 22 22 2 11 1sin cos22 2H mq m h m q mgh t mgq t ω ωω − + −  ω =− (14) Solving (13) for and substituting gives q 2 22 1sin cos22pH hp m q mgh t mgq tmω ωω − + − ω =+ (15) The Hamiltonian is therefore different from the total energy, T + U. The energy is not conserved in this problem since the Hamiltonian contains ti me explicitly. (The particle gains energy from the gravitational field.) HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 201 7-18. θ θy S ROA B Cmx –yx From the figure, we have the following relation: ACsR θ =−=−AA (1) where θ is the generalized coordinate. In terms of θ, the x,y coordinates of the mass are () ()cos sin cos sin cos sin cos sinxA C R R R yR A C R Rθ θθ θθ θ θθ θ =+ = −+ θ =− =− − A A (2) from which sin sin cos cosxR yRθθθ θθ θθθ θθ =−  =−   A   A (3) Therefore, the kinetic energy becomes ( )2 2 22 222 2 11222x y m R R Tm θ θθ θ θ   =+ = + −   A A (4) The potential energy is ( ) cos sin g ym g R R Um θ θ == − − θ     A (5) Then, the Lagrangian is ()22 222 2 12c os2LTU m R R m g R R sin θ θθ θ θ θ θ θ  =−= + − − −−      AA A (6) Lagrange’s equation for θ is ( )2cos 0 RR gθθ θ θ− −− =  A (7) Now let us expand about some angle 0θ, and assume the deviations are small. Defining 0 εθθ≡− , we obtain 0 00sin cosgg RR0 θ θεεθ θ+=−− AA (8) The solution to this differential equation is ()0 0cossinsinAtθεω δθ=+ + (9) where A and δ are constants of integration and 202 CHAPTER 7 0 0sing Rθωθ≡−A (10) is the frequency of small oscillati ons. It is clear from (9) that θ extends equally about 0θ when 0 2 θ π= . 7-19. PP m1m2 m2g m1gdθφ Because of the various constraints, only one gene ralized coordinate is needed to describe the system. We will use φ, the angle between a plane through P perpendicular to the direction of the gravitational force vector, and one of the extensionless strings, e.g., A, as our generalized coordinate. 2 The, the kinetic energy of the system is () ()2 11 2211 22Tm m2φ φ =+ A A (1) The potential energy is given by ( ) ( ) 11 22 sin sing m g Um πφθ φ =− − + − AA (2) from which the Lagrangian has the form ( ) ()22 2 11 22 1 1 2 21sin sin2U m m m g m g LT φ φθ φ =−= + + + +  AA A A (3) The Lagrangian equation for φ is ( )( )22 22 11 1 1 2 2 cos cos 0 mg m mφφ θ ++ − +  AA A A mg (4) φ= This is the equation which describes the motion in the plane . 12,,mmP To find the frequency of small oscillations around the equilibrium position (defined by 0 φφ=), we expand the potential energy U about 0φ: () () () () ()2 00 0 2 01 2 1 2UU U U Uφ φ φφ φφ φφ=+ + + ′′ ′ = ′′… (5) where the last equality follows because we can take ()00 φU = and because ()00 φ= ′U . From (4) and (5), the frequency of small osc illations around the equilibrium position is HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 203 ()0 2 2 11 22U mmφω′′=+AA2 (6) The condition ()00 φ= ′U gives 22 11 0 11costansinmm mθφθ+=AA A (7) or, ( )22 11 0 1222 22 11 22 1 212cosn 2c osmm mm m msiθφ θ+= ++AA AA A A (8) Then from (2), (7), and (8), ()0 Uφ′′ is found to be () ( ) () ( ) ( )0 0 22 11 11 0 22 2 22 11 11 22 11 1222 2222 1111 22 2 121 1222 22 11 22 2 121sin cos sin cot cos sincoscos 2c os 2c osUg m m m gm m mmmmm mm m m gm m m mφφ θ θ φ θ θθθ θ θ=+ + ′′ +  =+ +  + ++   =+ +AA A AA AAAAA AA A A AA A A (9) Finally, from (6) and (9), we have ( ) ( )1222 22 11 22 1 212 2 22 11 222c os mm m m g mmθ ++ = +AA A A AAω (10) which, using the relation, 22 12 12cos2dθ2+−=AA AA (11) can be written as () ( ) ( )1222 2 12 1 12 2 1 2 22 11 22gm m m m d m m mm ++ −= +AA AA2ω (12)  Notice that 2ω degenerates to the value gA appropriate for a simple pendulum when d → 0 (so that ). 12=AA 204 CHAPTER 7 7-20. The x-y plane is horizontal, and A, B, C are the fixed points lying in a plane above the hoop. The hoop rotates about the vertical through its center. z AB B′y x RC C′A′θ′ θ The kinetic energy of the system is given by 2 2 22 2 11 1 22 2 2MR zMz M2TI ω θθ∂=+ = +∂  θ (1) For small θ, the second term can be neglected since ( )00 z θθ =∂∂= The potential energy is given by UM g z= (2) where we take U = 0 at z = –A. Since the system has only one degree of freedom we can write z in terms of θ. When θ = 0, z=−A. When the hoop is rotated thorough an angle θ, then ( )( )222cos sin zR R R2θ θ =−− −A (3) so that ( )12222c os1 zR θ   =− + − A (4) and the potential energy is given by ( )12222c os1 UM g R θ   =− + − A (5) for small θ, 2s 1 2θθ−≅ − co ; then, 1222 2 22 21 12RUM g RMgθ θ ≅− −     ≅− −   AA AA (6) From (1) and (6), the Lagrangian is 22 22 21122RU M R M g LTθθ =−= + −    AA, (7) HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 205 for small θ. The Lagrange equation for θ gives 0gθθ+ =  A (8) where gω=A (9) which is the frequency of small rotational oscilla tions about the vertical through the center of the hoop and is the same as that for a simple pendulum of length A. 7-21. ω θ From the figure, we can easily write down the Lagrangian for this system. ( )2 22 2sin2mRT θ ω =+  θ (1) cos Um gR θ =− (2) The resulting equation of motion for θ is 2sin cos sin 0g Rθω θ θ θ−+ = (3) The equilibrium positions are found by finding the values of θ for which 02 0 0c osg Rθθ0 sin θ ωθ === −  θ (4) Note first that 0 and π are equilibrium, and a third is defined by the condition 0 2cosg Rθω= (5) To investigate the stability of each of these, expand using 0 εθθ=− (2 00 0 2cos sin sin cosg R)0 εωθ ε θ θ εω− + θ =− (6) 206 CHAPTER 7 For 0θ π=, we have 2 21g Rεωω=+  ε  (7) indicating that it is unstable. For 00 θ=, we have 2 21g Rεωω=−  ε (8)  which is stable if 2gR ω< and unstable if 2gR ω> . When stable, the frequency of small oscillations is 2gR ω− . For the final candidate, 22 0 sin ε ωθ=− ε (9) with a frequency of oscillations of ()22gR ωω− , when it exists. Defini ng a critical frequency 2 cgR ω≡ , we have a stable equilibrium at 00 θ= when c ωω< , and a stable equilibrium at ( )12 2 0cosc θ ωω−= when c ωω≥ . The frequencies of small oscillations are then ()21c ω ωω− and ()41c ω ωω− , respectively. To construct the phase diagram, we need the Hamiltonian LH L θθ∂≡ −∂ (10) which is not the total energy in this case. A conv enient parameter that describes the trajectory for a particular value of H is 22 2 221sin cos2cc cH mRθωK θ θωω ω −  − ≡= (11) so that we’ll end up plotting ()22 22c os s in ccKθωθ θωω =+ +   (12) for a particular value of ω and for various values of K. The results for c ωω< are shown in figure (b), and those for c ωω> are shown in figure (c). Note how the origin turns from an attractor into a separatrix as ω increases through cω. As such, the system could exhibit chaotic behavior in the presence of damping. HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 207 3 2 101 200.511.52 3K θ (b) 3 2 101 200.511.52 3K θ (c) 7-22. The potential energy U which gives the force ()() 2,t kFxt exτ−= (1) must satisfy the relation UFx∂=−∂ (2) we find tkUexτ−= (3) 208 CHAPTER 7 Therefore, the Lagrangian is 2 1 2tkLTU m x exτ−=− = −  (4) The Hamiltonian is given by xLH px L x Lx∂= −= −∂ (5) so that 2 2t xp kHmxeτ−=+ (6) The Hamiltonian is equal to the total energy, T + U, because the potential does not depend on velocity, but the total energy of the system is not conserved because H contains the time explicitly. 7-23. The Hamiltonian function can be written as [see Eq. (7.153)] jj jH pq L = − ∑ (1) For a particle which moves freely in a conservative field with potential U, the Lagrangian in rectangular coordinates is ( )22 2 1 2Lm x y z U= ++ − and the linear momentum components in rectangular coordinates are x y zLp mxx pm y pm z∂ ==∂ = =     (2) ( ) ( ) ( )222 2 2 2 22 2 2221 2 11 22xyzH mx my mz m x y z U mx y z U p p pm =+ +− + + −    + += + +     =+ (3) which is just the total energy of the particle. Th e canonical equations are [from Eqs. (7.160) and (7.161)] HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 209 xx yy z zUp mx Fx Up my Fy Up mz Fz∂== − =∂ ∂== − =∂ ∂== − =∂      (4) These are simply Newton’s equations. 7-24. θ m The kinetic energy and the potential en ergy of the system are expressed as ( ) ( )22 2 22 2 11 22 cosTm m Um gθα θ θ=+ = +  =−  AA A A (1) so that the Lagrangian is ( )22 2 1cos2U m m g LT α θ =−= + + AA θ (2) The Hamiltonian is 2 2 21cos22LHp L L pmm gmθ θθθθ α θ∂=− = −∂ =− − AA (3) which is different from the total energy, T + U. The total energy is not conserved in this system because work is done on the system and we have () 0dTUdt+ ≠ (4) 210 CHAPTER 7 7-25. z mzr y xθ In cylindrical coordinates the kinetic energy and the potential energy of the spiraling particle are expressed by 22 22 1 2Tm r r z Um g zθ  =+ +   =   (1) Therefore, if we use the relations, i.e., const.zkz k rθ θ ==  =  (2) the Lagrangian becomes 2 22 21 2rLmz zm gk=+ −  z (3) Then the canonical momentum is 2 21zLrp mzkz  ∂== +  ∂   (4) or, 2 21zpz rmk= +   (5) The Hamiltonian is 2 22 2212 1zz zzppH pz L p m g z rrmmkk= − +++ =− (6) or, 2 2 21 21zpH mgz rmk=++ (7) Now, Hamilton’s equations of motion are HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 211 zHpz∂− =∂; zHzp∂=∂ (8) so that zHmg pz∂− =− =∂ (9) 2 21z zp Hzp rmk∂= =∂ + (10) Taking the time derivative of (10) and substituting (9) into that equation, we find the equation of motion of the particle: 2 21gz r k= +   (11) It can be easily shown that Lagrange’s equation, computed from (3), gives the same result as (11). 7-26. a) θ m 22 1 2LTU m m g y θ =−= − A 22 1cos2Lm m g θ θ =+ AA 2 Lpmθ θθ∂==∂A so 2p mθθ= A Since U is velocity-independent and the coordinate transformations are time-independent, the Hamiltonian is the total energy 2 2cos2pHTU m gmθθ =+ = − AA The equations of motion are 212 CHAPTER 7 2and sinp HHpm gpmθ θ θθ θθ∂∂= =− =−∂∂ AA= b) a m2m2 m1m1x 2 22 12 2111 222xT m xm xIa=++ where I = moment of inertia of the pulley ( )12 Um g x m g x =−− − A 12 2 xLT Ip mm xxx a∂ ∂  === ++ ∂∂   So 12 2xpxImma= ++   H = T + U ()2 12 12 22xpH mg x mg xImma=− −++A− The equations of motion are ()12 2 12 1 22x x x xp HxI pmma Hp mg mg g m mp∂==∂ ++ ∂=− = − = −∂  HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 213 7-27. a) m2 m1k,b θ , coordinates ofiixy m i = Using A, θ as polar coordinates 21 cos xx θ =+A 21 sin yy θ = +A 21 cos sin xx θ θ =+ −   AA θ (1) 21 sin cos yy θ θ =+ +   AA θ (2) If we substitute (1) and (2) into ( ) ( ) ()2 22 22 11 1 22 2111 222U m x y mx y k b =−= + + + − −   A LT the result is () ( ) ( ) () () ()22 2 2 2 12 1 1 2 2 21 1 2 1 111 22 1cos sin cos sin2Lm m x y m mx y m y x k bθ θθ θ θθ=+ + + + ++ + − −  AA  AA −A The equations of motion are ()11 2 1 2 2 11 22:c os sin 0 xdxm mxm mdt mx mx pθθ θ  ++ − = = +=  AA  So = constant xp ()11 2 1 2 2 11 22:s in cos 0 ydym mym m my my pθθ θ  ++ + =dt = +=  AA  So yp= constant () () ()2 22 1 1 2 2 1 1 : cos sin cos sindmm x y m kb m y xdtθ θθ θ θ  ++ = − −+ −    AA A A θ which reduces to 214 CHAPTER 7 ()2 11 2cos sin 0kxy bmθθ θ−+ + + − =  AA A () () (2 22 1 1 21 1 2 1 1:c os sin sin cos cos sindmm y xdt mx y m x yθθ θ θ ) θ θθ θ +− =− − + − −  AA  AA θ which reduces to 11cos sin 20 yxθθθθ++ − =     AAA A b) As was shown in (a) 1constantxLpx∂==∂ 1constantyLpy∂==∂ (total linear momentum) c) Using L from part (a) ()1 12 12 2 1cos sinxLpm mxm mxθ θθ∂==+ + −∂  AA ()1 12 12 2 1sin cosyLpm mym myθ θθ∂==+ + −∂  AA 21 21 2cos sinLp mx my m θθ∂== + +∂A  AA 2 21 21 2sin cos pm x m y mθ θ θθ =− + +  AAA Inverting these equations gives (after much algebra) 1 1 1sin 1cosx xp pmpθθθ=− +A A 1 1 1cos 1siny yp pmpθθθ=−−A A 1112 121cos sinxymmp ppmmθθ +=− − + AA 1112 121sin cosxymmp ppmmθ θθ θ +=− +  AA Since the coordinate transformations are time independent, and U is velocity independent, HTU=+ Substituting using the above equations for in terms of gives iqip HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 215 ( ) ( ) ()11 1 1 112 22 2 12 2 12 212c os s in2 12s inc os2xy x y xyp mmHp p p p p pmm pp pkθ θθθ θθ  +=+ + + − +    +− +AAA AAb− The equations of motion are 1 11 1sin 1cosx xHxp ppmpθθθ∂ == − +∂ A A 1 11 1cos 1siny yHyp ppmpθθθ∂ == − −∂ A A 1112 121cos sinxymm Hpp ppm mθ θ+ ∂== − − ∂ A AA 1112 121sin cosxymm Hpp ppm mθ θθ θθ+ ∂== + − ∂  AA 11 1100xyHHppxy∂∂=− = =− =∂∂ ( )( )()112 12 32 12 1sin cosxymm p p Hp pp kmm mθ θθθ+ ∂=− = + − − −∂A AAA Ab ( )11 1 1 11sin cos cos sinxy x ypp Hpp p p pmmθ θ θ θθθ∂θ   =− = − + − + ∂A A Note: This solution chooses as its generalized coordinates what the student would most likely choose at this point in the text. If one looks ahead to Section 8.2 and 8.3, however, it would show another choice of generalized coordinate s that lead to three cyclic coordinates ( , , and θ ), as shown in those sections. CMxCMy 7-28. so 2Fk r−=−1Uk r−=− ( )22 2 1 2kLTU m r rrθ =−= + +   sor rp Lpm r rrm∂== =∂ 2 2sop Lpm rmrθ θ θθθ∂== =∂ Since the coordinate transfor mations are independent of t, and the potential energy is velocity- independent, the Hamiltonian is the total energy. 216 CHAPTER 7 ( )22 2 22 2 22 4 22 21 2 1 2 22r rkHTU m r rr pp kmrmm r r pp kHmm rrθ θθ =+= + − =+ −  =+ −  Hamilton’s equations of motion are 2 2 32 0r r rp p HHrp mp p Hkprm r r Hpθ θ θ θθ θ∂∂== ==∂∂ ∂=− = −∂ ∂=− =∂   mr 7-29. a k m θ b = unextended length of spring A = variable length of spring a) sin sin cos xx θ θθ ==  AA θ +A 21cos cos sin2ya t y a t θ θθ =− = − +   AA θ A Substituting into ( ) ()22 21 2 1 2Tm x y Um g y k b=+ =+ − A gives ( ) ()2 2 22 22 2 12s in c os c os22at kLTU m a t a t m g b θθ θ θ θ =−= + + + − + − − −   AA A A A A2 Lagrange’s equations give: HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 217 ()2: cos sin cosdma m t m m a t m g k bdtθθ θ θ θ −= ++ −  AA A A − 2: sin sin sin cosdmm at m at m g m atdtθ θθ θ θ θ  += − +AA A A A θ  Upon simplifying, the equations of motion reduce to: () ()2cos 0 2sin 0kag bm agθθ θθ θ−− + + − = ++ += AA A    AAA b) cos or cosp Lpm mat a tmθ θ∂==− = +∂A AAAA (1) 2sinLpm m atθ θ θθ∂== +∂AA o r 2sin pa t mθ θθAA=− (2) Since the transformation equations relating the generalized coordi nates to rectangular coordinates are not time-independent, th e Hamiltonian is not the total energy. ii Hp q L p pθθL = −= + − ∑ A  A Substituting (1) and (2) for and A θ and simplifying gives ()22 2 2 211sin cos cos22 2 2pp atHp atp k b mgatmmmθ θ g θ θθ =+ − + + − + −A A AAAA The equations for θ and A are  2sin cos agreeing with (1) and (2)p Ha t pm p Hatpmθ θθθ θ∂== −∂ ∂== +∂A A AA A The equations for and pA pθ are ()2 23sin cosp Ha tk bmgmppθ θ θθ∂=− =− − − + +∂A AAA A or ()2 23sin cos 0p atk b m gmθ θ θθ ++ −− +A AAApp = cos sin sinHa tatp mgθθpp θ θθθ∂=− =− + −∂A AA 218 CHAPTER 7 or cos sin sin 0atatp m gθθ θθ θ −− +A AApp = c) 2 sin , cos 12θθθ θ −  Substitute into Lagrange’s equations of motion () ()2 2102 20kag bm ag atθθ θθθ θ−− + −+ − =  ++ +− AA A A   AAA A= For small oscillations, . Dropping all second-order terms gives 1, 1, 1θθ   A  0kkag bmm agθθ+= + + ++=AA  A For θ, 22 Tagθππω==+A The solution to the equation for A is ()homogeneous particular cos sinkk mA tB t ag bmm k=+ =+ + +AA A + So for A, 22mTkππω==A 7-30. a) From the definition of a total derivative, we can write k k kkdg g g q g p dt t q t p tk  ∂∂ ∂ ∂ ∂=+ +  ∂ ∂∂ ∂∂ ∑ (1) Using the canonical equations HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 219 k k k k k kq Hqtp p Hptq∂ ∂ ==∂∂ ∂ ∂== − ∂∂   (2) we can write (1) as k kk kkdg g g g H H dt t q p p q  ∂∂ ∂ ∂∂=+ −  ∂∂ ∂ ∂ ∂  ∑ (3) or ,dg ggHdt t∂=+    ∂ (4) b) j j jq Hqtp∂ ∂==∂ ∂ (5) According to the definition of the Poisson brackets, ,jj j k kk kkqgH HqHqp pq∂∂  ∂∂ =−   ∂∂ ∂∂ ∑ (6) but j jk kq qδ∂ =∂ and 0j kq p∂ =∂ for any j,k (7) then (6) can be expressed as ,jj jHqH qp∂ = =∂ (8) In the same way, from the canonical equations, j jHpq∂=−∂ (9) so that ,jj j k kk kkppH HpHqp pq∂∂  ∂∂ =−   ∂∂ ∂∂ ∑ (10) but j jk kp pδ∂ =∂ and 0j kp q∂ =∂ for any j,k (11) then, 220 CHAPTER 7 ,jj jHp pHq∂  =− = ∂ (12) c) ,jj kk kjp p ppppqp pq∂∂   ∂∂ =−   ∂∂ ∂∂ ∑ A AA AA (13) since, 0kp q∂=∂A for any k,A (14) the right-hand side of (13) vanishes, and ,kjpp 0= (15) In the same way, ,jj kk kjqqqqqqqp pq∂∂   ∂∂ =−   ∂∂ ∂∂ ∑ A AA AA (16) since 0jq p∂ =∂A for any j,A (17) the right-hand side of (16) vanishes and ,kjqq 0= (18) d) ,jj kk kj kjp p qqqpqp pq δδ∂∂   ∂∂ =−   ∂∂ ∂∂  =∑ ∑A AA AA AA A (19) or, ,k j kj qp δ  = (20) e) Let ( ),kkgp q be a quantity that does not depend explicitly on the time. If ( ),kkgp q commutes with the Hamiltonian, i.e., if , gH 0=     (21) then, according to the result in a) above, 0dg dt= (22) and g is a constant of motion. HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 221 7-31. A spherical pendulum can be described in terms of the motion of a point mass m on the surface of a sphere of radius A, where A corresponds to the length of the pendulum support rod. The coordinates are as indicated below. θ φymg xz The kinetic energy of the pendulum is ( )222 2 2 12111sin222I m2TI φ θφ θ =+= +   A θ (1) and the potential energy is cos Um g θ = A (2) The Lagrangian is ( )22 2 2 1sin cos2Lm m g φ θθ θ =+ − AA (3) so that the momenta are 2 Lpmθ θθ∂==∂A (4) 22sinLpmφ φ θφ∂==∂A (5) The Hamiltonian then becomes ( ) ()22 2 2 2 21sin cos2 ,2Hp p m m g pVpmθφ θ φθ φφ θ θ θ=+− + + =+  AA Aθ (6) which is just the total energy of the system and where the effective potential is ()2 22,2s inpVp m gmθ φ cos θ θθ=+ AA (7) When 0 pφ=, V(θ,0) is finite for all θ, with a maximum at θ = 0 (top of the sphere) and a minimum at θ = π (bottom of the sphere); this is just the case of the ordinary pendulum. For different values of pφ, the V–θ diagram has the appearance below: 222 CHAPTER 7 V Pφ = 00 ππ 2θ When 0 pφ>, the pendulum never reaches θ = 0 or θ = π because V is infinite at these points. The V–θ curve has a single minimum and the motion is oscillatory about this point. If the total energy (and therefore V) is a minimum for a given pφ,θ is a constant, and we have the case of a conical pendulum. For further details, see J. C. Slater and N. H. Frank, Mechanics , McGraw-Hill, New York, 1947, pp. 79–86. 7-32. The Lagrangian for this case is ( )22 22 2 2 1sin2kU m r r rrθθ φ =−= + + +   LT (1) where spherical coordinates have been used due to the symmetry of U. The generalized coordinates are r, θ, and φ, and the generalized momenta are rLp mrr∂==∂ (2) 2 Lp mrθ θθ∂==∂ (3) 22sinLpm rφ φ θφ∂==∂ (4) The Hamiltonian can be constructed as in Eq. (7.155): ( )22 22 2 2 2 22 22 21sin2 1 2s inr rHp rp p L kmr r rr p pp k mm r m r rθφ φ θθφ θφ θ θ=+ + − =+ +  =+ + −       −  (5) Eqs. (7.160) applied to H as given in (5) reproduce equation s (2), (3), and (4). The canonical equations of motion are obtain ed applying Eq. (7.161) to H: 2 2 233 2sinrp p Hkprr m r m rφ θ θ∂=− =− + +∂ (6) HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 223 2 22cot sinp Hpmrφ θθ θ θ∂=− =∂ (7) 0Hpφφ∂=− =∂ (8) The last equation implies that const pφ= , which reduces the number of variables on which H depends to four: , , ,r rp pθ θ : 2 2 22 21c onst 2s inrp kHpmr rθ θ=+ +   r− (9) For motion with constant energy, (9) fixes the va lue of any of the four variables when the other three are given. From (9), for a given constant value of H = E, we obtain 1222 22sin 22sinrpc onst mkErrθ θ θ +=− + pm (10)  and so the projection of the phase path on the rrp− plane are as shown below. pr θ3 θ2 θ1 θ1 < θ2 < θ32mE 7-33. m1 m2 m3x′x Neglect the masses of the pulleys () ()22 2 12 311 1 22 2x m x x m x x =+ − + − − Tm ′ ′     ( ) ( )12 3mg x mg x x mg x x=− − − + − − + −U ′ ′′ AA A 224 CHAPTER 7 () () ( () ( )22 123 23 32 123 2311 constantm m x m m x x x m m gm m m x gm m x=+ + ++ + − ′′ +− −+− + ′  )22Lm We redefine the zero in U such that the constant in L = 0. () (123 32 xL) p mmm xmm xx∂== ++ + − ′∂ (1) () (32 23 xL) p mm xmm xx′∂==− ++ ′∂′ (2) Solving (1) and (2) for and gives xpxp′ ( ) ( )1 23 23 xx xD m m p m m p− ′   =+ + −  ( ) ( )1 23 123 xx xD mm p mmm p− ′   =+ + + + ′  where Dm 13 12 23 4 m m m m m=++ () () () () (22 123 23 32 123 2311 22 )HTU mmm x mm x mm x x gm m m x gm m x=+ + ++ + − ′ −− − − − ′ =+ Substituting for and and simplifying gives x x′ () () () ( ) ()12 12 23 123 1 23 123 23 13 12 2311 22 where 4xx xxHm m D p m m m D p mm D p p g mmm x g mm x Dm m m m m m−− ′ − ′=+ ++ + − − + − − ′ =+++− The equations of motion are () () () ( ) () ()11 23 23 11 23 123 123 23xx x xx x x xHx m mD p m mD pp Hxm mDp m m mDp Hpg m m mx Hpg m mx−− ′ −−p′ ′ ′ ′∂== + + −∂ ∂== − + + +′∂ ∂=− = − −∂ ∂=− = −∂′    HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 225 7-34. x Rr m Mθ The coordinates of the wedge and the particle are cos 0sMm Mmxx x r yy rinxθ θ== == −+ (1) The Lagrangian is then ( )22 2 22c o s 2 s i n s i n2Mm mL x r r xr xr mgrrθ θθ θ+=+ + + − +     θ (2) Note that we do not take r to be constant since we want the reaction of the wedge on the particle. The constraint equation is ( ),, 0fx r r Rθ =−= . a) Right now, however, we may take r = R and 0 rr==  to get the equations of motion for x and θ. Using Lagrange’s equations, ( )2sin cos xa R θ θθ θ =+    (3) sin cosxg Rθ θθ+= (4) where ( ) am Mm≡+ . b) We can get the reaction of the wedge from the Lagrange equation for r 2cos sinmx mR mgλ θθ =− −   θ (5) We can use equations (3) and (4) to express in terms of θ and x θ, and substitute the resulting expression into (5) to obtain ( )2 21sin1s inaRgaλ θθ −=+− θ (6) To get an expression for θ, let us use the conservation of energy ( )22 2 0 2 sin sin sin22Mm mHx R x Rm gR m gR θ θθ θ θ+=+ − − = −   (7) where 0θ is defined by the initial position of the particle, and 0 sin mgR θ − is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain siR n xa θ θ =  , and substitute this into the energy eq uation to obtain an expression for θ ( ) ( )0 2 22s i n s i n 1s ing Raθ θθ θ−= − (8) 226 CHAPTER 7 Finally, we can solve for the reaction in terms of only θ and 0θ ( ) () ( )3 0 223 sin sin 2 sin 1s inmMg a Mm aθ θθ θ−− =− +−λ (9) 7-35. We use iz and as our generalized coordinates, the subscript i corresponding to the ith particle. For a uniform field in the z direction the trajectories z = z(t) and momenta p = p(t) are given by ip 2 00 01 2ii i iizzv t g t pp m g t=+ −  =−  (1) where 0iz, , and 0ip00iivp= m are the initial displacement, momentum, and velocity of the ith particle. Using the initial conditions given, we have 2 0 101 2ptzzm=+ − gt t (2a) 10ppm g= − (2b) 2 0 20 01 2ptz zz gm=+ ∆ + − t t (2c) 20ppm g= − (2d) ( ) 00 2 301 2pp tz zm+∆=+ − gt t (2e) 30 0pp pm g= +∆ − (2f) ( ) 00 2 40 01 2pp tz zZ gm+∆=+ ∆+ − t t (2g) 40 0pp pm g= +∆ − (2h) The Hamiltonian function corresponding to the ith particle is 2 2 const.22i i iii i ip mzV m g z m g zm=+= + = + =HT (3) From (3) the phase space diagram for any of the four particles is a parabola as shown below. HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 227 p0 + ∆p0 z0 + ∆z0p0Area at t = 0 Area at t = t1 ∆p0 z0∆z0p z1234 From this diagram (as well as from 2b, 2d, 2f, and 2h) it can be seen that for any time t , 1pp2= (4) 3pp4= (5) Then for a certain time t the shape of the area described by the representative points will be of the general form p z1234 (p1,z1)( p2,z2)(p3,z3) (p4,z4) where the base 1 must parallel to the top 2 3 4 . At time t = 0 the area is given by 00zp ∆∆ , since it corresponds to a rectangle of base 0z∆ and height 0p∆. At any other time the area will be given by () { () } () { () }1 1 1 1 1 121 43 0 31 42 0 00base of parallelogram height of parallelogram =tt tt tt tt tt ttAz z zzz xp p p pp pz= = = = = === − =− = ∆ =− =− = ∆ ∆∆ (6) Thus, the area occupied in the phase plane is constant in time. 7-36. The initial volume of phase spac e accessible to the beam is 2 00VR p2 0 ππ = (1) After focusing, the volume in phase space is 22 11VR p1 ππ = (2) 228 CHAPTER 7 where now is the resulting radius of the distribu tion of transverse momentum components of the beam with a circular cross section of radius . From Liouville’s theorem the phase space accessible to the ensemble is invariant; hence, 1p 1R 22 2 00 0 11VR p VR p2 1 πππ == = π (3) from which 00 1 1RppR= (4) If , then , which means that the resulting spre ad in the momentum distribution has increased . 1RR<0 0 1pp> This result means that when the beam is bette r focused, the transverse momentum components are increased and there is a subsequent diverg ence of the beam past the point of focus. 7-37. Let’s choose the coordinate system as shown: x1 m1 m2m3x3 x2 The Lagrangian of the system is ()2 22 3 12 12 3 1 1 221 2dx dx dxLTU m m m g m x m x m xdt dt dt   =−= + + + + +       33 1 2 with the constraints and 1xy l+=23xy xy l−+− = which imply 2 22 3 12 123 1 2 2222( 2 )0 2dx dx dxxxx lldt dt dt++− += ⇒ + + = 0 (1) The motion equations (with Lagrange multiplier λ) are HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 229 2 1 11 22dxmg mdtλ0 − += (2) 2 2 22 20dxmg mdtλ − += (3) 2 3 33 20dxmg mdtλ − += (4) Combining (1)–(4) we find 124 411g mmmλ 3−= ++ Finally, the string tension that acts on m is (see Eq. (2)) 1 2 1 11 1 2 1282411g dxg mdt mmmλ =− = − = ++ 3Tm 7-38. The Hamiltonian of the system is 2 2 24 21 22 4 2 2p dx kx bx kx bxU mdt m=+ = + + = + + 4 4HT The Hamiltonian motion equations that follow this Hamiltonian are p dx H dt p m∂= =∂ 3()dp Hkx bxdt x∂=− =− +∂ 7-39. z The Lagrangian of the rope is 22 211 22 2mgz dz mz z dzU m g mdt b dt b  =−= − − = +      2LT 230 CHAPTER 7 from which follows the equation of motion 2 2mgz Ld L d zmzdt z b dt∂∂=⇒ =∂∂  7-40. m m 2 m xθ2θ1 We choose the coordinates for the system as shown in the figure. The kinetic energy is 2 22 2 11 1 22 12 1 2 12 11122 cos22 1cos cos sin sin2dd dx dx dxTm m b bdt dt dt dt dt dd dd dxmb b b bdt dt dt dt dt  =+ + +        ++ + + +   θθθ θθ θθθθ θ2θ 2 The potential energy is Um11 cos ( cos cos )gb m g b b θ θθ =− − + And the Lagrangian is 22 2 22 11 2 1 2 21 2 21 2 1122 cos2 cos cos ( ) 2 cos cosdd d dx dxL T U m mb mb mbdt dt dt dt dt dd d dxmb mb mgb mgbdt dt dt dt  =−= + + +      ++ − + +θθ θθ θθ θ 2 θ θθ θ θ From this follow 3 equations of motion 22 2 12 12 22 2 22 12 1204 2 c o s c o s 2s in s indd Ld L d xbx dt x dt dt dt ddbdt dt  ∂∂=⇒ = + +  ∂∂    −+  θθθ θ θθθθ HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 231 22 2 12 11 22 2 1 1 2 2 122s i n 2 2 c o s c o s ( ) sin( )dd Ld L d xgb bdt dt dt dt dbdt∂∂=⇒ −= + + −∂ ∂ +−θθ 1 2 θ θθθ θ θθθθ 2 2 2 22 2 211 22 1 2 22 2sin cos cos ( ) sin ( )Ld L dt dd d dxgb b bdt dt dt dtθ θ θθθ 1 2 θ θθ θ θ∂∂=⇒∂ ∂ −= + + − − − θ 7-41. For small angle of oscillation θ we have 22 2 11 22ddb mdt dtθ =+   bTm and Um cosgb θ =− So the Lagrangian reads 22 2 11cos22dd bU m b m m g bdt dtθLT θ =−= + +    from which follow 2 equations of motion 2 2 2cosLd L d d b dbgbd t d t d t d t b∂∂ =⇒ + = =−  ∂ ∂θ αθ 22 22 22sin 2 2Ld L d b d d d dmgb mb mb mb mbdt dt dt dt dt dtθ θθθαθ θ∂∂=⇒ − = + = − +∂ ∂θ 232 CHAPTER7 CHAPTER 8 Central- Force Motion 8-1. x3 m2m1 x2r1 r2 x1 In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x direction; then the masses and have the gravitational potential energies: 1 1m2m () () () () () () () ()11 1 1 11 1 22 2 2 12 1g gUF xm x UF xm xα α =− =−  =− =−  (1) where () () ()( )111 11 2 3 ,,xxx=r and where α is the constant gravitational acceleration. Therefore, introducting the relative coordinate r and the center of mass coordinate R according to ()12 11 22 1 2mm m m=−    += +rr r rr R (2) we can express r and r in terms of r and R by 1 2 2 1 12 1 2 12m mm m mm=++   =− + + rr rrR R (3) 233 234 CHAPTER 8 which differ from Eq. (8.3) in the text by R. The Lagrangian of the two-particle system can now be expressed in terms of r and R: ()() () ()22 12 11 22 22 21 12 12 12 21 12 12 1211 22 11 22gg Lm m U U U mmmm U rmm mm mmmx X m xmm mmαα=+− − − + − + −++ X=+  ++ + −+  ++ rr r rR rR   (4) where x and X are the components of r and R, respectively. Then, (4) becomes 1x () () ()2 2 22 21 12 1 12 12 12 12 12 1211 1 2mmLm m m mmm mm mm mmUr x m m Xmmαα =+ +  ++  −−+ + ++rr  2+ R22 (5) Hence, we can write the Lagrangian in the form () () ()2 2 12 1211 22r m m m m LU X µ α =− + + + + rR   (6) where µ is the reduced mass: 12 12mm mmµ=+ (7) Therefore, this case is reducible to an equivalent one-body problem. 8-2. Setting 1u=r, Eq. (8.38) can be rewritten as 2 2222du Ekuuθ µµ=− +−∫ AA (1) where we have used the relation ()21 du r dr=− . Using the standard form of the integral [see Eq. (E.8c), Appendix E]: 1 2212sin const. 4dx ax b a ax bx c b ac− −+= − ++ − + ∫ (2) we have 21 2 2222 const. sin 28k r kEµ θ µµ−    −+ +=   +   A AA (3) CENTRAL-FORCE MOTION 235 or, equivalently, ()2 2 2222 sin const. 28k r kEµ θ µ µ−+ += +A AA (4) We can choose the point from which θ is measured so that the constant in (4) is 2π− . Then, 2 2 211 cos 21kr E kµθ µ− = +A A (5) which is the desired expression. 8-3. When 2 kk→ , the potential energy will decrease to ha lf its former value; but the kinetic energy will remain the same. Since the original or bit is circular, the instantaneous values of T and U are equal to the average values, T and U. For a 2r1 force, the virial theorem states 1 2T=− U (1) Hence, 11 22ETU UU U=+ = − + = (2) Now, consider the energy diagram E D C Ar –k/rB where CB E= original total energy CA U= original potential energy C CD U= original centrifugal energy The point B is obtained from CB CA CD=− . According to the virial theorem, ()12=EU or ()12 . CB CA= Therefore, CD CB BA== 236 CHAPTER 8 Hence, if U suddenly is halved, the total energy is raised from B by an amount equal to ()12CA or by CB. Thus, the total energy point is raised from B to C; i.e., E(final) = 0 and the orbit is parabolic . 8-4. Since the particle moves in a central, invers e-square law force field, the potential energy is kUr=− (1) so that the time average is 01kUrτ τ=− dt∫ (2) Since this motion is a central motion, the angu lar momentum is a constant of motion. Then, (3) 2. rc o µθ ≡=A nst from which 2rdt dµθ =A (4) Therefore, (2) becomes 22 2 001 kr kUdrππµµrd θ θττ=− =−∫ ∫AA (5) Now, substituting 1c os rα εθ =+ and ( )322 a τµ π α= A , (5) becomes 2 32 01 21 coskUaπαdθπε=−+∫θ (6) where a is the semimajor axis of the ellipse. Us ing the standard integral [see Eq. (E.15), Appendix E], 2 2 01 1c os 1dπ2πθεθ ε=+ −∫ (7) and the relation, ( )21aα ε =− (8) (6) becomes kUa=− (9) The kinetic energy is CENTRAL-FORCE MOTION 237 2 2 21 22Trrµµ=− +A (10) and the time average is 2 2 3 0011 2kTT dt T raτπµdθτπ=− =∫∫A (11) Part of this integral is trivial, ()2 2 2 3 01 22kTr rdaπµµ πθπ µ =+    ∫A A (12) To evaluate the integral above, substitute the expression for r and make a change of variable () () ()22 22 1 22 2 2 00 1sin 1 22 1 1c osdxrr d xππθθ εεθµµ ε εθ −−  ==   + +∫AA2dx 11∫∫ (13) The reader is invited to evaluate this integral in either form. The solution presented here is to integrate by parts twice, which gives a third integral that can be looked up in a table: () () ()111 2 2 22 1111 11 1 1 11xd x x d x x x x xx εε ε ε ε −−−− −=− −+ + −+∫ ∫ (14) ()11 11 2 1 1sin sin 1 1 1xx x d x x εε ε−− − −x     =− −+ +   ∫ (15) () () () ()1 11 22 111 12sin tan11 1xxxε εε ε−− − −−  =− +++   −  (16) 2211 1π ε ε =− +  −   (17) Substituting this into (13) and then in to (12), we obtain the desired answer, 2kTa= (18) This explicitly verifies the virial theorem, whic h states that for an inverse-square law force, 1 2T=− U (19) 238 CHAPTER 8 8-5. a m1m2 Suppose two particles with masses and move around one another in a circular orbit with radius a. We can consider this motion as the moti on of one particle with the reduced mass µ moving under the influence of a central force 1m2m 2 12Gmm a . Therefore, the equation of motion before the particles are stopped is 2 12 2mmaGaµω = (1) where 1211 1 mmµ=+ , 2πωτ= (2) The radius of circular motion is 132 12 24Gmmaτ πµ =   (3) After the circular motion is stopped, the particle with reduced mass µ starts to move toward the force center. We can find the equation of motion from the conservation of energy: 2 12 12 1 2mm mmGx Gaxµ −= −  (4) or, 12 12 2 11 Gmmxxa µ  =−     (5) Therefore, the time elapsed before the collision is 0 12 2 11adxtd t Gmm xa µ== −  −  ∫∫ (6) where the negative sign is due to the fact that the time increases as the distance decreases. Rearranging the integrand, we can write 0 12 2aaxtGmm axµ=− −∫dx (7) Setting ( )22 xyd x y d y≡= , the integral in (7) becomes 00 2 22 a ay xId xax ay==− −∫∫dy CENTRAL-FORCE MOTION 239 Using Eq. (E.7), Appendix E, we find 02 12s in22 ayay y aa a 2Iπ− −  + = −  (8)  =− Therefore, 12 22aatGmmµ π= or, 42t=τ (9) 8-6. m1 m2 x1 x2 O r rxx=−21 When two particles are initially at rest separated by a distance , the system has the total energy 0r 12 0 0mmEGr=− (1) The coordinates of the particles, and , are measured from the position of the center of mass. At any time the total energy is 1x2x 22 12 11 2211 22mmEm x m x Gr=+−  (2) and the linear momentum, at any time, is 11 22 0 pm x m x= += (3) From the conservation of energy we have E0E=, or 22 12 12 11 22 011 22mm mmGm x mxGrr−= + −  (4) Using (3) in (4), we find 11 2 0 12 1 021 1 21 1GxvmMr r Gxv mM rr == −     == − −     (5) 240 CHAPTER 8 8-7. Since ()Fr k r =− is a central force, angular momentum is conserved and the areal velocity, 2 dA dt µ =A , is trivially constant (see Se ction 8.3). In order to compute U, we start with 2 22 2drdt EUr µµ=  −−  A (1) and 2 2krU= (2) The time average of the potential energy becomes max min0 3 42 21 2 2 2 22r rUU dt kr dr krErτ τ τ µ µ= =  −−  ∫ ∫A (3) Substituting 2 1 2r x dr dxr== (4) (4) becomes 2 max 2 min2 222 22r rxd x kU kEx xµ τ µ= −+−∫A (5) Using the integrals in Eqs. (E.9) and (E.8c), Appendix E, 21 221sin 2 4xd x ba xax bx ca aa ax bx c b ac− 2 b   +=+ ++  − ++ −  ∫ (6) (5) becomes max min12 22 14 122 222sin22 2 2rrkE E kr kUrkk k kE µ τµ µ−  − =− − − + −     −   A A2Er (7) But and were originally defined as the roots of maxrminr2 22EUrµ−−A. Hence, the second term vanishes at both limits of in tegration. On the other hand, CENTRAL-FORCE MOTION 241 max min max min2 2 2 422 2 2 2 22r r r rdr EUr rdr krE rτ µµ µ µ= −− = −+ −∫ ∫A A (8) or, using (5), 2 max 2 min max min2 2 12 2 1 122 22 22 2sin2r r r rdx kxE x Ek r k kEµτ µ µ µ−= −+ −  − =−   −∫A A (9) Using (9) to substitute for τ in (7), we have 2EU= (10) Now, 2ETE U=−= (11) The virial theorem states: 1 2nTU+= when Uk1nr+= (12) In our case n = 1, therefore, 2ETU= = (13) 8-8. The general expression for θ (r) is [see Eq. (8.17)] ()()2 2 222rd r r EUrθ µµ=  −−  ∫A A (1) where 22 rdr k r =− =−∫Uk in the present case. Substituting 2xr= and dx = 2r dr into (1), we have 242 CHAPTER 8 () 2 221 2 21dxr kExx xθ µµ= + −∫ AA (2) Using Eq. (E.10b), Appendix E, 1 212sin 4dx bx c c xa x b x c x b a c− 2  +=  − ++ −    ∫ (3) and expressing again in terms of r, we find ()2 2 1 022 2 4211sin2Er r Ekrµ θ θ µµ−−  = + A AA+ (4) or, ()2 0 22 221n 2 11E r kk EEµθθ µµ1−= − ++A AA2si (5) In order to interpret this result, we set 2 2 21k E Eεµ αµ +≡ ′    ≡′ A A (6) and specifying 0 4 θ π= , (5) becomes 21c os2rαε θ′=+ ′ (7) or, ( )22 2 2cos sin rr α εθ =+ −′′ θ (8) Rewriting (8) in x-y coordinates, we find ( )22 22xy xy αε=++ −′′ (9) or, 2 2 1 11y x α α ε ε=+′ ′ + − ′ ′ (10) Since a′ > 0, ε′ > 1 from the definition, (10) is equivalent to CENTRAL-FORCE MOTION 243 2 2 1 11y x αα ε ε=+′ ′ +−′ ′ (11) which is the equation of a hyperbola. 8-9. (a) By the virial theorem, 2U=−T for a circular orbit. The firing of the rocket doesn’t change U, so f i UU= But ( )22 122f i Tm v v =+ =T So 20fi i i iET UU U= += − += 0f iE E= The firing of the rocket doesn’t change the angula r momentum since it fires in a radial direction. 1f i=A A (b) E = 0 means the orbit is parabolic. The satellite will be lost. () () 0esGM mEr Urr== − ()esGM mTr E Ur=− = () ()22 2222esGM mVr Urrrrµ µ=+ = − +AA Behavior of V(r) is determined by 222 for small for large esrr GM m r rµ   −A 244 CHAPTER 8 Energy V(r) U(r)T(r) E(r) = 0 r0 Minimum in V(r) is found by setting 0dV dr= at 0rr= 2 23 000esGM m rr µ=− +A 2 0 esrGM mµ=−A 8-10. For circular motion 22 1 2ee Tm ω = r se eGM mUr=− We can get 2ω by equating the gravitational force to the centripetal force 2 2se ee eGM mmrrω= or 2 3s eGM rω= So 2 311 22se s ee eeGM m GMTm rrr=⋅ = = −2U 1 2ETU U=+ = If the sun’s mass suddenly goes to 1 2 its original value, T remains unchanged but U is halved. CENTRAL-FORCE MOTION 245 11 1022 2UT U U U =+ = + = − + =′′′ET T or he energy is 0, so the orbit is a parabola. For a parabolic bic, the earth will escape the solar system. 8-11. For central-force motion the equation of orbit is [Eq. (8.21)] ()22 2211 drFrdr rµ θ+= − A (1) force centerra θ In our case the equation of orbit is 2c o s ra θ = (2) Therefore, (1) becomes () () ()22112 24cos cos cos22dFrad aµ 2a 11θ θθ−− += − Aθ (3) But we have ()21 22 2 3sincoscos 2s i n 1 cos cosdd ddθθθθ θ θ θ θ−  =  =+ (4) Therefore, we have ()23 2 322s i n 8 11coscos cos cosaFrθµθθθ θ++ = −A (5) or, ()22 3528 8c osaFrarµθ µ=− =−A2 51A (6) so that ()5kFrr=− (7) 246 CHAPTER 8 8-12. βrere The orbit of the comet is a parabola ( ε = 1), so that the equation of the orbit is 1c o srαθ =+ (1) We choose to measure θ from perihelion; hence ( )0E rrθ β == (2) Therefore, 2 2Erkα βµ==A (3) Since the total energy is zero (the orbit is parabolic) and the potential energy is Uk=− r, the time spent within the orbit of the Earth is ()2 22 2 2 2 22 2 3E E E E E Er r r r E r E E rdrT k rr rd r k rr rrrrkβ βµµ µ β β µβΒ= − = − −−  =− −   ∫ ∫A (4) from which ()32 2221 13E Trkµβ β =+ −   (5) Now, the period and the radius of the Earth are related by 2 24E E rkπµτ=3 E′ (6) or, CENTRAL-FORCE MOTION 247 32 2E E Ekrτ µ π′= (7) Substituting (7) into (5), we find ()2221 132E EkTkτ µβ βµπ′=+ − (8) where s kG M µ = and sE kG M µ =′ . Therefore, () ()121 1 23E T β βτπ=− + (9) where 1 yearEτ= . Now, Mercury 0.387E rrβ== . Therefore, () ()12 1 0.387 1 2 0.387 365 days3π+× × T=− so that 76 daysT= (10) 8-13. Setting 1≡u we can write the force as r 2 23k3F ku urrλλ =− − =− − (1) Then, the equation of orbit becomes [cf. Eq. (8.20)] ( )2 2 22 21 duuk uduµλθ+= − − −A3u (2) from which 2 221du kud2µλµ θ+− =AA (3) or, 2 222 211 1du kudµλ µ µλ θ  0 +−−  − AA A= (4) If we make the change of variable, 2 21 1kvuµ µλ=− −A A (5) we have 2 221dvvdµλ θ0 +−A= (6) 248 CHAPTER 8 or, 2 2 20dvvdβθ+ = (7) where 2 21 βµ λ =− A. This equation gives different so lutions according to the value of λ. Let us consider the following three cases: i) 2λ µ <A : For this case and the solution of (7) is 20 β> ( ) cos vA βθδ = − By proper choice of the position θ = 0, the integration constant δ can be made to equal zero. Therefore, we can write 21coskArµβθµλ=+−A (9) When β = 1 ( λ = 0), this equation describes a conic sect ion. Since we do not know the value of the constant A, we need to use what we have learned from Kepler’s problem to describe the motion. We know that for λ = 0, ()211c osk rµε θ =+A and that we have an ellipse or circle (0 ≤ ε < 1) when E < 1, a parabola ( ε = 1) when E = 0, and a hyperbola otherwise. It is clea r that for this problem, if E ≥ 0, we will have some sort of parabolic or hyperbolic orbit. An ellipse should result when E < 0, this being the only bound orbit. When β ≠ 1, the orbit, whatever it is, precesses. This is most easily seen in the case of the ellipse, where the two turning points do not have an angular separation of π. One may obtain most constants of integration (in particular A) by using Equation (8.17) as a starting point, a more formal approach that confirms the statements made here. ii) 2λ µ =A For this case and (3) becomes 20 β= 2 2du k d2µ θ=A (10) so that 2 21 2kurµθθ A B ==+ +A (11) from which we see that r continuously decreases as θ increases; that is, the particle spirals in toward the force center. CENTRAL-FORCE MOTION 249 iii) 2λ µ >A For this case and the solution (7) is 20 β< ( )2cosh vA βθδ = −− (12) δ may be set equal to zero by the prop er choice of the position at which θ = 0. Then, ( )2 21coshkArµβθµλ=− +−A (13) Again, the particle spirals in toward the force center. 8-14. The orbit equation for the central-force field is [see Eq. (8.17)] 2 42 22 2dr rEUdrµ θµ2  =− −     A A (1) But we are given the orbit equation: 2rkθ= (2) from which 2 224drkdθθ= (3) Substituting (2) into (3), we have 2 244dr rkdkθ==kr (4) From (1) and (4), we find th e equation for the potential U: 42 2242rkr E Urµ µ2 =− −   A A (5) from which 22 321 2kUErrµµ=− −AA 21 (6) and ()Fr U r =−∂ ∂ . Therefore, ()2 4361kFrrrµ =− +  A (7) 250 CHAPTER 8 8-15. AB r Pm θb2 Let us denote by v the velocity of the particle when it is infinitely far from P and traveling along the line AB. The angular momentum is 2kmvb vbb== =A 2 (1) where we have used m = 1. Therefore, 22kv b= (2) The total energy E of the particle is equal to the initial kinetic energy: 2 41 24kEvb== (3) The general orbit equation for a force, ()5Frk r =− , is 22 42242drdr kErrθ=  +−  A A (4) Substituting for A and E from (1) and (3), we have ( )2 44 2 42 2 4 2221 22 2 2 2kd rdbr kk k br b r drb rb r b drb rbθ= +− = −+ = −2 or, 22drdbrbθ=−2− (5) where we have taken the negative square root because r decreases as θ increases (see the diagram). We can now use the integral [see Eq. (E.4b), Appendix E] CENTRAL-FORCE MOTION 251 1 22 21cothdx ax ax b a b b−=−− ∫ (6) from which we obtain 1 0 2c o t hr bθ θ−= + (7) or, 0coth 2rbθθ− =   (8) Now, coth φ → ∞ as φ → 0, since r → ∞ as θ → 0, we must have 00 θ=. Thus, ( ) coth 2 rb θ = (9) Notice that r is always greater than b (because coth φ → 1 as φ → ∞), so that the denominator in (5) never equals zero nor changes sign. Thus, r always decreases as θ increases. This is, the particle spirals in toward P but never approaches closer than a distance b. 8-16. The total energy of the particle is ETU=+ (1) a principle that by no means pushes the philosoph ical envelope of physical interpretation. The impulse that causes v → v + δv changes the kinetic energy, not the potential energy. We therefore have 2 1 2E T mv mv vδ δδ δ== =  (2) By the virial theorem, for a nearly circular orbit we have 2 1 2Em=− v (3) so that 2Ev Evδ δ=− (4) where we have written – E since E < 0. The major and minor axes of the orbit are given by 2 2kabE Eµ=− = −A (5) Now let us compute the changes in these quantities. For a we have 222kk E EaEE Eaδ δδδ =− = =    − (6) and for b we have 252 CHAPTER 8 31 22 22 2EbE EE EE bδ δδδ µµ µδδ  == + − = −    −−    − AA A A A (7) Easily enough, we can show that vv δ δ=AA and therefore 2 abE ab Evv δ δδ δ== =− (8) 8-17. The equation of the orbit is 1c osrαε θ =+ (1) from which 1c osrα ε θ=+ (2) where 2k α µ =A and 2 221E mkε=+A. Therefore, the radial distance r can vary from the maximum value ( ) 1α ε− to the minimum value ( ) 1α ε+. Now, the angular velocity of the particle is given by 2rωµ=A (3) Thus, the maximum and minimum values of ω become max 2 2 min min 2 2 max1 1r rωµ αµε ωµ αµε==  +   ==    − AA AA (4) Thus, 2 max min1 1nω ε ωε+= =− (5) from which we find 1 1n nε−= + (6) 8-18. Kepler’s second law states that the areal velocity is constant, and this implies that the angular momentum L is conserved. If a body is acted upon by a force and if the angular momentum of the body is not altered, then the force has imparted no torque to the body; thus, CENTRAL-FORCE MOTION 253 the force must have acted only along the line conn ecting the force center and the body. That is, the force is central. Kepler’s first law states that planets move in e lliptical orbits with the sun at one focus. This means the orbit can be described by Eq. (8.41): 1c os w ith 0r1αεθ ε =+ << (1) On the other hand, for central forces, Eq. (8.21) holds: ()22 2211 drFrdr rµ θ+= − A (2) Substituting 1 from (1) into the left-hand side of (2), we find r ()2 21rF raµ=−A (3) which implies, that ()2 2Frrαµ=−A (4) 8-19. The semimajor axis of an orbit is defined as one-half the sum of the two apsidal distances, and [see Eq. (8.44)], so maxrminr []max min 211 22 1 1rr1α αα ε εε+= + =+ −−  (1) This is the same as the semimajo r axis defined by Eq. (8.42). Ther efore, by using Kepler’s Third Law, we can find the semimajor axis of Ceres in astronomical units: 2 2 2 24 4C C CC E EE Ek a k aτπµ τπµ      =      (2) where cskM mc γ= , and 111 csMm µ=+ c Here, sM and are the masses of the sun and Ceres, respectively. Therefore, (2) becomes cm 132 Cs c c Es e EaM m aM mτ τ   +=  +      (3) from which 254 CHAPTER 8 ()13 21333, 4808,0004.6035333, 480 1C Ea a +  = +       (4) so that 2.767C Ea a≅ (5) The period of Jupiter can also be ca lculated using Kepler’s Third Law: 122 3123 2 34 4J J JJ J sE E Es E Eaka Mm Mmaakπµ τ πµ τ  J E   + ==     +        (6) from which ()12 3 333, 480 15.2028333, 480 318.35J Eτ τ+  = +   (7) Therefore, 11.862J Eτ τ≅ (8) The mass of Saturn can also be calculated from Kepler’s Third law, with the result 95.3s em m≅ (9) 8-20. Using Eqs. (8.42) and (8.41) for a and r, we have 4 4 2 01c os 1cos cos1adtrτεθθ θτε+ =    − ∫ (1) From Kepler’s Second Law, we can find the relation between t and θ: ()2 21 21c osdt dA dab abττ αθππ εθ== + (2) since ()212 dA r d θ = . Therefore, (1) becomes ( )(4 222 42 011cos cos 1 cos2 1aadra bπτ) θ θεθτπ ε= −∫θ + (3) It is easily shown that the value of the integral is 2 πε. Therefore, CENTRAL-FORCE MOTION 255 ( )4 2 4211cos 1a ra bθ αε ε= − (4) After substituting a and b in terms of ε and α [see Eqs. (8.42) and (8.43)], we obtain ( )4 522cos 1a rεθ ε= − (5) 8-21. If we denote the total energy and the potential of the family of orbits by E and U(r), we have the relation ()2 2 21 22rU rrµµE + +=A (1) from which ()22 122rE U r rµ=− −  A2µ (2) Here, E and U(r) are same for all orbits, an d the different values of A result from different values of ()212 rµ. For stable circular motion, 0r= , but for all other motions, 0r≠ . Therefore, for non- circular motions, and A is smaller than for the circu lar case. That is, the angular momentum of the circular orbit is the largest among the family. 20r> 8-22. For the given force, ()3Fr kr =− , the potential is ()22kUrr=− (1) and the effective potential is ()2 211 2Vr kr µ =−  A (2) The equation of the orbit is [cf. Eq. (8.20)] ( )2 3 22 2duuduµ θ+= − −Aku (3) or, 2 221du kudµ θ0 +−A= (4) Let us consider the motion for various values of A. 256 CHAPTER 8 i) 2kµ=A : In this case the effective potential V(r) vanishes and the orbit equation is 2 20du dθ= (5) with the solution 1uArθB ==+ (6) and the particle spirals toward the force center. ii) 2kµ>A : In this case the effective potential is positi ve and decreases monotonically with increasing r. For any value of the total energy E, the particle will approach the force center and will undergo a reversal of its motion at ; the particle will then proceed again to an infinite distance. 0rr= V(r) E rr0 Setting 22kµβ−≡ A 1 , (4) becomes 0> 2 2 20duudβθ+ = (7) with the solution ()1cos uArβθδ == − (8) Since the minimum value of u is zero, this solution corresponds to unbounded motion, as expected from the form of the effective potential V(r). iii) 2kµ<A : For this case we set 221 kGµ −≡ >A 0, and the orbit equation becomes 2 2 20duGudθ− = (9) with the solution ()1cosh uArβθδ == − (10) so that the particle spirals in toward the force center. CENTRAL-FORCE MOTION 257 In order to investigate the stability of a circular orbit in a 3r1 force field, we return to Eq. (8.83) and use ()3gr k r µ= . Then, we have () ()2 323 311k xx µρ ρ µ ρ ρ−= −3 x   ++  A  (11) or, ()2 3310 1xk x µµρ ρ+− ⋅ =  +A (12) Since 0rpr== , Eq. (8.87) shows that 2k µ =A . Therefore, (12) reduces to 0 x= (13) so that the perturbation x increases uniformly with the time. The circular orbit is therefore not stable. We can also reach the same conclusion by exam ining the basic criterion for stability, namely, that 2 20 and 0 r rVV rrρ ρ = =∂∂= >∂∂ The first of these relations requires 2k µ =A while the second requires 2kµ> A . Since these requirements cannot be met simultaneously, no stable circular orbits are allowed. 8-23. Start with the equation of the orbit: 1c osrαε θ =+ (1) and take its time derivative 2sin sinr rr2ε εθ θαα µ== A θ (2) Now from Equation (8.45) and (8.43) we have 222 1aabµ πµ ατπ ε=⋅ = − A A (3) so that from (2) max22 1arε πε µα τ ε=⋅= −A (4) as desired. 258 CHAPTER 8 8-24. r(b)(a) θb ra rp a) With the center of the earth as the origin, the equation for the orbit is 1 coxrαε θ =+ (1) Also we know ( )min 1 ra ε = − (2) ( )max 1 ra ε = + 6 min 300 km 6.67 10 mpe rr r== += × rr 6 max 3500 km 9.87 10 mae r == += × ( )6 18.27 10 m2ap ar r=+ =× Substituting (2) gives ε = 0.193. When θ = 0, min1a rε =+ which gives . So the equation of the orbit is 67.96 10 mα=× 67.96 10 m1 0.193 cosrθ×=+ When θ = 90°, 67.96 10 m rα== × T he satellite is 1590 km above the earth. b) b a – rminβθ CENTRAL-FORCE MOTION 259 1 min 1 mintan Using tan 101b ar ba a arθπβ π α αθπ− −=− =−− = =−°− Substituting into (1) gives ; which is 68.27 10 mr=× 1900 km above the earth 8-25. Let us obtain the major axis a by exploiting its relationship to the total energy. In the following, let M be the mass of the Earth and m be the mass of the satellite. 2 1 22p pGMm GMmEm var=− = = (1) where pr and pv are the radius and velocity of the satellite’s orbit at perigee. We can solve for a and use it to determine the radius at apogee by 1 222ap p ppGMra r rrv1−  =− = −       (2) Inserting the values 11 2 26.67 10 N m kgG− −=× ⋅ ⋅ (3) (4) 245.976 10 kg M=× (5) 66.59 10 mpr=× 37.797 10 m spv1−= ×⋅ (6) we obtain , or 288 km above the earth’s surface. We may get the speed at apogee from the conservation of angular momentum, 61.010 6.658 10 maprr =×  aa pp mr v mr v = (7) giving . The period can be found from Kepler’s third law 127,780 km hrav−=⋅ 23 24atGMπ= (8) Substitution of the value of a found from (1) gives τ = 1.49 hours. 260 CHAPTER 8 8-26. r ra rp First, consider a velocity kick applied along the direction of tr avel at an arbitrary place in the orbit. We seek the optimum location to apply the kick. v∆ ()1 2 2 2initial energy 1 2 final energy 1 2E GMmmvr E GMmmv vr= =− = =+ ∆ − We seek to maximize the energy gain EE21−: ( )2 21122EE m v v v−= ∆ + ∆ For a given , this quantity is clearly a maximum when v is a maximum; i.e., at perigee. v∆ Now consider a velocity kick applied at perigee in an arbitrary direction: V∆ ∆v v2v1 The final energy is 2 21 2pGMmmvr= This will be a maximum for a maximum 2v; which clearly occurs when and are along the same direction. 1v ∆v Thus, the most efficient way to change the energy of elliptical orbit (for a single engine thrust) is by firing along the direction of travel at perigee.an CENTRAL-FORCE MOTION 261 8-27. By conservation of angular momentum aa pp pp a amr v mr v rv or vr= = Substituting gives 1608 m/sav= 8-28. Use the conservation of energy for a spacecra ft leaving the surface of the moon with just enough velocity to reach r = ∞: escv ii ff TU T U+ =+ 2 m escGM 1002mmmvr− =+ esc2m mGMvr= where M 22mass of the moon 7.36 10 kgm== × × r 6radius of the moon 1.74 10 mm== Substituting gives esc 2380 m/s v= 8-29. max 0 min 0 , vv v vv v =+= − From conservation of angular momentum we know aa bbmv r mv r = or max max max min min max min min;rvvr vrrv== (1) Also we know ( )min 1 ra e= − (2) ( )max 1 ra e= + (3) Dividing (3) by (2) and setting the result equal to (1) gives 262 CHAPTER 8 () () () ()max max min min min max min max max min 01 1 11 22rv e re v ve v ev v v v ev ve+==− += − += − = 0vev= 8-30. To just escape from Earth, a velocity kick must be applied such that the total energy E is zero. Thus 2 2102eGM mmvr− = (1) where 2 24 11 2 2 6 6velocity after kick 5.98 10 kg 6.67 10 Nm /kg 200 km 200 km 6.37 10 m 6.57 10 me ev M G rr−= =× =× =+ =+ × =× Substituting into (1) gives v . 211.02 km/sec= For a circular orbit, the initial velocity v is given by Eq. (8.51) 1 1 7.79 km/seceGMvr== Thus, to escape from the earth, a velocity ck of 3.23 km/sec must be applied.ki CENTRAL-FORCE MOTION 263 Since E = 0, the trajectory is a parabola. parabolic escape orbit circular orbitEarth 8-31. From the given force, we find ()()324 dF r kkFrdr r r5′== +′ (1) Therefore, the condition of stability becomes [see Eq. (8.93)] () ()( ) ( )2 5 2 422301kkF Fkkρρ ρ ρρ ρρρ+′′ 3+= −+ ′+> (2) or, ( )2 20kk kkρ ρρ−′> +′ (3) Therefore, if , the orbit is stable. 2kkρ>′ 8-32. For this force, we have ()()32 32 2ra ra radF r kkFre edr r ar krera−− −== +′ =+ (1) Therefore, the condition of stability [see Eq. (8.93)] becomes () ()2330r Fr a Fr r r−+ + ′ +=> (2) This condition is satisfied if r < a. 264 CHAPTER 8 8-33. The Lagrangian of the particle subject to a gr avitational force is written in terms of the cylindrical coordinates as ( )22 22 1 2LTU m r r z m g z θ =−= + + −   (1) From the constraint , we have 24 ra=z 2rrza= (2) Therefore, (1) becomes 2 22 2 21124 4mg rLm r r raaθ=++ −  2 (3) Lagrange’s equation for θ is ( )20Ld L dmrdt dtθθ θ∂∂− =− =∂ ∂ (4) This equation shows that the angular momentum of the system is constant (as expected): (5) 2const. mrθ==A Lagrange’s equation for r is 2 22 2142 4mg Ld L m d rrr mr r m rrd t r a a d t aθ ∂∂−= + − − + = ∂∂  20 (6) from which 2 22 2 22142 4 2mg mrrr mr r m r rraa a aθ+−− + − =  20m  (7) After rearranging, th is equation becomes 22 2 22 31144 2mg rmrr raa a m r++ + −A  0= mr (8) For a circular orbit, we must have 0 rr==  or, r = ρ = constant. Then, 2 32mg amρ ρ=A (9) or, 2 2 2mg a4ρ =A (10) Equating this with 22 4m2ρθ =  A , we have 2 242 4 2mgmaρθ= ρ (11) CENTRAL-FORCE MOTION 265 or, 2 2g aθ= (12) Applying a perturbation to the circular orbit, we can write where 1xrxρρ→+  (13) This causes the following changes: 22 332 131rx x r rx rxρ ρ ρ ρ →+   −  →  → →    (14) from which, we have ( ) ( )22 22 20, in lowest order 2, in lowest ordrr x x rr xx xρ ρρ ρ →+ ≅ er →+ ≅     (15) Thus, (8) becomes ()2 2 231142mg xxaa mρρρρ ++ + − −  A 13 0= mx (16) But 2 32mg amρ ρ=A (17) so that (16) becomes 22 23142mgmx x xaa mρ ρ 40 + ++ = A (18) Substituting (17) into (18), we find 2 2214mgmx xaaρ 0 + +  = (19) or, 220 4gxx aaρ+ = + (20) 266 CHAPTER 8 Therefore, the frequency of small oscillations is 02g azω=+ (21) where 2 04zaρ= 8-34. The total energy of the system is ( )22 22 2 1cot cot2r r r m gr Em θ α =+ + +   α (1) or, ( )22 2 2 111c o t c o t22r m r m gr Em α θ =+ + +   α (2) Substituting 2mrθ =  A , we have ( )2 22 211c o t c o t22r m grmrEm α α =+ + +A (3) Therefore, the effective potential is ()2 2cot2Vr m g rmrα =+A (4) At the turning point we have , and (3) becomes a cubic equation in r: 0r= 2 32cot 02mgr Ermα− +=A (5) Energy mgr cot a r2 r1E rV(r) This cubic equation has three roots. If we attempt to find these roots graphically from the intersections of E = const. and ()222cm r m g r ot Vr α =+A , we discover that only two of the roots are real. (The third root is imaginary.) These two roots specify the planes between which the motion takes place. CENTRAL-FORCE MOTION 267 8-35. If we write the radial distance r as ,c o rx nst. ρ ρ =+= (1) then x obeys the oscillatory equation [see Eqs. (8.88) and (8.89)] 2 0 0 xxω+ =  (2) where ()()03ggρω ρρ=+ ′ (3) The time required for the radius vector to go from any maximum value to the succeeding minimum value is 0 2tτ∆= (4) where 0 02πτω= , the period of x. Thus, 0tπ ω∆= (5) The angle through which the particle mo ves during this time interval is 0tπωφωω=∆ = (6) where ω is the angular velocity of the orbital mo tion which we approximate by a circular motion. Now, under the force () () Fr gr µ=− , ω satisfies the equation () ()2Fr g µρω µ ρ =− = (7) Substituting (3) and (7) into (6), we find for the apsidal angle () ()()() ()0 33g ggggρπρ πω π ωφ ρ ρρρρ ρ== = ′++′ (8) Using ()1 nkgrrµ= , we have () ()g n gρ ρ ρ′=− (9) Therefore, (8) becomes 3n φπ= − (10) 268 CHAPTER 8 In order to have the closed orbits, the apsi dal angle must be a rational fraction of 2 π. Thus, n must be 2, 1, 6, n=−− … n = 2 corresponds to the inverse-square-force and n = –1 corresponds to the harmonic oscillator force. 8-36. The radius of a circular orbit in a force field described by ()2rakFrer−=− (1) is determined by equating F(r) to the centrifugal force: 2 2a kermρ 3ρ−=A (2) Hence, the radius ρ of the circular orbit must satisfy the relation 2 aemkρρ−=A (3) Since the orbit in which we are intere sted is almost circular, we write () () [ ] 1 rθ ρδ θ=+ (4) where ()1 δθ for all values of θ. (With this description, the apsides correspond to the maximum and minimum values of δ.) We can express the following quantities in terms of δ by using (4): (111 ur)δρ==− (5a) 22 211 dd dr d2δ θ ρθ=− (5b) () ()(1 2 2 21 1au aFu k ue keaρ ρδ ρδ− −=− ≅− − +) (5c) Then, substitution into Eq. (8.20) yields () (2 221111adm ke) padρδδρθ ρ− −= −Aδ −+ (6) Multiplying by ρ, using (3) and simplifying, (6) reduces to ()2 21dadδρδθ0 +−= (7) CENTRAL-FORCE MOTION 269 This equation obviously has two type s of solution depending on whether aρ is larger than or smaller than 1; we consider only ρ < a. (In fact, there is no stable circular orbit for ρ > a.) For the initial condition, we choose 0 δδ= to be a maximum (i.e., an apside) at θ = 0. Then, we have ( )12 0cos 1 , for a δδ ρ θ ρ=− a< (8) This solution describes an orbit with well-defi ned apsides. The advance of the apsides can be found from (8) by computing for what value of θ is δ again a maximum. Thus, 2 1 aπθ ρ= − (9) The advance of the apside is given by ( )1222 1 1 a θππ ρ−  ∆= − = − −  (10) In the particular case in which ρ  a we obtain, by extending (10), 22 12aρππ ∆≅ − +   (11) so that aπρ∆≅ (12) 8-37. From the equations in Section 8.8 regarding Hohmann transfers: ∆= 12 vv v∆ + ∆ ∆= 12 12 tt vv v v v − + − 2 11 2 1 2 21 22 r kk k kvmr r r mr mr mr r r − + −  ++ 1 2 r  ∆= (1) Substituting ( )( )11 2 2 24 1 2 66.67 10 Nm /kg 5.98 10 kg initial height above center of Earth 2 final height above center of Earth 3 radius of the Earth 6.37 10 me e e ekGMm rr rr r−==× × == == == × gives 1020 m/sv∆ 270 CHAPTER 8 8-38. Substitute the following into Eq. (1) of problem 8-37: ( )( )11 2 2 30 11 1 11 26.67 10 Nm /kg 1.99 10 kg mean Earth-sun distance 1.50 10 m mean Venus-sun distance 1.08 10 mskGMm r−==× × =× r=× The result is . The answer is negative because 5275 m/s v∆= −2rr1<; so the rocket must be fired in the direction opposite to the motion (the satellite must be slowed down). 5275 m/s; opposite to direction of motion.v∆= From Eq. (8.58), the time is given by 32 32 12 2trr mmTakkππ+ ==   (1) Substituting gives 148 daysτ 8-39. We must calculate the quantity 1v∆ for transfers to Venus and Mars. From Eqs. (8.54), (8.53), and (8.51): 1 11 2 11 2 12t vv v r kk mr r r mr∆= − =− + where ( )( )11 3 2 30 9 1 9 26.67 10 m /s -kg 1.99 10 kg mean Earth-sun distance 150 10 m Venus 108mean sun distance 10 mMars 228skGMm r−==× × ×   =− =    r== × Substituting gives Venus Mars2.53 km/sec 2.92 km/secvv ∆ =− ∆= where the negative sign for Venus means the velo city kick is opposite to the Earth’s orbital motion. Th us, a Mars flyby requires a larger than a Venus flyby. v∆ CENTRAL-FORCE MOTION 271 8-40. To crash into the sun, we calculate 1v∆ from Eq. (8.54) with r = mean distance from sun to Earth, and = radius of the sun. Using Eqs. (8.54), (8.53), and (8.51) we have 1 2r ()2 1sun 11 2 12ssGM GM rvrr r r∆= − + Substituting 11 2 2 30 11 1 8 2s u n6.67 10 Nm /kg 1.99 10 kg 1.5 10 m 6.96 10 ms seG M rr rr−=× =× == × == × gives ()1sun26.9 km/sec v∆= − To escape from the solar system, we must overcome the gravitational pull of both the sun and Earth. From conservation of energy (final 0 E =) we have: 2 102se se eGM m GM mmvrr−−+ = Substituting values gives 43500 m/s v= Now () ()escape43500 29700 m/s 13.8 km/si s sevvv GMvr v∆=− =− =− ∆= To les send the waste out of the solar system requires s energy than crashing it into the sun. 8-41. From the equations in Section 8.8 regarding Hohmann transfers 1212 12 ttvv v vv v v∆=∆ +∆ = −+− where 272 CHAPTER 8 12 1 11 2 12;tr kkvvmr r r mr== + 21 1 21 2 22;tr kkvvmr r r mr== + Substituting ( )( )11 2 2 24 65 1 8 26.67 10 Nm /kg 5.98 10 kg 200 km 6.37 10 m 2 10 m mean Earth-moon distance 3.84 10 me ekGMm r−==× × =+ = × + × ==rr × gives 3966 m/sv∆= From Eq. (8.58), the time of transfer is given by 32 32 12 2trr mmTakkτπ+ ==   Substituting gives 429,000 sec. 5 daysτ=  8-42. r=× 11 2 2 24 56 1 2 8 36.67 10 Nm /kg 5.98 10 kg 2 1 0 m 6 . 3 7 1 0 m ? mean Earth-moon distance 3.84 10 meG M r r−=× =× + × = == × We can get from Kepler’s Third Law (with τ = 1 day) 2r 132 7 2 24.225 10 m4eGMrτ π== ×  We know 2 EG Mm =− r CENTRAL-FORCE MOTION 273 So () () ()11 1 1 10 2 9 33.04 10 J2 4.72 10 J 5.19 10 JeGM mErr Er Er=− =− × =− × =− × To place the satellite in a synchronous orbit would require a minimum energy of ()()21Er Er − = 112.57 10 J × 8-43. In a circular orbit, the velocity v of satellite is given by 0 2 0 0 2mv GMm GMvRR R=⇒ = where M is the Earth’s mass. Conservation of energy implies 22 12 22mv mv GMm GMm RR−= −2 Conservation of angular momentum gives 12 2 mRv m Rv = From these equations, we find 14 3GMvR= so the velocity need to be increased by a factor 4 3 to change the orbit. 8-44. The bound motion means that 2 02mvEV= +< where rak r−=−Ve . The orbit of particle moving in this central force potential is given by 274 CHAPTER 8 ()( ) min min2 2 2 22 2/ 22 1 2 2r r r ra rrd rr EVr dr r keErrθ µµ µ µ−= − = +−∫ ∫A A A A In first order of , this is ( / )ra ()22 22 2222 22rdr dr kk k krE r Err a a rrµµθ µ µ+− − − +− ∫∫AA AA≈= Now effectively, this is the orbit of particle of total energy kEa− moving in potential k r−. It is well known that this orbit is given by (see Chapter 8) 1c osrαε θ =+ where2 kαµ=A and 2 221kEkaεµ=+ − A If 0 1 ε<< , the orbit is ellipsoid; if 0 ε=, the orbit is circular. 8-45. a) In equilibrium, for a circ ular orbit of radius r0, 2 0 00 0FFm rmrφφωω=⇒ = b) The angular momentum (which is conserved) of a particle in circular orbit is 2 0 Lm rφω =3 0mFr= The force acting on a particle , which is placed a distance r (r is very close to equilibrium position ) from the center of force is or () ()()3 2 00 3 32 2 00 0 0 32 4 00 033LFmrF Fmr LL Lrr F rr k rrmr mr mrφω=− = − ≈− − − = − − = − − CENTRAL-FORCE MOTION 275 where . So the frequency of oscillation is 2 0 3/kL m r ≡4 2 0 24 003 3 rF kL mm r m rω== = 8-46. In equilibrium circular orbit, 22 2244MvG M G MRRR=⇒ =v where M is the Sun’s mass. The period is 32 7 24 291 0 y rRR R D v GM GMππ π== = ≈ × T where D is the separation distance of 2 stars. 2=R 8-47. In equilibrium circular orbit of 1st star 2 11 1 2 12MvG M M LL= where 2 1 12LMLMM=+ is the distance from 1st star to the common center of mass. The corresponding velocity is 2 21 2 1 2 12()GM L GMvLL M M==+ Finally, the period is 32 8 1 1 12221.2 10 yr. ()LL v GM Mππ== = × +T 276 CHAPTER8 CHAPTER 9 Dynamics o f a System o f Particles 9-1. Put the shell in the z > 0 region, with the base in the x-y plane. By symmetry, 0 xy== . 2 1 2 1222 00 222 00sin sinr rr r rrzrd r dd z rd r ddππ φθ ππ φθρ θθφ ρ θθφ== = ====∫∫∫ ∫∫∫ Using z = r cos θ and doing the integrals gives ( ) ( )44 21 33 213 8rr z rr− = − 9-2. z h xyazh ah =− + ρ By symmetry, 0 xy== . Use cylindrical coordinates ρ, φ, z. 0mass density ρ= 277 278 CHAPTER 9 2 0000 24 00004haha z hah zzd d d zhz ddd zπρ φρ πρ φρρρ ρ φ ρρρφ−+ === −+ =====∫∫∫ ∫∫∫ The center of mass is on the axis 3of the cone from the vertex.4h 9-3. z h xya By symmetry, 0 xy== . From problem 9-2, the center of mass of the cone is at 1 4z h= . From problem 9-1, the center of mass of the hemisphere is at ()213,08za r a r =−= = So the problem reduces to 2 11 111;43z hm a h ρπ == i 3 22 232;83z am a ρ π =− = i ()22 11 22 1 2 12 1 23 42mz mz h azmm h aρρ ρ ρ+−==++ for 12ρ ρ= ()233 42hazah−=+ DYNAMICS OF A SYSTEM OF PARTICLES 279 9-4. θ′ θ/2θ/2y x aa By symmetry, 0y=. If mass lengthσ= then M aσθ= So 2 21x xdmMθ θθ=−′=∫ 2 21xxMθ θθadσθ =−′= ′ ∫ Using M aσθ= and cos xa θ = ′, 2 21cos sin sin22 2s i n2axa d aθ θθ θθθθθ θ θ−  == − ′′ −    =∫ 2sin2 0ax yθ θ= = 9-5. ri – r0 r0rimi 280 CHAPTER 9 th th 1position of the particle mass of the particle total mass constant gravitational fieldi ii mi Mm= = == =∑r g Calculate the torque about 0r () () ( ) ( ) ( )00 10 0 0i ii ii ii ii i iim mm mm mMττ= =− × =− × =×−× =× − =× − ××∑ ∑ ∑ ∑∑ ∑∑ ∑rr F rr g rg rg rg r rg r gg Now if the total torque is zero, we must have 0 iimM= ∑ rr or 01 iimM=∑ rr which is the definition of the center of mass. So 0C M 0 about or center of gravity center of mass.τ== =rr 9-6. Since particle 1 has F = 0, , then 00 0 ==rv10=r . For particle 2 0 20ˆˆ then FFrm==Fx  x Integrating twice with gives 00 0 ==rv 2 0 2ˆ 2Frtm= x 2 0 11 22 CM 12ˆ 4F mmtmm m+==+rrrx DYNAMICS OF A SYSTEM OF PARTICLES 281 2 0 CM 0 CM 0 CMˆ 4 ˆ 2 ˆ 2Ftm Ftm F m= = =rx vx ax 9-7. Oy a a HH x52˚ 52˚ By symmetry 0y= 016H mm= Let 0,1 6Hmm m== m Then 3 11 ii xmM=∑ x ()cos 52 12c o s 5 218 9axm am°=° = 0.068xa= 9-8. By symmetry, 0x=. Also, by symmetry, we may integrate over the x > 0 half of the triangle to get y. σ = mass/area 22 00 22 0032aax xy aax xyyd yd xay dy dxσ σ− == − ====∫∫ ∫∫ 32ay= 282 CHAPTER 9 9-9. POW! 45˚z ym1m2 v1vyvz Let the axes be as shown with the projectile in the y-z plane. At the top just before the explosion, the velocity is in the y direction and has magnitude 0 02yvv= . 0 12 00 0 122 22yE mm vEvmm+== =+ where and are the masses of the fragme nts. The initial momentum is 1m2m ()0 12 120, , 0iEpm mmm =+  +  The final momentum is 12 fp pp=+ ( )11 1 0,0, pmv= ( ) 22 ,,x y z p mvvv= The conservation of momentum equations are 2 :0xx v orv == 0x pm () ()01 2 2 01 2 21:yy y p Em m m v o r v Em mm+= = + 2 11 2 1 1 or :0z zzmp mv mv v vm=+ = − The energy equation is () ( )22 0 12 0 1 1 2 121 22 22 11 y zEmm E m v m vvmm++ = +++ or ( )22 01 1232 y z Em vm vv=+ + Substituting for yv and v gives 1 DYNAMICS OF A SYSTEM OF PARTICLES 283 () ()01 2 1 2 21 22 zEm m mvmm m−=+ 2 1 1zmvvm=− gives () ()021 1 11 22Emmvmm m−=−+ So travels straight down with speed =1m1v travels in the y-z plane 2m ( )() () () ()12012 22 2 21 2 121 11 124 2tan tanyz z yEmmvv vmm m mmm v vmθ−−+=+ =+ −==+m The mass is the largest it can be when 1m10v=, meaning 22mm1= and the mass ratio is 1 21 2m m= 9-10. x2x1xy B Aθ First, we find the time required to go from A to B by examining the motion. The equation for the y-component of velocity is 0sinyvv g t θ = − (1) At B, v ; thus 0y=0sinBtv g θ = . The shell explodes giving m and horizontal velocities v and (in the original direction). We solve for and using conservation of momentum and energy. 1 2 m1 2v1v2v ( )12 0 1 12 osx 2 :cp mm v m vm v θ += + (2) ()22 2 12 0 1 1 211:c os22mv E mv θ ++ = +2 21 2mv Em (3) Solving for in (2) and substituting into (3 ) gives an equation quadratic in v. The solution is 2v1 284 CHAPTER 9 ()2 10 11 2cosmEvvmm mθ2=±+ (4) and therefore we also must have ()1 20 21 2cosmEvvmm mθ2=+∓ (5) Now we need the positions where m and m land. The time to fall to the ocean is the same as the time it took to go from A to B. Calling the location where the shell explodes x = 0 gives for the positions of and upon landing: 1 2 1m2m 11 22 ;B xv t xv tB = = (6) Thus 0 12 12sinvxx vvgθ−= − (7) Using (4) and (5) and simplifying gives 0 1 12 12 2 1sin 2 v mm E gm m mmθ2xx −= +  +  (8) 9-11. The term in question is bαβ αβ α≠∑∑f For n = 3, this becomes ( )( )( ) 12 13 21 23 31 32 12 21 13 31 23 32+ + + + + =+++++ ffff ff ff ff ff But by Eq. (9.1), each quantity in parentheses is zero. Thus 33 110αβ αβ αβ== ≠= ∑∑ f 9-12. a) 0 0 lnmvv um=+ Assuming , we have 00 v= m1 0100 lns9v=0 8 2.02 m/s; yes, he runs out of gas.v= DYNAMICS OF A SYSTEM OF PARTICLES 285 b) Relative to Stumblebum’s original frame of reference we have: Before throwing tank 98 kg 2.02 m/s→ After throwing the tank we want Stumblebum’s ve locity to be slightly greater than 3 m/s (so that he will catch up to the orbiter). 8 kg 90 kg 3m/s V←→ Conservation of momentum gives ( )( )( )( )( ) 98 kg 2.02 m/s 90 kg 3 m/s 8 kg 9 m/sv v=− = (This velocity is relative to Stumblebum’s origin al reference frame; i.e., before he fires his pressurized tank.) Since Stumblebum is travelin g towards the orbiter at 2.02 m/s, he must throw the tank at v = 9 m/s + 2.02 m/s 11 m/sv= 9-13. From Eq. (9.9), the total force is given by ()e α αβ αα β αβ≠+∑ ∑∑ Ff As shown in Section 9.3, the second term is zero. So the total force is ()e α α∑F It is given that this quantity is zero. Now consider two coordinate systems with origins at 0 and 0 ′ OO′mα rα′rα r0 where is a vector from 0 to 0 ′ 0r αr is the position vector of mα in 0 286 CHAPTER 9 α′r is the position vector of mα in 0′ We see that 0 a α =+ ′ rrr The torque in 0 is given by ()e α α ατ=×∑rF The torque in 0 ′ is ( )() () 0 () () 0 () 0e e ee eαα α αα α α αα αα α ατ τ=×′′ =− × =× − × =− ×∑ ∑ ∑∑ ∑rF rr F rF rF rF But it is given that ()0e α α= ∑F Thus ττ=′ 9-14. Neither Eq. (9.11) or Eq. (9.31) is valid for a system of particles interacting by magnetic forces. The derivations leading to both of these equations assumes the weak statement of Newton’s Third Law [Eq. (9.31) assumes the strong statement of the Third Law also], which is αββ α =− ff That this is not valid for a system with magnet ic interactions can be seen by considering two particles of charge and moving with velocities and : 1q2q1v2v q1q2v2 f21v1 f12 Now ij ii i j q= × fv B where ijB is the magnetic field at due to the motion of iqjq. DYNAMICS OF A SYSTEM OF PARTICLES 287 Since ijf is perpendicular to both and ivijB (which is either in or out of the paper), ijf can only be parallel to jif if and ivjv are parallel, which is not true in general. Thus, equations (9.11) and (9.31) are not valid for ystem of particles with magnetic interactions.a s 9-15. σ = mass/length dpFdt= becomes mg mv mv= + where m is the mass of length x of the rope. So ; mx mxσ σ = = 2 2dvxg x xvdt dv dxxg x vdx dt dvxg x v vdxσσ σ =+ =+ =+ Try a power law solution: 1;nndvva x n a xdx−== Substituting, ()( )12 2 nnxg x a x n a x a x−=+n or ( )221nxg a n x=+ Since this must be true for all x, the exponent and coefficient of x must be the same on both sides of the equation. Thus we have: 1 = 2 n or 1 2n= ()2 21 o r 3gga n a=+ = So 2 3gxv= 288 CHAPTER 9 12 1222 33 3gx g gx dv dv dx dv dt dx dt dx− av  == = =     3ga= TU 00 ( 0 on tii y == = able) 2 2 11 22 3fgL mgLTm vm== = 3 2fLUm ghm g == − So 0;6ifmgLEE== − Energy lost6mgL= 9-16. T1 T2T x The equation of motion for the falling si de of the chain is, from the figure, ( ) ( ) 222bx bxxgρρ−−T = +  (1) From Example 9.2, we have for the energy conservation case ( ) () ()22 22 2 2gb xx xxg gbx bx− =− =+− − (2) Substitution gives us 2 24xTρ= (3) To find the tension on the other side of the bend, change to a moving coordinate system in which the bend is instantaneously at rest. This frame moves downward at a speed 2 ux= with respect to the fixed frame. The change in momentum at the bend is DYNAMICS OF A SYSTEM OF PARTICLES 289 () ()2 2222xp xu u tρρρ ∆= ∆ ⋅ = ∆ = ∆t (4) Equating this with the net force gives 2 122xTTρ+= (5) Using equation (3), we obtain 2 14xTρ= (6) as required. Note that equation (5) holds true for both the free fall and energy conservation cases. 9-17. As the problem states, we need to perform the following integral ()1212 21d εατ ααα−=− ∫ (1) Our choice of ε is 10 for this calculation, and the results are shown in the figure. We plot the natural velocity 4− 2 dd x g bατ = vs. the natural time τ. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.800.511.52 τd dα τ 9-18. Once we have solved Problem 9-17, it beco mes an easy matter to write the expression for the tension (Equation 9.18): ()212 6 21 2T mgα α α+−=− (1) This is plotted vs. the natural time using the solution of Problem 9-17. 290 CHAPTER 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.805101520 τT mg 9-19. released at t = 0x at time tceiling tableb The force that the tabletop exerts on the chain counteracts the force due to gravity, so that we may write the change in momentum of the center of the mass of the chain as dpbg Fdtρ= − (1) We can write out what the momentum is, though: ( ) p bx xρ= − (2) which has a time derivative () (23dpxb x x b g g xdtρρ=− + − = −  ) (3) where we have used and xg= 2 xg= x 3. Setting this last expression equal to (1) gives us F gxρ= (4) Although M. G. Calkin (personal communication) has found that experimentally the time of fall for this problem is consistently less than the va lue one would obtain in the above treatment by about 1.5%, he also finds evidence that suggests the free fall treatment is more valid if the table is energy absorbing. DYNAMICS OF A SYSTEM OF PARTICLES 291 9-20. a – x a + x Let ρ = mass/length The force on the rope is due to gravity ( ) ( ) 2F ax g axg xgρ ρ ρ=+− − = 2dp dv dvmadt dt dtρ == So dpFdt= becomes dvxg adt= Now dv dv dx dvvdt dx dt dx=⋅= So dvxg avdx= or gvdv xdxa= Integrating yields 221 22gvxac = + Since v = 0 when x = 0, c = 0. Thus 22gvxa= When the rope clears the nail, x = a. Thus vg a = 292 CHAPTER 9 9-21. Let us call the length of rope hanging over the edge of the table, and the total length of the rope. The equation of motion is x L 2 2mgx gx dxmxLd t=⇒ = L Let us look for solution of the form ttxA e B eω ω−=+ Putting this into equation of motion, we find g Lω= Initial conditions are ; (0 ) 0 0.3 mtxx===(0 ) 0m / stv== . From these we find 02 ABx== . Finally 0cosh ( ) xx t ω = . When , the corresponding time is xL= 1 01cosh 0.59 s.Ltx ω−== 9-22. Let us denote (see figure) m and 2 m mass of neutron and deuteron respectively 0v velocity of deuteron before collision 1v and v velocity of neutron and deuteron, respectively after collision 2 2m2m ν0ν2 ν1m αψ a) Conservation of energy: 2 22 2 2 0 12 1 02 222 2mv mv mv v vv =+ ⇒ = +2 2 22 Conservation of momentum is 2222 2 02 1 2 0 1 0 2 4 4 8 c os mv mv v v v v v v ψ =+ ⇒+ = +GG G Solving these equations, we obtain 2 sets of solutions 22 0 126 4 cos 2 cos 4 cos 33vψψ ψ+ v=− − 22 00 22 cos 4 cos 3 3vvvψψ+−=2 0v DYNAMICS OF A SYSTEM OF PARTICLES 293 or numerically 15.18 km/sv=214.44 km/s v= and 119.79 km/sv=25.12 km/s v= b) Let us call α the lab scattering angle of the neutron, then from the sine theorem we have 12 2 12sin 2 sinsin sinmv mv v vα ψψα=⇒ = 74.84 and 5.16α α ⇒= ° = ° c) From a) we see that 22 02 0244cos8vv v vvψ2 1 + −= 22 0 02 max 0226 330 3082vv vvψψ+⇒ ≤ ⇒ = =≥ ° 9-23. Conservation of momentum requires fv to be in the same direction as u (component of 1 fv ⊥ to must be zero). 1u 11 ipm u= ( )12 ffp mm v=+ 1 1 12if fmppvmm=→=+u The fraction of original kinetic energy lost is () ()22 2 11 11 1 2 2 12 2 11 2 1 1 12 1 2 1211 22 1 2if imumu m m KK mm Kmu mmmm m m mm−+ − += −+= =+ 294 CHAPTER 9 9-24. ω0 bv0x ωv maO θ θ The energy of the system is, of course, conserved, and so we have the following relation involving the instantaneous velocity of the particle: 2 011 22mv mv =2 (1) The angular momentum about the center of the cylinder is not conserved since the tension in the string causes a torque. Note that although th e velocity of the particle has both radial and angular components, there is only one inde pendent variable, which we chose to be θ. Here ωθ= is the angular velocity of the particle about the point of contact, which also happens to be the rate at which the point of contact is rotating about the center of the cylinder. Hence we may write ( )00 0 ; vb vb a ω ωθ = =− (2) From (1) and (2), we can solve for the angu lar velocity after turning through an angle θ 0 1a bωω θ= − (3) The tension will then be (look at the point of contact) ( )2 0 Tm ba m b ω θω ω =− = (4) 9-25. The best elements are those that will slow do wn the neutrons as much as possible. In a collision between (the neutron) and (moderator atom), we would thus want to minimize (kinetic energy of the neutron after th e collision); or alternatively, maximize T (kinetic energy of the moderator atom after the collision). From Eq. (9.88) 1m2m 1T2 ()2 21 2 2 0 124cosTm m T mmζ = + Since one cannot control the angle ζ, we want to maximize the function ()12 2 12mmf mm= + DYNAMICS OF A SYSTEM OF PARTICLES 295 with respect to . (m = constant) 2m1 ( ) ()22 11 2 12 4 2 120 when mm m dfmmdm mm− == += By evaluating 122 2 2mmdf dm=  12mm= one can show that the equilibri um point is a maximum. Thus, is a maximum for . Back to reactors, one would want elements whose mass is as close as possible to the neutron mass (thus, as light as possible). Naturally, there are many other factors to consider besides mass, but in general, the lower the mass of the moderator, the more energy is lost per collision by the neutrons. 2T 9-26. The internal torque for the system is 11 2 2 2 1 =×+ × Nr f r f (1) where f is the force acting on the first particle due to the second particle. Now 12 21 12=− ff (2) Then, ( ) ( )( ) ( ) ( )( )12 1 2 12 21 21 0 12 12 0rkv kr v=−×  =− × − − −    =− × −Nrr f rr rr rr rr rr  (3) This is not zero in general because ( ) 12−rr and ( ) 12−rr are not necessarily parallel. The internal torque vanishes only if the internal forc e is directed along the line joining two particles. The system is not conservative. 9-27. The equation for conservation of yp in the lab system is (see fig. 9-10c): 11 22 0s in s i mv mv n ψζ = − Thus 11 22sin sinmv mvζ ψ = or 11 22sin sinmT mTζ ψ = 296 CHAPTER 9 9-28. Using the notation from the chapter: 10:,if mT TT T1 = = 22:0 ;if mT T T = = Thus 1 012 001TTTTTo rTT=+ = +2 (1) If we want the kinetic energy loss for to be a maximum, we must minimize 1m1 0T T or, equivalently, maximize 2 0T T. From Eq. (9.88): ()2 21 2 2 0 124cosTm m T mmζ = + To maximize this, ζ = 0 (it can’t = 180°). ()21 2 0 124 Tm m T mm= +2 1 The kinetic energy loss for is T . The fraction of kinetic energy loss is thus 1m0T− 01 1 001TT TT TT2 0T−=− = (from (1)) ()01 12 2 0 12 max4 TT mm T mm−= + ζ = 0 implies ψ = 0, 180° (conservation of ). So the reaction is as follows vp Before: After:m1v1 m2 m1v1 m2v2 11 1 2 :xpm v m v m v2 = + 22 11 1111:222Em vm vm =+2 22v Solving for gives 1v12 1 12mmvvmm−=+ So travels in + x direction 2m DYNAMICS OF A SYSTEM OF PARTICLES 297 m travels in 112 12 direction if direction if xm xm+>  −<m m 9-29. From Eq. (9.69) ()12sintancos mmθψθ=+ From Eq. (9.74) 2 θπ ζ =− Substituting gives ( ) () ()12sin 2tancos 2 mmπζψπζ−=+− or () () ()12sin 2tancos 2 mmζψζ=− 9-30. Before: After:m1v1 m2 m1v1 m2v2 a) ( ) ( ) 0.06 kg 16 m/s cos 15 8 m/s cos 45 1.27 N secyp∆= ° − − °     =⋅ ( ) ( ) 0.06 kg 16 m/s cos 15 8 m/s sin 45 0.09 N secxp∆= ° − °     =− ⋅ The impulse P is the change in momentum. So ( ) 0.09 1.27 N sec=− + ⋅Px y b) dt t = =∆∫PF F So ( ) 91 2 7 N=− +Fxy 298 CHAPTER 9 9-31. From Eq. (9.69) 1 2sintan cosm mθψ θ= − From Eq. (9.74) θπφ=− Substituting gives 1 2sintan cosm mφψ φ= − 9-32. 1 ipm u= 122fp mv mv= + Conservation of momentum gives 11 2 11 22 uv vo r vu v2 =+= − ( )22 2 11 2 22 2 11 1 2 2 2 12 211 22 114422 23Tm u m v m v mu m u u v v mv mu v mv∆= − − =− − + − =−2 2 ()1 12 2 20 implies 2 6 or3dT uuv vdv∆== = ()2 2 20, so this is a maximumdT dv ∆<  1 11 2 23uvu v=− = 1 123uvv== DYNAMICS OF A SYSTEM OF PARTICLES 299 9-33. From Eq. (9.87b) in the text, we have ()222 2 11 2 2 01 12 22 22 22 2 112 1cos sin 1cos sin 2 cos sin 1Tm m Tm mm mm mm m mψψ 2ψ ψψ =+ −  +    =+ − + −      +ψ Substituting 21mm α≡ and cos ψ ≡ y we have ()2 22 2 2 1 012 1 2Tyy yTαα α−1   + − + + − =+  (1) α = 12 α = 4 α = 2 α = 1 π/2 ψπ1 0T T1 0 9-34. Before After 45˚ mm mxm v2v1 u1 θ Cons. of zp: 11 2 cos 45 cos mu mv mv θ =° + (1) Cons. of yp: 12 0s in45 s i mv mv nθ =° − (2) Cons. of energy (elastic collision) 22 11111 222mu mv mv =−2 2 (3) Solve (1) for cos θ : 11 22cosuv vθ−= Solve (2) for sin θ : 1 2sin 2v vθ= 300 CHAPTER 9 Substitute into , simplify, and the result is 22cos sin 1 θθ+= 222 12 1 1 2 uvv u v=−+1 Combining this with (3) gives 2 1122vu=1v We are told v , hence 10≠ 11 2 vu= Substitute into (3) and the result is 21 2 vu= Since v , (2) implies 1v=2 04 5=° 9-35. From the following two expressions for 10TT , 2 11 2 01Eq. (9.82)Tv Tu= ()222 2 11 2 2 1 12cos sin Eq. (9.87b)m Tm mmψψ =± −  + Tm we can find the expression for the final velocity of in the lab system in terms of the scattering angle ψ : 1v1m 2 2 11 2 1 12 1cos sinmu m mm mv ψ ψ   =± −   +   (1) If time is to be constant on a certain surface that is a distance r from the point of collision, we have 10 rv t= (2) Thus, 2 2 11 0 2 12 1cos sinmut m mm mr ψ ψ   =± −   +   (3) This is the equation of the required surf ace. Let us consider the following cases: DYNAMICS OF A SYSTEM OF PARTICLES 301 i) : 21mm= 2 10 10 cos 1 sin cos2utru t ψ ψ  =± − =ψ (4) (The possibility r = 0 is uninteresting.) iI) : 212 mm= 2 10cos 4 sin3utr ψ ψ   =± −  (5) iII) : Rewriting (3) as 2m=∞ 22 11 0 2 121 2 2cos sin 1 1mut m mm m mrψ ψ   =± −    +   (6) and taking the limit , we find 2m→∞ 10 ru t= (7) All three cases yield spherical surfac es, but with the centers displaced: v1t0 –v1t0m2 = 2m1m2 = ∞ m1 = m2 m2 m1O −vt10 3 This result is useful in the design of a certain type of nuclear detector. If a hydrogenous material is placed at 0 then for neutrons incident on the material, we have the case . Therefore, neutrons scattered from the hydrogenou s target will arrive on the surface A with the same time delay between scattering and arrival, independ ent of the scattering angle. Therefore, a coincidence experiment in which the time delay is measured can determine the energies of the incident neutrons. Since the entire surface A can be used, a very efficient detector can be constructed. 1mm=2 9-36. Since the initial kinetic energies of the two particles are equal, we have 22 2 11 22 21111 222mu mu mu α ==2 (1) or, 2 1 2m mα= (2) 302 CHAPTER 9 Now, the kinetic energy of the system is conserved because the collision is elastic . Therefore, 22 2 11 22 11 2211 1 22 2mu mu mu mv == =2 (3) since v . Momentum is also conserved, so we can write 10= ( )11 22 1 2 1 22mu mu m m u mv α += + = (4) Substituting the second equality in (4) into (3), we find 2 2 12 11 2 1 21 2mmmu m umα += 2 (5) or, 2 1 12 21 2mmmmα =+    (6) Using 2 12mm α= , (6) becomes ( )2222α αα=+ (7) solving for α, we obtain 212 ; 3 2 αα=− =∓ 2∓ (8) This gives us ()12 21:032 2 ; 1 2 w i t h :mu mu 0α α+ < =± = − ±− >  (9) 9-37. ( ) ( )361 072 0 7333Impulse 360 10 10360 6 10 6 10 N s3tFd t td t−× = −−= =− =⋅ × − ⋅ ×∫ ∫ ⋅ kg mImpulse 1.44s= Since the initial velocity is zero, fvv=∆ Impulse Pm v=∆= ∆ So DYNAMICS OF A SYSTEM OF PARTICLES 303 kg m1.44s 0.003 kgfv= muzzlem480sv = 9-38. v1′v1 Vθθ − ψ ψ 2 01 1 2 11 11 2 1 2Tm u Tm v=     = (1) Thus, 2 1 2 01Tv Tu=1 (2) Now, from the diagram above, we have ( )11 cos cos vV v ψ θψ =+− ′ (3) Using Eq. (9.68) in the text, this becomes (2 1 1cos cosmvVm) ψ θψ =+ −   (4) Thus, ()22 2 12 22 11 1cosvm V uu mψθ ψ =+ −    (5) Using Eq. (9.84) in the text, 1 11m V um m=+2 (6) Therefore, we find ()() ()22 11 2 01 12coscosTm Tm mmθψψ 2m  −=+  +     (7) If we define 304 CHAPTER 9 ( ) ()12coscosSmmθψψ−≡+ (8) we have ()2 2 11 2 0 12TmST mm= +× (9) as desired. 9-39. θ′θ u vy x As explained in Section 9.8, the component of ve locity parallel to the wall is unchanged. So sinxvu θ = yv is given by cosyy yvv u uεθ== or cosyvu ε θ = Thus 1222 2 2 2sin cos vu u θε θ   =+  1222 2sin cos vu θε θ   =+  sintancosu uθθε θ=′ or 11tan tan θ θε− =′   DYNAMICS OF A SYSTEM OF PARTICLES 305 9-40. Because of the string, is constrained to move in a circle of radius a. Thus, initially, will move straight up (taken to be the y direction). Newton’s rule applies to the velocity component along . The perpendicular component of veloci ty (which is zero) is unchanged. Thus will move in the original direction after the collision. 2m 2m 1u 1m From conservation of yp we have 11 11 22sin sin mu mv mv α α = + (1) From Newton’s rule we have ( )21 1cos 90vv uαε°− −= or 12 sin vv u1 αε = − (2) Substituting (2) into (1) and solving for gives 2v ()11 2 2 121s instraight upsinmu mmεα α+=+v (2) then gives ( )2 11 2 11 2 12sin along sinum m mmαε α− =+u v 9-41. Using 2 01 2yv t g t=− and vv , we can get the velocities before and after the collision: 0g t =− Before: 2 11 11where2ug t h g =− =1t So 1 1122hug g hg=− =− After: 02 2 0 0o r vg t tv=− = g 2 20 2 2 22 00 021 2 1or 22hv t g t vvvggg=− =− = h So 12 2 vg= h 306 CHAPTER 9 Thus 2 21 21 12 2gh vv uu ghε−==− 2 1h hε= lost i f TT T=− 22 11 12 2 11Fraction lost 1if iTT T uv hh h uh− = −−== = −2 1h 21if iTT Tε− =− 9-42. y x 30˚ 5 m/sθ As explained in Section 9.8, the velocity component in the y-direction is unchanged. m5 sin 30 2.5 m/ssyyvu== ° = For the x component we have 0.8m 3m5c os30 5s 2sx xx xv vv u== =° m23sxv= 1173 4.3 m/s2 2.5tan 36 23fv θ−= = °  DYNAMICS OF A SYSTEM OF PARTICLES 307 9-43. αT0v2 v145˚ 4m m m4m Conservation of gives xp 02124 c os 2mT mv mv α =+1 or 01 2122cos4mT mv mvα− = Conservation of yp gives 12104 2mv mv sinα =− or 1 2sin 42v vα= Substituting into si gives 22n cos 1αα+= 22 201 0 1 22 2 221222 32 16mT m v mT mvv vm v+− =+1 1 Simplifying gives 2 10 2 0 1 216 8 8vT m T vvmm=+ − (1) The equation for conservation of energy is ()22 0 0111462 2TTm v −= +2mv 2 2 or (2) 2 0153 1 2 Tm v m v =+ Substituting (1) into (2) gives a quadratic in : 1v 2 10 1 0 15 6 14 0mv T m v T − −= 308 CHAPTER 9 Using the quadratic formula (taking the positive sign since v ) gives: 10> 0 11.19Tvm= Substituting this into the previous expressions for cos α and sin α and dividing gives sintan 1.47cosααα== Thu s , the recoil angle of the helium, is 55.8 . α ° 9-44. v0 x grav impulse 0 0where mass/length , since , 0.Fm g x g F m vm vm v vvvµ µ == = =++ = =   We have ()0dd xmxdt dtv µ µµ == = So the total force is ()2 0 Fxx g vµ µ =+ We want F(x = a) ()2 0 Faa g vµ µ =+ or 2 01vFa gagµ =+     9-45. Since the total number of particles scattered into a unit solid angle must be the same in the lab system as in the CM system [cf. Eq. (9.124) in the text], () () 2s i n 2s i n dd σθπ θ θσ ψ π ψ ψ =⋅ (1) Thus, () ()sin sind dψψσθ σψθθ= (2) DYNAMICS OF A SYSTEM OF PARTICLES 309 The relation between θ and ψ is given by Eq. (9.69), which is sintancos xθψθ=+ (3) where 1 xm m=2. Using this relation, we can eliminate ψ from (2): ()2 22sin 11n 1 12 c o s cos1 1tan sinxx x2siθψ== (4) θ θ ψ θ= ++ ++ + ()( ) ( ) ()2 2 2tan cos cos sincostan cosdx dd d xdd ψ θθ ψψψθψ θ θ++== +θ (5) Since 2 21s1t a nψ coψ=+, (5) becomes ()()22 2 21c os 1c os 1 sin 12 c o s cos1 cosxx dx x xd xθ θ ψ θ θθ θ θ+ +=++ ++ += (6) Substituting (4) and (6) into (2), we find () () ( )3221c os 12 c o sx xxθσθ σψ θ+= ++ (7) 9-46. The change in angle for a particle of mass µ moving in a central-force field is [cf. Eq. (9.121)]. Let ψ = capital θ here. () ( )max min2 2222r rrd r EU rψ µµ∆= −−∫A A (1) b arminθ ψ In the scattering from an impenetrable sphere, is the radius of that sphere. Also, we can see from the figure that minr 2 θπψ=− . For , U = 0. Thus (1) becomes minrr> ()2 222ard r Erψ µ∞∆= −∫A A (2) Substituting 02;bT E T µ0 = = ′ ′ A (3) 310 CHAPTER 9 (2) becomes 2 21dr rrbαψ∞∆= −∫ (4) This integral can be solved by using Eq. (E. 10b), Appendix E: 1 22sin 4 arbψ∞ − − =      (5) Thus, sinb a=ψ (6) Therefore, we can find the relation between θ and b by substituting ( )2 θπψ=− into (6). We have cos2baθ= (7) Now, the differential cross section is given by Eq. (9.120): ()sinbd b dσθθθ= (8) From (7), we have ()21cos sinsin 2 2 2 4aaθθσθθ=×a= (9) Total cross section is given by ()2 44tad σ σθ π= Ω= ⋅ ∫ (10) so that 2 t a σ π= (11) 9-47. The number of recoil particles scattered into unit solid angle in each of the two systems, lab and CM, are the same. Therefore, () () sin sin dd σφφ φ σ ζζ ζ= (1) where φ and ζ are the CM and lab angles, respectively, of the recoil particle. From (1) we can write [cf. Eq. (9.125) in the text] () ()sin sind dσφ ζζ σζ φφ= (2) DYNAMICS OF A SYSTEM OF PARTICLES 311 Now, in general, 2 φζ= [see Eq. (9.74)]. Hence, sin sin 1 sin sin 2 2 cosζζ φζ ζ== (3) and 1 2dd ζ φ= (4) Using (3) and (4) in (2), we have () ()1 4c o sσφ σζ ζ= (5) For , the Rutherford scattering cross section is [Eq. (9.141)] 1mm=2 ()()2 24 01 4s ink Tσθθ=×2 (6) Also for this case, we have [Eqs. (9.71) and (9.75)] 2 2θψ πψζ=    =− (7) Hence, sin sin sin cos22θ πψζ ζ== − =  (8) and since the CM recoil cross section σ (φ) is the same as the CM scattering cross section ( ) σθ, (6) becomes ()2 24 01 4c osk Tσφζ=× (9) Using (5) to express σ (ζ ), we obtain () ()4c o s σζ σφ ζ =× (10) or, ()2 23 01 cosk Tσζζ=× (11) 312 CHAPTER 9 9-48. In the case , the scattering angle ψ for the incident particle measured in the lab system is very small for all energies. We can then anticipate that σ (ψ) will rapidly approach zero as ψ increases. 1mm2 Eq. (9.140) gives the Rutherford cross section in terms of the scattering angle in the CM system: () () ()2 CM 2 4 01 sin 2 4k Tσθθ= ′ (1) From Eq. (9.79) we see that for , 12mm 2 00 12 1mmTTmm m=≅′+2 0T (2) Furthermore, from Eq. (9.69), 2 11 2sintan sin cosm m m mθψ θ θ=≅ + (3) and therefore, since ψ is expected to be small for all cases of interest, 1 22sin tanmm mm1θ ψ ≅≅ ψ (4) Then, 2 1 2cos 1m mθψ =−    (5) and ()2 2 1 21sin 2 1 12m mθψ    =− −       (6) (Notice that 1 ψ, but since m , the quantity 1m2 1mm2 ψ is not necessarily small compared to unity.) With the help of (2) and (6), we can rewrite the CM cross section in terms of ψ as ()2 1 CM 2220 1 21 2 11mk mTm mψ= σψ     −−     (7) According to Eq. (9.129), DYNAMICS OF A SYSTEM OF PARTICLES 313 () ()22 11 22 LAB CM2 1 2cos 1 sin 1s inmm mm m m   +−    = −ψψ σ θ ψσψ (8) We can compute ()LABσ ψ with the help of (7) and the simplifications introduced in the right- hand side of (8) by the fact that 1 ψ: ()22 11 222 1 LAB 22220 11 221 2 11 1mm mmmk mTmm mm +−    ≅   −− −   ψ ψψ σψ (9) and so, ()( )222 12 0 LAB 222 11 222 11 1mk mT mm mm≅  −− −   ψψ σψ (10) This expression shows that the cross section has a second-order divergence at ψ = 0. For values of 2mm1 ψ> , (9) gives complex values for labσ. This result is due to the approximations involved in its derivation, making our result invalid for angles larger than 21mm . 9-49. The differential cross section for Rutherford scattering in the CM system is [cf. Eq. (9.140) in the text] ()2 2401 16sin2k Tσθθ=′ (1) where [cf. Eq. (9.79)] 2 0 12mTmm=′+0T (2) 314 CHAPTER 9 Thus, ()22 12 2402 22 1 24021 16sin2 1116sin2mm k Tm m k Tmσθθ θ  +=     =+    (3) Since 12 1 mm , we expand 2 11 2211 2mm mm+ ≅+ + … (4) Thus, to the first order in 1mm2, we have ()2 1 240211216sin2m k Tmσθθ =+    (5) This result is the same as Eq. (9.140) exce pt for the correction term proportional to 12mm . 9-50. The potential for the given force law is ()22kUrr= (1) First, we make a change of variable, 1z r= . Then, from Eq. (9.123), we can write max 12 2 max 2 00 22 2 0 1 2 max 0 2 0 2 2 01 sin 2 2z kzb mubd z kbzmu bz z kbmu b kbmuθ π− =+ −= −+ = + = +∫ (2) Solving (2) for b = b(θ ), 222 02 4kbmuθθ π θ= − (3) DYNAMICS OF A SYSTEM OF PARTICLES 315 According to Fig. 9-22 and Eq. (9.122), (1 2) θ πθ = − (4) so that b(θ ) can be rewritten as ( θ): () ()2 0 2kbmuπθθ θπθ−= − (5) The differential cross section can now be computed from Eq. (9.120): ()sinbd b dσθθθ= (6) with the result ()() ()2 2 22 02s ink muππθσθ θ πθ θ−= − (7) 9-51. θ φ = π − θ pn In the CM system, whenever the neutron is scattered through the angle θ, the proton recoils at the angle φ = π – θ. Thus, the neutron scattering cross sect ion is equal to the recoil cross section at the corresponding angles: () ()p ndN dN ddθ φ=ΩΩ (1) Thus, () ()pp n pdN dT dN dd Td θ φ=ΩΩ (2) where ppdN dT is the energy distribution of the re coil protons. According to experiment, ppdN dT = const. Since p n mm≅ , pT is expressed in terms of the angle ψ by using Eq. (9.89b): 2 0sinpTT ψ = (3) We also have 2θψ= for the case p n mm≅ . Thus, ()( )2 01sin2s i npdT dTddψφπ φ φ=Ω (4) 316 CHAPTER 9 or, ()2 0 0sin2s i n 2 sin cos2s i n 2 2 4pdT T d dd TT0πφ φπ φ φ φφ π φπ− = Ω   == (5) Therefore, we find for the angular di stribution of the scattered neutron, ()0 4p n pdN dN T dd T θ π=⋅Ω (6) Since ppdN dT = const., ndN d Ω is also constant. That is, the scattering of neutrons by protons is isotropic in the CM system. 9-52. Defining the differential cross section σ (θ ) in the CM system as in Eq. (9.116), the number of particles scattered into the interval from θ to θ + dθ is proportional to () ()( ) sin cos dN d d σθθ θ σ θ ∝= − θ 2 (1) From Eq. (9.87a) and the assumpti on of elastic collisions (i.e., TT01 T =+), we obtain ()(21 2 2 0 1221c o sTm m T mm)θ =− + (2) or, solving for cos θ, 22cosm mTT Tθ−= (3) where ()12 0 2 124 mmm mm= +T is the maximum energy attainable by the recoil particle in the lab system. Then, (1) can be rewritten as T ()22 mdTdNTσθ∝ (4) and consequently, we obtain the desire d result for the energy distribution: () 2dN dTσθ ∝ (5) DYNAMICS OF A SYSTEM OF PARTICLES 317 9-53. 1m=mass of particle1α, mass of 2m=238U 1,uu ′1: velocity of particle α in LAB and CM before collision 1,vv ′1 2 2: velocity of particle in LAB and CM after collision ∝ 2,uu ′: velocity of in LAB and CM before collision 238U 2,vv ′” velocity of in LAB and CM after collision 238U 20 u=, 2 11 1 7.7MeV2muT== 90ψ=° is angle through which particle α is deflected in LAB θ is angle through which particle α and are deflected in CM 238U ζ is recoil angle of in LAB 238U m1 u1v1 v v2′ –vCMm1 m2ζθ 2 a) Conservation of momentum in LAB: 11 22 11 11 22costansinmu mvumv mvvζζζ= = = ⇒ Conservation of energy in LAB: 22 11 11 22 222mu mv mv=+2 From these equations we obtain the recoil scattering angle of 238U tan 21 2144.52mm mm−= ⇒=+° ζζ b) The velocity of CM of system is 11 CM 12muvmm=+GG The velocity of in CM after collision is vv238U' 22C v =−MGGG. From the above figure we can obtain the scattering angle of particle in CM to be 238U 318 CHAPTER 9 22 2 21 2 21sinn 89.04coscmv mm vvmζ ζ−== ⇒ =−taθθ ° In CM, clearly after collision, particle α moves in opposite direction of that of 23. 8U c) The kinetic energy of particle after collision in LAB is 238U 222 2 10 22 2 11 1 1 21 2 1 220.25 MeV22c osmv m mu mTmm m m m ζ== = =++ mv Evidently, conservation of energy is satisfied. d) The impact parameter in CM is given in Section 9.10. 0cot22kbTθ=  ′ where 12 09qqkπε= and ( )2 01 1 21 2u m u =+′′2 2′ Tm is the total energy of system in CM, so 14 12 12 2 01 2 1cot 1.8 10 m42qq mmbmmuθ πε− + == × We note that b is the impact parameter of particle αwith respect to CM, so the impact parameter of particle α with respect to i s 238U( )12 14 21.83 10 m.mm b m− +=× e) In CM system, the orbit equation of particle α is ()1c osrαε θ =+′ where 0 θ= corresponds to rrmin= min1rα ε⇒=′+ is closest distance from particle α to the center of mass, and ( )22 11 0 11 2 14 mub mk qq mπεα′==A and () ()22 11 0 11 2 2 11 2 0 11 12 1412 12 41mub EEmk qq m mubmuqq mπεε πε1′=+ =+ ′=+ ′A But the actual minimum dist ance between particles is 14 12 min min 20.93 10 m.mmrrm− +== × ′ DYNAMICS OF A SYSTEM OF PARTICLES 319 f) Using formula () ()( )222 LAB CM22cos 1 sin 1s inxx xψ ψ σ θ ψ+− = −σψ where 1 2mxm= , 90ψ=° , () ()2 CM 2 401 4sin2k Tσθθ= ′ We find this differential cross section in LAB at ψ = 90°: ( )28 2 LAB 90 3.16 10 m σψ−=° = × g) Since d() s i n d dN Nσθψ ψ= φ we see that the ratio of probability is () ()sin11.1's i nσψ ψ σψ ψ=′ 9-54. Equation 9.152 gives the velocity of the rocket as a function of mass: ()00 00 0ln ln 0 lnmmvv u vmm mpm vm umµ =+ = = == To maximize p, set 0dp dm= 00l ndp mudm m1 = =−   1 00 01mm meo r emm mln−== = To check that we have a maximum, examine 1 022 22 mm edp dp u dm dm m −==−  1 02 2 00 mm edp uedm m −==−<  , so we have a maximum. 1 0mem−= 320 CHAPTER 9 9-55. The velocity equation (9.165) gives us: ()()0lnmvt g t umt =− +    (1) where ()0 mt m t α =− , the burn rate 091 0 m α τ = , the burn time 300 s τ= , and the exhaust velocity u . These equations are good only from t = 0 to t = τ. First, let us check that the rocket does indeed lift off at t = 0: the thrust -14500 m s=⋅ 2 0091 0 s uu m mm ατ−== ⋅ ⋅>0g 135 m. , as required. To find the maximum velocity of th e rocket, we need to check it at the times t = 0 and t = τ, and also check for the presence of any extrema in the region 0 < t < τ. We have v(0) = 0, , and calculate 1( ) 0 m s−=− ⋅ ln 10 vg uττ + = 740 () ()10dv u uggd t mt mtgαα =− + = − >   (2) The inequality follows since ()0 um gm t gα>> 100 m s−⋅. Therefore the maximum velocity occurs at t = τ, where vg . A similar single-stage rocket cannot reach the moon since ln 10 74uτ =− + = () ( )110 4 m s0 ln ln 10final u m m u vt−<= . ⋅  , which is less than escape velocity and independent of fuel burn rate. 9-56. a) Since the rate of change of mass of the droplet is proportional to its cross-sectional area, we have 2 dmkrdtπ= (1) If the density of the droplet is ρ, 3 4 3mrπρ = (2) so that 24dm dm dr drrdt dr dt dtπρ π == =2kr (3) Therefore, 4dr k dt ρ= (4) or, 04krr tρ=+ (5) as required. DYNAMICS OF A SYSTEM OF PARTICLES 321 b) The mass changes with time, so the equation of motion is ()dd v dmFm v m v mdt dt dt= =+= g (6) Using (1) and (2) this becomes 32 44 33dvrk rvdt3rgπ πρπ ρ += (7) or, 3 4dv kvgdt r ρ+ = (8) Using (5) this becomes 03 4 4dv k vgk dtrtρ ρ+ = + (9) If we set 3 4kAρ= and 4kBρ= , this equation becomes 0dv Avgdt r Bt+ =+ (10) and we recognize a standard form fo r a first-order differential equation: () ()dvPtv Qtdt+= (11) in which we identify () 0APtrB t=+; Q(t) = g (12) The solution of (11) is vt()() ()constantPt d t Pt d te e Q d t−   ∫∫=+   ∫ (13) Now, () () ()0 0 3 0ln lnAAP t dt dt r BtrB t B rB t==+ =+∫∫+ (14) since 3A B=. Therefore, (15) (3 0Pdter B∫=+ )t 322 CHAPTER 9 Thus, ()() () () ()33 00 34 00constant 4vt r B t r B t g d t grB t rB t CB− −  =+ + +  =+ + +∫ (16) The constant C can be evaluated by setting ( )0 0 vt v==: ( )4 00 3 01 4gvrrBC =+   (17) so that 3 00 04gCv r rB=−4 (18) We then have () ()()4 3 00 0 3 01 44ggr B t vr rBB rB t4 0 vt =+ + −   + (19) or, () ()() ()4 3 0 3104gvt B t rB Bt =+   (20) where ()3 0r0 means “terms of order and higher.” If is sufficiently small so that we can neglect these terms, we have 3 0r0r ()vt t∝ (21) as required. 9-57. Start from our definition of work: dpW F dx dx v dpdt== =∫ ∫∫ (1) We know that for constant acceleration we must have v = at (zero initial velocity). From Equation (9.152) this means 0at umm e−= (2) We can then compute dp: () ( )0 1at u atd mv d mat ma dt at dm m ae dtu−dp = + = − ==   (3) This makes our expression for the work done on the rocket () ( )0 0tat u rmaWa tuateu−=−∫dt (4) DYNAMICS OF A SYSTEM OF PARTICLES 323 The work done on the exhaust, on the other hand, is given with v → (v – u) and , so that (exhaust dp dm v u→ )− ()2 0 0tat u emaWa tueu−=−∫dt (5) The upper limit on the integrals is the burnout time, which we can take to be the final velocity divided by the acceleration. The total work done by the rocket engines is the sum of these two quantities, so that ( ) ()22 0 0001va vuat u x mau uat e dt m u x e dxu− −=− ∫∫W=− (6) where the obvious substitution wa s made in the last expression. Upon evaluating the integral we find 0vuW m uve muv−== (7) where m is the mass of the rocket after its engines have turned off and v is its final velocity. 9-58. From Eq. (9.165) the velocity is 0lndy mvg tudt m== − + 0lnmdy gt u dtm =− +  ∫∫ Since m dtα=− , dt dm α =− 2 0 1ln2m uyC g t d mm α+= − − ∫ ln 1 lnaadx xxx=+ ∫, so we have 2 0 1ln2m uyC g t mmm α += − − +   Evaluate C using y = 0 when t = 0, 0 mm= 0umCα=− ( ) 0 2 0 1ln2um m m mutm αα−=− − yg ; 0mm t α −= 2 0 1ln2m muyu t g tm α=− − At burnout, , B yy=Btt= 324 CHAPTER 9 2 0 1ln2BB Bm muyu t g tm α=− − After burnout, the equations are 2 001 2yy v t g t=+ − and vv0g t = − Calling the top of the path the final point 0f B f vv== − gt or fBtv g= 22 022BBvv vyygg−= − =2 B g; 00 y= 2 2Bvyg= 9-59. In order to immediately lift off, the thrust mu st be equal in magnitude to the weight of the rocket. From Eq. (9.157): Thru0 stvα= 0velocity of fuel v= So 0vm g α= or 0vm g α = 9-60. The rocket will lift off when the thrust just exceeds the weight of the rocket. Thrustdmuudtα =− = ( ) 0 Weight mg m t g α == − Set thrust = weight and solve for t: ( ) 0 umtαα=− g; 0m utg α= − With , 070000 kg m= 250 kg/s α= , 2500 m/su= , 29.8 m/sg= 25 sect The design problem is that there is too much fuel on board. The rocket sits on the ground burning off fuel until the thrust equals the weight. This is not what happens in an actual launch. A real rocket will lift off as soon as the engines reach full thrust. The time the rocket sits on the ground with the engines on is spent building up to full thrust, not burning off excess fuel. DYNAMICS OF A SYSTEM OF PARTICLES 325 9-61. From Eq. (9.153), the velocity after the first state is: 10 ln vvuk= + After the second stage: 21 0 ln 2 ln vvuk v uk=+= + After the third stage: 32 0 ln 3 ln vvuk v uk= += + After the n stages: 10 ln lnnnv v uk vn uk−= += + 0 lnnvvn u k=+ 9-62. To hover above the surface requires the thrust to counteract the gravitational force of the moon. Thus: 1 6 6dmumdt ud mdtgm−= −=g Integrate from m to 0.8 and t = 0 to T: 0m=0m ( ) 26 2000 m/s 6ln 0.8 ln 0.89.8 m/suTg=− =− 273 secT= 9-63. a) With no air resistance and constant gravity, the problem is simple: 2 01 2mv mgh = (1) giving the maximum height of the object as 2 021 800 km hv g=  . The time it takes to do this is 0 610 s vg  . b) When we add the expression for air resistance, the differential equation that describes the projectile’s ascent is 2 2 112W tdv vm g c A v m gdt vρ Fm       == − − = − + (2) 326 CHAPTER 9 where 122 5 ktW gc Aρ−=.  13 k gms⋅ vm would be the terminal velocity if the object were falling from a sufficient height (using 3m−.⋅ as the density of air). Solu tion of this differential equation gives ()1 0tan tant ttgt vvt vvv     −         =− (3) This gives v = 0 at time ( ) ( )1 0 tan 300 sttvg vvτ−=  . The velocity can in turn be integrated to give the y-coordinate of the projectile on the ascent. The height it reaches is the y-coordinate at time τ: 22 0ln 12t tvvhgv       =+ (4) which is  600 km. c) Changing the acceleration due to gravity from – g to ( ) ( )2 2 ee eeG MRy g R Ry  −+ =−+ changes our differential equation for y to 2 2 e tey RygvR y       =− ++    (5) Using the usual numerical techniques, we find that the projectile reaches a height of  630 km in a flight time of  330 s. d) Now we must replace the ρ in the air resistance equation with ( ) yρ . Given the dependence of on ρ, we may write the differential equation tv ()2 2 0e teyy RygvR yρ ρ       =− ++    (6) where we use log and 5 10( ) 0 11 (5 10 )yρ−=. − × y013 ρ=., with the ρ’s in and y in meters. The projectile then reaches a height of  2500 km in a flight time of  940 s. This is close to the height to which the projectile rises when there is no air resistance, which is  2600 km. 3kg m−⋅ DYNAMICS OF A SYSTEM OF PARTICLES 327 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.511.522.5 (a) (b) (c) (d)t (1000 s)height (1000 km) 9-64. We start with the equation of motion for a rocket influenced by an external force, Eq. (9.160), with including gravity, and later, air resistance. extF a) There is only constant acceleration due to gravity to worry about, so the problem can be solved analytically. From Eq. (9.166), we can obtain the rocket’s height at burnout 2 0 1ln2bb b bm muyu t g tm α=− −  (1) where is the mass of the rocket at burnout and bm ( ) 0 b mm t α=−b. Substitution of the given values gives . After burnout, the rocket travels an additional 250 kmby22bv , where is the rocket velocity at burnout. The final height the rocket ends up being  3700 km, after everything is taken into account. gbv b) The situation, and hence the differential eq uation, becomes more complicated when air resistance is added. Substituting 22ext WFm g c A vρ =− − (with ) into Equation (9.160), we obtain 313 k g mρ−=. ⋅ 2 2WcA v dv ugdt m mρ α=− − (2) We must remember that the mass m is also a function of time, an d we must therefore include it also in the system of equations. To be specific, the system of equations we must use to do this by computer are 2 2Wv y cA v uvgmmmρ α α      =− −      −    (3) These must be integrated from the beginning until the burnout time, and therefore must be integrated with the substitution α = 0. Firstly, we get the velocity and height at burnout to be and . We can numerically integrate to get the second part of the 17000 m sbv−⋅  230 kmby 328 CHAPTER 9 journey, or use the results of Problem 9-63(b) to help us get the additional distance travelled with air resistance, analytically. The tota l height to which the rocket rises is  890 km in a total flight time of  410 s. c) The variation in the acceleration of gravit y is taken into account by substituting ( ) ( )2 2 ee eeG MRy g R Ry  += + 16900 m sbv−⋅  230 kmby for g in the differential equation in part (b). This gives , , with total height  950 km and time-of-flight  460 s. d) Now one simply substitutes the give n expression for the air density, ρ(y) for ρ, into the differential equation from part (c). This gives v18200 m sb−⋅  , , and total height  8900 km with time-of-flight  2900 s. 250 kmby 0 0.5 1 1.5 2 2.5 30246810height (km) (a) (b) (c) (d)t (1000 s) 9-65. Total impulse 8.5 N sP=⋅ Total mass 00.054 kg m= Burn time 1.5 sft= Rocket cross section area 2 424.5 10 m4dSπ−== × Drag coefficient 0.75wc= Drag force 22 4 2 121 0 N2w Dc Sv K v v ρ−== =× where 1.2ρ= is density of air 3 kg/m and v is rocket’s speed Rocket exhaust speed u = 800 m/s a) The total mass of propellant is 0.0106 kgPmu∆= = DYNAMICS OF A SYSTEM OF PARTICLES 329 Since + , we will assume that the rocket’s ma ss is approximately constant in this problem. The equation of motion of rocket is 0 ~2 0 %mm 2 0 odvmm gudtα =− + − Kv (where37.1 10 kg/s fm tα− ∆== × is fuel burn rate) ()0 2 0md vdtum g K v α⇒=−− Using the initial condition at t = 0, v = 0, we find () () 00 2 0() t a n hum g K um g Kmααvt  −−  =    At burn-out, , we find 1.5 sftt== ( ) 114.3 m/sffvv t= = The height accordingly is given by () 0 0 2 0 0() () d l n c o s hku mg mvt t tkmαht   −  ==     ∫ At burn out, ftt=, we find the burn-out height ( ) 95.53 mffhh t = = b) After the burn-out, the equation of motion is 2 0 00 2 0d dddmv vgKvtm g=− − ⇒ =−+mm tKv with solution 0 0() t a nmg K gvt t CKm=− − Using the initial condition at ftt=, () f tv v=, we find the constant C = –1.43 rad, so 0 0() t a n 1 . 4 3mg K gvt tKm=− − and the corresponding height 0 0( ) ( ) d 0.88 ln cos 1.43 ft ff tKg mh vt t h tKmht      =+ + − =+             ∫ 330 CHAPTER 9 When the rocket reaches its maximum height (at ttmax= ) the time t can be found by setting We then find And the maximum height the rocket can reach is max max()vt =0.max 7.52 s. t= max max () 3 34 m hh t= = c) Acceleration in burn-out process is (see v(t) in a)) ()0 00 2 2 0d1()d coshum gv tm ku mgtmatα α−==  −     Evidently, the acceleration is maximum when t = 0 and 2 0 0(0 ) 9 5 . 4 m / smaxum gaa tmα−== = = d) In the fall-down process, the equation of motion is 2 0 0d dmvmg K vt=− , With the initial condition tt , v = 0, we find max= ( ) maxtt≥ ()0 max 0() t a n hmg K gvt t tKm =− −     (v(t) is negative for , because then the rocket falls downward) maxtt≥ The height of the rocket is () max0 max max max 0() l n c o s ht tKg mv t d th ttKmhh     =+ =− −          ∫ To find the total flight-time, we set h = 0 and solve for t. We find total 17.56 s t = e) Putting into the expression of V in part d), we find the speed at ground impact to be totaltt=()t ()0 max 0tanh 49.2 m/sgt otalmg K gtKm=− − =−  vt 9-66. If we take into account the change of the rocket’s mass with time 0 mm t α =− , where α is the fuel burn rate, 37.1 10 kg/sα−=× as calculated in problem 9-65. The equation of motion for the rocket during boost phase is DYNAMICS OF A SYSTEM OF PARTICLES 331 ()2 0 2 0dd dvvmt u K vtK v uαααα−= − ⇒ =−−dt tm Integrating both sides we obtain finally () ()2 0 2 01() 1Ku KuCt m uvtKCt mα αα α α+−= −− where C is a constant. Using the initial condition ( ) 0 at 0 vt t= =, we can find C and the velocity is 2 0 2 011 () 11Ku Kut m uvtKt mα α α∝ ∝−−  =  +−   a) The rocket speed at burn-out is (note t1.5 sf= ). ( ) 131.3 m/sffvv t= = b) The distance the rocket has traveled to the burn-out is ( ) 108.5 mft fhv tdt θ==∫ 9-67. From Equation (9.167) we have ( )2 0 0 2 0ln2f f bo ffgm m m uH mmm αα− m  + + − =−     Using numerical values from Example 9.12 41.42 10 kg/s α=− × 6 02.8 10 kg m=× 60.7 10 kgfm=× 2.600 m/su= we find 97.47 kmboH= . From Equation (9.168) we find 0() lnf o bo fgm m mvum α − =− +  2125 m/sbov = 332 CHAPTER9 CHAPTER 10 Motion in a Noninertial Re ference Frame 10-1. The accelerations which we feel at the surface of the Earth are the following: (1) Gravitational : 2980 cm/sec (2) Due to the Earth’s rotation on its own axis: ( ) ( )( )2 28 2852 rad/day6.4 10 cm86400 sec/day 6.4 10 7.3 10 3.4 cm/secrπω −=× ×   =× ×× =2 (3) Due to the rotation about the sun: ( ) ( )2 21 3 25 13 22 r a d / y e a r1.5 10 cm86400 365 sec/day 7.3 101.5 10 0.6 cm/sec365rπω − =× ×  ×    ×=× × =  10-2. The fixed frame is the ground. y a θx The rotating frame has the origin at the center of the tire and is the frame in which the tire is at rest. From Eqs. (10.24), (10.25): ( )2ff rr = + +× +× ×+ × a r r v  ω ω ω ω aR 333 334 CHAPTER 10 Now we have 0 00cos sin 0f rraa r Va rrθθ =− + = == ==Ri ri v a kk  ω ωj Substituting gives 2 0cos sinfvaaarθθ =− + + −ai j j i (2 0cos sin 1fvarθθ=− + + +ai j )a (1) We want to maximize fa, or alternatively, we maximize 2 fa: 42222 2 2 2 2 2 00 42 22 2 2 002cos cos 2 sin sin 22c os s infva vaa arr va vaarra θ θθ θθ=+ + + + + =+ + +a θ 2 2 2 0 0 22cos 2 cos 0 when tanfd avadr ar vθ θθ θ=− + ==a (Taking a second derivative shows this point to be a maximum.) 2 0 222 4 0n implies cosar v v ar vθθ== +ta and 0 22 4 0sinar ar vθ= + Substituting into (1) 22 0 22 4 22 4 0001far va var ar v ar v    =− + + +   ++ j  ai This may be written as 24 2 0 faa v r=+ +a MOTION IN A NONINERTIAL REFERENCE FRAME 335 θA This is the maximum acceleration. The point which experiences this acceleration is at A: where 0 2ar vtanθ= 10-3. We desire . From Eq. (10.25) we have eff0=F ( )eff 2f r mm m m = − −× −× × − × r r v  ω ω ω ω FFR r0ω The only forces acting are ce ntrifugal and friction, thus 2 smg m rµ ω= , or 2sgrµ ω= 10-4. Given an initial position of (–0.5 R,0) the initial velocity (0,0.5 ωR) will make the puck motionless in the fixed system. In the rotating sy stem, the puck will appear to travel clockwise in a circle of radius 0.5 R. Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the pr oblem, we will forgo such a solution here. 10-5. The effective acceleration in the merry -go-round is given by Equation 10.27: 22 xxyω ω =+  (1) 22 yyxω ω =−  (2) These coupled differential equations must be solved with the initial conditions ()0 0 0.5 m xx≡= − , ()0 00 m yy≡= , and ()()1 0 00 2 ms xyv−= =  0v⋅, since we are given in the problem that the initial velocity is at an angle of 45° to the x-axis. We will vary over some range that we know satisfies the condition that the path cross over . We can start by looking at Figures 10-4e and 10-4f, which indicate that we want . Trial and error can find a trajectory that does loop but doesn’t cross its path at all, such as 0v 1⋅ 0.00(,)xy 0.47 m> s− 0v153 m s−=⋅ . From here, one may continue to solve for different values of v until the wanted crossing is eyeball-suitable. This may be an entirely satisf actory answer, depending on the inclinations of the instructor. An interpolation over several trajec tories would show that an accurate answer to the problem is , which exits the merry-go-round at 3.746 s. The figure shows this solution, which was numerically integrat ed with 200 steps over the time interval. 0 10.512 m s−=⋅0v 336 CHAPTER 10 –0.5 0 0.5 10.500.5 x (m)y (m) 1 –1 –1 10-6. z m rz = f(r) Consider a small mass m on the surface of the water. From Eq. (10.25) ( )eff 2f r mm m r m = − −× −× × − × r v  ω ω ω ω FFR In the rotating frame, the mass is at rest; thus, eff0=F . The force F will consist of gravity and the force due to the pressure gradient, which is normal to the surface in equilibrium. Since , we now have 0fr== =Rv ω ( ) 0p mm= +− × ×gF r ω ω where pF is due to the pressure gradient. Fp mgmω2r θ′θ Since F , the sum of the gravitational and centrifugal forces must also be normal to the surface. eff0= Thus θ′ = θ. 2 tan tanr gωθθ==′ MOTION IN A NONINERTIAL REFERENCE FRAME 337 but tandz drθ= Thus 2 2constant2 The shape is a circular paraboloid.zrgω=+ 10-7. For a spherical Earth, the difference in the gr avitational field strength between the poles and the equator is only the centrifugal term: 2 poles equatorgg R ω −= For and R = 6370 km, this difference is only 3457.3 10 rad sω−=× ⋅1− 2 mm s−⋅. The disagreement with the true result can be explained by the fact that the Earth is really an oblate spheroid, another consequence of rotation. To qualitatively describe this effect, approximate the real Earth as a somewhat smaller sphere with a massive belt about the equator. It can be shown with more detailed analysis that the belt pu lls inward at the poles more than it does at the equator. The next level of analysis for the undaunted is th e “quadrupole” correction to the gravitational potential of the Earth, which is beyond the scope of the text. 10-8. xyz λω Choose the coordinates x, y, z as in the diagram. Then, the velocity of the particle and the rotation frequency of the Earth are expressed as ( ) ()0,0, cos , 0, sinz ω λω λ=   =−v  ω (1) so that the acceleration due to the Coriolis force is ( ) 22 0, c osz ω =− × = −ar  ω ,0λ (2) 338 CHAPTER 10 This acceleration is directed along the y axis. Hence, as the particle moves along the z axis, it will be accelerated along the y axis: 2c o s yz ω λ =−  (3) Now, the equation of motion for the particle along the z axis is 0 zvg t= −  (4) 2 01 2zvt g t=− (5) where v is the initial velocity and is equal to 0 2gh if the highest point the particle can reach is h: 0 2 v= gh c (6) From (3), we have 2c o s yz ω λ =− + (7) but the initial condition ( )0 yz== 0 implies c = 0. Substituting (5) into (7) we find ( )2 0 22 012c o s2 cos 2yv t gt v tωλ ωλ=− −  =− gt (8) Integrating (8) and usin g the initial condition y(t = 0) = 0, we find 22 01cos3yg t ωλ vt =−   (9) From (5), the time the particle strikes the ground ( z = 0) is 0102vg t=−t so that 02vtg= (10) Substituting this value into (9), we have 32 00 0 32 3 0 2841cos3 4cos3vvyg vgg v gωλ ωλ =−     =− (11) If we use (6), (11) becomes MOTION IN A NONINERTIAL REFERENCE FRAME 339 34cos3hygωλ =−8 (12) The negative sign of the displacement show s that the particle is displaced to the west. 10-9. Choosing the same coordinate system as in Example 10.3 (see Fig. 10-9), we see that the lateral deflection of the projectile is in the x direction and that the acceleration is ( )( ) 0 22 sin c osxz y ax v V ω ωλ α == = (1) Integrating this expression twice and using the initial conditions, ()0x 0= and ()0x =0, we obtain ()2 0cos sin xt V t ω α = λ (2) Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have ()2 01sin2ztV t g t α =− (3) from which the time T of impact is obtained by setting z = 0: 02s i nVTgα= (4) Substituting this value for T into (2), we find the lateral deflection at impact to be ()3 2 0 24sin cos sinVxTgωλ α = α (5) 10-10. In the previous problem we assumed the z motion to be unaffected by the Coriolis force. Actually, of course, there is an upward acceleration given by 2xyvω− so that 0 2c os c os z Vgω αλ = −  (1) from which the time of flight is obtained by integrating twice, using th e initial conditions, and then setting z = 0: 0 02s i n 2c os c osVTgVα ω αλ=′− (2) Now, the acceleration in the y direction is () (02 2c os s inyx z ay v V )gtω ωλ α== =− − (3) Integrating twice and using the initial conditions, ()0 0c os yV α = and ()00y =, we have 340 CHAPTER 10 ()32 01cos cos sin cos3t g t Vt Vt0 y ω λω λ α α =− + (4) Substituting (2) into (4), the range R ′ is () ()33 3 3 2 00 0 32 0 00sin cos 4 sin cos 2 cos cos 8 32 2 cos cos 2 cos cosVg V VRgV gV gV coscosω αλ ω αλ αλ ω αλ ωα λ ωα λ=− +′− −− (5) We now expand each of these three terms, retaining quantities up to order ω but neglecting all quantities proportional to 2ω and higher powers of ω. In the first two terms, this amounts to neglecting 02c os c V os ω αλ compared to g in the denominator. But in the third term we must use 2 2 0 00 0 3 2 0 0 22c o s s i n 22cos sin 1 cos cos 21c oscos 4sin cos cosV VV gg Vgg VRgαα ωα αα ωαλ ωαα λλ ≅+    − =+′ (6) where is the range when Coriolis effects are neglected [see Example 2.7]: 0R′ 2 0 02cos sinVRgα α =′ (7) The range difference, , now becomes 0 RRR∆= −′′′ 3 2 0 24 1cos sin cos sin3VRgω3λ αα α− ′ ∆= (8) Substituting for in terms of from (7), we have, finally, 0V0R′ 12 32 02 1cos cot tan3RRgω λα′ − ′ α∆= (9) MOTION IN A NONINERTIAL REFERENCE FRAME 341 10-11. θR sin θd = R θ This problem is most easily done in the fixed frame, not the rotating frame. Here we take the Earth to be fixed in space but rotating about its axis. The missile is fired from the North Pole at some point on the Earth’s surface, a direction th at will always be due south. As the missile travels towards its intended destination, the Eart h will rotate underneath it, thus causing it to miss. This distance is: ∆ = (transverse velocity of Earth at current latitude) × (missile’s time of flight) sin RTω θ =× (1) sindR d vRω =  (2) Note that the actual distance d traveled by the missile (that distance measured in the fixed frame) is less than the flight distance one woul d measure from the Earth. The error this causes in ∆ will be small as long as the miss distance is small. Using R = 6370 km, 57.27 10ω−=× rad⋅, we obtain for the 4800 km, T = 600 s flight a miss distance of 190 km. For a 19300 km flight the missile misses by only 125 km because there isn’t enough Earth to get around, or rather there is less of the Earth to miss. For a fixed velocity, the miss distance actually peaks somewhere around d = 12900 km. 1s− Doing this problem in the rotating frame is tric ky because the missile is constrained to be in a path that lies close to the Earth. Although a perturbative treatment would yield an order of magnitude estimate on the first part, it is enti rely wrong on the second part. Correct treatment in the rotating frame would at mi nimum require numerical methods. 10-12. z Fs r0 xλε 342 CHAPTER 10 Using the formula ( )eff 2f r mm m=− × × − ×Fa r v ω ω ω (1) we try to find the direction of when effFfma (which is the true force) is in the direction of the z axis. Choosing the coordinate system as in the diagram, we can express each of the quantities in (1) as 00 (c os , 0 , sin ) (0,0, ) (0,0, )r fR mm gω λω λ=    =− = =− v r aω (2) Hence, we have cosy Rω λ ×=reω (3) and (1) becomes eff 0 cos 0 sin 0c os 0xy z mg m Rz ω λω ωλ=− − −ee e λ 2 2Fe (4) from which, we have 2 eff 0 sin cos cosz x mg mR mRz ω λλ ω λ =− + + e Fe (5) e Therefore, 2 22 0() s i nc o s () c o sfx fzFm R Fm g m Rωλ λ ω λ =  =− +  (6) The angular deviation is given by 2 22 0() sin costancos ()fx fzF R gR Fω λλεω λ==− (7) Since ε is very small, we can put εε≅. Then, we have 2 22 0sin cos cosR gRω λλεω λ=− (8) It is easily shown that ε is a maximum for 45λ ° . Using , , , the maximum deviation is 86.4 10 cmR=×517.3 10 secω−−=×2980 cm/secg= 1.70.002 rad980ε≅≅ (9) MOTION IN A NONINERTIAL REFERENCE FRAME 343 10-13. ω λε z′z x x′Earth The small parameters which govern the approximat ions that need to be made to find the southerly deflection of a falling particle are: height of fall radius of Earthh Rδ≡= (1) and 2 0centrifugal force purely gravitational forceR gωα≡= (2) The purely gravitational component is define d the same as in Problem 10-12. Note that although both δ and α are small, the product 2 0 hg δα ω= is still of order 2ω and therefore expected to contribute to the final answer. Since the plumb line, which defines our vertical di rection, is not in the same direction as the outward radial from the Earth, we will use two coordinate systems to facilitate our analysis. The unprimed coordinates for the Northe rn Hemisphere-centric will have its x-axis towards the south, its y-axis towards the east, and its z-axis in the direction of the plumb line. The primed coordinates will share both its origin and its y′-axis with its unprimed counterpart, with the z′- and x′-axes rotated to make the z′-axis an outward radial (see figure). The rotation can be described mathematically by the transformation cos sin xx z ε ε =+′′ (3) (4) yy=′ sin cos z xz ε ε =− + ′ ′ (5) where 2 sin cosR gωε λ ≡ λ (6) as found from Problem 10-12. a) The acceleration due to the Coriolis force is given by 2X≡−× ′ av ω (7) Since the angle between ω and the z′-axis is π – λ, (7) is most appropriately calculated in the primed coordinates: 344 CHAPTER 10 2s in xy ω λ =′′  (8) ( ) 2c os s in yz x ω λ =− +′′ ′   λ (9) 2c os zy ω λ =′′  (10) In the unprimed coordinates, the interesting component is ( ) 2s in c osc os sin xy ω λε λ =+  ε (11) At our level approximation this becomes 2s i n xy ω λ   (12) Using the results for and y z, which is correct to order ω (also found from Example 10.3), 222s incos xg tω λ  λ (13) Integrating twice and using the zero th order result for the time-of-fall, 2h =t , we obtain for the deflection g 2 2 2sin cos3Xhdgω λ = λ (14) b) The centrifugal force gives us an acceleration of ( )c≡− × × ′ ar ω ω (15) The component equations are then ( )2sin sin cos xx R z ω λλ =+ + ′′ ′ λ     (16) (17) 2yω=′ y′ ( )2 0 cos sin cos z xR z ωλ λ λ=+ + ′′′  g− (18) where we have included the pure gravitational component of force as well. Now transform to the unprimed coordinates and approximate ( )2 0 sin cos sin xR z g ω λλ +−  ε (19) We can use Problem 10-12 to obtain sin ε to our level of approximation 2 0sin sin cosR gωεελ λ (20) The prompts a cancellation in equation (19), which becomes simply 2sin cos xzω λ  λ (21) Using the zeroth order result for the height, 22 zhg t=− , and for the time-of-fall estimates the deflection due to the centrifugal force 2 2 5sin cos6chdgω λ  λ (22) MOTION IN A NONINERTIAL REFERENCE FRAME 345 c) Variation in gravity causes the acceleration 0 3 gGMgr≡− +ar k (23) where ( ) xy R z=++ +′′ ′ j ri is the vector pointing to the particle from the center of the spherical Earth. Near the surface k ( )2 222 22 rx y z R R R z =++ + +′ ′′  ′ (24) so that (23) becomes, with the help of the binomial theorem, (02ggxy zR−+ − ) ′ ′′ ai j  k (25) Transform and get the x component ()0cos 2 singxx zRε ε −+′′  (26) () (0cos sin cos 2 sin cos singxz xzR) ε εε ε ε  =− − + +ε (27) (03s i ngxzR)ε −+  (28) Using (20), 23s in c os xz ω λ  λ (29) where we have neglected the x term. This is just thrice the part (b) result, R 2 2 5sin cos2ghdgω λ  λ (30) Thus the total deflection, correct to order 2ω, is 2 24s incoshdgω λ  λ (31) (The solution to this and the next problem fo llow a personal communication of Paul Stevenson, Rice University.) 10-14. The solution to part (c) of the Problem 10-13 is modified when the particle is dropped down a mineshaft. The force due to the variation of gravity is now 0 0 gggR≡− +ar k (1) As before, we approximate r for near the surface and (1) becomes (0 ggxyzR−+ + ) ′ ′′ ai j  k (2) In the unprimed coordinates, 346 CHAPTER 10 0xxgR− (3) To estimate the order of this term, as we probably should have done in part (c) of Problem 10-13, we can take 22~xh g ω , so that 2~hxhRω×  (4) which is reduced by a factor hR from the accelerations obtained previously. We therefore have no southerly deflection in this order due to the variation of gravity. The Coriolis and centrifugal forces still deflect the particle, however, so that th e total deflection in this approximation is 2 2 3sin cos2hdgω λ  λ (5) 10-15. The Lagrangian in the fixed frame is ()2 1 2ff Lm v U r =− (1) where fv and fr are the velocity and the position, respec tively, in the fixed frame. Assuming we have common origins, we have the following relation f rr= +× vv r ω (2) where v and are measured in the rotating frame. The Lagrangian becomes r r r () () (2 222rr r rmU  =+ ⋅ × + × −vr rω ω )rr Lv (3) The canonical momentum is (rr rLmm )r∂≡ =+ ×∂pvvωr (4) The Hamiltonian is then () (2 2 11 22rr r r r mvUr m ≡⋅− = − − ×vp r ω ) HL (5) H is a constant of the motion since 0 Lt∂∂ = , but H ≠ E since the coordina te transformation equations depend on time (see Section 7.9). We can identify (2 1 2cUm=− × rω )r (6) as the centrifugal potential energy because we may find, with the use of some vector identities, ()2 22 2crmUr ωr  −∇ = ∇ − ⋅ rω (7) ( )2 rr mω  =− ⋅ rr ω ω (8) MOTION IN A NONINERTIAL REFERENCE FRAME 347 ( )r m=−× ⋅ r ω ω (9) which is the centrifugal force. Computing th e derivatives of (3) required in Lagrange’s equations r rdmmdt∂ r = +×∂Lavωv (10) () (rr c rm∂ =∇ × ⋅ − ∇ +∂Lvrrω ) U U (11) ( ) ( )rr mm U=−× − × × − ∇ vrω ω ω (12) The equation of motion we obtain is then ( ) ( ) 2rr m m =− ∇ − × × − ×ar ω ω ωrv amU (13) If we identify F and F , then we do indeed reproduce the equations of motion given in Equation 10.25, withou t the second and third terms. eff rm= U =− ∇ 10-16. The details of the forces involved, save the Coriolis force, and numerical integrations in the solution of this problem are best explained in the solution to Problem 9-63. The only thing we do here is add an acceleration caused by the Coriolis force, and re-work every part of the problem over again. This is conceptually simp le but in practice makes the computation three times more difficult, since we now also must include the transverse coordinates in our integrations. The acceleration we add is ( ) 2s in s in c os c oscy x z y vv v vω λλ λ  =− + +j λk ai (1) where we have chosen the usual coordinates as shown in Figure 10-9 of the text. a) Our acceleration is C g=−+ak a (2) As a check, we find that the height reached is  1800 km, in good agreement with the result of Problem 9-63(a). The deflection at this height is found to be  77 km, to the west. b) This is mildly tricky. The correct treatment says that the equation of motion with air resistance is (cf. equation (2) of Problem 9-63 solution) 2 C tvgv=−+ +ak v a (3) The deflection is calculated to be  8.9 km. c) Adding the vaiation due to grav ity gives us a deflection of  10 km. d) Adding the variation of air dens ity gives us a deflection of  160 km. 348 CHAPTER 10 Of general note is that the deflection in a ll cases was essentially westward. The usual small deflection to the north did not contribute signific antly to the total transverse deflection at this precision. All of the heights obtained agr eed well with the answers from Problem 9-63. Inclusion of the centrifugal force also does not change the deflections to a significant degree at our precision. 10-17. Due to the centrifugal force, the water surface of the lake is not exactly perpendicular to the Earth’s radius (see figure). mg CB ββWater surfaceTangent to Earth surface α A The length BC is (using cosine theorem) 22()2 c o s α =+ −AC mg ACmgBC where AC is the centrifugal force 2cos ω α =AC m R with α = 47° and Earth’s radius , 6400 kmR≅ The angle β that the water surface is deviated from the direction tangential to the Earth’s surface is 5 sinsin 4.3 10sin sinαβαβ−=⇒ = = ×BC AC AC BC So the distance the lake falls at its center is sinβ =hr where r = 162 km is the lake’s radius. So finally we find h = 7 m. MOTION IN A NONINERTIAL REFERENCE FRAME 349 10-18. Let us choose the coordinate system Oxyz as shown in the figure. O νxνy νβα xy The projectile’s velocity is vv where β = 37° 0 0cos sin 00β β   == −   Gx yvv v gt The Earth’s angular velocity is cos sin 0ω α ω ωα− =−G   where α = 50° So the Coriolis acceleration is ( ) ( ) 00 2 2 cos sin 2 sin cosωω β α β ω α =× = − + −G GG c z v v g te av The velocity generated by Coriolis force is ()2 0 02 cos sin sin cos cosω βα β α ω − − ∫t cca dt v t gt α ==v And the distance of deviation due to the Coriolis force is ()3 2 0 0cossin3ω αωα β == − − − ∫t ccgtdtv tzv The flight time of the projectile is 02s i n 2β=vt . If we put this into , we find the deviation distance due to Coriolis force to be cz ~ 260 mcz 350 CHAPTER 10 10-19. The Coriolis force acting on the car is 22 sin ω ωα =× ⇒ =GG GG ccFmv Fm v where α = 65°, m = 1300 kg, v = 100 km/hr. So 4.76 N.cF=G 10-20. Given the Earth’s mass, M , the magnitude of the gravitational field vector at the poles is 245.976 10 kg=× 2 29.866 m/s ==pole poleGMgR The magnitude of the gravitational field vector at the equator is 22 e 2R 9.768 m/sω =− =eq q eqGMgR where ω is the angular velocity of the Earth about itself. If one use the book’s formula, we have at the poles 2( 90 ) 9.832 m/sλ=° =g and g at the equator 2( 0 ) 9.780 m/sλ=° = 10-21. The Coriolis acceleration acting on flowing water is 22 sin ω ωα =× ⇒ =G GG G ccav av Due to this force, the water is higher on the west bank. As in problem 10-17, the angle β that the water surface is deviated from the dire ction tangential to Earth’s surface is 5 22 2 2 222s i n2.5 10 4s inωα ωαsinβ−== = × ++ ∝c ca v ga g v The difference in heights of the two banks is 3sin 1.2 10 mβ−∆= = × Ah where A m is the river’s width. 47= 10-22. The Coriolis acceleration is 2cav ω = ×G GG. This accelerationcaG pushes lead bullets eastward with the magnitude 2 coscos 2ω αωG c gt α ==av , where α = 42°. The velocity generated by the Coriolis force is MOTION IN A NONINERTIAL REFERENCE FRAME 351 2() c o s ω α ==∫cvt a d t g t and the deviation distance is 3 () c o s3ω α ∆= =∫ccgtxv t dt The falling time of the bullet is 2=ht . So finally g 3 3 8cos 2.26 10 m3ωα−∆= = ×chxg 352 CHAPTER 10 CHAPTER 11 Dynamics of Rigid Bodies 11-1. The calculation will be simplified if we use spherical coordinates: sin cos sin sin cosxr yr zrθ φ θ φ θ=    = =  (1) z y x Using the definition of the moment of inertia, ()2 ij ij k i j kIr x xx ρδ dv =−   ∑ ∫ (2) we have ( ) ( ) ()22 33 22 2 2cos cosIr zdv rr r d r d dρ ρ θθ=− =−∫ ∫φ (3) or, ( )()12 42 33 01 0 51c o s c o s 4253R Ir dr d R+ −=− =⋅∫∫ ∫π d ρ θθ πρφ (4) 353 354 CHAPTER 11 The mass of the sphere is 3 4 3M= Rπρ (5) Therefore, 2 335IM R2= (6) Since the sphere is symmetrical around the origin, the diagonal elements of { I} are equal: 2 11 22 332 5III M R === (7) A typical off-diagonal element is () ()12 22 2sin sin cos cosIx ydv rr drdρ d ρ θφφ θ=− =−∫ ∫φ (8) This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vani sh. Therefore, the secular equation is 11 22 3300 00 00II II II− 0 − = − (9) From (9) and (7), we have 2 1232 5III M R=== (10) 11-2. a) Moments of inertia with respect to the x axes: i x3 = x3′ Rh CM x1x1′ x2x2′ It is easily seen that for i ≠ j. Then the diagonal elements become the principal moments I, which we now calculate. 0ijI=iiI i The computation can be simplified by noting that because of the symmetry, . Then, 12III=≠3 ( )222 12 12 3 1 2 222IIxxx d II vρ +== = ++ ∫ (1) DYNAMICS OF RIGID BODIES 355 which, in cylindrical coordinates, can be written as ( )222 1200 022hR zhd d z r z r dπII rρφ == + ∫∫ ∫ (2) where 23 MM VRρπ==h (3) Performing the integratio n and substituting for ρ, we find ( )2 123420II M R h== +2 (4) 3I is given by ( )22 2 31 2 Ix xdv r rdrddz ρ ρ =+= ⋅∫∫φ (5) from which 2 33 10IM R = (6) b) Moments of inertia with respect to the xi′ axes: Because of the symmetry of the body, the center of mass lies on the 3x′ axis. The coordinates of the center of mass are (00,0, )z, where 3 03 4xd v z h dv′ ==∫ ∫ (7) Then, using Eq. (11.49), 2 ij ij ij ij IIM a a a δ  =− −′  (8) In the present case, and 12 0 aa== () 334 a= h, so that 22 11 22 22 2 3393 1 16 20 4 93 1 16 20 4 3 10II M h M R h II M h M R h II M R=− = +′  =− = +′  =−′2 2 11-3. The equation of an ellipsoid is 2 22 3 12 2221x xx abc+ += (1) 356 CHAPTER 11 which can be written in normalized form if we make the following substitutions: 12 3 ,, xa xb xc ξ η ζ = == (2) Then, Eq. (1) reduces to 2221 ξηζ+ += (3) This is the equation of a sphere in the ( ξ,η,ζ) system. If we denote by dv the volume element in the system and by dτ the volume element in the (ξ,η,ζ ) system, we notice that the volume of the ellipsoid is ix 123 4 3Vd vd x dx dx a bc d d d abc d abcξηζ τπ== = ==∫∫ ∫ ∫ (4) because dτ∫ is just the volume of a sphere of unit radius. The rotational inertia with respect to the passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by 3-axisx ( ) (22 31 2 2222MIx xdvV Mabc a b dV=+ =+∫ ∫ ) ξ ητ (5) In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ : ( )22 22 2122 00 0 2 2sin cos sin 21235 4 15Razd v az r d r r dd ad d r d a a== = =× × × =∫∫ ∫∫ ∫ππθθφ φθ θθ π π4r (6) Therefore, () ( )2222 2 4 15ab d a b2 πξη τ+= + ∫ (7) and ( )22 31 5IM a b =+ (8) Since the same analysis can be applied for any axis, the other moments of inertia are DYNAMICS OF RIGID BODIES 357 ( ) ( )22 1 22 21 5 1 5IM b c IM a c=+ =+ (9) 11-4. The linear density of the rod is mρ=AA (1) For the origin at one end of the rod, the moment of inertia is 3 2 033mmIx dx ρ== = ∫A AAAA2 (2) If all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be (3) 2Im a= Equating (2) and (3), we find 3a=A (4) This is the radius of gyration . 11-5. JM a Qz – az a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short interval of time τ so that ()td t = ′′ ∫JF (1) The equations of motion are d dt=pF (2) d dt=×LrF (3) 358 CHAPTER 11 which, for this case, yield ()td t ∆== ′′ ∫pF J (4) ()td t ∆=× = × ′′ ∫Lr F r J (5) Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have ()CM 0 ; Iz aJω== = = ×=− V J L rJ pMωω (6) so that CM=JVM (7) and () 0JzaI ω=−ωω (8) where ()2 025 IM= a. The velocity of any point a on the ball is given by Eq. (11.1): CM α α = +× vV r ω (9) For the point of contact Q, this becomes ( )CM 512Q aJ J za Ma=− −  =−  JvV ω (10) Then, for rolling without slipping, 0Q=v , and we have ( ) 25az a= − (11) so that 7 5z a= (12) b) Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinni ng motion (“English”). For the table not to introduce spurious effects, the rail must be at such a height that the ball will be “reflected” upon collision. Consider the case in which the ball is incident norm ally on the rail, as in the diagram. We have the following relationships: DYNAMICS OF RIGID BODIES 359 y xVCM Before Collision After Collision Linear Momentum CM 0x yp MV p=− = CM 0x yp MV p=+′ =′ Angular Momentum 0 * 0x y zL L L= = = 0 0x y zL LL Ly=′ =−′ =′ * The relation between and depends on whether or not slipping occurs. yLCMV Then, we have CM 22x pp J M V ∆=− = = (13) ( )0 22y LLI J z a ω ∆= − = = − (14) so that ( )0C M 22IM V z a ω= − (15) from which 22 0 CM CM CM22 55I MaazaMV MV Vω ω ω−= = = (16) If we assume that the ball rolls without s lipping before it contacts the rail, then VCM aω= , and we obtain the same result as before, namely, 2 5za−= a (17) or, 7 5z a= (18) Thus, the height of the rail must be at a height of ()25a above the center of the ball. 11-6. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have 2 2(see Problem 11-1)5sIM R = (1) 360 CHAPTER 11 For the hollow sphere, θR sin θ ()2 22 00 43 0 4sin sin 2s in 8 3hId R R Rd Rππ πd σ φθ θ πσ θ θ πσ= = =∫∫ ∫θ or, using , we have 24RMπσ = 2 2 3hIM= R (2) Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is 2 11 22TM y I2θ =+   (3) where y is the measure of the distance along the plane. The potential energy is ( )sin UM g y α =− A (4) where A is the length of the plane and α is the angle of inclination of the plane. Now, y = Rθ, so that the Lagrangian can be expressed as 22 211sin22ILM y y M gyRα =+ + (5) where the constant term in U has been suppressed. The equation of motion for y is obtained in the usual way and we find 2 2sin gMRyMRIα=+ (6) Therefore, the sphere with the smaller moment of inertia (the solid sphere) will have the greater acceleration down the plane. DYNAMICS OF RIGID BODIES 361 11-7. R dx r φθ The force between the force center and the disk is, from the figure k=−Fr (1) Only the component along x does any work, so that the effective force is sinxF kr kx φ =− =− . This corresponds to a potential 22x=Uk . The kinetic energy of the disk is 22 11 3 22 4TM x I M x θ =+ =  2 (2) where we use the result 22 IM R= for a disk and dx = R dθ. Lagrange’s equations give us 302Mx kx+ =  (3) This is simple harmonic motion about x = 0 with an angular frequency of oscillations 2 3k mω= (4) 11-8. x3′ x1′x2′dh w We let be the vertical axis in the fixed system. Th is would be the axis (i.e., the hinge line) of the door if it were properly hung (no self-rotat ion), as indicated in the diagram. The mass of the door is M=ρwhd. 3x′ The moment of inertia of the door around the x3′ axis is 2 3 001 3hwmId hwddwwhd= ′′ ′∫∫2Mw = (1) where the door is considered to be a thin plate, i.e., d  w,h. 362 CHAPTER 11 The initial position of the self-closing door can be expressed as a two-step transformation, starting with the position in the diagram above. The first rotation is around the through an angle θ and the second rotation is around the 1-axisx′ 1-axisx′′ through an angle ψ: w hx3′ x2′ x1′x3′x3″ x2′ x2″ x1′ = x1″θ θx3′x3 = x3″ w3 = w x2′ x2 x2″x1″ x1ψψ The -axes are the fixed-system axes and the are the body system (or rotating) axes which are attached to the door. Here, the Euler angle φ is zero. 1x′ -axesix The rotation matrix that transforms the fixed axes into the body axes ( ) ixx→′i is just Eq. (11.99) with φ = 0 and θ → – θ since this rotation is performed cl ockwise rather than counterclockwise as in the derivation of Eq. (11.99): cos cos sin sin sin sin cos cos sin cos 0s in c osψ θψ θψ ψ θψ θψ θθ−  =− −  λ (2) The procedure is to find the torque acting on th e door expressed in the fixed coordinate system and then to obtain the component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle 3x 0ψ, the rotation is in the direction to decrease ψ. According to Eq. (11.119), 33 3 3IN Iω ψ = =   (3) where 12 0 ω ω== since . 0 φθ== In the body ( system the coordinates of the center of mass of the door are )ix 0 1 2Rw h =  (4) where we have set the thickness equal to zero. In the fixed ( )ix′ system, these coordinates are obtained by applying the inverse transformation 1λ− to R; but 1 tλ λ−=, so that sin 1cos cos sin2sin cos costw wh whψ θ ψθ θ ψθ−    == +′    −+ λ RR (5) Now, the gravitational force acting on the door is downward, and in the coordinate system is ix′ 3 Mg =−′ ′ Fe (6) DYNAMICS OF RIGID BODIES 363 There the torque on the door, expressed in the fixed system, is 12 31sin cos cos sin sin cos cos200 1 cos cos sin 1sin20Mg w w h w h wh Mg w=×′′ ′ ′′ ′  =− − + − +  +  =− NRF ee e ψ θψ θ θψ θ θψ θ ψ (7) so that in the body system we have 22cos cos sin cos cos sin 1sin sin2sin sinwh w Mg h wθ ψθ ψ θ θψ θψ ++ == − − ′′ λψ NN (8) Thus, 31sin sin2NM gw θ ψ −= (9) and substituting this expression into Eq. (3), we have 2 311sin sin23Mgw I Mw θ ψψ= =  ψ − (10) where we have used Eq. (1) for . Solving for 3I ψ, 3sin sin2g wψ θ =− ψ (11) This equation can be integrated by first multiplying by ψ: 213sin sin22 3sin cos2gdt dtw g w== − =∫∫   ψψ ψ θ ψψ θψ (12) where the integration constant is zero since cos 0 ψ= when 0 ψ= . Thus, 3sin cosg wψ θ =± ψ (13) We must choose the negative sign for the radical since 0 ψ< when cos 0 ψ>. Integrating again, from ψ = 90° to ψ = 0°, 0 0 23sin cosTg ddtwπψθ ψ=− ∫∫ (14) 364 CHAPTER 11 where T = 2 sec. Rewriting Eq. (14), 2 03sin cosg dTwπ ψθ ψ= ∫ (15) Using Eq. (E. 27a), Appendix E, we find 2 12 01 4cos3 2 4dπ πψψ−Γ=Γ∫ (16) From Eqs. (E.20) and (E.23), 11 11 0.90644 4 13.6244  Γ= Γ =     Γ= (17) And from Eqs. (E.20) and (E.24), 33 31 0.91944 4 31.2254  Γ= Γ =     Γ= (18) Therefore, 2 03.6242.6221 . 2 2 5 cosdπ ψπ ψ== ∫ (19) Returning to Eq. (15) and solving for sin θ, (2 2sin 2.623w gTθ=× ) (20) Inserting the values for g, w(= 1m), and T(= 2 sec), we find ( )1sin 0.058θ−= or, 3.33θ≅ ° (21) DYNAMICS OF RIGID BODIES 365 11-9. a ORQPC C′ θy x The diagram shows the slab rotated through an angle θ from its equilibrium position. At equilibrium the contact point is Q and after rotation the contact point is P. At equilibrium the position of the center of mass of the slab is C and after rotation the position is C′. Because we are considering on ly small departures from θ = 0, we can write QP R θ≅ (1) Therefore, the coordinates of C ′ are (see enlarged diagram below) = + ′ rO AA C (2) so that sin cos2 cos sin2axR R ayR Rθ θθ θ θθ =+ −    =+ +  (3) C QC′ RθRθ P OA θθ Consequently, 366 CHAPTER 11 cos cos sin2 cos sin2 sin cos sin2 sin cos2axR R R aR ayR R R aRθ θθθ θ θθθ θ θ θθ θ θθ θ θ=+ − + =+ =− + + +  =− +     θ from which 2 22 2 2 4axy R2θθ+= +  (4) The kinetic energy is ( )22 11 22TM x y I2θ =+ +   (5) where I is the moment of inertia of the slab with respect to an axis passing through the center of mass and parallel to the z-axis: ( )22 1 12IMa=+ A (6) Therefore, ()2 11 2Tf θθ =  (7) where ()2 22 14af MR θI θ = ++ (8) The potential energy is ()2 UM g y f θ = =− (9) where ()2 cos sin2afM gR R θ θθθ  =− + +     (10) and where Eq. (3) has been used for y. The Lagrangian is () ()2 11 2Lf f2 θθ =+  θ (11) The Lagrange equation for θ is DYNAMICS OF RIGID BODIES 367 0dL L dt θ θ∂ ∂− =∂ ∂ (12) Now, ()1Lfθθθ∂=∂ () ()11 2 22 2 224dLffdt aMR IM R∂=+∂ =+ + +    θθ θθθ θ θθ θ (13) () ()2 12 221 2 sin cos sin2Lff aMR Mg R R R∂=+ ′′∂   + − − =+    θθ θθ θθθ θ θ θ (14) Combining, we find 2 22 2 2sin cos sin 042aaMR I M R M gR R R θθ θ θ θ θ θ θ ++ + − + − −         = (15) For the case of small oscillations, 2θ θ and 2θ θ, so that Eq. (15) reduces to 220 4aMg R MaIθ−θ + = + (16) The system is stable for oscillations around θ = 0 only if 2 220 4aMg R MaIω−= > + (17) This condition is satisfied if 20 Ra−> , i.e., 2aR> (18) Then, the frequency is ( )2 222 1 41 2aMg R MaMaω−= ++ A (19) Simplifying, we have 368 CHAPTER 11 ( )22122 4agR aω−= +A (20) According to Eqs. (9) and (10), the potential energy is () cos sin2aUM g R R θ θθθ  =+ +     (21) This function has the following forms for 2 Ra> and 2 Ra< : U(θ) U(θ) θ−π/2 π/2Ra>2Ra<2Mg Ra+ 2 θ To verify that a stable condition exists only for 2 Ra> , we need to evaluate 2U2θ ∂∂ at θ = 0: sin cos2UaMg R θ θθθ∂  =− + ∂   (22) 2 2cos cos sin2UaMg R R θ θθ θθ∂  =− + − ∂   (23) and 2 2 02UMg R θθ=∂ =−  ∂a (24) so that 2 20 if 2URθ∂>>∂a (25) 11-10. z m Rθ When the mass m is at one pole, the z component of the angular momentum of the system is 2 2 5zLI M R ω ω == (1) After the mass has moved a distance vt = Rθ along a great circle on the surface of the sphere, the z component of the angular momentum of the system is DYNAMICS OF RIGID BODIES 369 22 2 2sin5zLM R m R θφ =+   (2) where φ is the new angular velocity. Since there is no external force acting on the system, angular momentum must be conserved. Therefore, equating (1) and (2), we have 2 22 22 5 2sin5MR MR mRω φ θ= + (3) Substituting vt Rθ= and integrating over the time interv al during which th e mass travels from one pole to the other, we have ()2 22 202 5 2sin5RtV tMR dt MR mR vt Rπω φ= == +∫ (4) Making the substitutions, () , vt R u dt R v du ≡= (5) we can rewrite (4) as 2 22 20 2 2 02 5 2sin5 2 1s inMRRduvMR mR u Rd u vu= + =+∫ ∫π πω φ ω β (6) where 52mM β≡ and where we have used the fact that the integrand is symmetric around 2 uπ= to write φ as twice the value of the integral ov er half the range. Using the identity (2 1sin 1 cos 22u=− )u (7) we express (6) as 2 02 111c22Rd u vuπωφ ββ=+−∫ os2 (8) or, changing the variable to x = 2u, 0 111c22Rd x vxπ osωφ ββ=+−∫ (9) Now, we can use Eq. (E.15), Appendix E, to obtain 370 CHAPTER 11 () ()1 01t an 2 2tan 11 2 25 2 25x R v RM vM m MTMm−  +=  ++     =+ =+π β ωφ ββ πω ω (10) where TR vπ= is the time required for the particle to move from one pole to the other. If m = 0, (10) becomes ( )0mTφ ω == (11) Therefore, the angle of retardation is ( )() 0mm αφφ= =− (12) or, 2125MTMmαω =−  +   (13) 11-11. a) No sliding: P P22 From energy conservation, we have 2 C.M.11 22 2 2mg mg mv I2ω =+ +AA (1) where v is the velocity of the center of mass when one face strikes the plane; v is related to ω by CM C.M. CM2v=Aω (2) I is the moment of inertia of the cube with respec t to the axis which is perpendicular to one face and passes the center: 2 1 6Im= A (3) Then, (1) becomes DYNAMICS OF RIGID BODIES 371 ()2 2 22 11 12122 2 6 3 2mg mmω2m ω ω −= + =   A AAA (4) from which, we have ()23212gω= −A (5) b) Sliding without friction: In this case there is no external force along the horizontal direction; therefore, the cube slides so that the center of mass falls directly downward along a vertical line. Pθ P While the cube is falling, the distance between the center of mass and the plane is given by cos 2y θ =A (6) Therefore, the velocity of center of mass when one face strikes the plane is 04 411sin22 2 = ==− =− =−A  A π θπy A θθθ ω (7) From conservation of energy, we have 2 2 11 1 1 22 2 2 6 2mg m m2mg ω ω  =+ − +     AAAA (8) from which we have ()212215gω= −A (9) 11-12. According to the definition of th e principal moments of inertia, ( ) ( ) ( )22 22 22 2 22 2jk i k i j jk i iiI I x x dv x x dv xx d v x d v Ix dv+= + + + =++ =+∫∫ ∫∫ ∫ρρ ρρ ρ (1) since 20ixd vρ > ∫ we have 372 CHAPTER 11 j ki III+≥ (2) 11-13. We get the elements of the inertia tensor from Eq. 11.13a: ( ) () () ()22 11 ,2 ,3 22 234 2 21 3Im x x mb m b mb m bαα α α=+ =+ +=∑ 2 b b 2 Likewise and 2 2216 Im=2 3315 Im= () ()12 21 ,1 ,2 2242 2II m x x mb m b m bαα α α== − =− − − =−∑ Likewise 2 13 31IIm b == and II 2 23 32 4mb == Thus the inertia tensor is {}213 2 1 21 6 4 14 1 5Im b−    =−     The principal moments of inertia are gotten by solving 213 2 1 21 6 4 14 1 5mbλ λ λ−− 0 − −=  −  Expanding the determinant gives a cubic equation in λ: 3244 622 2820 0 λλ λ− +−= Solving numerically gives 1 2 310.00 14.35 19.65λ λ λ= = = 2 1 2 2 2 3Thus the principal moments of inertia are 10 14.35 19.65Im b Im b Im b= = = To find the principal axes, we subs titute into (see example 11.3): DYNAMICS OF RIGID BODIES 373 ( ) () ()12 3 12 12 313 2 0 26 4 41 5 0ii i i ii i i ii i i−− + = −+ 1 − + = ++ − =λω ω ω ωλ ω ω ωω λ ω30 For i = 1, we have ( )110 λ= 11 21 31 11 21 31 11 21 3132 0 264 4500ω ωω ωωω ωωω− += − −+= −+= Solving the first for 31ω and substituting into the second gives 11 21ω ω= Substituting into the third now gives 31 21ω ω=− or 11 21 31: : 1:1: 1ω ωω = − So, the principal axis associated with is 1I ()1 3xyz+− Proceeding in the same way gives the other two principal axes: 2 : .81 .29 .52 3 : .14 .77 .63ii=− +− =− ++xy xyz z We note that the principal axes are mu tually orthogonal, as they must be. 11-14. z y x Let the surface of the hemisphere lie in the x-y plane as shown. The mass density is given by 333 2 2 3M MM Vbbρππ== = First, we calculate the center of mass of the hemisphere. By symmetry 374 CHAPTER 11 CM CM CM0 1 vxy z zd vMρ= = =∫ Using spherical coordinates 2(c os , s in ) zrd vrd rdd θ θθ == φ we have ()2 2 3 CM 00 0 4 3sin cos 31 1 3222 4 8b rz ddM bbbπ π φθρφθ θ θ ππ== ==   ==    ∫∫ ∫rdr We now calculate the inertia tensor with respec t to axes passing through the center of mass: z′ = z x′y′ 3 8b By symmetry, . Thus the axes shown are the principal axes. 12 21 13 31 23 32 0 IIIIII====== Also, by symmetry I . We calculate I using Eq. 11.49: 11 22I=11 2 11 113 8IJM v=− (1) where 11J= the moment of inertia with respect to the original axes ( ) ( ) ( ) ( )22 11 22 2 2 2 2 22 42 2 2 3 00 0 2 2 32 0 2sin sin cos sin 3sin sin cos sin2 3sin 2 cos sin10 2 5v v b rJy zdv rr r drdd Mrd r d db Mbd Mbπ π θφ π θρ θφ θ θ θ φ θ φθ φ θπ πθ πθ θ θπ== = ==+ =+    =+ =∫ ∫ ∫∫ ∫ ∫ θ =+ Thus, from (1) 22 11 2229 8 3 56 43 20I I Mb Mb Mb== − =2 Also, from Eq. 11.49 DYNAMICS OF RIGID BODIES 375 ()33 33 33 0 IJM J= −= ( should be obvious physically) 33 33IJ= So ( )22 33 43 2sin5v vIx ydv rd rdd Mρ ρθ θφ=+ ==∫ ∫2b 2 11 22 2 33Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are 83I320 2I5IM b Mb== = 11-15. θP g We suspend the pendulum from a point P which is a distance A from the center of mass. The rotational inertia with respect to an axis through P is (1) 2 0 IM R M=+ A2 where is the radius of gyration about the center of mass. Then, the Lagrangian of the system is 0R (2 1c o s2ILTU M gθ)θ =−= − − A (2) Lagrange’s equation for θ gives sin 0 IM g θθ+ =  A (3) For small oscillation, sin θθ≅. Then, 0Mg Iθθ+ =A (4) or, 22 00g Rθθ+ =+A A (5) 376 CHAPTER 11 from which the period of oscillation is 22 0 22R gπτπω+==A A (6) If we locate another point P′ which is a distance A′ from the center of mass such that the period of oscillation is also τ, we can write 22 2 00RR gg++2′=′A AAA (7) from which . Then, the period must be 2 0R=′AA 2 2gτπ+′=AA A A (8) or, 2gτπ+′=AA (9) This is the same as the period of a simple pendulum of the length A + A′ 1. Using this method, one does not have to measure the rotational inertia of the pendulum used; nor is one faced with the problem of approximating a simple pendulum phys ically. On the other hand, it is necessary to locate the two points for which τ is the same. 11-16. The rotation matrix is ()cos sin 0 sin cos 0 00θθ θθ   =−    λ (1) The moment of inertia tensor transforms according to ()()()()t=′Ι λΙλ (2) That is ()() () () ()11022cos sin 0 cos sin 0 11sin cos 0 0 sin cos 02200 1 0000AB AB IA B AB Cθθ θ θ θθ θ θ+−−      =− − +′       1 DYNAMICS OF RIGID BODIES 377 () () () () () () () ()11 11cos sin sin cos 022 22cos sin 0 11 1 1sin cos 0 cos sin sin cos 022 2 200 100AB AB AB AB AB AB AB AB Cθθ θ θθ θθ θ θ θ θ++ − − ++ −   =− − + + − − + +    θ () () () () () () () () () ()22 22 22 2211cos cos sin sin22 11sin cos22 0 11cos sin 022 11sin sin cos cos 022 0AB AB AB AB AB AB AB AB AB AB C++ − + +  =− − + −    −− −   +− − + +   θθ θ θ θθ θθ θθ θ θ or ()() () () () () () () ()22 2211 1cos sin cos sin 022 2 11 1sin cos cos sin 022 2 00AB AB AB AB AB AB AB AB Cθθ θ θ θθ θ θ++ − − − −  =− − + − + − −′   I (3) If 4 θπ= , sin cos 1 2θθ== . Then, ()00 00 00A B C   =′    I (4) 11-17. x3 x2 x1 378 CHAPTER 11 The plate is assumed to have negligible thickness and the mass per unit area is sρ. Then, the inertia tensor elements are ( )22 11 1 1 2 s xdx ρ=−∫Ir dx ( )22 2 231 2 2 1 2 ssxx d x d x x d x d xAρρ = ≡ ∫∫=+ (1) ( )22 2 22 2 1 2 1 1 2 ss xdxdx x dxdx ρρ=− =∫∫Ir (2) B≡ ( ) ( )22 22 33 31 2 121 ssr xd x d x x xd x d x ρρ=− =+∫∫ 2 BI (3) Defining A and B as above, becomes 33I 33IA=+ (4) Also, ( )12 1 2 1 2 s Ix xdxdx ρ C = −∫≡− (5) ( )21 2 1 1 2 s Ix xdxdx ρ C = −∫=− (6) ( ) 13 1 3 1 2 31 0s Ix xdxdx ρ I = −=∫= (7) ( ) 23 2 3 1 2 32 0s Ix xdxdx ρ I = −=∫= (8) Therefore, the inertia tensor has the form {}0 0 00AC CB AB−    =−   +  I (9) 11-18. The new inertia tensor {}′I is obtained from {}I by a similarity transformation [see Eq. (11.63)]. Since we are concerned only with a rotation around the , the transformation matrix is just 3-axisx φλ, as defined in Eq. (11.91). Then, 1 φ φ−=′IIλλ (1) where 1 t φ φ−=λ λ (2) Therefore, the similarity transformation is =−I cos sin 0 0 cos sin 0 sin cos 0 0 sin cos 0 00 1 0 0 00AC CB ABθθ θ θ θθ θ θ−−        − ′        +     1    Carrying out the operations and simplifying, we find DYNAMICS OF RIGID BODIES 379 {}() ()22 221cos sin 2 sin cos 2 sin 2 02 1cos 2 sin 2 sin sin 2 cos 02 00ACB C B A CB A A C B ABθθ θ θ θ θθ θ θ θ−+ −+ −  =− + − + +′ + I (3) Making the identifications stipulated in the statement of the problem, we see that 11 22 12 21, IA IB II C==′ ′′ ′    == −′′ ′  (4) and 33IA B A B=+= + ′ ′′ (5) Therefore {}0 0 00AC CB AB−′′   =−′′ ′   +′′  I (6) In order that and be principal axes, we require C′ = 0: 1x2x ()1cos 2 sin 2 02CB A θθ− −= (7) or, 2tan 2C BAθ=− (8) from which 1 12tan2C BAθ− = −  (9) Notice that this result is still valid if A = B. Why? (What does A = B mean?) 11-19. θ = 0x2 xηθθ = πθ = π /2 1 The boundary of the plate is given by rk eαθ= . Any point ( η,θ ) has the components 1 2cos sinx xη θ η θ=    =  (1) 380 CHAPTER 11 The moments of inertia are 2 1200 23 00sinke keIA xd d ddαθ αθπ πρ ηηθ ρ θθ η η== =∫∫ ∫∫ The integral over θ can be performed by using Eq. (E.18a), Appendix E, with the result 4 12kIA P==ρ α (2) where ( )4 21 16 1 4ePπα α−= + (3) In the same way, 2 2100 2 00coske keIB xd d d== =∫∫ ∫∫αθ αθπ π3dρ ηηθ ρ θθ η η (4) Again, we use Eq. (E.18a) by writing co2s 1 sin2θ θ =− , and we find ( )4 2 2 182kIB Pραα== + (5) Also 12 1 200 3 00cos sinke keIC x x d d ddαθ αθπ πρ ηηθ ρ θθ θ η=− =− =−∫∫ ∫∫η (6) In order to evaluate the integral over θ in this case we write () cos sin 1 2 sin 2θ θθ= and use Eq. (E.18), Appendix E. We find (7) 4 12IC k ρ =− = P Using the results of problem 11-17, the entire inertia tensor is now known. According to the result of Problem 11-18, the angle through which the coordinates must be rotated in order to make {}I diagonal is 1 12tan2C BAθ− = −  (8) Using Eqs. (2), (5), and (7) for A, B, C, we find 21 2C BA α=− (9) DYNAMICS OF RIGID BODIES 381 so that 1tan 22θα= (10) Therefore, we also have 2θ 2α1142+α 2 21sin 2 14 2cos 2 14θ α αθ α=+  =+  (11) Then, according to the relations specified in Problem 11-18, 2 1 cos sin 2 sin IAA C B2θ θ == − +′′ θ (12) Using ()( )2cos 1 2 1 cos 2 θ θ =+ and ()( )2sin 1 2 1 cos 2 θ θ =− , we have () ()111cos 2 sin 222A B A B C IA θ θ == + + − −′′ (13) Now, ( )4 2 414 4kPAB AB k Pραα αρ+= +   −= −  (14) Thus, ( )4 24 4 1222114 22 14 14kPk P k Pρααα ρ ραIA α α== + − × − ×′′ ++ (15) or, ()4 1IA k P Q R ρ == −′′ (16) where 2 214 2 14Q Raα α +=    =+  (17) Similarly, ()4 2IB k P Q R ρ == +′′ (18) and, of course, 382 CHAPTER 11 3IA B I I1 2 = += + ′ ′′ ′ ′ (19) We can also easily verify, for example, that I12 0 C=−=′ ′ . 11-20. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy 12bEU m g== (2b is the height of the center of mass above the floor.) When the rod hits the floor, the potential energy is zero. Thus 2 21 2ET I ω == where I is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length σ: 23 end 011 33b Ix dx b σσ== =∫2mb Thus 22 21 6Tm bω = By conservation of energy 12UT= 22 1 26bmg mb ω = 3g bω= 11-21. Using I to denote the matrix whose elements are those of {}I, we can write =LI ω (11.54) =′′′ LI ω (11.54a) We also have and and therefore we can express L and ω as =′xλx ′xt=′xλ t= ′ LLλ (11.55a) t= ′ ωλω (11.55b) substituting these expression s into Eq. (11.54), we have DYNAMICS OF RIGID BODIES 383 tt=′ ′ LIλ λω and multiplying on the left by λ, tt=′ ′ LIλλ λλω or ( )t=′ ′ LI λλω by virture of Eq. (11.54a), we identify t=′IIλλ (11.61) 11-22. According to Eq. (11.61), 1 ,ij ik k j klII λ λ−=′∑ AA (1) Then, {}1 , 1 , ,ii ik k i ii k ki i k ki kk k k kktr I I I II− −==′′ = ==∑ ∑∑ ∑∑ ∑∑IAA A AA A AA Aλ λ λλ δ (2) so that {} { }tr tr =′II (3) This relation can be verified for the examples in the text by straightforward calculations. Note: A translational transformation is not a similarity transformation and, in general, {}tr is not invariant under translation. (For example, I {}trI will be different for inertia tensors expressed in coordinate system with different origins.) 11-23. We have 1−=′IIλλ (1) Then, 384 CHAPTER 11 1 1 1− − −=′ =× × =×II I Iλλ λ λ λλ so that, =′II (2) This result is easy to verify for th e various examples involving the cube. 11-24. –a/2 a/2x2 x1 −3 2a The area of the triangle is 234 A a = , so that the density is 24 3MM A aρ== (1) a) The rotational inertia with respect to an axis through the point of suspension (the origin) is x2 x1 θ ( ) () () 122 31 2 1 2 2 0 22 11 2 0 322 422 31 24 6a axIx xdxdx dx x x dx aM a−−=+ =+ ==∫ ∫∫ρ ρ ρ2 (2) When the triangle is susp ended as shown and when θ = 0, the coordinates of the center of mass are 2(0, ,0)x , where DYNAMICS OF RIGID BODIES 385 () 12 212 2 0 12 0 321 2 23a axx x dx dxM dx x dxM a−−= = =−∫ ∫∫ρ ρ 2 (3) The kinetic energy is 2 311 21 2TI M a22θ θ ==  (4) and the potential energy is (1c o s 23MgaU )θ =− (5) Therefore, 22 1cos12 23MgaLM aθ θ =+  (6) where the constant term has been su ppressed. The Lagrange equation for θ is 3s i n 0g aθθ+ =  (7) and for oscillations with small amplitude, the frequency is 3g aω= (8) b) The rotational inertia for an axis through the point of suspension for this case is x2 x1 −3 2a ( )23 0 22 32 1 2 0 3 2 22 5 12x aId x x x Ma− −=+′ =∫∫ρ1dx (9) The Lagrangian is now 22 5cos24 3MgaLM aθ θ =+  (10) 386 CHAPTER 11 and the equation of motion is 12sin 0 53g aθθ+ =  (11) so that the frequency of small oscillations is 12 53g aω= (12) which is slightly smaller than the previous result. 11-25. x2 x1R r 2ρρ θ The center of mass of the disk is 2(0, )x, where () ()2 212 212 lower upper semicircle semicircle 2 00 0 32 sin 2 sin 2 3RRxx dxdx xdxdxM r rdrd r rdrdM R M =+  =⋅ +∫∫ ∫∫ ∫∫ππ πρ ρ⋅ =−θ θθ ρθ (1) Now, the mass of the disk is 22 211222 3 2MR R=⋅ + ⋅ =R ρ πρ π ρπ (2) so that 24 9xRπ=− (3) The direct calculation of the rotational inertia wi th respect to an axis through the center of mass is tedious, so we first compute I with respect to the and then use Steiner’s theorem. 3-axisx DYNAMICS OF RIGID BODIES 387 222 300 0 422 31 42RRIr rdrdr rd RM R=⋅ +⋅ ==∫∫ ∫∫ππ πrd ρ θθ πρ (4) Then, 2 03 2 22 2 2 211 28 1 13 2128 1IIM x 6MRM R MR=− =− ⋅  =−  π π (5) When the disk rolls without slipping, the veloci ty of the center of mass can be obtained as follows: Thus RCMθ θ CM 2 CM 2sin cosxR x yR xθ θ θ=− =− 2ρρ x2 CM 2 CM 2cos sinxR x yxθ θθ θθ=− =    ( )22 2 2 2 2 2 2 CM CM 2 2 2c o xy V R x R x s θ θθ += =+ −    θ 2 22Va θ= (6) where 22 222c os aR x R x θ =+ − (7) Using (3), a can be written as 216 8181 9aR cosθππ=+ − (8) The kinetic energy is 388 CHAPTER 11 trans rot 2 011 22TT T Mv I2= + =+ θ (9) Substituting and simplifying yields 22 13 8cos22 9TM Rθ θπ =−   (10) The potential energy is 21cos2 181c os29UM g R x MgR =+    =−  θ θπ (11) Thus the Lagrangian is 2 13 8 8cos 1 cos22 9 9R R g LM θ θππθ  =− − −       (12) 11-26. Since = φφ ω lies along the fixed 3-axisx′ , the components of φω along the body axes are given by the application of the transformation matrix λ [Eqs. (11.98) and (11.99)]: ()ix () () ()1 1 22 3 30 0φ φ φωφ ωφ φ φω  ==     λ (1) Carrying out the matrix multiplication, we find () () ()1 2 3sin sin cos sin cosφ φ φωψ θ ω φψ θω θ    =       (2) which is just Eq. (11.101a). The direction of = θθ ω coincides with the line of nodes and lies along the axis. The components of 1x′′′ θω along the body axes are therefore obtained by the application of the transformation matrix ψλ which carries the xi′′′ system into the system: ix () () ()1 2 3cos 0s in 00θ θψ θω θψ ω θ ω  ψ     == −            λ (3) DYNAMICS OF RIGID BODIES 389 which is just Eq. (11.101b). Finally, since ψω lies along the body , no transformation is required: 3-axisx () () ()1 2 30 0 1ψ ψ ψω ωψ ω  =    (4) which is just Eq. (11.101c). Combining these results, we obtain ()() () ()() () ()() ()1 11 2 22 3 33 sin sin cos cos sin sin cos  ++   ++   ++      +  =−   +      φθψ φθψ φθψωωω ωωω ωωω φ ψθ θψ φφψθ θ ψ θθ ψω= (5) which is just Eq. (11.02). 11-27. L α θωx3 x2x3′ Initially: 11 1 21 2 1 33 3 30 sin sin cosLI LL I I LL x o s I Iω θ ωω α θ ωω α== == = == = Thus 21 33tan tanLI LIθ α == (1) From Eq. (11.102) 3 cos ω φθ ψ= +   Since 3 cos ω ω= α, we have 390 CHAPTER 11 cos cos φ θω αψ= −   (2) From Eq. (11.131) 31 3 1II Iψ ω−=− Ω=− (2) becomes 3 1cos cosI I= φ θω α (3) From (1), we may construct the following triangle I3I3 tan α θ from which 3 1222 2 31s tanI IIθ α= +co Substituting into (3) gives 22 2 2 13 1sin cosIIIωφ αα =+ 11-28. From Fig. 11-7c we see that = φφ ω is along the 3-axisx′ , = θθ ω is along the line of nodes, and =ψψ ω is along the . Then, 3-axx is 3 φφ=′ ′e ω (1) whee is the unit vector in the direction. 3′e3x′ Projecting the lines of nodes into the 1-x′ and 2-axesx′ , we obtain ( ) 12cos sinθθ φ =+′′ ′ ee ω φ (2) ψω′ has components along all three of the xi′ axes. First, we write ψ′ω in terms of a component along the x and a component normal to this axis: 3-axis′ ( ) 12 3sin cosψψ θ =+′′ ′ ee ω θ (3) where 12 1 2 sin cosφ φ = − ′ ′′ ee e (4) Then, ( ) 12 3sin sin sin cos cosψψ θφ θφ θ =− +′′ ′ ′ ee e ω (5) DYNAMICS OF RIGID BODIES 391 Collecting the various components, we have 1 2 3cos sin sin sin sin cos cosω θφ ψθφ ω θφ ψθ ωψ θ φ=+′ =−′ =+′     φ (6) 11-29. When the motion is vertical θ = 0. Then, according to Eqs. (11.153) and (11.154), ( ) 3 PI Pφ ψ φψ = += (1) and using Eq. (11.159), we see that 33 PP Iψφ ω = = (2) Also, when θ = 0 (and ), the energy is [see Eq. (11.158)] 0θ= 2 331 2EI M g ω =+ h (3) Furthermore, referring to Eq. (11.160), 2 331 2EE I M g h ω =− =′ (3) If we wish to examine the behavior of the system near θ = 0 in order to determine the conditions for stability, we can use the values of Pψ, Pφ, and E′ for θ = 0 in Eq. (11.161). Thus, ( )222 33 2 12 2 121c o s 1cos22 sinII MghIωθMgh θ θθ−=+ +  (5) Changing the variable to z = cos θ and rearranging, Eq. (5) becomes ()()2 2 12 3 3 2 12121z2 2z Mgh I z I wI−  =+ −  (6) The questions concerning stability can be answered by examining th is expression. First, we note that for physically real motion we must have 20 z≥ . Now, suppose that the top is spinning very rapidly, i.e., that 3ω is large. Then, the term in the square brackets will be negative. In such a case, the only way to maintain the condition 20 z≥ is to have z = 1, i.e., θ = 0. Thus, the motion at θ = 0 will be stable as long as 22 12 3 3 4Mgh I I ω 0 − < (7) or, 12 22 3341Mgh I Iω< (8) 392 CHAPTER 11 Suppose now that the top is set spinning with θ = 0 but with 3ω sufficiently small that the condition in Eq. (8) is not met. Any small disturbance away from θ = 0 will then give z a negative value and θ will continue to increase; i.e., the motion is unstable. In fact, θ will continue until z reaches a value 0z that again makes the square brackets equal to zero. This is a turning point for the motion and nutation between z = 1 and 0 zz= will result. From this discussion it is evid ent that there exists a critical value for the angular velocity, cω, such that for 3 c ω ω> the motion is stable and for c ωω< there is nutation: 12 32 cMgh I Iω= (9) If the top is set spinning with 3 c ω ω> and θ = 0, the motion will be stable. But as friction slows the top, the critical angular velocity will eventua lly be reached and nutation will set in. This is the case of the “sleeping top.” 11-30. If we set , Eq. (1.162) becomes 0θ= ()( ) ( )2 2 12cos cos 21 c osPP M g h Iφψ θ EV θ θ θ− == +′ − (1) Re-arranging, this equation can be written as ( ) ( ) ( ) ( )32 2 12 12 12 12 2 cos 2 cos 2 cos 2 0Mgh I E I P P P Mgh I E I Pψφ ψ θθ θ−+ + − +− ′′2 φ= (2) which is cubic in cos θ. V(θ ) has the form shown in the diagram. Two of the roots occur in the region 1c o s 1 θ −≤ , and one root lies outside this range and is therefore imaginary. ≤ V(θ) cos θ –1+1 DYNAMICS OF RIGID BODIES 393 11-31. The moments of inertia of the plate are ()12 2 312 2 2 2cos 2 1c o s 2 2c o sII I III I Iα α α=    =+   =+   = = (1) We also note that ( ) 12 2 2 21c o s 2 2s i nII I Iα α−= − − =− (2) Since the plate moves in a force-free manner, the Euler equations are [see Eq. (11.114)] ( ) () ()12 1 23 3 23 2 31 1 31 3 12 20 0 0II I II I II Iωω ω ωω ω ωω ω −− =  −− =   −− =   (3) Substituting (1) and (2) into (3), we find ( ) ( ) () ( )22 21 2 2 3 22 3 21 23 1 2 22s i n 2c o s 0 cos 2 cos 2 0 0II II IIαω ω αω αω ω αω ωω ω −− =  −− =    −=   (4) These equations simplify to 2 31 2 12 3 23 1tan  =−   =− =  ω ωω α ωω ω ωω ω (5) From which we can write 2 123 2 21 13 3cot ωωω ωω ωω ωω α == − = −  (6) Integrating, we find () () ()22 22 2 2 2 2 22 11 3 3 0 0 cot 0 cot ω ωω ωω α ω −= − += − + α (7) Now, the initial conditions are 394 CHAPTER 11 () () ()1 2 30c os 00 0s inω α ω ω α=Ω   = =Ω  (8) Therefore, the equations in (7) become 2 22 2 2 2 2 2 21 3 cos cot cosαωω (9) α ω α =− +Ω =− +Ω From (5), we can write 22 232 1 ω ωω= (10) and from (9), we have 22 2 13 cot ω ω= α. Therefore, (10) becomes 2 23 cot ω ω= α (11) and using 22 2 2 2 32 sin tan ω αω α =Ω − from (9), we can write (11) as 2 22 2 2 2cottan sinωαωα α=−−Ω (12) Since 22dd t ω ω= , we can express this equation in terms of integrals as 2 22 2 2 2cottan sinddtωαωα α=−−Ω ∫∫ (13) Using Eq. (E.4c), Appendix E, we find () ( )1 2tan 1tanh cottan sin sintωαααα α− ΩΩ −= − (14) Solving for 2ω, () ( ) 2 cos tanh sin ttω α =Ω Ω α (15) 11-32. a) The exact equation of motion of the physical pendulum is sin 0 IM gL + = θθ where , so we have 2IM k= 2singL k=−θ θ or ()( ) 2dc o s d ddgL tk=θθθ or DYNAMICS OF RIGID BODIES 395 () ()2dd cogL k=  s θθθ so 2 22cosgLak= + θθ where a is a constant determined by the initial conditions. Suppose that at t , 0=0=θθ and at that initial position the angular veloci ty of the pendulum is zero, we find 0 22ak−= cosglθ. So finally () 0 22cos cosgl k=−θ θθ b) One could use the conservation of energy to fi nd the angular velocity of the pendulum at any angle θ, but it is exactly the result we obtained in a), so at 01=θ , we have ()1 0 22cos cos 53.7 sgL k−== − = ωθ θ θ 11-33. Cats are known to have a very flexible body that they can manage to twist around to a feet-first descent while falling with conserved ze ro angular momentum. First they thrust their back legs straight out behind their body and at the same time they tuck their front legs in. Extending their back legs helps to resist spinning , since rotation velocity evidently is inversely proportional to inertia momentum. This allows the cat to twist their body differently to preserve zero angular momentum: the front part of the bo dy twisting more than the back. Tucking the front legs encourages spinning to a downward direction preparing for touchdown and as this happens, cats can easily twist the rear half of their body around to catch up with the front. However, whether or not cats land on their feet depends on several factors, notably the distance they fall, because the twist maneuver takes a cert ain time, apparently around 0.3 sec. Thus the minimum height required for cats falling is about 0.5m. 11-34. The Euler equation, which describes the rotati on of an object about its symmetry axis, say 0x, is ( ) xx y z yzx II I −− =ωω ω N x where xNb=− ω is the component of torque along Ox. Because the object is symmetric about Ox, we have y zII= , and the above equation becomes 0d dx x xx xbtIIb et− =− ⇒ =ω x ω ωω 396 CHAPTER 11 CHAPTER 12 Coupled Oscillations 12-1. m1 = M k1 x1k12 k2m2 = M x2 The equations of motion are ( ) ()11 1 2 1 1 2 2 22 1 2 2 1 2 10 0Mx x x Mx x xκκ κ κκ κ ++ − =  ++ − =  (1) We attempt a solution of the form () ()11 22it itxt B e xt B eω ω =  =  (2) Substitution of (2) into (1) yields ( ) ( )2 11 2 11 2 2 2 12 1 2 12 20 0MB B BM Bκκ ω κ κκ κ ω +− − =  −+ + − =  (3) In order for a non-trivial solution to ex ist, the determinant of coefficients of and must vanish. This yields 1B2B ( ) ( )22 1 12 2 12 12 MM κκ ωκκ ω κ  +− +− =2 (4) from which we obtain ()212 1 2 12 1 22 1422MMκκ κ22ω κκ κ++=± − + (5) This result reduces to ( )2 12 12 M ωκ κκ=+ ± for the case 12κ κκ= = (compare Eq. (12.7)]. 397 398 CHAPTER 12 If were held fixed, the frequency of oscillation of m would be 2m1 (2 01 1 121 M) ω κκ =+ (6) while in the reverse case, would oscillate with the frequency 2m (2 02 2 121 M) ω κκ =+ (7) Comparing (6) and (7) with the two frequencies, ω+ and ω−, given by (5), we find ()2 22 12 1 2 12 1 21242Mκ κ κ κ κ+=+ + + − +ωκ () ()2 12 1 2 12 11 2 01122MMκκ κ κκ κκ ω + + − = + = 1 >+ (8) so that 01 ω ω+> (9) Similarly, ()2 22 12 1 2 12 1 21242Mκ κ κ κ κ−=+ + − − +ωκ () ()2 12 1 2 12 21 2 01122MMκκ κ κκ κκ ω + − − = + = 2 <+ (10) so that 02 ω ω−< (11) If , then the ordering of the frequencies is 1κκ>2 01 02 ω ωωω+ − >>> (12) 12-2. From the preceding problem we find that for 12 1 2 , κ κκ 11 2 21 2 12 ;MMκκκωω++≅≅κ (1) If we use 1 01 02 ;MM2 κ κωω== (2) then the frequencies in (1) can be expressed as COUPLED OSCILLATIONS 399 () ()12 10 1 0 1 1 1 12 20 2 0 2 2 211 11κωω ω εκ κωω ω εκ=+ ≅ +    =+ ≅ +   (3) where 12 12 12 12;22κ κεεκ κ== (4) For the initial conditions [Eq. 12.22)], () () () ()1 2120 , 00 , 00 , 00 x x x == =  ,= xD (5) the solution for ()1xt is just Eq. (12.24): ()12 12 1 cos cos22D t t xtω ωω ω + −  =      (6) Using (3), we can write ( )( ) 1 2 01 02 1 01 2 02 22ωω ω ω ε ω ε ω ε+++= + + + ≡Ω+ (7) ( )( ) 1 2 01 02 1 01 2 02 22ω ωω ω ε ωε ω ε−−− =−+ − ≡Ω+ (8) Then, () ( ) ( )1 cos cos xt D t t t t ε ε++ −− =Ω + Ω + (9) Similarly, () () (12 12 2 sin sin22 sin sinxt D t t Dt t t++ −−+− )t =     =Ω + Ω +ωω ωω ε ε (10) Expanding the cosine and sine fu nctions in (9) and (10) and taking account of the fact that ε+ and ε− are small quantities, we find, to first order in the ε’s, ()1 cos cos sin cos cos sinD t t t t t t t t εε+ − ++ − − + ≅Ω Ω −Ω Ω −Ω Ω  xt (11) − ()2 sin sin cos sin sin cosD t t t t t t t t εε+ − ++ − − + ≅Ω Ω + Ω Ω +Ω Ω  xt (12) − When either ()1xt or ()2xt reaches a maximum, the other is at a minimum which is greater than zero. Thus, the energy is never transfe rred completely to one of the oscillators. 400 CHAPTER 12 12-3. The equations of motion are 2 12 0 1 2 22 0 20 0mxxxM mxxxMω ω++=    ++=    (1) We try solutions of the form () ()11 22 ;it itxt B e xt B eω ω== (2) We require a non-trivial solution (i.e., the determinant of the coefficients of B and equal to zero), and obtain 1 2 B ( )2 222 4 0 0m Mωω ω− −= (3) so that 22 2 0m Mωω ω−= ± (4) and then 2 2 0 1m Mωω= ± (5) Therefore, the frequencie s of the normal modes are 2 0 1 2 0 21 1m M m Mωω ωω= + = − (6) where 1ω corresponds to the symmetric mode and 2ω to the antisymmetric mode. By inspection, one can see that the normal coordi nates for this problem are the same as those for the example of Section 12.2 [i.e., Eq. (12.11)]. 12-4. The total energy of the system is given by ( ) ( ) ()2 22 22 12 12 1 2 2 111 1 22 2ETU Mxx xx xx κκ=+ =+ + + + −  (1) Therefore, COUPLED OSCILLATIONS 401 () () ( )( ) () () () ()11 22 11 22 1 2 2 1 2 1 11 1 2 2 1 1 22 1 2 2 1 11 2 1 122 1 2 1211 2 2 2dEM x xx x x xx x xx xxdt Mx x x x x Mx x x x x Mx x x x Mx x x x=+ + + + − −     =+ − − ++ + −      + − +− + +                κκ κκ κκ κκ κ κ κκ2 =+ (2) which exactly vanishes because the coefficients of and are the left-hand sides of Eqs. (12.1a) and (12.1b). 1x2x An analogous result is obtained when T and U are expressed in terms of the generalized coordinates 1η and 2η defined by Eq. (12.11): ( )22 121 4TM η η =+  (3) ( )22 12 1 211 42U2 1 κηη κ η =+ + (4) Therefore, () [ 1 1 21 1 2 22dEMMdt] 22 η κκ η η η κ η η  ⋅= + + + +    (5) which exactly vanishes by virtue of Eqs. (12.14). When expressed explicitly in terms of the generalized coordinates, it is evident that there is only one term in the energy that has as a coefficient (namely, 12κ2 12 1κη 1), and through Eq. (12.15) we see that this implies that such a term depends on the ’s and 1C ω, but not on the ’s and 2C2ω. To understand why this is so, it is sufficient to recall that 1η is associated with the anitsymmetrical mode of oscillati on, which obviously must have 12κ as a parameter. On the other hand, 2η is associated to the symmetric mode, () ()12xt xt = , () ()12xt xt =  , in which both masses move as if linked together with a rigid, massless rod. For this mode, therefore, if the spring connecting the masses is changed, the motion is not affected. 12-5. We set . Then, the equations of motion are 121 2 κκκ== ≡ κ 11 1 2 22 2 120 20mx x x mx x xκκ κκ+− =    +− =   (1) Assuming solutions of the form () ()11 22it itxt B e xt B eω ω =  =  (2) we find that the equations in (1) become 402 CHAPTER 12 ( ) ( )2 112 2 12 220 20mBB Bm Bκω κ κκ ω −− =  −+− =  (3) which lead to the secular equation for 2ω: ( )( )22 12 22 mm2κ ωκ ωκ− −= (4) Therefore, 2 12311mmκµωµ =± − +  (5) where ( )12 1 2mm m mµ=+ 12= is the reduced mass of the system. Notice that (5) agrees with Eq. (12.8) for the case mm and M=12κ κ=. Notice also that 2ω is always real and positive since the maximum value of ( )1m+2 3 mµ is 3 4 . (Show this.) Inserting the values for 1ω and 2ω into either of the equations in (3), we find1 1 11 21 12321 1maammµ µ −+ − = +    (6) and 1 12 22 12321 1mammµ µ =− −− +   a (7) Using the orthonormality condition produces 11 11a D= (8) ()12 1 2 2 2 12 21 1321 1mm m m m mm Da +  −+ − +   = (9) where () ( )()2 22 1 12 1 1 2 2 22 123 222 1mmm m mmm mm≡− ++ − − +12mDm (10) The second eigenvector has the components ()12 1 2 2 2 12 12 2321 1mm m m m mm Da +  −− − +   = (11) 1 Recall that when we use 1 ωω= , we call the coefficients ( )11 a βω ω11 = = and ( )21 a βω ω ==21, etc. COUPLED OSCILLATIONS 403 22 21a D= (12) where ()222 111 21 2 1 2 22 2 22 2 123321 1 21 1mmm mm mmm m mm  ≡+ + − +− −   +   12mDm (13) The normal coordinates for the case in which ()0jq 0=  are ()() ()()11 1110 2 2120 21 1210 2 2220cos costm a xm a x t tm a xm a x1 2tη ω η ω=+ =+ (14) 12-6. mm mmkk x1 x2 If the frictional force acting on mass 1 due to mass 2 is ( )12 f xxβ=− −  (1) then the equations of motion are ( ) ()11 2 1 22 1 20 0mx x x x mx x x x +− + =  +− + =     βκ βκ (2) Since the system is not conservative, the eigenfre quencies will not be entirely real as in the previous cases. Therefore, we attempt a solution of the form () ()11 22 ;txt B e xt B et α α== (3) where i αλω=+ is a complex quantity to be determined. Substituting (3) into (1), we obtain the following secular equation by setting the determinant of the coefficients of the B’s equal to zero: ( )22m22α βα κ β α++ = (4) from which we find the two solutions ( )11 2 2; 1imm mmκ καω αβ β κ=± =±    =− ± − (5) The general solution is therefore 404 CHAPTER 12 () ( )22 11 1 1 1 1 2 1 2im t im t mt m mt m tmBe Be e Be Beκκ βκ βκ β −− +− − + −=++ + xt (6) and similarly for ()2xt . The first two terms in the expression for ()1xt are purely oscillatory, whereas the last two terms contain the damping factor teβ−. (Notice that the term ( )2 12expBm t β κ+− increases with time if 2m β κ> , but is not required to vanish in order to produce physically realizable motion because the damping term, exp(– βt), decreases with time at a more rapid rate; that is 12B+ 2ββ−+ 0κ<m− .) To what modes do 1α and 2α apply? In Mode 1 there is purely oscillating motion without friction. This can happen only if the two masses have no relative motion. Thus, Mode 1 is the symmetric mode in which the masses move in phase . Mode 2 is the antisymmetric mode in which the masses move out of phase and produce frictional damping. If 2m β κ< , the motion is one of damped oscillations, whereas if 2m β κ> , the motion proceeds monotonically to zero amplitude. 12-7. mk mkx1 x2 We define the coordinates and as in the diagram. Including the constant downward gravitational force on the masses results only in a displacement of the eq uilibrium positions and does not affect the eigenfrequencies or the norm al modes. Therefore, we write the equations of motion without the gravitational terms: 1x2x 11 2 22120 0mx x x mx x xκκ κκ+− =    +−=   (1) Assuming a harmonic time dependence for ()1tx and ()2xt in the usual way, we obtain ( ) ( )2 12 2 1220 0mB B Bm Bκω κ κκ ω −− =  −+ − =  (2) Solving the secular equation, we fi nd the eigenfrequencies to be COUPLED OSCILLATIONS 405 2 1 2 235 2 35 2m mκω κω+= −= (3) Substituting these frequencie s into (2), we obtain for the eigenvector components 11 21 12 2215 2 15 2aa aa −=   + = (4) For the initial conditions () ()1200xx ==  0, the normal coordinates are () ()11 1 10 2 0 21 2 10 2 015cos2 15cos2tm ax x t tm ax x tηω ηω −=+  +=+1 2 (5) Therefore, when , 10 20 1.6180 xx=− ()2 0tη = and the system oscillates in Mode 1, the antisymmetrical mode. When , 10 20.6180 xx=0 ()1 0tη = and the system oscillates in Mode 2, the symmetrical mode. When mass 2 is held fixed, the equation of motion of mass 1 is 112 mx x 0 κ+ =  (6) and the frequency of oscillation is 102 mκω= (7) When mass 1 is held fixed, the equation of motion of mass 2 is 22 0 mx x κ+ =  (8) and the frequency of oscillation is 20mκω= (9) Comparing these frequencies with 1ω and 2ω we find 11 2235 2 21.14414 350.61804mm mmκκωω κκωω+==0 0>    − == <  406 CHAPTER 12 Thus, the coupling of the oscillators produces a shift of the frequencies away from the uncoupled frequencies, in agreement with th e discussion at the end of Section 12.2. 12-8. The kinetic and potential energies for the doub le pendulum are given in Problem 7-7. If we specialize these results to the ca se of small oscillations, we have (222 12 1 21222Tm ) φ φφ φ =+ +   A (1) ( )22 12122Um g φ φ = A + (2) where 1φ refers to the angular displace ment of the upper pendulum and 2φ to the lower pendulum, as in Problem 7-7. (We have also discarded the constant term in the expression for the potential energy.) Now, according to Eqs. (12.34), ,1 2jkjk jkTm q =∑q (3) ,1 2jkjk jkUA q =∑ q (4) Therefore, identifying the elements of {}m and {}A, we find {}221 11m =   m A (5) {}20 01mg =   A A (6) and the secular determinant is 22 2222 0g gωω ωω−− = −−A A (7) or, 22 422ggωω ω  0 − −− =     AA (8) Expanding, we find 2 4242ggωω0 − + AA= (9) which yields COUPLED OSCILLATIONS 407 ()222gω=±A (10) and the eigenfrequencies are 1 222 1 .848 22 0 .765gg ggω ω=+ = =− =AA AA (11) To get the normal modes, we must solve ( )20jk r jk jr jAm a ω− = ∑ For k = 1, this becomes: ( ) ( )22 11 11 1 21 21 2 0rr r r Am a Am a ωω−+ −= For r = 1: () ()22 11 21 22 2 2 2 2ggmg m a m a −+ −+ = AA AAA0  Upon simplifying, the result is 21 11 2 aa=− Similarly, for r = 2, the result is 22 12 2 aa= The equations 11 1 11 2 22 1 1 2 2xa a xa a2 2η η η η= + =+ can thus be written as 11 1 1 2 21 2xa a2 η η =+ 1 11 1 22 2 2 xa a η η =− + Solving for 1η and 2η: 12 1 12 22 1122;2 22xx xx a aηη−+==2 408 CHAPTER 12 2 12 1 2 211 occurs when 0; i.e. when 2 occurs when 0; i.e. when 2xx xxηη ηη== ==− Mode 2 is therefore the symmetrical mode in wh ich both pendula are always deflected in the same direction; and Mode 1 is the antisymmetric al mode in which the pendula are always deflected in opposite directions. Notice that Mode 1 (the antisymmetrical mode), has the higher frequency, in agreement with th e discussion in Section 12.2. 12-9. The general solutions for ()1xt and ()2xt are given by Eqs. (12.10). For the initial conditions we choose oscillator 1 to be displaced a distance D from its equilibrium position, while oscillator 2 is held at , and both are released from rest: 20 x= () () () ()12 1 20 , 00 , 00 , 0 x x x == =  0= xD (1) Substitution of (1) into Eq. (12.10) determines the constants, and we obtain () ( 11 cos cos2Dxt t t )2 ω ω =+ (2) () ( 22 cos cos2Dxt t t )1 ω ω =− (3) where 12 122 MMκκ κωω+>= (4) As an example, take 121.2 ω ω = ; ()1xt vs. ()2xt is plotted below for this case. It is possible to find a rotation in configuratio n space such that the projection of the system point onto each of the new axes is simple harmonic. By inspection, from (2) and (3), the new coordinates must be 112 1 cos xxxD t ω ≡−= ′ (5) 212 2 cos xxxD t ω ≡+= ′ (6) These new normal axes correspond to the description by the normal modes. They are represented by dashed lines in the graph of the figure. COUPLED OSCILLATIONS 409 0.20.2 –0.50.40.60.8 1.0 0.4 0.6 0.8 1.0ωπ 25 2t= ωπ 23 2t=ωπ 23t=ωπ 22t= 7π 2x2(t)/Dx2′ x1′x1(t)/D ω2t = π 3π4π ω2t = 2πω1 = 1.2 ω2 ω2t = 0 12-10. The equations of motion are ( ) ()11 1 2 1 1 2 2 0 22 1 2 2 1 2 1cos 0mx bx x x F t mx bx x xκκκ ω κκ κ ++ + − =  ++ + − =    (1) The normal coordinates are the same as those for the undamped case [see Eqs. (12.11)]: 112 21 ; xx xx2 η η =−= + (2) Expressed in terms of these coordinates, the equations of motion become ( )( )( )( ) ( ) () () () () ()21 21 1 2 21 1 2 21 0 21 21 1 2 21 1 2 212c o s 0mb F mbt η ηη η κ κ η η κ η η ω ηη ηη κ κ ηη κηη ++ ++ + +− −=  −+ −+ + −− +=        (3) By adding and subtracting these equatio ns, we obtain the uncoupled equations: 0 12 11 1 0 2222cos cosF btmmm F btmmmκκηη η ω κηηη ω+ ++ =    ++=    (4) With the following definitions, 410 CHAPTER 12 2 12 1 2 2 02 2b m m m FAmβ κκω κω=  + =   =    = (5) the equations become 2 11 1 1 2 22 2 22c 2cos osA t A tη βη ω η ω η βη ω η ω++= ++=    (6) Referring to Section 3.6, we see that the solutions for ()1tη and ()2tη are exactly the same as that given for x(t) in Eq. (3.62). As a result ()1tη exhibits a resonance at 1 ωω= and ()2tη exhibits a resonance at 2 ωω= . 12-11. Taking a time derivative of the equations gives ( ) qI= 1 12 0ILI MIC++=   2 21 0ILI MIC+ +=   Assume 11itIB eω= , 22itIB eω= ; and substitute into the previous equations. The result is 22 11 210it it itLB e B e M B eCωω ωωω−+ − = 22 22 110it it itLB e B e M B eCωω ωωω−+ − = These reduce to ( )22 1210 BL B MCωω− +− = ( )22 1210 BM B LCωω− +− = This implies that the determ inant of coefficients of B and must vanish (for a non-trivial solution). Thus 1 2 B COUPLED OSCILLATIONS 411 22 221 01LMC MLCωω ωω−− = −− ( )2 222 10 LMCωω− −= 22 1LMCω ω −= ± or ()2 1 CL Mω=± Thus () ()1 21 1CL M CL Mω ω=+ =− 12-12. From problem 12-11: 11210 LI I MIC+ +=   (1) 22 110 LI I MIC+ +=   (2) Solving for  in (1) and substituting into (2) and similarly for I, we have 1I2 2 11 2 2 22 110 10MMLI I ILC C L MMLI I ILC C L −+ − =     −+ − =     (3) If we identify 2 12 11MmLL M LC M CLκ κ=−     =   =−    (4) 412 CHAPTER 12 then the equations in (3) become ( ) ()11 2 1 122 21 2 2 1210 0mI I I mI I Iκκ κ κκ κ ++ − =  ++ − =  (5) which are identical in form to Eqs. (12.1). Then, using Eqs. (12.8) for the characteristic frequencies, we can write () ()1 2 2 211 11M L M CL MCLL M L M CL MCLLω ω+ == −− − == +− (6) which agree with the results of the previous problem. 12-13. I1C1q1 I2C2q2 L1 L12 L2 The Kirchhoff circuit equations are ( ) ( )1 11 1 2 1 2 1 2 22 1 2 2 1 20 0qLI L I IC qLI L I IC++ −=   ++ − =    (1) Differentiating these equations using qI= , we can write () ()11 2 1 11 2 2 1 21 2 2 21 2 1 210 10LL I IL IC LL I IL IC++ − =    ++ − =    (2) As usual, we try solutions of the form () ()11 22 ;it itIt B e It B eω ω== (3) which lead to COUPLED OSCILLATIONS 413 () ()22 11 2 11 2 2 1 22 12 1 2 12 2 110 10LL BL BC LB L L BCωω ωω +− − =      −+ + − =     (4) Setting the determinant of the coefficients of the B’s equal to zero, we obtain () ()22 11 2 21 2 1 1211LL LL LCCωω +− +− =  42 2 ω  (5) with the solution () ()() () () ()22 11 2 1 21 2 2 11 2 1 21 2 2 1 2 1 22 2 12 1 1 2 2 1 2 1 24 2LL C LL C LL C LL C L C C CC L L L L Lω ++ +± +− + +=++ − (6) We observe that in the limit of weak coupling ( )12 0 L→ and 12LLL= =, CC , the frequency reduces to 12 C == 1 LCω= (7) which is just the frequency of un coupled oscillations [Eq. (3.78)]. 12-14. I1C1 C12I2C2L1 L2 The Kirchhoff circuit equations are (after differentiating and using qI= ) 11 1 2 11 2 1 2 22 2 1 21 2 1 211 10 11 10LI I ICC C LI I ICC C ++ − =      ++ − =      (1) Using a harmonic time dependence for ()1It and ()2tI , the secular equation is found to be 22 11 2 21 2 12 2 11 2 21 2 1 21 CC CCLLCC CC Cωω  ++−−   =  (2) Solving for the frequency, () () () ()222 11 2 1 2 22 1 1 2 11 2 1 2 22 1 1 2 1 2 122 12 1 2 1 24 2CL C C CL C C CL C C CL C C CCLL LLCCCω ++ +± +− + += (3) 414 CHAPTER 12 Because the characteristic frequencies are given by this complicated expression, we examine the normal modes for the special case in which 12LLL= = and CC12 C = =. Then, 2 12 1 12 2 22 1CC LCC LCω ω+= = (4) Observe that 2ω corresponds to the case of uncoupled oscillations. The equations for this simplified circuit can be set in the same form as Eq. (12.1), and consequently the normal modes can be found in the same way as in Section 12.2. There will be two possible modes of oscillation: (1) out of phase , with frequency 1ω, and (2) in phase , with frequency 2ω. Mode 1 corresponds to the currents and oscillating always out of phase : 1I2I ;I1 I2 I1 I2 Mode 2 corresponds to the currents and oscillating always in phase : 1I2I ;I1 I2 I1 I2 (The analogy with two oscillators coupled by a sp ring can be seen by associating case 1 with Fig. 12-2 for 1 ωω= and case 2 with Fig. 12-2 for 2 ωω= .) If we now let and , we do not have pure symmetrical and antisymmetric al symmetrical modes, but we can associate 1LL≠2 2 1CC≠ 2ω with the mode of highest degree of symmetry and 1ω with that of lowest degree of symmetry. 12-15. I1C1 I2C2 L1 RL2 Setting up the Kirchhoff circuit equa tions, differentiating, and using qI= , we find ( ) ( )11 1 2 1 1 22 2 1 2 210 10LI R I I IC LI RI I IC+− + =    +− + =      (1) Using a harmonic time dependence for ()1It and ()2tI , the secular equation is 22 12 12110 Li R Li R RCCωω ωω ω −− −− + = 22  (2) COUPLED OSCILLATIONS 415 From this expression it is clear that the oscillations will be damped because ω will have an imaginary part. (The resistor in the circuit dissi pates energy.) In order to simplify the analysis, we choose the special case in which 12LLL= = and 12CCC= =. Then, (2) reduces to 2 2 10 Li R RCωω ω 22−−+ = (3) which can be solved as in Problem 12-6. We find 1 2 21 LC iLRRLCω ω=±  =± −    (4) The general solution for is ()1It ()2211 11 1 1 1 1 2 1 2i L Ct i L Ct R LCtL R LCtL Rt LB e B e e B e B e−− − − +− + −It−  =+ + +    (5) and similarly for . The implications of these results follow closely the arguments presented in Problem 12-6. ()2It Mode 1 is purely oscillatory with no damping. Sinc e there is a resistor in the circuit, this means that and flow in opposite senses in the two parts of the circuit and cancel in R. Mode 2 is the mode in which both currents flow in the same direction through R and energy is dissipated. If 1I2I 2C<RL , there will be damped oscillations of and , whereas if 1I2I2RL C> , the currents will decrease monotonically without oscillation. 12-16. Oy x R P R MgQ(x,y) Mgθ φ Let O be the fixed point on the hoop and the origin of the coordinate system. P is the center of mass of the hoop and Q(x,y) is the position of the mass M. The coordinates of Q are ( ) ()sin sin cos cosxR yRθφ θ φ =+  =− + (1) The rotational inertia of the hoop through O is 416 CHAPTER 12 (2) 2 OC M 2 II M R M R =+ =2 The potential energy of the system is ()hoop mass 2c o s c o sUU U MgR θ φ= + =− + (3) Since θ and φ are small angles, we can use 2s 1 2xx≅− co . Then, discarding the constant term in U, we have ( )22 122UM gRθ φ =+ (4) The kinetic energy of the system is ( )hoop mass 22 2 O 22 2 2 211 22 122TT T IM x y MR MR=+ =+ +   =+ + +    θ  θ θφ θ φ (5) where we have again used the small-angle approximations for θ and φ. Thus, 22 2 1322TM R θ φθ φ   =+ +   (6) Using Eqs. (12.34), ,1 2jk jk jkTm q =∑q (7) ,1 2jk jk jkUA q =∑ q (8) we identify the elements of {}m and {}A: {}231 11MR =   m (9) {}20 01MgR =   A (10) The secular determinant is 22 2223 0g R g Rωω ωω−− = −− (11) from which COUPLED OSCILLATIONS 417 22 423gg RRωω ω  0 − −− =      (12) Solving for the eigenfrequencies, we find 1 22 2 2g R g Rω ω= = (13) To get the normal modes, we must solve: ( )20jk r jk jr jAm a ω− = ∑ For k = r = 1, this becomes: 22 11 21 22 3 2ggmgR mR a mR aRR0− −=  or 21 11 2 aa=− For k = 1, r = 2, the result is 12 22aa= Thus the equations 11 1 11 2 22 1 1 2 2xa a xa a2 2η η η η= + =+ can be written as 11 1 12 2 2 21 1 1 2 2 2xa a xa a2η η η η= + =− + Solving for 1η, 2η 12 1 11 11 222;33xx xx aaηη2 − +== 1η occurs when the initial conditions are such that 20 η=; i.e., 10 201 2xx=− This is the antisymmetrical mode in which the CM of the hoop and the mass are on opposite sides of the vertical through the pivot point. 2η occurs when the initial conditions are such that 10 η=; i.e., 10 20xx= 418 CHAPTER 12 This is the symmetrical mode in which the pi vot point, the CM of the hoop, and the mass always lie on a straight line. 12-17. k k m k m k m x1 x2 x3 Following the procedure outlined in section 12.6: 22 12111 222T m xm xm =++ 2 3x () ()2 2 22 12 1 3 2 222 1231 22 311 1 1 22 2 2Uk x k x x k x x k x kx x x x x x x=+ − + − + =+ + − −3 Thus 00 00 00m m m   =    m 20 2 02kk kkk kk−    =−−    −  A Thus we must solve 2 2 220 20 02km k kk mk kk mω ω ω−− −− − −−= This reduces to ( ) ( )322 222 2 km k kmωω 0 − −− = or ( )( )222 222 2km km kωω  −− −0= If the first term is zero, then we have 12k mω= If the second term is zero, then 222kmω−= ± k COUPLED OSCILLATIONS 419 which leads to () () 2322 22 ;kk mm+− ==ωω To get the normal modes, we must solve ( )20jk r jk jr jAm a ω− = ∑ For k = 1 this gives: ( ) ()2 12 20rr r km a k a ω− +− = Substituting for each value of r gives ( ) () ()11 21 21 12 22 22 12 13 23 23 131: 2 2 0 0 2: 2 0 2 3: 2 0 2rk ka kaa a ka a rk a ka a a=− − =→ = =− − = → = − =− =→ =rk a Doing the same for k = 2 and 3 yields 11 31 21 12 32 22 32 13 33 23 330 2 2aa a aa a a aa a a=−= == − == The equations 11 1 11 2 21 3 22 1 1 2 2 22 3 33 1 1 3 2 23 3xa a a xa a a xa a a3 3 3η ηη η ηη η ηη= ++ =++ =++ can thus be written as 11 1 1 2 2 23 3 3 22 2 2 3 3 3 31 1 1 2 2 2 3 31 2 2 1 2xa a a xa a xa a a3η ηη ηη η ηη=− + =+ =− − + We get the normal modes by solving these three equations for 1η, 2η, 3η: 420 CHAPTER 12 13 1 11 12 2 222 2 22xx a xx aη η3x−= −+ −= and 12 3 332 4xx aη3x + += The normal mode motion is as follows 11 22 3 213: :2 :2xx xx xxxη η•→ • ←• = − ←• •→←• = − = − •→ •→ •→ = =3 1 32 2x η 12-18. x1y1mbx MM θ 11 11sin ; cos cos ; sinxx b xx b yb b ybθ θθ θθ θ=+ =+ =− =    Thus ( ) ( ) ()22 2 11 22 2 2 111 22 112c os22 1c o sTM x m x y Mx m x b b x Um g y m g bθ θθ θ=+ + =+ + + == −     For small θ, 2 cos 12θθ − . Substituting and neglecting the term of order 2θθ gives COUPLED OSCILLATIONS 421 () ( )22 2 211222 2TM m xm b b x mgbUθ θ θ=++ + =  Thus 2M mm b mb mb+ =  m 00 0mgb =  A We must solve ( )22 220Mm m b mb mgb mbωω ωω−+ − 2=−− which gives ( )( ) ()22 24 2 22 20 0M m mb mgb m b Mb mgb m Mωω ω ωω+− −  −+ =2= Thus ()1 20 gMmmbω ω= =+ ( )20jk r jk jr jAm a ω− = ∑ Substituting into this equation gives ( ) ()()21 12 2202 2, 2ak bmaa kmM== =− = =+,1rr= Thus the equations 11 1 12 2 21 1 22 2xa a aaη η θ ηη= + =+ become ()11 1 22 2mbxa amMη η =−+ 422 CHAPTER 12 22 2aθ η = Solving for 1η, 2η: ()2 22 1 11a bmxmMnaθη θ= ++= ()12 21 occurs when 0; or 0 occurs when 0; or nn bmnn xmMθ θ== == −+ 12-19. With the given expression for U, we see that {}A has the form {}12 13 12 23 13 231 1 1ε ε ε ε εε−−    =− −    −− A (1) The kinetic energy is ( )222 1231 2T θ θθ =+ +  (2) so that {}m is {}100 010 001   =    m (3) The secular determinant is 2 12 13 2 12 23 2 13 231 1 1ωε ε εω ε εε ω−− − 0 − −− = −− − (4) Thus, ( )( )( )32 2 222 12 13 23 12 13 23 11 2ωω ε ε ε ε εε −− − + + − = 0 (5) This equation is of the form (with 12x − ≡ω ) 32 232 xx 0 − −=αβ (6) which has a double root if and only if ()3222α β= (7) COUPLED OSCILLATIONS 423 Therefore, (5) will have a double root if and only if 32222 12 13 23 12 13 233εεεεεε++=  (8) This equation is satisfied only if 12 13 23ε εε== (9) Consequently, there will be no degeneracy unless the three coupling coefficients are identical. 12-20. If we require , then Eq. (12.122) gives 11 212 aa=31 21 3 aa=− , and from Eq. (12.126) we obtain 2111 4 a= . Therefore, 121 3,, 14 14 14 =−  a (1) The components of can be readily found by substituting the components of above into Eq. (12.125) and using Eqs. (12.123) and (12.127): 2a1a 245 1,, 42 42 42−  =  a (2) These eigenvectors correspond to the following cases: a1 a2 12-21. The tensors {}A and {}m are: {}13 32 3 31102 1 22 102κκ 1κ κκ κκ        =       A (1) {}00 0 00m m m0   =    m (2) thus, the secular determinant is 424 CHAPTER 12 2 13 2 32 3 2 31102 11022 102m m mκω κ κκ ω κ κκ ω− − = − (3) from which ( )( ) ( )222 2 2 12 3 1102mm m κω κ κω −− − − κω (4) = In order to find the roots of this equation, we first set ()2 31 122 κ κκ= and then factor: ( )( )( ) (κω (5) ) () ( ) ()22 2 11 21 2 22 4 2 11 2 22 2 11 20 0 0mm m mm m mmmκω κω κω κ κ ω κ κ ω κω ω ω κ κ −− − −  −− + =  −− + == Therefore, the roots are 1 1 12 2 30m mκω κκω ω= += = (6) Consider the case 30 ω=. The equation of motion is 2 33 3 0 + = ηω η (7) so that 30 η= (8) with the solution ()3ta t bη =+ (9) That is, the zero-frequency mode corresponds to a translation of the system with oscillation. COUPLED OSCILLATIONS 425 12-22. The equilibrium configuration is shown in diagram (a) below, and the non- equilibrium configurations are shown in diagrams (b) and (c). 1 24 3 x1x2x3 O 2A2B(a) 12 x3xx3 OO′ AAAθ θ 2  }(b) 14 x3xx3 OO′B BBφ φ 2  {(c) The kinetic energy of the system is 22 31 211 1 22 2TM x I I2θ φ =+ +    (1) where ()2 113 IM= A and ()2 213 IM= B. The potential energy is () () () () ( )2222 3333 22 2 2 2 31 2 144 42Ux A B x A B x A B x A B xA B  =− − + + − + + + + − +  =+ +κ θ φθ φθ φθ κθ φφ (2) Therefore, the tensors {}m and {}A are {}2 200 103 1003M MA 0 MB        =       m (3) {}2 240 0 04 0 00 4A Bκ κ κ   =    A (4) The secular equation is 426 CHAPTER 12 ( )22 2 22 2 2 1144 433MA M A B M B κωκ ωκ ω −− −   0= (5) Hence, the characteristic frequencies are 1 2 322 32 32M M Mκω κω κω ω= = == (6) We see that 23ω ω= , so the system is degenerate . The eigenvector components are found from the equation ( )20jk r jk jr jAm a ω− = ∑ (7) Setting to remove the indeterminacy, we find 320 a= 2 12 3 2010 0; 3 ; 0 003M MA MB        == =          a aa (8) The normal coordinates are (for ()()()3000x θφ 0 = ==   ) () () ()13 0 1 0 2 0 33cos cos 3 cos 3tx M t AMt BMttηω θ 2t η ω φηω= = = (9) Mode 1 corresponds to the simple vertical osci llations of the plate (without tipping). Mode 2 corresponds to rotational oscillations around the axis, and Mode 3 corresponds to rotational oscillations around the -axis. 1x 2x The degeneracy of the system can be removed if the symmetry is broken. For example, if we place a bar of mass m and length 2 A along the of the plate, then the moment of inertia around the is changed: 2-axisx 1-axisx ()2 11 3IM m =+′ A (10) The new eigenfrequencies are COUPLED OSCILLATIONS 427 1 2 32 32 32M M m Mκω κω κω= =+ = (11) and there is no longer any degeneracy. 12-23. The total energy of the r-th normal mode is 2211 22rr r rrET U 2 r η ωη=+ =+ (1) where rit rr e=ωηβ (2) Thus, rit rr rie=ωηω β (3) In order to calculate T and U, we must take the squares of the real parts of r r rη and rη: () ( ) ()2 2 2 2Re Re cos sin cos sinrr r r rr r rr r r r rii t i ttt   == + +  − ηη ω µ ν ω ω ων ω ωµ ω=− (4) so that 221cos sin2rr rr r rTt ων ω µ ω=+ t     (5) Also () ( ) ()2 2 2 2Re Re cos sin cos sinrr r rr r rr r rit i ttt   == + +  ηη µ ν ω ω µω νω=− (6) so that 221cos2rr rr r rUt ωµ ω ν ω=− xint     (7) Expanding the squares in T and U, and then adding, we find r r 428 CHAPTER 12 ( )22 21 2rr r rr rET U ωµν=+ =+ Thus, 2 21 2rr r E ωβ= (8) So that the total energy associated with ea ch normal mode is separately conserved. For the case of Example 12.3, we have for Mode 1 () 11 0 20cos2Mxx1t η ω =− (9) Thus, () 11 1 0 2 0sin2Mxx1t η ω=− − ω (10) Therefore, 22 11 111 22E2 1 η ωη =+ (11) But 2 12 12 Mκ κω+= (12) so that () () () ()22 22 12 12 11 0 20 1 1 0 20 2 12 10 2022 11sin cos22 22 124MMEx x t x xMM xx++=− +− =+ −κκ κκ 1t ω ω κκ (13) which is recognized as the value of the potential energy at t = 0. [At t = 0, , so that the total energy is 12 0 xx== ( )1 0= Ut .] 12-24. Refer to Fig. 12-9. If the particles move along the line of the string, the equation of motion of the j-th particle is ( )( ) 1 jj j j m x xx xx κκ− =− − − −1j+ (1) Rearranging, we find ( ) 12jj j jxx x xm1κ − =− + + (2) which is just Eq. (12. 131) if we identify mdτ with mκ . COUPLED OSCILLATIONS 429 12-25. The initial conditions are () () () () () ()123 123000 000 0qqqa qqq===   ===  (1) Since the initial velocities are zero, all of the rν [see Eq. (12.161b)] vanish, and the rµ are given by [see Eq. (12.161a)] 3sin sin sin24 2 4rar r r π ππµ =+ +   (2) so that 1 2 321 2 0 21 2a aµ µ µ +=   =   −=  (3) The quantities ( ) si are the same as in Example 12.7 and are given in Eq. (12.165). The displacements of the particles are n 1jr nπ +   () () () () () ( ()) () ()11 3 1 21 3 1 31 3 112cos cos cos cos24 21cos cos cos cos22 12cos cos cos cos24qt a t t a t t a t t a t t qt a t t a t tωω ωω ωω ωω ωω ωω=+ + − =− + + =− + +3 3 3qt (4) where the characteristic freq uencies are [see Eq. (12.152)] 2s in , 1 ,28rrrmdτπω=,3 = (5) Because all three particles were initially displaced, there can exist no normal modes in which any one of the particles is located at a node. For three particles on a string, there is only one normal mode in which a particle is located at a node. This is the mode 2 ωω= (see Figure 12-11) and so this mode is absent. 12-26. Kinetic energy ( ) []2 2 222 2 123 2 2mb mb mbTm mb   =+ + ⇒ =    + θθθ Potential energy 430 CHAPTER 12 () () () () () ( ) ( ) []222 123 2 1 3 2 222 2 22 123 1 23 1 2 2 3 22 22 2 221c o s 1c o s 1c o s s i n s i n s i n s i n2 22 222 0 0kUm g b b mgb kb mgb kb kb A kb mgb kb kb kb mgb kb2   = − +− +− + − + −   ≈+ + + + + − −  +− ⇒= − + −  −+θθθ θ θ θ θθθ θ θθ θ θ θ θθ The proper frequencies are solutions of the equation [] []( )( ) ( ) ( )22 2 2 22 2 22 220 0D e t D e t 2 0mgb kb mb kb Am k b m gb k b m b k b kb mgb kb mb +− −  =− = − + − −  −+ − ω ωω ω2 22 We obtain 3 different proper frequencies 2 11 2 22 2 334.64 rad/s 334.81 rad/s 4.57 rad/smg kb mg kb mb mb mg kb mg kb mb mb mg g mb f++=⇒ == ++=⇒ == =⇒ = =ωω ωωωω Actually those values are very close to one another, because k is very small. 12-27. The coordinates of the system are given in the figure: L1 m1 m2θ1 θ2L2 Kinetic energy: () ( ) ( )22 2 22 2 2 1 1 1 2 1 1 2 2 1212 1 2 22 2 2 2 1 11 21 222 21212112c os22 11 1 22 2jk jk jkTmL m L L L L mL mL mL mLL m=+ + − − ≈+ + − = ∑      θθ θ θθ θ θ θ θθ θ θ θ COUPLED OSCILLATIONS 431 ( )2 12 1 2 1 2 2 212 22jkmm L m L LmmLL mL  +− ⇒=   −   Potential energy: ( ) ( ) ( ) ()11 1 2 1 1 2 2 22 12 12 1 2 21c o s 1c o s 1c o s 1 22 2jk j k jkUm g L m g L L mm g L m g L A  =− + − + −  ≈+ + = ∑θ θθ θθθθ ( )12 1 220 0jkmm g LAmg L  +⇒=     Proper oscillation frequencies are solutions of the equation [] [] ( )2Det 0 Am− = ω () ( ) () ( ) ( )22 2 12 1 2 12 1 1 2 2 1 2 1,2 1122mm g LL mm g m LL m LL mLL  ++ + + − + + ⇒= ω The eigenstate corresponding to 1ω is 11 21a a  where () ( ) () ( )() ()12 1 12 21 1122 21212 1 2 12 1 1 2 2 1 221mm L gm La amLmm g LL mm g m LL m LL +=− ++ + + − + +  × The eigenstate corresponding to 2ω is 12 22a a  where () () ( ) () ( ) ( )12 1 12 22 1222 21212 1 2 12 1 1 2 2 1 221mm L gm La amLmm g LL mm g m LL m LL +=− ++ − + − + +  × These expressions are rather complicat ed; we just need to note that and have the same sign 11a21a 11 210a a> while and have opposite sign 12a22a11 210a a<.  The relationship between coordinates ( ) 12,θθ and normal coordinates 12,ηη are 12 11 11 1 11 2 2 22 22 1 1 2 2 2 1 1 21 21~ ~a aa a aa a a−=+  ⇔ =+   −2 2η θθθηη θη ηη θθ 432 CHAPTER 12 To visualize the normal coordinate 1η, let 20=η . Then to visualize the normal coordinate 2η, we let 10=η . Because 11 210a a> and 12 22a a0<, we see that these normal coordinates describe two oscillation modes. In the first one, the two bobs move in opposite directions and in the second, the two bobs move in the same direction. 12-28. Kinetic energy: []2 2 22 22 11 22 2 20 11 22 0mbTm b m b mmb =+ ⇒ =    θθ Potential energy: () () () []11 22 1 22 1 22 21c o s 1c o s s i n s i n2kUm g b m g b b b mg b k b k bAkb m gb kb=− +− + −  +−⇒≈ −+2 θ θθ θ Solving the equation, [] [] ( )2Det 0 Am−ω =, gives us the proper frequencies of oscillation, 22 1ω 25 (rad/s)g b==22 2 1225.11(rad/s)g kk bm m=+ + =ω The eigenstate corresponding to 1ω is 11 22a a  with 21 117.44 aa= The eigenstate corresponding to 2ω is 12 22a a  with 22 128.55 aa= From the solution of problem 12-27 above, we see that the normal coordinates are 12 11 21 22~0a a−= +2 .12 η θθ θ θ 11 21 21 21~0a a−= +2 .13 η θθ θ θ Evidently 1η then characterizes the in-phase oscillation of two bobs, and 2η characterizes the out-of-phase oscillation of two bobs. Now to incorporate the initial conditions, let us write the most general oscillation form: ( ) ( ) ( )11 22 11 22 11 2211 1 1 2 22 1 2 2 11 12Re Re Re 7.44 8.35it i it i it i it i it i it iae ae ae ae ae ae−− −− −−=+ =+ =−ωδ ωδ ωδ ωδ ωδ ωδθα α α ααθα where α is a real normalization constant. The initia l conditions helps to determine parameters α’s, a’s, δ’s. COUPLED OSCILLATIONS 433 ()( ) ()()11 1 1 12 2 21 1 1 12Re 0 7 cos cos 0.122 rad Re 0 0 7.44 cos 8.35 cos 0ta a ta a° == − ⇒ + = − == ° ⇒ + = θα δα δ θα δ α2δ = 11 sin sin 0 ⇒= δ δ . Then ()11 1 1 1 2 2 1 2 21 1 1 1 2 2 2cos cos 0.065 cos 0.057 cos 7.44 cos 8.35 cos 0.48 cos cosat at t atat t=+= − − =+= −1tθ αω αω ω ω θ αω αω ω ω where 15.03 rad/s=ω , 24.98 rad/s=ω (found earlier) Approximately, the maximum angle 2θ is 0.096 rad and it happens when 2 1cos 1 cos 1t t=  =− ω ω which gives ()2 1 1 22 21 21 2tn k tk n=  +⇒==+ωπ ω ωπ ω because 1 2101 100=ω ω we finally find 50 kn== and 210063 s t==π ω. Note: 2m a x 0.96 rad= θ and at this value the small-angl e approximation breaks down, and the value 2maxθ we found is just a rough estimate. 434 CHAPTER 12 CHAPTER 13 Continuous S ystems; Waves 13-1. The initial velocities are zero and so all of the rν vanish [see Eq. (13.8b)]. The rµ are given by [see Eq. (13.8a)] 0 323sin sinL r rAx rxdxLL L Aππµ δ= =∫ (1) so that 3 0, 3rA rµ µ=    =≠  (2) The characteristic frequency 3ω is [see Eq. (13.11)] 33 Lπτωρ= (3) and therefore, ()33,c os s inxqxt A tLLπτ π ρ  =      (4) For the particular set of initial conditions us ed, only one normal mode is excited. Why? 13-2. L 3 h L 435 436 CHAPTER 13 The initial conditions are () ()3,03 ,0 3,23hLxxL qx hLLx xLL≤≤ =  − ≤≤ (1) (),0 0qx =  (2) Because (),0 0qx =  , all of the rν vanish. The rµ are given by ()3 22 03 2263sin sin 9sin3L L r Lhr x h r xxd xL xLL L L hr r=+ − ∫∫ππµ π πdx = (3) We see that 0rµ= for r = 3, 6, 9, etc. The displacement function is ()12 293 12 1 4, sin cos sin cos sin cos24 1 6h xx xqxt t t tLL L=+ × −…ππ πωωπ4−ω (4) where rr Lπτωρ= (5) The frequencies 3ω, 6ω, 9ω, etc. are absent because the initial displacement at 3L prevents that point from being a node. Thus, none of the harmonics with a node at 3L are excited. 13-3. The displacement function is () 13 2, 13 1 5sin cos sin cos sin cos89 25qxt xx xtthL L Lππ πωωπ+ …5tω+ =− (1) where 1 1 rL rπτωρ ωω=    =  (2) For t = 0, () 2,0 13 1 5sin sin sin89 25xx x hL L L=− + + …ππ π πqx (3) The figure below shows the first term, the fir st two terms, and the first three terms of this function. It is evident that the triangular shape is well represented by the first three terms. CONTINUOUS SYSTEMS; WAVES 437 1 term L1 0 2 terms L1 0 3 terms L1 0 The time development of q(x,t) is shown below at intervals of 1 of the fundamental period. 8 t = 0, T tT=1 87 8,T tT=1 434,T tT=3 85 8,T tT=1 2 13-4. The coefficients rν are all zero and the rµ are given by Eq. (13.8a): () ()2 0 338sin 1611L r rrxxL x d xLL r=−  =− −∫πµ π (1) so that 30, even 32, o dd 3rr r rµ π =  (2) Since 438 CHAPTER 13 (),s in cosr rrxqxt tLrπµ ω =∑ (3) the amplitude of the n-th mode is just nµ. The characteristic frequencie s are given by Eq. (13.11): nn Lπτωρ= (4) 13-5. The initial conditions are () ()0,0 0 1,2, 0, otherwiseqx vx s qxt=  −≤ =      (1) The rµ are all zero and the rν are given by [see Eq. (13.8b)] ()()2 0 2 02sin 4sin sin2Ls r r Ls rrxvdLL v rr s rL+ −=− =∫πνω ππ πωx (2) from which ()()12 00, even 41s in , o drr rr v rsrrLπ πω− =−− dν (3) (Notice that the even modes are all missing, as expected from the symmetrical nature of the initial conditions.) Now, from Eq. (13.11), 1Lπτωρ= (4) and 1 rr ω ω= . Therefore, ()()12 0 2 141s in , o dr rv rsrrLdπνπω− −=− (5) According to Eq. (13.5), CONTINUOUS SYSTEMS; WAVES 439 (),s in sin sinrit r r rr rrxqxt eL rxtL= =−∑ ∑ω πβ πνω (6) Therefore, ()0 13 14 13 3, sin sin sin sin sin sin9v sx s xt t tLL L Lππ π πωωπω=− … qx (7) + Notice that some of the odd modes—those for which ( ) sin 3 0 sLπ =—are absent. 13-6. The initial conditions are () ()0 0,0 0 404 4,024 02qx v LxxL vLLqx x xL LxL2L=  ≤≤       =−≤ ≤      ≤≤    (1) The velocity at t = 0 along the string, (),0qx , is shown in the diagram. v Lv0 L 4L 23 4L The rµ are identically zero and the rν are given by: () 10 0 32 12,0 sin 8sin 2 sin24L rrxqx d xrL L v r r=− r  =−  ∫πνω π π πω (2) Observe that for r = 4n, rν is zero. This happens because at t = 0 the string was struck at 4L, and none of the harmonics with modes at that point can be excited. Evaluation of the first few rν gives 440 CHAPTER 13 0 14 2 1 0 25 2 11 00 36 22 1180.414 0 812 .41 41 25 88 2.414 2 27 216v v vvννπω νπω πω ννπω πω=− ⋅ = 0 284 vν   =− ⋅ = ⋅    =− ⋅ = ⋅ (3) and so, ()0 12 2 1 358 12, 0.414 sin sin sin sin4 2.414 3 2.414 5sin sin sin sin27 135v xxqxt t tLL xxttLL=+  +− …ππωωπω ππωω − (4) From these amplitudes we can find how many db down the fundam ental are the various harmonics: Second harmonic: 20.25010 log 4.4 db0.414=− (5) Third harmonic: 22.414 2710 log 13.3 db0.414=− (6) These values are much smaller than those found fo r the case of example (13.1). Why is this so? (Compare the degree of symmetry of the initial conditions in each problem.) 13-7. 3 7L3 7L h hL O Since (),0 0qx =  , we know that all of the rν are zero and the rµ are given by Eq. (13.8a): () 02,0 sinL rrxqx d xLLπµ=∫ (1) The initial condition on (),qxt is CONTINUOUS SYSTEMS; WAVES 441 () () ()73,037 73,0 2 ,77 74,37hxxL hx L L x LL hLx LxLL4L qx −≤ ≤        =− ≤ ≤      −≤ ≤   (2) Evaluating the rµ we find 2298 4 3sin sin37rhr r r 7π πµπ =−   (3) Obviously, 0rµ= when 47r and 37r simultaneously are integers. This will occur when r is any multiple of 7 and so we conclude that th e modes with frequencies that are multiples of 17ω will be absent. 13-8. For the loaded string, we have [see Eq. (12.152)] ()2s in21rr md nτπω=+ (1) Using mdρ= and ( )1 Ln=+ d, we have () () ()2sin21 21sin21rr dn n r Ln=+ +=+τπωρ τ π ρ (2) The function ()()1s i n21 2r rnn Lω π τ ρ=++ (3) is plotted in the figure for n = 3, 5, and 10. For comparison, the characteristic frequency for a continuous string is also plotted: 2 2r r Lω π τ ρ= (4) 442 CHAPTER 13 00 2 4 6 8 1024681012 Continuous stringn = 10 n = 5 n = 3ω τ ρr L2 Of course, the curves have meaning only at the points for which r is an integer. 13-9. From Eq. (13.49), we have: 22 2 0 ;2Ds bπτβωρ ρ== (1) From section 3.5, we know that underdamped motion requires: 22 0 β ω< Using (1) this becomes 22 2 24Ds bπτ ρ ρ< 22 2 22 2 22 24or underdamped 4ise critically damped 4overdampedsDb sDb sDbρπ τ ρπ τ ρπ τ< = >Likew The complementary solution to Eq. (13.48) for underdamped motion can be written down using Eq. (3.40). The result is: () ( ) 1 cost sstC e tβ s η ωφ−=− CONTINUOUS SYSTEMS; WAVES 443 where 22 102ω ωβ=− , 0ω and β are as defined in (1), and C and s s φ are arbitrary constants depending on the initial conditions. The complete solution to Eq. (13.48) is the sum of the particular and complementary solutions (analogous to Eq. (13.50)): () ()() 0 122 222s i n c o s2coss t ss ssFt tC e t sDbbβπω δ φ πτηω ρ ωωρρ−−=− + −+  where 1 22 2tansD s bωδ πτρ ωρ− =   −     From Eq. (13.40): () ( ),s inr rrxqxt tbπη =∑ Thus ()() 22 2 0 222 222s i n c o s2,e xp c os sin24 (underdamped)r rr rrFtDt s D r xqxt C tbb rDbb  −    =− − − +       −+   ∑πωδπτπφρρ ρ πτρω ωρρ 13-10. From Eq. (13.44) the equation for the driving Fourier coefficient is: () ( ) 0,s i nb ssxft F x t d xbπ=∫ If the point x is a node for normal coordinate s, then where is an integerxnnsbs=≤ (This comes from the fact that normal mode s has s-half wavelengths in length b.) For xn bs=, () sin sin 0; hence 0ssxnfbt = ==ππ 444 CHAPTER 13 Thus, if the string is driven at an arbitrary point, al modes with nodes at the driving point will be excited.none of the norm 13-11. From Eq. (13.44) () ( ) 0,s i nb ssxft F x t d xbπ=∫ (1) where (),Fxt is the driving force, and ()sft is the Fourier coefficient of the Fourier expansion of (),Fxt . Eq. (13.45) shows that is the component of ()sft (),Fxt effective in driving normal coordinate s. Thus, we desire (),Fxt such that ()0f o r 0f o r sft s n sn= ≠ ≠ = From the form of (1), we are led to try a solution of the form () ( ),s innxFxt gtb=π where g(t) is a function of t only. Thus () () 0sin sinb snx sxft g t d xbb=∫ππ For n ≠ s, the integral is proportional to () 0sinb xnsx bπ =±  ; hence ()0sft = for s ≠ n. For n = s, we have () () ()2 0sin 02b snx bf t gt d x gtb==∫π≠ Only the n normal coordinate will be driven. th () ( )thThus, to drive the harmonic only, ,s inn nxFxt gtb=π CONTINUOUS SYSTEMS; WAVES 445 13-12. The equation to be solved is 22 0ss sDs bπτηη ηρρ+ +   = (1) Compare this equation to Eq. (3.35): 2 0 20xx x βω+ +=   The solution to Eq. (3.35) is Eq. (3.37): () ( ) ( )22 22 10 2 exp expte A t A t−+ −ββω βω0 xt=− − Thus, by analogy, the solution to (1) is ()22 22 22 2 12 22exp exp44Dt sDs Dsnt e A t A tbb−    =− + − −         ρ πτ πτ ρρ ρρ 13-13. Assuming k is real, while ω and v are complex, the wave function becomes ()( ) (),it i t k x tk x txt A e Aee+− − −= =αβ α βψ (1) whose real part is () ( ) ,c ostxt A e t k xβψ−= α− (2) and the wave is damped in time, with damping coefficient β. From the relation 2 2 2kvω= (3) we obtain (4) () (2 2 2ik u i w αβ+= + ) By equating the real and imaginary part of this equation we can solve for α and β in terms of u and w: 2ku wαβ= (5) and (6) kw ikuβ =  Since we have assumed β to be real, we choose the solution 446 CHAPTER 13 kwβ= (7) Substituting this into (5), we have kuα= (8) as expected. Then, the phase velocity is obtained from the oscillatory factor in (2) by its definition: ReVkkωα= = (9) That is, Vu= 13-14. Vn+2 Vn–2 L′ C′In Vn–1 L′In+1 Vn L′In+2 Vn+1 Qn–1 C′Qn C′Qn+1 C′Qn+2 C′Qn–2L′In–2 Consider the above circui t. The circuit in the inductor is I, and the voltage above ground at the point between the elements is V. Thus we have thnn thnn n nQVC=′ and 1 1n nn ndILV Vdt QQ CC− −=− ′ =−n ′ ′ (1) We may also write 1n nndQIIdt+ =− (2) Differentiating (1) with respect to time and using (2) gives [2 1 212n nn ndILI Idt C]1I− + =− + ′′ (3) or [2 1 212n nn ndIII Idt L C− =− + ]1+′′ (4) CONTINUOUS SYSTEMS; WAVES 447 Let us define a parameter x which increases by x∆ in going from one loop to the next (this will become the coordinate x in the continuous case), and let us also define ;LCLCxx′ ′≡≡∆ ∆ (5) which will become the inductance and the capacita nce, respectively, per unit length in the limit . 0 x∆→ From the above definitions and 1 rrII I+ r ∆=− (6) (4) becomes (2 1 210n nndIIIdt L C− ) + ∆− ∆ =′′ (7) or, ()2 20n nI dI dt L C∆∆− =′′ (8) Dividing by ()2x∆ , and multiplying by (– L′C′), we find () ()2 2 20n nI dILCdt x∆∆− = ∆ (9) But by virtue of the above definitions, we ca n now pass to the continuous limit expressed by () (),nIt I x t → (10) Then, () ( ) ()2 22, ,0Ix t Ix tLCxt∆∆ ∂− =∆∂ (11) and for , we obtain 0 x∆→ 22 22 210II xv t∂∂− =∂∂ (12) where 1v LC= (13) 13-15. Consider the wave functions () () () ()1 2exp expAi t k x Bi t k k xψω ψω ω =−    =+ ∆ −+∆    (1) where ; kk ω ω ∆∆ . A and B are complex constants: 448 CHAPTER 13 () ()exp expa bAA i BB iφ φ =  = (2) The superposition of 1ψ and 2ψ is given by () ()12 22exp exp exp22tk x tk xi abkit k x A ieB ieωiωψψψ ωωφ∆− ∆ ∆− ∆ φ −    =+    ∆∆   =+ − + × +            (3) which can be rewritten as 2exp22 2ab ba tk x tk xiiab kiw t k x A e B eωφ φ ωφ φφφ ωψ∆− ∆−+ ∆− ∆+− 2 −         + ∆∆   =+ − + + × +              (4) Define batx k ω δ φφα∆− ∆ ≡    −≡  (5) and ( ) ( ) 22 ii iAeB eδα δαeθ −+ ++= Γ (6) Therefore, ( )2222A B Γ= + (7) ()( ) 1222cos cos22a AB ABδθ+ +=  + (8) ()( ) 1222sin sin22a BA ABδθ+ −=  + (9) That is, θ is a function of ()() tk∆− ∆ x ω . Using (6) and (7) – (9), we can rewrite (4) as exp22 2i ab kit k xeθ φφ ω  + ∆∆   =Γ + − + +       ψω (10)  and then, Re cos cos22 2 sin sin22 2ab abktk x ktk x + ∆∆    =Γ + − + +           + ∆∆    −+ − + +         φφ ωψω θ φφ ωωθ (11) CONTINUOUS SYSTEMS; WAVES 449 From this expression we see that the wave func tion is modulated and that the phenomenon of beats occurs, but for A ≠ B, the waves never beat to zero amplitude; the minimum amplitude is, from Eq. (11), A B− , and the maximum amplitude is A B+. The wave function has the form shown in the figure. AB+ AB− Reψ wt – kx 13-16. As explained at the end of section 13.6, the wave will be reflected at and will then propagate in the – x direction. 0 xx= 13-17. We let ,2 ,2jmj n m mj n= ′ =  1 = + ′′ (1) where n is an integer. Following the procedure in Section 12.9, we write ( 22 2 1 2 22nn nn nFm q q qqd− == − +′ )1+τ (2a) ( 21 21 2 21 22 2nnn n Fm q qqqd++ +== − +′′ ) n+τ (2b) Assume solutions of the form ( ) 2 2it n k d nqA eω−= (3a) ()[ ] 21 21it n k d nqB eω−+ += (3b) Substituting (3a,b) into (2a,b), we obtain ( ) ( )2 22 2ikd ikd ikd ikdAB e A Bemd BA e B Aemdτω τω− −−= − +′   −= − + ′′ (4) from which we can write 450 CHAPTER 13 2 222cos 0 22cos 0AB kdmd md Ak dBmd mdττω ττω −− = ′′  −+ −= ′′ ′′   (5) The solution to this set of coupled equations is obtained by setting the determinant of the coefficients equal to zero. We th en obtain the secular equation 2 22 22 1 2cos 0kdmd m d mm dττ τωω  −− −  ′′ ′ ′ ′′    = (6) Solving for ω, we find 12211 11 4sinkddmm mm m mτ 2ω2   =+ ± + −      ′′ ′ ′′ ′ ′ ′ ′     (7) from which we find the two solutions 122 22 1 122 22 211 11 4sin 11 11 4sinkddmm mm m m kddmm mm m mτω τω   =+ + + −     ′′ ′ ′′ ′ ′ ′ ′       =+ − + −     ′′ ′ ′′ ′ ′ ′ ′    (8) If m′ < m″, and if we define 2 22,,ab cmd m dττ2 a b ω ωω ω ≡≡ =′′ ′ω+ (9) Then the ω vs. k curve has the form shown below in which two branches appear, the lower branch being similar to that for m′ = m″ (see Fig. 13-5). ωaωbωc 0 kω π/2d Using (9) we can write (6) as ( ) ()2 22 2 2 22sinab abkd Wωω ωω ωωω=+ − ≡ (10) From this expression and the figure above we see that for c ωω> and for abω ωω<< , the wave number k is complex. If we let , we then obtain from (10) kκ=+ iβ ( ) ()22 2 2 2sin sin cosh cos sinh 2 sin cos sinh cosh id d d d d i d d d dWκβ κ β κ β κ κ β β ω+= − + = (11) Equating the real and imaginary parts, we find CONTINUOUS SYSTEMS; WAVES 451 ()22 2 2sin cos sinh cosh 0 sin cosh cos sinhdd d d dddd Wκκ β β κβκβ=  ω −=  (12) We have the following possibilities that w ill satisfy the first of these equations: a) sin κd = 0, which gives κ = 0. This condition also means that cos κd = 1; then β is determined from the second equation in (12): ()2sinh dWβ ω −= (13) Thus, c ωω> , and κ is purely imaginary in this region. b) cos κd = 0, which gives κ = π/2d. Then, sin κd = 1, and ()2cosh dWβ ω = . Thus, abω ωω<< , and κ is constant at the value π/2d in this region. c) sinh βd = 0, which gives β = 0. Then, ()2ndW siκ ω = . Thus, a ωω< or bcω ωω<< , and κ is real in this region. Altogether we have the situatio n illustrated in the diagram. k κκ κ κ ωa ωbωc ωββπ 2d 13-18. The phase and group velocities for the propag ation of waves along a loaded string are ()( ) sin 2 22ckd dVkkk dωω== (1) () ()cos 22cd dUk k ddkωω== (2) where ( ) sin 2c kd ωω= (3) The phase and group velocities have the form shown below. V,U U(k)V(k) 0 k π/dωcd 2 When kdπ= , U = 0 but cd Vω π = . In this situation, the group (i.e., the wave envelope) is stationary, but the wavelets (i.e., the wave struct ure inside the envelope) move forward with the velocity V. 452 CHAPTER 13 13-19. The linear mass density of the string is described by 1 21if 0; if 0xx xLL < >  =  >< < ρ ρ ρρ I II III 0xρ1 ρ1ρ2 Consider the string to be divided in three different parts: I for x < 0, II for 0 < x < L, and III for x > L. Let be a wave train, oscillating with frequency ω, incident from the left on II. We can write for the different zones the corresponding wave functions as follows: ( 1 it k xAeωφ−=) IIψ() () () () ()11 11 22 22 1it k x it k x it i k x i k x I it k x it k x it i k x i k x it k x IIIAe Be e Ae Be Ce De e Ce De Eeωω ω ωω ω ωψ ψ−+ − −+ − −   =+ = +     = + =+    = (1) Where 12 1 2 12 1,, , V VVV2kkω ωτ τ ρ ρ=== = (2) and where τ is the tension in the string (constant throughout ). To solve the problem we need to state first the boundary conditions; these will be given by the co ntinuity of the wave functi on and its derivative at the boundaries x = 0 and x = L. For x = 0, we have ( ) ( )II I II I 0000 xxxx xx==== =   ∂∂ =∂∂ ψψ ψψ (3) and for x = L, the conditions are ( ) ( )II III II III0 xL xLxx xx==== = L  ∂∂ =∂∂ ψψ ψψ (4) Substituting ψ as given by (1) into (3) and (4), we have ()2 1ABCD kABC Dk+=+    −+= −+   (5) and CONTINUOUS SYSTEMS; WAVES 453 22 1 22 1 1 2ik L ik L ik L ik L ik L ik LCe De Ee kCe De Eek−− −− +=    −+= −  (6) From (6) we obtain () ()21 211 2 1 2112 112ik k L ik k LkCE ek kDE ek− −+ =+     =−   (7) Hence, 22 21 21ik L kkCekkD+=− (8) From (5) we have 2 11111122kk2A Ckk =+ +−  D (9) Using (7) and rearranging the above equation () ()()22 12 2 21 12 11 2ik L kkA ek kkk kD  +=− −  −     (10) In the same way () ()22 21 12 11 2ik LBk ke k kkD   =− + + +  (11) From (10) and (11) we obtain ( ) () ()2 222 2 12 2 2 12 121ik L ik Lkke B A kke kk −−= +− −2 (12) On the other hand, from (6) and (8) we have ()21 2 212 ik k L kDEekk+=− (13) which, together with (10) gives () () ()12 212 2 2 12 124ik k L ik Lkk e E A kke kk+ = +− −2 (14) Since the incident intensity is proportional to 0I2A, the reflected intensity is 2 rIB= , and the total transmitted intensity is 2 tEI= , we can write 454 CHAPTER 13 22 0 2,rtBEII II0 2A A== (15) Substituting (12) and (14) into (15), we have, for the reflected intensity, ( ) () ()2 2222 2 12 0 2 2 12 121ik L rik Lkke II kke kk −−= +− −2 (16) From which ( )() ( )222 12 2 0 244 2 2 22 12 1 2 12 21 cos 2 6 cos 2rkk k L kk k k kk k L −−=++ − −II (17)  and for the transmitted intensity, we have () () ()12 22 12 0 2 2 12 124ik k L tik Lkk eII kke kk+ = +− −2 (18) so that ( )22 12 0 244 2 2 22 12 1 2 12 28 62 c os2tkk kk k k kk k L= ++ − −II (19) We observe that II , as it must. 0 rt I += For maximum transmission we need minimum reflection; th at is, the case of best possible transmission is that in which 0 0t rII I=   = (20) In order that , (17) shows that L must satisfy the requirement 0rI= 2 1c o s 2 0 kL − = (21) so that we have 22,0 ,1,2,mmLmkππ τ ωρ== = … (22) The optical analog to the reflecti on and transmission of waves on a string is the behavior of light waves which are incident on a medium that consists of two parts of different optical densities (i.e., different indices of refraction). If a lens is given a coating of precisely the correct thickness of a material with the proper index of refrac tion, there will be almost no reflected wave. CONTINUOUS SYSTEMS; WAVES 455 13-20. II I M 0y We divide the string into two zones: I: 0 II: 0x x< > Then, () () ()I1 1 II 2it k x it k x it k xAe Be Ae−+ − =+  = ωω ωψ ψ (1) The boundary condition is ( ) ( )II I 0xx 0 === ψ ψ (2) That is, the string is continuous at x = 0. But because the mass M is attached at x = 0, the derivative of the wave function will not be cont inuous at this point. The condition on the derivative is obtained by inte grating the wave equation from x = –ε to x = +ε and then taking the limit ε → 0. Thus, 2 II I 2 0 0 x xMtx x= =∂∂ ∂ =−∂∂ ∂ ψψ ψτ  2 (3) Substituting the wave functions from (1), we find 11A BA+ = (4) ( )2 21 1 ik A A B MAτ−+− = −2 ω (5) which can be rewritten as ( )2 11 2ik M ABAik− −=τω τ (6) From (4) and (6) we obtain 11 2 11AB i k A Bi k Mτ τω+=−− (7) from which we write 2 2 1 22 12 21 2Mik B M A ik M M ikω τ ω τωω==−− τ (8) Define 456 CHAPTER 13 2 tan2MPkωθτ== (9) Then, we can rewrite (8) as 1 11Bi P A iP−=+ (10) And if we substitute this result in (4), we obtain a relation between 1A and 2A: 2 11 1A A iP=+ (11) The reflection coefficient , 2 1 1BRA= , will be, from (10), 22 2 1 2 1tan 11 tanBPRAP2θ θ== =++ (12) or, 2sinR θ = (13) and the transmission coefficient, 2 2 1ATA= , will be from (10) 2 2 2 111 11 tanATAP2θ== =++ (14) or, 2cosT θ = (15) The phase changes for the reflected and transmitted waves can be calculated directly from (10) and (11) if we substitute 11 1 11 1 22 2i i iBB e B AA e A A Ae Aφ φ φ =  =   = (16) Then, ( ) ()1 11 tan 1 1 1 2 11 1A i i B BPeAA Pφφ− −== +B Pe (17) and ( ) ()1 21 2 tan 2 2 111 1AAi i A AeAA Pφφ − − −== +Pe (18) CONTINUOUS SYSTEMS; WAVES 457 Hence, the phase changes are () () ()11 2111 111tan tan cot tan tan tanBA AAP Pφφ θ φ φθ−− −−−= = −= − = − = −θ (19) 13-21. The wave function can be wr itten as {see Eq. (13.111a)] (1) () ()(),it k xxt Ake d kωψ+∞ − −∞=∫ Since A(k) has a non-vanishing value only in the vicinity of 0 kk=, (1) becomes (2) ()()0 0,kk it k x kkxt e d kωψ+∆ − −∆=∫ According to Eq. (13.113), ( ) 00 0 kk ωω ω=+ − ′ (3) Therefore, (2) can now be expressed as ()() ( ) ()() ()() () () () () ()0 00 0 0 0 00 00 00 0 00 0 0 00 0, 2 2kk iw k t i t x k kk ik k t x ik k t x ik t i k t i tx k i tx kxt e e d k eeeit x ee e tx i+∆ −−′′ −∆ +∆ − −∆ − ′′ −′ −− ∆ − ∆ ′′ ′=   −=  −′      −= −′ ∫ωω ωω ωω ωω ω ωψ ω ω (4) and writing the term in the brackets as a sine, we have ()()()000 02s i n ,it k xtxk xt etxωω ψω− −∆′=−′ (5) The real part of the wave function at t = 0 is ()( ) 02s i nRe ,0 cosxkxxψ∆= kx (6) If ∆ , the cosine term will undergo many oscillati ons in one period of the sine term. That is, the sine term plays the role of a slowly va rying amplitude and we have the situation in the figure below. 0 kk 458 CHAPTER 13 xReΨ(x,0) 13-22. a) Using Eq. (13.111a), we can write (for t = 0) (1) () () () () ()2 0 2 00 0 2 0,0ikx kk ikx kk i kk x ik x ik x ui u xxA ke dk Be e d k Be e e dk Be e e du+∞ − −∞ +∞ −− − −∞ +∞ −− − − − −∞ +∞ − −− −∞= = = =∫ ∫ ∫ ∫σ σ σψ This integral can be evaluated by co mpleting the square in the exponent: 2 2 222 2 2 24 4 2 4bax xax bx a bbb ax xa a a b baxa aee d x e d x ee ee d x+∞ +∞ −− − −∞ −∞ +∞ −− +  −∞ +∞ −− −∞= = =∫∫ ∫ ∫dx (2) and letting 2 yxba=− , we have 2 2 24b ay ax bx ae e dx e e dy+∞ +∞ − − −∞ −∞= ∫∫ (3) Using Eq. (E.18c) in Appendix E, we have 2 24b ax bx aee d x eaπ+∞ − −∞= ∫ (4) Therefore, CONTINUOUS SYSTEMS; WAVES 459 ()2 0 4,0x ik xxB e eσ πψσ− −= (5) The form of (),0xψ (the wave packet) is Gaussian with a 1 width of e 4σ, as indicated in the diagram below. −2σ 2σ1 e⋅Bπ σBπ σΨx,0() x b) The frequency can be expressed as in Eq. (13.113a): () ( ) 00 0 kk k ωω ω = +− +′ … (6) and so, (7) () ()() ()() () ( ) () () () ( )00 0 2 00 0 00 0 2 00 0,it k x it k k t k x ik k t k k x it k x k k it k x it x u uxt Ak e d k Ak e d k Be e e dk Be e e du+∞ − −∞ +∞  +− −′ −∞ +∞  −− −′ −− −  −∞ +∞ −− ′ − −∞= = = =∫ ∫ ∫ ∫ω ωω ω ωσ ωω σψ Using the same integral as before, we find ()() ( )2 00 0 4,it k x t xxt B e e−− − ′=ω ω πψσσ (8) c) Retaining the second-order term in the Taylor expansion of ω (k), we have () () ()2 00 0 0 01 2kk k k k ω ω =+ −+ − + ′′ ′ … ωω (9) Then, ()()()() () () () ()2 00 00 000 2 0 00 01 2 2,ik k t k k t k k xit k x tiuit k x i w t x uxt e Ake d k Be e e du+∞ −+ − − −′′ ′ −  −∞ ′′  +∞−−−− ′  −∞= =∫ ∫ωωω ωσωψ (10) We notice that if we make the change 02 itσωσ− →′′ , then (10) becomes identical to (7). Therefore, 460 CHAPTER 13 ()() () 00 , 02,2it k x xtxt B i eiw tω α πψσ− −=−′′ (11) where ()() ()2 00 2 22 01 2, 4tx it xt tωσ ω α σω−+′ ′′ = +′′ (12) The 1 width of the wave packet will now be e ()()2 22 0 142etwtσω σ+′′= (13) or, ()2 0 1 412etWtωσσ′′ =+   (14) In first order, 1eW, shown in the figure above, does not depend upon the time, but in second order, 1eW depends upon t through the expression (14). But, as can be seen from (8) and (11), the group velocity is 0ω′, and is the same in both cases. Thus, the wave packet propagates with velocity 0ω′ but it spreads out as a function of time, as illustrated below. Oxt = t1t = 0Ψxt,() CHAPTER 14 The S pecial Theor y of Relativit y 14-1. Substitute Eq. (14.12) into Eqs. (14.9) and (14.10): 11vxx xcγ=−′ 1 (1) (11 1vxx xc=+ = ) ′ ′ ′ γ γγ (2) From (1) 1 11xv xcγ′  =−   From (2) 1 11 1x v x cγ′= +   So 11 1v v c cγ γ−=  +   or 221 1vcγ= − 461 462 CHAPTER 14 14-2. We introduce cosh , sinh yy vc α α ≅≅ and substitute these expressions into Eqs. (14.14); then 11 1 22 33cosh sinh cosh sinh ;xx c t xtt ac xx xxα α α= − ′   =−′   ==′′ (1) Now, if we use cosh α = cos ( iα) and i sinh α = sin ( iα), we can rewrite (1) as () () () ()11 1cos sin sin cosxx i i c t i ict x i ict iαα α α=+′   =− +′  (2) Comparing these equations with the relation between the rotated system and the original system in ordinary three-dimensional space, 11 2 21 2 33cos sin sin cosxx x xx x xxθθ θ θ=+′    =− +′ =′  (3) x2 x2′ x1x1′ θ We can see that (2) corresponds to a rotation of the 1xi c t− plane through the angle iα. 14-3. If the equation ()( )2 2 22, 1,xi c txi c tctψψ∂∇−∂0= (1) is Lorentz invariant, then in the transformed system we must have ()( )2 2 22, 1,xi c txi c tctψψ∂ ′′∇−′′ ′∂0=′ (2) where 22 2 22xyz2 2∂ ∂∂∇= + +′∂ ∂∂′ ′′ (3) We can rewrite (2) as ( )2 4 2 1,0xi c t xµ µψ =∂ ′′=∂′∑ (4) THE SPECIAL THEORY OF RELATIVITY 463 Now, we first determine how the operator 2 2xµ µ∂ ∂′∑ is related to the original operator 2 2xµ µ∂ ∂∑ . We know the following relations: xxµ µν ν νλ =′∑ (5) xxν µν µ µλ = ′ ∑ (6) µνµ λ ν λ µλλδ= ∑ (7) Then, x xx xν µν νν xµ νµλ ν∂ ∂∂==∂∂ ∂ ∂ ′′∑∑∂ (8) 2 2xx x xµν µλ µν µλ νλ ν λ xµ νλ νλλ λ λ∂∂ ∂ ∂ ∂∂ ∂ ∂′∑∑ ∑ ∑ (9) λ∂==∂ Therefore, 2 2 2 2xx xx x∂∂=∂∂′ ∂∂=∂∂ ∂=∂∑∑ ∑∑ ∑∑ ∑µν µλ µν λ µ x∂ ∂µ νλ νλ νλ νλ λλλ δ (10) Since µ and λ are dummy indices, we see that the operator 2x2 µ ∂∂∑ is invariant under a Lorentz transformation. So we have ( )2 2,0xi c t xµ µψ∂ ′′=∂′∑ (11) This equation means that the function ψ taken at the transformed point ( x′,ict′) satisfies the same equation as the original function ψ (x,ict) and therefore the equation is invariant. In a Galilean transformation, the coordinates become x y zxx v t yy v t zzv t tt=−′    =−′ =−′  =′  (12) Using these relations, we have 464 CHAPTER 14 1 1 1x y zxt xx xt xx v t yy v t zy v t tt∂∂ ∂∂ ∂ ∂ ∂ =+ = −∂∂ ∂∂ ∂∂ ∂′′ ′ ∂∂ ∂=− ∂∂ ∂′ ∂∂ ∂=− ∂∂ ∂′  ∂∂ =∂∂  ′ (13) Therefore, 222 2 2 2 2 2 22 2 2 2 2 2 2 2 2 2 2 22 21 1 111 2 1112xyz xyzx y z c t xyz c t v v v t vx t vy t vz t  ∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂++− = + + − + + +  ∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂′′ ′ ′   2 2   ∂∂∂−+ + ∂∂ ∂∂ ∂∂    (14) This means that the function ψ (x′,ict′) does not satisfy the same form of equation as does (, )xi c tψ , and the equation is not invarian t under a Galilean transformation. 14-4. In the K system the rod is at rest with its ends at and . The K′ system moves with a velocity v (along the x axis) relative to K. 1x2x KK′ x1 x2 If the observer measures the time for the ends of the rod to pass over a fixed point in the K′ system, we have 11 22 2 22 22 21 1 1 1vttc v c vttc v c =−′  1 2x x  −  =−′  − (1) where t and t are measured in the K′ system. From (1), we have 1′2′ THE SPECIAL THEORY OF RELATIVITY 465 () ( 12 12 1 2 22 21 1vtt xxc v c) tt −= − − −′′  − (2) We also have 12xx− =A (3) ( )12vt t− =A (4) ( )12vt t− = ′′A′ (5) Multiplying (2) by v and using (3), (4), and (5), we obtain the FitzGerald-Lorentz contraction: 2 21v c=−′AA (6) 14-5. The “apparent shape” of the cu be is that shape which would be recorded at a certain instant by the eye or by a camera (with an infini tesimally short shutter speed!). That is, we must find the positions that the various points of the cube occupy such that light emitted from these points arrives simultaneously at the eye of the observer. Those parts of the cube that are farther from the observer must then emit light earlier than those parts that are closer to the observer. An observer, looking directly at a cube at rest, would see just the front face, i.e., a square. When in motion, the edges of the cube are distor ted, as indicated in the figures below, where the observer is assumed to be on the line passing through the center of the cube. We also note that the face of the cube in (a) is actually bowed toward the observer (i.e., the face appears convex), and conversely in (b). (a) Cube moving toward the observer. (a) Cube moving away from the observer. 466 CHAPTER 14 14-6. K′ v x1 x2K We transform the time t at the points and in the K system into the K′ system. Then, 1x2x 1 1 2 2 2 2vxttc vxttc  =−′      =−′   γ γ (1) From these equations, we have ()12 12 21 xxttt v v xcc−− = − = − ∆ ′′′ γγ2∆= (2) 14-7. K K′ v x Suppose the origin of the K′ system is at a distance x from the origin of the K system after a time t measured in the K system. When the observer sees the clock in the K′ system at that time, he actually sees the clock as it was located at an earlier time because it takes a certain time for a light signal to travel to 0. Suppose we see the clock when it is a distance A from the origin of the K system and the time is t in K and 1 1 t′ in K′. Then we have ()11 2 1 1vttc ct t tv x tvγ =−′    −=    =  = A A A (1) We eliminate A, t, and x from these equations and we find 1 THE SPECIAL THEORY OF RELATIVITY 467 1 1vtcγ t =−′   (2) This is the time the observer reads by means of a telescope. 14-8. The velocity of a point on the surface of the Earth at the equator is ( )8 4 426 .38 1 0 cm 2 8.64 10 sec 4.65 10 cm/seceRvπ π τ×× ==× =× (1) which gives 4 6 104.65 10 cm/sec1.55 1031 0 c m / s e cv c− ×== = ××β (2) According to Eq. (14.20), the relationship between the polar and equatorial time intervals is 2 2112 1ttt β β∆ ∆= ≅ ∆ +′ − (3) so that the accumulated time difference is 21 2tt β t ∆=∆ −∆ = ∆′ (4) Supplying the values, we find ( )( )( )67 11.55 10 3.156 10 sec/yr 10 yr2−× × × ×2∆= × (5) Thus, 0.0038 sec∆= (6) 14-9. w dm′ m + dmv + dv The unsurprising part of the solution to the pro blem of the relativistic rocket requires that we apply conservation of momentum, as was done fo r the nonrelativistic case. The surprising, and key, part of the solution is that we not assume the mass of the ejected fuel is the same as the mass lost from the rocket. Hence ( )( )( )w p mv d m dm v dv dm wγγ γ γ + + + + == ′ (1) where – dm is the mass lost from the rocket, dm′ is the mass of the ejected fuel, ( )( )21 wv V v V c≡− − is the velocity of the exhaust with respect to the inertial frame, and 2211w wc γ≡− . One can easily calculate 3ddγγβ β= , ad after some algebra one obtains 468 CHAPTER 14 2 wdmmd v vd m wγγγ′++ (2) where we of course keep infinitesimals on ly to first order. The additional unknown dm′ is unalarming because of another conservation law ( )( )22 w c d m d m c d m c γγ γ γ == + + +2Em ′ (3) Subsequent substitution of dm′ into (2) gives, in one of its many intermediate forms ()21wmd v d m v wcβγ0 − +− =  (4) and will finally come to its desired form after dividing by dt ( )21dv dmmVdt dtβ 0 + −= (5) The quantity dt can be measured in any inertial frame, but would presumably only make sense for the particular one in which we measure v. Interestingly, it is not important for the ejected fuel to have an especially large kinetic energy but rather that it be near light speed, a nontrivial distinction. For such a case, a rocket can reach 0.6 c by ejecting half its mass. 14-10. From Eq. (14.14) ( )11xx vγ=−′ t (1) 1 2vtt xcγ=−′  (2) Solving (1) for and substituting into (2) gives 1x 1 2xvtt v tcγγ    ′=− +′        2 1 22vvtx t tccγγt γ+ =− = ′′ 1 2vtt xcγ =+ ′ ′   Solving (2) for t and substituting into (1) gives 11 1 2tvxx v xcγγ    ′=− +′        or ( )11xx v t γ=+ ′ ′ THE SPECIAL THEORY OF RELATIVITY 469 14-11. θ x1 From example 14.1 we know that, to an observer in motion relative to an object, the dimensions of objects are contracted by a factor of 221vc− in the direction of motion. Thus, the 1x′ component of the stick will be 22cos 1 vc θ − A while the perpendicular comp onent will be unchanged: sin θ A So, to the observer in K′, the length and orientation of the stick are ( )1222 2 2sin 1 cos vc θθ   =+ − ′ AA 1 22sintan cos 1 vcθθ θ−    =′   −  or 122 2 2cossin tan tanθθγ θγ θ =+′   =′AA 14-12. The ground observer measures the speed to be 8 100 m2.5 10 m/s.4 secvµ== × The length between the markers as measured by the racer is 22 21 2.5100 m 1 55.3 meters3vc =−′ =− =AA The time measured in the racer’s frame is given by 470 CHAPTER 14 ( )() ()12 8 2 22.5 10 m/s 100 m .4 sec 12 . 5 3 .22 secvtt xc cγ µ µ=−′   × −  = − = The speed observed by the racer is 82.5 10 m/s vtt′== = ×′AA 14-13. ( )1221.5 s 1 0.999 22.4tt tγ µ γ−∆= ∆′ ∆= =−  Therefore 34 st µ ∆′ . 14-14. KK ′ receiverv source In K, the energy and momentum of each photon emitted are 0 0andhEh pcνν== Using Eq. (14.92) to transform to K ′: ()0 11 00;hEh E v p pc vhhcννγ γν ν== − = −′  =+ THE SPECIAL THEORY OF RELATIVITY 471 So 0 0021 11 1 1v c=+  + +==− −νν γ β βννβ β which agrees with Eq. (14.31). 14-15. From Eq. (14.33) 01 1βν ν β−= + Since cλ ν= 01 1βλ λ β−= + or 01 1βλ λ β−= + With 0656.3 nm λ= and 4 841 0 31 0β×=×, λ = 656.4 nm. S o the shift is 0.1 nm toward the red (longer wavelength). 14-16. θθ′ EarthstarvK′ K Consider a photon sent from the star to the Earth. From Eq. (14.92) ( )1 EE v p γ=−′ also ( )1 EE v p γ=+ ′ ′ 472 CHAPTER 14 Now 0 01 1 , , cos , cosh hE h p pccEhν νν νθ = = =− =− ′′ θ′ Substituting yields ( ) 0 1c os ν νγ β θ=+ and ( ) 01c os νγν β θ=− ′ Thus ( )( )21 cos 1 cosβ θβ θ γ−+− ′= 122cos cos cos cos 1β θβ θ β θ θ β +− − = − ′′ cos cos cos cosθ θβ θ θ β − −=′′ − Solving for cos θ yields coscos1c osθβθβ θ−′=− ′ where angle in earth’s frame angle in star’s framevcβ θ θ= = =′ 14-17. From Eq. (14.33) 01 1βν νβ−=+ Since cν λ= , 01 1βλ λβ+=− We have 0 1.5λ λ = . This gives 5 13β= or 81.2 10 m/secv=× THE SPECIAL THEORY OF RELATIVITY 473 14-18. θ observersource lightvK′ K Proceeding as in example 14.11, we treat the light as a photon of energy hν. In 0 0 :,hhpcKEνν== ′′ ′ In ( )1 : h E pνγ ν== + ′ KE For the source approaching the observer at an early time we have 0 1hpcν= Thus 00 01 1v cβνγ ν ν νβ+ =+ = − For the source receding from the obse rver (at a much later time) we have 0 1hpcν=− and 01 1βννβ−=+ So 0 01source approaching observer1 1source receding from observer1βννβ βννβ+=− −=+ 474 CHAPTER 14 14-19. θ observersource vK K′ Proceeding as in the prev ious problem, we have In 122: cosr rtKE h hhpccν β ννθ β β= ′′ =− =−′ + In ( )10 :KE E p h γ νν =+= ′′ So 22 022 221 1r rt rt rthc cνβνν β β ββ ββhh        =− +     −− +    or ( ) 0221 1r rt−= −−ν βν β β 22 0 01 1rt rβ β λ ν νλ β−−==− For 0 λλ>, we have ( )2 2211rr t β ββ − >− − 2222tr rβ ββ>− ()221tr rβ ββ>− 14-20. As measured by observers on Earth, the entire trip takes 4 lightyears 802 y0.3 c 3=ears The people on earth age 80 3 years. The astronaut’s clock is ticking slower by a factor of γ. Thus, the astronaut ages THE SPECIAL THEORY OF RELATIVITY 475 2 80 801 0.3 0.95 years33−= So Those on Earth age 26.7 years. The astronaut ages 25.4 years. 14-21. () ( ) ( )0 0 3222 2 0 322 2122 11 1 1 1m dFmdt mββ ββ β ββ β β  −−    == +    −−   −       =+− −v vv vv    (1) If we take (this does not mean vv11v=ve23 0 = =  ), we have ( ) ( )11 1 0 1 10 1 32 322 221 11vvvm v ccvm v β ββ =+ == − −−  A 1 Fm (2) 0 2221tm 2 F vm v β== − (3) 0 3321tm 3 F vm v β== − (4) 14-22. The total energy output of the sun is ( )321.4 10 W m 4dERdt−=× ⋅ ×2π (1) where is the mean radius of the Earth’s orbit around the sun. Therefore, 111.50 10 mR=× 263.96 10 WdE dt×  (2) The corresponding rate of mass decrease is 9 214.4 10 kg sdm dE dt c dt1−= ×⋅  (3) The mass of the sun is approximately 1. , so this rate of mass decrease can continue for a time 3099 10 kg× 476 CHAPTER 14 30 13 911.99 10 yr1.4 10 yr4.4 10 kg sT−×=×⋅ × (4) Actually, the lifetime of the sun is limited by ot her factors and the sun is expected to expire about years from now. 94.5 10× 14-23. From Eq. (14.67) ()22 2 2 0 2 2 00 2 0 22 2 22 2pc E E ET E ET T pcT mc T=− = +− =+ =+ 14-24. The minimum energy will occur when the four pa rticles are all at rest in the center of the mass system after the collision. Conservation of energy gives (in the CM system) 224ppEm c = or 2 ,CM 0 22ppEm c == E which implies γ = 2 or 32 β= To find the energy required in the lab system (one proton at rest initially), we transform back to the lab ( )1 EE v p γ=+ ′ ′ (1) The velocity of K′(CM) with respect to K(lab) is just the velocity of the proton in the K′ system. So u = v. Then ()( )22 1C M v p v mu mv mc2vp γ γγ == = =′ β Since γ = 2, 32 β= , 103 2vp E=′ Substituting into (1) THE SPECIAL THEORY OF RELATIVITY 477 lab 0 0 0 0372222EE E E γ  =+ = =   7E 2 2 The minimum proton energy in the lab system 7 , of which 6 is kinetic energy.pp mc ismc 14-25. Let Bz 0B= x y vv= + vij Then 0 000 00xy yxqq v v B qvB vB×=   =− ij k vB ij () ()ddqmdt dtγ =× = =Fv B p v gives ( )0 yxqB dvvdt m=−viγj Define 0qB mω γ ≡ Thus and x yy x vv v v ω ω = =−  or 2 x y x vv ωω== −   v and 2 y x y vv ωω=− =−  v t t So cos sin cos sinx yvA t B vC t Dω ω ω ω= + =+ Take ()0xvv =, ()0yv =0. Then A = v, C = 0. Then () () 00xy 0 vv ω= =  () () 00 0,yxvv BD v=− =− →= = − v ω ω 478 CHAPTER 14 Thus cos sin vt vt= − vi j ω ω Then sin cosvvtt =+ri j ω ωωω The path is a circle of radius v ω 00 0p vm vrqB m qB qBγ γ== = From problem 14-22 122 22TpT mc =+   So 122 2 02TTmcrqB +  = 14-26. Suppose a photon traveling in the x-direction is converted into an e and as shown below −e+ before aftere+ e–θ θ Cons. of energy gives 2p e pcE= where momentum of the photon energy of energy of p ep Ee e+ −= == Cons. of gives xp ( ) 2 cos momentum of ,pe ep pp e θ e+ −== Dividing gives THE SPECIAL THEORY OF RELATIVITY 479 cosp e pepc Ecpp θ== or 22 2 2coseepc θE= (1) But , so (1) cannot be satisfied for 22 eeEp c>2 2cos 1 θ≤. An isolated photon cannot be converted to an electron-positron pair.in This result can also be seen by transforming to a frame where 0xp= after the collision. But, before the collision, 0xpp pc=≠ in any frame moving along the x-axis. So, without another object nearby, momentum cannot be conserved; thus, the process cannot take place. 14-27. The minimum energy required occurs when the p and p are at rest after the collision. By conservation of energy ( ) 022 938 MeV 938 MeVe eEET= E = =+ Since E , 0.5 MeVe= 937.5 MeVeeTT+−== 14-28. 2 classical1 2Tm = v ( )2 rel classical 1 Tm cT γ=− ≥ We desire rel classical rel0.01TT T−≤ ()2 21 2101mv mc γ−≤−.01 ()2 21 20.991v c γ≥− 2 1.981β γ≥− Putting ( )1221 γβ−=− and solving gives 480 CHAPTER 14 0.115 vc≤ 7The classical kinetic energy will be within 1% of the correct for 0 3.5 10 m/sec, independent of mass. v × value ≤≤ 14-29. 0 EEγ= For ( )9 6 0 4 122 2 10 230 10 eV 0.51 10 eV, 5.88 10 1or 1 1 111 1.4102E E γ γβ β βγ− −=× × × == − −= − ×  γ− ( )1011 . 41 0 c 0.99999999986 cv−=− × = 14-30. A neutron at rest has an energy of 939.6 Me V. Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV. Ot her than rest energies 0.8 MeV is available. 14-31. 0.98cθ θ Conservation of energy gives 2p EEπ= where E energy of each photon (Cons. of p=yp implies that the photons have the same energy). THE SPECIAL THEORY OF RELATIVITY 481 Thus 02p EEγ = 0 2135 MeV339 MeV2 2 1 0.98pEEγ== = − Th e energy of each photon is 339 MeV. Conservation of gives xp mv 2 cos where momentum of each photonppp p γ θ == ( )( ) ()2 2135 Mev/c 0.98 c s 0.98 2 1 0.98 339 MeV/c== −θco 1cos 0.98 11.3θ−= =° 14-32. From Eq. (14.67) we have 222 0 EEp c−=2 2With , this reduces to 0 EE T=+ 22 02ET T pc+= Using the quadratic formula (taking the + root since T ≥ 0) gives 22 2 00 TE p c E= +− Substituting pc = 1000 MeV ( )0electron 0.5 MeVE = ( ) 0proton 938 MeVE = gives electron proton999.5 MeV 433 MeVT T= = 482 CHAPTER 14 14-33. 120˚120˚ after beforepe n ν Conservation of yp gives sin 60 sin 60 or eepp p pν ν °= ° = Conservation of gives xp cos 60 cos 60p ee p ppν p = °+ °= So epp ppνp = =≡ Conservation of energy gives 0nepEE E Eν = ++ 22 2 22 2 00 0ne pEE p c E p c =+ ++ + pc (1) Substituting 0939.6 MeVnE= 0938.3 MeVpE= 00.5 MeVeE= and solving for pc gives p = 0.554 MeV/c 0.554 MeV/cpepppν === Substituting into 0 22 2 00TE E Ep cE=− = +− gives ( ) 00 Eν= 40.554 MeV 2 10 MeV, or 200 eV 0.25 MeVp eT T Tν −= =× = THE SPECIAL THEORY OF RELATIVITY 483 14-34. 2 2 2222 12 sc t x x x ∆ = − +++3 ′ ′′′′ Using the Lorentz transformation this becomes 22 22 122 21 222 11 23 22 22 22 2 22 2 2 1 1 22 22 23 22 2 2222 12322 11 1vxct xv txv t x v t csxvc vc vx vxc t tccxxvc ct x x x−− ++−∆= + ++′−−  −− −   +− =− + + +2x =+ So 22ss∆ =∆′ 14-35. Let the frame of Saturn be the unprimed fram e, and let the frame of the first spacecraft be the primed frame. From Eq. ( 14.17a) (switch primed and unprimed variables and change the sign of v) 1 1 1 2 1uvuuv c+′=′+ Substituting v = 0.9 c 10.2 uc=′ gives 10.93 uc= 14-36. Since ()123 and , , ,dd XX x xxidd==µµµ ττFm ct we have ()2 11 1 2 2 2 3 2 23 22 2 4 2dx d x dFm mdd d dx dxFm Fmdd di c t ddFm i cmdd dττ τ tτ τ τ ττ== == == 484 CHAPTER 14 Thus () () ()22 1 11 22 2 2 1 14 22 22 22 22 22 1 4 22 2 2 1 22 41;dx dFm m xv tdd dx dtmm v F idd dx dxFm m F FFdd vx dFi c m tdc dx dticm i mdd Fi F =− −′ ′== = + ′=== =′′ =−′  =− =−γττ γγ γττ ττ γτ γγ βττ γβ3 3Fβ Th us the required transformation equations are shown. 14-37. From the Lagrangian ( )22 1112Lm c k x β =− − −2 (1) we compute Lkxx∂=−∂ (2) 21LLmcvvβ β β β∂ ∂∂==∂∂ ∂ − (3) Then, from (2) and (3), the Lagrange equation of motion is 20 1dm ckxdtβ β + =  − (4) from which ( )3220 1mckxβ β+ = − (5) Using the relation dv dv dx dvcdt dx dt dxβ== =v (6) we can rewrite (4) as THE SPECIAL THEORY OF RELATIVITY 485 ( )2 3220 1mc dkxdxββ β+ = − (7) This is easily integrated to give 2 2 21 2 1mckx E β+ = − (8) where E is the constant of integration. The value of E is evaluated for some particular po int in phase space, the easiest being x = a; β = 0: 21 2Em c k a=+2 (9) From (8) and (9), 2 22 211 22 1mckx mc ka β+= + −2 (10) Eliminating 2β from (10), we have ( ) ( )( ) ( )24 2 2 22 2 22 2 22 2 22 21 1 2 4 2mc mc k a x kmc a x ka x kmc a x=− +−  +−  =−  +−  β (11) and, therefore, ( ) ( ) ( )22 2 22 22 24 1 2ka x m c ka x dx cd t mc k a x−+ − +−β== (12) The period will then be four times the integral of dt = dt(x) from x = 0 to x = a: ( ) ( )22 2 22 22 0 2124 14akaxm mcdxk kax axmc+− −+ −∫τ= (13)  Since x varies between 0 and a, the variable xa takes on values in the interval 0 to 1, and therefore, we can define sinx aφ= (14) from which 486 CHAPTER 14 22 cosax aφ−= (15) and 22dx a x d φ =− (16) We also define the dimensionless parameter, 22ak mc≡κ (17) Using (14) – (17), (13) transforms into ( )22 2 22 012 c o s 2 1c osadc+ = +∫πκφ τ φκ κφ (18) Since 221 ka mc  for the weakly relativistic case, we can expand the integrand of (18) in a series of powers of κ : ( ) ( )( )222 22 2 1222 22 2212 c o s 1 2 cos 1 cos2 1c os 112 c o s2 31c os2+ ≅+ − + ≅+ − =+κφ κκ φφ κφ κφ κφ (19) Substitution of (19) into (18) yields 2 22 0 2 0231c os2 31sin 222adc aa cc≅+  =+ +  ∫π πτ κφκ πκφφκφ (20) Evaluating (20) and substi tuting the expression for κ from (17), we obtain 2 2328ma kcπτπ=+k m (21) or, 2 0 23116ka mcττ =+    (22) THE SPECIAL THEORY OF RELATIVITY 487 14-38. () () (for constant)dp dFm udt dt dmu mdtγ γ== == 221dumdt uc    =   −  ( ) ( ) ( )12 1222 22 2 2211 1uuc u ucdu cmdt uc−  −− − −  −  = ( )32221dumu cdt−=− Thus ( )32221duFm u cdt−=− 14-39. The kinetic energy is 22 24 2 00 Tp c m c m c=+ − (1) For a momentum of 100 MeV/c, ()2 4 proton 10 931 931 936 931 5 MeV=+ − ≅ − = T (2) ()2 4 electron 10 0.51 0.51 100 0.5 99.5 MeV=+ − ≅ − = T (3) In order to obtain γ and β, we use the relation 2 22 0 021mcEm c m c γ β== = − (4) so that 2 0E mcγ= (5) and 211 βγ=− (6) electron1002000.51γ =≅ (7) 488 CHAPTER 14 2 electron110 .99200β=− ≅9988 (8) This is a relativistic velocity. proton9361.0054931γ =≅ (9) 2 proton111.0053β=− ≅0.1 (10) This is a nonrelativistic velocity. 14-40. If we write the velocity components of the center-of-mass system as jv, the transformation of ,jpα into the center-of-mass system becomes ,, 2j jjvE ppcα αα γ=−′ (1) where 2 21 1jv cγ= −. Since in the center-of-mass system, ,0jp=′∑ α α must be satisfied, we have ,, 20j jjvE ppc= −= ′ ∑∑α αα ααγ  (2) or, ,j jpcv c Eα α α α=∑ ∑ (3) 14-41. We want to compute 2 10 1 2 000Em c T TE m c−=− (1) where T and E represent the kinetic and total energy in the laboratory system, respectively, the subscripts 0 and 1 indicate th e initial and final states, and is the rest mass of the incident particle. 0m The expression for in terms of 0E1γ is 2 00Em c1γ − (2) 1E can be related to (total energy of particle 1 in the center of momentum reference frame after the collision) through the Lorentz transformati on [cf. Eq. (14.92)] (remembering that for the inverse transformation we switch the primed and unprimed variables and change the sign of v): 1E′ THE SPECIAL THEORY OF RELATIVITY 489 ( ) 11 1 1 1 cos EE cp γ β =+′′ ′ ′ θ 1 (3) where 10 1pm c βγ =′′ and Em2 10 c1γ =′′ : ( )22 2 10 1 1 1c os Em c γ β =+ ′′ θ (4) Then, from (1), (2), and (4), 22 2 11 1 1 01cos 1 1T Tγγ β θ γ+ − ′′ ′=− (5) For the case of collision between two particles of equal mass, we have, from Eq. (14.127), 2 1 11 2γγ+=′ (6) and, consequently, 22 2 1 11 1112γγβ γ−=− = ′′ ′ (7) Thus, with the help of (6) and (7), (5) becomes ( ) ()11 1 0111 co 21 1c o s 2T T−+ −=− +=s γ γθ γ θ (8) We must now relate the scattering angle θ in the center of momentum system to the angle ψ in the lab system. Squaring Eq. (14.128), which is valid only for 1mm2= , we obtain an equation quadratic in cos θ. Solving for cos θ in terms of tan2ψ, we obtain 2 1 2 11tan 12cos11t an2γψ θγψ+− ± =++ (9) One of the roots given in (9) corresponds to θ = π, i.e., the incident partic le reverses its path and is projected back along the incident direction. Substitution of the other root into (8) gives ()2 1 22 1012c o s 1 1 2c o s 1 s i n1t an2T Tψ γ2ψ γψψ+ +++== (10) An elementary manipulation with the denominator of (10), namely, 490 CHAPTER 14 ( ) ( ) () ()22 2 2 2 11 22 22 11 22 11 2 112 cos 1 sin 2 cos 1 cos sin sin cos cos cos 1c os c os 11 cos++ = + − + =+ + − + =+ − + =+ −−ψγ ψ ψ γ ψ ψ γ ψψ γ ψ γγψ ψ γγ ψψ (11) provides us with the desired result: () ()2 1 2 01 12c o s 11 cosT Tψ γ γψ=+− − (12) Notice that the shape of the curve changes when Tm , i.e., when 2 10 c >12 γ>. 0.2 0 0˚ 30˚ 60˚ 90˚0.40.60.81.0T T1 0 T1 = 0.1 GeV T1 = 1 GeV T1 = 10 GeV ψ 14-42. φ θhν hν′γmec2y x From conservation of energy, we have 22 ee hm c m chν γ ν += + ′ (1) Momentum conservation along the x axis gives cos cosehhmvccν νθγ′=+ φ (2) Momentum conservation along the y axis gives sin sinehmvcνγ φ′= θ (3) THE SPECIAL THEORY OF RELATIVITY 491 In order to eliminate φ, we use (2) and (3) to obtain 1cos cos sin sine ehh mv c c h mvννφ θγ νφθγ′   =−    ′ = (4) Then, 22 22 22 21s sin 1 2 cos ehh h h mv c c c cνν ν νco φ φθγ  ′′     += = + − +             (5) Since 2 21 1v cγ= − and 21cv γγ=− we have ( )22 2 21 vcγγ = − (6) Substituting γ from (1) into (6), we have () (2 2 22 222 eehhvmm c=− + − ) ′ ′ γ νν νν (7) From (5) and (7), we can find the equation for ν′: () (22 2 2 22c os 2ehh h h hhmcc c c cνν ν ν) θ νν νν′′    +− = − + −′ ′         (8) or, ()22221 c o semc mc hh2eν θν ν+− = ′  (9) Then, ()21 11 cos eh mcν ννθ      =′  +−   (10) or, ()1 211 cos eEEEmcθ− =+ −′    (11) The kinetic energy of the electron is 492 CHAPTER 14 ()22 211 11 cosee ec m c h h EE mcγν ν Tm θ =− = − = − ′ +−     ()2 2 21c o s 11 cose eETE mc mcθ θ−  =  +−    (12)