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Unformatted text preview: CHAPTER 0 Contents Preface v vii Problems Solved in Student Solutions Manual 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Matrices, Vectors, and Vector Calculus Newtonian MechanicsSingle Particle Oscillations 79 127 1 29 Nonlinear Oscillations and Chaos Gravitation 149 Some Methods in The Calculus of Variations 165 181 Hamilton's PrincipleLagrangian and Hamiltonian Dynamics CentralForce Motion 233 277 333 Dynamics of a System of Particles Motion in a Noninertial Reference Frame Dynamics of Rigid Bodies Coupled Oscillations 397 435 461 353 Continuous Systems; Waves Special Theory of Relativity iii iv CONTENTS CHAPTER 0 Preface This Instructor's Manual contains the solutions to all the endofchapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form. A Student Solutions Manual containing solutions to about 25% of the endofchapter 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 problems. There are now 509 problems, a significant number over the 4th edition. The instructor will find a large array of problems 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 provide 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 assigning 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 students to have access to this manual. Please do not give students solutions from this manual. Posting 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 assistance 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 suspected errors. Comments can be sent by email to stt@virginia.edu, the more detailed the better. Stephen T. Thornton Charlottesville, Virginia v vi PREFACE CHAPTER 1 Matrices, Vectors, and Vector Calculus 11. x2 = x2 x1 45 x1 45 x3 x3 Axes x and x lie in the x1 x3 plane. 1 3 The transformation equations are: x1 = x1 cos 45  x3 cos 45 x2 = x2 x3 = x3 cos 45 + x1 cos 45 x1 = 1 1 x1  x3 2 2 x2 = x2 x3 = So the transformation matrix is: 1 1 x1  x3 2 2 1 2 0 1 2 0  1 0 1 2 0 1 2 1 2 12. a) x3 CHAPTER 1 D E O A x1 B C x2 From this diagram, we have OE cos = OA OE cos = OB OE cos = OD Taking the square of each equation in (1) and adding, we find OE cos 2 + cos 2 + cos 2 = OA + OB + OD But OA + OB = OC 2 2 2 (1) 2 2 2 2 (2) (3) and OC + OD = OE 2 2 2 (4) 2 Therefore, OA + OB + OD = OE 2 2 2 (5) Thus, cos 2 + cos 2 + cos 2 = 1 b) x3 D E D E (6) O A A x1 B C C B x2 First, we have the following trigonometric relation: OE + OE  2OE OE cos = EE 2 2 2 (7) MATRICES, VECTORS, AND VECTOR CALCULUS 3 But, 2 EE = OB  OB + OA  OA + OD  OD 2 2 2 2 = OE cos  OE cos + OE cos  OE cos + OE cos  OE cos or, 2 2 (8) 2 2 2 EE = OE cos 2 + cos 2 + cos 2 + OE cos 2 + cos 2 + cos 2  2OE OE cos cos + cos cos + cos cos = OE 2 + OE2  2OE OE cos cos + cos cos + cos cos Comparing (9) with (7), we find cos = cos cos + cos cos + cos cos (10) (9) 13. x3 e3 O e1 x1 A e3 e2 e2 x2 e1 e2 e1 e3 Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 . Denote the new axes by x , x , x and the corresponding unit vectors by e1 , e2 , e3 . The effect of the 1 2 3 rotation is e1 e3 , e2 e1 , e3 e2 . Therefore, the transformation matrix is written as: cos ( e1 , e1 ) cos ( e1 , e2 ) cos ( e1 , e3 ) 0 1 0 = cos ( e , e1 ) cos ( e , e2 ) cos ( e , e3 ) = 0 0 1 2 2 2 1 0 0 3 3 3 cos ( e , e1 ) cos ( e , e2 ) cos ( e , e3 ) 14. a) Let C = AB where A, B, and C are matrices. Then, Cij = Aik Bkj k (1) (C ) t ij = C ji = Ajk Bki = Bki Ajk k k 4 Identifying Bki = Bt CHAPTER 1 ( ) ik and Ajk = At ( ) kj , (C ) = (B ) ( A ) t t t ij k ik kj (2) or, C t = ( AB) = Bt At t (3) b) To show that ( AB) = B1 A1 , 1 ( AB) B1 A1 = I = ( B1 A1 ) AB That is, (4) ( AB) B1 A1 = AIA1 = AA1 = I (5) (6) (B 15. 1 A1 ( AB) = B1 IB = B1B = I ) Take to be a twodimensional matrix: = Then, 11 12 = 11 22  12 21 21 22 (1) 2 2 2 2 2 2 2 2 2 2 2 2 = 11 22  211 22 12 21 + 12 21 + ( 11 21 + 12 22 )  ( 11 21 + 12 22 ) 2 2 2 2 2 2 2 2 2 2 2 = 22 11 + 12 + 21 11 + 12  11 21 + 211 2212 21 + 12 22 2 2 2 2 = 11 + 12 22 + 21  ( 11 21 + 12 22 ) ( ) ( ) ( ) (2) ( )( ) 2 But since is an orthogonal transformation matrix, Thus, ij j kj = ik . 2 2 2 2 11 + 12 = 21 + 22 = 1 11 21 + 12 22 = 0 Therefore, (2) becomes (3) =1 2 (4) 16. The lengths of line segments in the x j and x systems are j L= x j 2 j ; L = x i i 2 (1) MATRICES, VECTORS, AND VECTOR CALCULUS 5 If L = L , then x = x 2 j i j i 2 (2) The transformation is xi = ij x j j (3) Then, x = 2 j j i k ik xk i x (4) = xk x ik i i k, But this can be true only if i ik i = k (5) which is the desired result. 17. x3 (0,0,1) (0,1,1) (1,0,1) (1,1,1) (0,0,0) (0,1,0) x2 x1 (1,0,0) (1,1,0) There are 4 diagonals: D1 , from (0,0,0) to (1,1,1), so (1,1,1) (0,0,0) = (1,1,1) = D1 ; D 2 , from (1,0,0) to (0,1,1), so (0,1,1) (1,0,0) = (1,1,1) = D 2 ; D 3 , from (0,0,1) to (1,1,0), so (1,1,0) (0,0,1) = (1,1,1) = D 3 ; and D 4 , 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 D1 = D 2 = D 3 = D 4 = 3 The angle between any two of these diagonal vectors is, for example, D1 D 2 (1,1,1) ( 1,1,1) 1 = cos = = D1 D 2 3 3 6 so that 1 = cos1 = 70.5 3 CHAPTER 1 Similarly, D1 D 3 D D D D D D D D 1 = 1 4 = 2 3 = 2 4 = 3 4 = D1 D 3 D1 D 4 D2 D3 D2 D 4 D3 D 4 3 Let be the angle between A and r. Then, A r = A2 can be written as Ar cos = A2 18. or, r cos = A (1) This implies QPO = 2 (2) Therefore, the end point of r must be on a plane perpendicular to A and passing through P. 19. a) A = i + 2j  k B = 2i + 3j + k A  B = 3i  j  2k 2 2 A  B = ( 3) + ( 1) + (2)2 12 A  B = 14 b) B A component of B along A The length of the component of B along A is B cos . A B = AB cos B cos = A B 2 + 6  1 3 6 = = or A 2 6 6 The direction is, of course, along A. A unit vector in the A direction is 1 ( i + 2j  k ) 6 MATRICES, VECTORS, AND VECTOR CALCULUS 7 So the component of B along A is 1 ( i + 2j  k ) 2 c) cos = AB 3 3 3 = = ; = cos 1 AB 6 14 2 7 2 7 i j k 71 d) 2 1 1 1 1 2 j +k A B = 1 2 1 = i 3 1 2 1 2 3 2 3 1 A B = 5i + j + 7 k e) A  B = 3i  j  2k i 1 A + B = i + 5j j 5 k 0 ( A  B ) ( A + B ) = 3 1  2 ( A  B) ( A + B) = 10i + 2j + 14k 110. r = 2b sin t i + b cos t j v = r = 2b cos t i  b sin t j a) a = v = 2b 2 sin t i  b 2 cos t j =  2 r speed = v = 4b 2 2 cos 2 t + b 2 2 sin 2 t = b 4 cos 2 t + sin 2 t 12 12 speed = b 3 cos 2 t + 1 b) 12 At t = 2 , sin t = 1 , cos t = 0 So, at this time, v =  b j , a = 2b 2 i So, 90 8 111. a) Since ( A B) i = ijk Aj Bk , we have jk CHAPTER 1 (A B) C = ijk Aj Bk Ci i j,k = C1 ( A2 B3  A3 B2 )  C2 ( A1B3  A3 B1 ) + C3 ( A1B2  A2 B1 ) C1 = A1 B1 C2 A2 B2 C3 A1 A3 =  C1 B3 B1 A2 C2 B2 A3 A1 C3 = B1 B3 C1 A2 B2 C2 A3 B3 = A ( B C ) C3 (1) We can also write C1 C2 B2 A2 C3 B1 B2 C2 A2 (A B) C =  B1 A1 B3 = C1 A3 A1 C3 = B ( C A ) A3 B3 (2) We notice from this result that an even number of permutations leaves the determinant unchanged. b) Consider vectors A and B in the plane defined by e1 , e2 . Since the figure defined by A, B, C is a parallelepiped, A B = e3 area of the base, but e3 C = altitude of the parallelepiped. Then, C ( A B) = ( C e3 ) area of the base = altitude area of the base = volume of the parallelepiped 112. O a h c C ac b A ba cb B The distance h from the origin O to the plane defined by A, B, C is MATRICES, VECTORS, AND VECTOR CALCULUS 9 h= a ( b  a ) ( c  b) ( b  a ) ( c  b) = = a ( b c  a c + a b) bcac+ab ab c ab+bc+ca (1) The area of the triangle ABC is: A= 1 1 1 ( b  a ) ( c  b) = ( a  c ) ( b  a ) = ( c  b ) ( a  c ) 2 2 2 (2) 113. Using the Eq. (1.82) in the text, we have A B = A ( A X ) = ( X A ) A  ( A A ) X = A  A2 X from which X= ( B A ) + A A2 114. a) 1 2 1 2 1 0 1 2 1 AB = 0 3 1 0 1 2 = 1 2 9 2 0 1 1 1 3 5 3 3 2 9 1 9 1 2 +2 +1 3 3 5 3 5 3 AB = 104 Expand by the first row. AB = 1 b) 1 2 1 2 1 9 7 AC = 0 3 1 4 3 = 13 9 2 0 1 1 0 5 2 9 7 AC = 13 9 5 2 10 5 1 2 1 8 ABC = A ( BC ) = 0 3 1 2 3 2 0 1 9 4 5 5 ABC = 3 5 25 14 CHAPTER 1 c) d) AB  Bt At = ? 1 2 1 AB = 1 2 9 5 3 3 (from part a ) 2 0 1 1 0 2 1 1 5 B A = 1 1 1 2 3 0 = 2 2 3 0 2 3 1 1 1 1 9 3 t t 0 3 4 AB  B A = 3 0 6 4 6 0 t t 115. If A is an orthogonal matrix, then At A = 1 1 0 0 1 0 0 1 0 0 0 a a 0 a  a = 0 1 0 0  a a 0 a a 0 0 1 1 0 2 0 2a 0 0 a= 1 2 0 1 0 0 0 = 0 1 0 2a 2 0 0 1 MATRICES, VECTORS, AND VECTOR CALCULUS 11 116. x3 r a x2 x1 P r a r cos r a = constant ra cos = constant It is given that a is constant, so we know that r cos = constant But r cos is the magnitude of the component of r along a. The set of vectors that satisfy r a = constant all have the same component along a; however, the component perpendicular to a is arbitrary. Thus the surface represented by r a = constant is a plane perpendicular to a. 117. a B c A C b Consider the triangle a, b, c which is formed by the vectors A, B, C. Since C= AB C = ( A  B) ( A  B) 2 (1) = A 2  2A B + B 2 or, C = A2 + B2  2 AB cos 2 (2) which is the cosine law of plane trigonometry. 118. Consider the triangle a, b, c which is formed by the vectors A, B, C. a B c A C b 12 CHAPTER 1 C=AB so that (1) C B = ( A  B) B but the lefthand side and the righthand side of (2) are written as: (2) C B = BC sin e3 and (3) ( A  B) B = A B  B B = A B = AB sin e3 where e3 is the unit vector perpendicular to the triangle abc. Therefore, BC sin = AB sin or, C A = sin sin Similarly, C A B = = sin sin sin which is the sine law of plane trigonometry. 119. x2 a2 b2 a1 b1 a b x1 (4) (5) (6) a) We begin by noting that a  b = a 2 + b 2  2ab cos (  ) 2 (1) We can also write that a  b = ( a1  b1 ) + ( a2  b2 ) 2 2 2 2 = ( a cos  b cos ) + ( a sin  b sin ) 2 = a 2 sin 2 + cos 2 + b 2 sin 2 + cos 2  2 ab ( cos cos + sin sin ) = a 2 + b 2  2ab ( cos cos + sin sin ) (2) ( ) ( ) MATRICES, VECTORS, AND VECTOR CALCULUS 13 Thus, comparing (1) and (2), we conclude that cos (  ) = cos cos + sin sin b) (3) Using (3), we can find sin (  ) : sin (  ) = 1  cos 2 (  ) = 1  cos 2 cos 2  sin 2 sin 2  2cos sin cos sin = 1  cos 2 1  sin 2  sin 2 1  cos 2  2cos sin cos sin = sin 2 cos 2  2sin sin cos cos + cos 2 sin 2 = so that sin (  ) = sin cos  cos sin (5) ( ) ( ) ( sin cos  cos sin )2 (4) 120. a) Consider the following two cases: When i j When i = j Therefore, ij = 0 ij 0 but ijk 0 . but ijk = 0 . ij ijk ij = 0 (1) b) We proceed in the following way: When j = k, ijk = ijj = 0 . Terms such as j11 11 = 0 . Then, 12 jk ijk jk = i12 + i13 13 + i 21 21 + i 31 31 + i 32 32 + i 23 23 Now, suppose i = = 1 , then, = jk 123 123 + 132 132 = 1 + 1 = 2 14 CHAPTER 1 for i = = 2 , = 2 gives = jk jk 213 213 + 231 231 = 1 + 1 = 2 . For i = = 3 , = 1 ; i = 1, = jk 312 312 + 321 321 = 2 . But i = 1, = 3 ; i = 3, = 2. = 0 . Likewise for i = 2, = 3 ; i = 3, = 1 ; i = 2, Therefore, j,k ijk jk = 2 i (2) c) ijk ijk ijk = 123 123 + 312 312 + 321 321 + 132 132 + 213 213 + 231 231 = 1 1 + 1 1 + ( 1) ( 1) + ( 1) ( 1) + ( 1) ( 1) + (1) (1) or, ijk ijk ijk = 6 (3) 121. ( A B)i = ijk Aj Bk jk (1) ( A B) C = i jk ijk Aj Bk Ci (2) By an even permutation, we find ABC = ijk Ai Bj Ck ijk (3) 122. a) b) To evaluate k k ijk mk we consider the following cases: mk i= j: i= : k k ijk mk = iik = 0 for all i , , m ijk mk = ijk imk = 1 for j = m and k i , j k = 0 for j m c) i = m: k ijk mk = ijk k ik = 0 for j = 1 for j = and k i , j d) j= : k ijk mk = ijk jmk = 0 for m i k = 1 for m = i and k i , j MATRICES, VECTORS, AND VECTOR CALCULUS 15 e) j = m: k ijk mk = ijk k jk = 0 for i = 1 for i = and k i , j f) g) = m: k ijk mk = ijk k k = 0 for all i , j , k i or m : This implies that i = k or i = j or m = k. Then, h) k ijk mk = 0 for all i , j , , m j or m : k ijk mk = 0 for all i , j , , m Now, consider i jm  im j and examine it under the same conditions. If this quantity behaves in the same way as the sum above, we have verified the equation k ijk mk = i jm  im j a) b) i = j : i im  im i = 0 for all i , , m i = : ii jm  im ji = 1 if j = m, i j , m = 0 if j m c) i = m : i ji  ii j = 1 if j = , i j , = 0 if j d) j = : i m  im = 1 if i = m, i = 0 if i m e) j = m : i mm  im m = 1 if i = , m = 0 if i f) g) h) = m : i j  il j = 0 for all i , j , i , m : i jm  im j = 0 for all i , j , , m j , m : i jm  im i = 0 for all i , j , , m Therefore, k ijk mk = i jm  im j (1) Using this result we can prove that A ( B C ) = ( A C ) B  ( A B) C 16 First ( B C ) i = ijk Bj Ck . Then, jk CHAPTER 1 [ A (B C) ] = mn Am ( B C ) n = mn Am njk BjCk mn mn jk = jkmn mn njk Am Bj Ck = jkmn mn jkn Am Bj Ck = lmn jkn Am Bj Ck jkm n = jl km  k jm Am Bj Ck jkm ( ) = Am B Cm  Am BmC = B AmCm  C Am Bm m m m m = ( A C ) B  ( A B) C Therefore, A ( B C ) = ( A C ) B  ( A B) C (2) 123. Write ( A B) j = j m A Bm m ( C D) k = krs Cr Ds rs Then, MATRICES, VECTORS, AND VECTOR CALCULUS 17 [ ( A B) (C D)]i = ijk j m A Bm krs Cr Ds jk m rs = jk mrs ijk j m krs A BmCr Ds ijk rsk A BmCr Ds k ir = = j mrs j m j mrs ( j m js  is jr A BmCr Ds ) = j m A BmCi Dj  A Bm Di C j j m ( ) = j m Dj A Bm Ci  j m C j A Bm Di jm jm = (ABD)Ci  (ABC)Di Therefore, [( A B) (C D) ] = (ABD)C  (ABC)D 124. Expanding the triple vector product, we have e ( A e) = A ( e e)  e ( A e) (1) But, A ( e e) = A (2) Thus, A = e ( A e) + e ( A e) e(A e) is the component of A in the e direction, while e (A e) is the component of A perpendicular to e. (3) 18 125. er e e CHAPTER 1 The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by e = ( cos cos , cos sin ,  sin ) e = (  sin , cos , 0 ) er = ( sin cos , sin sin , cos ) (1) Thus, e =  cos sin  sin cos , cos cos  sin sin ,  cos =  er + cos e ( ) (2) Similarly, e =  cos ,  sin , 0 ( ) (3) (4) =  cos e  sin er er = sin e + e Now, let any position vector be x. Then, x = rer (5) x = rer + rer = r sin e + e + rer = r sin e + r e + rer x = r sin + r cos + r sin e + r sin e + r + r e + r e + rer + rer = 2r sin + 2r cos + r sin e + r  r 2 sin 2  r 2 er + 2r + r  r 2 sin cos e or, ( ) (6) ( ( ) ( ) ) ( ) ( ) (7) MATRICES, VECTORS, AND VECTOR CALCULUS 19 1 d 2 r  r 2 sin cos e x = a = r  r 2  r 2 sin 2 er + r dt 1 d 2 + r sin 2 e r sin dt ( ) ( ) (8) 126. When a particle moves along the curve r = k (1 + cos ) (1) we have 2 r =  k cos + sin Now, the velocity vector in polar coordinates is [see Eq. (1.97)] r =  k sin v = rer + r e so that v 2 = v = r 2 + r 2 2 2 (2) (3) = k 2 2 sin 2 + k 2 1 + 2 cos + cos 2 2 = k 2 2 2 + 2 cos and v 2 is, by hypothesis, constant. Therefore, (4) ( ) = Using (1), we find v2 2k (1 + cos ) 2 (5) = v 2kr (6) Differentiating (5) and using the expression for r , we obtain = v 2 sin v 2 sin = 2 2 4r 2 4 k (1 + cos ) (7) The acceleration vector is [see Eq. (1.98)] a = r  r 2 er + r + 2r e ( ) ( ) (8) so that 20 CHAPTER 1 a er = r  r 2 =  k 2 cos + sin  k (1 + cos ) 2 2 sin 2 =  k 2 cos + + (1 + cos ) 2 2 (1 + cos ) 1  cos 2 =  k 2 2 cos + + 1 2 (1 + cos ) 3 =  k 2 (1 + cos ) 2 or, a er =  ( ) (9) 3 v2 4 k (10) In a similar way, we find a e =  3 v 2 sin 4 k 1 + cos (11) From (10) and (11), we have a = ( a er ) 2 + ( a e ) 2 3 v2 4 k 2 1 + cos (12) or, a = (13) 127. Since r (v r) = (r r) v  (r v) r we have d d [ r ( v r ) ] = dt [ ( r r ) v  ( r v ) r ] dt = (r r) a + 2 (r v) v  (r v) v  ( v v) r  (r a) r = r 2a + ( r v ) v  r v2 + r a Thus, d [ r ( v r )] = r 2a + ( r v ) v  r r a + v 2 dt ( ) ( ) (1) (2) MATRICES, VECTORS, AND VECTOR CALCULUS 21 ( ln r ) ei x i 128. grad ( ln r ) = i (1) where r = Therefore, x i 2 i (2) 1 ( ln r ) = r x i = so that xi x i 2 xi 2 i r (3) grad ( ln r ) = or, 1 xe 2 i i r i (4) grad ( ln r ) = r r2 (5) 129. Let r 2 = 9 describe the surface S1 and x + y + z 2 = 1 describe the surface S2 . The angle between S1 and S2 at the point (2,2,1) is the angle between the normals to these surfaces at the point. The normal to S1 is grad ( S1 ) = grad r 2  9 = grad x 2 + y 2 + z 2  9 = ( 2xe1 + 2 ye2 + 2ze3 ) = 4e1  4e2 + 2e3 In S2 , the normal is: x = 2, y = 2, z = 1 ( ) ( ) (1) grad ( S2 ) = grad x + y + z 2  1 = ( e1 + e2 + 2ze3 ) = e1 + e2 + 2e3 Therefore, ( ) (2) x = 2, y =2, z = 1 22 CHAPTER 1 cos = grad ( S1 ) grad ( S2 ) grad ( S1 ) grad ( S2 ) = or, ( 4e1  4e2 + 2e3 ) (e1 + e2 + 2e3 ) 6 6 (3) cos = from which 4 6 6 (4) = cos 1 6 = 74.2 9 (5) 130. grad ( ) = ei i =1 3 ( ) = ei + x i i x i x i + ei x i x i i = ei i Thus, grad ( ) = grad + grad 131. a) 12 rn n 2 xj = ei grad r = ei x i x i j i =1 3 n 2 n = ei 2 xi x 2 j 2 j i 2 = ei x i n x 2 j j i = ei x i n r ( n  2) i n n 1 1 (1) Therefore, grad r n = nr ( n  2) r (2) MATRICES, VECTORS, AND VECTOR CALCULUS 23 b) grad f ( r ) = ei i =1 3 f ( r ) 3 f ( r ) r = ei x i r x i i =1 12 2 = ei xj x i j i = ei xi x 2 j j i = ei i f ( r ) r f ( r ) r (3) 1 2 x i f r dr Therefore, grad f ( r ) = c) 12 2 ln r 2 2 ( ln r ) = = 2 ln x j xi2 x i j i 2 1 2 1 2 xi x 2 j 2 j = 12 i x i 2 xj j r f ( r ) r r (4) = i x i 1 xi x 2 j j 2 = (  xi )( 2xi ) x 2 j j i = 2x 2 r 2 j i x + i x2 j i x i j 1 ( )( ) 2 1 + 3 2 r (5) = or, 2r 2 3 1 + 2 = 2 4 r r r 2 ( ln r ) = 1 r2 (6) 24 132. CHAPTER 1 Note that the integrand is a perfect differential: 2ar r + 2br r = a d d (r r) + b (r r) dt dt (1) Clearly, ( 2ar r + 2br r ) dt = ar 133. 2 + br 2 + const. (2) Since d r rr  rr r rr = =  2 dt r r2 r r we have (1) from which d r r rr r  r 2 dt = dt r dt (2) r r rr r  r 2 dt = r + C (3) where C is the integration constant (a vector). 134. First, we note that d AA =AA+AA dt But the first term on the righthand side vanishes. Thus, ( ) (1) ( A A) dt = dt ( A A) dt d so that (2) ( A A) dt = A A + C where C is a constant vector. (3) MATRICES, VECTORS, AND VECTOR CALCULUS 25 135. y x z 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: V1 = 2 a 2 a = 2 a 3 The rest of the common volume is formed by 8 equal parts from the other cylinder (the one along the xaxis). One of these parts extends from x = 0 to x = a, y = 0 to y = a 2  x 2 , z = a to ( ) (1) z = a 2  x 2 . The complementary volume is then V2 = 8 dx 0 a a2  x 2 0 dy a2  x 2 a dz = 8 dx a 2  x 2 a 2  x 2  a 0 a x 3 a3 x = 8 a2 x   sin 1 3 2 a 0 = Then, from (1) and (2): V = V1 + V2 = a 16 3 a  2 a 3 3 (2) 16 a 3 3 (3) 26 136. z CHAPTER 1 d y x c2 = x2 + y2 The form of the integral suggests the use of the divergence theorem. S A da = A dv V (1) Since A = 1 , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is 137. V dv = c 2 d (2) z R y x To do the integral directly, note that A = R3er , on the surface, and that da = daer . S A da = R 3 S da = R3 4 R 2 = 4 R 5 (1) To use the divergence theorem, we need to calculate A . This is best done in spherical coordinates, where A = r 3er . Using Appendix F, we see that A = Therefore, 1 2 r A r = 5r 2 r 2 r ( ) (2) V A dv = sin d 0 2 0 d r 2 5r 2 dr = 4 R5 0 R ( ) (3) Alternatively, one may simply set dv = 4 r 2 dr in this case. MATRICES, VECTORS, AND VECTOR CALCULUS 27 138. z z = 1 x2 y2 y C x x2 + y2 = 1 By Stoke's theorem, we have ( A) da = S C A ds (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 righthand rule. Now change to polar coordinates, so that we have ds = d e and A = sin i + cos k on the curve. Since e i =  sin and e k = 0 , we have 139. C A ds = 2 0 (  sin ) d =  2 (2) a) Let's denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB = (1, 2, 0) ; AC = (1, 0, 3) ; and AB AC = (6, 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB AC , so the unit vector perpendicular to plane (ABC) is n = (6 7 , 3 7 , 2 7 ) b) Let's denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then DH = ( x  1, y  1, z  1) is parallel to n given in a); this means x1 6 = =2 y1 3 and x 1 6 = =3 z1 2 Further, AH = ( x  1, y , z) is perpendicular to n so one has 6( x  1) + 3 y + 2 z = 0 . Solving these 3 equations one finds H = ( x , y , z) = (19 49 , 34 49 , 39 49) and DH = 5 7 140. a) At the top of the hill, z is maximum; 0= z = 2 y  6 x  18 x and 0= z = 2 x  8 y + 28 y 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 t1 = (1, 0, z ) = (1, 0, 8) x (1,1) and t2 = (0,1, z ) = (0,1, 22) y (1,1) Evidently t1 t2 = (8, 22,1) is perpendicular to the hill surface, and the angle between this and Oz axis is cos = (0, 0,1) (8, 22,1) 8 + 22 + 1 2 2 2 = 1 23.43 so = 87.55 degrees. c) Suppose that in the direction ( with respect to WE axis), at point A = (1,1,13) the hill is steepest. Evidently, dy = (tan )dx and dz = 2xdy + 2 ydx  6 xdx  8 ydy  18 dx + 28 dy = 22(tan  1)dx then tan = dx 2 + dy 2 dz = dx cos 1 = 22(tan  1)dx 22 2 cos ( + 45) The hill is steepest when tan is minimum, and this happens when = 45 degrees with respect to WE axis. (note that = 135 does not give a physical answer). 141. A B = 2a( a  1) then A B = 0 if only a = 1 or a = 0. CHAPTER 2 (1) Newtonian Mechanics Single Particle 21. The basic equation is F = mi xi a) b) F ( xi , t ) = f ( xi ) g ( t ) = mi xi : Not integrable (2) F ( xi , t ) = f ( xi ) g ( t ) = mi xi mi dxi = f ( xi ) g ( t ) dt g (t) dxi = dt : Integrable f ( xi ) mi c) (3) (4) F ( xi , xi ) = f ( xi ) g ( xi ) = mi xi : Not integrable 22. Using spherical coordinates, we can write the force applied to the particle as F = Fr er + F e + F e (1) But since the particle is constrained to move on the surface of a sphere, there must exist a reaction force  Fr er that acts on the particle. Therefore, the total force acting on the particle is Ftotal = F e + F e = mr (2) The position vector of the particle is r = Re r (3) where R is the radius of the sphere and is constant. The acceleration of the particle is a = r = Re r (4) 29 30 CHAPTER 2 We must now express er in terms of er , e , and e . Because the unit vectors in rectangular coordinates, e1 , e2 , e3 , do not change with time, it is convenient to make the calculation in terms of these quantities. Using Fig. F3, Appendix F, we see that er = e1 sin cos + e2 sin sin + e3 cos e = e1 cos cos + e2 cos sin  e3 sin e = e1 sin + e2 cos (5) Then er = e1  sin sin + cos cos + e2 cos sin + sin cos  e3 sin = e sin + e ( ) ( ) (6) Similarly, e = er + e cos e = er sin  e cos And, further, er = er 2 sin 2 + 2 + e  2 sin cos + e 2 cos + sin which is the only second time derivative needed. The total force acting on the particle is Ftotal = mr = mRer and the components are F = mR  2 sin cos (7) (8) ( ) ( ) ( ) (9) (10) ( ( ) ) (11) F = mR 2 cos + sin NEWTONIAN MECHANICSSINGLE PARTICLE 31 23. y v0 P x The equation of motion is F=ma The gravitational force is the only applied force; therefore, Fx = mx = 0 Fy = my =  mg Integrating these equations and using the initial conditions, x ( t = 0 ) = v0 cos y ( t = 0 ) = v0 sin We find x ( t ) = v0 cos y ( t ) = v0 sin  gt (1) (2) (3) (4) So the equations for x and y are x ( t ) = v0 t cos 1 2 y ( t ) = v0 t sin  gt 2 (5) Suppose it takes a time t0 to reach the point P. Then, cos = v0 t0 cos 1 2 sin = v0 t0 sin  gt0 2 Eliminating between these equations, 2v sin 2v0 1 + cos tan = 0 gt0 t0  0 2 g g from which (7) (6) 32 2v0 ( sin  cos tan ) g CHAPTER 2 t0 = (8) 24. One of the balls' height can be described by y = y0 + v0 t  gt 2 2 . The amount of time it takes to rise and fall to its initial height is therefore given by 2v0 g . If the time it takes to cycle the ball through the juggler's hands is = 0.9 s , then there must be 3 balls in the air during that time . A single ball must stay in the air for at least 3, so the condition is 2v0 g 3 , or v0 13.2 m s 1 . 25. flightpath N er plane point of maximum acceleration mg a) From the force diagram we have N  mg = mv 2 R er . The acceleration that the pilot feels is N m = g + mv R er , which has a maximum magnitude at the bottom of the maneuver. 2 ( ) ( ) b) If the acceleration felt by the pilot must be less than 9g, then we have 3 330 m s 1 v2 R = 8g 8 9.8 m s 2 ( ) 12.5 km (1) A circle smaller than this will result in pilot blackout. 26. Let the origin of our coordinate system be at the 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 x = x 0 + v0 t (1) NEWTONIAN MECHANICSSINGLE PARTICLE 33 y = y0  1 2 gt 2 (2) where y0 = 80 m , and we need to solve for x0 . From (2), the time the bale hits the ground is = 2 y0 g . If we want the bale to land at x ( ) = 30 m , then x0 = x ( )  v0 . Substituting v0 = 44.4 m s 1 and the other values, this gives x0 210 m behind the cattle. 210 m . The rancher should drop the bales 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 xdirection, then she will strike cattle. This time is given by ( 30 m ) v0 0.68 s . 27. Air resistance is always antiparallel to the velocity. The vector expression is: W= 1 1 v cw Av 2  =  cw Avv 2 2 v (1) Including gravity and setting Fnet = ma , we obtain the parametric equations x =  bx x 2 + y 2 y = by x 2 + y 2  g where b = cw A 2m . Solving with a computer using the given values and = 1.3 kg m 3 , we find that if the rancher drops the bale 210 m behind 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 allowed to make is the same as in the previous problem since it only depends on how fast the plane is moving. 80 60 (2) (3) y (m) 40 20 0 200 180 160 140 120 x (m) 100 80 60 40 With air resistance No air resistance 34 28. y v0 P h Q x CHAPTER 2 From problem 23 the equations for the coordinates are x = v0 t cos y = v0 t sin  1 2 gt 2 (1) (2) In order to calculate the time when a projective reaches the ground, we let y = 0 in (2): v0 t sin  t= 1 2 gt = 0 2 (3) (4) 2v0 sin g Substituting (4) into (1) we find the relation between the range and the angle as x= The range is maximum when 2 = 2 v0 sin 2 g (5) 2 , i.e., = 4 . For this value of the coordinates become v 1 x = 0 t  gt 2 2 2 x= v0 t 2 Eliminating t between these equations yields x2  2 v0 v2 x+ 0 y=0 g g (6) (7) We can find the xcoordinate of the projectile when it is at the height h by putting y = h in (7): x2  This equation has two solutions: x1 = 2 2 v0 v0 2  v0  4 gh 2g 2g 2 2 v0 v0 2 x2 = v0  4 gh + 2g 2g 2 v0 v2 h x+ 0 =0 g g (8) (9) NEWTONIAN MECHANICSSINGLE PARTICLE 35 where x1 corresponds to the point P and x2 to Q in the diagram. Therefore, d = x2  x1 = v0 g 2 v0  4 gh (10) 29. a) Zero resisting force ( Fr = 0 ): The equation of motion for the vertical motion is: F = ma = m Integration of (1) yields v =  gt + v0 where v0 is the initial velocity of the projectile and t = 0 is the initial time. The time tm required for the projectile to reach its maximum height is obtained from (2). Since tm corresponds to the point of zero velocity, v ( tm ) = 0 = v0  gtm , we obtain tm = v0 g (4) (3) (2) dv =  mg dt (1) b) Resisting force proportional to the velocity ( Fr =  kmv ) : The equation of motion for this case is: F=m dv =  mg  kmv dt (5) where kmv is a downward force for t < tm and is an upward force for t > tm . Integrating, we obtain v (t) =  For t = tm , v(t) = 0, then from (6), v0 = which can be rewritten as kv ktm = ln 1 + 0 g Since, for small z (z 1) the expansion (8) g ktm e 1 k g kv0 + g  kt e + k k (6) ( ) (7) 36 1 1 ln (1 + z ) = z  z 2 + z 3 2 3 is valid, (8) can be expressed approximately as CHAPTER 2 (9) v tm = 0 g 2 kv 1 kv 1  0 + 0  ... 2g 3 2g (10) which gives the correct result, as in (4) for the limit k 0. 210. The differential equation we are asked to solve is Equation (2.22), which is x =  kx . 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 10 8 of the analytical solution. 10 v (m/s) v vs t 5 0 5 10 15 t (s) x vs t 20 25 30 100 x (m) 50 0 0 5 10 15 t (s) 20 25 30 10 v (m/s) v vs x 5 0 0 20 40 x (m) 60 80 100 211. The equation of motion is m d2 x =  kmv 2 + mg dt 2 (1) This equation can be solved exactly in the same way as in problem 212 and we find NEWTONIAN MECHANICSSINGLE PARTICLE 37 x= 2 g  kv0 1 log 2 2k g  kv (2) where the origin is taken to be the point at which v = v0 so that the initial condition is x ( v = v0 ) = 0 . Thus, the distance from the point v = v0 to the point v = v1 is 2 g  kv0 1 s ( v0 v1 ) = log 2 2k g  kv1 (3) 212. The equation of motion for the upward motion is m d2 x =  mkv 2  mg 2 dt (1) Using the relation d 2 x dv dv dx dv = = =v 2 dt dt dx dt dx we can rewrite (1) as v dv =  dx kv 2 + g Integrating (3), we find 1 log kv 2 + g =  x + C 2k where the constant C can be computed by using the initial condition that v = v0 when x = 0: C= Therefore, x= kv 2 + g 1 log 0 kv 2 + g 2k 1 2 log kv0 + g 2k (3) (2) ( ) (4) ( ) (5) (6) Now, the equation of downward motion is m d2 x =  mkv 2 + mg dt 2 v dv = dx  kv 2 + g (7) This can be rewritten as (8) Integrating (8) and using the initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find 38 g 1 log g  kv 2 2k CHAPTER 2 x= (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): xh = kv 2 + g 1 log 0 g 2k (10) Then, substituting (10) into (9), kv 2 + g 1 g 1 = log 0 log 2k 2k g g  kv 2 (11) Solving for v, g 2 v0 k v= g 2 v0 + k (12) We can find the terminal velocity by putting x in(9). This gives vt = g k (13) Therefore, v= v0 vt 2 v0 + vt2 (14) 213. The equation of motion of the particle is m dv =  mk v 3 + a 2 v dt ( ) (1) Integrating, v(v and using Eq. (E.3), Appendix E, we find 2 dv =  k dt + a2 ) (2) v2 1 ln 2 =  kt + C 2 2 2a a + v (3) Therefore, we have v2 = C e  At a2 + v 2 (4) NEWTONIAN MECHANICSSINGLE PARTICLE 39 where A 2a 2 k and where C is a new constant. We can evaluate C by using the initial condition, v = v0 at t = 0: C = 2 v0 2 a 2 + v0 (5) Substituting (5) into (4) and rearranging, we have a 2C e  At v=  At 1  C e 12 = dx dt (6) Now, in order to integrate (6), we introduce u e  At so that du = Au dt. Then, x= a 2C e  At 1  C e  At 12 a dt = A C u 1  C u 12 du u = a C A du C u2 + u (7) Using Eq. (E.8c), Appendix E, we find x= a sin 1 (1  2C u) + C A (8) Again, the constant C can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1: C =  a sin 1 (1  2C ) A (9) Therefore, we have x= a sin 1 2C e  At + 1  sin 1 ( 2C + 1) A ( ) Using (4) and (5), we can write x= v 2 + a2 1 1 v 2 + a 2  sin 1 20 2 sin 2 2 2ak v +a v0 + a (10) From (6) we see that v 0 as t . Therefore, v 2 + a2 lim sin 1 2 = sin 1 (1) = 2 t 2 v +a Also, for very large initial velocities, v 2 + a2 lim sin 1 20 2 = sin 1 ( 1) =  v0 2 v0 + a Therefore, using (11) and (12) in (10), we have (12) (11) 40 CHAPTER 2 x (t ) = 2ka (13) and the particle can never move a distance greater than 2ka for any initial velocity. 214. d y x a) The equations for the projectile are x = v0 cos t y = v0 sin t  1 2 gt 2 Solving the first for t and substituting into the second gives y = x tan  Using x = d cos and y = d sin gives d sin = d cos tan  gd 2 cos 2 2 2v0 cos 2 gx 2 1 2 2 v0 cos 2 gd cos 2  cos tan + sin 0 = d 2 2 2v0 cos Since the root d = 0 is not of interest, we have d= 2 2 ( cos tan  sin ) v0 cos 2 g cos 2 2 2v0 cos ( sin cos  cos sin ) = g cos 2 d= 2 2v0 cos sin (  ) g cos 2 (1) NEWTONIAN MECHANICSSINGLE PARTICLE 41 b) Maximize d with respect to 2 2v0 d  sin sin (  ) + cos cos (  ) cos ( 2  ) ( d) = 0 = d g cos 2 cos ( 2  ) = 0 2  = 2 = c) Substitute (2) into (1) 4 + 2 dmax = Using the identity 2 2v0 2 cos 4 + 2 sin 4  2 g cos sin A  sin B = 2 cos we have 1 1 ( A + B) sin ( A  B) 2 2 dmax sin  sin v 2 1  sin 2 2v0 2 = = 0 2 g cos 2 g 1  sin 2 dmax = 2 v0 g (1 + sin ) 215. mg sin mg The equation of motion along the plane is m Rewriting this equation in the form 1 dv = dt k g sin  v 2 k (2) dv = mg sin  kmv 2 dt (1) 42 CHAPTER 2 We know that the velocity of the particle continues to increase with time (i.e., dv dt > 0 ), so that ( g k ) sin > v2 . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We 1 k v tanh 1 =t+C g g sin sin k k 1 find (3) The initial condition v(t = 0) = 0 implies C = 0. Therefore, v= g sin tanh k ( gk sin t = ) dx dt (4) We can integrate this equation to obtain the displacement x as a function of time: x= g sin tanh k ( gk sin t dt ) Using Eq. (E.17a), Appendix E, we obtain x= ln cosh gk sin t g sin + C k gk sin ( ) (5) The initial condition x(t = 0) = 0 implies C = 0. Therefore, the relation between d and t is d= From this equation, we can easily find t= cosh 1 e dk 1 ln cosh k ( gk sin t ) (6) ( ) gk sin (7) 216. The only force which is applied to the article is the component of the gravitational force along the slope: mg sin . So the acceleration is g sin . Therefore the velocity and displacement along the slope for upward motion are described by: v = v0  ( g sin ) t x = v0 t  1 ( g sin ) t 2 2 (1) (2) where the initial conditions v ( t = 0 ) = v0 and x ( t = 0 ) = 0 have been used. At the highest position the velocity becomes zero, so the time required to reach the highest position is, from (1), t0 = At that time, the displacement is v0 g sin (3) NEWTONIAN MECHANICSSINGLE PARTICLE 43 2 1 v0 2 g sin x0 = (4) For downward motion, the velocity and the displacement are described by v = ( g sin ) t x= 1 ( g sin ) t 2 2 (5) (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): t = Adding (3) and (7), we find t= 2v0 g sin (8) v0 g sin (7) for the total time required to return to the initial position. 217. v0 35 60 m 0.7 m Fence The setup for this problem is as follows: x = v0 t cos y = y0 + v0 t sin  1 2 gt 2 (1) (2) where = 35 and y0 = 0.7 m . The ball crosses the fence at a time = R ( v0 cos ) , where R = 60 m. It must be at least h = 2 m high, so we also need h  y0 = v0 sin  g 2 2 . Solving for v0 , we obtain 2 v0 = 2 cos R sin  ( h  y0 ) cos gR 2 (3) which gives v0 25.4 m s 1 . 44 218. CHAPTER 2 a) The differential equation here is the same as that used in Problem 27. It must be solved for many different values of v0 in order to find the minimum required to have the ball go over the fence. This can be a computerintensive and timeconsuming task, although if done correctly is easily tractable by a personal computer. This minimum v0 is 35.2 m s 1 , and the trajectory is shown in Figure (a). (We take the density of air as = 1.3 kg m 3 .) 15 10 y (m) 5 0 0 10 20 30 x (m) 40 50 60 With air resistance No air resistance fence height fence range b) The process here is the same as for part (a), but now we have v0 fixed at the result just obtained, and the elevation angle must be varied to give the 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). 20 15 y (m) 10 5 0 0 10 Flight Path fence height fence range 20 30 x (m) 40 50 60 NEWTONIAN MECHANICSSINGLE PARTICLE 45 219. The projectile's motion is described by x = ( v0 cos ) t 1 y = ( v0 sin ) t  gt 2 2 (1) where v0 is the initial velocity. The distance from the point of projection is r = x2 + y2 Since r must always increase with time, we must have r > 0 : (2) r= Using (1), we have xx + yy = xx + yy >0 r (3) 1 2 3 3 2 g t  g ( v0 sin ) t 2 + v0 t 2 2 (4) Let us now find the value of t which yields xx + yy = 0 (i.e., r = 0 ): t= 3 v0 sin v0 9 sin 2  8 2 g 2g (5) For small values of , the second term in (5) is imaginary. That is, r = 0 is never attained and the value of t resulting from the condition r = 0 is unphysical. Only for values of greater than the value for which the radicand is zero does t become a physical time at which r does in fact vanish. Therefore, the maximum value of that insures r > 0 for all values of t is obtained from 9 sin 2 max  8 = 0 or, sin max = so that 2 2 3 (7) (6) max 70.5 (8) 220. If there were no retardation, the range of the projectile would be given by Eq. (2.54): R= (1) 2 v0 sin 2 0 g The angle of elevation is therefore obtained from 46 Rg 2 v0 CHAPTER 2 sin 2 0 = (1000 m ) ( 9.8 m/sec 2 ) = (140 m/sec ) 2 = 0.50 so that (2) (3) 0 = 15 Now, the real range R, in the linear approximation, is given by Eq. (2.55): 4 kV R = R 1  3g = 2 v0 sin 2 4 kv0 sin 1  3g g (4) Since we expect the real angle to be not too different from the angle 0 calculated above, we can solve (4) for by substituting 0 for in the correction term in the parentheses. Thus, sin 2 = g R 4kv0 sin 0 v 1  3g 2 0 (5) Next, we need the value of k. From Fig. 23(c) we find the value of km by measuring the slope of the curve in the vicinity of v = 140 m/sec. We find km (110 N ) ( 500 m/s ) 0.22 kg/s . The curve is that appropriate for a projectile of mass 1 kg, so the value of k is k 0.022 sec 1 (6) Substituting the values of the various quantities into (5) we find = 17.1 . Since this angle is somewhat greater than 0 , we should iterate our solution by using this new value for 0 in (5). We then find = 17.4 . Further iteration does not substantially change the value, and so we conclude that = 17.4 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 R= (140 m/sec ) 2 sin 34.8 9.8 m/sec 2 1140 m NEWTONIAN MECHANICSSINGLE PARTICLE 47 221. x3 v0 x2 x1 Assume a coordinate system in which the projectile moves in the x2  x3 plane. Then, x2 = v0 t cos 1 x3 = v0 t sin  gt 2 2 or, r = x 2 e2 + x 3 e 3 (1) 1 = ( v0 t cos ) e2 + v0 t sin  gt 2 e3 2 The linear momentum of the projectile is p = mr = m ( v0 cos ) e2 + ( v0 sin  gt ) e3 and the angular momentum is (2) (3) L = r p = ( v0 t cos ) e2 + v0 t sin  gt 2 e3 m ( v0 cos ) e2 + ( v0 sin  gt ) e3 Using the property of the unit vectors that ei e j = e3 ijk , we find L= ( ) (4) 1 mg v0 t 2 cos e1 2 ( ) (5) This gives L =  ( mg v0 t cos ) e1 (6) Now, the force acting on the projectile is F =  mg e3 (7) so that the torque is 1 N = r F = ( v0 t cos ) e2 + v0 t sin  gt 2 e3 (  mg ) e3 2 =  ( mg v0 t cos ) e1 which is the same result as in (6). 48 222. z B E CHAPTER 2 y e x Our force equation is F = q ( E + 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 elementary means. In this case we have also v B so that v B = vB . macentripetal = Solving this for r r= with c qB/m . mv 2 = qvB r (2) mv v = qB c (3) 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 r = xi + yj + zk , with v = r and a = r . Let us calculate first the v B term. v B = x y z = B ( yi  xj) 0 0 B i j k (4) The Lorentz equation (1) becomes F = mr = qByi + q Ey  Bx j + qEz k ( ) (5) Rewriting this as component equations: x= qB y = c y m qEy Ey qB x+ =  c x  m m B (6) (7) (8) y= z= qEz m NEWTONIAN MECHANICSSINGLE PARTICLE 49 The zcomponent equation of motion (8) is easily integrable, with the constants of integration given by the initial conditions in the problem statement. z ( t ) = z0 + z0 t + qEz 2 t 2m (9) c) We are asked to find expressions for x and y , which we will call vx and vy , respectively. Differentiate (6) once with respect to time, and substitute (7) for v y Ey vx = c vy =  c2 vx  B or 2 vx + c2 vx = c (10) Ey B (11) This is an inhomogeneous differential equation that has both a homogeneous solution (the solution for the above equation with the right side set to zero) and a particular solution. The most general solution is the sum of both, which in this case is vx = C1 cos ( c t ) + C2 sin ( c t ) + Ey B (12) where C1 and C2 are constants of integration. This result may be substituted into (7) to get v y v y = C1 c cos ( c t )  C2 c sin ( c t ) vy = C1 sin ( c t ) + C2 cos ( c t ) + K (13) (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 T 2 c . x = Ey B , y =0 (15) d) We get the parametric equations by simply integrating the velocity equations. x= y= c C1 C1 sin ( c t )  cos ( c t ) + c C2 C2 cos ( c t ) + Ey B t + Dx (16) (17) c c sin ( c t ) + Dy where, indeed, Dx and Dy are constants of integration. We may now evaluate all the C's and D's using our initial conditions x ( 0 ) =  A c , x ( 0 ) = Ey B , y ( 0 ) = 0 , y ( 0 ) = A . This gives us C1 = Dx = Dy = 0 , C2 = A and gives the correct answer x (t) = A c cos ( c t ) + Ey B t (18) 50 CHAPTER 2 y(t) = A c sin ( c t ) (19) These cases are shown in the figure as (i) A > Ey B , (ii) A < Ey B , and (iii) A = Ey B . (i) (ii) (iii) 223. F ( t ) = ma ( t ) = kte  at (1) with the initial conditions x(t) = v(t) = 0. We integrate to get the velocity. Showing this explicitly, v(0)a (t ) dt = m 0te v( t ) t k  t dt (2) Integrating this by parts and using our initial conditions, we obtain v(t) = k 1 1 1  t + e  t m 2 (3) By similarly integrating v(t), and using the integral (2) we can obtain x(t). x (t) = k 2 1 1 2  t  3 + 2 + 2 t + e m x ( t ) = te  t 2 v ( t ) = 4  2 ( t + 2) e  t 2 (4) To make our graphs, substitute the given values of m = 1 kg, k = 1 N s 1 , and = 0.5 s 1 . (5) (6) (7) ( t ) = 16 + 4t + 4 ( t + 4 ) e  t 2 NEWTONIAN MECHANICSSINGLE PARTICLE 51 100 x(t) 50 0 0 5 10 t 15 20 4 v(t) 2 0 0 5 10 t 15 20 1 a(t) 0.5 0 0 5 10 t 15 20 224. Ff N d = length of incline s = distance skier travels along level ground s y N mg mg co mg x B y x Ff sin mg While on the plane: Fy = N  mg cos = my = 0 Fx = mg sin  Ff ; So the acceleration down the plane is: so N = mg cos Ff = N = mg cos mg sin  mg cos = mx a1 = g ( sin  cos ) = constant 52 CHAPTER 2 While on level ground: N = mg ; Ff =  mg So Fx = mx becomes  mg = mx a2 =  g = constant The acceleration while on level ground is For motion with constant acceleration, we can get the velocity and position by simple integration: x=a v = x = at + v0 x  x 0 = v0 t + Solving (1) for t and substituting into (2) gives: v  v0 =t a x  x0 = or 2 2a ( x  x0 ) = v 2  v0 (1) (2) 1 2 at 2 v0 ( v  v0 ) a 1 ( v  v0 ) + 2 a 2 Using this equation with the initial and final points being the top and bottom of the incline respectively, we get: 2a1d = VB2 VB = speed at bottom of incline 2a2 s = VB2 Thus a1d =  a2 s So gd ( sin  cos ) = gs Solving for gives d sin d cos + s a1 = g ( sin  cos ) a2 =  g Using the same equation for motion along the ground: (3) = Substituting = 17, d = 100 m, s = 70 m gives = 0.18 Substituting this value into (3): NEWTONIAN MECHANICSSINGLE PARTICLE 53 2 gs = VB2 VB = 2 gs VB = 15.6 m/sec 225. a) At A, the forces on the ball are: N mg The track counters the gravitational force and provides centripetal acceleration N  mg = mv 2 R Get v by conservation of energy: Etop = Ttop + Utop = 0 + mgh EA = TA + U A = 1 mv 2 + 0 2 Etop = EA v = 2 gh So N = mg + m2 gh R 2h N = mg 1 + R b) At B the forces are: N 45 mg N = mv 2 R + mg cos 45 = mv 2 R + mg Get v by conservation of energy. From a), Etotal = mgh . At B, E = 1 mv 2 + mgh 2 2 (1) 54 R 2 CHAPTER 2 R R 45 R cos 45 = R 2 h R= So Etotal = TB + UB becomes: R + h 2 or 1 h = R 1  2 1 1 2 mgh = mgR 1  + mv 2 2 Solving for v 2 1 2 2 gh  gR 1  =v 2 Substituting into (1): 2h 3 N = mg +  2 R 2 2 c) From b) vB = 2 g h  R + R 2 v = 2g h  R + R ( 2 ) 12 d) This is a projectile motion problem 45 45 B A Put the origin at A. The equations: x = x0 + vx 0 t y = y0 + v y 0 t  become x= R v + B t 2 2 vB 1 t  gt 2 2 2 (2) (3) 1 2 gt 2 y = h + Solve (3) for t when y = 0 (ball lands). NEWTONIAN MECHANICSSINGLE PARTICLE 55 gt 2  2 vBt  2h = 0 t= 2 2 vB 2vB + 8 gh 2g We discard the negative root since it gives a negative time. Substituting into (2): x= 2 R v 2 vB 2vB + 8 gh + B 2g 2 2 Using the previous expressions for vB and h yields x= e) ( 3 2  1 R + h + h 2  R2 + 2 R2 2 ) 12 U ( x) = mgy( x ) , with y(0) = h , so U ( x) has the shape of the track. 226. All of the kinetic energy of the block goes into compressing the spring, so that mv 2 2 = kx 2 2 , or x = v m k 2.3 m , where x is the maximum compression and the given values have been substituted. When there is a rough floor, it exerts a force k mg in a direction that opposes the block's velocity. It therefore does an amount of work k mgd in slowing the block down after traveling across the floor a distance d. After 2 m of floor, the block has energy mv 2 2  k mgd , which now goes into compressing the spring and still overcoming the friction on the floor, which is kx 2 2 + k mgx . Use of the quadratic formula gives x= mg k mv 2 2 mgd mg + +  k k k 2 (1) Upon substitution of the given values, the result is 1.12 m. 227. 0.6 m To lift a small mass dm of rope onto the table, an amount of work dW = ( dm) g ( z0  z ) must be done on it, where z0 = 0.6 m is the height of the table. The total amount of work that needs to be done is the integration over all the small segments of rope, giving W = ( dz) g( z0  z) = 0 z0 2 gz0 2 0.18 J . (1) When we substitute = m L = ( 0.4 kg ) ( 4 m ) , we obtain W 56 228. m v v M v3 v4 CHAPTER 2 before collision after collision The problem, as stated, is completely onedimensional. 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 m1v1 + m2 v2 = m1v3 + m2 v4 out of the energy conservation equation 1 1 1 1 2 2 2 2 m1v1 + m2 v2 = m1v3 + m2 v4 2 2 2 2 and get v1 + v3 = v2 + v4 (3) (2) (1) 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 v = 2gh , while the marble of mass m is falling at the same velocity. Conservation of momentum gives Mv + m ( v ) = Mv3 + mv4 and our result for elastic collisions in one dimension gives v + v3 = (  v ) + v 4 solving for v3 and v4 and setting them equal to hmarble 2 ghitem , we obtain 2 (4) (5) 3  h = 1+ 2 (6) 1  3 hsuperball = h 1+ (7) where m M . Note that if < 1 3 , the superball will bounce on the floor a second time after the collision. NEWTONIAN MECHANICSSINGLE PARTICLE 57 229. Ff mg cos mg mg sin x N y = tan 1 0.08 = 4.6 F y = N  mg cos = my = 0 N = mg cos F x = mg sin  Ff = mx Ff = N = mg cos so mx = mg sin  mg cos x = g ( sin  cos ) Integrate with respect to time x = gt ( sin  cos ) + x0 Integrate again: x = x0 + x0 t + 1 2 gt ( sin  cos ) 2 (2) (1) Now we calculate the time required for the driver to stop for a given x0 (initial speed) by solving Eq. (1) for t with x = 0 . t =  x0 ( sin  cos )1 g Substituting this time into Eq. (2) gives us the distance traveled before coming to a stop. ( x  x0 ) = x0t + 2 gt 2 ( sin  cos ) x =  x = 2 x0 1 x2 ( sin  cos )1 + 2 g g02 ( sin  cos )1 g 1 2 x0 ( cos  sin )1 2g 58 We have = 4.6 , = 0.45 , g = 9.8 m/sec 2 . For x0 = 25 mph = 11.2 m/sec , x = 17.4 meters . 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? x 30 meters gives x0 32.9 mph . T = t1 + t2 T = total time = 4.021 sec. t1 = the time required for the balloon to reach the ground. CHAPTER 2 230. (1) where t2 = the additional time required for the sound of the splash to reach the first student. We can get t1 from the equation y = y0 + y 0 t  1 2 gt ; 2 y0 = y 0 = 0 When t = t1 , y = h; so (h = height of building) h =  t= Substituting into (1): T= 2h h + g v or h 2h + T =0 v g h . Using the quadratic formula, we get: 2 gT 1 1 + V 1 2 gt1 2 or t1 = 2h g distance sound travels h = speed of sound v This is a quadratic equation in the variable  h= 2 g 2 v 2 4T + g v = v 2g Substituting V = 331 m/sec g = 9.8 m/sec 2 T = 4.021 sec and taking the positive root because it is the physically acceptable one, we get: NEWTONIAN MECHANICSSINGLE PARTICLE 59 h = 8.426 m1 2 h = 71 meters 231. For x0 0 , example 2.10 proceeds as is until the equations following Eq. (2.78). Proceeding from there we have B = x0 0 A = z0 so ( x  x0 ) = 0 cos t + 0 sin t z x ( y  y0 ) = y 0 t ( z  z0 ) =  0 cos t + 0 sin t Note that x z ( x  x 0 ) + ( z  z0 ) = 2 2 2 2 z0 + 2 2 x0 Thus the projection of the motion onto the xz plane is a circle of radius (x 1 2 0 2 + z0 ) 12 . So the motion is unchanged except for a change in the m 2 2 12 radius of the helix. The new radius is x 0 + z0 . qB0 ( ) 232. The forces on the hanging mass are T mg The equation of motion is (calling downward positive) mg  T = ma The forces on the other mass are or T = m ( g  a) (1) 60 CHAPTER 2 N y x T 2mg cos Ff 2mg 2mg sin The y equation of motion gives N  2mg cos = my = 0 or N = 2mg cos The x equation of motion gives Ff = k N = 2k mg cos ( ) (2) T  2mg sin  2 k mg cos = ma Substituting from (1) into (2) mg  2mg sin  2 k mg cos = 2ma When = 0 , a = 0. So g  2 g sin 0  2k g cos 0 = 0 1 = sin 0 + k cos 0 2 = sin 0 + k 1  sin 2 0 ( ) 12 Isolating the square root, squaring both sides and rearranging gives (1 + ) sin 2 k 2 2 0  sin 0  k = 0 1 4 Using the quadratic formula gives sin 0 = 2 1 k 3 + 4k 2 2 1 + k ( ) 233. The differential equation to solve is v 2 cW Av 2  g = g  1 y= 2m vt (1) where vt = 2mg cw A is the terminal velocity. The initial conditions are y0 = 100 m , and v0 = 0 . The computer integrations for parts (a), (b), and (c) are shown in the figure. NEWTONIAN MECHANICSSINGLE PARTICLE 61 100 100 100 y (m) 50 50 50 0 0 2 t (s) 4 6 0 0 5 t (s) 10 15 0 0 5 t (s) 10 0 0 0 5 v (m/s) 20 5 10 0 2 t (s) 0 0 4 6 0 5 t (s) 10 15 0 5 t (s) 10 0 a (m/s2) 5 5 5 0 2 t (s) 4 6 0 5 t (s) 10 15 0 5 t (s) 10 d) Taking = 1.3 kg m 3 as the density of air, the terminal velocities are 32.2, 8.0, and 11.0 (all m s 1 ) for the baseball, pingpong ball, and raindrop, respectively. Both the pingpong ball and the raindrop essentially reach their terminal velocities by the time they hit the ground. If we rewrite the mass as average density times volume, then we find that vt material R . 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 terminal velocity. Since the baseball has a higher terminal velocity than the pingpong ball, the magnitude of its deceleration is smaller for a given speed. If a person throws the two objects with the same initial velocity, the baseball goes farther because it has less drag. f) We have shown in part (d) that the terminal velocity of a raindrop of radius 0.004 m will be larger than for one with radius 0.002 m ( 9.0 m s 1 ) by a factor of 2 . 234. FR y mg Take the yaxis to be positive downwards. The initial conditions are y = y = 0 at t = 0. 62 a) CHAPTER 2 FR = v The equation of motion is my = m dv = mg  v dt m dv = dt mg  v Integrating gives:  m ln ( mg  v ) = t + C Evaluate C using the condition v = 0 at t = 0:  So or   m m ln ( mg ) = C ln ( mg  v ) + m ln ( mg ) = t t mg  v v = ln = ln 1  m mg mg Take the exponential of both sides and solve for v: e  r m = 1  v mg v mg v= dy = = 1  e  t m mg mg (1  e (1  e  t m ) (1)  t m ) dt Integrate again: y+C = y = 0 at t = 0, so: C= y= mg m = m2 g 2 mg m  t m t + e mg m m  t m  + t + e (2) Solve (3) for t and substitute into (4): 1 v mg = e  t m (3) NEWTONIAN MECHANICSSINGLE PARTICLE 63 t= v ln 1  mg m y= mg m m v m v mg v m v   ln 1    ln 1  + 1  = mg mg g mg y= mg v m ln 1  v + mg (4) b) FR = v 2 The equation of motion becomes: m dv = mg  v 2 dt m dv = dt mg  v 2 Integrate and apply the initial condition v = 0 at t = 0: mg From integral tables dv = 2 v m dt a 2 dx 1 x = tanh 1 ; so 2 x a a 1 v tanh 1 = t + C where a a a m 1 mg so: tanh 1 0 = 0 = 0 + C 1 v tanh 1 = t a a m Solving for v: v = a tanh a t m (5) dy at = a tanh dt m From integral tables tanh u du = ln cosh u So y + C = m ln cosh a t m 64 Apply the conditions at y = 0 and t = 0 C= m CHAPTER 2 ln ( cosh 0 ) = m ln 1 = 0 So y= m ln cosh a t m (6) Solving (5) for t: t= m tanh 1 v a Substituting into (6): y= m ln cosh tanh 1 , where u < 1 . v a Use the identity: tanh 1 u = cosh 1 1 1  u2 (In our case u < 1 as it should be because v v2 = ; and the condition that u < 1 just says that a mg gravity is stronger than the retarding force, which it must be.) So y= m ln cosh cosh 1 y= m 2 = ln 1  v mg 2 1  v mg 1 ( ) 1 2 m ln 1  v 2 mg 2 ( ) NEWTONIAN MECHANICSSINGLE PARTICLE 65 235. 14 12 10 y (km) 8 6 4 2 0 0 5 10 15 x (km) 20 25 30 35 30 Range (km) 20 10 0 0 0.02 0.04 0.06 0.08 k (1/s) 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 s1 ), with initial speed v0 = 600 m s 1 and angle of elevation = 60 . The first figure is produced by numerical solution of the differential equations, and agrees closely with Figure 28. Figure 29 can be most closely reproduced 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. 66 236. y h x R CHAPTER 2 Put the origin at the initial point. The equations for the x and y motion are then x = v0 ( cos ) t y = v0 ( sin ) t  1 2 gt 2 Call the time when the projectile lands on the valley floor. The y equation then gives  h = v0 ( sin )  Using the quadratic formula, we may find 2 v0 sin 2 + 2 gh v0 sin + = g g 1 2 g 2 (We take the positive since > 0 .) Substituting into the x equation gives the range R as a function of . R= 2 v0 cos sin + sin 2 + x 2 g (1) 2 where we have defined x 2 2 gh v0 . To maximize R for a given h and v0 , we set dR d = 0 . The equation we obtain is cos 2  sin 2  sin sin 2 + x 2 + sin cos 2 sin 2 + x 2 =0 (2) Although it can give x = x ( ) , the above equation cannot be solved to give = ( x ) in terms of the elementary functions. The optimum for a given x is plotted in the figure, along with its 2 respective range in units of v0 g . Note that x = 0, which among other things corresponds to 2 h = 0, gives the familiar result = 45 and R = v0 g . NEWTONIAN MECHANICSSINGLE PARTICLE 67 50 40 30 45 20 10 0 0 1 2 3 4 5 x 10 6 7 8 9 10 R/(v02/g) 5 1 0 0 1 2 3 4 5 x 6 7 8 9 10 237. v= x dv dx =  x 2 Since dv dv dx dv = = v then dt dx dt dx F=m dv dv = mv = m  2 dt dx x x F ( x ) =  m 2 x 3 v ( x ) = ax  n F=m dv dv dx dv =m = mv = m ax  n  nax  n 1 dt dx dt dx F ( x ) =  mna 2 x  ( 2 n + 1) 238. a) ( )( ) b) v ( x) = dx = ax  n dt x n dx = adt Integrate: x n+1 = at + C n+1 C = 0 using given initial conditions 68 x n + 1 = ( n + 1) at CHAPTER 2 x = [ ( n + 1) at ] c) Substitute x(t) into F(x): 1 ( n + 1) F ( t ) =  mna 2 {( n + 1) at} 1 ( n + 1)  ( 2 n + 1) F ( t ) =  mna 2 [ ( n + 1) at ]  ( 2 n + 1) ( n + 1) 239. a) F =  e v dv =  e v dt m e  v = v0 at t = 0, so  v dv =  m dt 1 e  v =  m t+C   Solving for v gives v (t) =  b) Solve for t when v = 0 1 e  v0 = C  e  v0 =  (e 1  v ) m t t  v0 ln +e m 1 t m t= c) From a) we have + e  v0 = 1 1  e  v0 m dx =  t  v0 ln +e dt m 1 NEWTONIAN MECHANICSSINGLE PARTICLE 69 Using ln ( ax + b) dx = ax + b ln ( ax + b )  x we obtain a t t  v0  v0 m + e ln m + e 1  t x + C =  m Evaluating C using x = 0 at t = 0 gives C= So x= mv0 v0 m e  v0 e  v0 + t  m t  v0 t  v0 +e ln m + e 2 m Substituting the time required to stop from b) gives the distance required to stop x= m 1  v0 e 1 v0 + 240. y (x(t),y(t)) an at x Write the velocity as v(t) = v(t)T(t). It follows that a (t) = dv dv dT T+v = = at T + an N dt dt dt (1) where N is the unit vector in the direction of dT dt . That N is normal to T follows from 0 = d dt ( T T ) . Note also an is positive definite. a) We have v = x 2 + y = A 5  4 cos t . Computing from the above equation, 2 at = dv 2 A 2 sin t = dt 5  4 cos t 2 (2) We can get an from knowing a in addition to at . Using a = x 2 + y = A 2 , we get an = a 2  at2 = A 2 2 cos t  1 5  4 cos t (3) b) Graphing an versus t shows that it has maxima at t = n , where an = A 2 . 70 241. a) CHAPTER 2 As measured on the train: Ti = 0 ; Tf = T = 1 mv 2 2 1 mv 2 2 b) As measured on the ground: Ti = 1 1 2 mu 2 ; T f = m ( v + u ) 2 2 T = 1 mv 2 + mvu 2 c) The woman does an amount of work equal to the kinetic energy gain of the ball as measured in her frame. W= 1 mv 2 2 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. W = mvu 242. R b R From the figure, we have h( ) = (R + b 2) cos + R sin , and the potential is U ( ) = mgh( ) . Now compute: dU b = mg  sin + R cos d 2 d 2U b = mg R  cos  R sin 2 2 d (1) (2) NEWTONIAN MECHANICSSINGLE PARTICLE 71 The equilibrium point (where dU d = 0 ) that we wish to look at is clearly = 0. At that point, we have d 2U d 2 = mg ( R  b 2) , which is stable for R > b 2 and unstable for R < b/ 2 . We can use the results of Problem 246 to obtain stability for the case R = b 2 , where we will find that the first nontrivial result is in fourth order and is negative. We therefore have an equilibrium at = 0 which is stable for R > b 2 and unstable for R b 2 . F =  kx + kx 3 2 U ( x ) =  F dx = 1 2 1 x4 kx  k 2 2 4 1 2 kx . For large x, the 2 243. To sketch U(x), we note that for small x, U(x) behaves like the parabola behavior is determined by  1 x4 k 4 2 U(x) E0 E1 E2 x4 x5 x1 x2 x3 x E3 = 0 E4 E= 1 mv 2 + U ( x ) 2 For E = E0 , the motion is unbounded; the particle may be anywhere. For E = E1 (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 slightly, it will move away from the equilibrium. What is the value of E1 ? We find the x values by setting 0 = kx  kx 3 2 x = 0, are the equilibrium points U ( ) = E1 = 1 2 1 2 1 2 k  k = k 2 4 4 dU =0. dx For E = E2 , the particle is either bounded and oscillates between  x2 and x2 ; or the particle comes in from to x3 and returns to . 72 CHAPTER 2 For E3 = 0 , the particle is either at the stable equilibrium point x = 0, or beyond x = x4 . For E4 , the particle comes in from to x5 and returns. 244. T T T m1 m1 g m2 m2 g From the figure, the forces acting on the masses give the equations of motion m1 x 1 = m1 g  T m2 x 2 = m2 g  2T cos where x2 is related to x1 by the relation x2 = (1) (2) ( b  x1 )2  d2 4 (3) and cos = d ( b  x1 ) 2 . At equilibrium, x 1 = x 2 = 0 and T = m1 g . This gives as the equilibrium values for the coordinates x10 = b  x20 = 4 m1 d 2 2 4 m1  m2 (4) m2 d 2 2 4 m1  m2 (5) We recognize that our expression x10 is identical to Equation (2.105), and has the same requirement that m2 m1 < 2 for the equilibrium to exist. When the system is in motion, the descriptive equations are obtained from the force laws: m1 ( x 1  g ) = m2 ( b  x1 ) 4 x2 (x 2  g) (6) To examine stability, let us expand the coordinates about their equilibrium values and look at their behavior for small displacements. Let 1 x1  x10 and 2 x2  x20 . In the calculations, take terms in 1 and 2 , and their time derivatives, only up to first order. Equation (3) then becomes 2 (m1 m2 )1 . When written in terms of these new coordinates, the equation of motion becomes 1 =  4 m1m2 ( m1 + m2 ) d 2 2 g 4 m1  m2 ( ) 32 1 (7) NEWTONIAN MECHANICSSINGLE PARTICLE 73 which is the equation for simple harmonic motion. The equilibrium is therefore stable, when it exists. 245. and 246. Expand the potential about the equilibrium point U ( x) = 1 di u i i x i = n + 1 i ! dx 0 (1) The leading term in the force is then F( x ) =  dU 1 d( n + 1)U =  ( n + 1) x n dx n! dx 0 (2) The force is restoring for a stable point, so we need F ( x > 0 ) < 0 and F ( x < 0 ) > 0 . This is never true when n is even (e.g., U = kx 3 ), and is only true for n odd when d( n + 1)U dx( n + 1) 0 < 0 . We are given U ( x) = U 0 ( a x + x a ) for x > 0 . Equilibrium points are defined by 247. dU dx = 0 , with stability determined by d 2U dx 2 at those points. Here we have dU a 1 = U0  2 + dx a x which vanishes at x = a. Now evaluate d 2U 2U0 dx 2 = a 3 > 0 a indicating that the equilibrium point is stable. 25 20 15 10 5 0 (1) (2) U(x)/U0 0 0.5 1 x/a 1.5 2 74 CHAPTER 2 248. In the equilibrium, the gravitational force and the eccentric force acting on each star must be equal Gm2 mv 2 = v= d2 d /2 mG d = = 2d v 2 d 3 2 mG 249. The distances from stars to the center of mass of the system are respectively r1 = dm2 m1 + m2 and r2 = dm1 m1 + m2 At equilibrium, like in previous problem, we have Gm1m2 m1v12 = v1 = d2 r1 Gm2 2 2 r1 2 d 3 2 = = d(m1 + m2 ) v1 G(m1 + m2 ) The result will be the same if we consider the equilibrium of forces acting on 2nd star. 250. a) d m0 v dt v2 1 2 c m0 v = Fd v2 0 1 2 c t m0 v = = Ft v(t) = v2 1 2 c t Ft 2 m0 + F 2t 2 c2 x(t) = v(t)dt = 0 c2 F 2t 2 m0 2 + 2  m0 F c b) v t c) From a) we find t= vm0 F 1 v2 c2 Now if F = 10 , then m0 NEWTONIAN MECHANICSSINGLE PARTICLE 75 when v = c 2 , we have t = c = 0.55 year 10 3 99c = 6.67 years 10 199 when v = 99% c, we have t = 251. a) m mv0 dv dv b =  bv 2 2 =  dt v(t) = dt v m btv0 + m 999m = 138.7 hours . v0 b Now let v(t) = v0/1000 , one finds t = v t b) x(t ) = vdt = 0 t m btv0 + m ln b m We use the value of t found in question a) to find the corresponding distance x(t ) = m ln(1000) = 6.9 km b 252. a) b) U F( x ) =  4U x dU x2 =  20 1  2 dx a a x When F = 0, there is equilibrium; further when U has a local minimum (i.e. dF dx < 0 ) it is stable, and when U has a local maximum (i.e. dF dx > 0 ) 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, F  4U 0 x  kx = a2 k = m 4U 0 ma 2 d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential, 2 2U 0 mvmin = U max = U 0 vmin = m 2 e) From energy conservation, we have 2 mv 2 U 0 x 2 mvmin dx + 2 = =v= a dt 2 2 2U 0 x2 1  2 m a We note that, in the ideal case, because the initial velocity is the escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression, t= a exp t m dx ma a+x ln = x(t) = 2U 0 8U 0 a  x x2 0 1  2 exp t a x 2 8U 0  1 ma 2 8U 0 + 1 ma 2 x t 253. F is a conservative force when there exists a nonsingular potential function U(x) satisfying F(x) = grad(U(x)). So if F is conservative, its components satisfy the following relations Fx Fy = y x and so on. a) In this case all relations above are satisfied, so F is indeed a conservative force. Fx =  bx 2 U = ayz + bx + c U =  ayzx   cx + f1 ( y , z) 2 x (1) where f1 ( y , z) is a function of only y and z Fy =  U = axz + bz U =  ayzx  byz + f 2 ( x , z) y (2) NEWTONIAN MECHANICSSINGLE PARTICLE 77 where f 2 ( x , z) is a function of only x and z Fz =  U = axy + by + c U =  ayzx  byz + f 3 ( y , z) z bx 2 +C 2 (3) then from (1), (2), (3) we find that U =  axyz  byz  cx 2  where C is a arbitrary constant. b) is Using the same method we find that F in this case is a conservative force, and its potential U =  z exp(  x)  y ln z + C c) Using the same method we find that F in this case is a conservative force, and its potential is ( using the result of problem 131b): U =  a ln r 254. 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 b) and consequently v = g k = 1000 m/s . dv F dx dv vdv = = ( g + kv) = dt =  dx =  dt m v g + kv g + kv 0 v0 xmax = g v0 g + 2 ln = 679.7 m k k g + kv0 where v0 is the initial velocity of the potato. x 0 255. Let's denote vx 0 and vy 0 the initial horizontal and vertical velocity of the pumpkin. vx 0  vxf dvx dv dx = Fx =  mkvx  =  dt = x x f = dt vx kvx k Evidently, vx 0 = vy 0 in this problem. m (1) where the suffix f always denote the final value. From the second equality of (1), we have  dt = Combining (1) and (2) we have xf = vx 0 (  kt 1 e f ) k (3) dvx  kt vxf = vx 0 e f kvx (2) 78 Do the same thing with the ycomponent, and we have m dvy dt = Fy =  mg  mkvy  dvy g + kvyf vy 0  vyf dy g =  dt = 0 = y f = 2 ln + vy g + kvy k g + kvy 0 k  dt = dvy g + kvy g + kvyf = ( g + kv0 f ) e  kt f CHAPTER 2 (4) and (5) From (4) and (5) with a little manipulation, we obtain 1 e  kt f = gkt f g + kvy 0 (6) (3) and (6) are 2 equations with 2 unknowns, t f and k. We can eliminate t f , and obtain an equation of single variable k. xf = Putting x f = 142 m and vx 0 = vy 0 = vx 0 (  kt g + kv ( gv ) 1  e f ( y0 ) x0 ) k v0 = 38.2 m/s we can numerically solve for k and obtain 2 K= 0.00246 s 1 . CHAPTER 3 Oscillations 31. a) 1 0 = 2 k 1 = m 2 10 4 dyne/cm 10 = 10 2 gram 2 gram cm sec 2 cm = 10 sec 1 gram 2 or, 0 1.6 Hz 0 = or, 1 (1) 0 = 2 sec 10 0 0.63 sec b) (2) E= 1 2 1 kA = 10 4 32 dynecm 2 2 so that E = 4.5 10 4 erg c) The maximum velocity is attained when the total energy of the oscillator is equal to the kinetic energy. Therefore, (3) 1 2 mvmax = 4.5 10 4 erg 2 v max = 2 4.5 10 4 100 79 80 CHAPTER 3 or, vmax = 30 cm/sec (4) 32. a) The statement that at a certain time t = t1 the maximum amplitude has decreased to onehalf the initial value means that xen = A0 e  t1 = or, e  t1 = so that 1 2 1 A0 2 (1) (2) = Since t1 = 10 sec , ln 2 0.69 = t1 t1 (3) = 6.9 10 2 sec 1 b) (4) According to Eq. (3.38), the angular frequency is 2 1 = 0  2 (5) where, from Problem 31, 0 = 10 sec 1 . Therefore, 1 = (10 )  ( 6.9 10 2 ) 2 2 2 1 10 1  ( 6.9) 10 6 sec 1 2 (6) so that 1 = which can be written as 10 (1  2.40 10 5 ) sec 1 2 (7) 1 = 0 (1  ) where (8) = 2.40 10 5 That is, 1 is only slightly different from 0 . (9) OSCILLATIONS 81 c) The decrement of the motion is defined to be e 1 where 1 = 1 1 . Then, e 1 1.0445 33. The initial kinetic energy (equal to the total energy) of the oscillator is m = 100 g and v0 = 1 cm/sec . 1 2 mv0 , where 2 a) Maximum displacement is achieved when the total energy is equal to the potential energy. Therefore, 1 1 2 2 mv0 = kx0 2 2 x0 = or, m 10 2 1 v0 = 1= cm 4 k 10 10 x0 = b) The maximum potential energy is 1 cm 10 (1) U max = or, 1 2 1 kx0 = 10 4 10 2 2 2 U max = 50 ergs (2) 34. a) Time average: The position and velocity for a simple harmonic oscillator are given by x = A sin 0 t x = 0 A cos 0 t where 0 = k m The time average of the kinetic energy is T = where = 2 is the period of oscillation. 0 1 t + (1) (2) t 1 mx 2 dt 2 (3) 82 CHAPTER 3 By inserting (2) into (3), we obtain T = or, T = 2 mA2 0 4 1 2 mA2 0 2 t+ cos t 2 0 t dt (4) (5) In the same way, the time average of the potential energy is U = 1 t + t 1 2 kx dt 2 t + = 1 kA2 2 kA2 4 sin t 2 0 t dt = 2 and since 0 = k m , (6) reduces to (6) U = From (5) and (7) we see that 2 mA2 0 4 (7) T = U The result stated in (8) is reasonable to expect from the conservation of the total energy. (8) E = T +U (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 exactly the same function, displaced by a time 2 : 2 mA2 0 cos 2 0 t 2 2 mA 0 t 2 sin 0 t U= 2 T= (10) Therefore, the time averages of T and U must be equal. Then, by taking time average of (9), we find T = U = b) Space average: E 2 (11) The space averages of the kinetic and potential energies are OSCILLATIONS 83 A T= and 1 1 2 2 mx dx A0 (12) 2 m 0 1 1 U = kx 2 dx = 2A A0 2 A x 0 A 2 dx (13) (13) is readily integrated to give U= 2 m 0 A2 6 (14) To integrate (12), we notice that from (1) and (2) we can write 2 2 x 2 = 0 A2 cos 2 0 t = 0 A2 1  sin 2 0 t 2 = 0 A2  x 2 ( ) (15) ( ) 0 A Then, substituting (15) into (12), we find T= 2 m 0 2A 2 m 0 2A A2  x 2 dx = or, 3 A3 A  3 (16) T=2 2 m 0 A2 6 (17) From the comparison of (14) and (17), we see that T = 2U To see that this result is reasonable, we plot T = T(x) and U = U(x): 1 x2 2 m 0 A2 1  2 2 A 1 2 2 U = m 0 x 2 (18) T= (19) 2 mA2 0 Energy E = const. = 1 2 mA2 0 2 U = U(x) T = T(x) A O A x And the area between T(x) and the xaxis is just twice that between U(x) and the xaxis. 84 35. CHAPTER 3 Differentiating the equation of motion for a simple harmonic oscillator, x = A sin 0 t (1) we obtain x = A 0 cos 0 t t (2) But from (1) sin 0 t = Therefore, cos 0 t = 1  ( x A) and substitution into (2) yields t = x 2 x A (3) (4) 0 A2  x 2 (5) Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval x at position x is given by t = x 0 A2  x 2 t/ = x 2 A2  x 2 (6) x A3 A2 A1 A1 A2 A3 This result implies that the harmonic oscillator spends most of its time near x = A, which is obviously true. On the other hand, we obtain a singularity for t at x = A. This occurs because at these points x = 0, and (2) is not valid. 36. k m1 x1 m2 x2 x Suppose the coordinates of m1 and m2 are x1 and x2 and the length of the spring at equilibrium is . Then the equations of motion for m1 and m2 are m1 x1 =  k ( x1  x2 + m2 x2 =  k ( x2  x1 + ) ) (1) (2) OSCILLATIONS 85 From (2), we have x1 = 1 ( m2 x2 + kx2  k k ) (3) Substituting this expression into (1), we find d2 m1 m2 x2 + ( m1 + m2 ) kx2 = 0 dt 2 from which x2 =  Therefore, x2 oscillates with the frequency m1 + m2 kx2 m1m2 (5) (4) = m1 + m2 k m1m2 (6) We obtain the same result for x1 . If we notice that the reduced mass of the system is defined as 1 we can rewrite (6) as = 1 1 + m1 m2 (7) = k k (8) This means the system oscillates in the same way as a system consisting of a single mass . Inserting the given values, we obtain 66.7 g and 2.74 rad s 1 . 37. A hb hs Let A be the crosssectional area of the floating body, hb its height, hs the height of its submerged part; and let and 0 denote the mass densities of the body and the fluid, respectively. The volume of displaced fluid is therefore V = Ahs . The mass of the body is M = Ahb . 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 (  0 gV =  0 ghs A ). The equilibrium situation occurs when the total force vanishes: 0 = Mg  0 gV = gAhb  0 ghs A which gives the relation between hs and hb : hs = hb (1) 0 (2) For a small displacement about the equilibrium position ( hs hs + x ), (1) becomes Mx = Ahb x = gAhb  0 g ( hs + x ) A Upon substitution of (1) into (3), we have (3) Ahb x =  0 gxA or, x+g (4) 0 x=0 hb (5) Thus, the motion is oscillatory, with an angular frequency 2 = g g gA 0 = = hb hs V (6) where use has been made of (2), and in the last step we have multiplied and divided by A. The period of the oscillations is, therefore, = Substituting the given values, 38. y 2 = 2 V gA (7) 0.18 s . 2a O x m s 2a The force responsible for the motion of the pendulum 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 F = ma = mv =  mg cos (1) OSCILLATIONS 87 where is the angle between the vertical and the tangent to the cycloidal path at the position of m. The cosine of is expressed in terms of the differentials shown in the figure (b) as cos = where ds = dx 2 + dy 2 m F mg dy dx dy ds (2) (3) ds S (a) (b) The differentials, dx and dy, can be computed from the defining equations for x() and y() above: dx = a (1  cos ) d dy =  a sin d Therefore, ds2 = dx 2 + dy 2 2 = a 2 (1  cos ) + sin 2 d 2 = 2a 2 (1  cos ) d 2 (4) = 4 a 2 sin 2 so that 2 d 2 (5) ds = 2a sin Thus, 2 d (6) dy  a sin d = ds 2a sin d 2 =  cos The velocity of the pendulum bob is 2 = cos (7) 88 CHAPTER 3 v= ds d = 2a sin dt 2 dt d dt = 4 a from which v = 4 a Letting z cos cos 2 cos 2 (8) d2 dt 2 (9) 2 be the new variable, and substituting (7) and (9) into (1), we have 4maz = mgz (10) or, z+ g z=0 4a (11) which is the standard equation for simple harmonic motion, 2 z + 0 z = 0 (12) If we identify 0 = where we have used the fact that g (13) = 4a . Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673). 39. The equation of motion for 0 t t0 is mx =  k ( x  x0 ) + F =  kx + ( F + kx0 ) while for t t0 , the equation is (1) mx =  k ( x  x0 ) =  kx + kx0 It is convenient to define (2) = x  x0 which transforms (1) and (2) into m =  k + F ; m =  k ; 0 t t0 (3) (4) t t0 OSCILLATIONS 89 The homogeneous solutions for both (3) and (4) are of familiar form ( t ) = Ae i t + Be  i t , where = k m . A particular solution for (3) is = F k . Then the general solutions for (3) and (4) are  = F + Ae i t + Be  i t ; k 0 t t0 (5) (6) + = Ce i t + De  i t ; t t0 To determine the constants, we use the initial conditions: x ( t = 0 ) = x0 and x(t = 0) = 0. Thus,  ( t = 0 ) =  ( t = 0 ) = 0 The conditions give two equations for A and B: F + A+B k 0 = i ( A  B) 0= Then A=B= and, from (5), we have F 2k (7) (8)  = x  x 0 = F (1  cos t ) ; k 0 t t0 (9) Since for any physical motion, x and x must be continuous, the values of  ( t = t0 ) and  ( t = t0 ) are the initial conditions for + ( t ) which are needed to determine C and D: + ( t = t0 ) = F (1  cos t0 ) = Ce i t0 + De  i t0 k F + ( t = t0 ) = sin t0 = i Ce i t0 k The equations in (10) can be rewritten as: Ce i t0 + De  i t0 = Ce i t0  De  i t0  De  i t0 (10) F (1  cos t0 ) k  iF = sin t0 k (11) Then, by adding and subtracting one from the other, we obtain F  i t0 e 1  e i t0 2k F i t0 D= e 1  e  i t0 2k C= ( ) ( ) (12) 90 CHAPTER 3 Substitution of (12) into (6) yields + = = = Thus, F e  i t0  1 e i t + e i t0  1 e  i t 2k F i (t  t0 ) i t  i t  t  e + e ( 0 )  e  i t e 2k F cos ( t  t0 )  cos t k (13) ( ) ( ) x  x0 = F cos ( t  t0 )  cos t ; t t0 k (14) 310. The amplitude of a damped oscillator is expressed by x ( t ) = Ae  t cos ( 1t + ) (1) Since the amplitude decreases to 1 e after n periods, we have nT = n 2 1 =1 (2) Substituting this relation into the equation connecting 1 and 0 (the frequency of undamped 2 2 oscillations), 1 = 0  2 , we have 2 2 0 = 1 + 1 1 2 = 1 1 + 4 2 n 2 2 n 2 (3) Therefore, 1 1 = 1 + 2 2 0 4 n so that 1 2 (4) 1 1 1 2 2 8 n 2 311. The total energy of a damped oscillator is E (t) = 1 1 2 2 mx ( t ) + kx ( t ) 2 2 (1) where x ( t ) = Ae  t cos ( 1t  ) x ( t ) = Ae  t  cos ( 1t  )  1 sin ( 1t  ) (2) (3) OSCILLATIONS 91 2 1 = 0  2 , 0 = k m Substituting (2) and (3) into (1), we have E (t) = A2 2 t 2 m 2 + k cos 2 ( 1t  ) + m 1 sin 2 ( 1t  ) e 2 ( ) + 2m 1 sin ( 1t  ) cos ( 1t  ) (4) Rewriting (4), we find the expression for E(t): E (t) = mA2 2 t 2 2 2 cos 2 ( 1t  ) + 0  2 sin 2 ( 1t  ) + 0 e 2 dE : dt (5) Taking the derivative of (5), we find the expression for dE mA2 2 t 2 2 0  4 3 cos 2 ( 1t  ) e = 2 dt 2  4 2 0  2 sin 2 ( 1t  ) 0  2 2 ( ) (6) The above formulas for E and dE dt reproduce the curves shown in Figure 37 of the text. To find the average rate of energy loss for a lightly damped oscillator, let us take 0 . This means that the oscillator has time to complete some number of periods before its amplitude decreases considerably, i.e. the term e 2 t 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 dE dt 312. mg sin mg 2  m 0 A2 e 2 t (7) The equation of motion is  m = mg sin (1) (2) = g sin If is sufficiently small, we can approximate sin , and (2) becomes 92 CHAPTER 3 = which has the oscillatory solution g (3) ( t ) = 0 cos 0 t where 0 = g and where 0 is the amplitude. If there is the retarding force 2m g , the equation of motion becomes  m = mg sin + 2m g (4) (5) or setting sin and rewriting, we have 2 + 2 0 + 0 = 0 (6) Comparing this equation with the standard equation for damped motion [Eq. (3.35)], 2 x + 2 x + 0 x = 0 (7) we identify 0 = . This is just the case of critical damping, so the solution for ( t) is [see Eq. (3.43)] ( t ) = ( A + Bt ) e 0t For the initial conditions ( 0 ) = 0 and (0) = 0, we find (8) ( t ) = 0 (1 + 0 t ) e  0t For the case of critical damping, = 0 . Therefore, the equation of motion becomes x + 2 x + 2 x = 0 313. (1) If we assume a solution of the form x (t) = y (t) e  t (2) we have x = ye  t  ye  t x = ye  t  2 ye  t + 2 ye  t Substituting (3) into (1), we find ye  t  2 ye  t + 2 ye  t + 2 ye  t  2 2 ye  t + 2 ye  t = 0 or, y=0 Therefore, y ( t ) = A + Bt (6) (5) (4) (3) OSCILLATIONS 93 and x ( t ) = ( A + Bt ) e  t which is just Eq. (3.43). 314. (7) For the case of overdamped oscillations, x(t) and x ( t ) are expressed by x ( t ) = e  t A1e 2t + A2 e  2t x ( t ) e  t  A1e 2t + + A2 e  2t + A1 2 e 2t  A2 2 e  2t (1) ( ) ( ) (2) 2 where 2 = 2  0 . Hyperbolic functions are defined as cosh y = or, ey + e y , 2 sinh y = ey  ey 2 (3) e y = cosh y + sinh y y e = cosh y  sinh y Using (4) to rewrite (1) and (2), we have x ( t ) = ( cosh t  sinh t ) ( A1 + A2 ) cosh 2t + ( A1  A2 ) sinh 2t and x ( t ) = ( cosh t  sinh t ) ( A1 2  A1 ) ( cosh 2t + sinh 2t )  ( A2 + A2 2 ) ( cosh 2t  sinh 2t ) (4) (5) (6) 315. We are asked to simply plot the following equations from Example 3.2: x ( t ) = Ae  t cos ( 1t  ) v(t) =  Ae  t cos ( 1t  ) + 1 sin ( 1t  ) (1) (2) with the values A = 1 cm , 0 = 1 rad s 1 , = 0.1 s 1 , and = 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 (b) 1 0.5 0 0.5 1 x(t) (cm) v(t) (cm/s) 0 5 10 15 20 25 30 t (s) 1 35 40 45 50 CHAPTER 3 (c) 0.5 v (cm/s) 0 0.5 1 1 0.5 0 x (cm) 0.5 1 (B) x (cm) 0.01 0 0.01 0 5 10 15 20 25 30 t (s) 35 40 45 50 55 316. If the damping resistance b is negative, the equation of motion is 2 x  2 x + 0 x = 0 (1) where  b 2m > 0 because b < 0. The general solution is just Eq. (3.40) with changed to : x ( t ) = e t A1 exp ( 2 2 2  0 t + A2 exp  2  0 t ) ( ) (2) From this equation, we see that the motion is not bounded, irrespective of the relative values of 2 2 and 0 . The three cases distinguished in Section 3.5 now become: 2 2 a) If 0 > 2 , the motion consists of an oscillatory solution of frequency 1 = 0  2 , multiplied by an everincreasing exponential: OSCILLATIONS 95 x ( t ) = e t A1e i1t + A2 e  i1t 2 b) If 0 = 2 , the solution is (3) x ( t ) = ( A + Bt ) e t which again is everincreasing. 2 c) If 0 < 2 , the solution is: (4) x ( t ) = e t A1e 2t + A2 e  2t where 2 2 = 2  0 (5) (6) This solution also increases continuously with time. The tree cases describe motions in which the particle 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 moves continually gives energy to the particle and the motion grows without bound. 317. For a damped, driven oscillator, the equation of motion is 2 x = 2 x + 0 x = A cos t (1) and the average kinetic energy is expressed as T = mA2 2 2 4 0  2 2 + 4 2 2 ( ) (2) Let the frequency n octaves above 0 be labeled 1 and let the frequency n octaves below 0 be labeled 2 ; that is 1 = 2n 0 2 = 2 n 0 The average kinetic energy for each case is T 1 2 22 n 0 mA2 2 2 2 4 0  22 n 0 2 + (4)22 n 0 2 (3) = ( ) (4) T 2
= 2 22 n 0 mA2 2 2 2 4 0  22 n 0 2 + (4)22 n 0 2 ( ) (5) Multiplying the numerator and denominator of (5) by 24 n , we have 96 2 22 n 0 mA2 2 2 2 4 0  22 n 0 2 + (4)22 n 0 2 CHAPTER 3 T Hence, we find 2 = ( ) T and the proposition is proven. 1 = T 2 (6) 318. Since we are near resonance and there is only light damping, we have 0 R , where is the driving frequency. This gives Q 0 2 . To obtain the total energy, we use the solution to the driven oscillator, neglecting the transients: x ( t ) = D cos ( t  ) We then have (1) E= 1 1 mD2 2 2 sin 2 ( t  ) + 0 cos 2 ( t  ) mx 2 + kx 2 = 2 2 2 1 2 m 0 D2 2 (2) The energy lost over one period is ( 2m x ) ( xdt ) = 2 m D T 0 2 (3) where T = 2 . Since 0 , we have E energy lost over one period 0 4 Q 2 (4) which proves the assertion. 319. The amplitude of a damped oscillator is [Eq. (3.59)] D= A ( 02  2 ) + 4 2 2 2 (1) 2 At the resonance frequency, = R = 0  2 , D becomes DR = A 2 2 0  2 (2) 1 DR : 2 A 2 2 Let us find the frequency, = , at which the amplitude is 1 1 A = DR = 2 2 2 2 0  2 Solving this equation for , we find ( 2 0  ) + 4 2 2 (3) OSCILLATIONS 97 2 =  2 2 0 1  2 0 2 2 0 2 12 (4) For a lightly damped oscillator, is small and the terms in 2 can be neglected. Therefore, 2 2 0 2 0 (5) or, 0 1 which gives 0 (6) = ( 0 + )  ( 0  ) = 2 We also can approximate R for a lightly damped oscillator: 2 R = 0  2 2 0 (7) (8) Therefore, Q for a lightly damped oscillator becomes Q 0 0 2 (9) 320. From Eq. (3.66), x=  A ( 02  2 ) + 4 2 2 2 sin ( t  ) (1) Therfore, the absolute value of the velocity amplitude v is given by v0 = A ( 2 0  2 2 ) + 4 2 (2) 2 The value of for v0 a maximum, which is labeled v , is obtained from v0 and the value is v = 0 . Since the Q of the oscillator is equal to 6, we can use Eqs. (3.63) and (3.64) to express in terms of 0 : =0 = v (3) 2 = 2 0 146 (4) 2 , where vmax = v0 ( = 0 ) . We need to find two frequencies, 1 and 2 , for which v0 = vmax We find 98 CHAPTER 3 vmax A = = 2 2 2 A ( 02  2 ) + 4 2 2 2 (5) Substituting for in terms of 0 from (4), and by squaring and rearranging terms in (5), we obtain ( from which 2 0 2  1,2 ) 2  2 2 2 1,2 0 = 0 73 ( ) (6) 2 2 0  1,2 = 2 1 1,2 0 1,2 0 73 6 (7) Solving for 1 , 2 we obtain 1,2 0 0 12 It is sufficient for our purposes to consider 1 , 2 positive: then (8) 1 so that 0 12 + 0 ; 2  0 12 + 0 (9) = 1  2 = A graph of v0 vs. for Q = 6 is shown. v0 0 6 (10) A vmax = 2 A 2 2 0 0 12 12 0 1 0 6 321. We want to plot Equation (3.43), and its derivative: x ( t ) = ( A + Bt ) e  t v ( t ) = [ B  ( A + Bt ) ] e  t (1) (2) where A and B can be found in terms of the initial conditions A = x0 B = v0 + x 0 (3) (4) OSCILLATIONS 99 The initial conditions used to produce figure (a) were ( x0 , v0 ) = ( 2, 4 ) , (1, 4) , (4,1) , (1,4) , (1,4) , and (4, 0) , where we take all x to be in cm, all v in cm s 1 , and = 1 s 1 . 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. lim v(t) =  Bte  t t (5) (6) lim x(t ) = Bte  t t so that in this limit, v = x, as required. (a) 4 3 2 1 v (cm/s) 0 1 2 3 4 4 2 0 x (cm) 2 4 (b) 0.4 0.2 v (cm/s) 0 0.2 0.4 0.5 0.25 0 x (cm) 0.25 0.5 322. For overdamped motion, the position is given by Equation (3.44) x ( t ) = A1e  1t + A2 e  2t (1) 100 CHAPTER 3 The time derivative of the above equation is, of course, the velocity: v ( t ) =  A11 e  1t  A2 2 e  2t a) (2) At t = 0: x0 = A1 + A2 v0 =  A11  A2 2 (3) (4) The initial conditions x0 and v0 can now be used to solve for the integration constants A1 and A2 . b) When A1 = 0 , we have v0 =  2 x0 and v ( t ) =  2 x ( t ) quite easily. For A1 0 , however, we have v ( t )  1 A1 e  1t =  1 x as t since 1 < 2 . 323. Firstly, we note that all the = solutions are just the negative of the = 0 solutions. The = 2 solutions don't make it all the way up to the initial "amplitude," A , due to the retarding force. Higher means more damping, as one might expect. When damping is high, less oscillation is observable. In particular, 2 = 0.9 would be much better for a kitchen door than a smaller , e.g. the door closing ( = 0), or the closed door being bumped by someone who then changes his/her mind and does not go through the door ( = 2 ). OSCILLATIONS 101 2 = 0.5, = 0 2 = 0.9, = 0 1 2 = 0.1, = 0 0.5 0 0.5 1 2 = 0.1, = /2 2 = 0.5, = /2 2 = 0.9, = /2 1 0.5 0 0.5 1 2 = 0.1, = 2 = 0.5, = 2 = 0.9, = 1 0.5 0 0.5 1 0 5 10 15 0 5 10 15 0 5 10 15 324. 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 , but the amplitude of the particular solution becomes larger as goes through the resonance near 0.96 rad s 1 , and decreases as is increased further. The plot closest to resonance here has 1 = 1.1 , which shows the least distortion due to transients. These figures are shown in figure (a). In figure (b), the 1 = 6 plot from figure (a) is reproduced along with a new plot with Ap = 20 m s 2 . 102 /1 = 1/9 1 1 /1 = 1/3 2 /1 = 1.1 CHAPTER 3 0 0 0 1 1 0 10 20 30 0 10 20 30 /1 = 3 /1 = 6 0.5 2 0 10 t (s) 20 30 0.5 0 0 0.5 0.5 Legend: xc xp x 1 0 10 t (s) 20 30 1 0 10 t (s) 20 30 (a) 0.5 0.5 0 0 0.5 0.5 Ap = 1 1 0 5 10 15 20 25 30 Ap = 20 1 0 5 10 15 20 25 30 (b) 325. This problem is nearly identical to the previous 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 increases, mostly because the complementary solution has a fixed amplitude whereas the amplitude due to the particular solution only decreases as increases. The latter fact is because there is no resonance in this case. OSCILLATIONS 103 /1 = 1/3 /1 = 1.1 /1 = 1/9 1 1 0.5 0 0 0 0.5 0.5 1 0 5 /1 = 3 10 1 0 5 /1 = 6 10 1 0 5 /1 = 6, Ap = 6 10 0 0 0 0.5 0.5 0.5 1 0 Legend: 5 xc 10 xp 1 0 x 5 10 1 0 5 10 326. The equations of motion of this system are m1 x1 =  kx1  b1 ( x1  x2 ) + F cos t m2 x2 =  b2 x2  b1 ( x2  x1 ) (1) The electrical analog of this system can be constructed if we substitute in (1) the following equivalent quantities: m1 L1 ; m2 L2 ; k 1 ; C b1 R1 ; x q b2 R2 F 0 ; Then the equations of the equivalent electrical circuit are given by 1 q1 = 0 cos t C L2 q2 + R2 q2 + R1 ( q2  q1 ) = 0 L1q1 + R1 ( q1  q2 ) + (2) Using the mathematical device of writing exp(i t) instead of cos t in (2), with the understanding that in the results only the real part is to be considered, and differentiating with respect to time, we have 104 CHAPTER 3 I1 = i 0 e i t C L2 I 2 + R2 I 2 + R1 I 2  I1 = 0 L1I1 + R1 I1  I 2 + ( ) ( ) ( ) (3) Then, the equivalent electrical circuit is as shown in the figure: L1 0 cos t I1(t) C 1 I2(t) R1 2 R2 L2 The impedance of the system Z is Z = i L1  i where Z1 is given by 1 1 1 = + Z1 R1 R2 + i L2 Then, Z1 = R1 R2 ( R2 + R1 ) + 2 L2 + i L2 R1 2 1 + Z1 C
(4) (5) ( R1 + R2 )2 + 2 L22 (6) and substituting (6) into (4), we obtain 1 2 2 2 R1 R2 ( R2 + R1 ) + 2 L2 + i R1 L2 + L1  2 ( R1 + R2 ) + L2 C Z= 2 2 2 ( R1 + R2 ) + L2 ( ) (7) 327. From Eq. (3.89), F (t) = 1 a0 + ( an cos n t + bn sin n t ) 2 n =1 (1) We write F (t) = 1 a0 + cn cos ( n t  n ) 2 n=1 (2) which can also be written using trigonometric relations as F (t) = 1 a0 + cn cos n t cos n + sin n t sin n 2 n =1 (3) Comparing (3) with (2), we notice that if there exists a set of coefficients cn such that OSCILLATIONS 105 cn cos n = an cn sin n = bn then (2) is equivalent to (1). In fact, from (4), 2 2 2 c n = a n + bn b tan n = n an (4) (5) with an and bn as given by Eqs. (3.91). 328. Since F(t) is an odd function, F(t) = F(t), according to Eq. (3.91) all the coefficients an vanish identically, and the bn are given by bn = =  F ( t ) sin n t dt 0   sin n t dt + 0 sin n t dt = 0 1 1 cos n t +  cos n t 0 n  n = 2 ( cos 0  cos n ) n 4 for n odd = n 0 for n even Thus, b( 2 n + 1) = b( 2 n) Then, we have F (t) = 4 4 ( 2n + 1) n = 0, 1, 2, ... =0 (1) (2) sin t + 4 4 sin 3 t + sin 5 t + ... 3 5 (3) 106 F(t) 1 / 1 / t CHAPTER 3 0.849 / Terms 1 + 2 / 0.849 t 1.099 / / 1.099 Terms 1 + 2 + 3 t 0.918 / / 0.918 Terms 1 + 2 + 3 + 4 t 329. In order to Fourier analyze a function of arbitrary period, say = 2P instead of 2 , proportional change of scale is necessary. Analytically, such a change of scale can be represented by the substitution x= t P or t= Px (1) for when t = 0, then x = 0, and when t = = 2P , then x = 2 . Thus, when the substitution t = Px is made in a function F(t) of period 2P , we obtain the function Px F = f ( x) and this, as a function of x, has a period of 2 . Now, f(x) can, of course, be expanded according to the standard formula, Eq. (3.91): (2) f ( x) = where 1 a0 + ( an cos n x + bn sin n x ) 2 n =1 (3) OSCILLATIONS 107 an = bn = f ( x ) cos n x dx 0 2 (4) 0 2 f ( x ) sin n x dx If, in the above expressions, we make the inverse substitutions x= the expansion becomes t P and dx = P dt (5) a t P t n t n t f = F = F ( t ) = 0 + an cos + bn sin P 2 n =1 P P P (6) and the coefficients in (4) become n t dt 0 P 2P n t bn = F ( t ) sin dt 0 P P an = F t cos ( ) P 2P (7) For the case corresponding to this problem, the period of F(t) is 4 , so that P = 2. Then, substituting into (7) and replacing the integral limits 0 and by the limits  obtain n t dt  2 2 n t bn =  2 F (t ) sin 2 dt 2 an = 2 and + 2 , we 2 F (t ) cos 2 2 (8) and substituting into (6), the expansion for F(t) is F (t) = Substituting F(t) into (8) yields n t dt 2 2 n t bn = 0 sin t sin 2 dt 2 an = a0 n t n t + an cos + bn sin 2 2 n =1 2 (9) 2 sin t cos 0 2 (10) Evaluation of the integrals gives 108 CHAPTER 3 b2 = 1 ; bn = 0 2 for n 2 a0 = a1 = 0 0 an ( n 2 ) = 4 2 n 4 ( ) n even n odd (11) and the resulting Fourier expansion is F (t) = t 4 3 t 5 t 7 t 1 4 4 4    +... sin t + cos cos cos cos 2 3 2 5 2 21 2 45 2 (12) 330. The output of a fullwave rectifier is a periodic function F(t) of the form  sin t ;  < t 0 F (t) = sin t ; 0<t< The coefficients in the Fourier representation are given by an = 0 (  sin t ) cos n t dt + sin t cos n t dt  0 (1) bn = 0  (  sin t ) sin n t dt + 0 sin t sin n t dt (2) Performing the integrations, we obtain 4 1  n2 ; if n even ( or 0 ) an = 0; if n odd ( ) bn = 0 The expansion for F(t) is F (t) = 2 for all n (3)  4 4 cos 2 t  cos 4 t ... 3 15 (4) 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) 1 .5 sin t / 2 /2 t 331. We can rewrite the forcing function so that it consists of two forcing functions for t > : 0 F (t) = a (t ) m a (t  ) a (t )  t<0 0<t< t > (1) During the interval 0 < t < , the differential equation which describes the motion is 2 x + 2 x + 0 x = at (2) The particular solution is x p = Ct + D , and substituting this into (2), we find 2 2 2 C + 0 Ct + 0 D = at (3) from which 2 2 C + 0 D = 0 a 2 C 0  = 0 (4) Therefore, we have D= which gives xp = Thus, the general solution for 0 < t < is x ( t ) = e  t A cos 1t + B sin 1t + and then, a a 2 a 4 0 , C= a 2 0 (5) 2 0 t 2 a 4 0 (6) 2 0 t 2 a 4 0 (7) 110 CHAPTER 3 x ( t ) =  e  t A cos 1t + B sin 1t + 1e  t  A sin 1t + B cos 1t + The initial conditions, x(0) = 0, x ( 0 ) = 0 , implies a 2 2 B=  1 2 2 1 0 0 A= 4 0 a 2 0 (8) 2 a (9) Therefore, the response function is x (t) = For the forcing function  a 2  t e t e cos 1t + 2 2 0 0 1 a (t  ) 2 2 2 2  1 sin 1t + t  2 0 0 (10) in (1), we have a response similar to (10). Thus, we add these two equations to obtain the total response function: x (t) = a 2  t e  t 2 2 cos 1t  e t cos 1 ( t  ) + 2 e 2  1 2 0 0 1 0 ( ) (11) sin 1t +  e sin 1 ( t  ) + t When 0, we can approximate e as 1 + , and also sin 1 1 , cos 1 1 . Then, x ( t ) 0 a 2  t e t e cos 1t  (1 + ) ( cos 1t + 1 sin 1t ) + 2 2 0 0 1 sin 1t  (1 + ) ( sin 1t  1 cos 1t ) + = 2 2 3 a + 1  e  t cos 1t  e  t 2 1  sin 1t 2 2 0 1 1 0 0 (12) 2 2 2  1 0 2 2 If we use 1 = 0  2 , the coefficient of e  t sin 1t becomes 1 . Therefore, x ( t ) 0 a 1  e  t cos 1t  e  t sin 1t 2 0 1 (13) This is just the response for a step function. 332. a) Response to a Step Function: From Eq. (3.100) H ( t0 ) is defined as OSCILLATIONS 111 0, t < t0 H ( t0 ) = a1 , t > t0 With initial conditions x ( t0 = 0 ) and x ( t0 = 0 ) , the general solution to Eq. (3.102) (equation of motion of a damped linear oscillator) is given by Eq. (3.105): a e  (t  t0 )  t t 1  e ( 0 ) cos 1 ( t  t0 )  sin 1 ( t  t0 ) for t > t0 2 0 1 for t < t0 x (t) = 0 (1) x (t) = (2) 2 where 1 = 0  2 . 2 2 For the case of overdamping, 0 < 2 , and consequently 1 = i 2  0 is a pure imaginary number. Hence, cos 1 ( t  t0 ) and sin 1 ( t  t0 ) are no longer oscillatory functions; instead, 2 they are transformed into hyperbolic functions. Thus, if we write 2 = 2  0 (where 2 is real), cos 1 ( t  t0 ) = cos i 2 ( t  t0 ) = cosh 2 ( t  t0 ) sin 1 ( t  t0 ) = sin i 2 ( t  t0 ) = i sinh 2 ( t  t0 ) The response is given by [see Eq. (3.105)] a e  (t  t0 )  t t 1  e ( 0 ) cosh 2 ( t  t0 )  sinh 1 ( t  t0 ) for t > t0 2 0 2 for t < t0 x (t) = 0 (3) x (t) = (4) For simplicity, we choose t0 = 0 , and the solution becomes x (t) = H ( 0) e t 1  e  t cosh 2t  sinh 2t 2 0 2 (5) This response is shown in (a) below for the case = 5 0 . b) Response to an Impulse Function (in the limit 0): From Eq. (3.101) the impulse function I ( t0 , t1 ) is defined as 0 I ( t0 , t1 ) = a 0 t < t0 t0 < t < t1 t > t1 (6) For t1  t2 = 0 in such a way that a is constant = b, the response function is given by Eq. (3.110): 112 CHAPTER 3 x (t) = b 1  t t e ( 0 ) sin 1 ( t  t0 ) for t > t0 (7) Again taking the "spike" to be at t = 0 for simplicity, we have x (t) = b 1 e  t sin 1 ( t ) for t > 0 (8) 2 For 1 = i 2 = i 2  0 (overdamped oscillator), the solution is x (t) = b 2 e  t sinh 2t ; t > 0 (9) This response is shown in (b) below for the case = 5 0 . (a) 2 x H ( 0) 0 1 0.5 0 0 1 2 3 4 5 0t 6 7 8 9 10 (b) x (b 2 ) 1 0.5 0 0 1 2 3 4 5 0t 6 7 8 9 10 333. a) In order to find the maximum amplitude of the response function shown in Fig. 322, we look for t1 such that x ( t ) given by Eq. (3.105) is maximum; that is, x (t) t From Eq. (3.106) we have x (t) t ( ) t = t1 =0 (1) ( ) = H ( 0) e 2 0  t 2 1 + sin 1t 1 (2) OSCILLATIONS 113 2 For = 0.2 0 , 1 = 0  2 = 0.98 0 . Evidently, t1 = 1 makes (2) vanish. (This is the absolute maximum, as can be seen from Fig. 322.) Then, substituting into Eq. (3.105), the maximum amplitude is given by x ( t ) max or, x ( t1 ) 1.53 a 2 0  a 1 = x ( t1 ) = 2 1 + e 0 (3) (4) b) In the same way we find the maximum amplitude of the response function shown in Fig. 324 by using x(t) given in Eq. (3.110); then, x (t) t ( ) t = t1  t t sin 1 ( t  t0 ) = be ( 0 ) cos 1 ( t  t0 )  1 t = t1 (5) If (5) is to vanish, t1 is given by t1  t0 = 1 1.37 tan 1 1 = tan 1 ( 4.9) = 1 1 1 1 (6) Substituting (6) into Eq. (3.110), we obtain (for = 0.2 0 ) x ( t ) max or, x ( t1 ) 0.76 a 1.37 b = x ( t1 ) = sin ( 1.37 ) e 1 0.98 0  (7) 0 (8) The response function of an undamped ( = 0) linear oscillator for an impulse function 2 I(0, ), with = , can be obtained from Eqs. (3.105) and (3.108) if we make the following 334. 0 substitutions: =0; t0 = 0 ; 1 = 0 2 t1 = = 0 (1) (For convenience we have assumed that the impulse forcing function is applied at t = 0.) Hence, after substituting we have 114 CHAPTER 3 x (t) = 0 x (t) = x (t) = 1  cos 0 t 2 0 t<0 0<t< t > = 2 a 0 2 cos ( w0 t  2 )  cos w0 t = 0 2 0 a 0 (2) This response function is shown below. Since the oscillator is undamped, and since the impulse lasts exactly one period of the oscillator, the oscillator is returned to its equilibrium condition at the termination of the impulse. 2a 2 0 a 2 0 2 0 0 t 335. The equation for a driven linear oscillator is 0 x + 2 x + w 2 x = f ( t ) where f(t) is the sinusoid shown in the diagram. f(t) a I II III t Region I: Region II: Region III: The solution of (2) is x=0 2 x + 2 x + 0 x = a sin t 2 x + 2 x + 0 x = 0 (1) (2) (3) x = e  t ( A sin 1t + B cos 1t ) + xP in which xP = Da sin ( t  ) where D= (4) (5) (  ) + 4 2 2 2 0 2 1 (6) OSCILLATIONS 115 = tan 1 Thus, 2 2 0  2 (7) x = e  t ( A sin 1t + B cos 1t ) + D a sin ( t  ) The initial condition x(0) = 0 gives (8) B = aD sin and x ( 0 ) = 0 (9)  B + 1 A + Da cos = 0 or, A = ( sin  cos ) The solution of (3) is aD 1 (10) x ( t ) = e  t A sin 1t + B cos 1t We require that x ( t ) and x ( t ) for regions II and III match at t = . The condition that (11) xII = xIII gives e  ( A sin + B cos ) + Da sin (  ) = e  ( A sin + B cos ) where = 1 or, A + B cot = A + B cot + aD sin e sin (12) The condition that xII = xIII gives  e  ( A sin + B cos ) + aD cos (  ) + e  ( A 1 cos  B 1 sin ) =  e  ( A sin + B cos ) + e  ( A 1 cos  B 1 sin ) or, A ( 1 cos  sin )  B ( 1 sin + cos ) = A (  sin + 1 cos )  B ( 1 sin + cos )  e aD cos or, sin + cos 1 sin + cos Da cos A  B 1 = A  B e 1 cos  sin 1 cos  sin 1 cos  sin (13) 116 CHAPTER 3 Substituting into (13) from (12), we have B (1 cos  sin ) cos + (1 sin + cos ) sin (1 cos  sin ) sin (1 cos  sin ) cos + (1 sin + cos ) sin + aDe sin + cos = (1 cos  sin ) sin sin 1 cos  sin (14) from which B = aD sin + Using (12), we can find A: a 1 De sin ( 1 cos  sin ) + cos sin (15) A = A + B cot + aD sin e  B cot sin (16) Substituting for A, B, and B from (10), (9), and (15), we have aD cos A = aD sin + e cos + sin  1 + e cos 1 1 1 ( ) (17) Thus, we obtained all constants giving us the response functions explicitly. 336. With the initial conditions, x ( t0 ) = x0 and x ( t0 ) = x0 , the solution for a step function for t > t0 given by Eq. (3.103) yields A1 = x0  a 2 0 ; A2 = 1 x0 + x0 a  2 1 01 (1) Therefore, the response to H ( t0 ) for the initial conditions above can be expressed as x x  t t x ( t ) = e ( 0 ) x0 cos 1 ( t  t0 ) + 0 + 0 sin 1 ( t  t0 ) 1 1 a  (t  t0 )  tt + 2 1  e ( 0 ) cos 1 ( t  t0 )  sin 1 ( t  t0 ) e 0 1 for t > t0 (2) given by (2) for t0 < t < t1 and by a superposition of solutions for H ( t0 ) and for H ( t1 ) taken individually for t > t1 . We must be careful, however, because the solution for t > t1 must be with initial conditions x ( 0 ) = 0 , x ( 0 ) = 0 , and using t1 instead of t0 in the expression. The solution for t > t1 is then The response to an impulse function I ( t0 , t1 ) = H ( t1 ) , for the above initial conditions will then be equal that given by (2) for t = t1 . This can be insured by using as a solution for H ( t1 ) Eq. (3.103) OSCILLATIONS 117 x x  t t x ( t ) = e ( 0 ) x0 cos 1 ( t  t0 ) + 0 + 0 sin 1 ( t  t0 ) + x1 ( t ) 1 1 where x1 ( t ) =  tt ae ( 0 ) e e cos 1 ( t  t0  )  cos 1 ( t  t0 ) + 2 0 1 (3) e + sin 1 ( t  t0  )  sin 1 ( t  t0 ) 1 1 (4) for t > t1 We now allow a as 0 in such a way that a = b = constant; expanding (3) for this particular case, we obtain x x b  tt x ( t ) = e ( 0 ) x0 cos 1 ( t  t0 ) + 0 + 0 + sin 1 ( t  t0 ) 1 1 1 which is analogous to Eq. (3.119) but for initial conditions given above. 337. Any function F ( t ) m can be expanded in terms of step functions, as shown in the figure below where the curve is the sum of the various (positive and negative) step functions. t > t0 (5) In general, we have 2 x + 2 x + 0 x = Fn ( t ) n = m = where an ( t ) Hn (t) = 0 n = H (t) n (1) t > tn = n t < tn = n (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 solutions for each of those functions. According to Eq. (3.105), the solution for H n ( t ) for t > tn is xn ( t ) = then, for F (t) = Hn (t) m n = an e  ( t  tn )  t t 1  e ( n ) cos 1 ( t  tn )  sin 1 ( t  tn ) 2 0 1 (3) (4) the solution is 118 CHAPTER 3 x (t) = e  ( t  tn )  t t H n ( t ) 1  e ( n ) cos 1 ( t  tn )  sin 1 ( t  tn ) 2 0 n = 1 1 = where n = mH (t ) G (t ) = F ( t ) G (t ) n n n = n n (5) 1 2 Gn ( t ) = m 0 0 or, comparing with (3) e  ( t  tn )  ( t  tn ) cos 1 ( t  tn )  sin 1 ( t  tn ) ; 1  e 1 t tn t < tn (6) xn ( t ) man , Gn ( t ) = 0 F (t ) m t tn t < tn (7) Therefore, the Green's function is the response to the unit step. tn tn+1 t Hn(t) 338. The solution for x(t) according to Green's method is x (t) = = t  F ( t ) G ( t , t ) dt F0 m 1 t 0 e  t sin t e  (t  t ) sin 1 ( t  t ) dt (1) Using the trigonometric identity, sin t sin 1 ( t  t ) = we have 1 cos ( 1 + ) t  1t  cos (  1 ) t + 1t 2 (2) OSCILLATIONS 119 x (t) = t t F0 e  t (  ) t cos ( + 1 ) t  1t  dt e (  )t cos (  1 ) t + 1t dt e 2m 1 0 0 (3) Making the change of variable, z = ( + 1 ) t  1t , for the first integral and y = (  1 ) t + 1t for the second integral, we find (  )1t F0 e  t e +1 x (t) = 2m 1 + 1 t  1t (  ) z dz e +1 cos z  e  (  )1t t 1  1 1t dy e (  ) y  1 cos y (4) After evaluating the integrals and rearranging terms, we obtain x (t) = F0 2 2 m (  ) + ( + 1 ) (  ) 2 + (  1 ) 2 sin t 2 2 e  t 2 (  ) cos t + [  ] + 1  2 ( ) (5) 2 2 sin 1t + e  t 2 (  ) cos 1t + [  ] + 2  1 ( ) 339. sin t F (t) = 0 From Equations 3.89, 3.90, and 3.91, we have F (t) = an = bn = an = a0 = a1 = 2 0<t< < t < 2 1 a0 + ( an cos n t + bn sin n t ) 2 n=1 F ( t ) cos n t dt 0 F ( t ) sin n t dt 0 2 sin t cos n t dt 0 2 0 sin t dt = sin t cos t dt = 0 0 120 CHAPTER 3 cos (1  n) t cos (1 + n) t an ( n 2) = sin t cos n t dt =   0 2 (1  n) 2 (1 + n) 0 Upon evaluating and simplifying, the result is 2 1  n2 an = 0 ( ) n even n = 0,1,2,... n odd b0 = 0 by inspection b1 = 0 sin 2 t dt = 1 2 sin (1  n) t sin (1 + n) t bn ( n 2) = sin t sin n t dt =  0 2 (1  n) 2 (1 + n) 0 So F (t) = or, letting n 2n F (t) = 1 1 1 2 sin 2 t + cos n t 2 2 n = 2,4,6,... 1  n =0 + ( ) + 1 2 sin t + cos 2n t 2 2 n = 1,2,... 1  4 n ( ) The following plot shows how well the first four terms in the series approximate the function. 1.0 Sum of first four terms 0.5 2 t 340. The equation describing the car's motion is m d2 y =  k ( y  a sin t ) dt 2 where y is the vertical displacement of the car from its equilibrium position on a flat road, a is the amplitude of sinecurve road, and k = elastic coefficient = dm g 100 9.8 = = 98000 N/m dy 0.01 OSCILLATIONS 121 = 2 v0 = 174 rad/s with v0 and being the car's speed and wavelength of sinecurve road. a 0 2 sin t 2 0  2 The solution of the motion equation can be cast in the form y ( t ) = B cos ( 0 t + ) + with 0 = k = 9.9 rad/s m We see that the oscillation with angular frequency has amplitude A= 2 a 0 = 0.16 mm 2 0  2 The minus sign just implies that the spring is compressed. 341. a) The general solution of the given differential equation is (see Equation (3.37)) 2 2 x ( t ) = exp (  t ) A1 exp t 2  0 + A2 exp t 2  0 ( ) ( ) and 2 2 v(t) = x ( t ) =  exp (  t ) A1 exp t 2  0 + A2 exp t 2  0 2 2 2 + exp (  t ) 2  0 A1 exp t 2  0  A2 exp t 2  0 ( ) ( ) ( ) ( ) at t = 0, x(t) = x0 , v(t) = v0 A1 = b) i) Underdamped, = v + x0 1 x + 0 2 0 2 2  0 and A1 = v + x0 1 x  0 2 0 2 2  0 (1) 0 2 In this case, instead of using above parameterization, it is more convenient to work with the following parameterization x(t ) = A exp (  t ) cos t 0 2  2  ( ) ( ) (2) (3) 2 2 2 v(t) =  A exp (  t ) cos t 0  2  + 0  2 sin t 0  2  ( ) Using initial conditions of x(t) and v(t), we find x0 v0 2 2 x + + 0  and tan ( ) = 2 0 0  2 2 A= v0 + x0  2 0 2 122 CHAPTER 3 In the case = 0 2 , and using (6) below we have tan = 2 v0 1 1 + = 3 x 0 0 3 3 2 x0 3 0 = 30 A= so finally x(t ) = 2 v0 v0 0 2 2 + + 0 = x0 2 x0 x0 3 3 2 1 + 30 x0 exp  0 t cos t 2 2 0 3 (4) ii) Critically damped, = 0 , using the same parameterization as in i) we have from (2) and (3): x(t ) = A exp (  t ) = x0 exp (  0 t ) and v(t) = x (t) =  0 x0 exp( 0 t v0 =  0 x0 iii) Overdamped, = 0 , returning to the original parameterization (1) we have (always using relation (6)), 2 2 x ( t ) = exp (  t ) A1 exp t 2  0 + A2 exp t 2  0 (5) (6) ( ) ( ) = ( 3 + 1) x0 exp 2 3 x (( 3  2) 0 t + ) ( 3  1) x0 exp  ( 3 + 2) 0 t 2 3 ( ) (7) Below we show sketches for equations (4), (5), (7) Underdamped Critically damped t Overdamped 342. a) m ( x + 0 2 x )  F0 sin t = 0 The most general solution is x(t ) = a sin 0 t + b cos 0 t + A sin t where the last term is a particular solution. (1) OSCILLATIONS 123 To find A we put this particular solution (the last term) into (1) and find A= F0 m (  2 ) 2 0 At t = 0, x = 0, so we find b = 0, and then we have x(t ) = a sin 0 t + A sin t v(t) = a 0 cos 0 t + A cos t At t = 0, v = 0 A=  a 0 F0 1 ( 0 sin t  sin 0t) m 0 ( 0 + )( 0  ) x(t ) = b) In the limit 0 one can see that x(t ) The sketch of this function is shown below. x F0 t 3 0 6m t 343. a) Potential energy is the elastic energy: U(r ) = 1 k(r  a)2 , 2 where m is moving in a central force field. Then the effective potential is (see for example, Chapter 2 and Equation (8.14)): U eff ( r ) = U (r ) + l2 1 l2 = k ( r  a )2 + 2mr 2 2 2mr 2 where l = mvr = m r 2 is the angular momentum of m and is a conserved quantity in this problem. The solid line below is U eff ( r ) ; at low values of r, the dashed line represents 1 l2 k(r  a)2 , and the solid line is dominated by . At large values of r, 2 2mr 2 1 U eff ( r ) U (r ) = k( r  a)2 . 2 U(r ) = 124 CHAPTER 3 Potential energy l2 2mr 2 U (r ) = 1 k ( r  a )2 2 U eff (r ) r b) In equilibrium circular motion of radius r0 , we have 2 k ( r0  a ) = m 0 r0 0 = k ( r0  a ) mr0 0 c) For given (and fixed) angular momentum l, V(r) is minimal at r0, because V (r ) r = r = 0 , so we make a Taylor expansion of V(r) about r0 ; 2 3m 0 (r  r0 )2 K (r  r0 )2 1 V (r ) = V (r0 ) + (r  r0 )V (r0 ) + (r  r0 )2 V ( r0 ) + ... = 2 2 2 2 where K = 3m 0 , so the frequency of oscillation is = K = 3 0 = m 3k(r0  a) mr0 344. This oscillation must be underdamped oscillation (otherwise no period is present). From Equation (3.40) we have x(t ) = A exp (  t ) cos ( 1t  ) so the initial amplitude (at t = 0) is A. Now at t = 4T = 8 1 8 x(4T ) = A exp  cos (8  ) 1 8 The amplitude now is A exp  , so we have 1 OSCILLATIONS 125 8 A exp  1 1 = A e 2 2 and because = 0  1 , we finally find 1 8 = 0 64 2 + 1 mgl 2 where is the amplitude. 2 345. Energy of a simple pendulum is For a slightly damped oscillation (t) exp(  t) . Initial energy of pendulum is mgl 2 . 2 l , is g Energy of pendulum after one period, T = 2 mgl mgl 2 (T )2 = exp (2 T ) 2 2 So energy lost in one period is mgl 2 mgl 2 ( 1  exp ( 2 T ) ) 2 T = mgl 2 T 2 2 So energy lost after 7 days is mgl 2 T (7 days) = mgl 2 (7 days) T This energy must be compensated by potential energy of the mass M as it falls h meters: Mgh = mgl 2 (7 days) = Mh = 0.01 s 1 ml (7 days) 2 Knowing we can easily find the coefficient Q (see Equation (3.64)) 0 2  2 2 = Q= R = 2 2 g  2 2 l = 178 2 126 CHAPTER 3 CHAPTER 4 Nonlinear Oscillations and Chaos 41. 0 = 0 +d m x s m d d 0 = 0 +d (a) (b) (c) The unextended length of each spring is 0 , 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 F = 2k ( s  where 0 ) sin (1) s= and sin = Then, 2 + x2 (2) x 2 + x2 (3) F ( x) =  2kx 2 +x 2 2 + x2  0 = 2kx 2 +x 2 2 + x2  (  d) = 2kx 1  1 2 d  d x2 = 2kx 1  1 + 2 2 + x2 (4) 127 128 CHAPTER 4 Expanding the radical in powers of x 2 2 and retaining only the first two terms, we have  d 1 x2 F ( x ) 2kx 1  1  2 2 d 1  d x2 = 2kx 1  1  + 2 2 = The potential is given by 2kd x k (  d) 3 x3 (5) U ( x ) =  F ( x ) dx so that (6) U ( x) = kd x2 + k (  d) 4 x 4 3 (7) 42. Using the general procedure explained in Section 4.3, the phase diagram is constructed as follows: U(x) E E2 1 E E4 3 E E6 5 x x E1 E2 E4 E5 E6 E3 x NONLINEAR OSCILLATIONS AND CHAOS 129 43. The potential U ( x ) =  ( 3x ) has the form shown in (a) below. The corresponding phase 3 diagram is given in (b): (a) U(x) E5 E E3 4 E E1 2 E1 x E2 (b) x x 44. Differentiation of Rayleigh's equation above yields 2 x  a  3bx 2 x + 0 x = 0 ( ) (1) The substitution, y = y0 implies that x= a y 3b y 0 a y 3b y 0 a y 3b y 0 3b x a (2) x= (3) x= When these expressions are substituted in (1), we find 3ba y 2 y a y a a y 2  + 0 =0 a  2 3b y 0 3b 3b y 0 b y0 y0 Multiplying by y0 3b and rearranging, we arrive at van der Pol's equation: a y a 2 2 y0  y 2 y + 0 y = 0 2 y0 (4) ( ) (5) 130 45. CHAPTER 4 a) A graph of the functions f1 ( x ) = x 2 + x + 1 and f 2 ( x ) = tan x in the region 0 x 2 shows that there is an intersection (i.e., a solution) for x1 3 8 . tan x x2 + x +1 O x /2 into f1 ( x ) and then solve for x = tan 1 f1 ( x1 ) . If the result is within some specified amount, The procedure is to use this approximate solution as a starting point and to substitute x1 = 3 8 say 10 4 , of 3 8 , 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 point and repeat the calculation. This procedure leads to the following values: x1 1.1781 1.2974 1.3247 1.3304 1.3316 1.3318 Thus, the solution is x = 1.3319. Parts b) and c) are solved in exactly the same way with the results: b) x = 1.9151 c) x = 0.9271 46. 2 f1 ( x1 ) = x1 + x1 + 1 tan 1 f1 ( x1 ) 1.2974 1.3247 1.3304 1.3316 1.3318 1.3319 Difference 0.11930 0.02728 0.00573 0.00118 0.00024 0.00005 3.5660 3.9806 4.0794 4.1004 4.1047 4.1056 For the plane pendulum, the potential energy is u = mg 1  cos (1) If the total energy is larger than 2mg , all values of are allowed, and the pendulum revolves continuously in a circular path. The potential energy as a function of is shown in (a) below. U 2mg mg /2 O /2 (a) Since T = E U( ), we can write NONLINEAR OSCILLATIONS AND CHAOS 131 T= 1 1 mv 2 = m 2 2 = E  mg (1  cos ) 2 2 (2) and, therefore, the phase paths are constructed by plotting = 12 2 E  mg (1  cos ) 2 m (3) versus . The phase diagram is shown in (b) below. E = 2mg 3 E = mg 2 E = 3mg E= 5 mg 2 E = mg  2 2 (b) 47. Let us start with the equation of motion for the simple pendulum: 2 =  0 sin (1) = sin . where 2 g . Put this in terms of the horizontal component by setting y x Solving for and taking time derivatives, we obtain = yy 2 y + 2 32 (1  y ) 1  y2 (2) 2 Since we are keeping terms to third order, we need to get a better handle on the y term. Help comes from the conservation of energy: 1 2 2 m  mg cos =  mg cos 0 2 (3) where 0 is the maximum angle the pendulum makes, and serves as a convenient parameter that describes the total energy. When written in terms of y , the above equation becomes (with the obvious definition for y0 ) y2 2 = 2 0 2 1 y ( 2 1  y 2  1  y0 ) (4) Substituting (4) into (2), and the result into (1) gives 132 2 2 y + 0 y 3 1  y 2  2 1  y0 = 0 CHAPTER 4 ( ) (5) Using the binomial expansion of the square roots and keeping terms up to third order, we can obtain for the x equation of motion 2 x0 3 g 2 x + 0 x 1 + 2  3 x 3 = 0 2 (6) 48. For x > 0, the equation of motion is mx =  F0 (1) If the initial conditions are x ( 0 ) = A , x ( 0 ) = 0 , the solution is x (t) = A  F0 2 t 2m (2) For the phase path we need x = x ( x ) , so we calculate x ( x) = 2F0 ( A  x) m (3) Thus, the phase path is a parabola with a vertex on the xaxis at x = A and symmetrical about both axes as shown below. x A t=0 t= 1 4 x Because of the symmetry, the period is equal to 4 times the time required to move from x = A to x = 0 (see diagram). Therefore, from (2) we have =4 2mA F0 (4) 49. The proposed force derives from a potential of the form 1 2 kx 2 U ( x) = 1 ( x + ) x  ax 2 2 x < a x > a which is plotted in (a) below. NONLINEAR OSCILLATIONS AND CHAOS 133 U(x) E7 E6 E5 E4 x E3 E2 E1 a O a (a) For small deviations from the equilibrium position (x = 0), the motion is just that of a harmonic oscillator. For energies E < E6 , the particle cannot reach regions with x < a, but it can reach regions of x > a if E > E4 . For E2 < E < E4 the possibility exists that the particle can be trapped near x = a. A phase diagram for the system is shown in (b) below. x E2 E1 E3 E4 x ... E7 (b) 410. The system of equations that we need to solve are y x y = 0.05 y  sin x + 0.7 cos t (1) The values of that give chaotic orbits are 0.6 and 0.7. Although we may appear to have chaos for other values, construction of a Poincar 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 than others. In particular, when = 0.8 the plot looks chaotic until it locks on to the point (2.50150, 0.236439) . The phase plot for = 0.3 shown in the figure was produced by numerical integration of the system of equations (1) with 100 points per drive cycle. The box encloses the point on the trajectory of the system at the start of a drive cycle. In addition, we also show Poincar plot for the case = 0.6 in figure, integrated over 8000 drive cycles with 100 points per cycle. 134 1 CHAPTER 4 0.5 0 0.5 1 1.5 1 0.5 0 0.5 1 1.5 3 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 411. The threecycle does indeed occur where indicated 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 threecycle near the 200th iteration, and stays that way until approximately the 270th iteration, although some may see it continue past the 300th. 1 0.8 0.6 x 0.4 0.2 0 100 200 300 400 500 iteration NONLINEAR OSCILLATIONS AND CHAOS 135 412. x1 = 0.4 0.25 0.2 0.15 0.1 0.05 0 0 0.2 0.4 0.6 0.8 1 x1 = 0.75 0.25 0.2 0.15 0.1 0.05 0 0 0.2 0.4 0.6 0.8 1 These plots are created in the manner described in the text. They are created with the logistic equation xn + 1 = 0.9 xn (1  xn ) The first plot has the seed value x1 = 0.4 as asked for in the text. Only one additional seed has been done here ( x1 = 0.75 ) as it is assumed that the reader could easily produce more of these plots after this small amount of practice. (1) 136 413. 1 0.8 0.6 0.4 0.2 0 30 32 34 36 38 40 42 iteration 44 46 48 50 CHAPTER 4 x1 = 0.7 x1 = 0.700000001 1 0.8 0.6 0.4 0.2 0 30 32 34 36 38 40 42 iteration 44 46 48 50 x1 = 0.7 x1 = 0.7000000001 The plots are created by iteration on the initial values of (i) 0.7, (ii) 0.700000001, and (iii) 0.7000000001, using the equation 2 xn + 1 = 2.5 xn 1  xn ( ) (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. 414. 1 0.8 0.6 0.4 0.2 0 20 22 24 26 28 iteration 30 32 34 x1 = 0.9 x1 = 0.9000001 fractional difference NONLINEAR OSCILLATIONS AND CHAOS 137 The given function with the given initial values are plotted in the figure. Here we use the notation x1 = 0.9 and y1 = 0.9000001 , with xn + 1 = f ( xn ) and yn + 1 = f ( yn ) where the function is f ( x ) = 2.5 x 1  x 2 ( ) (1) The fractional difference is defined as x  y x , and clearly exceeds 30% when n = 30. 415. A good way to start finding the bifurcations of the function f(,x) = sin x is to plot its bifurcation diagram. 1 0.8 0.6 x 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 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 n .Using Fractint for Windows, and Equation (4.47) to compute the Feigenbaum constant, we can obtain the following: n 1 2 3 4 5 6 7 8 0.71978 0.83324 0.85859 0.86409 0.86526 0.86551 0.865564 0.8655761 4.475 4.611 4.699 4.680 4.630 4.463 One can see that although we should obtain a better value of as n increases, numerical precision and human error quickly degrade the quality of the calculation. This is a perfectly acceptable answer to this question. One may compute the n 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 their algorithm, one obtains the following: 138 CHAPTER 4 n 1 2 3 4 5 6 7 0.719962 0.833266 0.858609 0.864084 0.865259 0.865511 0.865560 4.47089 4.62871 4.66198 4.65633 5.13450 Note that these are shown here only as reference, and the student may not necessarily be expected to perform to this degree of sophistication. The above values are only good to about 10 6 , but this time only limited by machine precision. Another alternative in computing the Feigenbaum constant, which is not requested in the problem, is to use the socalled "supercycles," or superstable points Rn , which are defined by f ( 2n1 ) R , 1 = 1 n 2 2 The values Rn obey the same scaling as the bifurcation points, and are much easier to compute since these points converge faster than for other (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, Jrgens and Saupe. As a result, the estimates for obtained in this way are more accurate than those obtained by calculating the bifurcation points. 416. where x0 = f ( x0 ) . Now expand f(x) in a Taylor series, so that near x0 we have f ( x) where f ( x0 ) + ( x  x0 ) = x 0 + ( x  x0 ) The function y = f(x) intersects the line y = x at x = x0 , i.e. x0 is defined as the point (1) df dx (2) x0 Now define n xn  x0 . If we have x1 very close to x0 , then 1 should be very small, and we may use the Taylor expansion. The equation of iteration xn + 1 = f ( xn ) becomes n+1 n (3) If the approximation (1) remains valid from the initial value, we have n +1 a) The values xn  x0 = n form the geometric sequence 1 , 1 , 2 1 , ... . b) Clearly, when < 1 we have stability since n1 . lim n = 0 n Similarly we have a divergent sequence when > 1 , although it will not really be exponentially divergent since the approximation (1) becomes invalid after some number of iterations, and normally the range of allowable xn is restricted to some subset of the real numbers. NONLINEAR OSCILLATIONS AND CHAOS 139 417. 0.6 0.4 0.2 0 0 2 4 6 8 10 iteration 12 14 16 18 20 = 0.4 = 0.7 The first plot (with = 0.4) converges rather rapidly to zero, but the second (with = 0.7) does appear to be chaotic. 418. 1 0.8 0.6 x 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 The tent map always converges to zero for < 0.5. Near = 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 < < 0.7 (0.7 is approximate), where the iterations are chaotic but oscillate between an upper and lower range of values. For > 0.7, there is only a single range of chaos, which becomes larger until it fills the range (0,1) at = 1. 419. From the definition in Equation (4.52) the Lyapunov exponent is given by = lim n df 1 n 1 ln dx n i=0 (1) xi The tent map is defined as for 0 < x < 1 2 2 x f ( x) = 2 (1  x ) for 1 2 < x < 1 This gives df dx = 2 , so we have (2) 140 CHAPTER 4 = lim n  1 ln ( 2 ) = ln ( 2 ) n n (3) As indicated in the discussion below Equation (4.52), chaos occurs when is positive: > 1 2 for the tent map. 420. 0.4 0.2 y 0 0.2 0.4 1.5 1 0.5 0 x 0.5 1 1.5 421. 0.4 0.2 y 0 0.2 0.4 1.5 1 0.5 0 x 0.5 1 1.5 The shape of this plot (the attractor) is nearly identical to that obtained in the previous problem. In Problem 420, 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. 422. The following. system of differential equations were integrated numerically y x y = 0.1 y  x 3 + B cos t (1) 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 drive 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 Poincar section (boxes) for this case are overlaid and shown in figure (a). All integrations are done here with 100 points per drive 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 that it has one period per cycle. The second plot is a Poincar 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 regular 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 B1 , B2 , and B3 . Circumventing the actual task of computing where these transition points are, we do know that 9.8 < B1 < 9.9 , 11.6 < B2 < 11.7 , and 13.3 < B3 < 13.4 . We can then describe the behavior of the system by region. B < B1 : one period per three drive cycles B1 < B < B2 : chaotic B2 < B < B3 : mixed chaotic/one period per drive cycle (depending on initial conditions) B3 < B : one period per drive cycle 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 (a) 5 0 y 5 3 2 1 0 x 1 2 3 (b) 10 5 y 0 5 2.4 2.6 2.8 x 3 3.2 3.4 3.6 (c) 5 y 0 5 3 2 1 0 x 1 2 3 (d) 10 5 y 0 5 10 2.4 2.6 2.8 3 3.2 x 3.4 3.6 3.8 NONLINEAR OSCILLATIONS AND CHAOS 143 423. The Chirikov map is defined by pn + 1 = pn  K sin qn qn + 1 = qn  pn + 1 (1) (2) The results one should get from doing this problem 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 notsorandom 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 orbits (the ones in the left are separate from the ones in the right). (a) 1 0.5 p/ 0 0.5 1 1 0.5 0 q/ 0.5 1 (b) 1 0.5 p/ 0 0.5 1 1 0.5 0 q/ 0.5 1 144 (c) 1 CHAPTER 4 0.5 p/ 0 0.5 1 1 0.5 0 q/ 0.5 1 424. a) The Van de Pol equation is d2 x dx + 0 2 x = ( a2  x 2 ) 2 dt dt Now look for solution in the form x(t ) = b cos 0 t + u(t) we have dx du =  b 0 sin 0 t + dt dt and d2 x d2u =  b 0 2 cos 0 t + 2 dt 2 dt Putting these into the Van de Pol equation, we obtain d 2 u(t) + 0 u(t) =  b 2 cos 2 0 t + u2 (t) + 2bu(t ) cos 0 t  a 2 dt 2 (1) { } {  b 0 sin 0 t + du(t ) dt } From this one can see that u(t) is of order (i.e. u ~ O( ) ), which is assumed to be small here. Keeping only terms up to order , the above equation reads d 2 u(t) + 0 u(t ) =   b 3 0 sin 0 t cos 2 0 t + a 2 b 0 sin 0 t 2 dt { } b2 b2 =  b 0 a 2  sin 0 t  sin 3 0 t 4 4 (where we have used the identity 4 sin 0 t cos 2 0 t = sin 0 t + sin 3 0 t ) This equation has 2 frequencies ( 0 and 3 0 ), and is complicated. However, if b = 2 a then the term sin 0 t disappears and the above equation becomes NONLINEAR OSCILLATIONS AND CHAOS 145 d 2 u(t) b3 sin 3 0 t + 0 u(t) = 0 4 dt 2 We let b = 2 a , and the solution for this equation is u(t ) =  b 3 a3 sin 3 0 t =  sin 3 0 t 32 0 4 0 So, finally putting this form of u(t) into (1), we obtain one of the exact solutions of Van de Pol equation: u(t ) = 2 a cos 0 t  a3 sin 3 0 t 4 0 b) See phase diagram below. Since = 0.05 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 1 2 1 1 2 x 1 2 425. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because = 0.07 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 424 for exact expression of stable solution). 146 . x 2 CHAPTER 4 1 2 1 1 2 x 1 2 b) 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 424 for exact expression of stable solution). . x 2 1 2 1 1 1 2 3 x 2 426. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because = 0.5 is not a small value, the limit cycle is NOT close to a circle (see problem 424 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 424 for exact expression of stable solution). NONLINEAR OSCILLATIONS AND CHAOS 147 . x 2 1 2 1 1 1 2 x 2 b) 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 424 for exact expression of stable solution). . x 2 1 2 1 1 2 3 x 1 2 148 CHAPTER 4 CHAPTER 5 Gravitation 51. a) Two identical masses: The lines of force (dashed lines) and the equipotential surfaces (solid lines) are as follows: b) Two masses, +M and M: In this case the lines of force do not continue outward 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 electrical lines of force run from +q to q. 149 150 52. Inside the sphere the gravitational potential satisfies CHAPTER 5 2 = 4 G ( r ) Since (r) is spherically symmetric, is also spherically symmetric. Thus, 1 2 = 4 G ( r ) r r 2 r r The field vector is independent of the radial distance. This fact implies = constant C r (1) (2) (3) Therefore, (2) becomes 2C = 4 G r or, (4) = C 2 Gr (5) 53. In order to remove a particle from the surface of the Earth and transport it infinitely far away, the initial kinetic energy must equal the work required to move the particle from r = Re to r = against the attractive gravitational force: Re G Me m 1 2 dr = mv0 2 r 2 (1) where Me and Re are the mass and the radius of the Earth, respectively, and v0 is the initial velocity of the particle at r = Re . Solving (1), we have the expression for v0 : v0 = 2G Me Re (2) Substituting G = 6.67 10 11 m 3/kg s 2 , Me = 5.98 10 24 kg , Re = 6.38 10 6 m , we have v0 11.2 km/sec (3) 54. The potential energy corresponding to the force is U =  F dx = mk 2 dx mk 2 = 2 x 3 2x (1) The central force is conservative and so the total energy is constant and equal to the potential energy at the initial position, x = d: GRAVITATION 151 E = constant = 1 1 k2 1 k2 mx 2  m 2 =  m 2 2 2 x 2 d 0 (2) Rewriting this equation in integrable form, dt =  d 0 dx 1 1 k 2  2 d x 2 = d x dx k d2  x2 d (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 t= or t= d2 k d d2  x2 k 0 (4) d 55. The equation of motion is mx = G Mm x2 (1) Using conservation of energy, we find 1 2 1 1 x  G M = E = G M 2 x x 1 1 dx =  2GM  dt x x (2) (3) where x is some fixed large distance. Therefore, the time for the particle to travel from x to x is t = x x 1 = GM 1 1 2GM  x x dx x x xx dx 2 ( x  x ) Making the change of variable, x y 2 , and using Eq. (E.7), Appendix E, we obtain t= x x 1 x ( x  x )  x sin 2GM x x x (4) If we set x = 0 and x = x 2 in (4), we can obtain the time for the particle to travel the total distance and the first half of the distance. 152 0 CHAPTER 5 T0 = x 2 x dt = 1 x GM 2 x 2 32 32 (5) T1 2 = Hence, x dt = 1 GM 1 + 2 (6) T1 2 T0 Evaluating the expression, T1 2 T0 or = 1+ 2 = 0.818 (7) T1 2 T0 9 11 (8) 56. z P x s r2drd(cos )d r y Since the problem has symmetry around the zaxis, the force at the point P has only a zcomponent. The contribution to the force from a small volume element is dg z = G where is the density. Using cos = a s2 r 2 dr d ( cos ) d cos (1) z  r cos and integrating over the entire sphere, we have s +1 1 2 g z = G r 2 dr d ( cos ) d 0 0 (r z  r cos 2 + z 2  2rz cos ) 32 (2) Now, we can obtain the integral of cos as follows: GRAVITATION 153 +1 1 I= ( z  r cos r + z 2  2rz cos 2 ) 32 d ( cos ) = z +1 1 (r 2 + z 2  2rz cos ) 12 d ( cos ) Using Eq. (E.5), Appendix E, we find I= z 1 2 r + z 2  2rz cos  rz ( ) 12 +1 1 (3) = 2 2 = z z z 2 Therefore, substituting (3) into (2) and performing the integral with respect to r and , we have g z = G = G But a3 2 2 3 z2 (4) 4 3 1 a 2 z 3 4 3 a is equal to the mass of the sphere. Thus, 3 g z = GM 1 z2 (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. 57. dx x s R P The contribution to the potential at P from a small line element is d = G where = M s dx (1) is the linear mass density. Integrating over the whole rod, we find the potential = G M 2  2 1 x 2 + R2 dx (2) 154 CHAPTER 5 Using Eq. (E.6), Appendix E, we have M 2 2 2 + R2 + GM 2 4 = ln 2 + R2  + 4 2 2 = G ln x + x 2 + R2  (3) = GM ln + 4 R2 + 2 + 4 R2  58. z z0 a z rdrddz y r 2 + ( z0  z ) 2 x r Since the system is symmetric about the zaxis, the x and y components of the force vanish and we need to consider only the zcomponent of the force. The contribution to the force from a small element of volume at the point (r,,z) for a unit mass at (0,0, z0 ) is dg z = G r + ( z0  z ) 2 rdrd dz 2 cos = G ( z0  z) rdrd dz 32 r 2 + ( z0  z ) 2 (1) where is the density of the cylinder and where we have used cos = ( z0  z ) 2 r 2 + ( z0  z ) . We can find the net gravitational force by integrating (1) over the entire volume of the cylinder. We find g z = G rdr 0 a 2 d dz 0 0 a z0  z0  z 32 2 r 2 + ( z0  z ) Changing the variable to x = z0  z , we have g z = 2 G rdr 0 z0 xdx r + x2 2 32 (2) Using the standard integral, GRAVITATION 155 we obtain xdx ( a2 x 2 3 ) = 1 a x2 2 (3) a r g z = 2 G dr r2 + z  0 (0 Next, using Eq. (E.9), Appendix E, we obtain ) 2  2 r 2 + z0 r (4) g z = 2 G a 2 + ( z0  )2  2 a 2 + z0 + (5) Now, let us find the force by first computing the potential. The contribution from a small element of volume is d = G Integrating over the entire volume, we have d = G dz 0 2 rdrd dz r + ( z0  z ) 2 2 (6) d dr 0 0 a r r 2 + ( z0  z ) 2 (7) Using Eq. (E.9), Appendix E, again, we find 2 d = 2 G dz a 2 + ( z0  z )  ( z0  z ) 0 (8) Now, we use Eqs. (E.11) and (E.8a), Appendix E, and obtain (z  = 2 G  0 2 + Thus, the force is z0 2 ) a 2 + ( z0  )2 + a2 ln 2 ( z0  2 ) + 2 ( z0  ) 2 + a 2 2 2 a 2 + z0  a2 1 2 ln 2z0 + 2 z0 + a 2  z0 + 2 2 156 CHAPTER 5 1 2 = 2 G gz =  a + ( z0  z 0 2 ) 2 1 + 2 ( z0  ) 2 2 a 2 + ( z0  ) 1 z0 1 a2 + 2 (z  0 z0  ( z0  ) 2 + a 2 ) + ( z0  ) 2 + a 2  1 2 1 2 a + z0  2 2 z +a z a  + 2 2 2  z0 + z0 + a 2 a 2 + z0 2 0 2 2 0 2 (9) or, g z = 2 G a 2 + ( z0  and we obtain the same result as in (5). In this case, it is clear that it is considerably easier to compute the force directly. (See the remarks in Section 5.4.) 59. a R r P )2  2 a 2 + z0 + (10 The contribution to the potential at the point P from a small line element d is = G d r M . Using 2 a (1) where is the linear mass density which is expressed as = r = R2 + a 2  2aR cos and d = ad, we can write (1) as = GM 2 2 0 d R + a  2aR cos 2 2 (2) This is the general expression for the potential. If R is much greater than a, we can expand the integrand in (2) using the binomial expansion: 1 2 a2 1 a = 1  2 cos  2 R R2 + a 2  2aR cos R R 1 2 1 a a2 3 a a2 1 = 1 + 2 cos  2 + 2 cos  2 + ... R 8 R R R 2 R (3) GRAVITATION 157 3 a If we neglect terms of order and higher in (3), the potential becomes R = GM 2 GM 2 2 0 3 a2 a a2 1 + cos  + cos 2 d R 2 2 2R 2R (4) = or, a2 3 a2 2  2 + 2 2 R R ( R)  GM 1 a 2 1+ R 4 R2 (5) We notice that the first term in (5) is the potential when mass M is concentrated in the center of GMa 2 . the ring. Of course this is a very rough approximation and the first correction term is  4 R3 510. P a x R r R cos R sin Using the relations x= ( R sin ) 2 + a2  2aR sin cos (1) (2) (3) r = x 2 + R2 cos 2 = R 2 + a 2  2 aR sin cos = the potential is expressed by = G M (the linear mass density), 2 a GM 2 R 2 d r = 0 d a a2 1  2 sin cos  2 R R 3 (4) If we expand the integrand and neglect terms of order ( a R) and higher, we have a a2 1  2 sin cos  2 R R Then, (4) becomes 1 2 1+ a 1 a2 3 a2 sin cos  sin 2 cos 2 + R 2 R2 2 R2 (5) 158 CHAPTER 5  Thus, GM 1 a2 3 a2 2  2 + sin 2 2 2 2 R 2R 2R ( R)  GM R 1 a2 3 2 1  2 R 2 1  2 sin (6) 511. P r dm z a The potential at P due to a small mass element dm inside the body is d = G dm dm = G 2 2 r z + a  2 za cos (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: d ave = 1 4 a 2 2 a 2 sin d G dm z 2 + a 2  2za cos 0 (2) Making the variable change cos = x, we have d ave Using Eq. (E.5), Appendix E, we find G =  dm 2 1 +1 (z ) dx 2 + a 2  2 zax ) (3) d ave =  = = G 1 dm  2 za (z 2 + a 2  2za + 1 za (z 2 + a 2 + 2 za ) G  ( z  a) + ( z + a) dm 2 za G dm z (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 159 512. dm r' O r R P Let P be a point on the spherical surface. The potential d due to a small amount of mass dm inside the surface at P is d =  Gdm r (1) The average value over the entire surface due to dm is the integral of (1) over d divided by 4. Writing this out with the help of the figure, we have d ave =  2 sin d Gdm 0 r 2 + R2  2r R cos 4 (2) Making the obvious change of variable and performing the integration, we obtain d ave =  Gdm 1 du Gdm 1 r 2 + R2  2r Ru =  R 4 (3) We can now integrate over all of the mass and get ave =  Gm R . This is a mathematical statement equivalent to the problem's assertion. 513. R1 R0 R2 1 2 R0 = position of particle. For R1 < R0 < R2 , we calculate the force by assuming that all mass for which r < R0 is at r = 0, and neglect mass for which r > R0 . The force is in the radially inward direction ( er ). The magnitude of the force is F= where M = mass for which r < R0 GMm 2 R0 160 CHAPTER 5 M= So F =  4 Gm 3 3 3 1R1 + 2 R0  2 R1 er 2 3R0 4 4 3 3 3 R1 1 + R0  R1 2 3 3 ( ) ( ) (  ) R3 4 F =  Gm 1 22 1 + 2 R0 er 3 R0 514. 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 where is the potential due to the mass already assembled, and dm is the mass of the shell. So 3 Mr 2 dr 3M dm = 4 r 2 dr = 4 r 2 dr = 3 R3 4 R = So Gm r3 where m = M 3 r R U = du = 3 Mr 2 dr GMr 2 R3  R3 r =0 R = 3GM 2 4 r dr R6 0 U= 3 GM 2 5 R R 515. 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 Fr : m r Fr = Gm 4 3 r where is the mass density of the Earth. The equation of motion is r2 3 GRAVITATION 161 Fr = mr =  or Gm 4 3 r r2 3 4 G 3 r + 2 r = 0 where 2 = This is the equation for simple harmonic motion. The period is T= 2 = 3 G Substituting in values gives a period of about 84 minutes. 516. z y x h r M r 2 + h2 For points external to the sphere, we may consider the sphere to be a point mass of mass M. Put the sheet in the xy plane. Consider force on M due to the sheet. By symmetry, Fx = Fy = 0 Fz = dFz = With dm = s 2 rdr and cos = we have Fz = 2sGMh h r + h2 2 r =0 (r GMdm cos 2 + h2 ) r =0 (r rdr 2 + h2 ) 32 1 Fz = 2sGMh 2 r + h2 ( ) 12 0 Fz = 2sGM The sphere attracts the sheet in the z direction with a force of magnitude 2sGM 162 517. y CHAPTER 5 water x moon (not to scale) Earth Start with the hint given to us. The expression for g x and g y are given by gx = GMm y GMe y 2GMm x GMe x   ; gy =  3 3 D3 R3 D R (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 origin of the coordinate system is at the center of the Earth. Evaluating the integrals: xmax 0 2 2GMm GMe xmax ; g x dx =  3 3 R 2 D ymax 0 2 GMm GMe ymax  3 g y dy =  3 R 2 D (2) To connect this result with Example 5.5, let us write (1) in the following way 2 ymax GMe 2 GMm 2 2 xmax  ymax xmax + = 3 3 2 2R D ( ) (3) The righthand side can be factored as GMe GMe x + ymax )( xmax  ymax ) = ( 2 R ) ( h ) = gh 3 ( max 2R 2R 3 2 If we make the approximation on the lefthand side of (3) that xmax Equation (5.55). Turning to the exact solution of (3), we obtain 2 ymax (4) R 2 , we get exactly h = 2R Me M m Me 2 M m + 3   3 R D R3 D3 Me Mm Me 2 M m + 3 +  3 R D R3 D3 (5) Upon substitution of the proper values, the answer is 0.54 m, the same as for Example 5.5. Inclusion of the centrifugal term in g x does not change this answer significantly. GRAVITATION 163 518. From Equation (5.55), we have with the appropriate substitutions hmoon hsun Substitution of the known values gives 3GMm r 2 3 2 gD3 M R = = m es 3GMs r 2 Ms D 3 2 gRes (1) hmoon 7.350 10 22 kg 1.495 1011 m = hsun 1.993 10 30 kg 3.84 10 8 m 3 2. 2 (2) 519. earth moon 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 observed frequency at which we see high tides. Thus 1 1 1 =  2Ttides Tearth Tmoon which gives Ttides 12 hours, 27 minutes. (1) 520. The differential potential created by a thin loop of thickness dr at the point (0,0,z) is 2 rdrM GM d ( r 2 ) 2GM ( z 2 + R2  z ) = ( z) = d( z) = 2 2 2 2 2 2 R R2 R z +r z +r G d( z) = Then one can find the gravity acceleration, 2 2 ^ d =  k 2GM z + R  z ^ g( z ) =  k dz R2 z 2 + R2 ^ where k is the unit vector in the zdirection. 164 521. CHAPTER 5 (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 dF = G ( M L) mdx x 2 GMm F = dF = L L+ D D dx GMm = 2 x D(L + D) And that is the total gravitational force acting on m by the rod. CHAPTER 6 (1) Some Methods in the Calculus of Variations 61. If we use the varied function y ( , x ) = x + sin (1  x ) Then dy = 1  cos (1  x ) dx Thus, the total length of the path is (2) S= 0 1 1 dy 1 + dx dx 12 2 = 2  2 cos (1  x ) + 2 2 cos 2 (1  x ) 0 dx (3) Setting (1  x ) u , the expression for S becomes S= 1 0 1 2 1  cos u + 2 2 cos 2 2 u 12 du (4) The integral cannot be performed directly since it is, in fact, an elliptic integral. Because is a small quantity, we can expand the integrand and obtain S= 2 0 2 1 1 2 2 1 1 2 2 2 2 1  cos u  cos u  cos u  cos u + ... du 8 2 2 2 (5) If we keep the terms up to cos 2 u and perform the integration, we find S= 2+ which gives 2 2 2 16 (6) 165 166 CHAPTER 6 S 2 2 = 8 Therefore S and S is a minimum when = 0. 62. (7) =0 a=0 (8) The element of length on a plane is dS = dx 2 + dy 2 (1) from which the total length is ( x2 , y2 ) S= ( x1 , y1 ) dx 2 + dy 2 = x2 x1 dy 1 + dx dx 2 (2) If S is to be minimum, f is identified as dy f = 1+ dx Then, the Euler equation becomes d d 1 + y2 = 0 dx dy where y = dy . (4) becomes dx d y dx 1 + y 2 or, y 1 + y2 from which we have = constant C (6) =0 (5) (4) 2 (3) y = Then, C2 = constant a 1  C2 (7) y = ax + b This is the equation of a straight line. (8) SOME METHODS IN THE CALCULUS OF VARIATIONS 167 63. The element of distance in threedimensional space is dS = dx 2 + dy 2 + dz 2 (1) Suppose x, y, z depends on the parameter t and that the end points are expressed by ( x1 (t1 ) , y1 (t1 ) , z1 (t1 )) , ( x2 (t2 ) , y2 (t2 ) , z2 (t2 )) . Then the total distance is S= t1 t2 dx dy dz dt + dt + dt dt 2 2 2 (2) The function f is identified as f = x2 + y2 + z2 Since (3) f f f = = = 0 , the Euler equations become x y z d f =0 dt x d f =0 dt y d f =0 dt z (4) from which we have = constant C1 x2 + y2 + z2 y = constant C2 x2 + y2 + z2 z = constant C3 x2 + y2 + z2 x From the combination of these equations, we have (5) y x = C1 C2 y z = C2 C3 If we integrate (6) from t1 to the arbitrary t, we have (6) 168 CHAPTER 6 x  x1 y  y1 = C1 C2 y  y1 z  z1 = C2 C3 On the other hand, the integration of (6) from t1 to t2 gives (7) x2  x1 y2  y1 = C1 C2 y2  y1 z2  z1 = C2 C3 (8) from which we find the constants C1 , C2 , and C3 . Substituting these constants into (7), we find y  y1 x  x1 z  z1 = = x2  x1 y2  y1 z2  z1 This is the equation expressing a straight line in threedimensional space passing through the two points ( x1 , y1 , z1 ) , ( x2 , y2 , z2 ) . (9) 64. z 1 dS y 2 x The element of distance along the surface is dS = dx 2 + dy 2 + dz 2 In cylindrical coordinates (x,y,z) are related to (,,z) by x = cos y = sin z=z from which dx =  sin d dy = cos d dz = dz (1) (2) (3) SOME METHODS IN THE CALCULUS OF VARIATIONS 169 Substituting (3) into (1) and integrating along the entire path, we find S = d + dz = 2 2 2 1 2 2 1 2 + z 2 d (4) where z = dz . If S is to be minimum, f 2 + z 2 must satisfy the Euler equation: d f f  =0 z z (5) Since f = 0 , the Euler equation becomes z z + z2 2 =0 (6) from which z + z2 2 = constant C (7) or, z= Since is constant, (8) means dz = constant d and for any point along the path, z and change at the same rate. The curve described by this condition is a helix. 65. z (x2,y2) ds (x1,y1) y C2 1  C2 (8) x The area of a strip of a surface of revolution is dA = 2 ds = 2 dx 2 + dy 2 Thus, the total area is (1) 170 x2 CHAPTER 6 A = 2 where y = x1 x 1 + y 2 dx (2) dy . In order to make A a minimum, f x 1 + y 2 must satisfy equation (6.39). Now dx f = 1 + y2 x xy f = y 1 + y2 Substituting into equation (6.39) gives 1 + y2 = xy 2 d x 1 + y2  dx 1 + y2 d x = dx 1 + y 2 = Multiplying by 1 + y 2  xy ( dy dx ) 1 + y 2 1 + y2 ( ) 1 2 1 + y 2 and rearranging gives  dy dx = x y 1 + y2 ( ) Integration gives  ln x + ln a = y2 1 ln 2 1 + y2 where ln a is a constant of integration. Rearranging gives y2 = Integrating gives y = b + a cosh 1 or x = a cosh which is the equation of a catenary. yb a x a ( 1 x a2  1 2 ) SOME METHODS IN THE CALCULUS OF VARIATIONS 171 66. 0 = y =0 2a (x2,y2) x (x1,y1) If we use coordinates with the same orientation as in Example 6.2 and if we place the minimum point of the cycloid at (2a,0) the parametric equations are x = a (1 + cos ) y = a ( + sin ) Since the particle starts from rest at the point ( x1 , y1 ) , the velocity at any elevation x is [cf. Eq. 6.19] v = 2 g ( x  x1 ) Then, the time required to reach the point ( x2 , y2 ) is [cf. Eq. 6.20] t= x2 (1) (2) x1 1 + y2 2 g ( x  x1 ) 12 dx (3) Using (1) and the derivatives obtained therefrom, (3) can be written as t= a g 1 + cos cos  cos1 2 = 0 1 12 d (4) Now, using the trigonometric identity, 1 + cos = 2 cos 2 2 , we have a g 1 t= 0 cos cos 2 2 d 2  cos 2 1 2 = a g 1 0 cos sin 2 2 d 2 1 2  sin 2 (5) Making the change of variable, z = sin 2 , the expression for t becomes sin 1 2 t=2 a g 0 dz sin 2 1 2 (6)  z2 The integral is now in standard form: 172 CHAPTER 6 Evaluating, we find x = sin 1 a a =x dx 2 2 (7) t= a g (8) Thus, the time of transit from ( x1 , y1 ) to the minimum point does not depend on the position of the starting point. 67. a 1 1 n1 n2 b 2 (n2 > n1) c n1 c v2 = n2 v1 = x d The time to travel the path shown is (cf. Example 6.2) 1 + y2 ds t= = dx v v Although we have v = v(y), we only have dv dy 0 when y = 0. The Euler equation tells us y d dx v 1 + y 2 Now use v = c n and y = tan to obtain n sin = const. This proves the assertion. Alternatively, Fermat's principle can be proven by the method introduced in the solution of Problem 68. 68. (1) =0 (2) (3) To find the extremum of the following integral (cf. Equation 6.1) J = f ( y , x ) dx we know that we must have from Euler's equation f =0 y This implies that we also have SOME METHODS IN THE CALCULUS OF VARIATIONS 173 f J = dx = 0 y y giving us a modified form of Euler's equation. This may be extended to several variables and to include the imposition of auxiliary conditions similar to the derivation in Sections 6.5 and 6.6. The result is g j J + j ( x) =0 y i y i j when there are constraint equations of the form g j ( yi , x ) = 0 a) The volume of a parallelepiped with sides of lengths a1 , b1 , c1 is given by V = a1b1c1 We wish to maximize such a volume under the condition that the parallelepiped is circumscribed by a sphere of radius R; that is, 2 2 2 a1 + b1 + c1 = 4 R 2 (1) (2) We consider a1 , b1 , c1 as variables and V is the function that we want to maximize; (2) is the constraint condition: g { a1 , b1 , c1 } = 0 Then, the equations for the solution are g V + =0 a1 a1 g V + =0 b1 b1 g V + =0 c1 c1 from which we obtain b1c1 + 2 a1 = 0 a1c1 + 2 b1 = 0 a1b1 + 2 c1 = 0 Together with (2), these equations yield a1 = b1 = c1 = 2 R 3 2 R. 3 (6) (3) (4) (5) Thus, the inscribed parallelepiped is a cube with side 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 2 a1 b2 c2 = 12 = 1 2 = 1 4 a 2 4b 4z (7) where a1 , b1 , c1 are the lengths of the sides of the parallelepiped. Combining (7) with (1) and (4) gives 2 2 2 a1 b1 c1 = 2 = 2 a2 b c (8) Then, a1 = a 2 2 2 , b1 = b , c1 = c 3 3 3 (9) 69. The average value of the square of the gradient of ( x1 , x2 , x3 ) within a certain volume V is expressed as I= 1 V ( ) 2 dx1 dx2 dx3 1 v= V In order to make I a minimum, 2 2 2 dx1dx2 dx3 + + x x x 1 2 3 (1) f = + + x1 x2 x3 must satisfy the Euler equation: 3 f f  =0 i = 1 x i x i 2 2 2 (2) If we substitute f into (2), we have x i =1 3 =0 i x i (3) which is just Laplace'sequation: 2 = 0 Therefore, must satisfy Laplace's equation in order that I have a minimum value. (4) SOME METHODS IN THE CALCULUS OF VARIATIONS 175 610. This problem lends itself to the method of solution suggested in the solution of Problem 68. The volume of a right cylinder is given by V = R2 H The total surface area A of the cylinder is given by (1) A = Abases + Aside = 2 R 2 + 2 RH = 2 R ( R + H ) We wish A to be a minimum. (1) is the constraint condition, and the other equations are (2) g A + =0 R R g A + =0 H H where g = V  R 2 H = 0 . The solution of these equations is R= 1 H 2 (3) (4) 611. y 1 2a R ds } The constraint condition can be found from the relation ds = Rd (see the diagram), where ds is the differential arc length of the path: ds = dx 2 + dy 2 which, using y = ax 2 , yields 1 + 4a 2 x 2 dx = Rd If we want the equation of constraint in other than a differential form, (2) can be integrated to yield A + R = x 2 4 ax 2 + 1 + 1 ln 2ax + 4 a 2 x 2 + 1 4a (2) ( ) 12 = Rd (1) ( ) (3) where A is a constant obtained from the initial conditions. The radius of curvature of a parabola, y = ax 2 , is given at any point (x,y) by r0 1 2a . The condition for the disk to roll with one and only one point of contact with the parabola is R < r0 ; that is, 176 CHAPTER 6 R< 1 2a (4) 612. The path length is given by s = ds = 1 + y 2 + z 2 dx (1) and our equation of constraint is g ( x, y, z) = x2 + y 2 + z2  2 = 0 (2) The Euler equations with undetermined multipliers (6.69) tell us that y d dx 1 + y 2 + z 2 dg = = 2 y dy (3) with a similar equation for z. Eliminating the factor , we obtain y 1 d y dx 1 + y 2 + z 2 1 d z  z dx 1 + y 2 + z 2 =0 (4) This simplifies to z y 1 + y 2 + z 2  y ( y y + z z )  y z 1 + y 2 + z 2  z ( y y + z z ) = 0 zy + ( yy + zz ) z y  yz  ( yy + zz ) y z = 0 ( ) ( ) (5) (6) and using the derivative of (2), ( z  xz ) y = ( y  xy ) z 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: Ax + By = z (7) (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 plane passing through the origin, as required. 613. 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 y1 ( x ) and y2 ( x ) respectively. (note that y1 > 0 and y2 < 0 ) The enclosed area is J ( y1 , y2 ) = a a y1 ( x)dx  a a y2 ( x)dx = a ( y1 (x)  y2 (x)) dx = a a a f ( y1 , y2 )dx SOME METHODS IN THE CALCULUS OF VARIATIONS 177 The total length of closed curve is K ( y1 , y2 ) = a a 1 + ( y1 ) dx + 2 a a 1 + ( y2 ) dx = 2 a { a 1 + ( y1 ) + 1 + ( y1 ) 2 2 } dx = a g ( y1 , y2 ) dx a Then the generalized versions of Eq. (6.78) (see textbook) for this case are g y1 f d f d g d  +  =0 = 0 1 2 y1 dx y1 dx 1 + ( y1 ) y1 dx y1 g y2 f d f d g d  +  =0 = 0 1 2 y2 dx y2 dx 1 + ( y2 ) y2 dx y2 Analogously to Eq. (6.85); from (1) we obtain from (2) we obtain (1) (2) ( x  A1 ) 2 + ( y1  A2 ) 2 = 2 ( x  B1 ) 2 + ( y2  B2 ) 2 = 2 (3) (4) where constants A's, B's can be determined from 4 initial conditions ( x = a, y1 = 0) and ( x = a, y2 = 0) We note that y1 < 0 and y2 > 0 , so actually (3) and (4) altogether describe a circular path of radius . And this is the sought configuration that renders maximum enclosed area for a given path length. 614. It is more convenient to work with cylindrical coordinates (r, ,z) in this problem. The constraint here is z = 1 r , then dz = dr ds2 = dr 2 + r 2 d 2 + dz 2 = 2 ( dr 2 + r 2 d 2 ) where we have introduced a new angular coordinate = 2 In this form of ds2 , we clearly see that the space is 2dimensional Euclidean flat, so the shortest line connecting two given points is a straight line given by: r= cos (  0 ) r0 =  0 cos 2 r0 this line passes through the endpoints (r = 1, = the shortest path equation 2 ) , then we can determine unambiguously r( ) = 2 2 cos 2 cos and z = 1 r 178 CHAPTER 6 Accordingly, the shortest connecting length is l= dr 2 d 2 d + r = 2 2 sin 2 2  2 2 2 615. dy 2 I[ y] =  y 2 dx dx 0 1 a) Treating I[y] as a mechanical action, we find the corresponding EulerLagrange equation y( x ) =  d2 y dx 2 Combining with the boundary conditions (x = 0, y = 0) and (x = 1, y = 1), we can determine unambiguously the functional form of y( x ) = (sin x) (sin 1) . b) The corresponding minimum value of the integral is 1 dy 2 1 I[ y] =  y 2 dx = dx cos 2 x = cot (1) = 0.642 sin 2 1 0 0 dx 1 c) If x = y then I[y] = ( 2 3 ) = 0.667. 616. a) S is arc length 9 dy dz dy S = dx 2 + dy 2 + dz 2 = dx 1 + + = dx 1 + + x = L dx dx dx dx 4 Treating S and L like a mechanical action and Lagrangian respectively, we find the canonical momentum associated with coordinate y p= 2 2 2 L dy dx = 1+ dy dx 9 dy x+ dx 4 2 Because L does not depend on y explicitly, then EL equation implies that p is constant (i.e. dp dx = 0 ), then the above equation becomes 9 32 1+ x dy 9 9 4 y = p =p +B dx 1 + x = A 1 + x dx 1  p2 1  p2 4 4 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 9 y( x ) = 3 2 1 + x  1 13  8 4 8 and z = x3 2 b) z 1 0.75 0.5 0.25 y 0 1 0.75 0.5 0.25 x 1 0. 0.75 0.5 0.25 00 617. a) Equation of a ellipse x2 y2 + =1 a2 b2 which implies xy ab 2 because 2xy x 2 y 2 2 + 2 ab a b so the maximal area of the rectangle, whose corners lie on that ellipse, is Max[A] = Max[4xy] = 2 iab. This happens when x= a 2 and y= b 2 b) The area of the ellipse is A0 = ab ; so the fraction of rectangle area to ellipse area is then Max[ A] 2 = A0 618. One can see that the surface xy = z is "locally" symmetric with respect to the line x =  y = z where x > 0, y < 0, z < 0. This line is a parabola. This 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. v( z) = 2 g( z + 1) 180 CHAPTER 6 CHAPTER 7 Hamilton's Principle Lagrangian and Hamiltonian Dynamics 71. Four coordinates are necessary to completely describe the disk. These are the x and y coordinates, the angle that measures the rolling, and the angle that describes the spinning (see figure). y x Since the disk may only roll in one direction, we must have the following conditions: dx cos + dy sin = R d (1) dy = tan dx (2) These equations are not integrable, and because we cannot obtain an equation relating the coordinates, the constraints are nonholonomic. This means that although 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 or (pitch and yaw), and vice versa. 72. Start with the Lagrangian L= = m v + at + cos 0 2 ( ) + ( sin ) + mg 2 2 cos (1) (2) m ( v0 + at ) 2 + 2 ( v0 + at ) cos + 2 2 + mg cos 2 181 182 CHAPTER 7 Now let us just compute d L d 2 = m ( v0 + at ) cos + m dt dt (3) = ma cos  m ( v0  at ) sin + m 2 L =  m ( v0 + at ) sin  mg sin According to Lagrange's equations, (4) is equal to (5). This gives Equation (7.36) (4) (5) = g sin + a cos = 0 (6) To get Equation (7.41), start with Equation (7.40) = and use Equation (7.38) g cos e  a sin e (7) tan e =  a g (8) to obtain, either through a trigonometric identity or a figure such as the one shown here, g 2 + a2 e g a cos e = Inserting this into (7), we obtain g g +a 2 2 sin e = a g + a2 2 (9) = as desired. a2 + g 2 (10) We know intuitively that the period of the pendulum 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 pendulum will be in equilibrium when it is in line with the effective acceleration. Since the acceleration is sideways and gravity is down, and the period can only depend on the magnitude of the effective acceleration, the correct form is clearly a2 + g 2 . HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 183 73. R If we take angles and as our generalized coordinates, the kinetic energy and the potential energy of the system are T= 2 1 1 m ( R  ) + I 2 2 2 (1) (2) U = R  ( R  ) cos mg 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 I = ( 2 5) m 2 , the Lagrangian becomes L = T U = 1 1 2 m ( R  ) 2 + m 2 2  R  ( R  ) cos mg 2 5 (3) When the sphere is at its lowest position, the points A and B coincide. The condition A0 = B0 gives the equation of constraint: f ( , ) = ( R  )  = 0 Therefore, we have two Lagrange's equations with one undetermined multiplier: L d  dt f L + = 0 f L d L  + =0 dt (4) (5) After substituting (3) and f = R  and f =  into (5), we find  ( R  ) mg sin  m ( R  ) + ( R  ) = 0 2 (6) (7)  From (7) we find : 2 m 2  = 0 5 =  m or, if we use (4), we have 2 5 (8) =  m ( R  ) Substituting (9) into (6), we find the equation of motion with respect to : 2 5 (9) 184 CHAPTER 7 =  2 sin where is the frequency of small oscillations, defined by (10) = 5g 7 ( R  ) (11) 74. y m r x If we choose (r, ) as the generalized coordinates, the kinetic energy of the particle is T= 1 1 m x 2 + y 2 = m r 2 + r 2 2 2 2 U r ( ) ( ) (1) Since the force is related to the potential by f = we find U= A (2) r (3) where we let U(r = 0) = 0. Therefore, the Lagrangian becomes L= 1 A m r 2 + r 2 2  r 2 ( ) (4) Lagrange's equation for the coordinate r leads to mr  mr 2 + Ar 1 = 0 Lagrange's equation for the coordinate leads to d mr 2 = 0 dt Since mr 2 = is identified as the angular momentum, (6) implies that angular momentum is conserved. Now, if we use , we can write (5) as mr  Multiplying (7) by r , we have mrr  r 2 + Ar r 1 = 0 mr 3 (8) 2 3 (5) ( ) (6) mr + Ar 1 = 0 (7) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 185 which is equivalent to 2 d A d 1 d mr 2 + r =0 + 2 dt 2 dt 2mr dt (9) Therefore, d (T + U ) = 0 dt (10) and the total energy is conserved. 75. y r m x Let us choose the coordinate system so that the xy 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,) as T= U= 1 m r 2 + r 2 2 2 ( ) (1) (2) A r + mgr sin from which the Lagrangian is L = T U = 1 A m r 2 + r 2 2  r  mgr sin 2 ( ) (3) Therefore, Lagrange's equation for the coordinate r is mr  mr 2 + Ar 1 + mg sin = 0 Lagrange's equation for the coordinate is d mr 2 + mgr cos = 0 dt (4) ( ) (5) Since mr 2 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 76. y M S m x CHAPTER 7 Let us choose ,S as our generalized coordinates. The x,y coordinates of the center of the hoop are expressed by x = + S cos + r sin y = r cos + ( Therefore, the kinetic energy of the hoop is  S) sin (1) Thoop = = 1 1 m x 2 + y 2 + I 2 2 2 1 m + S cos 2 ( ) ( ) + ( S sin ) + 1 I 2 2 2 2 (2) Using I = mr 2 and S = r , (2) becomes Thoop = 1 m 2S 2 + 2 + 2S cos 2 (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 xaxis which is Tplane = Therefore, the total kinetic energy becomes T = mS2 + The potential energy is U = mgy = mg r cos + (  S) sin Hence, the Lagrangian is l = mS 2 + 1 ( m + M ) 2 + mS cos  mg r cos + (  S) sin 2 (7) (6) 1 ( m + M ) 2 + mS cos 2 (5) 1 M2 2 (4) from which the Lagrange equations for and S are easily found to be 2mS + m cos  mg sin = 0 (8) (9) ( m + M ) + mS cos = 0 HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 187 or, if we rewrite these equations in the form of uncoupled equations by substituting for and S , we have m cos 2 2  S  g sin = 0 m+ M mg sin cos = 2 2 ( m + M )  m cos Now, we can rewrite (9) as d ( m + M ) + mS cos = 0 dt where we can interpret ( m + M ) as the x component of the linear momentum of the total system and mS cos 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 xaxis. 77. y1 1 x1 m 2 x2 m x y2 y (10) (11) If we take (1 , 2 ) as our generalized coordinates, the x,y coordinates of the two masses are x1 = cos 1 y1 = sin 1 x2 = cos 1 + cos 2 y2 = sin 1 + sin 2 Using (1) and (2), we find the kinetic energy of the system to be T= = = m 2 m 2 2 2 x1 + y1 + x2 + y2 2 2 m 2 m 2 2 (1) (2) ( ) ( ) 2 2 2 1 + 1 + 2 + 212 ( sin 1 sin 2 + cos 1 cos 2 ) 2 2 21 + 2 + 21 2 cos (1  2 ) 2 (3) 188 CHAPTER 7 The potential energy is U =  mgx1  mgx2 =  mg 2 cos 1 + cos 2 Therefore, the Lagrangian is (4) 2 1 2 L = m 2 1 + 2 + 12 cos (1  2 ) + mg 2 cos 1 + cos 2 2 from which L = m 21 2 sin (1  2 )  2mg sin 1 1 L 2 2 = 2m 1 + m 2 cos (1  2 ) 1 L =  m 21 2 sin (1  2 )  mg sin 2 2 L 2 2 = m 2 + m 1 cos (1  2 ) 2 The Lagrange equations for 1 and 2 are 2 21 + 2 cos (1  2 ) + 2 sin (1  2 ) + 2 (5) (6) g sin 1 = 0 (7) 2 2 + 1 cos (1  2 )  1 sin (1  2 ) + g sin 2 = 0 (8) 78. U1 y v2 2 1 v1 U2 x Let us choose the x,y coordinates so that the two regions are divided by the y axis: U1 U ( x) = U2 x<0 x>0 If we consider the potential energy as a function of x as above, the Lagrangian of the particle is L= 1 m x2 + y2  U ( x) 2 ( ) (1) Therefore, Lagrange's equations for the coordinates x and y are HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 189 mx + dU ( x ) =0 dx (2) (3) my = 0 Using the relation mx = dP dP dx Px dpx d mx = x = x = dt dt dx dt m dx Px dPx dU ( x ) + =0 m dx dx (4) (2) becomes (5) Integrating (5) from any point in the region 1 to any point in the region 2, we find 1 2 dU ( x ) Px dPx dx + dx = 0 m dx dx 1 2 (6) Px22 2m or, equivalently,  Px21 2m + U 2  U1 = 0 (7) 1 1 2 2 mx1 + U1 = mx2 + U 2 2 2 Now, from (3) we have d my = 0 dt (8) and my is constant. Therefore, my1 = my2 (9) From (9) we have 1 1 2 2 my1 = my2 2 2 Adding (8) and (10), we have 1 1 2 2 mv1 + U1 = mv2 + U 2 2 2 From (9) we also have mv1 sin 1 = mv2 sin 2 (10) (11) (12) Substituting (11) into (12), we find sin 1 v2 U1  U 2 = = 1 + T1 sin 2 v1 12 (13) 190 CHAPTER 7 This problem is the mechanical analog of the refraction of light upon passing from a medium of a certain optical density into a medium with a different optical density. 79. y O x m M Using the generalized coordinates given in the figure, the Cartesian coordinates for the disk are ( cos , sin ), and for the bob they are ( sin + cos , cos sin ). The kinetic energy is given by 1 1 1 2 2 T = Tdisk + Tbob = M 2 + I 2 + m x bob + y bob 2 2 2 ( ) (1) Substituting the coordinates for the bob, we obtain T= 1 1 1 ( M + m) 2 + I 2 + m 2 2 + m cos ( + a) 2 2 2 (2) The potential energy is given by U = Udisk + U bob = Mgydisk + mgy bob =  ( M + m) g sin  mg cos (3) Now let us use the relation = R to reduce the degrees of freedom to two, and in addition substitute I = MR2 2 for the disk. The Lagrangian becomes 1 1 3 L = T  U = M + m 2 + m 2 2 + m cos ( + a) + ( M + m) g sin + mg cos 4 2 2 The resulting equations of motion for our two generalized coordinates are 3 2 M + m  ( M + m) g sin + m cos ( + )  sin ( + ) = 0 2 (5) (6) (4) + cos ( + ) + 1 g sin = 0 710. y x M S y x M HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 191 Let the length of the string be so that (S  x)  y = Then, x = y a) The Lagrangian of the system is (1) (2) L= 1 1 Mx 2 + My 2  Mgy = My 2  Mgy 2 2 d L L  = 2 My + Mg = 0 dt y y (3) Therefore, Lagrange's equation for y is (4) from which y= g 2 (5) Then, the general solution for y becomes y (t) =  g 2 t + C1t + C2 4 (6) If we assign the initial conditions y ( t = 0 ) = 0 and y ( t = 0 ) = 0 , we find y (t) =  g 2 t 4 (7) b) If the string has a mass m, we must consider its kinetic energy and potential energy. These are Tstring = U string =  m 1 my 2 2 y mg 2 y = 2 2 mg 2 1 my 2 + y 2 2 (8) (9) yg Adding (8) and (9) to (3), the total Lagrangian becomes L = My 2  Mgy + (10) Therefore, Lagrange's equation for y now becomes ( 2 M + m) y  mg y + Mg = 0 (11) In order to solve (11), we arrange this equation into the form ( 2 M + m) y = mg M y  m (12) 192 CHAPTER 7 Since d2 dt 2 M d2 = y y , (12) is equivalent to m dt 2 d2 dt 2 M y  m = mg ( 2 M + m) M y  m (13) which is solved to give y M = Ae t + Be  t m (14) where = mg ( 2 M + m) (15) If we assign the initial condition y ( t = 0 ) = 0 ; y ( t = 0 ) = 0 , we have A = +B =  M 2m Then, y (t) = M (1  cosh t ) m (16) 711. x y er m x y The x,y coordinates of the particle are x = R cos t + R cos ( + t ) y = R sin t + R sin ( + t ) (1) Then, x =  R sin t  R + sin ( + t ) y = R cos t + R + cos ( + t ) ( ) ( ) (2) Since there is no external force, the potential energy is constant and can be set equal to zero. The Lagrangian becomes HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 193 L= 1 m x2 + y2 2 ( ) (3) = from which m 2 2 2 2 2 R + R ( + ) + 2R ( + ) cos 2 L =  mR2 + sin ( ) (4) d L d = mR 2 + + cos dt dt ( ) (5) Therefore, Lagrange's equation for becomes + 2 sin = 0 (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 a = 2 R ( i + er ) (7) where the primed unit vectors are as indicated in the figure. The part proportional to e does r not affect the motion since it has no contribution 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. 712. m r Put the origin at the bottom of the plane L = T U = 1 m r 2 + r 2 2  mgr sin 2 (
) = t; = L= 1 m r 2 + 2 r 2  mgr sin t 2 mr = m 2 r  mg sin t ( ) Lagrange's equation for r gives or r  2 r =  g sin t (1) 194 CHAPTER 7 The general solution is of the form r = rp + rh where rh is the general solution of the homogeneous equation r  2 r = 0 and rp is a particular solution of Eq. (1). So rh = Ae t + Be  t For rp , try a solution of the form rp = C sin t . Then rp = C 2 sin t . Substituting into (1) gives C 2 sin t  C 2 sin t =  g sin t C= g 2 2 g So r ( t ) = Ae t + Be  t + 2 2 sin t We can determine A and B from the initial conditions: r ( 0 ) = r0 r ( 0) = 0 (2) (3) (2) implies r0 = A + B (3) implies 0 = A  B + g 2 2 g 1 r0  2 2 2 g 1 r0 + 2 2 2 Solving for A and B gives: A= r (t) = B= g t 1 g  t g 1 r0  2 2 e + 2 r0 + 2 2 e + 2 2 sin t 2 or r ( t ) = r0 cosh t + g 2 2 ( sin t  sinh t ) 713. a) a b m HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 195 x= 1 2 at  b sin 2 y =  b cos x = at  b cos y = b sin L= = 1 m x 2 + y 2  mgy 2 1 m a 2t 2  2 at b cos + b 2 2 + mgb cos 2 d L L = dt gives ( ) ( ) d  mat b cos + mb 2 = mat b sin  mgb sin dt This gives the equation of motion + g a sin  cos = 0 b b b) To find the period for small oscillations, we must expand sin and cos about the equilibrium point 0 . We find 0 by setting = 0 . For equilibrium, g sin 0 = a cos 0 or tan 0 = a2 + g 2 0 g a g a Using the first two terms in a Taylor series expansion for sin and cos gives f ( ) sin cos tan 0 = f ( 0 ) + f ( ) = (  0 ) 0 sin 0 + (  0 ) cos 0 cos 0  (  0 ) sin 0 a implies sin 0 = g cos 0 = a a + g2 2 , g a2 + g 2 196 Thus sin cos 1 CHAPTER 7 a + g2 2 ( a + g  g 0 ) ( g  a + a 0 ) 1 a2 + g 2 Substituting into the equation of motion gives 0 = + This reduces to g b a2 + g ( a + g  g 0 )  2 g 2 + a2 b a b a2 + g 2 ( g  a + a 0 ) + = g 2 + a2 b 0 The solution to this inhomogeneous differential equation is = 0 + A cos + B sin where (g = Thus 2 2 + a2 b1 2 ) 14 T= = (g 2 b1 2 2 + a2 ) 14 714. a b m x = b sin y= 1 2 at  b cos 2 x = b cos y = at + b sin T= 1 1 m x 2 + y 2 = m b 2 2 + a 2t 2 + 2abt sin 2 2 ( ) ( ) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 197 1 U = mgy = mg at 2  b cos 2 L = T U = Lagrange's equation for gives d mb 2 + mabt sin = mabt cos  mgb sin dt b 2 + ab sin + abt cos = abt cos  gb sin 1 1 m b 2 2 + a 2t 2 + 2 abt sin + mg b cos  at 2 2 2 ( ) + For small oscillations, sin a+ g sin = 0 b + a+ g =0. b Comparing with + 2 = 0 gives T= 2 = 2 b a+ g 715. k m b = unextended length of spring = variable length of spring T= U= 1 m 2 ( 2 + 2 2) 1 1 2 2 k (  b ) + mgy = k (  b )  mg cos 2 2 1 m 2 L = T U = ( 2 + 2 2)  1 (  b ) 2 + mg cos 2 Taking Lagrange's equations for and gives : d m = m 2  k (  b ) + mg cos dt 198 CHAPTER 7 : This reduces to d m 2 =  mg sin dt  2 + k (  b )  g cos = 0 m g sin = 0 + 2 + 716. x = a sin t b m For mass m: x = a sin t + b sin y =  b cos x = a cos t + b cos y = b sin Substitute into T= 1 m x2 + y2 2 U = mgy and the result is L = T U = 1 m a 2 2 cos 2 t + 2ab cos t cos + b 2 2 + mgb cos 2 ( ) ( ) Lagrange's equation for gives d mab cos t cos + mb 2 =  mabw cos t sin  mgb sin dt  ab 2 sin t cos  ab cos t sin + b 2 =  ab cos t sin  gb sin or ( ) + g a sin  2 sin t cos = 0 b b HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 199 717. y A h C q mg B Using q and (= t since (0) = 0), the x,y coordinates of the particle are expressed as x = h cos + q sin = h cos t + q ( t ) sin t y = h sin  q cos = h sin t  q ( t ) cos t from which x =  h sin t + q cos t + q sin t y = h cos t + q sin t  q cos t Therefore, the kinetic energy of the particle is T= = The potential energy is U = mgy = mg ( h sin t  q cos t ) Then, the Lagrangian for the particle is L= 1 1 1 mh 2 2 + mq2 2 + mq2  mgh sin t + mgq cos t  mh q 2 2 2 q  2 q = g cos t The complementary solution and the particular solution for (6) are written as (5) (4) 1 m x2 + y2 2 (2) (1) ( ) ) (3) 1 m h 2 2 + q2 2 + q2  mh q 2 ( Lagrange's equation for the coordinate is (6) qc ( t ) = A cos ( i t + ) g qP ( t ) =  2 cos t 2 so that the general solution is q ( t ) = A cos ( i t + )  Using the initial conditions, we have g 2 2 cos t (7) (8) 200 =0 q ( 0 ) =  i A sin = 0 q ( 0 ) = A cos  g 2 2 Therefore, CHAPTER 7 (9) = 0, A = and q (t) = or, q (t) = g 2 2 q(t) g 2 2 (10) g 2 2 ( cos it  cos t ) ( cosh t  cos t ) (11) (12) g 2 2 t In order to compute the Hamiltonian, we first find the canonical momentum of q. This is obtained by p= Therefore, the Hamiltonian becomes L = mq  m h q (13) H = pq  L = mq2  m hq  so that H= 1 2 1 1 mq  m 2 h 2  m 2 q2 + mgh sin t  mgq cos t 2 2 2 (14) 1 1 1 m 2 h 2  m 2 q2  mq2 + mgh sin t  mgq cos t + m qh 2 2 2 Solving (13) for q and substituting gives p2 1 H= + hp  m 2 q2 + mgh sin t  mgq cos t 2m 2 (15) The Hamiltonian is therefore different from the total energy, T + U. The energy is not conserved in this problem since the Hamiltonian contains time explicitly. (The particle gains energy from the gravitational field.) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 201 718. y S R O A B x y m C x From the figure, we have the following relation: AC =  s =  R where is the generalized coordinate. In terms of , the x,y coordinates of the mass are (1) x = AC cos + R sin = (  R ) cos + R sin y = R cos  AC sin = R cos  (  R ) sin from which x = R sin  sin y = R cos  cos Therefore, the kinetic energy becomes T= The potential energy is U = mgy = mg R cos  (  R ) sin Then, the Lagrangian is L = T U = 1 m 2 2 + R 2 2 2  2R 2  mg R cos  (  R ) sin 2 1 1 m x 2 + y 2 = m 2 2 + R 2 2 2  2R 2 2 2 (2) (3) ( ) (4) (5) (6) Lagrange's equation for is (  R )  R 2  g cos = 0 Now let us expand about some angle 0 , and assume the deviations are small. Defining  0 , we obtain (7) + g sin 0 g cos 0 =  R 0  R 0 cos 0 sin 0 (8) The solution to this differential equation is = A sin ( t + ) + where A and are constants of integration and (9) 202 CHAPTER 7 g sin 0  R 0 (10) is the frequency of small oscillations. It is clear from (9) that extends equally about 0 when 0 = 2 . 719. P d m1 m1g m2 m2 g Because of the various constraints, only one generalized coordinate is needed to describe the system. We will use , the angle between a plane through P perpendicular to the direction of the gravitational force vector, and one of the extensionless strings, e.g., 2 , as our generalized coordinate. The, the kinetic energy of the system is T= The potential energy is given by U =  m1 g 1 1 m1 2 ( ) 1 2 + 1 m2 2 ( ) 2 2 (1) sin (  ( + ) )  m2 g 2 sin (2) from which the Lagrangian has the form L = T U = 1 m1 2 ( 2 1 + m2 2 2 ) 2 + m1 g 1 sin ( + ) + m2 g 2 sin (3) The Lagrangian equation for is m2 g 2 cos + m1 g 1 cos ( + )  m1 ( 2 1 + m2 2 2 ) = 0 (4) This is the equation which describes the motion in the plane m1 , m2 , P . To find the frequency of small oscillations around the equilibrium position (defined by = 0 ), we expand the potential energy U about 0 : U ( ) = U ( 0 ) + U ( 0 ) + = 1 U (0 ) 2 2 1 U (0 ) 2 + ... 2 (5) where the last equality follows because we can take U (0 ) = 0 and because U (0 ) = 0 . From (4) and (5), the frequency of small oscillations around the equilibrium position is HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 203 2 = The condition U (0 ) = 0 gives tan 0 = or, sin 0 = U (0 ) m1 2 1 + m2 2 2 (6) m2 + m1 1 cos m1 1 sin 2 (7) (m m2 2 + m2 2 2 2 + m1 1 cos 1 2 2 2 1 1 + 2m1m2 cos ) 12 (8) Then from (2), (7), and (8), U (0 ) is found to be U (0 ) = g sin 0 ( m2 = 2 + m1 + m1 1 cos + m1 cos ) 2 1 1 sin cot 0 ) m2 2 + m1 1 cos + 2 m1 2 sin 2 1 m2 2 + m1 1 cos (m ( g ( m2 2 + m2 2 1 2 2 2 1 2 2 1 1 + 2m2 m1 2 2 cos ) 12 2 = g m1 2 + m2 + 2m2 m1 2 1 cos ) 12 (9) Finally, from (6) and (9), we have which, using the relation, 2 (m =g 2 2 1 1 2 + m2 (m 2 2 + 2m1m2 + m2 1 2 2 2 2 1 1 ) cos ) 12 (10) cos = can be written as 2 1 + 2 2 2  d2 (11) 1 2 = 2 g ( m1 + m2 ) m1 ( (m 2 1 + m2 + m2 2 2 2 2 2 1 1 )d m m ) 2 1 2 12 (12) Notice that 2 degenerates to the value g (so that 1 = 2 ). appropriate for a simple pendulum when d 0 204 CHAPTER 7 720. The xy 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 A C B y x R A C B The kinetic energy of the system is given by T= 1 2 1 MR2 2 1 z 2 + M I + Mz 2 = 2 2 2 2 =0 2 (1) For small , the second term can be neglected since ( z ) The potential energy is given by U = Mgz =0 (2) where we take U = 0 at z = . Since the system has only one degree of freedom we can write z in terms of . When = 0, z =  . When the hoop is rotated thorough an angle , then z2 = so that 2  ( R  R cos )  ( R sin ) 2 2 (3) z =  and the potential energy is given by 2 + 2R 2 ( cos  1) 12 (4) U =  Mg for small , cos  1  2 2 ; then, 2 + 2R2 ( cos  1) 12 (5) R 2 2 U  Mg 1  2 R 2 2  Mg 1  2 2 From (1) and (6), the Lagrangian is L = T U = 12 (6) R 2 2 1 MR 2 2 + Mg 1  , 2 2 2 (7) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 205 for small . The Lagrange equation for gives + where g =0 (8) = g (9) which is the frequency of small rotational oscillations about the vertical through the center of the hoop and is the same as that for a simple pendulum of length . 721. From the figure, we can easily write down the Lagrangian for this system. T= mR 2 2 + 2 sin 2 2 U =  mgR cos ( ) (1) (2) The resulting equation of motion for is  2 sin cos + g sin = 0 R (3) The equilibrium positions are found by finding the values of for which 0 = = 0 g = 2 cos 0  sin 0 R (4) Note first that 0 and are equilibrium, and a third is defined by the condition cos 0 = g 2R (5) To investigate the stability of each of these, expand using =  0 = 2 cos 0  g  sin 0 ( sin 0 + cos 0 ) 2R (6) 206 For 0 = , we have CHAPTER 7 = 2 1 + indicating that it is unstable. For 0 = 0 , we have g 2R (7) = 2 1  g 2R (8) which is stable if 2 < g R and unstable if 2 > g R . When stable, the frequency of small oscillations is 2  g R . For the final candidate, =  2 sin 2 0 (9) with a frequency of oscillations of 2  ( g R) , when it exists. Defining a critical frequency 2 2 c2 g R , we have a stable equilibrium at 0 = 0 when < c , and a stable equilibrium at 0 = cos 1 ( c2 2 ) when c . The frequencies of small oscillations are then 1  ( c ) and 1  ( c ) , respectively. 4 To construct the phase diagram, we need the Hamiltonian H L L (10) which is not the total energy in this case. A convenient parameter that describes the trajectory for a particular value of H is 1 K = 2 2 m c R 2 so that we'll end up plotting H 2 2  sin 2  cos c c (11) 2 = 2 ( K + cos ) + sin c c 2 2 (12) for a particular value of and for various values of K. The results for < c are shown in figure (b), and those for > c are shown in figure (c). Note how the origin turns from an attractor into a separatrix as increases through c . As such, the system could exhibit chaotic behavior in the presence of damping. HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 207 2 1.5 K 1 0.5 0 3 2 1 0 1 2 3 (b) 2 1.5 K 1 0.5 0 3 2 1 0 1 2 3 (c) 722. The potential energy U which gives the force F ( x, t) = k ( t ) e x2 U x (1) must satisfy the relation F= we find U= k t e x (3) (2) 208 CHAPTER 7 Therefore, the Lagrangian is L = T U = The Hamiltonian is given by H = px x  L = x so that H= 2 px k  t + e 2m x 1 k mx 2  e  t x 2 L L x (4) (5) (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. 723. The Hamiltonian function can be written as [see Eq. (7.153)] H = p j qj  L j (1) For a particle which moves freely in a conservative field with potential U, the Lagrangian in rectangular coordinates is L= 1 m x2 + y2 + z2  U 2 ( ) and the linear momentum components in rectangular coordinates are px = L = mx x py = my pz = mz (2) 1 H = mx 2 + my 2 + mz 2  m x 2 + y 2 + z 2  U 2 ( ) = 1 1 2 2 2 m x2 + y2 + z2 + U = px + py + pz 2 2m ( ) ( ) (3) which is just the total energy of the particle. The canonical equations are [from Eqs. (7.160) and (7.161)] HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 209 px = mx =  py = my =  pz = mz =  These are simply Newton's equations. 724. U = Fx x U = Fy y U = Fz z (4) m The kinetic energy and the potential energy of the system are expressed as T= 1 m 2 ( 2 + 2 2 = ) U =  mg cos so that the Lagrangian is L = T U = The Hamiltonian is H = p  L = = 1 m 2 + 2 2 2 ( ) (1) 1 m 2 + 2 2 + mg cos 2 L L ( ) (2) 2 p 1  m 2  mg cos 2 2m 2 (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 d (T + U ) 0 dt (4) 210 725. z z r CHAPTER 7 m y x In cylindrical coordinates the kinetic energy and the potential energy of the spiraling particle are expressed by 1 m r 2 + r 2 2 + z 2 2 U = mgz T= Therefore, if we use the relations, i.e., z = k r = const. z = k the Lagrangian becomes L= Then the canonical momentum is pz = or, z= pz r2 m 2 + 1 k (5) r2 L = m 2 + 1 z z k (4) 1 r2 2 m 2 z + z 2  mgz 2 k (3) (2) (1) The Hamiltonian is H = pz z  L = p z pz r2 m 2 + 1 k  2 pz r2 2m 2 + 1 k + mgz (6) or, H= 1 2 2 pz r2 m 2 + 1 k + mgz (7) Now, Hamilton's equations of motion are HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 211  so that H = pz ; z H =z p z (8)  H =  mg = pz z pz r2 m 2 + 1 k =z (9) (10) H = p z Taking the time derivative of (10) and substituting (9) into that equation, we find the equation of motion of the particle: z= g r k 2 + 1 2 (11) It can be easily shown that Lagrange's equation, computed from (3), gives the same result as (11). 726. a) m L = T U = L= 1 2 2 m  mgy 2 1 2 2 m + mg cos 2 L = m 2 p = so = p m 2 Since U is velocityindependent and the coordinate transformations are timeindependent, the Hamiltonian is the total energy H = T +U = The equations of motion are 2 p  mg cos 2m 2 212 p H H = 2 and p =  =  mg sin p m CHAPTER 7 = b) a x m2 m1 T= 1 1 1 x2 m1 x 2 + m2 x 2 + I 2 2 2 2 a where I = moment of inertia of the pulley U =  m1 gx  m2 g (  x ) px = So x= px I m1 + m2 + a 2 H=T+U H= 2 px L T I = = m1 + m2 + 2 x x x a I 2 m1 + m2 + 2 a  m1 gx  m2 g (  x ) The equations of motion are x= px H = I p x 2 m1 + m2 + 2 a H = m1 g  m2 g = g ( m1  m2 ) p x px =  HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 213 727. a) k,b m1 m2 xi , yi = coordinates of mi Using , as polar coordinates x2 = x1 + cos y2 = y1 + sin x2 = x1 + cos  sin y2 = y1 + sin + cos If we substitute (1) and (2) into L = T U = the result is L= 1 1 ( m1 + m2 ) x12 + y12 + m2 2 2 1 1 1 2 2 2 2 2 m1 x1 + y1 + m2 x2 + y2  k (  b ) 2 2 2 (1) (2) ( ) ( ) ( ) ( 2 + 2 2 ) 1 2 + m2 ( x1 cos + y1 sin ) + m2 ( y1 cos  x1 sin )  k (  b ) 2 The equations of motion are x1 : d ( m1 + m2 ) x1 + m2 cos  m2 sin = 0 dt = m1 x1 + m2 x2 = px So px = constant y1 : d ( m1 + m2 ) y1 + m2 sin + m2 cos = 0 dt = m1 y1 + m2 y2 = py So py = constant : d m2 + m2 ( x1 cos + y1 sin ) = m2 2  k (  b ) + m2 ( y1 cos  x1 sin ) dt which reduces to 214 CHAPTER 7  2 + x1 cos + y1 sin + k (  b) = 0 m2 : d m2 2 + m2 ( y1 cos  x1 sin ) dt =  m2 ( x1 sin  y1 cos ) + m2 (  x1 cos  y1 sin ) which reduces to + b) 2 + cos y1  sin x1 = 0 As was shown in (a) L = px = constant x1 L = py = constant y1 (total linear momentum) c) Using L from part (a) px1 = py1 = p = L = ( m1 + m2 ) x1 + m2 cos  m2 sin x1 L = ( m1 + m2 ) y1 + m2 sin  m2 cos y1 L = m2 x1 cos + m2 y1 sin + m2 p =  m2 x1 sin + m2 y1 cos + m2 2 Inverting these equations gives (after much algebra) x1 = y1 = = sin 1 p px1  p cos + m1 cos 1 p py1  p sin  m1 1 m1 + m2 p  px1 cos  py1 sin + m1 m2 1 m1 + m2 p px1 sin  py1 cos + m1 m2 H = T +U = Since the coordinate transformations are time independent, and U is velocity independent, Substituting using the above equations for qi in terms of pi gives HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 215 H= m1 + m2 1 2 2 px1 + py1 + m2 2m1 2 2 p p + 2  2 p px1 cos + py1 sin p 1 2 +2 px1 sin  py1 cos + k (  b ) 2 ( ) ( ) The equations of motion are x1 = y1 = = sin 1 H = p px1  p cos + px1 m1 cos 1 H = p py1  p sin  py1 m1 1 m1 + m2 H p  px1 cos  py1 sin = m1 m2 p 1 m1 + m2 H p + px1 sin  py1 cos = p m1 m2 H =0 x1 py1 =  H =0 y1 = px1 =  2 p H ( m1 + m2 ) p p = = + 2 px1 sin  py1 cos  k (  b ) 3 m1m2 m1 ( ) p =  p H p =  px1 sin + py1 cos  px1 cos + py1 sin m1 m1 ( ) 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 coordinates that lead to three cyclic coordinates ( xCM , yCM , and ), as shown in those sections. F =  kr 2 U =  kr 1 L = T U = pr = p = 1 k m r 2 + r 2 2 + 2 r 728. so ( ) p L = mr so r = r m r so = p mr 2 L = mr 2 Since the coordinate transformations are independent of t, and the potential energy is velocityindependent, the Hamiltonian is the total energy. 216 1 k m r 2 + r 2 2  2 r CHAPTER 7 H = T +U = = ( ) 2 p2 k 1 pr m 2 + r 2 2 4  mr r 2 m 2 pr p2 k + 2 r 2m 2mr H= Hamilton's equations of motion are r= H p r = p r m p2 k H = 3  2 r r mr H =0 = p H = p mr 2 pr =  p =  729. a k m b = unextended length of spring = variable length of spring a) x = sin y= 1 2 at  cos 2 1 m x2 + y2 2 x = sin + cos y = at  cos + sin Substituting into T = ( ) U = mgy + gives L = T U = 1 m 2 2 1 2 k (  b) 2 at 2 k 2 + 2 2 + a 2t 2 + 2at sin  cos + mg cos   2 (  b) 2 ( ) Lagrange's equations give: HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 217 : d m  amt cos = m 2 + mat sin + mg cos  k (  b ) dt d m 2 + mat sin = mat sin  mg sin + mat cos dt : Upon simplifying, the equations of motion reduce to:  2  ( a + g ) cos + k (  b) = 0 m sin = 0 = p + at cos m (1) + b) 2 + a+ g p = p = or L = m  mat cos or L = m 2 + mat sin = p at sin  2 m (2) Since the transformation equations relating the generalized coordinates to rectangular coordinates are not timeindependent, the Hamiltonian is not the total energy. H = pi qi  L = p + p  L Substituting (1) and (2) for H= and and simplifying gives p2 p2 1 1 at 2 + 2  p sin + atp cos + k (  b ) + mgat 2  mg cos 2 m 2m 2 2 are p at H = 2  sin p m H p = + at cos m p agreeing with (1) and (2) The equations for and = = The equations for p and p are 2 p H at p = =  2 p sin  k (  b ) + mg cos + m 3 or p + p =  at 2 p sin + k (  b )  mg cos + 2 p =0 m 3 at H =  p cos + at p sin  mg sin 218 CHAPTER 7 or p  c) at p cos  at p sin + mg sin = 0 sin , cos 1  2 2 2 k  2  ( a + g ) 1  + (  b ) = 0 2 m Substitute into Lagrange's equations of motion + For small oscillations, 1, 1, 2 + a+ g  at =0 1 . Dropping all secondorder terms gives + k k = a+ g+ b m m + For , a+ g =0 T = The solution to the equation for is 2 = 2 a+ g = homogeneous + particular = A cos So for , k k m t + B sin t + (a + g) + b m m k T = 2 = 2 m k 730. a) From the definition of a total derivative, we can write g qk g pk dg g = + + dt t p k t k qk t Using the canonical equations (1) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 219 qk H = qk = t p k pk H = pk =  t qk (2) we can write (1) as g H g H dg g = +  dt t p k qk k qk pk (3) or dg g = + g , H dt t b) (4) qj = q j t = H p j (5) According to the definition of the Poisson brackets, q H g j H qj , H = j  pk qk k qk p k (6) but q j qk = jk and q j pk = 0 for any j,k (7) then (6) can be expressed as H q j , H = p = q j j In the same way, from the canonical equations, pj =  so that p H p j H pj , H = j  p k qk k qk pk but p j pk then, = jk and p j qk = 0 for any j,k (11) (10) H q j (9) (8) 220 H = pj , H q j CHAPTER 7 pj =  (12) c) p p j p k p j pk , p j = k  p q q p pk = 0 for any k, q (13) since, (14) the righthand side of (13) vanishes, and pk , p j = 0 In the same way, q q j qk q j qk , q j = k  q p p q since q j p the righthand side of (16) vanishes and qk , q j = 0 d) (15) (16) = 0 for any j, (17) (18) q p j qk p j qk , p j = k  q p p q = k j or, qk , p j = kj e) Let g ( pk , qk ) be a quantity that does not depend explicitly on the time. If g ( pk , qk ) commutes with the Hamiltonian, i.e., if (19) (20) g , H = 0 then, according to the result in a) above, dg =0 dt and g is a constant of motion. (21) (22) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 221 731. A spherical pendulum can be described in terms of the motion of a point mass m on the surface of a sphere of radius , where corresponds to the length of the pendulum support rod. The coordinates are as indicated below. z mg x y The kinetic energy of the pendulum is T= and the potential energy is U = mg cos The Lagrangian is L= so that the momenta are p = p = The Hamiltonian then becomes H = p + p  1 m 2 2 1 1 1 I1 2 + I 2 2 = m 2 2 2 2 ( 2 sin 2 + 2 ) (1) (2) 1 m 2 2 ( 2 sin 2 + 2  mg cos ) (3) L = m 2 (4) (5) L = m 2 sin 2 ( 2 sin 2 + 2 + mg cos (6) ) p2 = 2 + V , p 2m ( ) which is just the total energy of the system and where the effective potential is 2 p V , p = + mg cos 2m 2 sin 2 ( ) (7) When p = 0 , V(,0) is finite for all , with a maximum at = 0 (top of the sphere) and a minimum at = (bottom of the sphere); this is just the case of the ordinary pendulum. For different values of p , the V diagram has the appearance below: 222 CHAPTER 7 V 0 2 P = 0 When p > 0 , the pendulum never reaches = 0 or = because V is infinite at these points. The V 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 p , is a constant, and we have the case of a conical pendulum. For further details, see J. C. Slater and N. H. Frank, Mechanics, McGrawHill, New York, 1947, pp. 7986. 732. The Lagrangian for this case is L = T U = 1 k m r 2 + r 2 2 + r 2 sin 2 2 + 2 r ( ) (1) where spherical coordinates have been used due to the symmetry of U. The generalized coordinates are r, , and , and the generalized momenta are pr = p = p = L = mr r L = mr 2 L = mr 2 sin 2 (2) (3) (4) The Hamiltonian can be constructed as in Eq. (7.155): H = pr r + p + p  L = = 1 k m r 2 + r 2 2 + r 2 2 sin 2  2 r 2 2 2 k p p 1 pr + +  2 2 2 mr sin r 2 m mr ( ) (5) Eqs. (7.160) applied to H as given in (5) reproduce equations (2), (3), and (4). The canonical equations of motion are obtained applying Eq. (7.161) to H: pr =  p p2 H k =  2 + 3 + 3 r r mr mr sin 2 2 (6) HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 223 2 p cot H p =  = mr 2 sin 2 (7) (8) p =  H =0 The last equation implies that p = const , which reduces the number of variables on which H depends to four: r , , pr , p : H= 1 2m 2 2 p const k pr + 2 + 2  r r sin 2 r (9) For motion with constant energy, (9) fixes the value of any of the four variables when the other three are given. From (9), for a given constant value of H = E, we obtain p 2 sin 2 + const 2mk + pr = 2 mE  2 r sin 2 r 12 (10) and so the projection of the phase path on the r  pr plane are as shown below. pr 2mE 3 2 1 1 < 2 < 3 733. x m1 x m2 m3 Neglect the masses of the pulleys T= 1 1 1 2 2 m1 x 2 + m2 ( x  x ) + m3 (  x  x ) 2 2 2 U =  m1 gx  m2 g (  x + x )  m3 g (  x +  x ) 224 1 1 ( m1 + m2 + m3 ) x 2 + 2 ( m2 + m3 ) x 2 + xx ( m3  m2 ) 2 + g ( m1  m2  m3 ) x + g ( m2  m3 ) x + constant L = ( m1 + m2 + m3 ) x + ( m3  m2 ) x x L = ( m3  m2 ) x + ( m2 + m3 ) x x CHAPTER 7 L= We redefine the zero in U such that the constant in L = 0. px = px = (1) (2) Solving (1) and (2) for px and px gives x = D1 ( m2 + m3 ) px + ( m2  m3 ) px x = D1 ( m2 + m3 ) px + ( m1 + m2 + m3 ) px where D = m1m3 + m1m2 + 4 m2 m3 H =T +U = 1 1 ( m1 + m2 + m3 ) x 2 + 2 ( m2 + m3 ) x 2 + ( m3  m2 ) xx 2  g ( m1  m2  m3 ) x  g ( m2  m3 ) x Substituting for x and x and simplifying gives H= 1 1 ( m2 + m3 ) D1 px2 + 2 ( m1 + m2 + m3 ) D1 px2 2 + ( m2  m3 ) D1 px px  g ( m1  m2 + m3 ) x  g ( m2  m3 ) x where D = m1m3 + m1 m2 + 4 m2 m3 The equations of motion are x= x = H = ( m2 + m3 ) D1 px + ( m2  m3 ) D1 px p x H = ( m2  m3 ) D1 px + ( m1 + m2 + m3 ) D1 px p x H = g ( m1  m2  m3 ) x H = g ( m2  m3 ) x px =  px =  HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 225 734. x r R m M The coordinates of the wedge and the particle are xM = x yM = 0 The Lagrangian is then L= M+m 2 m 2 x + r + r 2 2 + 2 xr cos  2 xr sin + mgr sin 2 r xm = r cos + x ym =  r sin (1) ( ) (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 f ( x , , r ) = r  R = 0 . a) Right now, however, we may take r = R and r = r = 0 to get the equations of motion for x and . Using Lagrange's equations, x = aR sin + 2 cos ( ) (3) (4) = where a m ( M + m) . x sin + g cos R b) We can get the reaction of the wedge from the Lagrange equation for r = mx cos  mR 2  mg sin (5) We can use equations (3) and (4) to express x in terms of and , and substitute the resulting expression into (5) to obtain = a1 2 R + g sin 2 1  a sin ( ) (6) To get an expression for , let us use the conservation of energy H= M+m 2 m 2 2 x + R  2xR sin  mgR sin =  mgR sin 0 2 2 ( ) (7) where 0 is defined by the initial position of the particle, and  mgR sin 0 is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain x = aR sin , and substitute this into the energy equation to obtain an expression for 2 = 2 g ( sin  sin 0 ) R 1  a sin 2 ( ) (8) 226 CHAPTER 7 Finally, we can solve for the reaction in terms of only and 0 = mMg 3 sin  a sin 3  2 sin 0 ( ( M + m) ( 1  a sin ) 2 ) 2 (9) 735. We use zi and pi 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 z i = z i 0 + vi 0 t  1 2 gt 2 pi = pi 0  mgt (1) where zi 0 , pi 0 , and vi 0 = pi 0 m are the initial displacement, momentum, and velocity of the ith particle. Using the initial conditions given, we have z1 = z0 + p0 t 1 2  gt m 2 (2a) (2b) p1 = p0  mgt z2 = z0 + z0 + p2 = p0  mgt z 3 = z0 + p0 t 1 2  gt m 2 (2c) (2d) (2e) (2f) (2g) (2h) ( p0 + p0 ) t  1 gt 2 m 2 p3 = p0 + p0  mgt z4 = z0 + Z0 + ( p0 + p0 ) t  1 gt 2 m 2 p4 = p0 + p0  mgt The Hamiltonian function corresponding to the ith particle is H i = Ti + Vi = p2 mzi2 + mgzi = i + mgzi = const. 2 2m (3) From (3) the phase space diagram for any of the four particles is a parabola as shown below. HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 227 p0 + p0 p0 Area at t = 0 1 p 2 3 4 Area at t = t1 p0 z0 z0 z0 + z0 z From this diagram (as well as from 2b, 2d, 2f, and 2h) it can be seen that for any time t, p1 = p2 p3 = p 4 (4) (5) Then for a certain time t the shape of the area described by the representative points will be of the general form p (p3,z3) 3 (p4,z4) 4 1 (p1,z1) 2 (p2,z2) z where the base 1 2 must parallel to the top 3 4 . At time t = 0 the area is given by z0 p0 , since it corresponds to a rectangle of base z0 and height p0 . At any other time the area will be given by A = base of parallelogram t = t = ( z2  z1 ) t = t 1 { 1 = ( z 4  z 3 ) t = t = z 0 1 } x height of parallelogram t = t = ( p3  p1 ) 1 { t = t1 = ( p 4  p2 ) =p0 z0 Thus, the area occupied in the phase plane is constant in time. 736. t = t1 = p0 } (6) The initial volume of phase space accessible to the beam is 2 2 V0 = R0 p0 (1) After focusing, the volume in phase space is 2 2 V1 = R1 p1 (2) 228 CHAPTER 7 where now p1 is the resulting radius of the distribution of transverse momentum components of the beam with a circular cross section of radius R1 . From Liouville's theorem the phase space accessible to the ensemble is invariant; hence, 2 2 2 2 V0 = R0 p0 = V1 = R1 p1 (3) from which p1 = R0 p0 R1 (4) If R1 < R0 , then p1 > p0 , which means that the resulting spread in the momentum distribution has increased. This result means that when the beam is better focused, the transverse momentum components are increased and there is a subsequent divergence of the beam past the point of focus. 737. Let's choose the coordinate system as shown: x1 x2 x3 m1 m3 m2 The Lagrangian of the system is 2 2 2 1 dx dx dx L = T  U = m1 1 + m2 2 + m3 3 + g ( m1 x1 + m2 x2 + m3 x3 ) dt dt 2 dt with the constraints x1 + y = l1 and x2  y + x3  y = l2 (1) which imply 2x1 + x2 + x3  (2l1 + l2 ) = 0 2 d 2 x1 d 2 x2 d 2 x3 + 2 + 2 =0 dt 2 dt dt The motion equations (with Lagrange multiplier ) are HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 229 m1 g  m1 m2 g  m2 m3 g  m3 Combining (1)(4) we find d 2 x1 + 2 = 0 dt 2 d 2 x2 + =0 dt 2 d 2 x3 + =0 dt 2 (2) (3) (4) = 4 g 4 1 1 + + m1 m2 m3 Finally, the string tension that acts on m1 is (see Eq. (2)) T1 = m1 g  m1 8g d 2 x1 = 2 = 2 4 1 1 dt + + m1 m2 m3 738. The Hamiltonian of the system is H =T +U = kx 2 bx 4 p 2 kx 2 bx 4 1 dx m + + = + + 2 dt 2 4 2m 2 4 dx H p = = dt p m dp H = = (kx + bx 3 ) dt x 2 The Hamiltonian motion equations that follow this Hamiltonian are 739. z The Lagrangian of the rope is mgz 2 1 dz mz z 1 dz L = T  U = m   g = m + 2 dt b 2 2 dt 2b 2 2 230 CHAPTER 7 from which follows the equation of motion mgz d2 z L d L = =m 2 z dt z b dt 740. 1 m 2 m 2m x We choose the coordinates for the system as shown in the figure. The kinetic energy is 2 2 2 d 1 dx 1 1 2 d 1 dx dx T = 2m + m b cos 1 + + 2b dt dt dt 2 2 dt dt 2 2 d 1 d 2 d 2 1 dx d 1 cos 1 + b cos 2 + b sin 1 + b sin 2 + m +b dt dt dt dt 2 dt The potential energy is U =  mgb cos 1  mg(b cos 1 + b cos 2 ) And the Lagrangian is 1 dx d 1 dx d d cos 1 + mb 2 2 L = T  U = 2m + mb 2 1 + 2mb dt dt dt 2 dt dt 2 2 2 + mb dx d 2 d d cos 2 + mb 2 1 2 cos ( 1  2 ) + 2mgb cos 1 + mgb cos 2 dt dt dt dt d 2 d2 x d 2 + b 2 21 cos 1 + 22 cos 2 2 dt dt dt From this follow 3 equations of motion L d L = x dt x 0=4 2 d 2 d  b 2 1 sin 1 + 2 sin 2 dt dt HAMILTON'S PRINCIPLELAGRANGIAN AND HAMILTONIAN DYNAMICS 231 L d L = 1 dt 1  2 g sin 1 = 2b 2 d 2 1 d 2 d2 x + 2 2 cos 1 + b 22 cos( 1  2 ) dt 2 dt dt d + b 2 sin( 1  2 ) dt L d L = 2 dt 2  g sin 2 = b d 2 2 d 2 x d 2 d + 2 cos 2 + b 21 cos ( 1  2 )  b 1 sin ( 1  2 ) 2 dt dt dt dt 2 741. For small angle of oscillation we have T= So the Lagrangian reads 1 2 d 1 db mb + m dt 2 2 dt 2 2 and U =  mgb cos L = T U = 1 2 d 1 db mb + m + mgb cos dt 2 2 dt 2 2 from which follow 2 equations of motion L d L = b dt b L d L = dt d2b d d b + g cos = 2 =  dt dt dt 2  mgb sin = 2mb db d d 2 d d 2 + mb 2 2 = 2mb + mb 2 2 dt dt dt dt dt 232 CHAPTER 7 CHAPTER 8 CentralForce Motion 81. x3 m1 r1 r2 x2 m2 x1 In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x1 direction; then the masses m1 and m2 have the gravitational potential energies: ( ( ( U g1) =  F (1) x11) =  m1 x11) ( 2) ( 2) ( 2 ) ( 2) U g =  F x1 =  m2 x1 (1) ( ( ( where r1 = x11) , x21) , x31) and where is the constant gravitational acceleration. Therefore, ( ) introducting the relative coordinate r and the center of mass coordinate R according to r = r1  r2 m1r1 + m2 r2 = ( m1 + m2 ) R (2) we can express r1 and r2 in terms of r and R by m1 r2 =  r+R m1 + m2 r1 = m2 r+R m1 + m2 (3) 233 234 CHAPTER 8 which differ from Eq. (8.3) in the text by R. The Lagrangian of the twoparticle system can now be expressed in terms of r and R: L= = 1 1 2 2 ( ( m1 r1 + m2 r2  U ( r )  U g1)  U g2) 2 2 m2 m1 1 1 r + R + m2  r + R  U (r) m1 2 2 m1 + m2 m1 + m2 m2 m1 x + X + m2  x + X + m1 m1 + m2 m1 + m2 2 2 (4) where x and X are the x1 components of r and R, respectively. Then, (4) becomes 2 m2 2 1 m1 2 1 1 L = m1 r + m2 r + ( m1 + m2 ) R 2 2 2 m1 + m2 m1 + m2 m1m2  m1m2 x + ( m1 + m2 ) X  U (r) + m1 + m2 2 (5) Hence, we can write the Lagrangian in the form L= 2 1 1 2 r  U ( r ) + ( m1 + m2 ) R + ( m1 + m2 ) X 2 2 (6) where is the reduced mass: = m1m2 m1 + m2 (7) Therefore, this case is reducible to an equivalent onebody problem. 82. Setting u = 1 r , Eq. (8.38) can be rewritten as =  2E 2 + du 2 k 2 (1) uu 2 where we have used the relation du =  1 r 2 dr . Using the standard form of the integral [see Eq. (E.8c), Appendix E]: 2ax + b 1 sin 1 + const. 2 a b  4 ac ( ) we have dx ax 2 + bx + c = (2) 2 2 k  + 2 1 r + const. = sin 2 2 k + 8 E 2 2 (3) CENTRALFORCE MOTION 235 or, equivalently, sin ( + const.) = 2 2 k  + 2 r 2 E 2 k 2 +8 2 (4) We can choose the point from which is measured so that the constant in (4) is  2 . Then, cos = 1 1 k r 2E 2 1+ k 2 2 (5) which is the desired expression. 83. When k k 2 , the potential energy will decrease to half its former value; but the kinetic energy will remain the same. Since the original orbit is circular, the instantaneous values of T and U are equal to the average values, T and U . For a 1 r 2 force, the virial theorem states T = Hence, 1 U 2 (1) 1 1 E = T +U =  U +U = U 2 2 Now, consider the energy diagram E D (2) C r k/r B A where CB = E CA = U CD = UC original total energy original potential energy original centrifugal energy The point B is obtained from CB = CA  CD . According to the virial theorem, E = (1 2) U or CB = (1 2) 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 (1 2) CA or by CB . Thus, the total energy point is raised from B to C; i.e., E(final) = 0 and the orbit is parabolic. 84. Since the particle moves in a central, inversesquare law force field, the potential energy is U= so that the time average is U = k r (1) 0 1 k dt r (2) Since this motion is a central motion, the angular momentum is a constant of motion. Then, r 2 = const. from which dt = Therefore, (2) becomes U = 1 2 (3) r 2 d (4) 0 k r 2 k d =  r 2 rd 0 (5) Now, substituting r = 1 + cos and = ( 2 U = k 2 a 3 2 ) 2 0 a 3 2 , (5) becomes 1 (6) 1 + cos d where a is the semimajor axis of the ellipse. Using the standard integral [see Eq. (E.15), Appendix E], 2 1 + cos d = 0 1 2 1 2 (7) and the relation, = a (1  2 ) (6) becomes U = The kinetic energy is k a (8) (9) CENTRALFORCE MOTION 237 2 1 T =  r 2 + 2 2 r 2 (10) and the time average is T = Part of this integral is trivial, 1 T dt = 0 1 2 k a3 2 Tr 0 2 d (11) T = 1 2 2 k 2 ( rr ) 2 d + a3 2 0 (12) To evaluate the integral above, substitute the expression for r and make a change of variable 1 2 2 ( rr ) 0 2 1 d = 2 2 2 (1 + cos ) 0 sin 2 d 2 = 2 1 1 (1 + x ) 1  x 2 dx 2 (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: 1 (1 + x ) 1 1  x 2 dx 2 1  x2 = (1 + x ) 1  1 1 1 x dx 1  x 2 (1 + x ) (14) 1 1 x sin 1 x = 1 + x 1 1  sin 1 x dx 2 1 ( 1 + x ) 1 (15) 1 1 2 tan 1 =  2 sin 1 x + 2 1 (1  )(1  x ) (1 + )(1 + x ) 1 (16) = 1 1 + 2 1  2 k 2a (17) Substituting this into (13) and then into (12), we obtain the desired answer, T = (18) This explicitly verifies the virial theorem, which states that for an inversesquare law force, T = 1 U 2 (19) 238 85. a m1 m2 CHAPTER 8 Suppose two particles with masses m1 and m2 move around one another in a circular orbit with radius a. We can consider this motion as the motion of one particle with the reduced mass moving under the influence of a central force G m1m2 a 2 . Therefore, the equation of motion before the particles are stopped is a 2 = G where 1 m1m2 a2 2 (1) The radius of circular motion is = 1 1 + , m1 m2 = (2) G m1m2 2 a= 2 4 13 (3) After the circular motion is stopped, the particle with reduced mass starts to move toward the force center. We can find the equation of motion from the conservation of energy: G or, 2G m1m2 x= Therefore, the time elapsed before the collision is t = dt =  a m1m2 1 2 mm = x  G 1 2 a 2 x 12 (4) 1 1  x a (5) 0 dx 2G m1m2 1 1 x  a (6) where the negative sign is due to the fact that the time increases as the distance decreases. Rearranging the integrand, we can write t= a 2G m1m2 a 0 x dx ax (7) Setting x y 2 ( dx = 2 y dy ) , the integral in (7) becomes I= a 0 x dx = 2 ax a 0 y2 a  y2 dy CENTRALFORCE MOTION 239 Using Eq. (E.7), Appendix E, we find y a  y2 a y + sin 1 I = 2  2 2 a Therefore, t= or, t= a a 2G m1m2 2 0 = a a 2 (8) 4 2 (9) 86. m1 x1 O r r = x2  x1 m2 x2 When two particles are initially at rest separated by a distance r0 , the system has the total energy E0 = G m1m2 r0 (1) The coordinates of the particles, x1 and x2 , are measured from the position of the center of mass. At any time the total energy is E= 1 1 mm 2 2 m1 x1 + m2 x2  G 1 2 2 2 r p = m1 x1 + m2 x2 = 0 From the conservation of energy we have E = E0 , or G Using (3) in (4), we find x1 = v1 = m2 2G 1 1  M r r0 (5) x1 = v2 =  m1 2G M 1 1  r r0 m1m2 1 1 mm 2 2 = m1 x1 + m2 x2  G 1 2 r0 2 2 r (4) (2) and the linear momentum, at any time, is (3) 240 CHAPTER 8 87. Since F ( r ) =  kr is a central force, angular momentum is conserved and the areal velocity, dA dt = 2 , is trivially constant (see Section 8.3). In order to compute U , we start with dt = dr 2 2 EU  2 r 2 (1) and U= kr 2 2 (2) The time average of the potential energy becomes U = 1 2 U dt 0 rmax rmin = kr 3 2 dr 2 2 2 kr 4 Er   2 2 (3) Substituting r2 = x (4) becomes dr = 1 dx 2r (4) U = k 2 2 2 rmax 2 rmin x dx k  + Ex  x 2 2 2 2 (5) Using the integrals in Eqs. (E.9) and (E.8c), Appendix E, (5) becomes k U = 2 x dx ax 2 + bx + c = 2ax + b b 1 ax 2 + bx + c + sin 1 2 a 2a  a b  4 ac (6) 12 E 2  2 k k 2 E  kr 2 2 k 4 2   r + Er  sin 2 12 2 2 E2  k k rmin 1 rmax (7) But rmax and rmin were originally defined as the roots of vanishes at both limits of integration. On the other hand, EU  2 2 r 2 . Hence, the second term CENTRALFORCE MOTION 241 rmax rmin =2 dr 2 2 EU  2 2 r rmax rmin = 2 rdr 2 k  r 4 + Er 2  2 2 (8) or, using (5), = 2 2 rmax dx  12 2 k 2 x + Ex  2 2 2 rmin 2 1 = sin 2 k 2 12 E2  k E  kr 2 rmax (9) rmin Using (9) to substitute for in (7), we have U = Now, T =E U = The virial theorem states: T = In our case n = 1, therefore, T = U = E 2 (13) n+1 U 2 when U = kr n +1 (12) E 2 (11) E 2 (10) 88. The general expression for (r) is [see Eq. (8.17)] (r) = ( r 2 dr ) 2 2 E  U  2 2 r (1) where U =  kr dr =  kr 2 2 in the present case. Substituting x = r 2 and dx = 2r dr into (1), we have 242 CHAPTER 8 (r) = 1 2 dx k 2E x 2 x2 + 2 x  1 (2) Using Eq. (E.10b), Appendix E, x dx ax + bx + c 2 = bx + 2c 1 sin 1 c x b 2  4 ac (3) and expressing again in terms of r, we find E 2 2 r  1 1 1 ( r ) = sin 2 2 2 E 2 k + 2 r 4 or, sin 2 (  0 ) = 1 k 1+ E2 2 + 0 (4)  1 r2 2 E 2 k 1+ E2 (5) In order to interpret this result, we set 1+ k 2 E 2 E 2 (6) and specifying 0 = 4 , (5) becomes r2 or, = 1 + cos 2 (7) = r 2 + r 2 ( cos 2  sin 2 ) Rewriting (8) in xy coordinates, we find (8) = x2 + y2 + ( x2  y2 ) or, y2 x2 1= + 1 + 1  Since a > 0, > 1 from the definition, (10) is equivalent to (9) (10) CENTRALFORCE MOTION 243 1= y2 x2 + 1 + 1 (11) which is the equation of a hyperbola. 89. (a) By the virial theorem, T =  U 2 for a circular orbit. The firing of the rocket doesn't change U, so U f = U i But Tf = So E f = 2Ti + U i = U i + U i = 0 Ef Ei =0 1 m v 2 + v 2 = 2Ti 2 ( ) The firing of the rocket doesn't change the angular momentum since it fires in a radial direction. f i =1 (b) E = 0 means the orbit is parabolic. The satellite will be lost. E(r) = 0 T (r) = E  U = V (r) = U (r) + Behavior of V(r) is determined by 2 2 r 2  GM m r e s U (r) =  GMe ms r 2 GMe ms r 2 r 2 = 2 GMe ms + 2 r 2 r for small r for large r 244 CHAPTER 8 Energy V(r) T(r) r0 E(r) = 0 U(r) Minimum in V(r) is found by setting dV = 0 at r = r0 dr 2 GMe ms 0= + 3 r0 r02 r0 =  2 GMe ms 810. For circular motion T= 1 me 2 re2 2 GMs me re U= We can get 2 by equating the gravitational force to the centripetal force GMs me = me 2 re re2 or 2 = So T= GMs re3 GM GMs me 1 1 me re2 3 s = = U re 2 2re 2 E=T +U = 1 U 2 If the sun's mass suddenly goes to 1 its original value, T remains unchanged but U is halved. 2 CENTRALFORCE MOTION 245 1 1 1 U= U+ U=0 2 2 2 E = T + U = T + The energy is 0, so the orbit is a parabola. For a parabolic orbic, the earth will escape the solar system. 811. For centralforce motion the equation of orbit is [Eq. (8.21)] d2 1 1 r 2 + =  2 F (r) 2 d r r (1) r force center a In our case the equation of orbit is r = 2 a cos (2) Therefore, (1) becomes 1 d2 1 4 a2 ( cos )1 + 2a ( cos )1 =  2 F ( r ) cos2 2 a d 2 But we have d2 d sin 1 cos ) = 2 ( d cos 2 d = Therefore, we have 2 sin 2 8 a3 1 1 + + =  2 F ( r ) cos 2 cos cos 3 cos or, (5) 2 sin 2 1 + cos cos 3 (4) (3) F (r) =  so that 2 8 a2 = 8 a 3 cos 5 2 2 1 r5 (6) F (r) =  k r5 (7) 246 812. CHAPTER 8 re re The orbit of the comet is a parabola ( = 1), so that the equation of the orbit is r = 1 + cos (1) We choose to measure from perihelion; hence r ( = 0 ) = rE Therefore, (2) = 2 k = 2 rE (3) Since the total energy is zero (the orbit is parabolic) and the potential energy is U =  k r , the time spent within the orbit of the Earth is T = 2 rE dr 2 2 k r  2 r 2 rE rE rE = 2 k r dr r  rE rE (4) 2 2 ( 2 rE  r ) = r  rE  3 k rE from which T= 2 k 2 3 2 3 rE ( 2 + 1) 1  4 2 E 3 rE k (5) Now, the period and the radius of the Earth are related by 2 E = (6) or, CENTRALFORCE MOTION 247 3 rE 2 = k E E 2 (7) Substituting (7) into (5), we find T= 2 2 k 3 k E ( 2 + 1) 1  E 2 (8) where k = GMs and k = GMs E . Therefore, T= 1 3 2 (1  ) (1 + 2 ) E (9) where E = 1 year . Now, = rMercury rE = 0.387 . Therefore, T= so that 1 3 2 (1  0.387 ) (1 + 2 0.387 ) 365 days T = 76 days (10) 813. Setting u 1 r we can write the force as F= k  =  ku 2  u3 r2 r3 (1) Then, the equation of orbit becomes [cf. Eq. (8.20)] d2u 1 + u =  2 2  ku2  u3 2 d u ( ) (2) from which d2u k + u 1  2 = 2 2 d or, (3) d 2 u k 1 + 1  2 u  2 =0 2 d 1 2 If we make the change of variable, (4) v=u k 2 1 1 2 (5) we have d 2 v + 1 2 v = 0 d 2 (6) 248 or, CHAPTER 8 d2v + 2v = 0 d 2 (7) where 2 = 1  2 . This equation gives different solutions according to the value of . Let us consider the following three cases: i) < 2 : For this case 2 > 0 and the solution of (7) is v = A cos (  ) By proper choice of the position = 0, the integration constant can be made to equal zero. Therefore, we can write 1 = A cos + r 2 k  (9) When = 1 ( = 0), this equation describes a conic section. 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 = 0, 1 k = 2 (1 + cos ) r and that we have an ellipse or circle (0 < 1) when E < 1, a parabola ( = 1) when E = 0, and a hyperbola otherwise. It is clear 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 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 . 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) = 2 For this case 2 = 0 and (3) becomes d2u k = 2 d 2 so that 1 k 2 = + A + B r 2 2 (10) u= (11) from which we see that r continuously decreases as increases; that is, the particle spirals in toward the force center. CENTRALFORCE MOTION 249 iii) > 2 For this case 2 < 0 and the solution (7) is v = A cosh (  2  ) (12) may be set equal to zero by the proper choice of the position at which = 0. Then, 1 = A cosh r (  2 + ) 2 k  (13) Again, the particle spirals in toward the force center. 814. The orbit equation for the centralforce field is [see Eq. (8.17)] 2 r 4 dr = 2 d 2 2 EU  2 2 r (1) But we are given the orbit equation: r = k 2 from which dr 2 2 d = 4 k Substituting (2) into (3), we have dr 2 r d = 4 k k = 4 kr From (1) and (4), we find the equation for the potential U: 4 kr = from which U = E and F ( r ) =  U r . Therefore, F (r) =  6k 1 + r4 r3 2 2 2 r 4 EU  2 2 2 r 2 2 (2) (3) (4) (5) 2k 2 2 1 1  3 r 2 r 2 (6) (7) 250 815. A m r P b 2 CHAPTER 8 B 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 = where we have used m = 1. Therefore, k = mvb 2 = vb 2 b (1) v= k 2 b2 (2) The total energy E of the particle is equal to the initial kinetic energy: E= 1 2 k v = 4 2 4b (3) The general orbit equation for a force, F ( r ) =  k r 5 , is d = dr r 2 2 k 2 E + 4  2 4r 2r (4) Substituting for and E from (1) and (3), we have d = k 1 b r2 dr k k k + 4 2 2 4 b r 2b 2r dr r  2b 2 r 2 + b 4 4 =b 2 =b 2 or, dr (r 2  b2 ) 2 d =  b 2 dr r  b2 2 (5) where we have taken the negative square root because r decreases as increases (see the diagram). We can now use the integral [see Eq. (E.4b), Appendix E] CENTRALFORCE MOTION 251 a x 2 dx 1 =  coth 1 2 2 b ab ax b (6) from which we obtain r = 2 coth 1 + 0 b or,  0 r = b coth 2 Now, coth as 0, since r as 0, we must have 0 = 0 . Thus, r = b coth (7) (8) ( 2 ) (9) Notice that r is always greater than b (because coth 1 as ), so that the denominator in (5) never equals zero nor changes sign. Thus, r always decreases as increases. This is, the particle spirals in toward P but never approaches closer than a distance b. 816. The total energy of the particle is E = T +U (1) a principle that by no means pushes the philosophical envelope of physical interpretation. The impulse that causes v v + v changes the kinetic energy, not the potential energy. We therefore have E = T = mv 2 = mv v 1 2 By the virial theorem, for a nearly circular orbit we have 1 E =  mv 2 2 so that (2) (3) E E = 2 v v (4) where we have written E since E < 0. The major and minor axes of the orbit are given by a= k 2E b= 2E (5) Now let us compute the changes in these quantities. For a we have a =  and for b we have k k E E = 2 = a 2E 2E E (6) 252 CHAPTER 8 b = + = 2E 2E3 2E E 1  2 E = b  2E (7) Easily enough, we can show that = v v and therefore a a 817. = b b = E E = 2 v v (8) The equation of the orbit is r from which = 1 + cos (1) r= where = 1 + cos (2) 2E 2 . Therefore, the radial distance r can vary from the mk 2 maximum value (1  ) to the minimum value (1 + ) . Now, the angular velocity of the particle is given by 2 k and = 1 + = r 2 (3) Thus, the maximum and minimum values of become max = r 2 min = 1 + 1  2 2 min = 2 rmax = 2 (4) Thus, max 1 + = =n min 1  from which we find (5) = n 1 n +1 (6) 818. 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, CENTRALFORCE MOTION 253 the force must have acted only along the line connecting the force center and the body. That is, the force is central. Kepler's first law states that planets move in elliptical orbits with the sun at one focus. This means the orbit can be described by Eq. (8.41): r = 1 + cos with 0 < < 1 (1) On the other hand, for central forces, Eq. (8.21) holds: d2 d 2 r 1 1 r + r =  2 F (r) 2 (2) Substituting 1 r from (1) into the lefthand side of (2), we find 1 =  2 r2 F (r) a which implies, that (3) F (r) =  2 r 2 (4) 819. The semimajor axis of an orbit is defined as onehalf the sum of the two apsidal distances, rmax and rmin [see Eq. (8.44)], so 1 1 [ rmax + rmin ] = 2 1 + + 1  = 1  2 2 (1) This is the same as the semimajor axis defined by Eq. (8.42). Therefore, by using Kepler's Third Law, we can find the semimajor axis of Ceres in astronomical units: kC 2 C aC 4 2 C = aE kE 2 4 2 E E where kc = Ms mc , and 1 (2) c = 1 1 + Ms mc Here, Ms and mc are the masses of the sun and Ceres, respectively. Therefore, (2) becomes aC Ms + mc = aE Ms + me from which c E 2 13 (3) 254 1 333, 480 + aC 8, 000 = ( 4.6035) 2 aE 333, 480 + 1 so that 13 CHAPTER 8 (4) aC 2.767 aE The period of Jupiter can also be calculated using Kepler's Third Law: (5) 4 2 J 3 aJ J kJ = E 4 2 E 3 aE kE from which 12 M + m aJ 3 E = s Ms + mJ aE 12 (6) 12 J 333, 480 + 1 = ( 5.2028 ) 3 E 333, 480 + 318.35 (7) Therefore, J 11.862 E The mass of Saturn can also be calculated from Kepler's Third law, with the result (8) ms 95.3 me (9) 820. Using Eqs. (8.42) and (8.41) for a and r, we have 1 1 + cos a cos cos = dt r 0 1 2 4 4 (1) From Kepler's Second Law, we can find the relation between t and : dt = 1 2 dA = d ab ab 2 (1 + cos ) 2 (2) since dA = (1 2) r 2 d . Therefore, (1) becomes 1 1 a cos = r 1 2 4 ( ) 4 a2 ab 2 2 cos (1 + cos ) 0 2 d (3) It is easily shown that the value of the integral is 2. Therefore, CENTRALFORCE MOTION 255 1 a cos = r 1  2 4 ( ) 4 1 2 ab (4) After substituting a and b in terms of and [see Eqs. (8.42) and (8.43)], we obtain a cos = r 1  2 4 ( ) 52 (5) 821. If we denote the total energy and the potential of the family of orbits by E and U(r), we have the relation 2 1 2 r + + U (r) = E 2 2 r 2 (1) from which 2 1 = 2 r 2 E  U ( r )  r 2 2 (2) Here, E and U(r) are same for all orbits, and the different values of result from different values of (1 2) r 2 . For stable circular motion, r = 0 , but for all other motions, r 0 . Therefore, for noncircular motions, r 2 > 0 and is smaller than for the circular case. That is, the angular momentum of the circular orbit is the largest among the family. 822. For the given force, F ( r ) =  k r 3 , the potential is U (r) =  k 2r 2 (1) and the effective potential is V (r) = The equation of the orbit is [cf. Eq. (8.20)] d2u + u =  2 2  ku3 2 d u 1 2 2 1  k r2 (2) ( ) (3) or, d2u k + 1 2 u = 0 d 2 Let us consider the motion for various values of . (4) 256 i) 2 CHAPTER 8 = k : In this case the effective potential V(r) vanishes and the orbit equation is d2u =0 d 2 with the solution u= 1 = A + B r (6) (5) and the particle spirals toward the force center. ii) 2 > k : In this case the effective potential is positive 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 r = r0 ; the particle will then proceed again to an infinite distance. V(r) E r0 r Setting 1  k 2 2 > 0 , (4) becomes d2u + 2u = 0 d 2 (7) with the solution u= 1 = A cos (  ) r (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) 2 < k : 2 For this case we set k  1 G 2 > 0 , and the orbit equation becomes d2u  G2u = 0 2 d (9) with the solution u= 1 = A cosh (  ) r (10) so that the particle spirals in toward the force center. CENTRALFORCE MOTION 257 In order to investigate the stability of a circular orbit in a 1 r 3 force field, we return to Eq. (8.83) and use g ( r ) = k r 3 . Then, we have x or, 2 1 =0 x + k  3 1 + ( x ) 3 2 1 + ( x ) 2 3 3 = k 1 + ( x ) 3 3 (11) (12) Since r r = p = 0 , Eq. (8.87) shows that k = 2 . Therefore, (12) reduces to x=0 (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 examining the basic criterion for stability, namely, that V r = 0 and r= 2V r 2 >0 r= The first of these relations requires k = 2 while the second requires 2 > k . Since these requirements cannot be met simultaneously, no stable circular orbits are allowed. 823. Start with the equation of the orbit: r and take its time derivative = 1 + cos (1) r = sin = sin 2 r r 2 Now from Equation (8.45) and (8.43) we have 2 2 a 1 2 (2) = so that from (2) ab = (3) r max = as desired. 2 a = 1  2 (4) 258 824. (b) b (a) r CHAPTER 8 ra rp a) With the center of the earth as the origin, the equation for the orbit is r Also we know = 1 + cox (1) rmin = a (1  ) rmax = a (1 + ) rmin = rp = 300 km + re = 6.67 10 6 m rmax = ra = 3500 km + re = 9.87 10 6 m a= 1 ra + rp = 8.27 106 m 2 (2) ( ) Substituting (2) gives = 0.193. When = 0, a = 1+ rmin which gives = 7.96 106 m . So the equation of the orbit is 7.96 10 6 m = 1 + 0.193 cos r When = 90, r = = 7.96 10 6 m The satellite is 1590 km above the earth. b) b a rmin CENTRALFORCE MOTION 259 =  =  tan 1 Using b = a b a  rmin =  tan 1 Substituting into (1) gives a a  rmin 101 r = 8.27 10 6 m ; which is 1900 km above the earth 825. 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. E= GMm 1 GMm 2 = mvp = 2a 2 rp (1) where rp and vp 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 2GM ra = 2 a  rp = rp  1 2 rp vp Inserting the values G = 6.67 10 11 N m 2 kg 2 M = 5.976 10 24 kg rp = 6.59 106 m vp = 7.797 10 3 m s 1 we obtain ra 1.010 rp = 6.658 106 m , or 288 km above the earth's surface. We may get the mra va = mrp vp giving va = 27,780 km hr 1 . The period can be found from Kepler's third law t2 = 4 2 a 3 GM (8) (7) (3) (4) (5) (6) 1 (2) speed at apogee from the conservation of angular momentum, Substitution of the value of a found from (1) gives = 1.49 hours. 260 826. r ra rp CHAPTER 8 First, consider a velocity kick v applied along the direction of travel at an arbitrary place in the orbit. We seek the optimum location to apply the kick. E1 = initial energy = 1 GMm mv 2  2 r E2 = final energy = 1 GMm 2 m ( v + v )  2 r We seek to maximize the energy gain E2  E1 : E2  E1 = 1 m 2v v + v 2 2 ( ) For a given v , this quantity is clearly a maximum when v is a maximum; i.e., at perigee. Now consider a velocity kick V applied at perigee in an arbitrary direction: v v2 v1 The final energy is 1 GMm 2 mv2 = 2 rp This will be a maximum for a maximum v 2 ; which clearly occurs when v 1 and v are along the same direction. Thus, the most efficient way to change the energy of an elliptical orbit (for a single engine thrust) is by firing along the direction of travel at perigee. CENTRALFORCE MOTION 261 827. By conservation of angular momentum mra va = mrp vp or Substituting gives va = rp vp ra va = 1608 m/s 828. Use the conservation of energy for a spacecraft leaving the surface of the moon with just enough velocity vesc to reach r = : Ti + U i = Tf + U f 1 GM m m 2 =0+0 mvesc  2 rm vesc = where 2GMm rm Mm = mass of the moon = 7.36 10 22 kg rm = radius of the moon = 1.74 10 6 m Substituting gives vesc = 2380 m/s vmax = v + v0 , vmin = v  v0 mva ra = mvb rb or 829. From conservation of angular momentum we know vmax rmin = vmin rmax ; Also we know rmax vmax = rmin vmin (1) rmin = a (1  e ) rmax = a (1 + e ) Dividing (3) by (2) and setting the result equal to (1) gives (2) (3) 262 CHAPTER 8 rmax 1 + e vmax = = rmin 1  e vmin vmin (1 + e ) = vmax (1  e ) e ( vmin + vmax ) = vmax  vmin e ( 2v ) = 2v0 e= v0 v 830. To just escape from Earth, a velocity kick must be applied such that the total energy E is zero. Thus GMe m 1 2 =0 mv2  2 r where (1) v2 = velocity after kick Me = 5.98 10 24 kg G = 6.67 10 11 Nm 2/kg 2 r = 200 km + re = 200 km + 6.37 10 6 m = 6.57 106 m Substituting into (1) gives v2 = 11.02 km/sec . For a circular orbit, the initial velocity v1 is given by Eq. (8.51) v1 = GMe = 7.79 km/sec r Thus, to escape from the earth, a velocity kick of 3.23 km/sec must be applied. CENTRALFORCE MOTION 263 Since E = 0, the trajectory is a parabola. parabolic escape orbit Earth circular orbit 831. From the given force, we find dF ( r ) 2k 4 k = F (r) = 3 + 5 dr r r (1) Therefore, the condition of stability becomes [see Eq. (8.93)] k 2 + 2k F ( ) 3 5 3 + = + >0 1 F ( )  k 2 + k 4 2 ( ) ( ) (2) or, k2  k >0 k2 + k ( ) (3) Therefore, if 2 k > k , the orbit is stable. 832. For this force, we have dF ( r ) k 2k = F ( r ) = 3 er a + 2 er a dr r ar k = 3 er a r r 2 + a (1) Therefore, the condition of stability [see Eq. (8.93)] becomes F (r) 3 + = F (r) r This condition is satisfied if r < a. r  2 + + 3 a >0 r (2) 264 CHAPTER 8 833. The Lagrangian of the particle subject to a gravitational force is written in terms of the cylindrical coordinates as L = T U = From the constraint r 2 = 4 az , we have 1 m r 2 + r 2 2 + z 2  mgz 2 ( ) (1) z= Therefore, (1) becomes L= Lagrange's equation for is rr 2a (2) mg 2 r2 1 m 1 + 2 r 2 + r 2 2  r 2 4a 4a d L d L  = mr 2 = 0 dt dt (3) ( ) (4) This equation shows that the angular momentum of the system is constant (as expected): mr 2 = = const. Lagrange's equation for r is mg m d r2 L d L r  m 1 + 2 r = 0  = 2 rr 2 + mr 2  dt 4 a 2a r dt r 4 a from which mg m 2 r2 m rr + mr 2  r  m 1 + 2 r  2 rr 2 = 0 2 4a 2a 2a 4a After rearranging, this equation becomes 2 mg r2 m 1 m 1 + 2 r + 2 rr 2 + r =0 m r3 4a 2a 4a (5) (6) (7) (8) For a circular orbit, we must have r = r = 0 or, r = = constant. Then, 2 mg = 2a m 3 (9) or, 2 = m2 g 4 2a (10) Equating this with 2 = m2 4 2 , we have m2 4 2 = m2 g 4 2a (11) CENTRALFORCE MOTION 265 or, 2 = g 2a (12) Applying a perturbation to the circular orbit, we can write r + x where This causes the following changes: r 2 2 + 2 x x 1 3 1 3 r3 rx rx from which, we have rr 2 ( + x ) x 2 0, in lowest order 2 2 2 r r + 2 x x x , in lowest order (15) x 1 (13) (14) ( ) Thus, (8) becomes 2 mg 1 + x)  m 1 + 2 2 x + ( m 3 2a 4a x 1  3 = 0 (16) But 2 mg = 2a m 3 (17) so that (16) becomes mg 2 3 2 m 1 + 2 x + x+ x=0 2a m 4 4a Substituting (17) into (18), we find 2mg 2 m 1 + 2 x + x=0 a 4a or, x+ 2g a+ (19) (18) 2 4a x=0 (20) 266 Therefore, the frequency of small oscillations is CHAPTER 8 = where 2g a + z0 (21) z0 = 2 4a 834. The total energy of the system is E= 1 m r 2 + r 2 2 + r 2 cot 2 + mgr cot 2 ( ) (1) or, E= Substituting 1 1 m 1 + cot 2 r 2 + mr 2 2 + mgr cot 2 2 ( ) (2) = mr 2 , we have E= 2 1 + mgr cot m 1 + cot 2 r 2 + 2 2mr 2 ( ) (3) Therefore, the effective potential is V (r) = 2 2mr 2 + mgr cot (4) At the turning point we have r = 0 , and (3) becomes a cubic equation in r: mgr 3 cot  Er 2 + 2 2m =0 (5) V(r ) E Energy mg ta r co r1 r2 r This cubic equation has three roots. If we attempt to find these roots graphically from the intersections of E = const. and V ( r ) = 2 2mr 2 + mgr cot , 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. CENTRALFORCE MOTION 267 835. If we write the radial distance r as r = + x, = const. (1) then x obeys the oscillatory equation [see Eqs. (8.88) and (8.89)] 2 x + 0 x = 0 (2) where 0 = 3g ( ) + g ( ) (3) The time required for the radius vector to go from any maximum value to the succeeding minimum value is t = where 0 = 2 , the period of x. Thus, t = 0 2 (4) 0 0 0 (5) The angle through which the particle moves during this time interval is = t = (6) where is the angular velocity of the orbital motion which we approximate by a circular motion. Now, under the force F ( r ) =  g ( r ) , satisfies the equation 2 =  F ( r ) = g ( ) Substituting (3) and (7) into (6), we find for the apsidal angle g ( ) (7) = = 0 + g ( ) 3 g ( ) = g ( ) 3+ g ( ) (8) Using g ( r ) = k 1 , we have rn g ( ) n = g ( ) (9) Therefore, (8) becomes = 3n (10) 268 CHAPTER 8 In order to have the closed orbits, the apsidal angle must be a rational fraction of 2. Thus, n must be n = 2,  1,  6, ... n = 2 corresponds to the inversesquareforce and n = 1 corresponds to the harmonic oscillator force. 836. The radius of a circular orbit in a force field described by F (r) =  k r a e r2 (1) is determined by equating F(r) to the centrifugal force: 2 k  a = e r2 m 3 (2) Hence, the radius of the circular orbit must satisfy the relation e  a = 2 mk (3) Since the orbit in which we are interested is almost circular, we write r ( ) = [1 + ( ) ] where ( ) 1 for all values of . (With this description, the apsides correspond to the maximum and minimum values of .) We can express the following quantities in terms of by using (4): u= 1 1 = (1  ) r (5a) (4) d2 1 1 d 2 = d 2 r d 2 (5b) F ( u) =  ku2 e 1 au  Then, substitution into Eq. (8.20) yields  1 d 2 1 mke  a + (1  ) = (1  p a) 2 d 2 (6) k e a 2 (1 + ) 2 (1  a) (5c) Multiplying by , using (3) and simplifying, (6) reduces to d 2 + (1  a ) = 0 d 2 (7) CENTRALFORCE MOTION 269 This equation obviously has two types of solution depending on whether a is larger than or smaller than 1; we consider only < a. (In fact, there is no stable circular orbit for > a.) For the initial condition, we choose = 0 to be a maximum (i.e., an apside) at = 0. Then, we have = 0 cos (1  a) , for < a 12 (8) This solution describes an orbit with welldefined apsides. The advance of the apsides can be found from (8) by computing for what value of is again a maximum. Thus, = The advance of the apside is given by 2 1 a (9) =  2 = 2 1  (1  a ) In the particular case in which 1 2 (10) a we obtain, by extending (10), 2  2 1 + 2a so that (11) a (12) 837. From the equations in Section 8.8 regarding Hohmann transfers: v = v1 + v2 v = vt1  v1 + v2  vt2 v = 2k r2 k k 2k r1 +   mr1 r1 + r2 mr1 mr2 mr2 r1 + r2 (1) Substituting k = GMe = 6.67 10 11 Nm 2/kg 2 5.98 10 24 kg m r1 = initial height above center of Earth = 2re r2 = final height above center of Earth = 3re re = radius of the Earth = 6.37 10 6 m gives v 1020 m/s ( )( ) 270 838. CHAPTER 8 Substitute the following into Eq. (1) of problem 837: k = GMs = 6.67 10 11 Nm 2/kg 2 1.99 10 30 kg m r1 = mean Earthsun distance 1.50 1011 m r2 = mean Venussun distance 1.08 1011 m ( )( ) The result is v = 5275 m/s . The answer is negative because r2 < r1 ; so the rocket must be fired in the direction opposite to the motion (the satellite must be slowed down). v = 5275 m/s; opposite to direction of motion. From Eq. (8.58), the time is given by T = Substituting gives m 32 at = k m k r1 + r2 2 32 (1) 148 days 839. We must calculate the quantity v1 for transfers to Venus and Mars. From Eqs. (8.54), (8.53), and (8.51): v1 = vt1  v1 = where k = GMs = 6.67 10 11 m 3/s 2 kg 1.99 10 30 kg m r1 = mean Earthsun distance = 150 10 9 m Venus 108  sun distance = 109 m r2 = mean Mars 228 Substituting gives vVenus = 2.53 km/sec vMars = 2.92 km/sec where the negative sign for Venus means the velocity kick is opposite to the Earth's orbital motion. Thus, a Mars flyby requires a larger v than a Venus flyby. 2k r2 k  mr1 r1 + r2 mr1 ( )( ) CENTRALFORCE MOTION 271 840. To crash into the sun, we calculate v1 from Eq. (8.54) with r1 = mean distance from sun to Earth, and r2 = radius of the sun. Using Eqs. (8.54), (8.53), and (8.51) we have ( v1 )sun = Substituting 2GMs r2 GMs  r1 r1 + r2 r1 G = 6.67 10 11 Nm 2/kg 2 Ms = 1.99 10 30 kg r1 = rse = 1.5 1011 m r2 = rsun = 6.96 10 8 m gives ( v1 )sun = 26.9 km/sec To escape from the solar system, we must overcome the gravitational pull of both the sun and Earth. From conservation of energy ( Efinal = 0 ) we have:  Substituting values gives v = 43500 m/s Now v = v  v i =v GMs rse GMs m GMe m 1  + mv 2 = 0 rse re 2 = ( 43500  29700 ) m/s ( v) escape = 13.8 km/s To send the waste out of the solar system requires less energy than crashing it into the sun. 841. From the equations in Section 8.8 regarding Hohmann transfers v = v1 + v2 = vt1  v1 + v2  vt2 where 272 2k r2 ; v1 = mr1 r1 + r2 2k mr2 r1 ; v1 = r1 + r2 CHAPTER 8 vt1 = vt2 = Substituting k mr1 k mr2 k = GMe = 6.67 10 11 Nm 2/kg 2 5.98 10 24 kg m r1 = 200 km + re = 6.37 10 6 m + 2 10 5 m r2 = mean Earthmoon distance = 3.84 10 8 m gives v = 3966 m/s From Eq. (8.58), the time of transfer is given by T = Substituting gives m 32 at = k m r1 + r2 k 2 32 ( )( ) = 429, 000 sec. 5 days 842. G = 6.67 10 11 Nm 2/kg 2 Me = 5.98 10 24 kg r1 = 2 10 5 m + 6.37 10 6 m r2 = ? r3 = mean Earthmoon distance = 3.84 10 8 m We can get r2 from Kepler's Third Law (with = 1 day) GMe 2 r2 = 2 4 We know E =  GMm 2r 13 = 4.225 107 m CENTRALFORCE MOTION 273 So E ( r1 ) =  GMe m = 3.04 1011 J 2r1 E ( r2 ) = 4.72 1010 J E ( r3 ) = 5.19 10 9 J To place the satellite in a synchronous orbit would require a minimum energy of E ( r2 )  E ( r1 ) = 2.57 1011 J 843. In a circular orbit, the velocity v0 of satellite is given by 2 mv0 GMm GM = v0 = 2 R R R where M is the Earth's mass. Conservation of energy implies 2 2 mv1 GMm mv2 GMm  =  2 2 2R R Conservation of angular momentum gives mRv1 = m2Rv2 From these equations, we find v1 = so the velocity need to be increased by a factor 4GM 3R 4 3 to change the orbit. 844. The bound motion means that E = k r a e . r mv 2 +V <0 2 where V =  The orbit of particle moving in this central force potential is given by 274 r CHAPTER 8 (r) = rmin ( / r 2 ) dr 2 2 E  V 2 2 r r = 1 2 rmin r 2 dr E+ 2 ke  ra  2 r 2 r In first order of (r / a) , this is ( r ) 2 dr r2 E + 2 k k   2 r 2 r a = 2 dr 2 k k r2 E  +  a r 2 r 2 k k Now effectively, this is the orbit of particle of total energy E  moving in potential  . It is r a well known that this orbit is given by (see Chapter 8) r where = 2 = 1 + cos k and = 1+ k E  a k 2 2 2 If 0 < < 1 , the orbit is ellipsoid; if = 0 , the orbit is circular. 845. a) In equilibrium, for a circular orbit of radius r0, 2 F0 = m r0 = F0 mr0 b) The angular momentum (which is conserved) of a particle in circular orbit is L = mr02 = mFr03 The force acting on a particle, which is placed a distance r (r is very close to equilibrium position ro ) from the center of force is 2 F = m r  F0 = L3  F0 mr 3 L3 3L2 3L2  ( r  r0 )  F0 =  4 ( r  r0 ) =  k ( r  r0 ) mr03 mr02 mr0 CENTRALFORCE MOTION 275 where k 3L2 / mr04 . So the frequency of oscillation is r = 3L2 k = = m m2 r04 3F0 mr0 846. In equilibrium circular orbit, Mv 2 GM 2 = R 4 R2 where M is the Sun's mass. The period is 2 R 4 R R = = v GM R= GM 4v 2 T= 2 D3 2 9 107 yr GM where D = 2R is the separation distance of 2 stars. 847. In equilibrium circular orbit of 1st star where 2 M1v1 GM1 M2 = L1 L2 mass. L1 = LM2 is the distance from 1st star to the common center of M1 + M2 The corresponding velocity is v1 = Finally, the period is GM2 L1 = L2 2 GM2 L( M1 + M2 ) T= 2 L1 2 L3 2 = = 1.2 10 8 yr. v1 G( M1 + M2 ) 276 CHAPTER 8 CHAPTER 9 Dynamics of a System of Particles 91. Put the shell in the z > 0 region, with the base in the xy plane. By symmetry, x = y = 0 . z= =0 2 =0 =0 2 2 r2 r = r1 2 r2 zr 2 dr sin d d r 2 dr sin d d = 0 r = r1 Using z = r cos and doing the integrals gives z= ( 8(r 3 r24  r14 3 2 r 3 1 ) ) 92. z z= h +h a h y a x By symmetry, x = y = 0 . Use cylindrical coordinates , , z. 0 = mass density 277 278 h  + h a 0 =0 =0 z=0 h a  +h 2 4 0 =0 =0 z =0 2 a CHAPTER 9 z= z d d dz d d dz = h 4 The center of mass is on the axis 3 of the cone h from the vertex. 4 93. z h y a x By symmetry, x = y = 0 . From problem 92, the center of mass of the cone is at z = 1 h. 4 From problem 91, the center of mass of the hemisphere is at z= So the problem reduces to i z1 = 1 1 h ; m1 = 1 a 2 h 4 3 3 2 a ; m2 = 2 a 3 8 3 3 a ( r2 = a, r1 = 0 ) 8 i z2 =  z= for 1 = 2 m1 z1 + m2 z2 1 h 2  32 a 2 = m1 + m2 4 ( 1 h + 22 a ) z= h 2  3a3 4 ( 2a + h) DYNAMICS OF A SYSTEM OF PARTICLES 279 94. y a /2 /2 a x By symmetry, y = 0 . If = mass length then M = a So x= x= Using M = a and x = a cos , 1 M 2 = 2 xdm 1 M 2 = 2 x ad x= = a 1 2  2 a cos d = a sin 2  sin  2 2 sin 2 x= 2a sin 2 y=0 95.
mi ri ri r0 r0 280 CHAPTER 9 ri = position of the i th particle m1 = mass of the i th particle M = mi = total mass g = constant gravitational field Calculate the torque about r0 = i = ( ri  r0 ) Fi = ( ri  r0 ) mi g = r1 mi g  r0 mi g = = ( m r ) g  ( m ) r i i i 0 g ( m r ) g  Mr i i 0 g Now if the total torque is zero, we must have m r or r0 = i i = Mr0 1 M m r i i which is the definition of the center of mass. So or center of gravity = center of mass. = 0 about r0 = rCM 96. Since particle 1 has F = 0, r0 = v 0 = 0 , then r1 = 0 . For particle 2 ^ F2 = F0 x then r = F0 ^ x m Integrating twice with r0 = v 0 = 0 gives r2 = F0 2 ^ t x 2m rCM = F m1r1 + m2 r2 ^ = 0 t2 x m1 + m2 4m DYNAMICS OF A SYSTEM OF PARTICLES 281 rCM = v CM = a CM = F0 2 ^ t x 4m F0 ^ tx 2m F0 ^ x 2m 97. y H a O 52 52 a H x By symmetry y = 0 m0 = 16 mH Let mH = m, m0 = 16 m Then x= x= 1 M m 1 3 i xi a cos 52 1 ( 2ma cos 52) = 9 18 m x = 0.068 a 98. By symmetry, x = 0 . Also, by symmetry, we may integrate over the x > 0 half of the triangle to get y . = mass/area y= a 2 x=0 a 2 x=0 a x 2 y =0 a x 2 y =0 y dy dx dy dx a 3 2 = a 3 2 y= 282 99. z vz POW! m1 v1 45 y m 2 vy CHAPTER 9 Let the axes be as shown with the projectile in the yz plane. At the top just before the explosion, v the velocity is in the y direction and has magnitude v0 y = 0 . 2 v0 = 2 2E0 m1 + m2 2 E0 m1 + m2 v0 y = = where m1 and m2 are the masses of the fragments. The initial momentum is E0 , 0 pi = ( m1 + m2 ) 0, m1 + m2 The final momentum is p f = p1 + p2 p1 = m1 ( 0, 0, v1 ) p2 = m2 vx , vy , vz The conservation of momentum equations are px : py : pz : The energy equation is 0 = m2 vx or vx = 0 or ( ) E0 ( m1 + m2 ) = m2 vy vy = 1 E0 ( m1 + m2 ) m2 0 = m1v1 + m2 vz or v1 =  m2 vz m1 E0 1 1 1 2 2 2 + E0 = m1v1 + m2 vy + vz ( m1 + m2 ) 2 m1 + m2 2 2 or 2 2 2 3E0 = m1v1 + m2 vy + vz ( ) ( ) Substituting for vy and v1 gives DYNAMICS OF A SYSTEM OF PARTICLES 283 E0 m1 ( 2m2  m1 ) 2 m2 ( m1 + m2 ) m2 vz gives m1 E0 ( 2m2  m1 ) m1 ( m1 + m2 ) vz = v1 =  v1 =  So m1 travels straight down with speed = v1 m2 travels in the yz plane 2 2 v2 = v y + v z ( ) 12 = E0 ( 4 m1 + m2 ) m2 ( m1 + m2 ) m1 ( 2m2  m1 ) ( m1 + m2 ) = tan 1 vz = tan 1 vy The mass m1 is the largest it can be when v1 = 0 , meaning 2m2 = m1 and the mass ratio is m1 1 = m2 2 910. y B x A x1 x2 First, we find the time required to go from A to B by examining the motion. The equation for the ycomponent of velocity is vy = v0 sin  gt (1) At B, vy = 0 ; thus tB = v0 sin g . The shell explodes giving m1 and m2 horizontal velocities v1 and v2 (in the original direction). We solve for v1 and v2 using conservation of momentum and energy. px : E: ( m1 + m2 ) v0 cos = m1v1 + m2v2 1 1 1 ( m1 + m2 ) v02 cos2 + E = m1v12 + m2v22 2 2 2 (2) (3) Solving for v2 in (2) and substituting into (3) gives an equation quadratic in v1 . The solution is 284 CHAPTER 9 v1 = v0 cos and therefore we also must have 2 m2 E m1 ( m1 + m2 ) (4) v2 = v0 cos 2 m1E m2 ( m1 + m2 ) (5) Now we need the positions where m1 and m2 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 m1 and m2 upon landing: x1 = v1tB ; Thus x2 = v2tB v0 sin v1  v2 g (6) x1  x2 = Using (4) and (5) and simplifying gives (7) x1  x2 = v0 sin g m2 2E m1 + m1 + m2 m2 m1 (8) 911. The term in question is f b For n = 3, this becomes f12 + f13 + f 21 + f 23 + f 31 + f 32 = f12 + f 21 + f13 + f 31 + f 23 + f 32 But by Eq. (9.1), each quantity in parentheses is zero. Thus ( ) ( ) ( ) =1 =1 f = 0 3 3 912. a) v = v0 + u ln m0 m Assuming v0 = 0 , we have m 100 v = 100 ln s 98 v = 2.02 m/s; yes, he runs out of gas. 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 velocity to be slightly greater than 3 m/s (so that he will catch up to the orbiter). 8 kg V Conservation of momentum gives 90 kg 3m/s ( 98 kg ) ( 2.02 m/s) = ( 90 kg ) ( 3 m/s)  ( 8 kg ) v v = 9 m/s (This velocity is relative to Stumblebum's original reference frame; i.e., before he fires his pressurized tank.) Since Stumblebum is traveling towards the orbiter at 2.02 m/s, he must throw the tank at v = 9 m/s + 2.02 m/s v = 11 m/s 913. From Eq. (9.9), the total force is given by F (e) + f As shown in Section 9.3, the second term is zero. So the total force is F (e) It is given that this quantity is zero. Now consider two coordinate systems with origins at 0 and 0 m r r r0 O O where r 0 is a vector from 0 to 0 r is the position vector of m in 0 286 r is the position vector of m in 0 CHAPTER 9 We see that ra = r0 + r The torque in 0 is given by ( = r Fe ) The torque in 0 is ( = r Fe ) ( = r  r 0 Fe ) ( ) ( ( = r Fe )  r 0 Fe ) ( =  r 0 Fe ) But it is given that Thus F (e) =0 = 914. 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 f =  f That this is not valid for a system with magnetic interactions can be seen by considering two particles of charge q1 and q2 moving with velocities v1 and v2 : v2 f21 q2 f12 q1 v1 Now f ij = qi v i Bij where Bij is the magnetic field at qi due to the motion of q j . DYNAMICS OF A SYSTEM OF PARTICLES 287 Since f ij is perpendicular to both v i and Bij (which is either in or out of the paper), f ij can only be parallel to f ji if v i and v j are parallel, which is not true in general. Thus, equations (9.11) and (9.31) are not valid for a system of particles with magnetic interactions. 915. = mass/length F= dp becomes dt mg = mv + mv where m is the mass of length x of the rope. So m = x; m = x x g = x xg=x dv + x v dt dv dx + v2 dx dt dv + v2 dx x g = xv Try a power law solution: v = ax n ; Substituting, dv = nax n 1 dx x g = x ax n nax n 1 + a 2 x 2 n or ( )( ) x g = a 2 ( n + 1) x 2 n 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 = 2n or n = 1 2 g = a 2 ( n + 1) or a = So 2g 3 v= 2 gx 3 288 CHAPTER 9 dv dv dx dv 2 gx = =v = a= dt dx dt dx 3 a= Ti = 0 Tf = Ui = 0 g 3 12 g 2 gx 3 3 1 2 ( y = 0 on table) 1 1 2 gL mgL mv 2 = m = 2 2 3 3 L 2 U f = mgh =  mg So Ei = 0; E f =  mgL 6 Energy lost = mgL 6 916. T x T1 T2 The equation of motion for the falling side of the chain is, from the figure, (b  x) 2 x= (b  x) 2 g + T2 (1) From Example 9.2, we have for the energy conservation case x=g Substitution gives us T2 = g 2bx  x 2 2 (b  x) 2 ( ) =g+ x2 2 (b  x) (2) x 2 4 (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 u = x 2 with respect to the fixed frame. The change in momentum at the bend is DYNAMICS OF A SYSTEM OF PARTICLES 289 p = ( x ) ( 2u) = 2u2 t = Equating this with the net force gives T1 + T2 = Using equation (3), we obtain T1 = x 2 2 t (4) x 2 2 (5) x 2 4 (6) as required. Note that equation (5) holds true for both the free fall and energy conservation cases. 917. As the problem states, we need to perform the following integral 12 = 1  2 d 2 (1  ) (1) Our choice of is 10 4 for this calculation, and the results are shown in the figure. We plot the natural velocity d d = x 2 gb vs. the natural time . 2 1.5 d d 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 918. Once we have solved Problem 917, it becomes an easy matter to write the expression for the tension (Equation 9.18): T 1 + 2  6 2 = mg 2 (1  2 ) This is plotted vs. the natural time using the solution of Problem 917. (1) 290 20 CHAPTER 9 15 T mg 10 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 919. ceiling x released at t=0 b at time t table 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 dp = bg  F dt We can write out what the momentum is, though: p = (b  x) x which has a time derivative dp =  x 2 + ( b  x ) x = ( bg  3 gx ) dt where we have used x = g and x = 2 gx . Setting this last expression equal to (1) gives us F = 3 gx (4) (3) (2) (1) Although M. G. Calkin (personal communication) has found that experimentally the time of fall for this problem is consistently less than the value 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 291 920. ax a+x Let = mass/length The force on the rope is due to gravity F = ( a + x) g  ( a  x) g = 2 x g dp dv dv =m = 2 a dt dt dt So F = dp becomes dt xg = a Now dv dv dx dv = =v dt dx dt dx So xg = av or vdv = Integrating yields 1 2 g 2 v = x +c 2 2a Since v = 0 when x = 0, c = 0. Thus v2 = When the rope clears the nail, x = a. Thus v = ga g 2 x a g xdx a dv dx dv dt 292 CHAPTER 9 921. Let us call x the length of rope hanging over the edge of the table, and L the total length of the rope. The equation of motion is mgx gx dx 2 =m 2 x= L dt L Let us look for solution of the form x = Aet + Be t Putting this into equation of motion, we find = g L Initial conditions are x(t = 0) = x0 = 0.3 m ; v(t = 0) = 0 m/s . From these we find A = B = x0 2 . Finally x = x0 cosh ( t) . When x = L , the corresponding time is t= L cosh 1 = 0.59 s. x0 1 922. Let us denote (see figure) mass of neutron and deuteron respectively velocity of deuteron before collision velocity of neutron and deuteron, respectively after collision 2 0 2m 2m m 1 m and 2m v0 v1 and v2 a) Conservation of energy: 2 2 2 2mv0 mv1 2mv2 v2 v2 2 = + v0 = 1 + 2 2 2 2 2 2 Conservation of momentum is 2mv0 = 2mv2 + v1 2 2 2 4v2 + 4v0 = v1 + 8v0 v2 cos Solving these equations, we obtain 2 sets of solutions v1 = 2v0 6  4 cos 2 + 2 cos 4 cos 2  3 3 v2 = 2 2 2v0 cos + 4v0 cos 2  3v0 3 DYNAMICS OF A SYSTEM OF PARTICLES 293 or numerically v1 = 5.18 km/s v1 = 19.79 km/s v2 = 14.44 km/s and v2 = 5.12 km/s b) Let us call the lab scattering angle of the neutron, then from the sine theorem we have mv1 2mv2 = sin sin = 74.84 and = 5.16 c) From a) we see that cos = sin = 2 v2 sin v1 2 2 2 4v0 + 4v2  v1 8v0 v2 = 2 2 2v0 + 6v2 3 8 v0 v 2 2 300 max = 30 923. Conservation of momentum requires v f to be in the same direction as u1 (component pi = m1u1 p f = ( m1 + m2 ) v f pi = p f v f = m1 u1 m1 + m2 of v f to u1 must be zero). The fraction of original kinetic energy lost is 2 2 m1 u1 1 1 2 m1u1  ( m1 + m2 ) 2 Ki  K f 2 ( m1 + m2 )2 = 1 Ki 2 m1u1 2 = m1  2 m1 m1 + m2 m1 = m2 m1 + m2 294 924. v CHAPTER 9 x O a 0 b v0 m The energy of the system is, of course, conserved, and so we have the following relation involving the instantaneous velocity of the particle: 1 1 2 mv 2 = mv0 2 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 the velocity of the particle has both radial and angular components, there is only one independent variable, which we chose to be . Here = 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 v0 = 0 b ; v = 0 ( b  a ) (2) From (1) and (2), we can solve for the angular velocity after turning through an angle = 0 a 1 b (3) The tension will then be (look at the point of contact) T = m 2 ( b  a ) = m 0 b (4) 925. The best elements are those that will slow down the neutrons as much as possible. In a collision between m1 (the neutron) and m2 (moderator atom), we would thus want to minimize T1 (kinetic energy of the neutron after the collision); or alternatively, maximize T2 (kinetic energy of the moderator atom after the collision). From Eq. (9.88) T2 4 m1 m2 cos 2 = T0 ( m1 + m2 ) 2 Since one cannot control the angle , we want to maximize the function f= ( m1 + m2 ) 2 m1m2 DYNAMICS OF A SYSTEM OF PARTICLES 295 with respect to m2 . ( m1 = constant) 2 2 m1 m1  m2 df = = 0 when m1 = m2 dm2 ( m1 + m2 ) 4 ( ) By evaluating d2 f one can show that the equilibrium point is a maximum. Thus, T2 is a 2 dm2 m = m 1 2 maximum for m1 = m2 . 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. 926. The internal torque for the system is N = r1 f 12 + r 2 f 21 (1) where f12 is the force acting on the first particle due to the second particle. Now f 21= f12 (2) Then, N = r1  r 2 f12 ( ) r r 2  r1 = k r1  r 2 r 2  r1  v0 ( ) ( ) ( ) = kr r1  r 2 r1  r 2 v0 ( ) ( ) (3) This is not zero in general because r1  r 2 and r1  r 2 are not necessarily parallel. The internal torque vanishes only if the internal force is directed along the line joining two particles. The system is not conservative. 927. ( ) ( ) The equation for conservation of py in the lab system is (see fig. 910c): 0 = m1v1 sin  m2 v2 sin Thus sin = or sin = m1T1 sin m2T2 m1v1 sin m2 v2 296 928. CHAPTER 9 Using the notation from the chapter: m1 : Ti = T0 , T f = T1 m2 : Ti = 0; Tf = T2 Thus T0 = T1 + T2 or 1 = T1 T2 + T0 T0 T1 or, T0 (1) If we want the kinetic energy loss for m1 to be a maximum, we must minimize equivalently, maximize From Eq. (9.88): T2 4 m1 m2 = cos 2 T0 ( m1 + m2 ) 2 To maximize this, = 0 (it can't = 180). T2 4 m1 m2 = T0 ( m1 + m2 ) 2 T2 . T0 The kinetic energy loss for m1 is T0  T1 . The fraction of kinetic energy loss is thus T0  T1 T T = 1  1 = 2 (from (1)) T0 T0 T0 T0  T1 T0 = max ( m1 + m2 )2 v1 4 m1 m2 = 0 implies = 0, 180 (conservation of pv ). So the reaction is as follows Before: m1 m2 v2 m2 After: m1 v1 px : m1v = m1v1 + m2 v2 E: Solving for v1 gives v1 = So m2 travels in + x direction m1  m2 v m1 + m2 1 1 1 2 2 m1v 2 = m1v1 + m2 v2 2 2 2 DYNAMICS OF A SYSTEM OF PARTICLES 297 + x direction if m1 > m2 m1 travels in  x direction if m1 < m2 929. From Eq. (9.69) tan = sin cos + ( m1 m2 ) From Eq. (9.74) =  2 Substituting gives tan = or tan = sin ( 2 ) ( m1 m2 )  cos ( 2 ) sin (  2 ) ( m1 m2 ) + cos (  2 ) 930. Before: m1 After: m1 v1 m2 v1 m2 v2 a) py = ( 0.06 kg ) 16 m/s cos 15  ( 8 m/s ) cos 45 = 1.27 N sec px = ( 0.06 kg ) 16 m/s cos 15 ( 8 m/s ) sin 45 = 0.09 N sec The impulse P is the change in momentum. So P = ( 0.09x + 1.27 y ) N sec b) P = F dt = F t F =  ( 9x + 127 y ) N So 298 931. CHAPTER 9 From Eq. (9.69) tan = sin m1  cos m2 From Eq. (9.74) =  Substituting gives tan = sin m1  cos m2 932. pi = mu1 p f = mv1 + 2mv2 Conservation of momentum gives u1 = v1 + 2v2 T = = or v1 = u1  2v2 1 1 2 2 2 mu1  mv1  mv2 2 2 1 1 2 2 2 2 mu1  m u1  4u1v2 + 4v2  mv2 2 2 ( ) 2 = 2mu1v2  3mv2 d ( T ) = 0 implies 2u1 = 6v2 dv2 or v2 = u1 3 d 2 ( T ) < 0, so this is a maximum 2 dv2 v1 = u1  2v2 = v1 = v2 = u1 3 u1 3 DYNAMICS OF A SYSTEM OF PARTICLES 299 933. From Eq. (9.87b) in the text, we have T1 m = T0 ( m1 + m2 ) 2 1 2 1 2 cos + m2  sin 2 m1 2 2 2 cos 2 + m2  sin 2 + 2 cos m2  sin 2 = 2 m2 m1 m1 1+ m1 Substituting m2 m1 and cos y we have T1 2 = (1 + ) 2 y 2 + 2  1 + 2 y 2 + y 2  1 T0 1 = 12 =4 0 =1 =2 /2 (1) T1 T0 934. Before After v1 u1 m m 45 m m v2 x Cons. of pz : Cons. of py : mu1 = mv1 cos 45 + mv2 cos 0 = mv1 sin 45  mv2 sin (1) (2) Cons. of energy (elastic collision) 1 1 1 2 2 2 mu1 = mv1  mv2 2 2 2 Solve (1) for cos : cos = Solve (2) for sin : sin = v1 2 v2 u1  v1 v2 2 (3) 300 CHAPTER 9 Substitute into cos 2 + sin 2 = 1 , simplify, and the result is 2 2 2 u1 = v2  v1 + 2 u1v1 Combining this with (3) gives 2 2v1 = 2 u1v1 We are told v1 0 , hence v1 = u1 2 Substitute into (3) and the result is v2 = u1 2 Since v1 = v2 , (2) implies 0 = 45 935. From the following two expressions for T1 T0 , 2 T1 v1 = 2 T0 u1 Eq. (9.82) 2 T1 m = T ( m1 + m2 ) 2 2 1 2 cos m2  sin 2 m1 Eq. (9.87b) we can find the expression for the final velocity v1 of m1 in the lab system in terms of the scattering angle : v1 = 2 m m1u1 cos 2  sin 2 m1 + m2 m1 (1) If time is to be constant on a certain surface that is a distance r from the point of collision, we have r = v1t0 (2) Thus, 2 m2 m1u1t0 cos  sin 2 r= m1 + m2 m1 (3) This is the equation of the required surface. Let us consider the following cases: DYNAMICS OF A SYSTEM OF PARTICLES 301 i) m2 = m1 : r= u1t0 cos 1  sin 2 = u1t0 cos 2 (4) (The possibility r = 0 is uninteresting.) iI) m2 = 2m1 : r= u1t0 cos 4  sin 2 3 (5) iII) m2 = : Rewriting (3) as mut r= 1 10 m 1+ 1 m2 and taking the limit m2 , we find 2 cos 1 sin 2  2 m2 m2 m1 (6) r = u1t0 All three cases yield spherical surfaces, but with the centers displaced: m2 = m2 = 2m1 m1 = m2 O (7) m1 v1t0  v1t0 3 m2 v1t0 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 m1 = m2 . Therefore, neutrons scattered from the hydrogenous target will arrive on the surface A with the same time delay between scattering and arrival, independent 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. 936. Since the initial kinetic energies of the two particles are equal, we have 1 1 1 2 2 2 m1u1 = m2 u2 = 2 m2 u1 2 2 2 (1) or, m1 = 2 m2 (2) 302 CHAPTER 9 Now, the kinetic energy of the system is conserved because the collision is elastic. Therefore, 1 1 1 2 2 2 2 m1u1 = m2 u2 = m1u1 = m2 v2 2 2 2 since v1 = 0 . Momentum is also conserved, so we can write m1u1 + m2 u2 = ( m1 + m2 ) u1 = m2 v2 Substituting the second equality in (4) into (3), we find m + m2 2 1 m u = m2 1 u1 2 m2 2 1 1 2 (3) (4) (5) or, 1 m1 = m2 2 Using m1 m2 = 2 , (6) becomes 2 2 = 2 + solving for , we obtain m1 + m2 2 (6) ( ) 2 (7) = 1 2; This gives us m1 = 3 2 2; m2 2 = 3 2 2 (8) u2 + : < 0 =  1 2 with u1  : > 0 ( ) (9) 937. Impulse = F dt = 6 10 3 t=0 ( 360  10 t ) dt 7 2 3 107 = 360 6 10 3  6 10 3 N s 3 ( ) Impulse = 1.44 Since the initial velocity is zero, v f = v kg m s Impulse = P = mv So DYNAMICS OF A SYSTEM OF PARTICLES 303 kg m s vf = 0.003 kg 1.44 vmuzzle = 480 m s 938. v1 v1  V 1 2 m1u1 2 1 2 T1 = m1v1 2 T0 = Thus, 2 T1 v1 = 2 T0 u1 (1) (2) Now, from the diagram above, we have v1 = V cos + v1 cos (  ) Using Eq. (9.68) in the text, this becomes m v1 = V cos + 2 cos (  ) m1 Thus, 2 v1 V 2 m = 2 cos + 2 (  ) 2 u1 u1 m1 2 (3) (4) (5) Using Eq. (9.84) in the text, V m1 = u1 m1 + m2 Therefore, we find 2 T1 m1 = T0 ( m1 + m2 ) 2 (6) cos (  ) cos + ( m1 m2 ) 2 (7) If we define 304 CHAPTER 9 S cos + we have cos (  ) ( m1 m2 ) (8) 2 T1 m1 = S2 T0 ( m1 + m2 ) 2 (9) as desired. 939. u v y x As explained in Section 9.8, the component of velocity parallel to the wall is unchanged. So vx = u sin vy is given by = or vy uy = vy u cos vy = u cos Thus v = u2 sin 2 + 2 u2 cos 2 v = u sin 2 + 2 cos 2 tan = or u sin u cos 12 12 = tan 1 tan 1 DYNAMICS OF A SYSTEM OF PARTICLES 305 940. Because of the string, m2 is constrained to move in a circle of radius a. Thus, initially, m2 will move straight up (taken to be the y direction). Newton's rule applies to the velocity component along u1 . The perpendicular component of velocity (which is zero) is unchanged. Thus m1 will move in the original direction after the collision. From conservation of py we have m1u1 sin = m1v1 sin + m2 v2 From Newton's rule we have v2 cos ( 90  )  v1 u1 (1) = or v1 = v2 sin  u1 Substituting (2) into (1) and solving for v2 gives v2 = (2) then gives v1 = u1 m1 sin 2  m2 m1 sin + m2 2 (2) ( + 1) m1u1 sin m1 sin 2 + m2 straight up ( ) along u1 941. Using y = v0 t  1 2 gt and v = v0  gt , we can get the velocities before and after the 2 where h1 = 1 2 gt1 2 2 h1 =  2 gh1 g collision: Before: u1 =  gt1 So After: 0 = v0  gt2 h2 = v0 t2  = So u1 =  g or t2 = v0 g 1 2 gt2 2 or v0 = 2 gh2 v1 = 2 gh2 2 2 v0 1 v0  g 2 g 306 CHAPTER 9 Thus = v2  v1 = u2  u1 2 gh2 2 gh1 = h2 h1 Tlost = Ti  Tf Fraction lost = = Ti  T f Ti 2 2 u1  v1 h1  h2 h = = 1 2 2 u1 h1 h1 Ti  Tf Ti = 1 2 942. 30 5 m/s y x As explained in Section 9.8, the velocity component in the ydirection is unchanged. m vy = uy = 5 sin 30 = 2.5 m/s s For the x component we have 0.8 = vx ux = vx vx = m 3 m 5 5 s cos 30 2 s vx = 2 3 vf = 1 73 2 m s 4.3 m/s 36 = tan 1 2.5 2 3 DYNAMICS OF A SYSTEM OF PARTICLES 307 943. v2 4m T0 m 4m 45 m v1 Conservation of px gives 2mT0 = 4 mv2 cos + or cos = Conservation of py gives 0= or sin = Substituting into sin 2 + cos 2 = 1 gives v2 1= 1 2 + 32v2 Simplifying gives 2 v2 = 2 v1 T0 v1 T0 m +  16 8 m 8m 1 mv1 2 2mT0  1 mv1 2 4 mv2 1 mv1  4 mv2 sin 2 v1 4 2 v2 2mT0 + 1 2 2 m v1  2 mT0 mv1 2 2 16 m2 v2 (1) The equation for conservation of energy is T0  or 2 2 5T0 = 3mv1 + 12mv2 T0 1 1 2 2 = mv1 + ( 4 m) v2 6 2 2 (2) Substituting (1) into (2) gives a quadratic in v1 : 2 15mv1  6 T0 m v1  14T0 = 0 308 CHAPTER 9 Using the quadratic formula (taking the positive sign since v1 > 0 ) gives: v1 = 1.19 T0 m Substituting this into the previous expressions for cos and sin and dividing gives tan = sin = 1.47 cos Thus , the recoil angle of the helium, is 55.8. 944. v0 x Fgrav = mg = xg where = mass/length Fimpulse = mv + mv + mv0 , since v = v0 , v = 0. We have m= So the total force is 2 F ( x ) = xg + v0 d dx ( x ) = = v0 dt dt We want F(x = a) 2 F ( a ) = ag + v0 or v2 F = ag 1 + 0 ag 945. 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], ( ) 2 sin d = ( ) 2 sin d Thus, sin d sin d (1) ( ) = ( ) (2) DYNAMICS OF A SYSTEM OF PARTICLES 309 The relation between and is given by Eq. (9.69), which is tan = sin cos + x (3) where x = m1 m2 . Using this relation, we can eliminate from (2): sin = 1 1 1+ tan 2 = 1 ( cos + x ) 2 1+ sin 2 = sin 1 + 2 x cos + x 2 (4) d ( tan ) cos ( cos + x ) + sin 2 d d 2 = = cos d d ( tan ) d ( cos + x )2 Since cos 2 = 1 , (5) becomes 1 + tan 2 d = d 1 + x cos 1 + x cos 1 = 2 2 sin ( cos + x ) 1 + 2x cos + x 2 1+ ( cos + x ) 2 (5) (6) Substituting (4) and (6) into (2), we find ( ) = ( ) (1 + 2x cos + x ) 1 + x cos 2 32 (7) 946. The change in angle for a particle of mass moving in a centralforce field is [cf. Eq. (9.121)]. Let = capital here. = rmax rmin 2 E  U  ( ( r 2 dr 2 ) 2 r 2 ) (1) b a rmin In the scattering from an impenetrable sphere, rmin is the radius of that sphere. Also, we can see from the figure that =  2 . For r > rmin , U = 0. Thus (1) becomes = Substituting b 2T0 = ; E = T0 (3) a ( r 2 dr 2 ) 2E  r2 (2) 310 CHAPTER 9 (2) becomes = dr r2 r 2 1 b (4) This integral can be solved by using Eq. (E. 10b), Appendix E: 2 = sin 4 r b2 a 1 (5) Thus, sin = b a (6) Therefore, we can find the relation between and b by substituting ( =  2 ) into (6). We have b = a cos 2 (7) Now, the differential cross section is given by Eq. (9.120): ( ) = From (7), we have b db sin d (8) ( ) = Total cross section is given by 1 a a2 a cos sin = sin 2 2 2 4 (9) t = ( ) d = so that a2 4 4 (10) t = a2 (11) 947. The number of recoil particles scattered into unit solid angle in each of the two systems, lab and CM, are the same. Therefore, ( ) sin d = ( ) sin d (1) where and are the CM and lab angles, respectively, of the recoil particle. From (1) we can write [cf. Eq. (9.125) in the text] ( ) sin d = ( ) sin d (2) DYNAMICS OF A SYSTEM OF PARTICLES 311 Now, in general, = 2 [see Eq. (9.74)]. Hence, sin sin 1 = = sin sin 2 2 cos and d 1 = d 2 Using (3) and (4) in (2), we have (4) (3) ( ) 1 = ( ) 4 cos For m1 = m2 , the Rutherford scattering cross section is [Eq. (9.141)] (5) ( ) = k2 1 2 4 4T0 sin ( 2) (6) Also for this case, we have [Eqs. (9.71) and (9.75)] =  2 = 2 (7) Hence, sin = sin = sin  = cos 2 2 (8) and since the CM recoil cross section () is the same as the CM scattering cross section ( ) , (6) becomes ( ) = Using (5) to express ( ), we obtain k2 1 2 4T0 cos 4 (9) ( ) = ( ) 4 cos or, (10) ( ) = k2 1 2 T0 cos 3 (11) 312 CHAPTER 9 948. In the case m1 m2 , the scattering angle for the incident particle measured in the lab system is very small for all energies. We can then anticipate that () will rapidly approach zero as increases. Eq. (9.140) gives the Rutherford cross section in terms of the scattering angle in the CM system: CM ( ) = From Eq. (9.79) we see that for m1 m2 , T0 = Furthermore, from Eq. (9.69), tan = ( 4T0) k2 2 sin ( 2) 4 1 (1) m2 m T0 2 T0 m1 + m2 m1 sin m1 + cos m2 (2) m2 sin m1 (3) and therefore, since is expected to be small for all cases of interest, sin Then, m cos = 1  1 m2 and 2 m1 1 sin ( 2) = 1  1  2 m2 2 2 m1 m tan 1 m2 m2 (4) (5) (6) (Notice that to unity.) 1 , but since m1 m2 , the quantity m1 m2 is not necessarily small compared With the help of (2) and (6), we can rewrite the CM cross section in terms of as mk 1 CM ( ) = 1 2 2 2m2T0 1  1  m1 m 2 According to Eq. (9.129), 2 (7) DYNAMICS OF A SYSTEM OF PARTICLES 313 2 2 m1 cos + 1  m1 sin m2 m2 LAB ( ) = CM ( ) 2 m 1  1 sin m2 (8) We can compute LAB ( ) with the help of (7) and the simplifications introduced in the righthand side of (8) by the fact that 1 : 2 m1 + 1  m1 2 m2 m2 m1k LAB ( ) 2 2 2 2m2T0 m1 m1 1  1  1 m2 m2 2 (9) and so, LAB ( ) (m k 2 1 2 2m2 T0 2 ) 2 2 1  1  m1 m2 (10) 2 m 1 1 m2 This expression shows that the cross section has a secondorder divergence at = 0. For values of > m2 m1 , (9) gives complex values for lab . This result is due to the approximations involved in its derivation, making our result invalid for angles larger than m2 m1 . 949. The differential cross section for Rutherford scattering in the CM system is [cf. Eq. (9.140) in the text] ( ) = where [cf. Eq. (9.79)] T0 = 1 k2 2 16T0 sin 4 2 m2 T0 m1 + m2 (1) (2) 314 CHAPTER 9 Thus, 1 m1 + m2 k2 ( ) = 2 16T0 sin 4 m2 2 1 k2 m = 1+ 1 2 m2 16T0 sin 4 2 Since m1 m2 1 , we expand m1 m1 +... 1 + 1+ 2 m2 m2 Thus, to the first order in m1 m2 , we have 2 2 2 (3) (4) ( ) = m k2 1 1+ 2 1 2 16T0 sin 4 m2 2 (5) This result is the same as Eq. (9.140) except for the correction term proportional to m1 m2 . 950. The potential for the given force law is U (r) = k 2r 2 (1) First, we make a change of variable, z = 1 r . Then, from Eq. (9.123), we can write = zmax b dz k 2 z 1  b 2 + 2 mu0 b z zmax k zmax = b 2 + 2 mu0 0 1 2 0 = k 2 b + 2 mu0 2 sin 1 (2) = b 2 b2 + k 2 mu0 Solving (2) for b = b( ), b = k 2 mu0 2  4 2 2 (3) DYNAMICS OF A SYSTEM OF PARTICLES 315 According to Fig. 922 and Eq. (9.122), = so that b( ) can be rewritten as (): b ( ) = 1 (  ) 2 (4) k 2 mu0  ( 2  ) (5) The differential cross section can now be computed from Eq. (9.120): ( ) = with the result b db sin d (6) ( ) = 2 mu0 2 ( 2  ) sin 2 k 2 (  ) (7) 951. n = p In the CM system, whenever the neutron is scattered through the angle , the proton recoils at the angle = . Thus, the neutron scattering cross section is equal to the recoil cross section at the corresponding angles: dN p dN n = d ( ) d ( ) Thus, dN p dTp dN n = d ( ) dTp d ( ) where dN p dTp is the energy distribution of the recoil protons. According to experiment, dN p dTp = const. Since mp mn , Tp is expressed in terms of the angle by using Eq. (9.89b): Tp = T0 sin 2 We also have = (3) (2) (1) 2 for the case mp mn . Thus, d ( ) dTp = d 1 T0 sin 2 2 sin d ( ) (4) 316 CHAPTER 9 or, d ( ) dTp = T0 d  sin 2 2 sin d 2 T0 T sin cos = 0 2 sin 2 2 4 (5) = Therefore, we find for the angular distribution of the scattered neutron, dN p T0 dN n = d ( ) dTp 4 (6) Since dN p dTp = const., dN n d is also constant. That is, the scattering of neutrons by protons is isotropic in the CM system. 952. Defining the differential cross section ( ) in the CM system as in Eq. (9.116), the number of particles scattered into the interval from to + d is proportional to dN ( ) sin d =  ( ) d ( cos ) From Eq. (9.87a) and the assumption of elastic collisions (i.e., T0 = T1 + T2 ), we obtain T2 2m1 m2 = (1  cos ) T0 ( m1 + m2 ) 2 or, solving for cos , cos = where Tm = 4 m1 m2 Tm  2T2 Tm (1) (2) (3) ( m1 + m2 )2 T0 is the maximum energy attainable by the recoil particle in the lab system. Then, (1) can be rewritten as dN 2 ( ) dT2 Tm (4) and consequently, we obtain the desired result for the energy distribution: dN ( ) dT2 (5) DYNAMICS OF A SYSTEM OF PARTICLES 317 953. m1 = mass of particle 1 , m2 = mass of 238 U u1 , u : velocity of particle in LAB and CM before collision 1 v1 , v : velocity of particle in LAB and CM after collision 1 u2 , u : velocity of 2 v2 , v " velocity of 2 u2 = 0 , T1 = 238 U in LAB and CM before collision U in LAB and CM after collision 238 2 m1u1 = 7.7MeV 2 = 90 is angle through which particle is deflected in LAB is angle through which particle and is recoil angle of 238 238 U are deflected in CM U in LAB v1 m1 u1 m2 m1 v2 vCM v2 a) Conservation of momentum in LAB: m1 u1 = m2 v2 cos v1 = u1 tan m1 v1 = m2 v2 sin Conservation of energy in LAB: 2 2 2 m1u1 m1v1 m2 v2 = + 2 2 2 From these equations we obtain the recoil scattering angle of tan = b) The velocity of CM of system is 238 U m2  m1 m2 + m1 = 44.52 vCM = The velocity of 238 m1u1 m1 + m2 ' U in CM after collision is v2 = v2  vCM . From the above figure we can obtain 238 the scattering angle of particle U in CM to be 318 CHAPTER 9 tan = v2 sin = v2 cos  vcm 2 2 m2  m1 2 m1 = 89.04 238 In CM, clearly after collision, particle moves in opposite direction of that of c) The kinetic energy of particle 238 U. U after collision in LAB is 2 2 2 2 m2 v2 m2 m1v0 m1 u1 2m1 T1 = 0.25 MeV = = = m cos 2 2 2 m1 + m2 m1 + m2 Evidently, conservation of energy is satisfied. d) The impact parameter in CM is given in Section 9.10. b= where k = so k cot 2 2T0 q1q2 1 and T0 = m1u1 2 + m2 u2 2 is the total energy of system in CM, 2 9 0 b= q1 q2 m1 + m2 cot = 1.8 10 14 m 2 2 4 0 m1 m2 u1 ( ) We note that b is the impact parameter of particle with respect to CM, so the impact ( m1 + m2 ) b = 1.83 1014 m. parameter of particle with respect to 238 U is m2 e) In CM system, the orbit equation of particle is r = (1 + ) cos where = 0 corresponds to r = rmin rmin = 1+ is closest distance from particle to the center of mass, and 4 0 ( m1u1b ) = = m1k q1q2 m1 2 2 and = 1+ 2 4 0 ( m1u1b ) E 2 = 1 + 2E m1k q1 q2 m1 4 0 ( m1u1b ) q1q2 m1 2 2 = 1 + m1u1 2 But the actual minimum distance between particles is rmin = m1 + m2 rmin = 0.93 10 14 m. m2 DYNAMICS OF A SYSTEM OF PARTICLES 319 f) Using formula LAB ( ) = CM ( ) where x = m1 , = 90 , m2 ( x cos + 1  x 2 sin 2 1  x sin 2 2 ) 2 CM ( ) = ( 4T0) k2 2 1 sin 4 2 We find this differential cross section in LAB at = 90: LAB ( = 90) = 3.16 10 28 m 2 g) Since dN = ( ) sin d d we see that the ratio of probability is N ( ) sin = 11.1 ( ') sin 954. Equation 9.152 gives the velocity of the rocket as a function of mass: v = v0 + u ln m0 m = ln 0 m m m0 m ( v0 = 0 ) p = mv = mu ln To maximize p, set dp =0 dm 0= ln m0 =1 m dp m = u ln 0  1 dm m m0 =e m or m = e 1 m0 To check that we have a maximum, examine d2 p dm2 m = m e1 0 d2 p u = 2 dm m d2 p u e < 0 , so we have a maximum. = 2 dm m = m e1 m0 0 m = e 1 m0 320 955. CHAPTER 9 The velocity equation (9.165) gives us: m v ( t ) =  gt + u ln 0 m (t) (1) where m ( t ) = m0  t , the burn rate = 9m0 10 , the burn time = 300 s , and the exhaust velocity u = 4500 m s 1 . These equations are good only from t = 0 to t = . First, let us check that the rocket does indeed lift off at t = 0: the thrust u = 9um0 10 = 13.5 m s 2 m0 > m0 g , as required. To find the maximum velocity of the rocket, we need to check it at the times t = 0 and t = , and also check for the presence of any extrema in the region 0 < t < . We have v(0) = 0, v( ) =  g + u ln 10 = 7400 m s 1 , and calculate u dv u = g + =g  1 > 0 dt m (t) m (t) g (2) The inequality follows since u > m0 g > m ( t ) g . Therefore the maximum velocity occurs at t = , where v =  g + u ln 10 = 7400 m s 1 . A similar singlestage rocket cannot reach the moon since v ( t ) < u ln m0 m final = u ln 10 10.4 m s 1 , which is less than escape velocity and independent of fuel burn rate. 956. a) Since the rate of change of mass of the droplet is proportional to its crosssectional area, we have ( ) dm = k r 2 dt If the density of the droplet is , m= so that dm dm dr dr = = 4r 2 = kr 2 dt dr dt dt Therefore, dr k = dt 4 or, r = r0 + as required. k t 4 4 r3 3 (1) (2) (3) (4) (5) DYNAMICS OF A SYSTEM OF PARTICLES 321 b) The mass changes with time, so the equation of motion is F= Using (1) and (2) this becomes d dv dm = mg ( mv) = m + v dt dt dt (6) 4 3 dv 4 3 + kr 2 v = r r g 3 3 dt or, dv 3k v=g + dt 4 r Using (5) this becomes dv 3k v + =g dt 4 r + k t 0 4 If we set A = 3k k and B = , this equation becomes 4 4 dv A + v=g dt r0 + Bt and we recognize a standard form for a firstorder differential equation: dv + P (t) v = Q (t) dt in which we identify P (t) = The solution of (11) is  P ( t ) dt P (t) dt Q dt + constant v (t) = e e (7) (8) (9) (10) (11) A ; Q(t) = g r0 + Bt (12) (13) Now, P (t ) dt = r A = 3 . Therefore, B e 0 A A dt = ln ( r0 + Bt ) B + Bt 3 = ln ( r0 + Bt ) since (14) Pdt = ( r0 + Bt ) 3 (15) 322 CHAPTER 9 Thus, 3 3 v ( t ) = ( r0 + Bt ) ( r0 + Bt ) g dt + constant 3 g 4 = ( r0 + Bt ) ( r0 + Bt ) + C 4B (16) The constant C can be evaluated by setting v ( t = 0 ) = v0 : v0 = so that C = v0 r03  We then have v (t) = or, v (t) = 1 ( Bt ) 3 g 4 3 4B ( Bt ) + 0 r0 g 4 r0 4B (18) 1 g 4 r0 + C r03 4B ( ) (17) ( r0 + Bt ) 1 3 g 4 4 g 3 4B ( r0 + Bt ) + v0 r0  4B r0 (19) ( ) (20) where 0 r03 means "terms of order r03 and higher." If r0 is sufficiently small so that we can neglect these terms, we have ( ) v (t) t as required. 957. (21) Start from our definition of work: W = F dx = dp dx = v dp dt (1) We know that for constant acceleration we must have v = at (zero initial velocity). From Equation (9.152) this means m = m0 e  at u We can then compute dp: (2) at dp = d ( mv ) = d ( mat ) = ma dt + at dm = m0 ae  at u 1  dt u This makes our expression for the work done on the rocket (3) Wr = m0 a t ( at )( u  at ) e  at u dt u 0 (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 dp dmexhaust (v  u) , so that We = m0 a t ( at  u) 2 e  at u dt u 0 (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 W= m0 a v a 2  at u 2 0 u  uat e dt = m0 u u ( )  0 (1  x ) e v u x dx (6) where the obvious substitution was made in the last expression. Upon evaluating the integral we find W = m0 uve  v u = muv where m is the mass of the rocket after its engines have turned off and v is its final velocity. 958. (7) From Eq. (9.165) the velocity is v= dy m =  gt + u ln 0 dt m m0  gt + u ln m dt dy = Since m =  , dt =  dm dt y+C = a 1 2 u gt  2 ln m0 dm m ln x dx = x 1 + ln x , so we have y+C = 1 2 u gt  2 a m0 m + m ln m Evaluate C using y = 0 when t = 0, m = m0 C= y= u ( m0  m)  um0 1 2 mu m0 ln gt  ; m0  m = t m 2 1 2 mu m0 ln gt  2 m y = ut  At burnout, y = yB , t = tB 324 CHAPTER 9 yB = utB  After burnout, the equations are 1 2 mu m0 gtB  ln m 2 y = y 0 + v0 t  Calling the top of the path the final point 1 2 gt and v = v0  gt 2 or v f = 0 = vB  gt f y  y0 = t f = vB g y0 = 0 2 2 2 vB vB vB  = ; g 2g 2g 2 vB 2g y= 959. In order to immediately lift off, the thrust must be equal in magnitude to the weight of the rocket. From Eq. (9.157): Thrust = v0 So v0 = velocity of fuel v0 = mg or v0 = mg 960. The rocket will lift off when the thrust just exceeds the weight of the rocket. Thrust =  u dm = u dt Weight = mg = ( m0  t ) g Set thrust = weight and solve for t: u = ( m0  t ) g ; t = m0  u g With m0 = 70000 kg , = 250 kg/s , u = 2500 m/s , g = 9.8 m/s 2 t 25 sec 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 325 961. From Eq. (9.153), the velocity after the first state is: v1 = v0 + u ln k After the second stage: v2 = v1 + u ln k = v0 + 2u ln k After the third stage: v3 = v2 + u ln k = v0 + 3u ln k After the n stages: vn = vn 1 + u ln k = v0 + nu ln k vn = v0 + nu ln k 962. To hover above the surface requires the thrust to counteract the gravitational force of the moon. Thus: u  dm 1 = mg dt 6 6u dm = dt g m Integrate from m = m0 to 0.8 m0 and t = 0 to T: T= 6 ( 2000 m/s ) 6u ln 0.8 =  ln 0.8 9.8 m/s 2 g T = 273 sec 963. a) With no air resistance and constant gravity, the problem is simple: 1 2 mv0 = mgh 2 2 giving the maximum height of the object as h = v0 2 g v0 g 610 s . (1) 1800 km . The time it takes to do this is b) When we add the expression for air resistance, the differential equation that describes the projectile's ascent is F=m dv 1 v =  mg  cW Av 2 =  mg 1 + v dt 2 t 2 (2) 326 CHAPTER 9 where vt = 2mg cW A 2.5 km s 1 would be the terminal velocity if the object were falling from a sufficient height (using 1.3 kg m 3 as the density of air). Solution of this differential equation gives v ( t ) = vt tan tan 1 v0 gt  v vt t (3) This gives v = 0 at time = ( vt g ) tan 1 ( v0 vt ) 300 s . The velocity can in turn be integrated to give the ycoordinate of the projectile on the ascent. The height it reaches is the ycoordinate at time : v2 v h = t ln 1 + 0 2 g vt 2 (4) which is 600 km. c) Changing the acceleration due to gravity from g to  GMe ( Re + y ) 2 =  g Re ( Re + y ) 2 changes our differential equation for y to y 2 R 2 e y =  g + v t Re + y Using the usual numerical techniques, we find that the projectile reaches a height of a flight time of 330 s. (5) 630 km in d) Now we must replace the in the air resistance equation with ( y ) . Given the dependence of vt on , we may write the differential equation ( y) y 2 R 2 e y = g + 0 vt Re + y (6) where we use log 10 ( y ) = 0.11  (5 10 5 )y and 0 = 1.3 , with the 's in kg m 3 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 2600 km. height to which the projectile rises when there is no air resistance, which is DYNAMICS OF A SYSTEM OF PARTICLES 327 2.5 2 height (1000 km) 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 t (1000 s) 0.6 0.7 0.8 0.9 1 (a) (b) (c) (d) 964. We start with the equation of motion for a rocket influenced by an external force, Eq. (9.160), with Fext including gravity, and later, air resistance. 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 yb = utb  1 2 mu m0 gtb  ln 2 mb (1) where mb is the mass of the rocket at burnout and = ( m0  mb ) tb . Substitution of the given values gives yb 2 250 km . After burnout, the rocket travels an additional vb 2 g , where vb is the rocket velocity at burnout. The final height the rocket ends up being is taken into account. 3700 km, after everything b) The situation, and hence the differential equation, becomes more complicated when air resistance is added. Substituting Fext =  mg  cW Av 2 2 (with = 1.3 kg m 3 ) into Equation (9.160), we obtain c Av 2 dv u = g W dt m 2m (2) We must remember that the mass m is also a function of time, and 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 v y 2 cW Av u v = m  g  2m m  These must be integrated from the beginning until the burnout time, and therefore must be integrated with the substitution = 0. Firstly, we get the velocity and height at burnout to be vb 7000 m s 1 and yb 230 km . We can numerically integrate to get the second part of the (3) 328 CHAPTER 9 journey, or use the results of Problem 963(b) to help us get the additional distance travelled with air resistance, analytically. The total height to which the rocket rises is 890 km in a total flight time of 410 s. 2 c) The variation in the acceleration of gravity is taken into account by substituting GMe vb ( Re + y ) = g Re ( Re + y ) for g in the differential equation in part (b). This gives 2 6900 m s 1 , yb 230 km , with total height 950 km and timeofflight 460 s. d) Now one simply substitutes the given expression for the air density, (y) for , into the differential equation from part (c). This gives vb 8200 m s 1 , yb 250 km , and total height 8900 km with timeofflight 10 8 2900 s. height (km) 6 4 2 0 0 0.5 1 1.5 t (1000 s) 2 2.5 3 (a) (b) (c) (d) 965. Total impulse Total mass Burn time Rocket cross section area Drag coefficient Drag force P = 8.5 N s m0 = 0.054 kg t f = 1.5 s S= d2 4 = 4.5 10 4 m 2 cw = 0.75 D= 1 cw Sv 2 = Kv 2 = 2 10 4 v 2 N 2 where = 1.2 kg/m 3 is density of air and v is rocket's speed Rocket exhaust speed u = 800 m/s a) The total mass of propellant is m = P = 0.0106 kg u DYNAMICS OF A SYSTEM OF PARTICLES 329 Since m ~ 20% m0 , we will assume that the rocket's mass is approximately constant in this problem. The equation of motion of rocket is m0 (where = dv =  mo g + u  Kv 2 dt m = 7.1 10 3 kg/s is fuel burn rate) tf m0 dv = dt ( u  m0 g )  Kv2 Using the initial condition at t = 0, v = 0, we find v(t) = At burnout, t = t f = 1.5 s , we find v f = v(t f ) = 114.3 m/s The height accordingly is given by h(t) = v(t) dt = 0 ( u  m0 g ) tanh K K ( u  m0 g ) m0 2 k ( u  m g ) m0 0 ln cosh t 2 k m0 At burn out, t = t f , we find the burnout height h f = h(t f ) = 95.53 m b) After the burnout, the equation of motion is m0 dv =  m0 g  Kv 2 dt m0 dv =  dt m0 g + Kv 2 with solution v(t) =  Kg m0 g tan t  C K m0 Using the initial condition at t = t f , v(t) = v f , we find the constant C = 1.43 rad, so v(t) =  and the corresponding height h(t) = h f + v(t) dt = h f + tf t Kg m0 g tan t  1.43 K m0 m0 K Kg  1.43 0.88 + ln cos t m0 330 CHAPTER 9 When the rocket reaches its maximum height (at t = tmax ) the time tmax can be found by setting v(tmax ) = 0. We then find tmax = 7.52 s. And the maximum height the rocket can reach is hmax = h(tmax ) = 334 m c) Acceleration in burnout process is (see v(t) in a)) a(t ) = dv u  m0 g = dt m0 1 k ( u  m g ) 0 cosh 2 t 2 m0 Evidently, the acceleration is maximum when t = 0 and amax = a(t = 0) = u  m0 g = 95.4 m/s 2 m0 d) In the falldown process, the equation of motion is m0 dv = m0 g  Kv 2 , dt With the initial condition t = tmax , v = 0, we find ( t tmax ) v(t) =  m0 g Kg tanh ( t  tmax ) K m0 (v(t) is negative for t tmax , because then the rocket falls downward) The height of the rocket is h = hmax + Kg m0 ln cosh ( t  tmax ) K m0 tmax t v(t) dt = hmax  To find the total flighttime, we set h = 0 and solve for t. We find ttotal = 17.56 s e) Putting t = ttotal into the expression of V(t) in part d), we find the speed at ground impact to be vg =  m0 g Kg tanh ( ttotal  tmax ) = 49.2 m/s K m0 966. If we take into account the change of the rocket's mass with time m = m0  t , where is the fuel burn rate, = 7.1 10 3 kg/s as calculated in problem 965. The equation of motion for the rocket during boost phase is DYNAMICS OF A SYSTEM OF PARTICLES 331 ( m0  t ) dt = u  Kv2 Integrating both sides we obtain finally dv dv dt = Kv  u t  m0 2 1 + C t  m 2 ( 0) v(t) = 1  C ( t  m ) 2 0 u Ku K Ku where C is a constant. Using the initial condition v(t ) = 0 at t = 0 , we can find C and the velocity is 2 1  1  t m0 u v(t) = K 2 1 + 1  t m0 a) The rocket speed at burnout is (note t f = 1.5 s ). Ku Ku v f = v(t f ) = 131.3 m/s b) The distance the rocket has traveled to the burnout is h f = v(t) dt = 108.5 m tf 967. From Equation (9.167) we have H bo =  g m0  m2 f 2 2 ( ) + u m f mf ln + m0  m f m0 Using numerical values from Example 9.12 = 1.42 10 4 kg/s m0 = 2.8 106 kg m f = 0.7 106 kg u = 2.600 m/s we find H bo = 97.47 km . From Equation (9.168) we find vbo =  g(m0  m f ) m + u ln o mf vbo = 2125 m/s 332 CHAPTER 9 CHAPTER 10 Motion in a Noninertial Reference Frame 101. (1) (2) The accelerations which we feel at the surface of the Earth are the following: : Gravitational 980 cm/sec 2 Due to the Earth's rotation on its own axis: 2 rad/day r = 6.4 10 cm 86400 sec/day 2 ( ( 8 ) 2 = 6.4 10 8 7.3 10 5 (3) Due to the rotation about the sun: ) ( ) 2 = 3.4 cm/sec 2 r = 1.5 10 2 ( ( 13 2 rad/year cm 86400 365 sec/day ) 2 7.3 10 5 2 = 1.5 1013 = 0.6 cm/sec 365 ) 2 102. The fixed frame is the ground. y a x 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): a f = R f + a r + r + ( r ) + 2 v r 333 334 Now we have R f =  a cos i + a sin j r = r0 i = Substituting gives V k r0 vr = ar = 0 = a k r0 CHAPTER 10 a f =  a cos i + a sin j + a j  v2 i r0 (1) v2 a f = i + a cos + j ( sin + 1) a r0 We want to maximize a f , or alternatively, we maximize a f : 2 af 2 = v4 2 av 2 cos + a 2 + 2 a 2 sin + a 2 sin 2 + a 2 cos 2 + 2 r0 r0 v4 2av 2 + 2a2 + cos + a 2 sin 2 r02 r0 = d af d 2 = 2av 2 cos + 2a 2 cos r0 ar0 v2 v2 a 2 r02 + v 4 = 0 when tan = (Taking a second derivative shows this point to be a maximum.) tan = and sin = Substituting into (1) ar0 a r + v4 2 2 0 ar0 implies cos = v2 v2 av 2 a f = i + r0 a 2 r02 + v 4 This may be written as ar0 + j + 1 a a 2 r02 + v 4 a f = a + a 2 + v 4 r02 MOTION IN A NONINERTIAL REFERENCE FRAME 335 A This is the maximum acceleration. The point which experiences this acceleration is at A: where tan = ar0 v2 103. We desire Feff = 0 . From Eq. (10.25) we have Feff = F  mR f  m r  m ( r )  2m v r r 0 The only forces acting are centrifugal and friction, thus s mg = m 2 r , or r= s g 2 104. Given an initial position of (0.5R,0) the initial velocity (0,0.5R) will make the puck motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise in a circle of radius 0.5R. Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the problem, we will forgo such a solution here. 105. The effective acceleration in the merrygoround is given by Equation 10.27: x = 2 x + 2 y y = 2 y  2 x These coupled differential equations must be solved with the initial conditions x0 x ( 0 ) = 0.5 m , y0 y ( 0 ) = 0 m , and x ( 0 ) = y ( 0 ) = v0 2 m s 1 , since we are given in the problem that the initial velocity is at an angle of 45 to the xaxis. We will vary v0 over some range that we know satisfies the condition that the path cross over ( x0 , y0 ) . We can start by (1) (2) looking at Figures 104e and 104f, which indicate that we want v0 > 0.47 m s 1 . Trial and error can find a trajectory that does loop but doesn't cross its path at all, such as v0 = 0.53 m s 1 . From here, one may continue to solve for different values of v0 until the wanted crossing is eyeballsuitable. This may be an entirely satisfactory answer, depending on the inclinations of the instructor. An interpolation over several trajectories would show that an accurate answer to the problem is v0 = 0.512 m s 1 , which exits the merrygoround at 3.746 s. The figure shows this solution, which was numerically integrated with 200 steps over the time interval. 336 1 CHAPTER 10 0.5 y (m) 0 0.5 1 1 0.5 0 x (m) 0.5 1 106. z z = f(r) m r Consider a small mass m on the surface of the water. From Eq. (10.25) Feff = F  mR f  m r  m ( r )  2m v r In the rotating frame, the mass is at rest; thus, Feff = 0 . The force F will consist of gravity and the force due to the pressure gradient, which is normal to the surface in equilibrium. Since R f = = v r = 0 , we now have 0 = mg + Fp  m ( r ) where Fp is due to the pressure gradient. Fp m2r mg Since Feff = 0 , the sum of the gravitational and centrifugal forces must also be normal to the surface. Thus = . tan = tan = 2r g MOTION IN A NONINERTIAL REFERENCE FRAME 337 but tan = Thus dz dr z= 2 2g r 2 + constant The shape is a circular paraboloid. 107. For a spherical Earth, the difference in the gravitational field strength between the poles and the equator is only the centrifugal term: g poles  gequator = 2 R For = 7.3 10 5 rad s 1 and R = 6370 km, this difference is only 34 mm s 2 . 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 pulls inward at the poles more than it does at the equator. The next level of analysis for the undaunted is the "quadrupole" correction to the gravitational potential of the Earth, which is beyond the scope of the text. 108. z y x 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 = (  cos , 0, sin ) so that the acceleration due to the Coriolis force is a = 2 r = 2 ( 0,  z cos , 0 ) v = ( 0, 0, z ) (1) (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: y = 2 z cos Now, the equation of motion for the particle along the z axis is z = v0  gt z = v0 t  where v0 is the initial velocity and is equal to h: 1 2 gt 2 (4) (5) (3) 2gh if the highest point the particle can reach is (6) v0 = 2 gh From (3), we have y = 2 z cos + c but the initial condition y ( z = 0 ) = 0 implies c = 0. Substituting (5) into (7) we find 1 y = 2 cos v0 t  gt 2 2 = cos gt 2  2v0 t 2 (7) ( ) (8) Integrating (8) and using the initial condition y(t = 0) = 0, we find 1 y = cos gt 2  v0 t 2 3 From (5), the time the particle strikes the ground (z = 0) is 1 0 = v0  gt t 2 so that t= Substituting this value into (9), we have 2v0 g (9) (10) 1 8v 3 4v 2 y = cos g 30  v0 20 g 3 g v3 4 =  cos 0 3 g2 If we use (6), (11) becomes (11) MOTION IN A NONINERTIAL REFERENCE FRAME 339 4 8h3 y =  cos 3 g The negative sign of the displacement shows that the particle is displaced to the west. (12) 109. Choosing the same coordinate system as in Example 10.3 (see Fig. 109), we see that the lateral deflection of the projectile is in the x direction and that the acceleration is ax = x = 2 z vy = 2 ( sin )(V0 cos ) Integrating this expression twice and using the initial conditions, x ( 0 ) = 0 and x ( 0 ) = 0 , we obtain x ( t ) = V0 t 2 cos sin Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have z ( t ) = V0 t sin  1 2 gt 2 (1) (2) (3) from which the time T of impact is obtained by setting z = 0: T= 2V0 sin g (4) Substituting this value for T into (2), we find the lateral deflection at impact to be x (T ) = 4V03 sin cos sin 2 2 g (5) 1010. 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 2 x vy so that z = 2 V0 cos cos  g (1) from which the time of flight is obtained by integrating twice, using the initial conditions, and then setting z = 0: T = 2V0 sin g  2V0 cos cos (2) Now, the acceleration in the y direction is ay = y = 2 x vz = 2 (  cos )(V0 sin  gt ) Integrating twice and using the initial conditions, y ( 0 ) = V0 cos and y ( 0 ) = 0 , we have (3) 340 CHAPTER 10 y (t) = 1 gt 3 cos  V0 t 2 cos sin + V0 t cos 3 (4) Substituting (2) into (4), the range R is R = 4V03 sin 3 cos 2V02 cos cos 8 V03 g sin 3 cos  + 3 ( g  2V0 cos cos ) 3 ( g  2V0 cos cos ) 2 g  2V0 cos cos (5) We now expand each of these three terms, retaining quantities up to order but neglecting all quantities proportional to 2 and higher powers of . In the first two terms, this amounts to neglecting 2V0 cos cos compared to g in the denominator. But in the third term we must use 2V02 cos sin 2V0 cos cos g 1  g 2V0 2V02 cos sin 1 + cos cos g g 4V03 sin cos 2 cos 2 g = R0 + (6) where R is the range when Coriolis effects are neglected [see Example 2.7]: 0 R0 = 2V02 cos sin g (7) The range difference, R = R  R0 , now becomes R = 4V03 1 cos sin cos 2  sin 3 2 g 3 (8) Substituting for V0 in terms of R from (7), we have, finally, 0 R = 2R0 1 cos cot1 2  tan 3 2 g 3 (9) MOTION IN A NONINERTIAL REFERENCE FRAME 341 1011. d= R R sin 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 that will always be due south. As the missile travels towards its intended destination, the Earth will rotate underneath it, thus causing it to miss. This distance is: = (transverse velocity of Earth at current latitude) (missile's time of flight) = R sin T = d R d sin R v (1) (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 would measure from the Earth. The error this causes in will be small as long as the miss distance is small. Using R = 6370 km, = 7.27 10 5 rad s 1 , 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. Doing this problem in the rotating frame is tricky 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 entirely wrong on the second part. Correct treatment in the rotating frame would at minimum require numerical methods. 1012. r0 Fs z x 342 CHAPTER 10 Using the formula Feff = ma f  m ( r )  2m v r (1) we try to find the direction of Feff when ma f (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 = ( cos , 0, sin ) r = (0, 0, R) ma f = (0, 0,  mg0 ) Hence, we have r = R cos e y and (1) becomes (3) vr = 0 (2) ex Feff =  mg0 ez  m  cos 0 from which, we have ey ez (4) 0 sin R cos 0 Feff =  mg0 ez + mR 2 sin cos ex + mR 2 cos 2 e z Therefore, ( Ff )x = mR 2 sin cos 2 2 ( Ff )z =  mg 0 + mR cos (5) (6) The angular deviation is given by tan = ( Ff )x ( Ff )z = R 2 sin cos g 0  R 2 cos 2 (7) Since is very small, we can put . Then, we have = R 2 sin cos g 0  R 2 cos 2 45 . (8) It is easily shown that is a maximum for Using R = 6.4 10 8 cm , = 7.3 10 5 sec 1 , g = 980 cm/sec 2 , the maximum deviation is 1.7 0.002 rad 980 (9) MOTION IN A NONINERTIAL REFERENCE FRAME 343 1013. z z Earth x x The small parameters which govern the approximations that need to be made to find the southerly deflection of a falling particle are: and height of fall h = R radius of Earth (1) centrifugal force R 2 = g0 purely gravitational force (2) The purely gravitational component is defined the same as in Problem 1012. Note that although both and are small, the product = h 2 g 0 is still of order 2 and therefore expected to contribute to the final answer. Since the plumb line, which defines our vertical direction, 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 Northern Hemispherecentric will have its xaxis towards the south, its yaxis towards the east, and its zaxis in the direction of the plumb line. The primed coordinates will share both its origin and its yaxis with its unprimed counterpart, with the zand xaxes rotated to make the zaxis an outward radial (see figure). The rotation can be described mathematically by the transformation x = x cos + z sin y = y z =  x sin + z cos where (3) (4) (5) as found from Problem 1012. R 2 sin cos g (6) a) The acceleration due to the Coriolis force is given by a X 2 v Since the angle between and the zaxis is , (7) is most appropriately calculated in the primed coordinates: (7) 344 x = 2 y sin y = 2 ( z cos + x sin ) z = 2 y cos In the unprimed coordinates, the interesting component is CHAPTER 10 (8) (9) (10) x = 2 y ( sin cos + cos sin ) At our level approximation this becomes 2 y sin (11) x (12) Using the results for y and z , which is correct to order (also found from Example 10.3), x 2 2 gt 2 sin cos (13) Integrating twice and using the zeroth order result for the timeoffall, t = 2 h g , we obtain for the deflection dX = 2 h2 2 sin cos 3 g (14) b) The centrifugal force gives us an acceleration of a c  ( r ) The component equations are then (15) x = 2 sin x sin + ( R + z ) cos y = 2 y z = 2 cos x sin + ( R + z ) cos  g0 (16) (17) (18) where we have included the pure gravitational component of force as well. Now transform to the unprimed coordinates and approximate x 2 ( R + z ) sin cos  g0 sin We can use Problem 1012 to obtain sin to our level of approximation sin (19) R 2 sin cos g0 x 2 z sin cos (20) The prompts a cancellation in equation (19), which becomes simply (21) Using the zeroth order result for the height, z = h  gt 2 2 , and for the timeoffall estimates the deflection due to the centrifugal force dc 5 h2 2 sin cos 6 g (22) MOTION IN A NONINERTIAL REFERENCE FRAME 345 c) Variation in gravity causes the acceleration ag  GM r + g0 k r3 (23) where r = x i + y j + ( R + z ) k is the vector pointing to the particle from the center of the spherical Earth. Near the surface r 2 = x 2 + y 2 + ( z + R) 2 R2 + 2Rz (24) so that (23) becomes, with the help of the binomial theorem, ag Transform and get the x component  g0 ( x i + y j  2 z k ) R (25) x = g0 (  x cos + 2z sin ) R g0  ( x cos  z sin ) cos + 2 ( x sin + z cos ) sin R g0 (  x + 3z sin ) R (26) (27) (28) Using (20), x 3 2 z sin cos (29) where we have neglected the x R term. This is just thrice the part (b) result, dg 5 h2 2 sin cos 2 g (30) Thus the total deflection, correct to order 2 , is d 4 h2 2 sin cos g (31) (The solution to this and the next problem follow a personal communication of Paul Stevenson, Rice University.) 1014. The solution to part (c) of the Problem 1013 is modified when the particle is dropped down a mineshaft. The force due to the variation of gravity is now ag  g0 r + g0 k R (1) As before, we approximate r for near the surface and (1) becomes ag In the unprimed coordinates,  g0 ( x i + y j + z k ) R (2) 346 CHAPTER 10 x  g0 x R (3) To estimate the order of this term, as we probably should have done in part (c) of Problem 1013, we can take x ~ h 2 2 g , so that x ~ 2h h R (4) which is reduced by a factor h R 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 the total deflection in this approximation is d 3 h2 2 sin cos 2 g (5) 1015. The Lagrangian in the fixed frame is L= 1 mv 2  U rf f 2 ( ) (1) where v f and rf are the velocity and the position, respectively, in the fixed frame. Assuming we have common origins, we have the following relation v f = v r + rr where vr and rr are measured in the rotating frame. The Lagrangian becomes (2) L= The canonical momentum is m 2 2 v + 2v r ( rr ) + ( rr )  U ( rr ) r 2 L = mv r + m ( rr ) v r (3) pr The Hamiltonian is then (4) H v r pr  L = 1 1 2 mvr2  U ( rr )  m ( rr ) 2 2 (5) H is a constant of the motion since L t = 0 , but H E since the coordinate transformation equations depend on time (see Section 7.9). We can identify Uc =  1 2 m ( rr ) 2 (6) as the centrifugal potential energy because we may find, with the use of some vector identities, U c = m 2 2 2 rr  ( rr ) 2 (7) (8) = m 2 rr  ( rr ) MOTION IN A NONINERTIAL REFERENCE FRAME 347 (9) =  m ( rr ) which is the centrifugal force. Computing the derivatives of (3) required in Lagrange's equations d L = ma r + m v r dt v r L = m ( v r ) rr  (U c + U ) rr =  m ( v r )  m ( rr )  U The equation of motion we obtain is then (10) (11) (12) ma r = U  m ( rr )  2m ( v r ) (13) If we identify Feff = ma r and F = U , then we do indeed reproduce the equations of motion given in Equation 10.25, without the second and third terms. 1016. 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 963. The only thing we do here is add an acceleration caused by the Coriolis force, and rework every part of the problem over again. This is conceptually simple 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 a c = 2 vy sin i  ( vx sin + vz cos ) j + vy cos k where we have chosen the usual coordinates as shown in Figure 109 of the text. a) Our acceleration is (1) a =  gk + a C As a check, we find that the height reached is 1800 km, in good agreement with the result of 77 km, to the west. (2) Problem 963(a). The deflection at this height is found to be b) This is mildly tricky. The correct treatment says that the equation of motion with air resistance is (cf. equation (2) of Problem 963 solution) v a =  g k + 2 vt The deflection is calculated to be c) d) v + aC (3) 8.9 km. 10 km. 160 km. Adding the vaiation due to gravity gives us a deflection of Adding the variation of air density gives us a deflection of 348 CHAPTER 10 Of general note is that the deflection in all cases was essentially westward. The usual small deflection to the north did not contribute significantly to the total transverse deflection at this precision. All of the heights obtained agreed well with the answers from Problem 963. Inclusion of the centrifugal force also does not change the deflections to a significant degree at our precision. 1017. Due to the centrifugal force, the water surface of the lake is not exactly perpendicular to the Earth's radius (see figure). B m Ta n g ge nt Wa ter sur to fac e Ea rth A su r C fa ce The length BC is (using cosine theorem) BC = AC 2 + (mg ) 2  2 ACmg cos where AC is the centrifugal force AC = m 2 R cos with = 47 and Earth's radius R 6400 km , The angle that the water surface is deviated from the direction tangential to the Earth's surface is BC AC = sin sin sin = AC sin = 4.3 105 BC So the distance the lake falls at its center is h = r sin where r = 162 km is the lake's radius. So finally we find h = 7 m. MOTION IN A NONINERTIAL REFERENCE FRAME 349 1018. Let us choose the coordinate system Oxyz as shown in the figure. O y x y x The projectile's velocity is vx v0 cos v = v y = v0 sin  gt 0 0 The Earth's angular velocity is where = 37  cos =  sin 0 So the Coriolis acceleration is where = 50 ac = 2v = 2v0 cos sin + 2 ( v0 sin  gt ) cos e z The velocity generated by Coriolis force is ( ) vc = ac dt = 2v0 t ( cos sin  sin cos )  gt 2 cos 0 t And the distance of deviation due to the Coriolis force is gt 3 cos zc = vc dt = v0 t sin (  )  3 0 t 2 The flight time of the projectile is t = distance due to Coriolis force to be 2v0 sin . If we put this into zc , we find the deviation 2 zc ~ 260 m 350 1019. CHAPTER 10 The Coriolis force acting on the car is Fc = 2m v Fc = 2mv sin where = 65, m = 1300 kg, v = 100 km/hr. So Fc = 4.76 N. 1020. Given the Earth's mass, M = 5.976 1024 kg , the magnitude of the gravitational field vector at the poles is g pole = GM = 9.866 m/s 2 2 R pole The magnitude of the gravitational field vector at the equator is g eq = GM  2 R e q = 9.768 m/s 2 2 Req where is the angular velocity of the Earth about itself. If one use the book's formula, we have g ( = 90) = 9.832 m/s 2 at the poles and g ( = 0) = 9.780 m/s 2 at the equator 1021. The Coriolis acceleration acting on flowing water is ac = 2v ac = 2v sin Due to this force, the water is higher on the west bank. As in problem 1017, the angle that the water surface is deviated from the direction tangential to Earth's surface is sin = ac g +a 2 2 c = 2v sin g + 4v sin 2 2 2 2 = 2.5 105 The difference in heights of the two banks is h = sin = 1.2 103 m where = 47 m is the river's width. The Coriolis acceleration is ac = 2v . This acceleration ac pushes lead bullets 1022. eastward with the magnitude ac = 2v cos = 2 gt cos , where = 42. The velocity generated by the Coriolis force is MOTION IN A NONINERTIAL REFERENCE FRAME 351 vc (t ) = a dt = gt 2 cos and the deviation distance is xc = vc (t ) dt = The falling time of the bullet is t = gt 3 cos 3 2h g . So finally xc = 8h3 3 g cos = 2.26 103 m 352 CHAPTER 10 CHAPTER 11 Dynamics of Rigid Bodies 111. The calculation will be simplified if we use spherical coordinates: x = r sin cos y = r sin sin z = r cos z (1) y x Using the definition of the moment of inertia, I ij = ( r ) ij we have x k 2 k  xi x j dv (2) I 33 = r 2  z 2 dv = r  r cos r dr d ( cos ) d 2 2 2 2 ( ( ) ) (3) or, I 33 = r 4 dr 0 R +1 1 (1  cos ) d ( cos ) d 2 0 2 = 2 R5 4 5 3 (4) 353 354 CHAPTER 11 The mass of the sphere is M= Therefore, I 33 = 2 MR 2 5 (6) 4 R3 3 (5) Since the sphere is symmetrical around the origin, the diagonal elements of {I} are equal: I11 = I 22 = I 33 =
2 MR2 5 (7) A typical offdiagonal element is I12 = (  xy ) dv =  r sin sin cos r dr d ( cos ) d 2 2 2 (8) This vanishes because the integral with respect to is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is I11  I 0 I 22  I 0 0 I 33  I 0 0 From (9) and (7), we have =0 (9) 0 I1 = I 2 = I 3 = 2 MR2 5 (10) 112. a) Moments of inertia with respect to the xi axes: x3 = x3 R h CM x1 x1 x2 x2 It is easily seen that I ij = 0 for i j. Then the diagonal elements I ii become the principal moments I i , which we now calculate. The computation can be simplified by noting that because of the symmetry, I1 = I 2 I 3 . Then, I1 = I 2 = I1 + I 2 = 2 2 ( 2x 2 3 2 2 + x1 + x2 dv ) (1) DYNAMICS OF RIGID BODIES 355 which, in cylindrical coordinates, can be written as I1 = I 2 = where 2 2 0 d dz 0 h Rz h 0 (r 2 + 2z 2 rdr ) (2) = M 3M = V R2 h (3) Performing the integration and substituting for , we find I1 = I 2 = I 3 is given by 2 2 I 3 = x1 + x2 dv = r 2 rdr d dz 3 M R2 + 4 h 2 20 ( ) (4) ( ) (5) from which I3 = 3 MR 2 10 (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 x axis. The coordinates of 3 the center of mass are (0 , 0, z0 ) , where z0 = Then, using Eq. (11.49), x dv = 3 h dv 4 3 (7) I ij = I ij  M a 2 ij  ai a j In the present case, a1 = a2 = 0 and a3 = ( 3 4 ) h , so that (8) I1 = I1  I2 = I2  I3 = I3  9 3 1 Mh 2 = M R2 + h2 16 20 4 9 3 1 Mh 2 = M R2 + h 2 16 20 4 3 MR2 10 113. The equation of an ellipsoid is 2 2 2 x1 x2 x3 + 2 + 2 =1 a2 b c (1) 356 CHAPTER 11 which can be written in normalized form if we make the following substitutions: x1 = a , x2 = b , x3 = c Then, Eq. (1) reduces to (2) 2 +2 + 2 = 1 This is the equation of a sphere in the (,,) system. If we denote by dv the volume element in the xi system and by d the volume element in the (,, ) system, we notice that the volume of the ellipsoid is (3) V = dv = dx1 dx2 dx3 = abc d d d 4 = abc d = abc 3 because d is just the volume of a sphere of unit radius. The rotational inertia with respect to the x3 axis passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by I3 = = M V (4) (x 2 1 2 + x2 dv ) M abc a 2 2 + b 2 2 d V ( ) (5) In order to evaluate this integral, consider the following equivalent integral in which z = r cos : a 2 z 2 dv = a 2 z 2 ( r dr r sin d d ) = a2 2 0 d cos 2 sin d 0 R =1 0 r 4 dr = a 2 2 = Therefore, 4 a 2 15 2 1 3 5 (6) (a 2 2 + b 2 2 d = ) 4 2 a + b2 15 ( ) (7) and I3 = 1 M a2 + b 2 5 ( ) (8) Since the same analysis can be applied for any axis, the other moments of inertia are DYNAMICS OF RIGID BODIES 357 I1 = 1 M b2 + c2 5 ( ) ) (9) 1 I 2 = M a2 + c2 5 ( 114. The linear density of the rod is = m (1) For the origin at one end of the rod, the moment of inertia is I = x 2 dx = 0 m 3 3 = m 3 2 (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 I = ma 2 Equating (2) and (3), we find a= This is the radius of gyration. 115. z J a Q M za (3) 3 (4) a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short interval of time so that J = F ( t ) dt The equations of motion are dp =F dt dL =rF dt (1) (2) (3) 358 CHAPTER 11 which, for this case, yield p = F ( t ) dt = J L = r F ( t ) dt = r J Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have p = MVCM = J; L = I 0 = r J = ( z  a ) J (4) (5) (6) so that VCM = J M (7) and = J ( z  a) I0 (8) where I 0 = ( 2 5) Ma 2 . The velocity of any point a on the ball is given by Eq. (11.1): v = VCM + r For the point of contact Q, this becomes (9) v Q = VCM  a = J J (10) J 5 ( z  a) 1 2a M Then, for rolling without slipping, v Q = 0 , and we have 2a = 5 ( z  a) so that z= 7 a 5 (11) (12) b) Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinning 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 normally on the rail, as in the diagram. We have the following relationships: DYNAMICS OF RIGID BODIES 359 y VCM x Before Collision Linear Momentum px =  MVCM py = 0 Angular Momentum Lx = 0 Ly = * Lz = 0 After Collision px = + MVCM py = 0 Lx = 0 Ly =  Ly Lz = 0 * The relation between Ly and VCM depends on whether or not slipping occurs. Then, we have p = 2 px = J = 2 MVCM L = 2Ly = 2I 0 = J ( z  a) so that 2I 0 = 2 MVCM ( z  a ) from which (13) (14) (15) za= I0 2 Ma 2 2 a2 = = MVCM 5 MVCM 5 VCM (16) If we assume that the ball rolls without slipping before it contacts the rail, then VCM = a , and we obtain the same result as before, namely, za= or, 2 a 5 (17) z= 7 a 5 (18) Thus, the height of the rail must be at a height of ( 2 5) a above the center of the ball. 116. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have Is = 2 MR 2 5 (see Problem 111) (1) 360 CHAPTER 11 For the hollow sphere, R sin I h = d ( R sin ) R2 sin d 2 0 0 2 = 2 R 4 sin 3 d 0 8 = R 4 3 or, using 4 R 2 = M , we have Ih = 2 MR 2 3 (2) Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is T= 1 1 M y 2 + I 2 2 2 (3) where y is the measure of the distance along the plane. The potential energy is U = Mg (  y ) sin (4) where is the length of the plane and is the angle of inclination of the plane. Now, y = R, so that the Lagrangian can be expressed as L= 1 1 I 2 M y2 + y + Mgy sin 2 2 R2 (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 y= gMR 2 sin MR2 + I (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 361 117. x R d r The force between the force center and the disk is, from the figure F =  kr (1) Only the component along x does any work, so that the effective force is Fx =  kr sin =  kx . This corresponds to a potential U = kx 2 2 . The kinetic energy of the disk is T= 1 1 3 Mx 2 + I 2 = Mx 2 2 2 4 (2) where we use the result I = MR2 2 for a disk and dx = R d. Lagrange's equations give us 3 Mx + kx = 0 2 This is simple harmonic motion about x = 0 with an angular frequency of oscillations (3) = 2k 3m (4) 118. x3 h w x1 x2 d We let x be the vertical axis in the fixed system. This would be the axis (i.e., the hinge line) of 3 the door if it were properly hung (no selfrotation), as indicated in the diagram. The mass of the door is M=whd. The moment of inertia of the door around the x3 axis is m 1 2 2 I3 = dh w d dw = 3 Mw whd 0 0 h w (1) where the door is considered to be a thin plate, i.e., d w,h. 362 CHAPTER 11 The initial position of the selfclosing door can be expressed as a twostep transformation, starting with the position in the diagram above. The first rotation is around the x axis through 1 an angle and the second rotation is around the xaxis through an angle : 1 x3 w h x3 x3 x3 x3 = x3 w3 = w x2 x1 x2 x1 = x1 x2 x1 x1 x2 x2 x2 The x axes are the fixedsystem axes and the xi axes are the body system (or rotating) axes 1 which are attached to the door. Here, the Euler angle is zero. The rotation matrix that transforms the fixed axes into the body axes ( xi xi ) is just Eq. (11.99) with = 0 and since this rotation is performed clockwise rather than counterclockwise as in the derivation of Eq. (11.99): cos =  sin 0 cos sin cos cos sin  sin sin  sin cos cos (2) The procedure is to find the torque acting on the door expressed in the fixed coordinate system and then to obtain the x3 component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle 0 , the rotation is in the direction to decrease . According to Eq. (11.119), I 3 3 = N 3 = I 3 (3) where 1 = 2 = 0 since = = 0 . In the body ( xi ) system the coordinates of the center of mass of the door are 0 1 R = w 2 h (4) where we have set the thickness equal to zero. In the fixed ( x) system, these coordinates are i obtained by applying the inverse transformation 1 to R; but 1 = t , so that  w sin 1 R = R = w cos cos + h sin 2  w sin cos + h cos t (5) Now, the gravitational force acting on the door is downward, and in the x coordinate system i is F =  Mg e 3 (6) DYNAMICS OF RIGID BODIES 363 There the torque on the door, expressed in the fixed system, is N = R F e1 1 =  Mg  w sin 2 0 e2 w cos cos + h sin 0 e 3  w sin cos + h cos 1 w cos cos + h sin 1 =  Mg w sin 2 0 (7) so that in the body system we have w cos cos 2 + h sin cos + w cos sin 2 1 N = N =  Mg  h sin sin 2 w sin sin (8) Thus, N3 = 1 Mgw sin sin 2 (9) and substituting this expression into Eq. (3), we have  1 1 Mgw sin sin = I 3 = Mw 2 2 3 (10) where we have used Eq. (1) for I 3 . Solving for , = 3 g sin sin 2w (11) This equation can be integrated by first multiplying by : dt = 2 = 1 2 = 3 g sin sin dt 2w (12) 3 g sin cos 2w where the integration constant is zero since cos = 0 when = 0 . Thus, = 3g sin cos w (13) We must choose the negative sign for the radical since < 0 when cos > 0 . Integrating again, from = 90 to = 0, 0 2 d = cos T 3g sin dt w 0 (14) 364 where T = 2 sec. Rewriting Eq. (14), CHAPTER 11 0 2 d =T cos 3g sin w (15) Using Eq. (E. 27a), Appendix E, we find 1 4 1 2 cos d = 2 3 0 4 2 (16) From Eqs. (E.20) and (E.23), 1 1 1 = 1 = 0.906 4 4 4 1 = 3.624 4 And from Eqs. (E.20) and (E.24), 3 3 3 = 1 = 0.919 4 4 4 3 = 1.225 4 Therefore, (17) (18) 0 2 d 3.624 = = 2.62 2 1.225 cos (19) Returning to Eq. (15) and solving for sin , sin = w 2 ( 2.62) 3 gT 2 (20) Inserting the values for g, w(= 1m), and T(= 2 sec), we find = sin 1 ( 0.058 ) or, 3.33 (21) DYNAMICS OF RIGID BODIES 365 119. y C Q C P R a O x The diagram shows the slab rotated through an angle 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 only small departures from = 0, we can write QP R Therefore, the coordinates of C are (see enlarged diagram below) (1) r = OA + AC so that a sin  R cos 2 a y = R + cos + R sin 2 (2) x = R + (3) C C R A Q R P O Consequently, 366 x = R + a cos  R cos + R sin 2 CHAPTER 11 a = cos + R sin 2 a y =  R + sin + R cos + R sin 2 a =  sin + R cos 2 from which a2 x 2 + y 2 = + R 2 2 2 4 (4) The kinetic energy is T= 1 1 M x 2 + y 2 + I 2 2 2 ( ) (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 zaxis: I= 1 M 12 ( 2 + a2 ) (6) Therefore, T= 1 f1 ( ) 2 2 (7) where a2 f1 ( ) = M + R 2 2 + I 4 (8) The potential energy is U = Mgy =  f 2 ( ) (9) where a f 2 ( ) =  Mg R + cos + R sin 2 (10) and where Eq. (3) has been used for y. The Lagrangian is L= 1 f1 ( ) 2 + f 2 ( ) 2 (11) The Lagrange equation for is DYNAMICS OF RIGID BODIES 367 d L L  =0 dt (12) Now, L = f1 ( ) d L = f1 ( ) + f1 ( ) dt a2 = M + R 2 2 + I + 2 MR 2 2 4 L 1 = f1( ) 2 + f 2 ( ) 2 a = MR 2 2 + Mg R + sin  R cos  R sin 2 (13) (14) Combining, we find a2 2 2 2 2 M + R + I + MR  Mg R + 4 a sin  R cos  R sin = 0 2 (15) For the case of small oscillations, 2 and 2 , so that Eq. (15) reduces to a Mg R  2 + =0 2 Ma +I 4 The system is stable for oscillations around = 0 only if a Mg R  2 = 2 > 0 2 Ma +I 4 This condition is satisfied if R  a 2 > 0 , i.e., R> Then, the frequency is a Mg R  2 2 Ma 1 + M 2 + a2 4 12 a 2 (16) (17) (18) = ( ) (19) Simplifying, we have 368 CHAPTER 11 = a 12 g R  2 ( 2 + 4a2 ) (20) According to Eqs. (9) and (10), the potential energy is a U ( ) = Mg R + cos + R sin 2 This function has the following forms for R > a 2 and R < a 2 : R> a 2 U() (21) Mg R + a 2 U() R< a 2 /2 /2 To verify that a stable condition exists only for R > a 2 , we need to evaluate 2U 2 at = 0: U a = Mg  sin + R cos 2 2U a = Mg  cos + R cos  R sin 2 2 and a 2U = Mg R  2 = 0 2 so that 2U a > 0 if R > 2 2 1110. z m R (22) (23) (24) (25) When the mass m is at one pole, the z component of the angular momentum of the system is Lz = I = 2 MR 2 5 (1) After the mass has moved a distance vt = R 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 2 Lz = MR 2 + mR 2 sin 2 5 (2) where 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 MR 2 + mR2 sin 2 5 2 MR 2 5 (3) Substituting = vt R and integrating over the time interval during which the mass travels from one pole to the other, we have t= R V = t=0 2 MR 2 + mR 2 sin 2 ( vt R) 5 2 MR2 5 dt (4) Making the substitutions, vt R u, dt = ( R v ) du we can rewrite (4) as (5) = 0 R du 2 v 2 2 2 MR + mR sin u 5 2 2 MR 2 5 2R = v 1 + sin 0 du 2 u (6) where 5m 2 M and where we have used the fact that the integrand is symmetric around u = 2 to write as twice the value of the integral over half the range. Using the identity sin 2 u = we express (6) as 2R = v or, changing the variable to x = 2u, 2 1 (1  cos 2u) 2 (7) 0 du 1 1 1 +  cos 2u 2 2 (8) = R v 0 dx 1 1 1 +  cos x 2 2 (9) Now, we can use Eq. (E.15), Appendix E, to obtain 370 (1 + ) tan ( x 2) 2R = tan 1 v 1+ 1+ = CHAPTER 11 0 R v 2M 2 M + 5m 2M 2 M + 5m (10) = T where T = R v is the time required for the particle to move from one pole to the other. If m = 0, (10) becomes ( m = 0) = T Therefore, the angle of retardation is (11) = ( m = 0 )  ( m) or, (12) = T 1  2M 2 M + 5m (13) 1111. a) No sliding: 2 2 P P From energy conservation, we have mg 2 = mg 2 + 1 1 2 mvC.M. + I 2 2 2 (1) where vCM is the velocity of the center of mass when one face strikes the plane; vC.M. is related to by vCM = 2 (2) I is the moment of inertia of the cube with respect to the axis which is perpendicular to one face and passes the center: I= Then, (1) becomes 1 m 6 2 (3) DYNAMICS OF RIGID BODIES 371 mg 2 from which, we have ( 1 1 m 2 2 1 = m 2 2 2 1 = m + 2 2 2 6 3 ) 2 (4) 2 = b) Sliding without friction: 3 g 2 ( 2 1 ) (5) 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. P P While the cube is falling, the distance between the center of mass and the plane is given by y= 2 cos (6) Therefore, the velocity of center of mass when one face strikes the plane is = 0 = 4 y 2 sin = 4 = 1 1 = 2 2 (7) From conservation of energy, we have mg from which we have 1 1 11 = mg + m  + m 2 2 2 6 2 2 2 2 2 (8) 2 = 12 g 5 ( 2 1 ) (9) 1112. According to the definition of the principal moments of inertia, 2 I j + I k = xi2 + xk dv + xi2 + x 2 dv j 2 = x 2 + xk dv + 2 xi2 dv j ( ) ( ) ( ) = I i + 2 xi2 dv since (1) x we have 2 i dv > 0 372 I j + Ik Ii CHAPTER 11 (2) 1113. We get the elements of the inertia tensor from Eq. 11.13a: 2 2 I11 = m x ,2 + x ,3 ( ) ( ) = 3m b 2 + 4 m 2b 2 + 2m b 2 = 13mb 2 Likewise I 22 = 16 mb 2 and I 33 = 15mb 2 I12 = I 21 =  m x ,1 x ,2 ( ) ( ) = 4 m b 2  2m  b 2 = 2mb 2 Likewise I13 = I 31 = mb 2 and I 23 = I 32 = 4mb 2 Thus the inertia tensor is 13 2 1 {I } = mb 2 16 4 1 4 15 2 ( ) ( ) The principal moments of inertia are gotten by solving 2 1 13  mb 2 16  4 =0 1 4 15  2 Expanding the determinant gives a cubic equation in : 3  44 2 + 622  2820 = 0 Solving numerically gives 1 = 10.00 2 = 14.35 3 = 19.65 Thus the principal moments of inertia are I1 = 10 mb 2 I 2 = 14.35 mb 2 I 3 = 19.65 mb 2 To find the principal axes, we substitute into (see example 11.3): DYNAMICS OF RIGID BODIES 373 (13  i ) 1i  2 2i + 3i = 0 2 1i + (16  i ) 2 i + 4 3 i = 0 1i + 4 2 i + (15  i ) 3 i = 0 For i = 1, we have ( 1 = 10 ) 3 11  2 21 + 31 = 0 2 11  6 21 + 4 31 = 0 11  4 21 + 5 31 = 0 Solving the first for 31 and substituting into the second gives 11 = 21 Substituting into the third now gives 31 =  21 or 11 : 21 : 31 = 1 : 1 :  1 So, the principal axis associated with I1 is 1 ( x + y  z) 3 Proceeding in the same way gives the other two principal axes: i = 2: i = 3: .81x + .29y  .52z .14 x + .77 y + .63z We note that the principal axes are mutually orthogonal, as they must be. 1114. z y x Let the surface of the hemisphere lie in the xy plane as shown. The mass density is given by = M M 3M = = V 2 b 3 2 b 3 3 First, we calculate the center of mass of the hemisphere. By symmetry 374 xCM = yCM = 0 zCM = 1 M CHAPTER 11 z dv v b Using spherical coordinates ( z = r cos , dv = r 2 sin dr d d ) we have zCM = M 2 =0 2 d =0 sin cos d r =0 r 3 dr 3 1 1 3 2 ) b 4 = b = 3( 2 b 2 4 8 We now calculate the inertia tensor with respect to axes passing through the center of mass: z = z y 3 b 8 x By symmetry, I12 = I 21 = I13 = I 31 = I 23 = I 32 = 0 . Thus the axes shown are the principal axes. Also, by symmetry I11 = I 22 . We calculate I11 using Eq. 11.49: 3 I11 = J11  M v 8 2 (1) where J11 = the moment of inertia with respect to the original axes J11 = y 2 + z 2 dv v ( ) = r 2 sin 2 sin 2 + r 2 cos 2 r 2 sin dr d d v ( ) = 3M 2 b 3 r =0 b r 4 dr 2 2 2 2 sin sin + cos d sin d = 0 = 0 2 ( ) 3 Mb 2 = 10 = Thus, from (1) 2 Mb 2 5 2 =0 ( sin 3 + 2 cos 2 sin ) d I11 = I 22 = Also, from Eq. 11.49 2 9 83 Mb 2  Mb 2 = Mb 2 5 64 320 DYNAMICS OF RIGID BODIES 375 I 33 = J 33  M ( 0 ) = J 33 ( I 33 = J 33 should be obvious physically) So I 33 = x 2 + y 2 dv v (
) = r 4 sin 3 dr d d = v 2 Mb 2 5 Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are I11 = I 22 = I 33 = 83 Mb 2 320 2 Mb 2 5 1115. P g We suspend the pendulum from a point P which is a distance from the center of mass. The rotational inertia with respect to an axis through P is 2 I = MR0 + M 2 (1) where R0 is the radius of gyration about the center of mass. Then, the Lagrangian of the system is L = T U = Lagrange's equation for gives I + Mg sin = 0 For small oscillation, sin . Then, (3) I 2  Mg (1  cos ) 2 (2) + or, Mg =0 I (4) + g R + 2 0 2 =0 (5) 376 from which the period of oscillation is CHAPTER 11 = 2 = 2 2 R0 + g 2 (6) If we locate another point P which is a distance from the center of mass such that the period of oscillation is also , we can write 2 R0 + g 2 from which R0 = 2 2 R0 + 2 g = (7) . Then, the period must be = 2 or, + g 2 (8) = 2 + g (9) This is the same as the period of a simple pendulum of the length + . 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 physically. On the other hand, it is necessary to locate the two points for which is the same. 1116. The rotation matrix is cos ( ) =  sin 0 sin cos 0 0 0 1 (1) The moment of inertia tensor transforms according to ( ) = ( ) ( ) ( t ) That is cos ( I ) =  sin 0 sin cos 0 1 ( A + B) 0 2 1 0 ( A  B) 2 1 0 1 ( A  B) 0 2 cos 1 ( A + B) 0 sin 2 C 0 0  sin cos 0 0 0 1 (2) DYNAMICS OF RIGID BODIES 377 cos =  sin 0 sin cos 0 1 1 2 ( A + B) cos + 2 ( A  B) sin 0 1 1 0 ( A  B) cos + ( A + B) sin 2 2 1 0  1 1 ( A + B) sin + ( A  B) cos 2 2 1 1  ( A  B) sin + ( A + B) cos 2 2 0 0 0 C 1 1 2 2 2 ( A + B) cos + ( A  B) cos sin + 2 ( A + B) sin 1 1 =  ( A  B) sin 2 + ( A  B) cos 2 2 2 0 1 1 0 ( A  B) cos 2  ( A  B) sin 2 2 2 1 1 ( A + B) sin 2  ( A  B) sin cos + ( A + B) cos 2 0 2 2 C 0 or 1 2 ( A + B) + ( A  B) cos sin 1 1 ( I ) =  ( A  B) sin 2 + ( A  B) cos 2 2 2 0 If = 4 , sin = cos = 1 2 . Then, A 0 0 ( I ) = 0 B 0 0 0 C 1 1 ( A  B) cos 2  ( A  B) sin 2 0 2 2 1 0 ( A + B)  ( A  B) cos sin 2 C 0 (3) (4) 1117. x3 x2 x1 378 CHAPTER 11 The plate is assumed to have negligible thickness and the mass per unit area is s . Then, the inertia tensor elements are I11 = s (r 2 2  x1 dx1 dx2 ) 2 2 2 = s x2 + x3 dx1dx2 = s x2 dx1 dx2 A ( ) (1) (2) (3) I 22 = s I 33 = s (r (r 2 2 2  x2 dx1 dx2 = s x1 dx1 dx2 B 2 2 2  x3 dx1 dx2 = s x1 + x2 dx1 dx2 ) ) 2 ( ) Defining A and B as above, I 33 becomes I 33 = A + B Also, I12 = s I 21 = s I13 = s I 23 = s Therefore, the inertia tensor has the form A C {I} = C B 0 0 0 0 A + B (4) (  x x ) dx 1 2 1 dx2 C dx2 = C dx2 = 0 = I 31 dx2 = 0 = I 32 (5) (6) (7) (8) (  x x ) dx 2 1 1 3 1 (  x x ) dx 2 3 1 (  x x ) dx 1 (9) 1118. 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 x3 axis , the transformation matrix is just , as defined in Eq. (11.91). Then,  I = I 1 (1) where  t 1 = (2) Therefore, the similarity transformation is cos I =  sin 0 sin cos 0 0 A C 0 C B 1 0 0 0 cos 0 sin A + B 0  sin cos 0 0 0 1 Carrying out the operations and simplifying, we find DYNAMICS OF RIGID BODIES 379 1 ( B  A) sin 2 2 0 0 A + B 2 2 A cos  C sin 2 + B sin 1 {I} = C cos 2 + ( B  A) sin 2 2 0 C cos 2 + A sin 2 + C sin 2 + B cos 2 0 (3) Making the identifications stipulated in the statement of the problem, we see that I11 = A , I 22 = B I12 = I 21 = C and I 33 = A + B = A + B Therefore 0 A C 0 {I } =  C B 0 0 A + B In order that x1 and x2 be principal axes, we require C = 0: C cos 2  or, tan 2 = from which 2C B A (8) 1 ( B  A) sin 2 = 0 2 (7) (5) (4) (6) = 1 2C tan 1 2 B  A (9) Notice that this result is still valid if A = B. Why? (What does A = B mean?) 1119. x2 = /2 = =0 x1 The boundary of the plate is given by r = ke . Any point (, ) has the components x1 = cos x2 = sin (1) 380 The moments of inertia are CHAPTER 11 I1 = A = 0 ke 0 2 x2 d d = sin 2 d 0 ke 0 3 d The integral over can be performed by using Eq. (E.18a), Appendix E, with the result I1 = A = where k 4 P 2 (2) P= In the same way, e 4  1 16 1 + 4 2 ( ) (3) I2 = B = 0 ke 0 2 x1 d d = cos 2 d 0 ke 0 3 d (4) Again, we use Eq. (E.18a) by writing cos 2 = 1  sin 2 , and we find I2 = B = Also k 4 P ( 1 + 8 2 ) 2 0 (5) I12 = C =  ke 0 x1 x2 d d ke 0 (6) =  cos sin d 0 d 3 In order to evaluate the integral over in this case we write cos sin = (1 2) sin 2 and use Eq. (E.18), Appendix E. We find I12 = C = k 4 P Using the results of problem 1117, the entire inertia tensor is now known. According to the result of Problem 1118, the angle through which the coordinates must be rotated in order to make {I} diagonal is (7) = 1 2C tan 1 2 B  A (8) Using Eqs. (2), (5), and (7) for A, B, C, we find 2C 1 = B  A 2 (9) DYNAMICS OF RIGID BODIES 381 so that tan 2 = Therefore, we also have 1 + 4 2 2 2 1 1 2 (10) sin 2 = cos 2 = 1 1 + 4 2 1 + 4 2 2 (11) Then, according to the relations specified in Problem 1118, I1 = A = A cos 2  C sin 2 + B sin 2 Using cos 2 = (1 2)(1 + cos 2 ) and sin 2 = (1 2)(1  cos 2 ) , we have (12) I1 = A = Now, 1 1 ( A + B) + ( A  B) cos 2  C sin 2 2 2 k 4 P (1 + 4 2 ) (13) A+B= (14) A  B = 4k 4 P Thus, I1 = A = or, k 4 P (1 + 4 2 )  2k 4 P 2 2  k 4 P 1 2 2 1 + 4 1 + 4 I1 = A = k 4 P ( Q  R ) (15) (16) where Q= 1 + 4 2 2 R = 1 + 4a2 (17) Similarly, I 2 = B = k 4 P ( Q + R ) (18) and, of course, 382 I 3 = A + B = I1 + I 2 CHAPTER 11 (19) We can also easily verify, for example, that I12 = C = 0 . 1120. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy E = U1 = mg b 2 ( b is the height of the center of mass above the floor.) 2 When the rod hits the floor, the potential energy is zero. Thus E = T2 = 1 2 I 2 where I is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length : I end = x 2 dx = 0 b 1 3 1 b = mb 2 3 3 Thus T2 = 1 mb 2 2 6 By conservation of energy U1 = T2 mg b 1 = mb 2 2 2 6 = 3g b 1121. Using I to denote the matrix whose elements are those of {I} , we can write L = I L = I We also have x = x and x = t x and therefore we can express L and as (11.54) (11.54a) L = t L = t (11.55a) (11.55b) substituting these expressions into Eq. (11.54), we have DYNAMICS OF RIGID BODIES 383 t L = I t and multiplying on the left by , t L = I t or L = I t ( ) by virture of Eq. (11.54a), we identify I = I t (11.61) 1122. According to Eq. (11.61), I ij = ik I k j1 k ,l (1) Then, tr {I} = I ii = ik I k i1 i i k, = Ik k, i 1 i ik (2) = I k k = I kk k, k so that tr {I} = tr {I} (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 {I} is not invariant under translation. (For example, tr {I} will be different for inertia tensors expressed in coordinate system with different origins.) 1123. We have I = I 1 (1) Then, 384 CHAPTER 11 I = I 1 = I 1 = 1 I so that, I = I This result is easy to verify for the various examples involving the cube. 1124. x2 a/2 a/2 x1 (2)  3 a 2 The area of the triangle is A = 3 a 2 4 , so that the density is = M 4M = A 3 a2 (1) a) The rotational inertia with respect to an axis through the point of suspension (the origin) is x2 x1 2 2 I 3 = x1 + x2 dx1 dx2 a2 0 ( ) = 2 0 dx1  3 ( a  2 x1 ) 2 (x 2 1 2 + x2 ) dx2 = 3 4 1 a = Ma 2 24 6 (2) When the triangle is suspended as shown and when = 0, the coordinates of the center of mass are (0, x2 , 0) , where DYNAMICS OF RIGID BODIES 385 1 x2 dx1 dx2 M a2 x2 = 2 = M = 0 dx1  3 ( a  2 x1 ) 0 x2 dx2 a 2 3 (3) The kinetic energy is T= 1 1 I 3 2 = Ma 2 2 2 12 Mga (1  cos ) 2 3 (4) and the potential energy is U= (5) Therefore, L= Mga 1 cos Ma 2 2 + 12 2 3 (6) where the constant term has been suppressed. The Lagrange equation for is + 3 g sin = 0 a (7) and for oscillations with small amplitude, the frequency is = 3 g a (8) b) The rotational inertia for an axis through the point of suspension for this case is x2 x1  0  x2 3 a 2 I 3 = 2 3a  2 dx2 (x 0 3 2 1 2 + x2 dx1 ) = 5 Ma 2 12 Mga 5 cos Ma 2 2 + 24 3 (9) The Lagrangian is now L= (10) 386 and the equation of motion is CHAPTER 11 + 12 g sin = 0 5 3 a (11) so that the frequency of small oscillations is = 12 g 5 3 a (12) which is slightly smaller than the previous result. 1125. x2 R r x1 2 The center of mass of the disk is (0, x2 ) , where x2 = 2 x2 dx1 dx2 + x2 dx1 dx2 M lower upper semicircle semicircle = M 0 R ( r sin ) rdrd + 2 ( r sin ) rdrd R 2 0 0 = 2 R 3 3 M 1 1 M = R 2 + 2 R 2 2 2 = (1) Now, the mass of the disk is 3 R2 2 (2) so that x2 =  4 R 9 (3) The direct calculation of the rotational inertia with respect to an axis through the center of mass is tedious, so we first compute I with respect to the x3 axis and then use Steiner's theorem. DYNAMICS OF RIGID BODIES 387 I 3 = 0 = R 0 r 2 rdr d + 2 R 0 2 r 2 rdr d 3 1 R 4 = MR2 4 2 2 I 0 = I 3  Mx2 (4) Then, = = 1 16 MR2  M R2 2 81 2 1 32 MR2 1  2 81 2 (5) When the disk rolls without slipping, the velocity of the center of mass can be obtained as follows: Thus 2 x2 CM R xCM = R  x2 sin yCM = R  x2 cos xCM = R  x2 cos yCM = x2 sin (x where 2 CM 2 2 + yCM = V 2 = R 2 2 + x2 2  2 2 R x2 cos ) V 2 = a 2 2 (6) 2 a = R 2 + x2  2R x2 cos (7) Using (3), a can be written as a = R 1+ The kinetic energy is 16 8 cos  2 81 9 (8) 388 T = Ttrans + Trot = Substituting and simplifying yields T= The potential energy is 1 U = Mg R + x2 cos 2 = Thus the Lagrangian is L= 1 8 3 8 MR R 2  cos  g 1  cos 2 2 9 9 CHAPTER 11 1 1 Mv 2 + I 0 2 2 2 (9) 1 3 8 MR2 2  cos 2 2 9 (10) 1 8 MgR 1  cos 2 9 (11) (12) 1126. Since = lies along the fixed x axis , the components of along the body axes 3 ( xi ) are given by the application of the transformation matrix [Eqs. (11.98) and (11.99)]: ( ) 0 ( ) = = 0 ( ) 1 2 3 1 2 3 1 2 3 (1) Carrying out the matrix multiplication, we find which is just Eq. (11.101a). The direction of = coincides with the line of nodes and lies along the x axis. The 1 components of along the body axes are therefore obtained by the application of the transformation matrix which carries the xi system into the xi system: ( ) 1 ( ) 2 = ( ) 3 0 = 0 cos  sin 0 ( ) sin sin ( ) = cos sin cos ( ) (2) (3) DYNAMICS OF RIGID BODIES 389 which is just Eq. (11.101b). Finally, since lies along the body x3 axis , no transformation is required: which is just Eq. (11.101c). Combining these results, we obtain = ( ) 0 ( ) = 0 ( ) 1 1 2 3 (4) ( ) + ( ) + ( ) ( ) + ( ) + ( ) ( ) + ( ) + ( ) 1 1 1 2 3 2 3 2 3 sin sin + cos = cos sin  sin cos + which is just Eq. (11.02). 1127. x3 x3 (5) L x2 Initially: L1 = 0 = I1 1 L2 = L sin = I1 2 = I1 sin L3 = L xos = I 3 3 = I 3 cos Thus tan = From Eq. (11.102) L2 I1 = tan L3 I 3 (1) 3 = cos + Since 3 = cos , we have 390 CHAPTER 11 cos = cos  From Eq. (11.131) I 3  I1 3 I1 (2) =  =  (2) becomes cos = I3 cos I1 (3) From (1), we may construct the following triangle I3 tan I3 from which cos = I3 I + I tan 2 2 3 2 1 12 Substituting into (3) gives = I1 2 2 I1 sin 2 + I 3 cos 2 1128. From Fig. 117c we see that = is along the x3 axis , = is along the line of = e 3 nodes, and = is along the x3 axis . Then, (1) whee e is the unit vector in the x3 direction. 3 Projecting the lines of nodes into the x1  and x2 axes , we obtain = ( e1 cos + e sin ) 2 along the x3 axis and a component normal to this axis: = ( e12 sin + e cos ) 3 where (3) (2) has components along all three of the xi axes. First, we write in terms of a component e12 = e1 sin  e cos 2 Then, = ( e1 sin sin  e sin cos + e cos ) 2 3 (4) (5) DYNAMICS OF RIGID BODIES 391 Collecting the various components, we have 1 = cos + sin sin 2 = sin  sin cos 3 = cos + When the motion is vertical = 0. Then, according to Eqs. (11.153) and (11.154), P = I 3 + = P (6) 1129. ( ) (1) and using Eq. (11.159), we see that P = P = I 3 3 (2) Also, when = 0 (and = 0 ), the energy is [see Eq. (11.158)] E= 1 2 I 3 3 + Mgh 2 (3) Furthermore, referring to Eq. (11.160), E = E  1 2 I 3 3 = Mgh 2 (3) If we wish to examine the behavior of the system near = 0 in order to determine the conditions for stability, we can use the values of P , P , and E for = 0 in Eq. (11.161). Thus, 2 2 I 3 3 (1  cos ) 1 + Mgh cos Mgh = I12 2 + 2 2I12 sin 2 2 (5) Changing the variable to z = cos and rearranging, Eq. (5) becomes z2 = (1  z ) 2 2 2 2 2 Mgh I12 (1 + z )  I 3 w3 I12 (6) The questions concerning stability can be answered by examining this expression. First, we note that for physically real motion we must have z 2 0 . Now, suppose that the top is spinning very rapidly, i.e., that 3 is large. Then, the term in the square brackets will be negative. In such a case, the only way to maintain the condition z 2 0 is to have z = 1, i.e., = 0. Thus, the motion at = 0 will be stable as long as 2 2 4 Mgh I12  I 3 3 < 0 (7) or, 4 Mgh I12 <1 2 2 I3 3 (8) 392 CHAPTER 11 Suppose now that the top is set spinning with = 0 but with 3 sufficiently small that the condition in Eq. (8) is not met. Any small disturbance away from = 0 will then give z a negative value and will continue to increase; i.e., the motion is unstable. In fact, will continue until z reaches a value z0 that again makes the square brackets equal to zero. This is a turning point for the motion and nutation between z = 1 and z = z0 will result. From this discussion it is evident that there exists a critical value for the angular velocity, c , such that for 3 > c the motion is stable and for < c there is nutation: c = 2 Mgh I12 I3 (9) If the top is set spinning with 3 > c and = 0, the motion will be stable. But as friction slows the top, the critical angular velocity will eventually be reached and nutation will set in. This is the case of the "sleeping top." If we set = 0 , Eq. (1.162) becomes 1130. (P  P E = V ( ) = 2 I12 1  cos 2 ( cos ) 2 ) + Mgh cos (1) Rearranging, this equation can be written as ( 2 Mgh I12 ) cos3  ( 2E I12 + P2 ) cos2 + 2 ( P P  Mgh I12 ) cos + ( 2E I12  P2 ) = 0 which is cubic in cos . (2) V( ) has the form shown in the diagram. Two of the roots occur in the region 1 cos 1 , and one root lies outside this range and is therefore imaginary. V() +1 1 cos DYNAMICS OF RIGID BODIES 393 1131. The moments of inertia of the plate are I2 = I 3 = I1 + I 2 = I 2 (1 + cos 2 ) = 2I 2 cos 2 I1 = I 2 cos 2 (1) We also note that I1  I 2 =  I 2 (1  cos 2 ) = 2I 2 sin 2 (2) Since the plate moves in a forcefree manner, the Euler equations are [see Eq. (11.114)] ( I1  I 2 ) 1 2  I 3 3 = 0 ( I 2  I 3 ) 2 3  I11 = 0 ( I 3  I1 ) 31  I 2 2 = 0 Substituting (1) and (2) into (3), we find (3) ( 2I sin 2 1 2  2I 2 cos 2 3 = 0 (  I 2 cos 2 ) 2 3  ( I 2 cos 2 ) 1 = 0 I 2 3 1  I 2 2 = 0 2 ) ( ) (4) These equations simplify to 3 =  1 2 tan 2 1 =  2 3 2 = 3 1 From which we can write (5) 1 2 3 = 2 2 =  1 1 =  3 3 cot 2 Integrating, we find 2 2 2 2 2 2 2  2 ( 0 ) =  1 + 1 ( 0 ) =  3 cot 2 + 3 ( 0 ) cot 2 (6) (7) Now, the initial conditions are 394 CHAPTER 11 2 ( 0) = 0 3 ( 0 ) = sin Therefore, the equations in (7) become 2 2 2 2 =  1 + 2 cos 2 =  3 cot 2 + 2 cos 2 1 ( 0 ) = cos (8) (9) From (5), we can write 2 2 2 2 = 3 1 2 2 and from (9), we have 1 = 3 cot 2 . Therefore, (10) becomes 2 2 = 3 cot 2 2 and using 3 = 2 sin 2  2 tan 2 from (9), we can write (11) as (10) (11) 2 =  cot tan  2 sin 2 2 2 2 (12) Since 2 = d 2 dt , we can express this equation in terms of integrals as  Solving for 2 , 2 2 d 2 =  cot dt tan 2  2 sin 2 (13) Using Eq. (E.4c), Appendix E, we find tan 1 tanh 1 2 = t cot ( tan )( sin ) sin (14) 2 ( t ) = cos tanh ( t sin ) 1132. a) The exact equation of motion of the physical pendulum is (15) I + MgL sin = 0 where I = Mk 2 , so we have = or gL sin k2 gL d ( cos ) d = 2 dt k d () or DYNAMICS OF RIGID BODIES 395 d ( ) = so gL d ( cos ) k2 2 = 2 gL cos + a k2 where a is a constant determined by the initial conditions. Suppose that at t = 0 , = 0 and at 2 gl that initial position the angular velocity of the pendulum is zero, we find a = 2 cos 0 . So k finally = 2 gl ( cos  cos 0 ) k2 b) One could use the conservation of energy to find the angular velocity of the pendulum at any angle , but it is exactly the result we obtained in a), so at = 10 , we have = = 2 gL ( cos  cos 0 ) = 53.7 s1 k2 1133. Cats are known to have a very flexible body that they can manage to twist around to a feetfirst descent while falling with conserved zero 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 body 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 certain time, apparently around 0.3 sec. Thus the minimum height required for cats falling is about 0.5m. 1134. The Euler equation, which describes the rotation of an object about its symmetry axis, say 0x, is I x x  I y  I z y z = N x where N x =  b x is the component of torque along Ox. Because the object is symmetric about Ox, we have I y = I z , and the above equation becomes Ix d x = b x dt x = e  b t Ix ( ) x0 396 CHAPTER 11 CHAPTER 12 Coupled Oscillations 121. m1 = M k1 x1 k12 x2 m2 = M k2 The equations of motion are Mx1 + ( 1 + 12 ) x1  12 x2 = 0 Mx2 + ( 2 + 12 ) x2  12 x1 = 0 We attempt a solution of the form x1 ( t ) = B1 e it x2 ( t ) = B2 e it Substitution of (2) into (1) yields (1) (2) ( + 12  M 2 B1  12 B2 = 0  12 B1 + 2 + 12  M 2 B2 = 0 1 ) ( ) (3) In order for a nontrivial solution to exist, the determinant of coefficients of B1 and B2 must vanish. This yields 2 1 + 12  M 2 2 + 12  M 2 = 12 ( ) ( ) (4) from which we obtain 2 = 1 + 2 + 2 12 2M 1 2M 2 ( 1  2 )2 + 4 12 (5) This result reduces to 2 = ( + 12 12 ) M for the case 1 = 2 = (compare Eq. (12.7)]. 397 398 CHAPTER 12 If m2 were held fixed, the frequency of oscillation of m1 would be 2 01 = 1 ( 1 + 12 ) M (6) while in the reverse case, m2 would oscillate with the frequency 2 02 = 1 ( 2 + 12 ) M (7) Comparing (6) and (7) with the two frequencies, + and  , given by (5), we find 2 + = 1 + 2 + 2 12 + 1 2M 2 ( 1  2 )2 + 4 12 (8) > so that 1 1 2 1 + 2 + 2 12 + ( 1  2 ) = M ( 1 + 12 ) = 01 2M + > 01 Similarly, 2  = (9) 1 2M + + 2  2 12 1 2 ( 1  2 )2 + 4 12 (10) < so that 1 1 2 1 + 2 + 2 12  ( 1  2 ) = M ( 2 + 12 ) = 02 2M  < 02 If 1 > 2 , then the ordering of the frequencies is (11) + > 01 > 02 >  122. (12) From the preceding problem we find that for 12 1,2 M (1) 1 If we use 1 + 12 M ; 2 2 + 12 01 = 1 M ; 02 = 2 M (2) then the frequencies in (1) can be expressed as COUPLED OSCILLATIONS 399 1 = 01 1 + 2 = 02 where 12 01 (1 + 1 ) 1 1 + 12 02 (1 + 2 ) 2 (3) 1 = For the initial conditions [Eq. 12.22)], 12 ; 2 = 12 2 1 2 2 (4) x1 ( 0 ) = D, x2 ( 0 ) = 0, x1 ( 0 ) = 0, x2 ( 0 ) = 0 , the solution for x1 ( t ) is just Eq. (12.24): (5) + 2 x1 ( t ) = D cos 1 2 Using (3), we can write  2 t cos 1 2 t (6) 1 + 2 = ( 01 + 02 ) + ( 1 01 + 2 02 ) 2 + + 2 + (7) 1  2 = ( 01  02 ) + ( 1 01  2 02 ) 2  + 2  Then, x1 ( t ) = D cos ( + t + + t ) cos (  t +  t ) Similarly, + 2 x2 ( t ) = D sin 1 2  2 t sin 1 2 t (8) (9) = D sin ( + t + + t ) sin (  t +  t ) Expanding the cosine and sine functions in (9) and (10) and taking account of the fact that + and  are small quantities, we find, to first order in the 's, x1 ( t ) D cos + t cos  t  + t sin + t cos  t   t cos + t sin  t x2 ( t ) D sin + t sin  t + + t cos + t sin  t +  t sin + t cos  t When either x1 ( t ) or x2 ( t ) reaches a maximum, the other is at a minimum which is greater than zero. Thus, the energy is never transferred completely to one of the oscillators. (10) (11) (12) 400 123. CHAPTER 12 The equations of motion are m 2 x2 + 0 x1 = 0 M m 2 x2 + x2 + 0 x2 = 0 M x1 + (1) We try solutions of the form x1 ( t ) = B1 e it ; x2 ( t ) = B2 e it (2) We require a nontrivial solution (i.e., the determinant of the coefficients of B1 and B2 equal to zero), and obtain 2 ( 02  2 )  4 m = 0 2 M (3) so that 2 0  2 = 2 m M (4) and then 2 = 2 0 m 1 M (5) Therefore, the frequencies of the normal modes are 1 = 2 0 1+ m M 2 0 (6) 2 = 1 m M where 1 corresponds to the symmetric mode and 2 to the antisymmetric mode. By inspection, one can see that the normal coordinates for this problem are the same as those for the example of Section 12.2 [i.e., Eq. (12.11)]. 124. The total energy of the system is given by E = T +U = Therefore, 1 1 1 2 2 2 2 2 M x1 + x2 + x1 + x2 + 12 ( x2  x1 ) 2 2 2 ( ) ( ) (1) COUPLED OSCILLATIONS 401 dE = M ( x1 x1 + x2 x2 ) + ( x1 x1 + x2 x2 ) + 12 ( x2  x1 )( x2  x1 ) dt = Mx1 + x1  12 ( x2  x1 ) x1 + Mx2 + x2 + 12 ( x2  x1 ) x2 = Mx1 + ( + 12 ) x1  12 x2 x1 + Mx2  12 x1 + ( + 12 ) x2 x2 (2) which exactly vanishes because the coefficients of x1 and x2 are the lefthand sides of Eqs. (12.1a) and (12.1b). An analogous result is obtained when T and U are expressed in terms of the generalized coordinates 1 and 2 defined by Eq. (12.11): T= U= 1 2 2 M 1 + 2 4 ( ) (3) (4) 1 1 2 2 2 1 + 2 + 121 4 2 ( ) Therefore, 2 dE = M1 + ( + 2 12 ) 1 1 + [ M2 + 2 ] 2 dt (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 2 one term in the energy that has 12 as a coefficient (namely, 121 ), and through Eq. (12.15) we see that this implies that such a term depends on the C1 's and 1 , but not on the C2 's and 2 . To understand why this is so, it is sufficient to recall that 1 is associated with the anitsymmetrical mode of oscillation, which obviously must have 12 as a parameter. On the other hand, 2 is associated to the symmetric mode, x1 ( t ) = x2 ( t ) , x1 ( t ) = x2 ( t ) , 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. 125. We set 1 = 2 = 12 . Then, the equations of motion are m1 x1 + 2 x1  x2 = 0 m2 x2 + 2 x2  x1 = 0 (1) Assuming solutions of the form x1 ( t ) = B1e it i t x2 ( t ) = B2 e (2) we find that the equations in (1) become 402 CHAPTER 12 ( 2  m ) B 2 1  B2 = 0  B1 + 2  m2 2 B2 = 0 1 ( ) (3) which lead to the secular equation for 2 : ( 2  m )( 2  m ) = 2 2 1 2 2 (4) Therefore, 2 = 3 1 1  m1 + m2 (5) where = m1 m2 ( m1 + m2 ) is the reduced mass of the system. Notice that (5) agrees with Eq. the maximum value of 3 ( m1 + m2 ) is 3 4 . (Show this.) (12.8) for the case m1 = m2 = M and 12 = . Notice also that 2 is always real and positive since 1 Inserting the values for 1 and 2 into either of the equations in (3), we find m1 2  and 3 1 + 1  a11 = a21 m1 + m2 (6) m 3 a12 = 2  1 1  1  a22 m1 + m2 Using the orthonormality condition produces a11 = 1 D1 (7) (8) 2 a21 = where m1 + m2 3m1m2 1 + 1  m2 ( m1 + m2 ) 2 D1 (9) D1 2 ( m2  m1 ) + 2 2 m1 2 2 2 m1  m2 + m2 m2 ( ) 1 ( m1 + m2 ) 2 3m1m2 (10) The second eigenvector has the components 2 a12 = m1 + m2 3m1m2 1  1  m2 ( m1 + m2 ) 2 D2 (11) 1 Recall that when we use = 1 , we call the coefficients 1 ( = 1 ) = a11 and 2 ( = 1 ) = a21 , etc. COUPLED OSCILLATIONS 403 a22 = where m2 D2 2m1 1 + 1 + m2 2 m2 1 D2 (12) 2 3m1 1  2 + 2m1 m2 2 m1 3m1 m2 1 2 1 ( m1 + m2 )2 m2 (13) The normal coordinates for the case in which q j ( 0 ) = 0 are 1 ( t ) = ( m1 a11 x10 + m2 a21 x20 ) cos 1t 2 ( t ) = ( m1 a12 x10 + m2 a22 x20 ) cos 2t (14) 126. m m k k x1 x2 If the frictional force acting on mass 1 due to mass 2 is f =  ( x1  x2 ) then the equations of motion are mx1 + ( x1  x2 ) + x1 = 0 mx2 + ( x2  x1 ) + x2 = 0 Since the system is not conservative, the eigenfrequencies will not be entirely real as in the previous cases. Therefore, we attempt a solution of the form x1 ( t ) = B1e t ; x2 ( t ) = B2 e t (3) (2) (1) where = + i 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: ( m from which we find the two solutions 2 + + ) 2 = 2 2 (4) 1 = i m 1 2   m 2 = m m ; 1 = (5) ( ) The general solution is therefore 404 CHAPTER 12 + x1 ( t ) = B11 e i mt  + B11 e i mt + + e  t m B12 e ( 2  m t m  + B12 e 2  m t m ) (6) and similarly for x2 ( t ) . The first two terms in the expression for x1 ( t ) are purely oscillatory, whereas the last two terms + contain the damping factor e  t . (Notice that the term B12 exp ( 2  m t increases with time if ) + 2 > m , but B12 is not required to vanish in order to produce physically realizable motion because the damping term, exp(t), decreases with time at a more rapid rate; that is  + 2  m < 0 .) To what modes do 1 and 2 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 2 < m , the motion is one of damped oscillations, whereas if 2 > m , the motion proceeds monotonically to zero amplitude. 127. k m k m x1 x2 We define the coordinates x1 and x2 as in the diagram. Including the constant downward gravitational force on the masses results only in a displacement of the equilibrium positions and does not affect the eigenfrequencies or the normal modes. Therefore, we write the equations of motion without the gravitational terms: mx1 + 2 x1  x2 = 0 mx2 + x2  x1 = 0 Assuming a harmonic time dependence for x1 ( t ) and x2 ( t ) in the usual way, we obtain (1) ( 2  m ) B 2  B2 = 0  B1 +  m 2 B2 = 0 1 ( ) (2) Solving the secular equation, we find the eigenfrequencies to be COUPLED OSCILLATIONS 405 2 1 = 3+ 5 2 m 3 5 2 m (3) 2 2 = Substituting these frequencies into (2), we obtain for the eigenvector components 1 5 a11 = a21 2 1+ 5 a12 = a22 2 For the initial conditions x1 ( 0 ) = x2 ( 0 ) = 0 , the normal coordinates are (4) 1 ( t ) = ma11 x10 + 1 5 x20 cos 1t 2 (5) 1+ 5 x20 cos 2t 2 ( t ) = ma12 x10 + 2 Therefore, when x10 = 1.6180 x20 , 2 ( t ) = 0 and the system oscillates in Mode 1, the antisymmetrical mode. When x10 = 0.6180 x20 , 1 ( t ) = 0 and the system oscillates in Mode 2, the symmetrical mode. When mass 2 is held fixed, the equation of motion of mass 1 is mx1 + 2 x1 = 0 and the frequency of oscillation is 2 m (6) 10 = (7) When mass 1 is held fixed, the equation of motion of mass 2 is mx2 + x2 = 0 and the frequency of oscillation is (8) 20 = m (9) Comparing these frequencies with 1 and 2 we find 1 = 2 = 3+ 5 4 3 5 4 2 2 = 1.1441 > 10 m m = 0.6180 < 20 m m 406 CHAPTER 12 Thus, the coupling of the oscillators produces a shift of the frequencies away from the uncoupled frequencies, in agreement with the discussion at the end of Section 12.2. 128. The kinetic and potential energies for the double pendulum are given in Problem 77. If we specialize these results to the case of small oscillations, we have T= 1 m 2 U= 2 ( 2 2 1 2 + 2 + 21 2 ) (1) (2) 1 2 2 mg 21 + 2 2 ( ) where 1 refers to the angular displacement of the upper pendulum and 2 to the lower pendulum, as in Problem 77. (We have also discarded the constant term in the expression for the potential energy.) Now, according to Eqs. (12.34), T= U= 1 mjk qj qk 2 j,k 1 Ajk qj qk 2 j,k (3) (4) Therefore, identifying the elements of {m} and {A} , we find {m} = m 2 1 1 {A} = mg 0 1 and the secular determinant is 2 g  2 2 2 2 1 (5) 2 0 (6)  2 g =0  2 (7)  or, g 2 g 2 4 2  2   = 0 Expanding, we find (8) 4 4 g g +2 =0 2 2 (9) which yields COUPLED OSCILLATIONS 407 2 = 2 2 and the eigenfrequencies are ( )g g (10) 1 = 2 + 2 2 = 2  2 To get the normal modes, we must solve g = 1.848 = 0.765 (11) g g (A j jk  r2 m jk a jr = 0 ) For k = 1, this becomes: (A For r = 1: 11  r2 m11 a1r + A21  r2 m21 a2 r = 0 ) ( ) g g 2m 2 a11  2 + 2 m 2mg  2 + 2 Upon simplifying, the result is ( ) ( ) 2 a21 = 0 a21 =  2 a11 Similarly, for r = 2, the result is a22 = 2 a12 The equations x1 = a11 1 + a12 2 x2 = a21 1 + a22 2 can thus be written as x1 = a11 1 + 1 a22 2 2 x1 =  2 a11 1 + a22 2 Solving for 1 and 2 : 1 = 2 x1  x2 ; 2 = 2 2 a11 2 x1 + x2 2 a22 408 x2 2 CHAPTER 12 1 occurs when 2 = 0; i.e. when x1 =  2 occurs when 1 = 0; i.e. when x1 = x2 2 Mode 2 is therefore the symmetrical mode in which both pendula are always deflected in the same direction; and Mode 1 is the antisymmetrical 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 the discussion in Section 12.2. 129. The general solutions for x1 ( t ) and x2 ( t ) 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 x2 = 0 , and both are released from rest: x1 ( 0 ) = D, x2 ( 0 ) = 0, x1 ( 0 ) = 0, x2 ( 0 ) = 0 (1) Substitution of (1) into Eq. (12.10) determines the constants, and we obtain x1 ( t ) = x2 ( t ) = D ( cos 1t + cos 2t ) 2 D ( cos 2t  cos 1t ) 2 (2) (3) where 1 + 2 12 M > 2 = M (4) As an example, take 1 = 1.2 2 ; x1 ( t ) vs. x2 ( t ) is plotted below for this case. It is possible to find a rotation in configuration 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 x1 x1  x2 = D cos 1t x2 x1 + x2 = D cos 2t (5) (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 x2(t)/D 0.8 2t = 5 2 2t = 2 0.6 0.4 0.2 0.2 2t = 2 2t = 3 0.4 7 2 0.5 3 x1 2t = 0.6 0.8 2t = 0 1.0 x1(t)/D 4 x2 1 = 1.2 2 1.0 2t = 3 2 1210. The equations of motion are mx1 + bx1 + ( + 12 ) x1  12 x2 = F0 cos t mx2 + bx2 + ( + 12 ) x2  12 x1 = 0 (1) The normal coordinates are the same as those for the undamped case [see Eqs. (12.11)]: 1 = x1  x2 ; 2 = x1 + x2 (2) Expressed in terms of these coordinates, the equations of motion become m (2 + 1 ) + b (2 + 1 ) + ( + 12 )(2 + 1 )  12 (2  1 ) = 2F0 cos t m (2  1 ) + b (2  1 ) + ( + 12 )(2  1 )  12 (2 + 1 ) = 0 (3) By adding and subtracting these equations, we obtain the uncoupled equations: F b + 2 12 1 + 1 = 0 cos t m m m F b 2 + 2 + 2 = 0 cos t m m m 1 + (4) With the following definitions, 410 b m CHAPTER 12 2 = 2 1 = + 2 12 m = 2 2 m F0 m (5) A= the equations become 2 1 + 21 + 1 1 = A cos t 2 2 + 22 + 2 2 = A cos t (6) Referring to Section 3.6, we see that the solutions for 1 ( t ) and 2 ( t ) are exactly the same as that given for x(t) in Eq. (3.62). As a result 1 ( t ) exhibits a resonance at = 1 and 2 ( t ) exhibits a resonance at = 2 . Taking a time derivative of the equations gives ( q = I ) LI1 + LI 2 + I1 + MI 2 = 0 C I2 + MI1 = 0 C 1211. Assume I1 = B1e i t , I 2 = B2 e it ; and substitute into the previous equations. The result is  2 LB1e i t +  2 LB2 e it + These reduce to 1 B1  2 L + B2  M 2 = 0 C 1 B1 e it  M 2 B2 e it = 0 C 1 B2 e it  M 2 B1e it = 0 C ( ) 1 B1  M 2 + B2  2 L = 0 C ( ) This implies that the determinant of coefficients of B1 and B2 must vanish (for a nontrivial solution). Thus COUPLED OSCILLATIONS 411 1  2L C  M 2 2  M 2 1  2L C =0 1 2 2 C  L  M ( ) 2 =0 1  2 L = M 2 C or 2 = Thus 1 C (L M) 1 = 2 = 1 C (L + M) 1 C (L  M) 1212. From problem 1211: LI1 + LI 2 + 1 I1 + MI 2 = 0 C 1 I 2 + MI1 = 0 C (1) (2) Solving for I1 in (1) and substituting into (2) and similarly for I 2 , we have M2 M 1 L I1 + C I1  CL I 2 = 0 L M2 M 1 L  L I 2 + C I 2  CL I1 = 0 If we identify M 12 = LC M 1 = 1  C L m= L M2 L (3) (4) 412 then the equations in (3) become mI1 + ( + 12 ) I1  12 I 2 = 0 mI 2 + ( + 12 ) I 2  12 I1 = 0 which are identical in form to Eqs. (12.1). Then, using Eqs. (12.8) for the characteristic frequencies, we can write M 1 L = 1 = 2 M C (L  M) CL  L 1+ CHAPTER 12 (5) (6) M 1 1 L 2 = = 2 M C (L + M) CL  L which agree with the results of the previous problem. 1213. q1 C1 L1 q2 C2 L12 I2 L2 I1 The Kirchhoff circuit equations are q1 + L12 I1  I 2 = 0 C1 q L2 I 2 + 2 + L12 I 2  I1 = 0 C2 L1 I1 + ( ) ( ) (1) Differentiating these equations using q = I , we can write 1 I1  L12 I 2 = 0 C1 1 ( L2 + L12 ) I 2 + I 2  L12 I1 = 0 C2 ( L1 + L12 ) I1 + (2) As usual, we try solutions of the form I1 ( t ) = B1 e it ; which lead to I 2 ( t ) = B2 e it (3) COUPLED OSCILLATIONS 413 2 1 2 ( L1 + L12 )  B1  L12 B2 = 0 C1 2 1 2  L12 B1 + ( L2 + L12 )  B2 = 0 C1 (4) Setting the determinant of the coefficients of the B's equal to zero, we obtain 2 1 2 1 4 2 ( L1 + L12 )  ( L2 + L12 )  = L12 C1 C2 with the solution (5) 2 ( L1 + L12 ) C1 + ( L2 + L12 ) C2 ( L1 + L12 ) C1  ( L2 + L12 ) C2 = 2C1C2 ( L1 + L12 )( L2 + L12 )  L2 12 2 + 4 L2 C1C2 12 (6) We observe that in the limit of weak coupling ( L12 0 ) and L1 = L2 = L , C1 = C2 = C , the frequency reduces to = 1 LC (7) which is just the frequency of uncoupled oscillations [Eq. (3.78)]. 1214. L1 C1 C12 L2 C2 I1 I2 The Kirchhoff circuit equations are (after differentiating and using q = I ) 1 1 1 L1I1 + + I2 = 0 I1  C12 C1 C12 1 1 1 L2 I 2 + + I1 = 0 I2  C12 C2 C12 Using a harmonic time dependence for I1 ( t ) and I 2 ( t ) , the secular equation is found to be C1 + C12 C2 + C12 1 2 2 L1  L2  = 2 C1 C12 C2 C12 C12 Solving for the frequency, 2 2 C1L1 ( C2 + C12 ) + C2 L2 ( C1 + C12 ) C1L1 ( C2 + C12 )  C2 L2 ( C1 + C12 ) + 4C1 C2 L1L2 = 2L1L2C1C2C12 2 2 (1) (2) (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 L1 = L2 = L and C1 = C2 = C . Then, 2 1 = 2C + C12 LCC12 (4) 1 2 2 = LC Observe that 2 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 1 , and (2) in phase, with frequency 2 . Mode 1 corresponds to the currents I1 and I 2 oscillating always out of phase: ; I1 I2 I1 I2 Mode 2 corresponds to the currents I1 and I 2 oscillating always in phase: ; I1 I2 I1 I2 (The analogy with two oscillators coupled by a spring can be seen by associating case 1 with Fig. 122 for = 1 and case 2 with Fig. 122 for = 2 .) If we now let L1 L2 and C1 C2 , we do not have pure symmetrical and antisymmetrical symmetrical modes, but we can associate 2 with the mode of highest degree of symmetry and 1 with that of lowest degree of symmetry. 1215. C1 L1 R C2 L2 I1 I2 Setting up the Kirchhoff circuit equations, differentiating, and using q = I , we find 1 I1 = 0 C1 1 L2 I 2 + R I 2  I1 + I2 = 0 C2 L1I1 + R I1  I 2 + ( ) ( ) (1) Using a harmonic time dependence for I1 ( t ) and I 2 ( t ) , the secular equation is 2 2 1 1 2 2 L1  C  i R L2  C  i R + R = 0 1 2 (2) COUPLED OSCILLATIONS 415 From this expression it is clear that the oscillations will be damped because will have an imaginary part. (The resistor in the circuit dissipates energy.) In order to simplify the analysis, we choose the special case in which L1 = L2 = L and C1 = C2 = C . Then, (2) reduces to 1 2 2 2 L   i R + R = 0 C which can be solved as in Problem 126. We find 2 (3) 1 = i 2 = L The general solution for I1 ( t ) is + I1 ( t ) = B11e 1 LC L 2 R R  C (4) i 1 LC t  + B11 e  i 1 LC t + + e  Rt L B12 e R2  L C t L  + B12 e  R2  L C t L (5) and similarly for I 2 ( t ) . The implications of these results follow closely the arguments presented in Problem 126. Mode 1 is purely oscillatory with no damping. Since there is a resistor in the circuit, this means that I1 and I 2 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 R2 < L C , there will be damped oscillations of I1 and I 2 , whereas if R2 > L C , the currents will decrease monotonically without oscillation. 1216. y O R P Mg R Q(x,y) Mg x 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 x = R ( sin + sin ) y =  R ( cos + cos ) (1) The rotational inertia of the hoop through O is 416 I O = I CM + MR 2 = 2 MR 2 The potential energy of the system is CHAPTER 12 (2) U = U hoop + U mass =  MgR ( 2 cos + cos ) (3) Since and are small angles, we can use cos x 1  x 2 2 . Then, discarding the constant term in U, we have U= The kinetic energy of the system is T = Thoop + Tmass = 1 1 I O 2 + M x 2 + y 2 2 2 1 MgR 2 2 + 2 2 ( ) (4) ( ) (5) = MR 2 2 + 1 MR 2 2 + 2 + 2 2 where we have again used the smallangle approximations for and . Thus, T= Using Eqs. (12.34), T= U= we identify the elements of {m} and {A} : 1 mjk qj qk 2 j,k 1 Ajk qj qk 2 j,k (7) 1 MR 2 3 2 + 2 + 2 2 (6) (8) {m} = MR2 {A} = MgR The secular determinant is 2 g  3 2 R  from which 2 3 1 1 1 (9) 2 0 0 1 (10)  2 g  2 R =0 (11) COUPLED OSCILLATIONS 417 g 2 g 2 4 2  3   = 0 R R (12) Solving for the eigenfrequencies, we find 1 = 2 2 = To get the normal modes, we must solve: 2 2 g R g R (13) (A j jk  r2 m jk a jr = 0 ) For k = r = 1, this becomes: g g 2 2 2mgR  2 3mR a11  2 mR a21 = 0 R R or a21 = 2a11 For k = 1, r = 2, the result is a12 = a22 Thus the equations x1 = a11 1 + a12 2 x2 = a21 1 + a22 2 can be written as x1 = a11 1 + a22 2 x2 = 2 a11 1 + a22 2 Solving for 1 , 2 1 = x1  x2 ; 3a11 1 = 2x1 + x2 3a22 1 2 1 occurs when the initial conditions are such that 2 = 0 ; i.e., x10 =  x20 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. 2 occurs when the initial conditions are such that 1 = 0 ; i.e., x10 = x20 418 CHAPTER 12 This is the symmetrical mode in which the pivot point, the CM of the hoop, and the mass always lie on a straight line. 1217. k m x1 k m x2 k m k x3 Following the procedure outlined in section 12.6: T= 1 1 1 2 2 2 mx1 + mx2 + mx3 2 2 2 U= 1 2 1 1 1 2 2 2 kx1 + k ( x2  x1 ) + k ( x3  x2 ) + kx3 2 2 2 2 2 2 2 = k x1 + x2 + x3  x1 x2  x2 x3 Thus m 0 0 m= 0 m 0 0 0 m 2k  k 0 A =  k 2k  k 0  k 2k Thus we must solve 2k  2 m k 0 This reduces to k 2k  m 2 0 k 2k  m 2 =0 k ( 2 k  m) 2 3  2k 2 2k  2 m = 0 ( ) or ( 2k  m) ( 2k  m) 2 2 2  2k 2 = 0 If the first term is zero, then we have 1 = If the second term is zero, then 2k m 2k  2 m = 2 k COUPLED OSCILLATIONS 419 which leads to 2 = (2 + 2 ) k ; m 3 = (2  2) k m To get the normal modes, we must solve (A j jk  r2 m jk a jr = 0 ) For k = 1 this gives: ( 2 k  m) a 2 r 1r + (  k ) a2 r = 0 Substituting for each value of r gives r = 1: r = 2: r = 3: ( 2k  2k ) a11  ka21 = 0 a21 = 0 ( ( 2 k a12  ka22 = 0 a22 =  2 a12 2 k a13  ka23 = 0 a23 = 2 a13 ) ) Doing the same for k = 2 and 3 yields a11 =  a31 a12 = a32 a13 = a33 The equations a21 = 0 a22 =  2 a32 a23 = 2 a33 x1 = a11 1 + a12 2 + a13 3 x2 = a21 1 + a22 2 + a23 3 x3 = a31 1 + a32 2 + a33 3 can thus be written as x1 = a11 1  1 a22 2 + a33 3 2 x2 = a22 2 + 2 a33 3 x3 =  a11 1  1 a22 2 + a33 3 2 We get the normal modes by solving these three equations for 1 , 2 , 3 : 420 x1  x3 2a11  x1 + 2 x2  x3 2 2 a22 CHAPTER 12 1 = 2 = and 3 = The normal mode motion is as follows x1 + 2 x2 + x3 4 a33 1 : x1 =  x3 x2 =  2 x1 =  2 x3 2 : 3 : x2 = 2 x1 = 2 x3 1218. x M b m y1 x1 x1 = x + b sin ; x1 = x + b cos y1 = b  b cos ; Thus T= = 1 1 2 2 Mx 2 + m x1 + y1 2 2 y1 = b sin ( ) 1 1 Mx 2 + m x 2 + b 2 2 + 2b x cos 2 2 ( ) U = mgy1 = mgb (1  cos ) For small , cos 1 2 2 . Substituting and neglecting the term of order 2 gives COUPLED OSCILLATIONS 421 1 1 ( M + m) x 2 + m b 2 2 + 2b x 2 2 mgb 2 2 M + m mb m= mb 2 mb 0 0 A= 0 mgb T= U= Thus ( ) We must solve  2 ( M + m)  2 mb which gives  2 mb mgb  2 mb 2 =0 2 ( M + m) ( 2 mb 2  mgb )  4 m2 b 2 = 0 2 2 Mb 2  mgb ( m + M ) = 0 Thus 1 = 0 2 = g ( M + m) mb  r2 m jk a jr = 0 (A j jk ) Substituting into this equation gives a21 = 0 a12 =  Thus the equations x = a11 1 + a12 2 bm a ( m + M ) 22 ( k = 2, r = 1) ( k = 2, r = 2) = a21 1 + a22 2 become x = a11 1  mb a ( m + M ) 22 2 422 CHAPTER 12 = a22 2 Solving for 1 , 2 : 2 = a22 x+ bm (m + M) a11 n1 = n1 occurs when n2 = 0; or = 0 n2 occurs when n1 = 0; or x =  (m + M) bm 1219. With the given expression for U, we see that { A} has the form 1 {A} = 12  13  12 1  23  13  23 1 (1) The kinetic energy is T= so that {m} is 1 0 0 {m} = 0 1 0 0 0 1 The secular determinant is 1 2  12  13 Thus,  12 1  23 2 1 2 2 2 1 + 2 + 3 2 ( ) (2) (3)  13  23 = 0 1 2 (4) (1  )  (1  )( 2 3 2 2 12 2 2 + 13 + 23  2 12 13 23 = 0 ) (5) This equation is of the form (with 1  2 x ) x 3  3 2 x  2 2 = 0 which has a double root if and only if (6) ( ) 2 32 = 2 (7) COUPLED OSCILLATIONS 423 Therefore, (5) will have a double root if and only if 2 2 2 12 + 13 + 23 3 32 = 12 13 23 (8) This equation is satisfied only if 12 = 13 = 23 Consequently, there will be no degeneracy unless the three coupling coefficients are identical. 1220. (9) If we require a11 = 2a21 , then Eq. (12.122) gives a31 = 3a21 , and from Eq. (12.126) we 14 . Therefore, 1 3 2 , , a1 = 14 14 14 (1) obtain a21 = 1 The components of a2 can be readily found by substituting the components of a1 above into Eq. (12.125) and using Eqs. (12.123) and (12.127): 5 1 4 , , a2 = 42 42 42 These eigenvectors correspond to the following cases: (2) a1 a2 1221. The tensors {A} and {m} are: 1 1 {A} = 3 2 0 1 3 2 2 1 3 2 0 1 3 2 1 (1) m 0 0 {m} = 0 m 0 0 0 m (2) thus, the secular determinant is 424 1 3 2 CHAPTER 12 1  m 2 1 3 2 0 from which 0 1 3 2 =0 2  m 2 1 3 2 (3) 1  m 2 ( 1  m 2 ) ( 2 2 1 2  m 2  3 1  m 2 = 0 2 ) ( ) (4) 2 In order to find the roots of this equation, we first set (1 2) 3 = 1 2 and then factor: ( ( ( Therefore, the roots are 1  m 2 1  m 2 2  m 2  1 2 = 0  m 2 m2 4  ( 1 + 2 ) m 2 = 0  m 2 m 2 m 2  ( 1 + 2 ) = 0 )( ) ) )( ) 1 (5) 1 1 = 2 = 3 = 0 1 m 1 + 2 m (6) Consider the case 3 = 0 . The equation of motion is 2 3 + 3 3 = 0 (7) so that 3 = 0 with the solution (8) 3 ( t ) = at + b That is, the zerofrequency mode corresponds to a translation of the system with oscillation. (9) COUPLED OSCILLATIONS 425 1222. The equilibrium configuration is shown in diagram (a) below, and the nonequilibrium configurations are shown in diagrams (b) and (c). (a) x3 4 3 O 1 2A x1 2B 2 x2 (b) 1 x3 2 O A O A }A x3 x2 (c) 4 x3 1 B { B x 3 O B O x2 The kinetic energy of the system is T= 1 1 1 2 Mx3 + I1 2 + I 2 2 2 2 2 (1) where I1 = (1 3) MA2 and I 2 = (1 3) MB2 . The potential energy is 1 2 2 2 2 U = ( x3  A  B ) + ( x3 + A  B ) + ( x3 + A + B ) + ( x3  A + B ) 2 1 2 = 4 x3 + 4 A2 2 + 4B2 2 2 Therefore, the tensors {m} and {A} are M {m} = 0 0 4 {A} = 0 0 The secular equation is 0 1 MA2 3 0 0 4 A 0 2 ( ) 0 1 MB2 3 0 0 0 4 B2 (2) (3) (4) 426 CHAPTER 12 ( 4  M ) 4 A 2 2  1 1 MA2 2 4 B2  MB2 2 = 0 3 3 (5) Hence, the characteristic frequencies are 1 = 2 2 = 2 3 = 2 We see that 2 = 3 , so the system is degenerate. M 3 M 3 = 2 M (6) The eigenvector components are found from the equation (A j jk  r2 m jk a jr = 0 ) (7) Setting a32 = 0 to remove the indeterminacy, we find 0 1 M 0 a1 = 0 ; a 2 = 3 MA2 ; a 3 = 0 0 0 3 MB2 The normal coordinates are (for x3 ( 0 ) = ( 0 ) = ( 0 ) = 0 ) (8) 1 ( t ) = x30 M cos 1t 2 ( t ) = 3 ( t ) = 0 A M 3 cos 2t cos 3t (9) 0 B M 3 Mode 1 corresponds to the simple vertical oscillations of the plate (without tipping). Mode 2 corresponds to rotational oscillations around the x1 axis, and Mode 3 corresponds to rotational oscillations around the x2 axis. 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 2A along the x2 axis of the plate, then the moment of inertia around the x1 axis is changed: I1 = The new eigenfrequencies are 1 ( M + m) A2 3 (10) COUPLED OSCILLATIONS 427 1 = 2 2 = 2 3 = 2 and there is no longer any degeneracy. 1223. M 3 M+m 3 M (11) The total energy of the rth normal mode is Er = Tr + U r = 1 2 1 2 2 r + r r 2 2 (1) where r = r e ir t Thus, (2) r = i r r e ir t In order to calculate Tr and U r , we must take the squares of the real parts of r and r : (3) r2 = ( Re r ) = Re i r ( r + i r ) ( cos r t + i sin r t ) 2 2 =  r r cos r t  r r sin r t so that Tr = Also 2 (4) 1 2 2 r r cos r t + r sin r t 2 2 (5) r2 = ( Re r ) = Re ( r + i r ) ( cos r t + i sin r t ) 2 = r cos r t  r sin r t so that Ur = 2 (6) 1 2 2 r r cos r t  r xin r t 2 (7) Expanding the squares in Tr and U r , and then adding, we find 428 Er = Tr + U r = Thus, Er = 1 2 r r 2 2 CHAPTER 12 1 2 2 r r + r2 2 ( ) (8) So that the total energy associated with each normal mode is separately conserved. For the case of Example 12.3, we have for Mode 1 1 = Thus, M ( x10  x20 ) cos 1t 2 (9) 1 =  1 Therefore, E1 = But M ( x10  x20 ) sin 1t 2 (10) 1 2 1 2 2 1 + 1 1 2 2 (11) 2 1 = + 2 12 M (12) so that E1 = = 1 + 2 12 M 1 + 2 M ( x10  x20 ) 2 sin 2 1t + 2 M 12 2 ( x10  x20 ) 2 cos2 1t M 2 2 1 ( + 2 12 ) ( x10  x20 ) 2 4 (13) which is recognized as the value of the potential energy at t = 0. [At t = 0, x1 = x2 = 0 , so that the total energy is U1 ( t = 0 ) .] 1224. Refer to Fig. 129. If the particles move along the line of the string, the equation of motion of the jth particle is mx j =  x j  x j 1  x j  x j+ 1 Rearranging, we find xj = ( ) ( ) (1) m (x j 1  2x j + x j +1 ) (2) which is just Eq. (12.131) if we identify md with m . COUPLED OSCILLATIONS 429 1225. The initial conditions are q1 ( 0 ) = q2 ( 0 ) = q3 ( 0 ) = a q1 ( 0 ) = q2 ( 0 ) = q3 ( 0 ) = 0 (1) Since the initial velocities are zero, all of the r [see Eq. (12.161b)] vanish, and the r are given by [see Eq. (12.161a)] r = so that a 2 r r 3 r sin 4 + sin 2 + sin 4 2 +1 a 2 2 = 0 2 1 3 = a 2 (2) 1 = (3) The quantities sin jr ( n + 1) are the same as in Example 12.7 and are given in Eq. (12.165). The displacements of the particles are q1 ( t ) = q2 ( t ) = q3 ( t ) = 1 2 a ( cos 1t + cos 3t ) + a ( cos 1t  cos 3t ) 2 4 2 1 a ( cos 1t  cos 3t ) + a ( cos 1t + cos 3t ) 2 2 1 2 a ( cos 1t  cos 3t ) + a ( cos 1t + cos 3t ) 2 4 (4) where the characteristic frequencies are [see Eq. (12.152)] r = 2 r sin , r = 1, 2, 3 md 8 (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 (see Figure 1211) and so this mode is absent. mb 2 [ m] = + mb 2 1226. mb 2 2 2 2 Kinetic energy T = 1 + 2 + 3 2 ( ) mb 2 Potential energy 430 CHAPTER 12 k 2 2 U = mgb (1  cos 1 ) + (1  cos 2 ) + (1  cos 3 ) + b 2 ( sin 2  sin 1 ) + ( sin 3  sin 2 ) 2 mgb 2 kb 2 2 2 2 2 2 1 + 2 + 3 + 1 + 2 2 + 3  2 1 2  2 2 3 2 2 ( ) ( ) mgb + kb 2 [ A] =  kb 2 0  kb 2 mgb + kb  kb 2 2  kb mgb + kb 2 0 2 The proper frequencies are solutions of the equation mgb + kb 2  mb 2 2  kb 2 0 = Det [ A]  2 [ m] = D et 0 ( ) ( ) ( mgb + 2kb  kb 2 2  mb 2 2 )  kb 2 (  kb 2 mgb + kb 2  mb 2 2 0 ) We obtain 3 different proper frequencies 2 1 = mg + kb mb 1 = 2 = 3 = mg + kb = 4.64 rad/s mb mg + 3kb = 4.81 rad/s mb g = 4.57 rad/s f 2 2 = mg + 3kb mb 2 3 = mg mb Actually those values are very close to one another, because k is very small. 1227. The coordinates of the system are given in the figure: 1 L1 m1 2 L2 m2 Kinetic energy: T= 1 2 2 2 1 2 2 m1 1 L1 + m2 L2 1 + L2 2  2L1L2 1 2 cos ( 1  2 ) 1 2 2 2 ( ) 1 2 1 1 2 1 m1L2 + m2 L2 + m2 L2 2  m2 L1L2 1 2 = m jk j k 1 1 2 2 2 2 jk ( ) COUPLED OSCILLATIONS 431 ( m + m2 ) L2 1 m jk = 1  m2 L1 L2  m2 L1L2 m2 L2 2 Potential energy: U = m1 gL1 (1  cos 1 ) + m2 g L1 (1  cos 1 ) + L2 (1  cos 2 ) ( m1 + m2 ) gL1 2 1 2 + m2 gL2 2 2 2 = 1 Ajk j k 2 jk m2 gL2 0 ( m + m2 ) gL1 Ajk = 1 0 Proper oscillation frequencies are solutions of the equation Det [ A]  2 [ m] = 0 1,2 = ( ) ( m1 + m2 ) g ( L1 + L2 ) + ( m1 + m2 ) g 2 m1 ( L1  L2 )2 + m2 ( L1 + L2 ) 2 2m1 L1 L2 a11 The eigenstate corresponding to 1 is where a21 a21 ( m + m2 ) L1 1  = 1 m1L2 a 11 2 2 ( m1 + m2 ) g ( L1 + L2 ) + ( m1 + m2 ) g 2 m1 ( L1  L2 ) + m2 ( L1 + L2 ) 2 gm1 L2 a12 The eigenstate corresponding to 2 is where a22 a22 ( m + m2 ) L1 1  = 1 m1L2 a 12 2 2 ( m1 + m2 ) g ( L1 + L2 )  ( m1 + m2 ) g 2 m1 ( L1  L2 ) + m2 ( L1 + L2 ) 2 gm1 L2 These expressions are rather complicated; we just need to note that a11 and a21 have the same a a sign 11 > 0 while a12 and a22 have opposite sign 11 < 0 . a21 a21 The relationship between coordinates ( 1 , 2 ) and normal coordinates 1 , 2 are 1 = a11 1 + a12 2 2 = a21 1 + a22 2 a12 1 ~ 1  a 2 22 a11 2 ~ 1  2 a21 432 CHAPTER 12 To visualize the normal coordinate 1 , let 2 = 0 . Then to visualize the normal coordinate 2 , a a we let 1 = 0 . Because 11 > 0 and 12 < 0 , we see that these normal coordinates describe two a21 a22 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. m2 b 2 0 1228. Kinetic energy: T= 1 1 2 2 m1b 2 1 + m2 b 2 2 2 2 [ m] = 0 m2 b 2 Potential energy: U = m1 gb (1  cos 1 ) + m2 gb (1  cos 2 ) + k ( b sin 1  b sin 2 ) 2 [ A] m1 gb + kb 2  kb 2 m2 gb + kb  kb 2 2 Solving the equation, Det [ A]  2 [ m] = 0 , gives us the proper frequencies of oscillation, 2 1 = ( ) g = 25 (rad/s)2 b 2 2 = g k k + + = 25.11 (rad/s)2 b m1 m2 a11 The eigenstate corresponding to 1 is with a21 = 7.44 a11 a22 a12 The eigenstate corresponding to 2 is with a22 = 8.55 a12 a22 From the solution of problem 1227 above, we see that the normal coordinates are 1 ~ 1  2 ~ 1  a12 2 = 1 + 0.12 2 a22 a11 2 = 1 + 0.13 2 a21 Evidently 1 then characterizes the inphase oscillation of two bobs, and 2 characterizes the outofphase oscillation of two bobs. Now to incorporate the initial conditions, let us write the most general oscillation form: 1 = Re ( a11 e i1t  i1 + a12 e i2t  i 2 ) 2 = Re ( a21 e i1t  i1 + a22 e i 2t  i 2 ) = Re 7.44 a11 e i1t  i1  8.35 a12 e i 2t  i 2 ( ) where is a real normalization constant. The initial conditions helps to determine parameters 's, a's, 's. COUPLED OSCILLATIONS 433 Re 1 ( t = 0 ) = 7 a11 cos 1 + a12 cos 2 = 0.122 rad Re ( 2 ( t = 0 ) = 0 7.44 a11 cos 1 + 8.35 a12 cos 2 = 0 ) sin 1 = sin 1 = 0 . Then ( ) 1 = a11 cos 1t + a12 cos 2t = 0.065 cos 1t  0.057 cos 2 2 = 7.44 a11 cos 1t + 8.35 a12 cos 2t = 0.48 ( cos 2t  cos 1t ) where 1 = 5.03 rad/s , 2 = 4.98 rad/s (found earlier) Approximately, the maximum angle 2 is 0.096 rad and it happens when cos 2t = 1 cos 1t = 1 which gives 1t = ( 2k + 1) because 2 t = 2n 1 2k + 1 = 2n 2 1 101 100 = we finally find k = n = 50 and t = = 63 s . 2 100 2 Note: 2 max = 0.96 rad and at this value the smallangle approximation breaks down, and the value 2max we found is just a rough estimate. 434 CHAPTER 12 CHAPTER 13 Continuous Systems; Waves 131. The initial velocities are zero and so all of the r vanish [see Eq. (13.8b)]. The r are given by [see Eq. (13.8a)] r = 2A 3 x r x sin sin dx L L L 0 L (1) = A 3 r so that 3 = A r = 0, r 3 (2) The characteristic frequency 3 is [see Eq. (13.11)] 3 = and therefore, 3 L (3) 3 q ( x , t ) = A cos L 3 x t sin L (4) For the particular set of initial conditions used, only one normal mode is excited. Why? 132. L 3 h L 435 436 CHAPTER 13 The initial conditions are L 3h L x, 0 x 3 q ( x , 0) = 3h L ( L  x) , xL 3 2L q ( x , 0) = 0 Because q ( x , 0 ) = 0 , all of the r vanish. The r are given by (1) (2) 6h r = 2 L = r 2 L3 0 r x 3h x sin dx + 2 L L r 3 L3 ( L  x ) sin L r x dx L (3) 9h 2 sin We see that r = 0 for r = 3, 6, 9, etc. The displacement function is q ( x, t) = where 9 3h 2 2 1 2 x 1 4 x x sin L cos 1t + 4 sin L cos 2t  16 sin L cos 4 t  ... (4) r = r L (5) The frequencies 3 , 6 , 9 , etc. are absent because the initial displacement at L 3 prevents that point from being a node. Thus, none of the harmonics with a node at L 3 are excited. 133. The displacement function is q ( x, t) x 1 3 x 1 5 x = sin cos 1t  sin cos 3t + sin cos 5t + ... 8h 2 L 9 L 25 L (1) where 1 = L (2) r = r 1 For t = 0, q ( x, 0) x 1 3 x 1 5 x = sin  sin + +... sin 2 8h L 9 L 25 L The figure below shows the first term, the first 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. (3) CONTINUOUS SYSTEMS; WAVES 437 1 term 1 0 1 0 1 0 L 2 terms L 3 terms L The time development of q(x,t) is shown below at intervals of 1 8 of the fundamental period. t = 0, T t= 1 7 T, T 8 8 t= 1 3 T, T 4 4 t= 3 5 T, T 8 8 t= 1 T 2 134. The coefficients r are all zero and the r are given by Eq. (13.8a): 8 r x r = 2 x ( L  x ) sin dx L 0 L L = 1  ( 1) r r 3 3 16 (1) so that 0, r = 32 , 3 r 3 Since r even (2) r odd 438 CHAPTER 13 q ( x , t ) = r sin r r x cos r t L (3) the amplitude of the nth mode is just n . The characteristic frequencies are given by Eq. (13.11): n L n = (4) 135. The initial conditions are q ( x , 0) = 0 v0 , q ( x, t) = 0, ( L 2) + s 1 s 2 otherwise x (1) The r are all zero and the r are given by [see Eq. (13.8b)] r =  = from which 2 r x v sin dx r L ( L )  s 0 L 2 (2) 4 v0 r r s sin sin 2 r r L 0, r = 4v r s 0 , ( 1)( r 1) 2 sin  r r L r even r odd (3) (Notice that the even modes are all missing, as expected from the symmetrical nature of the initial conditions.) Now, from Eq. (13.11), 1 = and r = r 1 . Therefore, L (4) r = According to Eq. (13.5), 4v0 r s , r odd ( 1)( r 1) 2 sin r 2 1 L (5) CONTINUOUS SYSTEMS; WAVES 439 q ( x , t ) = r e i r t sin r r x L r x L (6) =  r sin r t sin r Therefore, q ( x, t) = 1 4 v0 sin 1t sin s L sin x L 1 3 s 3 x  sin 3t sin sin + ... 9 L L (7) Notice that some of the odd modesthose for which sin ( 3 s L) = 0 are absent. 136. The initial conditions are q ( x , 0) = 0 4 v0 L 4v q ( x , 0) = 0 L 0 x L 2  x 0x L 4 L L x 4 2 L xL 2 (1) The velocity at t = 0 along the string, q ( x , 0 ) , is shown in the diagram. v v0 L 4 L 2 3L 4 L The r are identically zero and the r are given by: r =  = 2 r x q ( x , 0) sin L dx rL 1 0 L r r sin 2  2 sin 4 r 1 8 v0 2 3 (2) Observe that for r = 4n, r is zero. This happens because at t = 0 the string was struck at L 4 , and none of the harmonics with modes at that point can be excited. Evaluation of the first few r gives 440 CHAPTER 13 1 = 0.414 2 =  3 =  and so, 1 2 8 v0 4 = 0 1 8 v0 4 2 1 2.414 8v0 27 2 1 2.414 8v0 5 = 125 2 1 2 8 v0 2 6 = 216 1 (3) q ( x, t) = 8 v0 2 x x 1 + sin 2t sin 0.414 sin 1t sin 2 L 4 L 1 2.414 3 x 2.414 5 x + sin 3t sin  sin 5t sin  ... L L 27 135 (4) From these amplitudes we can find how many db down the fundamental are the various harmonics: Second harmonic: 0.250 = 4.4 db 10 log 0.414 Third harmonic: 2.414 27 10 log = 13.3 db 0.414 2 2 (5) (6) These values are much smaller than those found for the case of example (13.1). Why is this so? (Compare the degree of symmetry of the initial conditions in each problem.) 137. 3 L 7 h O h L 3 L 7 Since q ( x , 0 ) = 0 , we know that all of the r are zero and the r are given by Eq. (13.8a): r = The initial condition on q ( x , t ) is 2 r x q ( x , 0) sin L dx L0 L (1) CONTINUOUS SYSTEMS; WAVES 441 7h  3L x , 7h q ( x , 0 ) = ( 2 x  L) , L 7h 3L ( L  x ) , Evaluating the r we find 0x 3 4 L x L 7 7 4 L xL 7 3 L 7 (2) r = 98 h 3 r 2 2 4 r 3r sin 7  sin 7 (3) Obviously, r = 0 when 4 r 7 and 3r 7 simultaneously are integers. This will occur when r is any multiple of 7 and so we conclude that the modes with frequencies that are multiples of 7 1 will be absent. 138. For the loaded string, we have [see Eq. (12.152)] r = 2 Using = m d and L = ( n + 1) d , we have md sin r 2 ( n + 1) (1) r = = The function 2 d r sin 2 ( n + 1) r sin 2 ( n + 1) (2) 2 ( n + 1) L r 2 L = ( n + 1) sin r 2 ( n + 1) (3) is plotted in the figure for n = 3, 5, and 10. For comparison, the characteristic frequency for a continuous string is also plotted: r 2 L = r 2 (4) 442 CHAPTER 13 12 n = 10 10 ss tri ng Co nti nu ou r 2 L 8 6 n=5 4 n=3 2 0 0 2 4 6 8 10 Of course, the curves have meaning only at the points for which r is an integer. 139. From Eq. (13.49), we have: = D s 2 2 2 ; 0 = 2 b (1) From section 3.5, we know that underdamped motion requires: 2 2 < 0 Using (1) this becomes D 2 s 2 2 < 4 2 b or D2 < 4 s 2 2 b 4 s 2 2 b 4 s 2 2 b underdamped critically damped overdamped Likewise D 2 = D2 > The complementary solution to Eq. (13.48) for underdamped motion can be written down using Eq. (3.40). The result is: s ( t ) = Cs e  t cos ( 1t  s ) CONTINUOUS SYSTEMS; WAVES 443 2 2 where 1 = 0  2 , 0 and are as defined in (1), and Cs and s 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)): s ( t ) = C s e  t s 2 F0 sin cos ( t  s ) 2 cos ( 1t  s ) + 2 2 s D  2 + 2 b b where s = tan 1 D 2 s 2  b 2 From Eq. (13.40): q ( x , t ) = r ( t ) sin r r x b Thus r s 2 2 D 2 2 F0 sin 2 cos ( t  r ) Dt r x q ( x , t ) = Cr exp  cos  2 t  r + sin b 2 2 b 2 4 r b r  2 + D 2 b (underdamped) 1310. From Eq. (13.44) the equation for the driving Fourier coefficient is: f s ( t ) = F ( x , t ) sin 0 b s x dx b If the point x is a node for normal coordinate s, then x n = where n is an integer s b s (This comes from the fact that normal mode s has shalf wavelengths in length b.) For x n = , b s sin s x = sin n = 0; hence f s ( t ) = 0 b 444 CHAPTER 13 Thus, if the string is driven at an arbitrary point, none of the normal modes with nodes at the driving point will be excited. 1311. From Eq. (13.44) f s ( t ) = F ( x , t ) sin 0 b s x dx b (1) where F ( x , t ) is the driving force, and f s ( t ) is the Fourier coefficient of the Fourier expansion of coordinate s. Thus, we desire F ( x , t ) such that F ( x , t ) . Eq. (13.45) shows that f s ( t ) is the component of F ( x , t ) effective in driving normal f s ( t ) = 0 for s n 0 for s = n From the form of (1), we are led to try a solution of the form F ( x , t ) = g ( t ) sin n x b where g(t) is a function of t only. Thus f s ( t ) = g ( t ) sin 0 b n x s x sin dx b b ; hence f s ( t ) = 0 for s n. x=0 b For n s, the integral is proportional to sin For n = s, we have ( n s) x b f s ( t ) = g ( t ) sin 2 0 b n x b dx = g ( t ) 0 2 b Only the nth normal coordinate will be driven. Thus, to drive the nth harmonic only, n x F ( x , t ) = g ( t ) sin b CONTINUOUS SYSTEMS; WAVES 445 1312. The equation to be solved is s + Compare this equation to Eq. (3.35): D s + s 2 2 =0 b s (1) 2 x + 2 x + 0 x = 0 The solution to Eq. (3.35) is Eq. (3.37): x ( t ) = e  t A1 exp ( 2 2 2  0 t + A2 exp  2  0 t ) ( ) Thus, by analogy, the solution to (1) is D 2 s 2 2 D 2 s 2 2   ns ( t ) = e  Dt 2 A1 exp t + A2 exp  t 2 2 b b 4 4 Assuming k is real, while and v are complex, the wave function becomes 1313. ( x , t ) = Ae i(t + it  kx) = Ae ( t  kx ) e  t whose real part is (1) ( x , t ) = Ae  t cos ( t  kx ) and the wave is damped in time, with damping coefficient . From the relation k2 = we obtain (2) 2 v2 (3) ( + i ) 2 = k 2 ( u + iw ) 2 (4) By equating the real and imaginary part of this equation we can solve for and in terms of u and w: = and k 2 uw (5) kw = iku (6) Since we have assumed to be real, we choose the solution 446 CHAPTER 13 = kw Substituting this into (5), we have (7) = ku as expected. Then, the phase velocity is obtained from the oscillatory factor in (2) by its definition: V= (8) Re = k k V=u (9) That is, 1314. Vn2 In2 L Vn1 In L Vn In+1 L Qn Vn+1 In+2 L Vn+2 Qn2 C Qn1 Qn+1 C C Qn+2 C C Consider the above circuit. The circuit in the nth inductor is I n , and the voltage above ground at the point between the nth elements is Vn . Thus we have Vn = Qn C and L dI n = Vn 1  Vn dt = Qn 1 Qn  C C (1) We may also write dQn = I n  I n+1 dt (2) Differentiating (1) with respect to time and using (2) gives L d2 I n 1 = [ I n 1  2I n + I n + 1 ] 2 dt C (3) or d2 In 1 = [ I n  1  2I n + I n + 1 ] 2 dt L C (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 L L ; x C C x (5) which will become the inductance and the capacitance, respectively, per unit length in the limit x 0 . From the above definitions and I r = I r + 1  I r (4) becomes d2 In 1 + ( I n1  I n ) = 0 2 dt LC or, d 2 I n ( I n )  =0 dt 2 LC Dividing by ( x ) , and multiplying by (LC), we find 2 (6) (7) (8) ( I n ) ( x ) 2  LC d2 In =0 dt 2 (9) But by virtue of the above definitions, we can now pass to the continuous limit expressed by In (t) I ( x, t) Then, ( I ( x , t ) ) x 2 and for x 0 , we obtain 2I 1 2I  =0 x 2 v 2 t 2 where v= 1 LC (13) (12)  LC 2 I ( x, t) =0 t 2 (11) (10) 1315. Consider the wave functions 1 = A exp i ( t  kx ) 2 = B exp i ( ( + ) t  ( k + k ) x ) (1) where ; k k . A and B are complex constants: 448 CHAPTER 13 A = A exp ( i a ) B = B exp ( ib ) The superposition of 1 and 2 is given by (2) = 1 + 2 t kx t kx i i k = exp i + t  k + x A exp ( i a ) e 2 + B exp ( ib ) e 2 2 2 (3) which can be rewritten as = exp i w + Define t kx  a + b t kx + b  a i i + b k 2 2 A e + B e t k + x + a 2 2 2 (4) t  xk b  a and A e  i( + ) 2 + B e i( + ) 2 = e i Therefore, (5) (6) =2 A + B 2 2 ( 2 ) ( + a ) 2 (7) (8) cos = 2 A + B 2 2 ( A+B ) 12 cos sin = 2 A 2 + B 2 ( B A ) 12 sin ( + a ) 2 (9) That is, is a function of ( ) t  ( k ) x . Using (6) and (7) (9), we can rewrite (4) as a + b i k = exp i + e t k + x + 2 2 2 and then, a + b k Re = cos + t  k + x cos + 2 2 2 k  sin + t  k + x sin + 2 2 (10) a + b 2 (11) CONTINUOUS SYSTEMS; WAVES 449 From this expression we see that the wave function 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. A+B Re AB wt kx 1316. As explained at the end of section 13.6, the wave will be reflected at x = x0 and will then propagate in the x direction. 1317. We let m , mj = m , j = 2n j = 2n + 1 (1) where n is an integer. Following the procedure in Section 12.9, we write F2 n = m q2 n = d ( q2 n1  2q2n + q2n+1 ) d (2a) (2b) F2 n +1 = m q2 n +1 = Assume solutions of the form ( q2n  2q2n+1 + q2n+ 2 ) q2 n = Ae i(t  2 nkd) q2 n + 1 = Be [ Substituting (3a,b) into (2a,b), we obtain  2 A =  B = 2 i t  ( 2 n + 1) kd ] (3a) (3b) m d ( Be ikd  2 A + Be  ikd  2B + Ae ) ) m d ( Ae ikd  ikd (4) from which we can write 450 CHAPTER 13 2 2  2  B A cos kd = 0 md m d 2 2 cos kd + B A  2 = 0 m d m d The solution to this set of coupled equations is obtained by setting the determinant of the coefficients equal to zero. We then obtain the secular equation 1 2 2 2 2 m d  m d   m m Solving for , we find 2 1 1 1 4 + = + sin 2 kd  d m m m m m m 2 (5) 2 d cos kd = 0 2 (6) 1 12 (7) from which we find the two solutions 2 1 1 1 4 + + sin 2 kd = +  d m m m m m m 2 1 1 12 12 2 1 1 1 1 4 2 2 + 2 = + sin kd   d m m m m m m (8) If m < m, and if we define a 2 , b m d 2 2 2 , c = a + b m d (9) Then the 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. 135). c b a /2d 0 k Using (9) we can write (6) as sin 2 kd = 2 2 2 + b  2 ) W ( ) 2 2 ( a ab (10) From this expression and the figure above we see that for > c and for a < < b , the wave number k is complex. If we let k = + i , we then obtain from (10) sin 2 ( + i ) d = sin 2 d cosh 2 d  cos 2 d sinh 2 d + 2i sin d cos d sinh d cosh d = W ( ) Equating the real and imaginary parts, we find (11) CONTINUOUS SYSTEMS; WAVES 451 2 2 2 2 sin d cosh d  cos d sinh d = W ( ) We have the following possibilities that will satisfy the first of these equations: a) sin d = 0, which gives = 0. This condition also means that cos d = 1; then is determined from the second equation in (12): sin d cos d sinh d cosh d = 0 (12)  sinh 2 d = W ( ) Thus, > c , and is purely imaginary in this region. (13) b) cos d = 0, which gives = /2d. Then, sin d = 1, and cosh 2 d = W ( ) . Thus, a < < b , and is constant at the value /2d in this region. c) sinh d = 0, which gives = 0. Then, sin 2 d = W ( ) . Thus, < a or b < < c , and is real in this region. Altogether we have the situation illustrated in the diagram. k 2d a b c 1318. The phase and group velocities for the propagation of waves along a loaded string are V (k) = U (k) = k = c d sin ( kd 2) 2 kd 2 (1) (2) d c d = cos ( kd 2) dk 2 where = c sin ( kd 2) The phase and group velocities have the form shown below. c d 2 (3) V(k) U(k) /d V,U 0 k When k = d , U = 0 but V = c d . In this situation, the group (i.e., the wave envelope) is stationary, but the wavelets (i.e., the wave structure inside the envelope) move forward with the velocity V. 452 1319. CHAPTER 13 The linear mass density of the string is described by 1 = 2 > 1 I 1 0 II 2 if x < 0; x > L if 0 < x < L III 1 x 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 = A e ( i t  k1 x ) be a wave train, oscillating with frequency , incident from the left on II. We can write for the different zones the corresponding wave functions as follows: I = Ae i(t  k1x ) + Be i(t + k1x) = e it Ae  ik1x + Be ik1x II = Ce i(t  k2 x) + De i(t + k2 x ) = e it Ce  ik2 x III = Ee i(t  k1x ) Where + De ik2 x (1) k1 = V1 , k2 = V2 , V1 = , V2 = 1 2 (2) and where 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 continuity of the wave function and its derivative at the boundaries x = 0 and x = L. For x = 0, we have I ( x = 0 ) = II ( x = 0 ) I x and for x = L, the conditions are = x=0 II x x=0 (3) II ( x = 0 ) = III ( x = L) II x = x=L III x x=L (4) Substituting as given by (1) into (3) and (4), we have k2  A + B = ( C + D) k1 and A+B=C+D (5) CONTINUOUS SYSTEMS; WAVES 453 C e  ik2 L + D e ik2 L = E e  ik1L C e From (6) we obtain  ik2 L +De ik2 L k =  1 E e  ik1L k2 (6) C= k 1 1 + 1 E e i( k2  k1 ) L 2 k2 1 k D = 1  1 E e  i( k2 + k1 ) L 2 k2 k 2 + k1 i 2 k 2 L e D k 2  k1 (7) Hence, C= From (5) we have (8) A= 1 k2 1 k2 1 + k C + 2 1  k D 2 1 1 (9) Using (7) and rearranging the above equation 2 1 ( k1 + k 2 ) i 2 k2 L A= e  ( k 2  k1 ) D 2k1 ( k 2  k1 ) (10) In the same way B= From (10) and (11) we obtain 1  ( k 2 + k1 ) e i 2 k2 L + ( k1 + k 2 ) D 2 k1 (11) 2 2 k1  k 2 e i 2 k2 L  1 B = A ( k1 + k 2 ) 2 e i 2 k 2 L  ( k1  k 2 ) 2 ( ) (12) On the other hand, from (6) and (8) we have E= which, together with (10) gives 2k 2 D i( k2 + k1 ) L e k 2  k1 (13) 4 k1 k 2 e i( k1 + k2 ) L E = A ( k1 + k 2 ) 2 e i 2 k 2 L  ( k1  k 2 ) 2 Since the incident intensity I 0 is proportional to A , the reflected intensity is I r = B , and the total 2 2 (14) transmitted intensity is I t = E , we can write 2 454 2 2 2 2 CHAPTER 13 Ir = I0 B A , It = I 0 E A (15) Substituting (12) and (14) into (15), we have, for the reflected intensity, 2  k 2 e i 2 k2 L  1 Ir = I0 ( k1 + k 2 ) 2 e i 2 k 2 L  ( k1  k 2 ) 2 2 1 (k ) 2 (16) From which 2 2 2 k1  k 2 (1  cos 2k 2 L) Ir = I0 4 4 2 2 2 2 k1 + k 2 + 6 k1 k 2  k1  k 2 2 cos 2k2 L ( ) ( ) (17) and for the transmitted intensity, we have It = I0 so that ( k1 + k 2 ) 2 e i 2 k L  ( k1  k 2 ) 2 2 4 k1k 2 e i( k1 + k2 ) L 2 (18) It = I0 2 2 8 k1 k 2 4 4 2 2 2 2 k1 + k 2 + 6 k1 k 2  2 k1  k 2 ( ) 2 (19) cos 2k 2 L We observe that I r + I t = I 0 , as it must. For maximum transmission we need minimum reflection; that is, the case of best possible transmission is that in which It = I0 Ir = 0 In order that I r = 0 , (17) shows that L must satisfy the requirement (20) 1  cos 2k2 L = 0 so that we have (21) L= m m = k2 , m = 0,1, 2,... 2 (22) The optical analog to the reflection 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 refraction, there will be almost no reflected wave. CONTINUOUS SYSTEMS; WAVES 455 1320. y I M 0 II We divide the string into two zones: I: x < 0 II: x > 0 Then, I = A1e i(t  kx) + B1e i(t + kx) II = A2 e The boundary condition is i ( t  kx ) (1) I ( x = 0 ) = II ( x = 0 ) 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 continuous at this point. The condition on the derivative is obtained by integrating the wave equation from x = to x = + and then taking the limit 0. Thus, M 2 t 2 I = II  x x (2) (3) x=0 x=0 Substituting the wave functions from (1), we find A1 + B1 = A2 ik (  A2 + A1  B1 ) =  2 MA2 which can be rewritten as A1  B1 = A2 From (4) and (6) we obtain A1 + B1 ik = A1  B1 ik  2 M from which we write (7) (4) (5) ( ik  M) 2 ik (6) 2 M 2ik B1 2M = = A1 2ik  2 M 1  2 M 2ik Define (8) 456 CHAPTER 13 2M = P = tan 2k Then, we can rewrite (8) as  iP B1 = A1 1 + iP And if we substitute this result in (4), we obtain a relation between A1 and A2 : A2 1 = A1 1 + iP B The reflection coefficient, R = 1 , will be, from (10), A1 tan 2 B P2 = R= 1 = 1 + P 2 1 + tan 2 A1 or, 2 2 (9) (10) (11) (12) R = sin 2 A and the transmission coefficient, T = 2 , will be from (10) A1 T= or, T = cos 2 A2 A1 2 2 (13) = 1 1 = 2 1+ P 1 + tan 2 (14) (15) The phase changes for the reflected and transmitted waves can be calculated directly from (10) and (11) if we substitute B1 = B1 e i B1 i A1 = A1 e A1 A2 = A2 e i A2 (16) Then, B i(  ) B1 P i tan 1 ( 1 P ) = 1 e B1 A1 = e A1 A1 1 + P2 and 1 A2 A2 i(A2 A1 ) 1 = = e e i tan (  P) A1 A1 1 + P2 (17) (18) CONTINUOUS SYSTEMS; WAVES 457 Hence, the phase changes are B1  A1 = tan 1 = tan 1 ( cot ) P A2  A1 =  tan ( P ) =  tan 1 1 1 (19) ( tan ) =  1321. The wave function can be written as {see Eq. (13.111a)] ( x, t) = +  A(k) e k0 +k k0 k i ( t  kx ) dk (1) Since A(k) has a nonvanishing value only in the vicinity of k = k0 , (1) becomes ( x, t) = According to Eq. (13.113), e i(t  kx ) dk (2) = 0 + 0 ( k  k0 ) Therefore, (2) can now be expressed as k0 +k k0 k (3) ( x , t ) = e i(0  w0 k0 )t =e( i 0  0 k0 ) t e( i 0t  x) k dk e i( k0 +k )(0 t  x )  e i( k0 k )( 0 t  x) i ( 0 t  x ) e i( 0 t  x) k  e i( 0 t  x) k 2i (4) = 2e ( i 0  0 k0 ) t 0t  x and writing the term in the brackets as a sine, we have ( x, t) = 2 sin ( 0 t  x ) k i( 0t  k0 x) e 0t  x (5) The real part of the wave function at t = 0 is Re ( x , 0 ) = 2 sin ( xk ) cos k0 x x (6) If k k0 , the cosine term will undergo many oscillations in one period of the sine term. That is, the sine term plays the role of a slowly varying amplitude and we have the situation in the figure below. 458 Re(x,0) CHAPTER 13 x 1322. a) Using Eq. (13.111a), we can write (for t = 0) ( x, 0) = +  A(k) e +   ikx dk =B e  ( k  k0 ) 2 e  ikx dk 2 = Be  ik0 x +  + e e  ( k  k0 ) i kk x e ( 0 ) dk = Be  ik0 x  u2 e  iux du (1)  This integral can be evaluated by completing the square in the exponent: +  e  ax 2 + e dx = bx  e e b  a x 2  x a dx b2 = 2 +  a x 2  b x + b a 4 a2 e 4 a dx  b b 2 +  a x  2a 4a  2 =e and letting y = x  b 2a , we have + e dx (2)  e  ax 2 e dx = e bx b 2 +  ay 2 4a  e a dy (3) Using Eq. (E.18c) in Appendix E, we have +  e  ax 2 e dx = bx e b2 4a (4) Therefore, CONTINUOUS SYSTEMS; WAVES 459 ( x , 0) = B  ik0 x  x2 4 e e (5) The form of ( x , 0 ) (the wave packet) is Gaussian with a 1 e width of 4 , as indicated in the diagram below. ( x, 0 ) B 2 2 1 B e x b) The frequency can be expressed as in Eq. (13.113a): ( k ) = 0 + 0 ( k  k0 ) + ... and so, + (6) ( x, t) = =  + A(k) e A(k) e i ( t  kx ) dk i 0 t + 0 ( k  k0 ) t  kx dk  = Be ( = Be ( i 0 t  k0 x ) +  i 0 t  k0 x ) + e e  ( k  k0 ) 2 e i 0 ( k  k0 ) t  ( k  k0 ) x dk  u2 e( i 0t  x) u du (7)  Using the same integral as before, we find ( x, t) = B i(0t  k0 x)  (0 t  x )2 e e 4 (8) c) Retaining the secondorder term in the Taylor expansion of (k), we have ( k ) = 0 + 0 ( k  k0 ) + 0 ( k  k0 ) + ... 2 1 2 (9) Then, ( x , t ) = e i ( 0 t  k 0 x ) = Be ( +  i 0 t  k0 x ) + A(k) e 1 2 i 0 ( k  k0 ) t + 0 ( k  k0 ) t  ( k  k0 ) x 2 dk  e t   i 0 u2 2 e( i w0 t  x ) u du (10) We notice that if we make the change  i 0 t 2 , then (10) becomes identical to (7). Therefore, 460 CHAPTER 13 ( x, t) = B where 2 i tk x i ( 0 0 ) e  ( x ,t) 2  iw0 t (11) ( x, t) = ( 0t  x ) 2 + 2 i 0t 1 4 2 + ( 0 ) t 2 2 (12) The 1 e width of the wave packet will now be w1 e ( t ) = 2 or, W1 e ( t ) = 4 t 1+ 0 2 2 4 2 + ( 0 ) t 2 2 (13) (14) In first order, W1 e , shown in the figure above, does not depend upon the time, but in second order, W1 e depends upon t through the expression (14). But, as can be seen from (8) and (11), the group velocity is 0 , and is the same in both cases. Thus, the wave packet propagates with velocity 0 but it spreads out as a function of time, as illustrated below. (x, t) t=0 t = t1 O x CHAPTER 14 (1) The Special Theory of Relativity 141. Substitute Eq. (14.12) into Eqs. (14.9) and (14.10): v x1 = x1  x1 c v x1 = x1 + x1 c From (1) ( = ) (2) x1 v = 1  x1 c From (2) x1 = x1 So 1 v 1 + c v 1 1  = c v 1+ c or = 1 1  v2 c2 461 462 CHAPTER 14 142. We introduce cosh y , sinh y v c and substitute these expressions into Eqs. (14.14); then x1 = x1 cosh  ct sinh x1 t = t cosh a  sinh c x2 = x2 ; x3 = x3 Now, if we use cosh = cos (i) and i sinh = sin (i), we can rewrite (1) as (1) ict =  x1 sin ( i ) + ict cos ( i ) Comparing these equations with the relation between the rotated system and the original system in ordinary threedimensional space, x1 = x1 cos + x2 sin x2 =  x1 sin + x2 cos x3 = x3 x2 x2 x1 = x1 cos ( i ) + ict sin ( i ) (2) (3) x1 x1 We can see that (2) corresponds to a rotation of the x1  ict plane through the angle i. If the equation 2 ( x , ict )  143. 1 2 ( x , ict ) =0 c2 t 2 1 2 ( x , ict ) =0 c2 t 2 (1) is Lorentz invariant, then in the transformed system we must have 2 ( x , ict )  (2) where 2 = 2 2 2 + + 2 2 x y z 2 (3) We can rewrite (2) as 2 ( x , ict ) x 2 = 0 =1 4 (4) THE SPECIAL THEORY OF RELATIVITY 463 Now, we first determine how the operator We know the following relations: 2 x 2 is related to the original operator 2 x 2 . (5) (6) (7) x = x x = x = Then, x = = x x x x 2 = 2 x x Therefore, 2 x 2 = x x = (8) = x x x (9) x x = 2 2 x (10) Since and are dummy indices, we see that the operator Lorentz transformation. So we have 2 ( x , ict ) x 2 = 0 2 2 x is invariant under a (11) This equation means that the function taken at the transformed point (x,ict) satisfies the same equation as the original function (x,ict) and therefore the equation is invariant. In a Galilean transformation, the coordinates become x = x  vx t y = y  vy t z = z  vz t t = t (12) Using these relations, we have 464 CHAPTER 14 x t 1 = + =  x x x t x x v x t 1 =  y y v y t 1 =  z y v z t = t t Therefore, 2 1 2 1 2 1 1 1 2 2 2 2 2 2 + +  2 = 2 + 2 + 2  2 2+ 2 + 2 + 2 2 x 2 y 2 z 2 c t 2 x y z c t v x v y v z t 1 1 1  2 + + v x x t v y y t v z z t 2 2 2 (13) (14) This means that the function (x,ict) does not satisfy the same form of equation as does ( x , ict) , and the equation is not invariant under a Galilean transformation. 144. In the K system the rod is at rest with its ends at x1 and x2 . The K system moves with a velocity v (along the x axis) relative to K. K K 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 v t1  2 x1 c v 1 2 c v 1 t2 = t  2 x2 2 2 c v 1 2 c t1 = 1 2 (1) where t1 and t2 are measured in the K system. From (1), we have THE SPECIAL THEORY OF RELATIVITY 465 t1  t2 = v ( t1  t2 )  c 2 ( x1  x2 ) v 1 2 c 1 2 (2) We also have x1  x2 = v ( t1  t2 ) = v ( t1  t2 ) = Multiplying (2) by v and using (3), (4), and (5), we obtain the FitzGeraldLorentz contraction: = 1 v2 c2 (6) (3) (4) (5) 145. The "apparent shape" of the cube is that shape which would be recorded at a certain instant by the eye or by a camera (with an infinitesimally 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 distorted, 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. 466 146. K CHAPTER 14 K v x1 x2 We transform the time t at the points x1 and x2 in the K system into the K system. Then, vx t1 = t  21 c vx t2 = t  22 c From these equations, we have t = t1  t2 =  v (1) ( x1  x2 ) =  vx c 2 1 c2 (2) 147. 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 from the origin of the K system and the time is t1 in K and t in K. Then we have 1 v t1 = t1  2 c c ( t  t1 ) = tv = x t1v = We eliminate , t1 , and x from these equations and we find (1) THE SPECIAL THEORY OF RELATIVITY 467 v t1 = 1  t c This is the time the observer reads by means of a telescope. 148. (2) The velocity of a point on the surface of the Earth at the equator is v= 2 Re = 2 6.38 108 cm 8.64 10 sec 4 ( ) (1) = 4.65 10 4 cm/sec which gives v 4.65 10 4 cm/sec = = 1.55 10 6 c 3 1010 cm/sec 1 t 1 + 2 2 1 2 = (2) According to Eq. (14.20), the relationship between the polar and equatorial time intervals is t = t (3) so that the accumulated time difference is = t  t = Supplying the values, we find = Thus, = 0.0038 sec (6) 1 1.55 10 6 3.156 107 sec/yr 10 2 yr 2 1 2 t 2 (4) ( ) ( ) ( ) (5) 149. w dm m + dm v + dv The unsurprising part of the solution to the problem of the relativistic rocket requires that we apply conservation of momentum, as was done for 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 p = mv = ( + d ) ( m + dm)( v + dv ) + w dm w where dm is the mass lost from the rocket, dm is the mass of the ejected fuel, w ( v  V ) 1  vV c 2 is the velocity of the exhaust with respect to the inertial frame, and (1) ( ) w 1 1  w 2 c 2 . One can easily calculate d = 3 d , ad after some algebra one obtains 468 CHAPTER 14 2 m dv + v dm + w dm w (2) where we of course keep infinitesimals only to first order. The additional unknown dm is unalarming because of another conservation law E = mc 2 = ( + d ) ( m + dm) c 2 + w dm c 2 Subsequent substitution of dm into (2) gives, in one of its many intermediate forms (3) 2 m dv 1  w + dm ( v  w ) = 0 c (4) and will finally come to its desired form after dividing by dt m dv dm +V 1 2 = 0 dt dt ( ) (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 0.6c by ejecting half its mass. 1410. From Eq. (14.14) x1 = ( x1  vt ) v t = t  2 x1 c Solving (1) for x1 and substituting into (2) gives (1) (2) v x t = t  2 1 + vt c t + v v2 t x1 = t  2 t = 2 c c v t = t + 2 x1 c Solving (2) for t and substituting into (1) gives t v x1 = x1  v + 2 x1 c or x1 = ( x1 + vt ) THE SPECIAL THEORY OF RELATIVITY 469 1411. 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 component of the stick will be 1  v 2 c 2 in the direction of motion. Thus, the x 1 cos 1  v 2 c 2 while the perpendicular component will be unchanged: sin So, to the observer in K, the length and orientation of the stick are = sin 2 + 1  v 2 c 2 cos 2 ( ) 12 = tan 1 or cos 1  v 2 c 2 sin cos 2 = sin 2 + 2 tan = tan 12 1412. The ground observer measures the speed to be v= 100 m = 2.5 108 m/s .4 sec The length between the markers as measured by the racer is = 1  v2 c2 2 2.5 = 100 m 1  = 55.3 meters 3 The time measured in the racer's frame is given by 470 CHAPTER 14 v t = t  2 x1 c 2.5 10 8 m/s (100 m ) .4 sec  c2 1  ( 2.5 3) 2 ( ) = = .22 sec The speed observed by the racer is v= = = 2.5 108 m/s t t 1413. t = t t = 1.5 s = ( 1  0.9992 ) Therefore t 34 s . 1 2 22.4 1414. K K v source receiver In K, the energy and momentum of each photon emitted are E = h 0 and p= h 0 c Using Eq. (14.92) to transform to K: h E = h = ( E  vp1 ) ; p1 =  0 c v = h 0 + h 0 c THE SPECIAL THEORY OF RELATIVITY 471 So v = 0 1 + c = 0 = 0 1+ 1 2 1+ 1 which agrees with Eq. (14.31). 1415. From Eq. (14.33) = Since = c 1 1+ 0 0 = or 1 1+ = With 0 = 656.3 nm and = 1 1+ 0 4 10 4 , = 656.4 nm. 3 10 8 So the shift is 0.1 nm toward the red (longer wavelength). 1416. K v K star Earth Consider a photon sent from the star to the Earth. From Eq. (14.92) E = ( E  vp1 ) also E = ( E + vp1 ) 472 CHAPTER 14 Now E = h , E = h 0 , Substituting yields p1 =  h cos , c p1 =  h 0 cos c 0 = (1 + cos ) and = 0 (1  cos ) Thus (1 + cos )(1  cos ) = 2 1 + cos  cos  2 cos cos = 1  2 cos  cos  cos cos =  Solving for cos yields cos = where cos  1  cos =v c = angle in earth's frame = angle in star's frame 1417. From Eq. (14.33) = Since 1 0 1+ =c , = We have = 1.5 0 . This gives = or 5 13 1+ 0 1 v = 1.2 10 8 m/sec THE SPECIAL THEORY OF RELATIVITY 473 1418. K v source K light observer Proceeding as in example 14.11, we treat the light as a photon of energy h. In K : E = h 0 , p = In K : h 0 c E = h = ( E + p1 ) h 0 c 1+ 1 For the source approaching the observer at an early time we have p1 = Thus = 0 + 0 = 0 v c For the source receding from the observer (at a much later time) we have p1 =  and h 0 c 1 1+ = 0 So = 0 = 0 1+ source approaching observer 1 1 source receding from observer 1+ 474 1419. K CHAPTER 14 source K v observer Proceeding as in the previous problem, we have In K : E = h p1 =  In K : So h h cos =  c c r + t2 2 r E = ( E + p1 ) = h 0 h 0 = or h r h  c r2 + t2 2 2 c 2 + 2 1  r  t r t 1 0 = (1  r ) 1  r2  t2 1  r2  t2 0 = = 0 1  r For > 0 , we have (1  r )2 > 1  r2  t2 t2 > 2 r  2 r2 t2 > 2 r (1  r ) 1420. As measured by observers on Earth, the entire trip takes 4 lightyears 80 2 = 3 years 0.3 c The people on earth age the astronaut ages 80 years. The astronaut's clock is ticking slower by a factor of . Thus, 3 THE SPECIAL THEORY OF RELATIVITY 475 80 80 1  0.32 = 0.95 years 3 3 So Those on Earth age 26.7 years. The astronaut ages 25.4 years. 1421. 1  ( 2 ) mv v 2 d 0 +v F= = m0 32 dt 1  2 1 2 1  2 ( ) (1) v v = m0 + 32 1  2 1  2 ( ) If we take v = v1e1 (this does not mean v2 = v3 = 0 ), we have v v v1 1 1 v1 m0 c c = F1 = m0 + 2 2 32 2 1 1 1 ( ) ( ) 32 v1 = m v1 (2) F2 = m0 1  2 m0 1  2 v2 = mt v2 (3) F3 = v3 = mt v3 (4) 1422. The total energy output of the sun is dE = 1.4 10 3 W m 2 4 R 2 dt ( ) (1) where R = 1.50 1011 m is the mean radius of the Earth's orbit around the sun. Therefore, dE dt The corresponding rate of mass decrease is dm 1 dE = dt c 2 dt 4.4 109 kg s 1 (3) 3.96 10 26 W (2) The mass of the sun is approximately 1.99 10 30 kg , so this rate of mass decrease can continue for a time 476 1.99 10 30 yr 4.4 109 kg s 1 CHAPTER 14 T= 1.4 1013 yr (4) Actually, the lifetime of the sun is limited by other factors and the sun is expected to expire about 4.5 10 9 years from now. 1423. From Eq. (14.67) 2 p 2 c 2 = E2  E0 2 = ( E0 + T )  E0 2 = 2E0 T + T 2 p 2 c 2 = 2T mc 2 + T 2 1424. The minimum energy will occur when the four particles are all at rest in the center of the mass system after the collision. Conservation of energy gives (in the CM system) 2Ep = 4 mp c 2 or Ep ,CM = 2mp c 2 = 2E0 which implies = 2 or = 3 2 To find the energy required in the lab system (one proton at rest initially), we transform back to the lab E = ( E + vp1 ) (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 vp1 = v ( pCM ) = v ( mu) = mv 2 = mc 2 2 Since = 2, = 3 2 , vp1 = Substituting into (1) 3 E0 2 THE SPECIAL THEORY OF RELATIVITY 477 3 7 Elab = 2E0 + E0 = 2 E0 = 7E0 2 2 The minimum proton energy in the lab system is 7 mp c 2 , of which 6 mp c 2 is kinetic energy. 1425. Let B = B0 z v = vx i + v y j Then i qv B = q v x 0 j vy 0 k 0 B0 = q vy B0 i  vx B0 j F = qv B = d d ( p) = m ( v ) gives dt dt qB d v = 0 v y i  vx j dt m ( ) Define q B0 m Thus vx = vy and vy =  vx or vx = vy =  2 vx and vy =  vx =  2 vy So vx = A cos t + B sin t vy = C cos t + D sin t Take vx ( 0 ) = v , vy ( 0 ) = 0 . Then A = v, C = 0. Then vx ( 0 ) = vy ( 0 ) = 0 vy ( 0 ) =  vx ( 0 ) =  v B = 0, D =  v 478 Thus CHAPTER 14 v = i v cos t  j v sin t Then r=i The path is a circle of radius v v sin t + j v cos t r= v q B0 m = mv q B0 = p q B0 From problem 1422 T2 p = 2Tm + 2 c 12 So T2 2Tm + 2 c r= q B0 12 1426. below Suppose a photon traveling in the xdirection is converted into an e  and e + as shown e+ e before after Cons. of energy gives pp c = 2Ee where pp = momentum of the photon Ee = energy of e + = energy of e  Cons. of px gives pp = 2 pe cos Dividing gives (p e = momentum of e + , e  ) THE SPECIAL THEORY OF RELATIVITY 479 pp c pp Ee pe cos =c= or 2 pe c 2 cos 2 = Ee2 2 But Ee2 > pe c 2 , so (1) cannot be satisfied for cos 2 1 . (1) An isolated photon cannot be converted into an electronpositron pair. This result can also be seen by transforming to a frame where px = 0 after the collision. But, before the collision, px = pp c 0 in any frame moving along the xaxis. So, without another object nearby, momentum cannot be conserved; thus, the process cannot take place. 1427. The minimum energy required occurs when the p and p are at rest after the collision. By conservation of energy 2Ee = 2 ( 938 MeV ) Ee = 938 MeV = T + E0 Since Ee = 0.5 MeV , Te+ = Te = 937.5 MeV 1428. Tclassical = 1 mv 2 2 Trel = (  1) mc 2 Tclassical We desire Trel  Tclassical 0.01 Trel 1 mv 2 2 1 0.01 (  1) mc2 1 2 v 2 0.99 (  1) c 2 2 1 Putting = 1  2 1.98 ( ) 1 2 and solving gives 480 v 0.115 c CHAPTER 14 The classical kinetic energy will be within 1% of the correct value for 0 v 3.5 107 m/sec, independent of mass. 1429. E = E0 E = 30 109 eV E0 For 0.51 106 eV, 5.88 10 4 1 1  2 1 2 2 = or = 1  2 = 1  1.4 10 10 ( ) 12 1 v = 1  1.4 10 10 c = 0.99999999986 c ( ) 1430. A neutron at rest has an energy of 939.6 MeV. Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV. Other than rest energies 0.8 MeV is available. 1431. 0.98c Conservation of energy gives E = 2Ep where Ep = energy of each photon (Cons. of py implies that the photons have the same energy). THE SPECIAL THEORY OF RELATIVITY 481 Thus E0 = 2Ep Ep = E0 2 = 135 MeV 2 1  0.98 2 = 339 MeV The energy of each photon is 339 MeV. Conservation of px gives mv = 2 pp cos where pp = momentum of each photon cos = (135 Mev/c ) ( 0.98 c) 2 2 1  0.98 2 ( 339 MeV/c ) = 0.98 = cos 1 0.98 = 11.3 1432. From Eq. (14.67) we have 2 E2  E0 = p 2 c 2 With E = E0 + T , this reduces to 2E0 T + T 2 = p 2 c 2 Using the quadratic formula (taking the + root since T 0) gives 2 T = E0 + p 2 c 2  E0 Substituting pc = 1000 MeV E0 ( electron ) = 0.5 MeV E0 ( proton ) = 938 MeV gives Telectron = 999.5 MeV Tproton = 433 MeV 482 1433. e 120 n before after p 120 CHAPTER 14 Conservation of py gives pe sin 60 = p sin 60 or pe = p Conservation of px gives pp = pe cos 60 + p cos 60 = pe So pe = pp = p p Conservation of energy gives E0 n = Ee + Ep + E 2 2 E0 n = E0 e + p 2 c 2 + E0 p + p 2 c 2 + pc (1) Substituting E0 n = 939.6 MeV E0 p = 938.3 MeV E0 e = 0.5 MeV and solving for pc gives p = 0.554 MeV/c pp = pe = p = 0.554 MeV/c Substituting into T = E  E0 2 = E0 + p 2 c 2  E0 gives ( E0 = 0 ) T = 0.554 MeV Tp = 2 10 4 MeV, or 200 eV Te = 0.25 MeV THE SPECIAL THEORY OF RELATIVITY 483 s 2 =  c 2t 2 + x1 2 + x2 2 + x3 2 1434. Using the Lorentz transformation this becomes s 2 = c 2t 2  2 v 2 x1 + 2 x1vt x 2 + v 2t 2  2x1vt 2 2 c2 + 1 + x2 + x3 1  v2 c2 1  v2 c2 2 2 v 2 x1 2 2 v 2 2 x1  c 2  c t  c 2 t + x2 + x2 = 2 3 2 2 1 v c 2 2 2 =  c 2t 2 + x1 + x2 + x3 So s 2 = s2 1435. Let the frame of Saturn be the unprimed frame, 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) u1 = Substituting v = 0.9 c u1 + v u v 1 + 12 c u1 = 0.2 c gives u1 = 0.93 c 1436. Since F = d d dX m d and X = ( x1 , x2 , x3 , ict ) we have F1 = d d d2 x dx1 = m 21 m d d d 2 x2 d 2 F3 = m d 2 x3 d 2 F2 = m F4 = d d d 2t d ( ict ) = icm 2 m d d 484 Thus F1 = m d 2 x1 d2  m 2 ( x1  vt ) d 2 d d 2 x1 d 2t = mv 2 = ( F1 + i F4 ) d 2 d CHAPTER 14 =m F2 = m d 2 x2 d 2 x2 =m = F2 ; F3 = F3 d 2 d 2 d d 2 vx1 t  c 2 F4 = icm = icm d 2t d2 x  i m 21 d 2 d = ( F4  i F1 ) Thus the required transformation equations are shown. 1437. From the Lagrangian L = mc 2 1  1  2  ( ) 1 2 kx 2 (1) we compute L =  kx x L L = = mc v v 1  2 Then, from (2) and (3), the Lagrange equation of motion is d dt from which mc + kx = 0 1  2 mc (4) (2) (3) (1  ) Using the relation c = we can rewrite (4) as 2 32 + kx = 0 (5) dv dv dx dv = =v dt dx dt dx (6) THE SPECIAL THEORY OF RELATIVITY 485 mc 2 2 32 (1  ) This is easily integrated to give mc 2 d + kx = 0 dx (7) 1 where E is the constant of integration. 2 + 1 2 kx = E 2 (8) = 0: The value of E is evaluated for some particular point in phase space, the easiest being x = a; 1 2 ka 2 E = mc 2 + From (8) and (9), mc 2 1 Eliminating 2 from (10), we have 2 (9) + 1 2 1 kx = mc 2 + ka 2 2 2 (10) 2 = 1  m2 c 4 2 2 1 mc + k a 2  x 2 2 ( ) = k a2  x 2 ( ) 2 k 2 2 mc + 4 a  x 2 2 k 2 mc + a  x 2 2 ( ) ( ) (11) and, therefore, 1 dx = = c dt k a2  x 2 ( ) mc 2 + k a 2  x 2 4 mc + k a  x 2 2 ( ( 2 ) ) 2 (12) The period will then be four times the integral of dt = dt(x) from x = 0 to x = a: m =4 k 0 a k 2 2 1 + 2mc 2 a  x dx k 2 2 2 2 a  x 1+ a x 4 mc 2 ( ) ( ) (13) Since x varies between 0 and a, the variable x a takes on values in the interval 0 to 1, and therefore, we can define sin = from which x a (14) 486 a2  x 2 a CHAPTER 14 cos = and (15) dx = a 2  x 2 d (16) We also define the dimensionless parameter, Using (14) (17), (13) transforms into a 2 k mc 2 (17) = Since ka 2 mc 2 2a c 2 0 (1 + 2 2 cos 2 1 + 2 cos 2 ) d (18) 1 for the weakly relativistic case, we can expand the integrand of (18) in a series of powers of : (1 + 2 cos ) 1 + 2 ( (1 + cos ) 2 2 2 2 12 2 2 cos 2 1  cos 2 2 ) 1 1 + 2  2 cos 2 2 3 = 1 + 2 cos 2 2 Substitution of (19) into (18) yields 2a c = 2 (19) 1 + 2 0 3 2 cos 2 d 2 a 3 a + c 2c 1 + 2 sin 2 0 (20) Evaluating (20) and substituting the expression for from (17), we obtain = 2 or, m 3 a 2 + k 8c 2 k m (21) = 0 1 + 3 ka 2 16 mc 2 (22) THE SPECIAL THEORY OF RELATIVITY 487 1438. F= dp d = ( mu) dt dt d ( u) dt d u dt 1  u 2 c 2 (for m = constant) =m =m 2 2 1 u c = m ( ) 12 u  u  2 1  u2 c 2 c 1  u2 c2 ( ( ) ) 1 2 du dt = m 1  u2 c 2 Thus F=m ( ) 3 2 du dt du 1  u2 c 2 dt ( ) 3 2 1439. The kinetic energy is 2 T = p 2 c 2 + m0 c 4  m0 c 2 (1) For a momentum of 100 MeV/c, Tproton = 10 4 + ( 931)  931 936  931 = 5 MeV 2 (2) (3) Telectron = 10 4 + ( 0.51)  0.51 100  0.5 = 99.5 MeV 2 In order to obtain and , we use the relation E = mc 2 = m0 c 2 = so that m0 c 2 1  2 (4) = and E m0 c 2 (5) = 1 electron = 1 2 (6) (7) 100 200 0.51 488 CHAPTER 14 electron = 1  200 0.999988 1 This is a relativistic velocity. 2 (8) proton = 936 1.0054 931 1 2 (9) proton = 1  1.0053 0.1 This is a nonrelativistic velocity. 1440. (10) If we write the velocity components of the centerofmass system as v j , the v j E p , j = p , j  2 c transformation of p , j into the centerofmass system becomes (1) = 0 must be satisfied, we have where = 1 1 v2 j c2 . Since in the centerofmass system, p ,j p ,j v j E = p , j  2 = 0 c (2) or, vj c p c = E ,j (3) 1441. We want to compute T1 E1  m0 c 2 = T0 E0  m0 c 2 (1) where T and E represent the kinetic and total energy in the laboratory system, respectively, the subscripts 0 and 1 indicate the initial and final states, and m0 is the rest mass of the incident particle. The expression for E0 in terms of 1 is E0  m0 c 2 1 (2) E1 can be related to E (total energy of particle 1 in the center of momentum reference frame 1 after the collision) through the Lorentz transformation [cf. Eq. (14.92)] (remembering that for the inverse transformation we switch the primed and unprimed variables and change the sign of v): THE SPECIAL THEORY OF RELATIVITY 489 E1 = 1 ( E1 + c1 p1 cos ) (3) where p1 = m0 c1 1 and E1 = m0 c 2 1 : E1 = m0 c 2 1 2 1 + 1 2 cos Then, from (1), (2), and (4), ( ) (4) T1 1 2 + 1 2 1 2 cos  1 = T0 1 1 For the case of collision between two particles of equal mass, we have, from Eq. (14.127), 1+ 1 2 (5) 12 = and, consequently, (6) 1 2 1 2 = 1 2  1 = Thus, with the help of (6) and (7), (5) becomes 1 1 2 (7) T1 1  1 + ( 1  1) cos = T0 2 ( 1  1) = 1 + cos 2 (8) We must now relate the scattering angle in the center of momentum system to the angle in the lab system. Squaring Eq. (14.128), which is valid only for m1 = m2 , we obtain an equation quadratic in cos . Solving for cos in terms of tan 2 , we obtain cos =  1 +1 2 2 tan 2 1 tan 2 1+ 1 +1 (9) One of the roots given in (9) corresponds to = , i.e., the incident particle reverses its path and is projected back along the incident direction. Substitution of the other root into (8) gives 2 cos 2 T1 1 = = T0 1 + 1 + 1 tan 2 2 cos 2 + ( 1 + 1) sin 2 2 An elementary manipulation with the denominator of (10), namely, (10) 490 2 cos 2 + ( 1 + 1) sin 2 = 2 cos 2 + 1 1  cos 2 + sin 2 = 1 + sin 2 + cos 2  1 cos 2 + cos 2 = 1 + 1  1 cos 2 + cos 2 = ( 1 + 1)  ( 1  1) cos 2 CHAPTER 14 ( ) (11) provides us with the desired result: 2 cos 2 T1 = T0 ( 1 + 1)  ( 1  1) cos 2 Notice that the shape of the curve changes when T1 > m0 c 2 , i.e., when 1 > 2 . T1 T0 (12) 1.0 T1 = 0.1 GeV 0.8 T1 = 1 GeV 0.6 T1 = 10 GeV 0.4 0.2 0 0 30 60 90 1442. y mec2 x h h From conservation of energy, we have h + me c 2 = me c 2 + h (1) Momentum conservation along the x axis gives h h cos + me v cos = c c (2) Momentum conservation along the y axis gives me v sin = h sin c (3) THE SPECIAL THEORY OF RELATIVITY 491 In order to eliminate , we use (2) and (3) to obtain cos = 1 h h c  c cos me v h sin = sin me v (4) Then, 1 cos + sin = 1 = 2 2 2 me v 2 2 h 2 h 2 h h cos  2 + + c c c c (5) Since = 1 1 v c2 2 and v = c 2  1 we have 2 v 2 = c 2 ( 2  1) (6) Substituting from (1) into (6), we have 2v2 = 2h h2 (  ) + 2 2 (  ) 2 me me c (7) From (5) and (7), we can find the equation for : h2 h h h h 2 cos = 2 hme (  ) + 2 (  ) + 2 c c c c c 2 2 (8) or, 2me c 2 2me c 2 + 2 (1  cos ) = h h (9) Then, 1 = 1 + h (1  cos ) me c 2 (10) or, E 1  cos ) E = E 1 + 2 ( me c 1 (11) The kinetic energy of the electron is 492 1 T = me c 2  me c 2 = h  h = E 1  E 1  cos ) 1+ 2 ( me c CHAPTER 14 T= 1  cos E2 2 me c 1 + E 1  cos ( ) me c 2 (12) ...
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This note was uploaded on 04/08/2008 for the course ENGR 2350 taught by Professor Borzova during the Spring '08 term at SMU.
 Spring '08
 BORZOVA
 Dynamics, Bifurcation, The Land

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