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Classical dynamics of particles and systems solutions pdf
Classical dynamics of particles and systems marion solutions. Classical dynamics of particles and systems 5th edition solutions. Classical-dynamics-of-particles-and-systems-marion-thornton solutions manual. Student solutions manual for thornton/marion classical dynamics of particles and systems 5th. Classical dynamics of particles and systems
solutions manual pdf.
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/wcsstore/CengageStoreFrontAssetStore/images/NoImageIcon.jpg "," ItemThumbnailImage ":" / wcsstore / CengageStoreFrontAssetStore / /NoImageIcon.jpg images "}] 0th chapter index v Prefácio problems solved in 1 Solutions Student manual vii matrices, vectors and vector Calculating Newtonian Mechanicsà ¢ 1 2 única partÃcula 29 3 4 79 nA
oscilaçÃμes £ linear chaos and oscilaçÃμes 127 5149 6 gravitaçà £ o Some mà © all of Calculating Variations of 165 7 Hamiltonà ¢ s and Hamiltonian Dynamics 181 8 Central-Force of Motion 233 9 Dining of a Particola System 277 10 Movement in an uninternial reference frame 333 11 Dynamics of Rigid Bodies 353 12 397 13 Coupled
Oscillations Container Systems ; Waves 435 14 Special theory of Relativity 461 III IV Content Chapter 1 Matrices, Vectors, and Vector Calculus 1-1. x2 = x2a â² x1a ² 45e x1 x3 x3a ² 45e axes and are on the plane. 1Xà ¢² 3x a 3x x Transform equations as the following: 1 1 3Cst 45 COS 45x xx =  ° AA  ° ,² 2 2x x = A ² 3 3 1st 45 Cos 45x xx =  °
à £² 1 1 1 1 2 2 xx = â € £² 3x 2 2x x = a ² 3 1 1 1 2 2 xx = â € £² 3x So, the matrix Transformation is: 1 1 0 2 2 0 1 0 1 1 0 2 2 yeah £ â € £ £ £ £ £ £ £ £ £ € € £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ 1 1-2 chapter 1-2. a) x1 abcd ct ± py oe x2 x3 from this diagram, we have cosoe oai  ± = cosoe obeseâ ² = (1) cosee odÞ³ = Taking the square of each of
the equations in (1) and Adding, we found 2 2 2 2COS Cos OA OB ODI  ± PY Registe £¹ + + = + + I £ £ 2 2 2 å ' ° ° ° ° ° (2), But 2 2 OA OB OC + = 2 (3) and 2 2 OC OB OE + = 2 (4) So 2 2 2 OA OD OE + + = 2 (5) So 2 2st 2 Cos Cos 1A  ± Py + + = (6) b) x3 aa ² x1 x2o edc bi c ² Bà ² and ² in first place, we have the following trigonometic
relationship: 2 2 2 COSEE OE OE OE Eii 2 â² q²² + a = a ² (7) matrices, vectors and vector carton 3 but, 2 2 2 2 2 2 2 cos cos cos cos cos cos and ee ob ob o oa od od oe oe Oe oe oe oe ppu  ± yy registry â € £ â € £ â € £ â € £ â € £ â € £ â € ¢ â € œa  ² + A + Ai £ 0. Register £ £ £ £ £ £ £ £ £ £ £ £ â € ° I  ° »I  °   ° I  °  ° »I   °   ± â € £²
= A + â € â € ¢ ²  € £ £  °  °  °  ° I  °  °  °  °  °   ± â € ° I  ± â € £ â € £ 2  ° I   ° »I  ±    ±  ± â € ƒâ € œThe £ £ £» (8) or, 2 2 22 2 2 2 2 2 2 2 2 Cos Cos Cos Cos Cos Cos Cos Cos Cos Cos Cos Cos 2 cos cos cos cos cos and oe oe oe oe oe oe oe oe ct ± py o  ± pp ± ct  ± ppyyy ± ct ct ± pp à  ² ® I  ±  ±  ± â € = + + +
+ + â € £ â € ¢ ² £²  ° ° ° ° °  ° â € £ £ ² â € ƒâ € ƒâ € ƒâ € ƒâ € œ £ â € £ £ £ £ £ £ £ £ »â € ƒâ € ƒ  € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒ² » ± PPYY = + + A ² â € ™2 (10) 3/1. X1 x2 x3 E3A ² E1 E1 E3 A E2A ² E1a ² E1 E1 E3 denote the original axes by ,,, and the corresponding unit vectors by and ,,. They
denote the new axes by, and the corresponding unit vectors per 1x 2x 3x 1 2e 3e 3E 1xà ² 2xà ¢² ² ² 1st, ², 2a ², and. The effect of rotation is and and, and. Therefore, the transformation matrix is written as: 3a â² 1 3 ² 1st, ² and 3 a2a ² () () () () () () () () () 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3 COS, COS, COS, 0 1 0 COS, COS, 0 0 1 1 0 0s, COS,
COS, Regista  ⠀ £ 1  € â € ¢ sa ² CT â € £ â € £ â € £ â € £ £ £ £ £ £ £ £ 0  € £ â € â € œQ² £ 1. Register £ £ = = I £ 1. Register £ â € £ £ £ £ £ 1. Register â € œa  ° I           ° a) Be c = AB in which A, B, and C are matrices. Then ij ik k kj ca = ba (1) () t ji jk ki ki jkij kk ccabba = = = AA 6 chapter 1, so that a cos 1 70.5 3 iai     €
¶ = =  ° I   € £ £ £ £ £ £ £ £ £ £  °¸ In the same way, 3 2 3 1 3 41 4 2 4 1 3 1 2 4 3 2 4 3 4 1 3 AA à ã, = = = = = ddddd dd ddddddddddddddd to  ± 1-8. Be the angle between A and R. Then 2-aa = a r can be written as 2 ° or, cosr ai = (1) this implies two qpo i = (2) thus, the end point of R has to be in a plane perpendicular to a and passing by P.
1-9. 2 = + AA IJK 2 3 = A + B + IJKA) 3 2à ¢ = ¢ ¢ AB IJK () () 1 22 2 23 1 (2) I   € ¢ Ai   ° I  ° »AB 14A = AB B) The B component along ABA IO Length of the B component along A is B COS I. COSAB IA = AB 2 6 1 3 6 COS or 26 6A IAA + A = = = ABB The direction is, of course, along A. A vector unit in the direction () 1 2 6 + the IJK matrices,
vectors and vector colon 7 thus, the of ba a aorto of air air () 1 2 + A IJKC) 3 3 3 Cos 6 14 2AB I A = = = = = =. 1 3 Cos 2 7 I = 71. O I â + ijke ab) 3 2 26 ¢. ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢. ¢+ = A + AB IJ ( () 3 1 1 5 0 A A + = à à à j i k A B A B 2 () () 10 2 14A = A + + + A B A B i j k 10/1. 2 sin cosb tb tà I = + Rija) 2 2 2 cos sen 2 sin cos btbtbtbt I I I I I I I 2I I = = A
= = à à = à vrijavijr 1 22 2 2 2 2 2 1 22 2 4 velocidade cos sin sin cos 4 btbbtt I I I I I I i ti £ ® I £ ¹ = = + i £ ° I £ »i £ ® I £ ¹ = + I £ £ ° I» v 1 22speed 3 cos 1b Ti i I £ ® I £ ¹ = + I £ £ ° I »b) para a 2t I I =, sin 1Ti =, cos 0tà = Assim, neste momento, Bi = VJ, 22bà = â ai Assim, 90i ° 8 CapÃtulo 1 1-11. a) Uma
vez que () IJK J ki jk Um BÃμà = A AB, temos () () () (), 1 2 3 3 2 2 1 3 3 1 3 1 2 2 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 () IJK jkiijk ABCCABABCABABCABABCCC AAAAAAAAACCCBBBBBBBBB CCC Ãμà â = = à à A + A = = A = = à à à ABCAB Ca (1) podemos também gravação (1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 ()
CCCBBBBBBCCCAAAAAA A A = A = = a a ABCBCA) (2) Notamos deste resultado que um mesmo número de permutações folhas o determinante inalterado. b) Considere vectores A e B no plano definido por e,. Uma vez que a figura definido por A, B, C é um paralelepÃpedo, a área da base, mas um 2e 3 3à = A A B e A = E C Altura do
paralelepÃpedo. Então, () () 3 área da base = altitude área da base a = volume do paralelepÃpedo A A = A A A C A B C E 12/1. O A B C ha B C A A C C A b b a a h A distância a partir da origem O para o plano definido por A, B, C é MATRIZES, vectores e do cálculo de vetor novembro 01-16. x3 P r i x2 x1 Ar I ar Porque eu Constanta = ra cos
constantra i = à dado que uma é constante, por isso sabemos que cos constantr i = mas eu COSR é a magnitude do componente de R ao longo de uma. O conjunto de vectores que satisfazem todos têm o mesmo ao longo de um componente; no entanto, o componente perpendicular a um é arbitrária. CONSTANÅ¢A = r uma Th é-nos Ã
superfÃcie representada pela constante de um plano perpendicular ao. à = r a um 17/01. um A i b B c C Considere o triângulo a, b, c, que é formado pelos vectores A, B, C. Desde () (2 2 22) AB = A = A à à = a a + CABCABABAB (1) ou, 2 2 2 2 B AB cosa i = + a C (2), que é a lei co-seno de trigonometria plana. 1-18. Considere o triângulo a, b,
c, que é formado pelos vectores A, B, C. A i ± CB y p bc um CapÃtulo 12 1 = A CAB (1), de modo que () A = A A CBABB (2), mas o lado esquerdo e o lado direito de (2) são escritos como: 3sinBC ct ± a = CB-e (3) e () 3sinAB óâ a = a a a = a = ABBABBBAB e (4) em que e é o vector de unidade perpendicular ao triângulo ABC. Portanto, 3
pecado sinBC ABI ± y = (5) ou, sin sin C Um y ± ct = Do mesmo modo, sin sin sin C A B y oc ± p = = (6), que é a lei sinusoidal de trigonometria plana. 1-19. x2 A i ± x1 a2 b2 a1 b1 b p a) Começamos notando que () 2 2 2 2 cosa b ab a- ± pA = + A AB (1) também pode escrever que ( ) () () () () () () () 2 22 1 1 2 2 2 2 2 2 2 2 2 2 2 2 cos cos
sin sin sin cos sin cos 2 cos cos sin sin 2 cos cos sin sin ababababab ab ab ab ct ct ± p ± p ± ct ct ± p p p ± ct ct ct ± ± ± u p p a = a + a = a + a = + + + A + = + A + ab p (2) MATRICES, vectores e do cálculo de vetor 13 Assim, comparando (1) e (2), conclui-se que () cos cos cos sin sini ± p ± p ct ct ± à = + p (3) b) Utilizando (3),
podemos encontrar () sin ± ct pA: () () () () () 2 2 2 2 2 2 2 2 2 2 2 2 2 2 sin 1 cos 1 cos cos sin sin 2cos sin cos pecado 1 cos 1 sin sin 1 cos 2cos sin cos cos sin sin 2sin sin cos cos cos sin sin cos cos pecado ct ± p ct ± p ct ± p ct ± p ct ± u ± p p p ± ct ct ct p ± ± ± ct ct p ± p ± p ct ct ct ± p ± p ± p ct ct ± p a = a a = a â a =
a a a a a = a + = a p (4), de modo que () sin sin cos cos sini ± p ± p ct ct ± â = j uma p (5) 1-20. a) Considerar os dois casos seguintes: Quando I A = 0ijÃ' mas 0ijkÃμ â. Quando i = j 0ijÃ' â mas 0ijkÃμ =. Portanto, 0ijk ij ij £ ô = a (1) b) Nós proceder da seguinte maneira: quando j = k, 0ijk ijjÃμ £ = = Termos como 11 and 11 0jà μ =.
Em seguida, 12 12 13 13 21 21 31 31 32 32 23 23ijk JK iiiiii JK  £  £  £  £  £  £  £  £  £  £  £  £  £  £ = + + + + + A = Agora, she suponha, entà £ o, 1 = = 123 123 132 132 1 1 2 JK  £  £  £  £ = + = + A 16 CapÃtulo 1 () primeiro IJK J ki JK B Repair μà A = BC. Entà £ o, () [] () () () () MNM MNM NJK j KN MN MN JK MM NJK
mjk MN jkn mjk jkmn jkmn LMN jkn mjk JKM n JL km k JM mjk JKM mmmmmmmmmmmm ABCABCABCABCABCABCABC ABCBACCABBC  £  £  £  £  £  £  £  £  £ vaccines vaccines vaccines vaccines c = a = = = n  £  «Â £ i = n ¶ ¬  £  £  Registe · I  £  £ Registe a = a i  £  «Â £ she ¶ a £ « a = a = a £ ¬ a £  · a £ ˆ ¯
 £  £ s a a £ = a a a a aaaa à ¢ à ¢ à ¢ ABCACAB Registe  £ ¶ Registe  £   · I a £ Assim, () () () = the time to study ¢ à ¢ ABCACBABC (2) 1-23. Write () JM MJ my Um ba μà AB = A () KRS sk rs r = C two μà um CD from £ EntÃ, matrizes, vectores to cálculo of vetor 17 () () [] ( ) () IJK JMM KRS r Si JK my rs IJK JM KRS SRA JK SRA JM IJK
RSK mrsj SRA kjm IR JS à © jr mrsj SRA jmmijmijjmjmjmijmjmj ABCDABCDABCDABCDABCDA BDCDABC  £  £  £  £  £  £  £  £  £  £ vaccines vaccines vaccines vaccines  £  £  £  £  Registe «she  £ ¶ a £ «  £ ¶ she AAA = a £ ¬ a £  · a £ ¬ a £  · a £ a £ £ a a a a a = £ a £  «Â £ i = a ¶ £ ¬ a £  · a £ a £ a = a = a a £ Â
«Â £ i = a ¶ she ¬  £  £ s · a £ £ a a a a a AA ABCD () () jmimii CABDCD Registe  £  «Â she ¶ Registe £  £  £ ¬ Registe I  ·  £  Registe a = £ à ¢ à ¢ Abd Portanto ABC, [() ()] () () = AAA ABCD ABC Abd C D 1-24. Expandindo from Produto triplo vector, temos () () () = the time to study ¢ à ¢ A A A A EEE (1) If, () A = A him (2) Assim, () ()
= a + um aaae him of the EU (3) (a study it) à © the componente of the direcçà £ o of it, he Tempo ao mesmo (aae) à © o componente of Uma perpendicular to it. CapÃtulo 18 de Janeiro 1-25. r Ei Ei ii Os vectores unitários em coordenadas esfà © ricas in £ from expressos em termos of coordenadas rectangulares por () () () cos cos, cos there, so
that, cos, 0 cos there, so that she COSR iiiiiiiiiiiiii  £ ¹ a = i º  £  £ à  ° a = i º  £  £ à  à º º £ = a £  »EEE (1) Assim, () cos cos there then, cos cos that there Cosi IIIIIIIIIIII ii = aaaae COSR iiii + a = a (2) is forma semelhante, () cos, that 0i iiii = to a cos pecado Rii ii IA is = (3) is SINR i ii II = + EEE (4) Agora, vamos qualquer vector
posiçà from £ à © x. Em seguida, eventually = xe (5) () so that rrrrrrrrrr IIIIIIIIII = + = + + = + + xeeeeeeeer (6) () () () () () 2 2 2 2 cos so that there are 2 2 cos cos 2 so that there rrrrrrrrrrrrrrrrrrrr iiiiiiii II iiiiiiiiii II iiiiiiiiiii = + + + + + + + + = + + + + a + a re xeeeeee him (7) OU, matrizes, vectores to cálculo of vetor 21 1-28. () () Full full ii ix
ha of um graduado rre (1) onde 2i ix = AR (2) Assim, () 2 2 1 full iiiiixxxx A = A = Uma RRR (3), of Modo que () 2 1 full IIIx Registe  £  «Â £ i = n ¶ •  £  £  £ Registe r er um formando Quero  £  · (4) OU, () 2ln r = r r graduado (5 ) 1-29. Let descrever to superfÃcie is 9r 2 = 1 + 2 + 1 = XYZ descrever to superfÃcie S. O à ¢ ngulo entre ie
ponto or (2, 2,1) from à © à ¢ ngulo entre estas the most perpendiculares superfÃcies, or ponto. From normal à © of 2 1s 2s 1s () () () () 2 2 2 1 1 2 3 2 2 1 1 2 3 9 9 2 2 2 4 4 2 XYZ is rxyzxyz = = = = A = + + + + = = a + graduado graduado graduado eeeeee 2-a (1), the normal time ©: (2s) () () 2 2 1 2 3 2 2 1 2 3 1 2 2 x 1a z E xyzz = = a = = + + a
+ = + = + + graduado graduado eeeeee (2) Assim, CapÃtulo 22 1 () () () () () () 1 2 1 2 1 2 3 1 2 3 cos 4 4 2 2 6 6 SSSS is a = a + a + + = graduado graduado graduado graduado eeeeee (3) OU, 4 cos i = 6 6 (4) of the qual partir um 74,2 9 6 cos i = a =  ° (5) 1-30. () () 3 1 iii ii ii iiii ixxxx iiiiii ix IIIIIII  £ = ® I  £ à ¢ à ¢ ¢ ¹à = = i +  £  £
Registe to ¢ à ¢ AI ºà  £  £  ° I  »= c + aaaa eeee out graduado Assim, () = + IIIIII graduado grad 1-31. a) () 1 2 3 2 1 1 2 2 1 2 2 2 2 2 nnniiji ji = iniijijniijijniiirrxx xnxxxnxxnr à  £ ® I I ≤ ¹ï £ £   " £ i £   ¶AAI Sign ¯  £  ° I =  £ ¬ Sign  £ à ¢  · A i  £  £  Sign ºï  £  £ Register i  £  £   ° i »i  £  'i  £  £ n = ¶
¬ Register ·  £ i £  Register Register £ £   " £ ¶ i = n £   £ ¬ Register £   · I Register £ à = aaaaaa graduated eeeee (1) therefore, (2) = NNR NR degree R (2) MATRICES, vectors and vector Calculating 23 b) () () () () () 3 3 1 1 1 2 2 1 2 2 ii iii IIJI ji iijijiiifrfrrfrxrfrxx rfrxxr fx r dr = A A x A = ¢ à £     'i  à £ ¶ A =  £  £ ¬ · A A
A A A A A A £ £ £     «i ¢ à  £ ¶ =   à £ ¬ ¯  £  £  · A i A A A A £ ¢ = AAA A A A graduated eeeee (3) Therefore, () () f RF = rrr grad R (4) c) ( ) () () () () 1 2 2 2 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 22 2 2 ln LN 1 2 2 2 1 2 3 ji ji ji iijjiijjiji iiijjiji ji jirrxxxxxxxxxxxxxxxxx rr aa £  ® AI I £  £   ¹ï 'i  £  £ Reg ¶AAI iste  £  ° I =  £ ¬ Sign
 £ à ¢  · AI  £  £  Sign ºï  £  £ Register i  £  °   £  »Sign  £ ® I  £  £  ¹ï 'i  £ ¶ Sign  £  £  Sign ºà ¢ Sign  £ ¬ Sign  £  £   · I Sign  £  £ ºï Sign  £ A Ui =  £  £  ° Register  £ Ia '¶ï to £  £ £ º A A A A A A A A £ £ £ ¬ à  ºï ¯  £  £  ·     £ i £    £  £  ° ºï to £  »Â £  £  ® The
¹ï  £ «  £ i £  ¶A Register Register  £  £ n ° = ¬ Register · A i £  £    Register ºï £ £ £ i Register £    ° i £  »i  £ Â"  £ i £ ¶ Register to ' £ i = a i + ¶A ¢  £  £ ¬ Register ·  £ i £  ¬ Register · A i   £ £ £ Register Register Register Register £ £  ®  = a + i £ i £   ° a ¢ à ¢ ¢ AAA aaaaaaaa 2 4 2 2 2 3 Register £
rrr 1r ¹ ºï to £  £  '= a + = (5) or () 2 = 2 1 ln rr to (6), 26 1 1-36 chapter. X d Z y c2 = x2 + y2 The integrally suggests the use of the divergência theorem. (1) V S = a ¢  ¢ à ¢ "à ¢ 'A A A A From the dv, only need to evaluate the total volume. Our cylinder has a radius c and time d and so the response to 1 = © (2) V 2 = dv c Di à ¢ '1-37. X Y Z
R to make the member directly, note that A, on the surface, and. RR = 3 and e = RD 5 3 3 24 4 d R RR SS IR = IA = A =  "à ¢ 'The one (1) To use divergência theorem, we need to calculate AA A. This à © best done in EFSA coordinates © rich, which one. Using Apêndice F, we see that 3 rr = E () 221 5rrr r 2 r à ¢ AA = AA (2) Therefore, () 2 2 2 0
0 0 5 4 sin RV DV DDRR Dr R ii 5i iiiaa = à ¢  "à ¢ Â"  A_ «Ã ¢ '(3) Alternatively, one can simply set this case dv. DRI = 24 r MATRIX, vectors and vector 27 Calculating 1-38. xzy C x2 + y2 = 1 z = 1 ¢ à ¢ x2 y2 At Stokeà theorem ¢ s we () R d C to A a = à ¢  "à ¢  'A A s (1) the curve C, which includes the Surface © s to the cÃrculo unitário
which is in the xy plane. Since área element in Surface à © chosen to be outside of the origin, the curve © anti-directed schedule, as required by the rule of poor £ right. Now switch to polar coordinates, so that we have d = D II and are sewed and sin I = A i + k on the curve. Since IA sini = EI = ¢ à 0i and k, we have () sin 2 2 0 C d I I i di = A = A a
A 'a ¢ Â' S (2) 1-39. a) Leta s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then, (1, 2,0) = AB; (1,0, 3) = CA; and (6 3.2) The AC = AB. Any vector perpendicular to the plane (ABC) should be parallel to an AB AC, so that the unit vector perpendicular to the plane (ABC) A © (6 7 3 7 2 7) = nb) Leta s is D = ( 1, 1,1) and M = (x, y, z) is the point on the
plane (ABC) nearest H. then (1, 1, 1 x yz = a a DH) a © n parallel data in a); Hereby 1 6 2 1 3 ¢ A and xy = 1 6 3 1 2 ¢ à Another xz = (A = 1 ,, XY) perpendicular to ZAH à © n 6 thus has (1) 3 2 xy 0 = Za +. Solving these equaçÃμes 3 is H (,,) (49.34 19 49, 39 49) X = Y = Z 5:07 DH = 1-40. a) On top of the hill, z © máximo; 0 2 and 6 x z y x = a ¢ Ã
¢ 0 18 2 8 x y y z A = 28 A + 1 A 28 well chapter x = 2; y = 3, and s hilla © height of one máximo [Z] = 72 m. In fact, this à © máximo the value of z, because given the equaçà £ z implies that for each given value of X (or Y) z discloses an upside down parábola term in y (or x) Variable. B) Point A: x = Y = 1, Z = 13. At this point, two of the tangent
vectors at the superfine of the hill are will be (1,0) (1.0, 8) ZX A = a t = and 2 (1,1) (0.1), (0.1, 22) ZY = ¢ = TT is evidently perpendicular to the hill surface, and the angle of a 2 (8, 22.1) of the axis a = AI between this and oz is 2 2 2 (0.0,1) (8 , 22.1) 1 23,438 s 22 1 a + + co I = So I = 87.55 degrees. = C) Suppose that in the I meaning  ± (in relation
to the W-e axis), at point A = (1,1,13) the hill is more ogreme. Of course, DY = (tan?  ±) DX ED 2 2 6 8 18 28 22 22 (TAN 1) Z XDY YDX XDX Ydy DX DY DXI  ± = + AAA + = A, then 2 2 COS 1 22 (tan 1) 22 2 COS (45) DX DX DY DZ DX CT CT   ± U + A = A + P = TAN = The hill is more accentuated when tanÞⲠis minimal, and when this
happens to = 45 degrees with respect to the WE axis. (Note that I  ± 135 does not give a physical response). 1-41. 2 (1) AA = a a b, then if only one or = a = 0. 0 = a b newtonian mechanicsà ¢ single partisicle 31 2-3. Y x v0 p P CT  ± The equation of the movement is M = F A (1) The gravitational force is the only applied force; therefore, 0x y f mx f
my mg = = i â € ƒâ € œ â € œ  ° = = A I  ± (2) integration of these equations and using the Condi Initial, () () () 0 0 0 0 COS sin xtvytv â €  ±  ± q £ £ £ £ £ = = I  € »(3) We found () () () 0 0 Cos Sin Xtvytv gt   ±  ±  € £ £ £ £ â € ƒâ € ƒâ € ƒâ € œA £ for x and y () () () () 0 2 0 1 COS SEN 2 XTVTYTVT GT CT CT ±   ±  ° ž  ° °  € »(5)
Suppose it takes a time t to reach Point P. So, 0 0 0 0 0 0 0 0 1 Cos Cos Sin Sin P 2 VTVT GT CT CT ±  ± = n £ ¹ º Regista  ° = A  ° º º  ° ž  € "(6), the elimination between these equations, 0 00 0 2 21 Sin Cos tan 0 2 VVTGG CT CT      ±   °   ° GT A + I  ±  ± ¯  ± Regista  ¸ = n £ · (7) from which 32 chapter 2 () 00 2 Sin
COS tan VTG CT CT    ± P = A (8) 4/2. One of the height Ballsà ¢ can be described by 20 0 2Y y v t gt = + a. The amount of time it takes to rise and descend to its initial height is, therefore, given by 02V g. If the time leading to the ball cycle through Jugglerà ¢ s's hands is 0.9 si =, so it should be 3 balls in the air for the time I. A single ball has to
remain in the air for at least 3a, so that the condition is 02 3v g i to ¥, or. 10 13.2 m SV ¢ â € ™ 2-5. Point Mg of N Max FlightPath Acceleration Plan A) From the force diagram that has () 2 RM MV r = n g and m. The acceleration that the pilot feels () 2 rm mv r = + n g and, which has a maximum amplitude at the bottom of the maneuver. b) If the
acceleration felt by the pilot should be less than 9 g, then we have () 12 2 3 330 ms 12.5 km 8 9 9.8 msv r gaaaa à ¢ â € = (1) a smaller circle than this will result in pilot darkening. 2-6. Let the origin of our coordinate system be at the end of the cattle tail (or the closest of cow / bull). a) The bales are initially moving at the velocity of the plane when it
fell. Describe one of these bales by the parametric equations 0 x x v = T + (1) Newtoniano Mechanicsà ¢ single Particle 33 20 1 2 y y gt = a (2) where, and we have to solve. (2), the time the burden reaches the soil is 0 80 mine = 0 x 02Y GI =. If we want the burden to land in () 30 mx i = a, then () 0 x x v0ã i = a. Replacing and other values, which
gives 44.4 -1S0 MV = to 0 210x mA. The farmer should drop the bales of 210 m for back the cattle. b) She could drop the burden before for any amount of time and did not hit the cattle. If she was late for the amount of time it takes the burden (or plan) to travel by 30 m in direction X, then she will attack the cattle. This time is given by () 030 m 0.68
SV. 2-7. Air resistance is always anti-parallel speed. The expression vector is: 2 1 1 2W C AV C AV VII Â ± â € £ â € = â € = Â ± â € ° ° ° ° ° I »V W WI VA (1) Including gravity and configuration, they obtison the parametic equations Mquidae M = F 2x BX XY = A + 2 (2) 2 2Y by XYG = A + (3) where 2W to MB CI =. Solving with a computer using the
data and, we discovered that if the farmer fall the burden 210 m behind the cattle (the response of the previous problem) then takes 4.44 s à earth 62,5 62.5 trás cattle. This means that the burden should be discarded in 178 m atrás cattle à land 30 m to trás. This soluçà £ â © © what plotted in FIG. The error since it à © à © allowed to do the
same as in the previous problem that already-Only depends on the Wed £ Fast Avia £ está the moving. -31.3 kg ¢ à ¢ MI = 200 A 180 A 160 A 140 A 120 A 100 A 80 A 60 A 40 0 A 20 40 60 80 Resistance With Without air the air Resistance x ( m) Y (m) 36 chapter 2 () 1 2 3 21 1ln 3z + zz z = a + (9) © válido (8) can be approximately expressed as 2 0
0 2011 3 kV v 2m Register KR tggg ® I £ £  £  ® ¹ï I ¹ £  £   à £   Sign ° Ai = A + £  £ Register º to £ The Sign ºï  £  £  £   ° I »i  £  £   ° A» ¢ | (10) Gives the correct result, as in (4) to limit the k 0. 2-10. The equaçà £ differential are invited to solve the à © Equaçà the £ (2.22), which à ©. Using the above values,
frames sà £ shown in the figure. Of course, the reader in the será £ able to distinguish between the results presented here and analÃticos results. The reader will have to take the word of the author that the graphs were obtained using a mà © all © rich in a computer. The results were within the 10 máximo x k = A x £ soluçà 8a of the analÃtica. 0
5 10 15 20 25 30 5 10 v vs TT (s) v (m / s) 0 20 40 60 80 0 5 10 100 V vs xx (m) v (m / s) 0 5 10 15 20 25 30 0 vs TT 50 100 x (s) x (m) 2-11. The equaçà the motion £ © 2 2 2 DXM KMV mg dt = A + (1) This equaçà the £ can be solved exactly the same way as in 2-12 problem and found NEWTONIAN MECHANICSà ¢ única partÃcula 37 2 0 1 2 log 2
g Register KR KR xkg I £ ®  £ ¢ ¹à n =  £  ¢ Register Register ºà £ £ £    ° I '(2) where the origin © taken as the point at which v = 0 v so that the initial condiçà £ â © () vv = 0x = 0. Thus, Dista ¢ 0v INSTANCE point to the point v = v 1 v = © () 1 0 0 2 1 2 1 2 g of KR KR svvkg log Register I £ ®  £ ¢ A = ¹à IU £  ¢ Register Register
ºà £ £ £    ° I '(3) 2-12. The £ equaçà the movement to the upward movement © 2 2 2 dt = DXM MKV mg to A (1) Using the Interface £ 2 2 dv dv dv dx dx dt v dt = dx dt = dx (2) you can rewrite (1) as dv dx 2 v kv g = a + (3) the Integraçà £ (3), we find () log 2 21 kv GX + C k = a + (4) where the constant C can be calculated using the initial £
condiçà that when x = 0: v = 0v () 201 g log2C KVK = + (5) therefore 1 2 0 2 2 log KR KR gxk = g + + (6) Now, the low equaçà £ © movement DXM 2 2 2 MKV mg dt = a + (7) This can be written as dv dx 2 v kv g = a + (8) Integrating (8) and using the initial condiçà £ x = 0 at v = 0 (w taking the highest point as the source for downward
movement) TITLE found Case 38 2 2 1 2 gxkg KR = log to (9) at the highest point the speed of partÃcula should be zero. Therefore, the point higher by substituting v = 0 (6): 2 01 log 2h kv gxkg + = (10) Then, substituting (10) into (9) 2 0 2 1 1 log log 2 2 kv ggkgkgk = a + v (11) Solving for v 2 0 2 0 kV = gv + GVK (12) we can find the terminal
velocity by putting x on (9). This Gives T g = v k (13) Therefore, 0 2 2 0 t t v v v v v = + (14) 2-13. The equaçà £ partÃcula of the motion © () 3 2dvm mk va dt = A + v (1) Integraçà £ the () 2 2 k dv dt = A + A vva  «Ã ¢ ' (2) and using eq. (E.3), Apêndice And, there are 2 2 2 2 1 2 ln kt C v AAV Register £ £  ® ¹ I = a + i £  £    ° Sign + i £
  ° I £   "(3) Therefore, aTV 2 2 2 a = EVA C a² + (4) NEWTONIAN MECHANICSà ¢ única partÃcula 41 b) Maximize d Interface for the to £ ± Ct () () () (2 0 0 2 2 2 cos sin sin cos vd DDG) Ct ± a ± a ± p Ct Ct p ± ± ± Ct p p = a + AAI £   ® R ¹ï £  £  £   ° I '() 0I cos 2 ± 2 ± 2 II A 4 = pA 2 I ²à  + c = ±) substitute (2) into (1) 0 Bad
2 ¡x 2 2 cos cos sin 2 4 4 2 pA = vgd The PPP Register I £ ®  £  £  ¹ï 'i  £  £ ¶ Register "= i +  £ ¬ Sign  £  £  · I sign ¬  £  £  Sign ¶A · I  £  £  Sign ºï  £  £  Sign Sign  £ sign  £  £  ° i i  £  »Using the identity () () sin 2 cos 1 1n 2 sin 2 a = ABAB Si + B 2 have máx 2 0 0 2 2 sin sin cos sin 2 1 2 1 2 sin vvggd  Â
ppppai  £ i £ ® = A = UI £ 1. Register  ° â €  °  ° I  ° »() 2 Max 1 SIN VDG P = + 2-15. MG I MG SIN I The equation of the movement along the plane is 2SIN DV M MG KM DT I = A V (1) (1) This equaçà as the £ 1 2 sin i = dv dtgk the VK (2) chapter 42 2 It is known that the speed of partÃcula continues to increase with time (ie 0dv dt>), so
that ( ) VI 2sing k>. Therefore, we must use Eq. (E.5a) Apêndice And to carry out the £ integraçÃ. Found 1 1 1 tanh vt sin sin C kggkk IIAI ®  £  £ ¹ I + i = £  £  Register º to £  £ Register º to £  £   ° I "(3) the initial condiçà £ v (t = 0) = 0 stands for C = 0. therefore, () sin tanh sing dxv gkk tt = i = di (4) that can integrate this £
equaçà for the displacement x as a £ funçà the time in. () sin = tanh singx ii gkk to the "t t Using equation (E.17a) Apêndice And, m is obta © () LN cosh sin sin sin tan k gk = gk IIIX the a² + C (5) condiçà £ x initial (t = 0) = 0 implies -C a² = 0. therefore, the DET between the Interface £ à © () 1 ln cosh sind Ti = gkk (6) from this equaçà the £
which can easily find () DKE 1cosh sin t ia = gk (7) 2-16. The única força that applied to the article to © © The strength of the component of gravity along the slope: mg ± sin Ct. Thus, the aceleraçà £ à ©  ± sin g ct. Therefore, the scroll speed and the slope for upward movement Sà £ o described by: () 0 SINV eg IT ± = A (1) (20 1 2 sin xvtgt) I
≤ ± = A (2) where initial Conditions () = 00 and v = vt () t = 0x = 0 were used. In posiçà £ the higher the speed becomes zero, so that Necessary time to reach the highest posiçà £ à ©, from (1) 00 ± sin VTG Ct = (3) the time, the scroll © NEWTONIAN MECHANICSà ¢ única partÃcula 43 2 0 0 1 2 sin vxg Ct ± = (4) for the downward movement,
the velocity and displacement sà £ o described by () SINV iT = ± g (5) (21 ug sin 2 Ct = ±) t (6) where dê a new origin for the XET £ posiçà the highest so that the initial Conditions sà £ ov (t = 0) = 0 and x (t = 0) = 0. We will find the Necessary time to move from the high posiçà £ £ posiçà to the initial, substituting (4) into ( 6): sin 0 = VTG Ct ±
à ¢ a² (7) of the Adiçà £ (3) and (7), there are 0 2 sin VTG ct = a ± (8) for the total time Necessary to return aa posiçà £ initial. 2-17. 35E v0 About 0.7 m to 60 m £ The configuraçà the problem for this à © as follows: 0 = cosx v i T (1) 20 0 1 2 sin y + v = t GTI A (2) where e. The ball crosses the about once 35i = 0 = 0 7 my. () I = 0 COSR VI,
where R = 60 m. Should be at least h = 2 m high, so © m tamba need SINV 20 0 i i A Gi = 2H Ya. For Solving we obtain 0v () 02 0 2 2 cos cos sin v gR Rh Yi I i = I ®  £  £ ¢ ¹à The I £  £    ° I '(3) Gives v. 10 25 m 4 S A. 46 chapter 2 () () () 0 2 0 2 2 2 sin 1000 m 9.8 m / sec 140 m / sec Rg iv = 0.50 A = (2) so 0 = 15i that a ° (3) Now, the true
range R a², in the linear aproximaçà £ â © given by equation (2.55):. 2 0 0 4 1 3 2 4 1 3 kV sin sin RR gv gg kv III ® I £ £  ¹ = à ¢ ¢  £ a² Register Register º £ £   à £   ° I 'A A A A £ £ ® = ¹ al  £ £ º a a a a a a a £ £  ° a "(4) since I to expect the real angle ¢ £ to be on the very different ¢ à 0i angle calculated above , we can solve (4) in
the case of i, for i 0I replacing the term in the parenthesis correcçà £. Thus, 2 0 0 0 2 4 1 3 g sin sin R kv ¢ IIA eg n = A²  £  ® I £ £ ai ¹ The Register º £  à  £  £  ° I '(5) Next, we need the value of k. From Fig. 2-3 (c) is the value of km per £ mediçà the curve slope in vizinhança v = 140 m / sec. Found () () C 110 500 m / s 0.22 kg / SKM
A. The curve A © appropriate for mass proja © ethyl 1 kg, so that the value of K © The seck 10.022 (6) Substituting the values of different quantities (5), we find a = 17.1à °. From this angle à ¢ à © slightly larger than 0i, we interact our soluçà the £ using this new value for 0i in (5). Then find a = 17.4à °. Beyond this iteraçà © £ m £ the sampler
substantially change the value, and therefore conclude that A = A 17.4à ° unless there £ the delay, one shot ethyl proja © à ¢ an angle of 17.4à a ° with an initial velocity of 140 m / sec would have a range of () 2 2 140 m / sec 34.8 9.8 m / sec 1140 m R ° = Newtoniano Mechanicsà ¢ single Particle 47 2-21. X3 CT  ± v0 x1 x2 Suppose a coordinate
system in which the project is moving in the X3A 2x plane. 2x. 2 0 2 3 0 1 COS SEN 2 XVTXVT GT CT CT   ±  ±  ° º  ° ž  ° º º º £  ± (1) Or, ( ) 2 2 3 3 2 0 2 0 1 COS SEN 2 XXVTVT GTI   ±  ± = + I £ £ £ £ £ £ 1. Register £  ° · I   ± and (2) the amount of linear movement of the project () () () 0 2 0Bond SIMM MVV GTI  ± 3I   Â
  € = = = + A I   ° I                                  € 3I  ± â € ° ° I  ° I  ° I  ° »I   °   °» Lpeeee (4) using the establishment of the unit vectors that 3I J = IJKà ± ee (), we find () 201 mg Cos2 VT ± CT = G 1E (5) This gives () 0 Cosg VT ± CT = L 1E ã, (6) Now the force acting on the project is 3mg = to
Fe (7), so that the binarium is () () () 2 0 2 0 3 0 1 1 2 Sin Cos Cos VTVT GT MG MG VT Â ± CT CT CT Â Â Â Â Â Â Â Â € £ Â Â ° £ £ · I £ 1. Register £ £ £ £ 1 Â ° I Â ° Â Â € = = A N R F Eeee 3, which is the same result as in (6). 48 Capan Tulo 02-22 February. x y z and b and the equation of the force is () q = + f and v b (1) a) Note that when e = 0, the
force is always perpendicular to the speed. This is a centriputous acceleration and can be analyzed â € â € R to be unstable Rb . We can use the results of Problem 2-46 for stability in case 2 2R BA unstable for ¤. 2-43. 3 2F kx kx ± Ct = A + () 4 2 2 1 1 2 4 x L x M dx = kx ki a = ± à ¢ 'For esboçar G (x), note that for small x, L (x) behaves like the
parábola 2 2 1 kx. For large x, the behavior © determined by 4 2 1 4 xk Ct ± a U (x) E0 E1 E2 E3 E4 = 0 x1 x2 x3 x4 x5 x () 2 1 2 E mv L x = + to E, the motion © unlimited; the partÃcula could be anywhere. For E = 0D (as in U máximos (x)) to the © partÃcula at a point equilÃbrio unstable. You can stay at rest where he está but was troubled a
little, it will move away from the balance. 1E = What à © the value of? We will find the values of x defining 1E dV dx = 0. 3 20 kx kx = Ct x = ± 0, ± U A A A £ ± sà equilÃbrio of the points (21) 2 1 1 1 2 4 4 EU kkk 2A Ct ± ± ± a ± a =  ± a = a to e, the partÃcula or © bounded and varies between 2E = ¢ and 2XA; or partÃcula reaches a ± a ±  ¢ a,
and returns to a ± a. 2x to 3x 2 and chapter 72, the partÃcula à © either at the point of equilÃbrio Stable x = 0, or m © Ala it. 3 x4 = 0 x = a ±  To And, the partÃcula comes from one to four Å ±  ± 5xà and returns. 2-44. m1 m2 m2 g TT I T M1G from the figure, the forças acting on the masses to the movement equaçÃμes m 11 1m to T = gx (1)
22 mg 2 2 cosmetic Tx i = a (2) where está associated with the Interface for £ 1x 2x () 21 2 2 4 bxx à ¢ d = a (3) and () b 1cos 2d Xi = Sign ®  £ i £  £   ° I ¹ï  £  ». In equilÃbrio, 1 x 2 x = 0 and T = M1G. This Gives as equilÃbrio values to the coordinates 1 2 110 2 4 4 AA = mdxbmm 2 (4) 1 24 220 2 2 = A mdxmm (5) We will recognize that
the expressed idêntica © £ â à £ equaçà the (2,105), and has the same requirement 10x 2 1 2 mm £ A A A A A A A £ £ º a a ºï £  £  £   ° a »Â £  £  ° a a '(2) indicates that the point of equilÃbrio à © Stable. 0 0.5 1 1.5 2 0 5 10 15 20 25 X / U (x) / G chapter 2 76 0 Thus, it can be seen that this problem x = x = £ to an AS of the standings
equilÃbrio instáveis, x = 0 à © £ posiçà one of the Stable balance. c) around the origin 0 02 2 4 x Uk 4U F kx am going ¡aa ¢ à ¢ m = d) to escape from the infinite x = 0, the partÃcula needs to obtain at least for the peak potential 0min 2 min 0 max 2 2 Umv UU vm = a = e) £ conservaçà the energy we 0min 2 22 2 0 2 2 2 1 2 2 U x dx XV Umv -dt
m £    " £ ¶ + i = ai = the ¬ £ £  · a a a a a a £ £ mV one we note that in the ideal case because initial speed © escape velocity found in d) in preferência x © always less than or equal to one, then from the expressed £ the above 0 22 2 0 00 0 2 2 8 exp 1 ln ( ) exp 81 2 8 1 x dx L in mam m AXT UU U xx rt £ ma   " £ i £   ¶ï 'i   £ i
£ ¶ A A A ¬ £  £ ¬ · A A A A A A ·  £ £ £    £ i £ A A = A + A = A A £  'i  à £ ¶ ¯  £  «i  £ « i  £  £ ¬ ¶A     · £ + A £ ¬  £  ·      £ ¬ £ £  · A A A A A A A £ £ £    £ i £      ¢ " £ ¶ TX TX 2-53. F © força one conservative nA when there is a potential unique funçà £ £ U (x) which satisfies F (x) = a
graduation (G (x)). Therefore, if M © conservative its components satisfy the following relaçÃμes FF yx x y a = A and so on. a) In this case, all the relaçÃμes above sà £ satisfied, so that f © indeed a conservative força. 2 1 (,) G bx Ayz 2x and L ayzx bx cx = fyzx à ¢ + à ¢ A = A + AF = a (1) where one funçà © £ only the y and z 1 (,) fyz 2 (,) L y L
AXZ bz ayzx Byz fxzy à ¢ = + a = a + AF = a (2) NEWTONIAN MECHANICSà ¢ única partÃcula 77, where one funçà © £ only the X and Z 2 ( ,) FXZ 3 (,) L Axy by c L ayzx Byz fy ZZF z ¢ = + a = à ¢ + a = a (3), then from (1) (2) ( 3) discovered that L 2 2 2 bx cx axyz Byz yyyy + C where C = A © arbitrária one constant. b) Using the same mà ©
whole found that M, in this case, a conservative força ©, and the potential à © exp () Lnu zxyz C = A A + C) Using the same © mà all we find in this case F à © força a conservative, and their potential à © (using the result of problem 1-31b): Lnu Ar = a 2-54. a) means terminal end speed constant speed (here assumed that the potato reaches this
speed before impact with the earth), when the total força acting on potato à © zero. mg = KMV and thus to 1000 m / g = skv. b) 0 0 0 () F XV VDV dv dx dt dx dt kv g mvg KR KR g = A + A = AA + AA + A «Â £ à 'dv = 0 0 2 ln máx 679, 7 mg xkkg gv = + kv = v + © where the initial speed of the potato. 2-55 0. And s is Leta 0xv 0yv the initial
horizontal and vertical speed abóbora. Of course, v = 0x 0yv this problem. XFX 0x xx x x x v f VDV DVDx MKV dt v dt kv x k m = a ¢ M = a a = a = a = (1) where the suffix f denotes the final value always. From the second equality (1) having fktx xf 0 xx dv dt kV VSL = a = a (2) £ the combination of (1) and (2) having (1 0 ¢ à fktxf vxk = a) and (3) 78
chapter 2 do the same with the y component, and have 0 2 0 0 lny y yf yfy yfyyy dv dv g kv v vdy gm F mg MKV dt dt y VG kV kg KR K + a = a = a = = = + + + (4) and () 0 fkty YF FY dv dt gg KR KR e.g. kv a + = + a = (5) from (4) and (5 ) with a little Handling the £, m is obta © 0 1 FKT fy GKT e.g. a = kV + (6) (3) and (6) SA £ 2 2 equaçÃμes
incógnitas, pA © if k. We can eliminate foot ©, and get a £ equaçà the a Single Variable k. () () () 1 0 00 fy xkt g kv gvxf eK + vx = a putting and MFX 142 0 0 0 = 38.2 m / S2X YV = vv = ke can solve numerically to obtain K = 0, 00246 1SA. ES OSCILAà 81 c) decrà © © scimo of motion defined as where E 1A ²à 1 1I V =. Then, 1 1,0445Eã² 3-3.
Initial initial energy (total energy equal to) the oscillator © 20 mV 1 2 where m = 100 g, and v. 0 1 cm / s = a) deslocaçà the máxima £ à © alcançada when the total energy equal to the potential energy ©. Therefore, 20 2 0 1 1 2 mv 2 = kx 2 0 0 1 4 10 10 1 c 10 = mxvk A = m or 0 = 10 x 1 cm (1) b) The potential energy máxima à © 2 4max 0 1
1 10 10 2 2 L = kx 2A ¢ a or in máximo 50 ergsU = (2) 4/3. a) Time audio mà ©: The posiçà o £ speed for a simple oscillator harmónico sà given by the £ 0sinx The Ti = (1) 0 i = cosx 0TA A (2) where k = 0 M © mà day time cinnamic optical energy © à © 2 1 2 1 T tt = mx + II à ¢ 'dt (3) where 0 = 2i IIA perÃodo of the © £ the oscilaçÃ. 82
chapter 3, with the Insertion £ (2) (3) is obta © m-1 20 2 2 cos 2 t tt = mA + Tiiii à ¢ '0 dt (4) or 2 2 0 4 mA t i = (5) likewise ma © day of the potential energy of the time © 2 2 2 0 2 1 1 2 1 2 4 sin tttt L kx dt kA t dt kA iiiii + + = = = a ¢  "à ¢ '(6) and once 20 ml k = (6) reduces to 2 2 0 4 mA ui = (7) from (5) and (7) we see TU = ( 8) the result given
in (8) © razoável expect from £ conservaçà the total energy. E = T L + (9) This equality © válido instantly, as well as on hand © day. On the other hand, when T and £ -U sa expressed by the (1) and (2), we note that they sà £ described by the exact same funçà £ o, shifted by a time 2I: 2 2 20 0 2 20 cos 0 sin 2 mA 2 mA t t U = ¹ tiiiii  £ £   Â
 £ º º A A A A º £ £ £ ä ä = º ºï   £ '(10) therefore, the mà © days time T and U must be equal. Enta £ the taking mà © day time (9) we find ETU = 2 (11) b) Day Spacebar mà ©: The day of spatial mà © © optical cinnamic and the potential energy sà £ à OSCILAà ES 83 2 0 1 1 2 a = mx aT the 'dx (12) and 2 2 0 0 0 1 1 2 a 2 Am x dx L kx i
= AA = after' 2 DXA '(13) (13 ) à © easily integrated to give 2 0 2 6 m AU à = (14) to integrate (12), we note that from (1) and (2) we can write () () 2 2 2 2 2 2 2 0 0 0 0 2 2 2 1 cos AA x t the sinx 0TA IIII = a = a (15) then, substituting (15) into (12) we will find 0 2 20 2 2 30 2 2 3 03 : TA 00 m x AAA DXI II ®  £  £ i = Ai ¹ £ £    ° I »i   £ i £
® ¹ = A i £   à Register £ º I £   °  £  »Ã ¢ '(16) or, May 6 2 2 02 = (17) from the comparaçà £ (14) and (17) we see that G = 2T ( 18) To see that this result © razoável, mark T = T (x) and G = G (x): 2 2 2 0 2 2 2 0 1 1 2 1 2 x mx Tm AAU III £ ¹à ¯   £ i £ ® ¹ Ai = £  £  £ ºï The Sign º to £  £   ° I »i  £  £ à º º to £ A ° = n
£  £   ºï '( 19) L = L (x) T = T (x) The energy of an X 2i is 0 mA mA2 = const. 1 2 2 0 2I And área between T (x) and the axis X © only twice that between L (x) and the x-axis. Chapter 3 86 There are two forças acting on the body: that due to gravity (Mg), and due to the fluid pushing up the body (SGV 0 0 gh = Ai i to one). The situaçà £
equilÃbrio occurs when the total força disappears 0 0 0 b Mg gV GAH = gh III as A = A (1), which gives the Interface between the £ e: SH 0 bh HI s i = bh (2) for a small displacement on the equilÃbrio posiçà £ (), (1) becomes s + sh hà ¢ x () 0B SMX b Ah x GAH gh XI I i = a + a (3) the aft substituiçà £ the (1) to (3) have 0bAh gxAà x i = a (4) or 0
0 II + = bxgxh (5) Thus, the motion © oscilatório with an angular frequência 2 0 BSG ga VIII GHH = = (6) in which use was made of (2), and último step têm multiplied and divided by A. the perÃodo the oscilaçÃμes à © therefore, 2 V 2 = IIII AG (7) Substituting the above values, 0 18 SI. . 3-8. The y x m s 2a 2a responsável The strength of the
pendulum movement à © The strength of the component of gravity acting on m £ porçà perpendicular to the rectilÃnea of the suspended rope £. This component © seen from the figure (a) below, to be cos f ma mv mg ±  = I = A (1) where i oscilaçÃμes 87 ± à  © à ¢ the angle between the vertical and the tangent to the path in the cycloidal
posiçà the £ m. The cosine? ± A ©  expressed in terms of Differences indicated in the figure (b) as dc cos dy ds = ± (2) dx dy where 2DS = + 2 (3) R u  ± ± m F DX DY S DS (A) (b) Differentials, DX and DY, can be calculated from the equations define by X (i) Ey (i) above: () 1 COS Sin DX DY Animance that announcement IIII  € â¹ = A º º  ° â
€ œ  ° I  ± £ »(4) So () () 2 2 2 22 2 2 2 2 2 2 1 2 1 COS Sin Cos in 2 DS DX DY ADAAD 2D 4 SIIIIII + I =   € = A + = AI  ° I  ± = (5), so that 2 SIN 2 DS AD II = (6) , Sin Sin 2 2 Cos 2 Dy Animancio DS AD IIIIII ± A = = A = (7) The Pendulum Speed is 88 Chapter 2 3 4 2 Sin Cos 2 DS DT DT DT DT DT DT = = I ® I â € £ â¹ = Ai £ 1. Register º
° ° ° I  ° »(8) from which two 24 COS 2 DVA DT II    € = A I £ £ £ £ â € œ  °  °  ° »(9) Leaving COS 2 Zia is the new variable, and starting from (7) and (9) in (1), which has 4maz mgzà ¢ = (10) or, 0 4 gzza + = (11) which is the standard equation for a simple harmonic movement, 20 0z zi + = (12) if identify 0 gi = (13) Where the fact that it
was used. 4a = Thus, the movement is exactly isochronous, regardless of the breadth of oscillations. This fact was discovered by Christian Huygene (1673). 3-9. The equation of the movement for 00 t ta ¤ ¤ () () () 0mx kxx f kx f = kx one + = + to + 0 (1) while for, the equation The à 0t ¥ () 0mx kxx kx kx = aa = a + 0 (2) is convenient to define 0x
xÞ⾠= q-transforming (1) and (2) into m kÞ⾠‡ f = a +; 00 t ta ¤ ¤ (3) m kÞ⾠ ‡ = A; (4) 0T ~ ~ ~ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒ1 1 1 1 1 1 2 2 2 2 2 2 () () ة Cos 2 Sin Sin Cos Tae Temktmttt P 2 P omâ € ƒâ € ƒâ € ƒâ € ƒâ € ƒ IAI   ® = + A + I   ° I  ±  ±  ° (4) Rewriting (4), we can find the expression of and (t): () () () 2 2 2 2 2 2 1
0 1COM2 Sin 22 TMA ET Tážâ² P \ Awe Awe? IAI       € = A + AI  ° I  ° »0 + and t (5) Taking the derivative of (5), we found the expression for DT: () () () 2 2 2 3 0 1 2 2 2 2 0 1 0 2 4 4 Cos2 Sin 2 2 TDE MA Ett P òÀ² Á ounce à £ â € ƒâ € ƒ ®   ° I  ° I  ° I   ° â € ¢ To find the rate of energy loss of a slightly moist oscillator, let's give DT
0â € ™2. This means that the oscillator has time to complete a certain number of perhaps before its amplitude decreases considerably, that is, the term 2 te pa does not change much in the time it takes for a complete period. The co-seno and terms sine, in mine, to almost zero in comparison with the constant term in the DT, and obtained at this limit
of 2 2 20 TDE A and DT ãž² Ž² A (7) 3-12. MG MG SIN II The equation of the movement is Sinm Mgi ia = (1) Sin GII = A (2) If I are small enough, we can approach sin ct ion, and (2) becomes 92 Chapter 3 GII = a (3) which has oscillatory solution () 0 cost 0TÞ⸠II = (4) where 0 = GI and where amplitude is 0i. If there is the retardant force 2m GI, the
motion equation becomes sin 2m mg of M Gi = IA + I (5) or the SIN I would and rewrite, we have 20 02 ¸ IIII 0+ + = (6) Comparing this equation with the standard equation for the damped movement [eq. (3.35)], 202x x xÞⲠto 0+ + = (7) that identify 0i p =. This is only the case of critical damping, so that the solution of I (t) is [see eq. (3,43)] () ()
0TT a BT EIIA = + (8) for The initial conditions () 00Þ  00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ii 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ii 00 00 00 00 ii ii 00 00 00 00 00 ii ii ii 00 ii ii ii ii ii = For the case of christian damping, 0² I =. Therefore, the equation of movement becomes 22x x
xãžâ² p 0+ + = (1) if we assume a solution of the form () () tx tyte pa = (2) which has 22 ttttx ye ye x ye ye ye pppppppppaaaaai £ £ £² Register º  ° ž = A + I  € »(3) Replacing (3) in (1), we find ye ( 4) 2 22 2 2t tttt taço ye ye ye yeÞⲠPPPPPPP â € ™22 pa aaaa à ¢ à ¢ + + a + 2 0 = 0 or = y, (5) therefore () yt bt = + (6) of oscillations at 93 and ()
() () tx t a bt and pa = + That's just eq. (3.43). 3-14. For the case of superamortected oscillations, x (t) and () xt are expressed by () 21 2t tx to and the eéžâ² I a 2ti registry â € £ â € = + I  ° I  ° »(1) () () () 2 2 21 2 1 2 2t T You and A And A And A Ei Iipia á ¢ à ¢ + + + A '2tà ¯ â € ° á ¯ ¯ â € q q ¯ ¯ ¯ ¯ ¯ ¯ ¯oving áceâ € ° áceâ ân ¯âaâ ân¯lands ½,
where 22 2 0 ¯ ° ¯ ¯ ounces, q ¯ ¯ ¯ ouncer. = The hyperbolic functions are defined as Cosh 2 y Ey à ¢ + =, Sinh 2 Y Ye Ey'à ¢ â € € = (3) or, Cosh Sinh Cosh Sinh Yyeye y from ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Àº º º º ech £ £ â € £ (4) To rewrite (1) and (2), we have () () () () () () () 1 2 2 1 2 2cosh Sin Cosh Singt T AA T AA Tážâ² PII xt Registe £ â € £ £ £ £ £ £
£ £ £ £ 5 (5) and () () () () () () () 1 2 1 2 2 2 2 2 2 2 Cosh Sinh Cosh Sinh Cosh Sinh XTTTT AA AA TT TPPP echar £ £ £ â € ƒâ € ƒâ € ƒâ € ƒâ € ƒâ € œH = We are requested to simply plot the following equations of Example 3.2: () () 1Costare Tá®â² Tis ¯ â € ° '' = ¢ (1) () () 1 1 1 () COS SINTAE TT 5 of Ác ¯ ~a Regista    € = AA + A I  ° I  ° »(2) with
the values of A = 1 cm 10 1 Rad SI A = A, 10 1 SÞⲠà ¢ =. And à '= N rad. The position passes by x = 0 a total of 15 times before dropping at 0.01 of its initial amplitude. An exploded (or reduced) seen from figure (b), shown here as figure (b), is the best to determine this number, as is easily demonstrated. showing.
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