Classical Mechanics Joel phần 1 pdf

... MoreGeneralMotion 10 0 4.4 Dynamics 10 7 4.4 .1 Euler’sEquations 10 7 4.4.2 Eulerangles 11 3 4.4.3 Thesymmetrictop 11 7 5 Small Oscillations 12 7 5 .1 Small oscillations about stable equilibrium 12 7 5 .1. 1 MolecularVibrations ... Externalandinternalforces 10 1. 3.2 Constraints 14 1. 3.3 Generalized Coordinates for Unconstrained Sys- tems 17 1. 3.4 Kineticenergyingeneralizedco...

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... 10 6

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... asym- metry of the potential, and the orbits become egg-shaped. 1 -1 x p 1. 210 .8 0.6 0.40.2-0.2-0.4 0 0.3 0.2 0 .1 0 -0 .1 -0.2 -0.3 x U U(x) Figure 1. 3. Motion in a cubic poten- tial. If the system has ... other than gravity. The Earth has a mass of 6.0 10 24 kg and aradiusof6.4 × 10 6 m. Newton’s gravitational constant is 6.67 × 10 11 N · m 2 /kg 2 . 1. 2 In the discussion...

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... energy is T = 1 2 m 1 ˙r 2 1 + 1 2 m 2 ˙r 2 2 = 1 2 m 1  ˙  R − m 2 M ˙ r  2 + 1 2 m 2  ˙  R + m 1 M ˙ r  2 = 1 2 (m 1 + m 2 ) ˙  R 2 + 1 2 m 1 m 2 M ˙ r 2 = 1 2 M ˙  R 2 + 1 2 µ ˙ r 2 , where µ ... invertible, 8 we also have ˙q = M 1 · P ,so H = P T · ˙q −L = P T · M 1 · P −  1 2 ˙q T · M · ˙q −U(q)  = P T · M 1 · P − 1 2 P T · M 1 · M ·M...

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... orthogonal and has determinant 1, det  (A − 1I ) A T  =det(1I−A T ) = det(1I − A) =det(A − 1I ) d e t ( A)=det(−(1I −A)) = ( 1) 3 det(1I − A) = −det(1I −A), so det(1I − A)=0and1I− A is a singular matrix. ... θ =0. Then 1 r = A cos θ + B = 1 r p  1 − e 1+ e (1 −cos θ)  = 1 r p 1+ e cos θ 1+ e where e = A/B. What is this orbit? Clearly r p just sets the scale of the whole or...

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... I 1 = I 2 ,sowehave I 1 ˙ω 1 =(I 1 −I 3 )ω 2 ω 3 , I 1 ˙ω 2 =(I 3 −I 1 )ω 1 ω 3 , I 3 ˙ω 3 =(I 1 −I 2 )ω 1 ω 2 =0. We see that ω 3 is a constant. Let Ω = ω 3 (I 3 −I 1 )/I 1 .Thenweseethat ˙ω 1 = ... the gravitational field 15 . T = 1 2 (ω 2 1 + ω 2 2 )I 1 + 1 2 ω 2 3 I 3 = 1 2  ˙ φ 2 sin 2 θ + ˙ θ 2  I 1 + 1 2  ˙ φ cos θ + ˙ ψ  2 I 3 , (4.34) U = Mg...

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... ˙η = 1 2 ˙η T ·  O 1 1 ·m ·O 1  · ˙η = 1 2  ˙η T ·O T 1  ·m · (O 1 · ˙η) = 1 2 (O 1 · ˙η) T · m · (O 1 · ˙η) = 1 2 ˙x T · m · ˙x. Similarly the potential energy becomes U = 1 2 x T ·O 1 · ... DYNAMICS 13 9 so U(y 1 , ,y i , ,y n ) =  y i 0 dy i τ a (2y i − y i +1 −y i 1 )+F (y 1 , ,y i 1 ,y i +1 , ,y n ) = τ a  y 2 i − (y i +1 + y i 1 )y i  + F (y 1...

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... 2-forms at a point, dx 1 ∧ dx 2 , dx 1 ∧ dx 3 ,anddx 2 ∧ dx 3 ,wheredx 1 ∧ dx 2 = dx 1 ⊗ dx 2 − dx 2 ⊗ dx 1 , which means that, acting on u and v, dx 1 ∧ dx 2 (u,v)=u 1 v 2 − u 2 v 1 . The product ... <i k ω i 1 i k dx i 1 ∧···∧dx i k over S by  S ω (k) =   i 1 ,i 2 , ,i k ω i 1 i k (x(u))  k   =1 ∂x i  ∂u   du 1 du 2 ···du k . Wehadbettergivesomeexa...

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... this difference, d(ω 1 − ω  1 )=ω 2 − ω  2 =0,soω 1 − ω  1 is an closed 1- form. Thus it is exact 11 , and there must be a function F on phase space such that ω 1 − ω  1 = dF .WecallF the ... =tan 1 Q/P as the new coordinate, and we might hope to have the radial coordinate related to the new momentum, P = −∂F 1 /∂θ.AsP = ∂F 1 /∂Q is also Q cot θ,wecantakeF 1 = 1 2...

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... symmetry point, q = 0, to the next such crossing. The -1 -0.5 0 0.5 1 -2 -1. 5 -1 -0.5 0.5 1 1.5 q p -1 -0.5 0 0.5 1 -2 -1. 5 -1 -0.5 0.5 1 1.5 q p Fig. 4. The trajectory in phase space of the system ... a circle. 0 10 20 30 -1. 5 -1 -0.5 0 0.5 1 1.5 -1. 5 -1 -0.5 0 0.5 1 1.5 Fig. 7. The surface Σ 1 for a harmonic oscillator with a spring constant which varies, for...

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