Using the relation. (2) becomes (1) (2) or, equivalently, CHAPTER 7 (4) (4) (5) (5) (6) This reduces to (7) This reduces to (6) (8) 7-9. (9) 7-16.

Σχετικά έγγραφα
2. Μηχανικό Μαύρο Κουτί: κύλινδρος με μια μπάλα μέσα σε αυτόν.

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

Section 8.3 Trigonometric Equations

Areas and Lengths in Polar Coordinates

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

CE 530 Molecular Simulation

Areas and Lengths in Polar Coordinates

derivation of the Laplacian from rectangular to spherical coordinates

Space-Time Symmetries

( )( ) ( ) ( )( ) ( )( ) β = Chapter 5 Exercise Problems EX α So 49 β 199 EX EX EX5.4 EX5.5. (a)

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

Srednicki Chapter 55

Finite Field Problems: Solutions

Similarly, we may define hyperbolic functions cosh α and sinh α from the unit hyperbola

Second Order RLC Filters

Inverse trigonometric functions & General Solution of Trigonometric Equations

Homework 3 Solutions

4.6 Autoregressive Moving Average Model ARMA(1,1)

1 String with massive end-points

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

2 Composition. Invertible Mappings

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

Jackson 2.25 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

( y) Partial Differential Equations

w o = R 1 p. (1) R = p =. = 1

Matrices and Determinants

EE512: Error Control Coding

Example Sheet 3 Solutions

Lanczos and biorthogonalization methods for eigenvalues and eigenvectors of matrices

D Alembert s Solution to the Wave Equation

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Forced Pendulum Numerical approach

Approximation of distance between locations on earth given by latitude and longitude

Partial Differential Equations in Biology The boundary element method. March 26, 2013

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Name: Math Homework Set # VI. April 2, 2010

The Simply Typed Lambda Calculus

6.3 Forecasting ARMA processes

Chapter 5. Exercise Solutions. Microelectronics: Circuit Analysis and Design, 4 th edition Chapter 5 EX5.1 = 1 I. = βi EX EX5.3 = = I V EX5.

Second Order Partial Differential Equations

Section 7.6 Double and Half Angle Formulas

Lecture 26: Circular domains

Homework 8 Model Solution Section

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

F19MC2 Solutions 9 Complex Analysis

Strain gauge and rosettes

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

CRASH COURSE IN PRECALCULUS

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Answer sheet: Third Midterm for Math 2339

TMA4115 Matematikk 3

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

Chap 8 Mapping by Elementary Functions

Section 8.2 Graphs of Polar Equations

ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω

Tridiagonal matrices. Gérard MEURANT. October, 2008

MathCity.org Merging man and maths

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

Section 9.2 Polar Equations and Graphs

( ) 2 and compare to M.

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

ECON 381 SC ASSIGNMENT 2

To find the relationships between the coefficients in the original equation and the roots, we have to use a different technique.

Notes on the Open Economy

CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD

Spin Precession in Electromagnetic Field

Solutions to Exercise Sheet 5

ST5224: Advanced Statistical Theory II

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

A Note on Intuitionistic Fuzzy. Equivalence Relation

C.S. 430 Assignment 6, Sample Solutions

Statistical Inference I Locally most powerful tests

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Geodesic Equations for the Wormhole Metric

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

PARTIAL NOTES for 6.1 Trigonometric Identities

Math221: HW# 1 solutions

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

Example of the Baum-Welch Algorithm

Part III - Pricing A Down-And-Out Call Option

If we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering

Parametrized Surfaces

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

Solution to Review Problems for Midterm III

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

If we restrict the domain of y = sin x to [ π 2, π 2

ECE 308 SIGNALS AND SYSTEMS FALL 2017 Answers to selected problems on prior years examinations

Differential equations

Spherical Coordinates

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

Fractional Colorings and Zykov Products of graphs

Transcript:

U 3 U the relation HAILTON S PRINCIPLE LAGRANGIAN AND HAILTONIAN DYNAICS 83 P P P p t t t 4 beoes 7-3. P P U R ρ φ R ρ Iφ U P P If we take anles an φ as our eneralize oorinates, the kineti ener an the potential ener of the sste are T 5 Interatin 5 fro an point in the reion to an point in the reion, we fin P U R R ρ os P 6 U U 7 or, equivalentl, U U where is the ass of the sphere an where U at the lowest position of the sphere. I is the oent of inertia of sphere with respet to an iaeter. Sine I 5 ρ, the Laranian beoes 8 Now, fro 3 we have L T U R ρ ρ φ R R ρ os 5 When the sphere is at its lowest position, the points A an B oinie. The onition A B ives the equation of onstraint: 88 f, φ R ρ ρφ 4 Therefore, we have two Larane s equations with one uneterine ultiplier: The potential ener is f L λ os φ os φ U t Therefore, the Laranian is f L λ φ φ L φ φ φφφt os φ φ os φ os φ After substitutin 3 an f R ρ an f φ ρ into 5, we fin fro whih R ρ R ρ λ R ρ φφ φ φ φ φ ρ φ λρ 5 φ φ os φ φ φ Fro 7 we fin λ: or, if we use 4, we have λ ρφ φφ 5φ φ φ φ φ φ os φ φ λ R ρ φ 5 84 The Larane9 equations for fin φ an Substitutin into 6, we the φequation of otion with respet to : are φ φ os φ φ φ φ φ ω φ where ω is the frequen of sall osillations, efine b φ φ os φ φ φ φ φ φ 5 ω 7 R ρ 7-8. 7-4. U v 5 5 6 7 v : t v Substitutin into, we fin : This reues to t 9 7-9. 7-6. 7-6. a ωt O a ωt b ξ 7 b φ 8 For ass : For ass : α a ω t b 3 4 aω os bos os aω ωt ωbt os T Tisk Tbob ξ I bob bob b b Substitutin the oorinates for the bob, we obtain Substituteinto into Substitute T ξ I φ φξ os φ a T T The potential ener is iven b U U Uisk U bob isk bobu ξ α os φ an the result is an the result is relation ξ R to reue the erees of freeo to two, an in aition Now let us use the substitute I R Laranian beoes L T for U the isk. b a ωthe os ω t abω os ω t os b os L T U a ω os ω t abω os ω t os b b os 3 L T U ξ φ φξ φ a ξ α os os φ Larane s equation for ives Larane s 4equation for ives eneralize oorinates are The resultin equations for our two abofωotion os ω t os b abw os ω t b t 3 t abω os ω t os b abw os ω t b ξ α φω b abω ω t os abω os ω t φos bφ α ab os φωt α U the relation Larane s equation for the oorinate r leas to P P P p 4 t r rt Ar α t 5 beoes Larane s equation for the oorinate leas to P P U 5 6 r t Interatin 5 fro an point in the reion to an point in the reion, we fin Sine r is ientifie as the anular oentu, 6 iplies that anular oentu is U P onserve. Now, if we use, we anpwrite 6 as 5 ultiplin 7 b r, we have or, equivalentl, 98 6 8 U A L r r r α α v U v U Ain 8 an, we have v k U U 3 b os T v k b os This proble is the ehanial analo of the of liht upon pas fro a eiu of refration withaifferent optial a ertain optial ensit into a eiu ensit. U the eneralize oorinates iven theωfiure, the in a t b Cartesian oorinates for the isk are ξ os α, ξ α, an for the bob the are φ ξ os α, os φ ξ α. The kineti b os b os ener is iven b This reues to 9 9 98 9 we also have Fro U < r r U U > Sine the fore is relate to the potential b If we onsier the potential ener as a funtion of as above, the Laranian of the partile is U f L r U we fin HAILTON S PRINCIPLE LAGRANGIAN AND HAILTONIAN DYNAICS 89 Therefore, Larane s equations for the oorinates an are A α U r 3 Uα beoes where we let Ur. Therefore, the Laranian 4 Let ushoose hooser, the so that the two reions are ivie bthe thepartile ais: is If we, as oorinates the eneralize oorinates, the kineti ener of T an is onstant. Therefore, Fro 9 we have r v t 3 α Pr P 3 Ar r U U 7 7 r α rr r 3 Ar r U U 8 8 or or 7-. 3 4 5 abω ω t os abω os ω t b abω os ω t b φ ξ os φ α φ 6 a ω ω t os b b a b ω ω t os b S

7-7. A C q h U q an ω t e, the, oorinates of the partile are epresse as fro whih Therefore, the kineti ener of the partile is The potential ener is hos q h os ωt q t ωt h q os h ωt q t os ωt hω ωt qω os ωt q ωt hω os ωt qω ωt qos ωt T Then, the Laranian for the partile is B h ω q ω q hω q 3 ω os ω U h t q t 4 L h ω q ω q h ωt q os ωt hω q 5 Larane s equation for the oorinate is q ω q os ωt 6 The opleentar solution an the partiular solution for 6 are written as so that the eneral solution is U the initial onitions, we have Therefore, an or, q t A os iωt δ 7 qp t os ωt ω qt A os iωt δ os ω t 8 ω q Aos δ ω q iωa δ δ, A ω qt os iωt os ωt ω qt osh ωt os ωt ω ω qt In orer to opute the Hailtonian, we first fin the anonial oentu of q. This is obtaine b Therefore, the Hailtonian beoes so that Solvin 3 for H pq L p q ω h q ω ω ω ω os ω ω q hq h q q h t q t qh H q ω h ω q h ωt q ωt t 9 3 os 4 an substitutin ives Solvin 3 for q an substitutin ives p os ω ω ω 5 H hp q h t q ω t The Hailtonian is therefore ifferent fro the total ener, T U. The ener is not onserve in this proble e the Hailtonian ontains tie epliitl. The partile ains ener fro the ravitational fiel. 7-. Fro the fiure, we an easil write own the Laranian for this sste. The resultin equation of otion for is ω R T ω U R os ω os R 3 The equilibriu positions are foun b finin the values of for whih ω R 4 os Note first that an π are equilibriu, an a thir is efine b the onition To investiate the stabilit of eah of these, epan u ε For π, we have os 5 ω R os ω R os 6 ε ω ε ε ε ω ε R iniatin that it is unstable. For, we have ω ε ω R ε 8 whih is stable if ω < R an unstable if ω > R. When stable, the frequen of sall osillations is ω R. For the final aniate, ω ε 9 ε ω with a frequen of osillations of ω ωr, when it eists. Definin a ritial frequen ω R, we have a stable equilibriu at when ω < ω, an a stable equilibriu at ω ω os when ω ω ω ω ω, respetivel. an 4. The frequenies of sall osillations are then ω ω ω To onstrut the phase iara, we nee the Hailtonian H L whih is not the total ener in this ase. A onvenient paraeter that esribes the trajetor for a partiular value of H is H ω K os ωr ω ω so that we ll en up plottin ω K os ω ω for a partiular value of ω an for various values of K. The results for ω < ω are shown in fiure b, an those for ω > ω are shown in fiure. Note how the oriin turns fro an attrator into a separatri as ω inreases throuh ω. As suh, the sste oul ehibit haoti behavior in the presene of apin. 7

.5 K.5 3 b 3.5 K.5 3 3