Physics Mechanics 110A notes UCSD

8/10/2019 Physics Mechanics 110A notes UCSD

1/305

Classical Mechanics

(UCSD Physics 110A-110B)

January 1, 2015


2/305

2

Contents

0 Introduction 9

1 Review of Newtonian Mechanics 101.1 Newtons Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3 Physical Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Motion with Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5.1 Example: Motion in a Constant Gravitational Field . . . . . . . 181.5.2 Example: Motion with Constant g and Drag Force proportional

to Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5.3 Example: Motion with Constant g and Drag Force proportional

to Velocity Squared . . . . . . . . . . . . . . . . . . . . . . . . . 221.6 Example: Rockets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.6.1 Example: Rockets in Constant Gravity . . . . . . . . . . . . . . . 271.6.2 Example: The Falcon Heavy Rocket . . . . . . . . . . . . . . . . 29

1.7 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.8 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2 Review of Energy and Conservative Forces 422.1 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.2 Conservative Forces and Potential Energy . . . . . . . . . . . . . . . . . 422.3 Total Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.3.1 One Dimensional Problems . . . . . . . . . . . . . . . . . . . . . 462.3.2 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.3 Example: Phase Portrait for Harmonic Oscillator . . . . . . . . . 48

2.4 Conservation of Momentum and Angular Momentum . . . . . . . . . . . 492.4.1 Example: Phase Portrait for Gravity . . . . . . . . . . . . . . . . 50

2.5 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.6 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3 Oscillations 53

3.1 Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Energy of the Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3 Phase Curves of the Simple Harmonic Oscillator . . . . . . . . . . . . . 553.4 Damped Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56


3/305

3

3.5 The Driven Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.5.1 Periodic Driving Forces . . . . . . . . . . . . . . . . . . . . . . . 593.5.2 Superposition of Driving Forces . . . . . . . . . . . . . . . . . . . 633.5.3 Fourier Series Solution of Driven Oscillator . . . . . . . . . . . . 64

3.5.4 Oscillator Driven by an Impulse . . . . . . . . . . . . . . . . . . 673.5.5 Greens Function Solution of the Driven Oscillator . . . . . . . . 713.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.6.1 Example: Sawtooth Driving Force . . . . . . . . . . . . . . . . . 723.6.2 Example: Exponentially Decreasing Force . . . . . . . . . . . . . 74

3.7 Derivations and Computations . . . . . . . . . . . . . . . . . . . . . . . 753.7.1 Orthonormality of Cosines in Fourier Series . . . . . . . . . . . . 75

3.7.2 Response to a Step Function at t0 . . . . . . . . . . . . . . . . . 763.7.3 Limit of Impulse Response for t 0 . . . . . . . . . . . . . . . 773.8 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 793.9 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4 Newtonian Gravity 854.1 Inverse Square Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.2 Gravitational Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.3 Using Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 894.4 Gravitational Field of a Spherical Shell . . . . . . . . . . . . . . . . . . . 904.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.5.1 Potential of a Uniform Density Spherical Shell . . . . . . . . . . 914.5.2 Gravitational Potential of a (Partial) Ring . . . . . . . . . . . . . 94

4.6 Derivations and Computations . . . . . . . . . . . . . . . . . . . . . . . 964.6.1 The Divergence of the Field for an Inverse Square Law . . . . . . 96


5 The Calculus of Variations 1005.1 Eulers Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.2 Second Eulers Equation if f x = 0 . . . . . . . . . . . . . . . . . . . . 1035.3 Functions of Several Dependent Variables . . . . . . . . . . . . . . . . . 104

5.4 Euler Equation with Constraints . . . . . . . . . . . . . . . . . . . . . . 1045.5 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5.6.1 The Brachistochrone Problem . . . . . . . . . . . . . . . . . . . . 108


4/305

4

5.6.2 Minimum Surface of Revolution . . . . . . . . . . . . . . . . . . . 1105.6.3 Geodesic on a Sphere . . . . . . . . . . . . . . . . . . . . . . . . 112


6 Hamiltons Principle and Lagrange Equations 1186.1 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.2 Lagranges Equations with Undetermined Multipliers . . . . . . . . . . . 1216.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

6.3.1 One Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . 1226.3.2 The Plane Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . 1226.3.3 Mass Sliding on Cone . . . . . . . . . . . . . . . . . . . . . . . . 123

6.3.4 Pendulum Hanging from a Rotating Support . . . . . . . . . . . 1256.3.5 Bead on Spinning Parabola . . . . . . . . . . . . . . . . . . . . . 1276.3.6 Atwoods Machine . . . . . . . . . . . . . . . . . . . . . . . . . . 1286.3.7 A Mass Sliding O ff a Sphere Using an Undetermined Multiplier . 130


7 Conserved Momenta and Noethers Theorem 137

7.1 Consider Symmetry Transformations Leaving Action Invariant . . . . . 1387.2 More General Noether Theorem . . . . . . . . . . . . . . . . . . . . . . . 1407.3 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1417.4 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

8 Central Force Motion and Planetary Motion 1428.1 Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.2 The Center of Mass Transformation . . . . . . . . . . . . . . . . . . . . 145

8.2.1 Keplers Area Law . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.3 Solution for the Gravitational Potential . . . . . . . . . . . . . . . . . . 147

8.3.1 Inverse Cube Force . . . . . . . . . . . . . . . . . . . . . . . . . . 1538.4 The Energy View of Orbits . . . . . . . . . . . . . . . . . . . . . . . . . 1538.5 Keplers Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1538.6 Keplers Equation for (t) . . . . . . . . . . . . . . . . . . . . . . . . . . 1538.7 Stability of Circular Orbits . . . . . . . . . . . . . . . . . . . . . . . . . 1548.8 Small Oscillations Around Circular Orbits . . . . . . . . . . . . . . . . . 1548.9 Hohmann Transfers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1548.10 The Slingshot E ff ect . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1548.11 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1558.12 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155


5/305

5

9 Coupled Oscillations and Normal Modes 1599.1 Two Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . 1599.2 Weak Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1639.3 The General Problem of Small Oscillations . . . . . . . . . . . . . . . . 164

9.4 Normal Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1669.5 Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1679.6 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1689.7 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

10 Nonlinear Mechanics and Chaos 16910.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16910.2 Homework Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16910.3 Sample Test Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

11 Vectors and Rotations 17011.1 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17111.2 Cross Products and Axial Vectors . . . . . . . . . . . . . . . . . . . . . . 17311.3 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17411.4 The Vector Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176


12 Noninertial Frames of Reference 17912.1 Rotating Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 179

12.1.1 Calculating the E ff ective Force (simple rotation) . . . . . . . . . 18112.2 Calculating the E ff ective Force (a more general derivation) . . . . . . . 18412.3 Motion Near the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

12.4 The Tides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18912.4.1 Example: Deection of a Falling Object . . . . . . . . . . . . . . 19312.4.2 Example: Deection of Cannon Balls . . . . . . . . . . . . . . . . 19412.4.3 Weather Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 19512.4.4 Example: The Foucault Pendulum . . . . . . . . . . . . . . . . . 196


13 Dynamics of Rigid Bodies 19913.1 Calculating the Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . 20013.2 The Inertia Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

13.2.1 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 20213.2.2 Simple Example: Inertia Tensor for Dumbbell . . . . . . . . . . . 202


6/305

6

13.2.3 Transforming the Inertia Tensor . . . . . . . . . . . . . . . . . . 20413.3 Parallel Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20413.4 Example: The Inertia Tensor for a Cube . . . . . . . . . . . . . . . . . . 20613.5 Principal Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

13.5.1 Example: Cube Rotating about a Corner . . . . . . . . . . . . . 20813.5.2 Proof: Principal Axes Orthogonal . . . . . . . . . . . . . . . . . 20913.5.3 Proof: Roots I (a ) are Real . . . . . . . . . . . . . . . . . . . . . . 210

13.6 Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21013.7 Body Frame Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 21213.8 Eulers Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

13.8.1 Example: Symmetric Top with no Torque . . . . . . . . . . . . . 213

13.9 Stability of Rigid Body Rotations . . . . . . . . . . . . . . . . . . . . . . 21413.10Lagrange Equations for Top with One Fixed Point . . . . . . . . . . . . 21613.11Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

14 Special Relativity 22114.1 Some History of Special Relativity . . . . . . . . . . . . . . . . . . . . . 22114.2 The Michelson Morley Experiment: Some Analysis . . . . . . . . . . . . 22514.3 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . . 22814.4 Checking Michelson Morley with Lorentz Transformation . . . . . . . . 232

14.4.1 Phenomena of the Lorentz Transformation . . . . . . . . . . . . . 23314.5 Minkowski Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

14.5.1 Proper Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23714.6 Causality and the Light Cone . . . . . . . . . . . . . . . . . . . . . . . . 23814.7 Symmetry Transformations in Minkowski Space . . . . . . . . . . . . . . 23814.8 Rotations in 4 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 239

14.8.1 Imaginary Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 24114.8.2 A Boost in an Arbitrary Direction . . . . . . . . . . . . . . . . . 242

14.9 Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24214.10The Momentum-Energy 4-Vector . . . . . . . . . . . . . . . . . . . . . . 243

14.10.1Deriving the Momentum-Energy 4-Vector . . . . . . . . . . . . . 24514.10.2The Force 4-Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 246

14.11Summary of 4-Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24714.12The 4D Gradient Operator x . . . . . . . . . . . . . . . . . . . . . . . 24914.13The Relativistic Doppler E ff ect . . . . . . . . . . . . . . . . . . . . . . . 24914.14The Twin Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25114.15Kinematics Problems in Electron Volts . . . . . . . . . . . . . . . . . . . 252


7/305

7

14.15.1 0 Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25214.15.2Neutron Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25314.15.3Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . 255

14.16Lagrange Equations in Special Relativity . . . . . . . . . . . . . . . . . . 257

14.17Covariant Electricity and Magnetism Equations . . . . . . . . . . . . . . 25914.17.1Rationalized Heaviside-Lorentz Units . . . . . . . . . . . . . . . . 25914.17.2The Electromagnetic Field Tensor . . . . . . . . . . . . . . . . . 26114.17.3Lorentz Transformation of the Fields . . . . . . . . . . . . . . . . 263

14.18E&M is a Vector Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 26414.19Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

15 A Little General Relativity 26715.1 Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26815.2 The Metric Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26815.3 The Schwarzschild Metric . . . . . . . . . . . . . . . . . . . . . . . . . . 26915.4 Gravitys E ff ect on Time and the Gravitational Red Shift . . . . . . . . 27015.5 The Singularity in Schwarzschild Coordinates . . . . . . . . . . . . . . . 27115.6 The Geodesic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 27215.7 Conserved Energy and Angular Momentum in the Schwarzschild Metric 27315.8 Orbits in the Schwarzschild Metric . . . . . . . . . . . . . . . . . . . . . 27415.9 Orbits of Photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

15.9.1 Deection of Light . . . . . . . . . . . . . . . . . . . . . . . . . . 27715.10Black Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27815.11Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

16 Hamiltonian Mechanics 280

16.1 Recalling Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . 28016.2 The Hamiltonian Formalism . . . . . . . . . . . . . . . . . . . . . . . . . 28216.3 Conserved Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28316.4 Simple Example: Particle on the Surface of a Cylinder . . . . . . . . . . 28316.5 Example: Particle in 3D Potential V (x,y,z ) . . . . . . . . . . . . . . . . 28516.6 Example: A spherical pendulum . . . . . . . . . . . . . . . . . . . . . . 28616.7 Example: Motion in a Central Potential V (r ) . . . . . . . . . . . . . . . 286

16.8 Phase Space and Liouvilles Theorem . . . . . . . . . . . . . . . . . . . . 28716.9 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

17 Continuum Mechanics 29017.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290


8/305

8

18 Appendices 29118.1 Review of Complex Numbers and Complex Exponentials . . . . . . . . . 29118.2 Review of the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 292

18.2.1 Hyperbolic Function Identities . . . . . . . . . . . . . . . . . . . 293

18.3 Solutions to Linear Diff

erential Equations with Constant Coeffi

cients . . 29518.3.1 Homogeneous equations with constant coe fficients . . . . . . . . 29518.3.2 The Inhomogeneous Equation . . . . . . . . . . . . . . . . . . . . 296

18.4 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29618.5 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 29918.6 Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29918.7 Study of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300

18.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30018.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30018.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

19 Formulas 302


9/305

9

0 Introduction

Classical Mechanics, is the science of the motion of macroscopic objects in the ap-proximation that Quantum Mechanical e ff ects can be neglected. Macroscopic objects

are substantially bigger and more massive than atoms where quantum eff

ects are im-portant. (More quantitatively, they are much bigger than their Debroglie wavelength.

We can learn a lot about Physics by studying the Classical Mechanics of simple visiblesystems where we have good intuition about how they will move. Much of what welearn can be applied to the Quantum realm as well as to the Physics of Fields wherethings are harder to visualize. Thus we will concentrate on simple mechanical systemsin this course. We will take a step away from our intuition when we study Relativitywhere objects move at velocities that are well above our own experience. but, this will

still be the study of the motion of simple objects.

In this course we will learn some of the more advanced techniques for solving me-chanical problems. These techniques are quite applicable to more advanced physicalproblems.

First we consider Single Objects or particles in this simple discussion of New-tons Laws.


10/305

10

1 Review of Newtonian Mechanics

Isaac Newton developed the Universal Law of Gravity (inverse square law) in1666. It was time to understand the orbits of planets (Keplers Laws of Planetary

Motion), Comets, and the Tides, since the measurements had been made and t. Helater (1686) developed Newtons Laws of motion in his book Philosophi NaturalisPrincipia Mathematica (Mathematical Principles of Natural Philosophy). In thisbook he used various mathematical tools (including calculus), developed by him, tosolve Physics problems of objects in motion. It will be a lot easier for us.

1.1 Newtons Laws

Newtons Laws: Original Text

Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiterin directum, nisi quatenus a viribus impressis cogitur statum illum mutare.

Lex II: Mutationem motus proportionalem esse vi motrici impressae, et eri se-cundum lineam rectam qua vis illa imprimitur.

Lex III: Actioni contrariam semper et qualem esse reactionem: sive corporum duo-rum actiones in se mutuo semper esse quales et in partes contrarias dirigi.

Newtons Laws: Translation

Law 1: Every body persists in its state of being at rest or of moving uniformly straightforward, except insofar as it is compelled to change its state by force impressed.

Law 2: The alteration of motion is ever proportional to the motive force impressed;and is made in the direction of the right line in which that force is impressed.

Law 3: To every action there is always opposed an equal reaction: or the mutual ac-tions of two bodies upon each other are always equal, and directed to contraryparts.

Newton proposed these as universal laws . They are independent of position .They are independent of time . They are independent of the velocity of theobject. They are independent of the direction of the force or the velocity. These

universalities are seen as Symmetries of the Laws of Physics and are perhaps moreimportant than the laws themselves.


11/305

11

Newtons Laws: Modern Statement

Law 1: When viewed in an inertial reference frame, an object either remainsat rest or continues to move at a constant velocity, unless acted upon

by an external force.Law 2: The vector sum of the forces on an object is equal to the change in

momentum of that object. (Denition of Force.)

Law 3: When one body exerts a force on a second body, the second body simul-taneously exerts a force equal in magnitude and opposite in directionon the rst body.

These laws can be summarized in one equation relating the change in momen-tum of two objects interacting with each other.

d~ p1dt

= d~ p2dt

Laws 1, 2, and 3 in any inertial frame.

~ p = m~ v Non-relativistic momentum of an object.

This single equation derived from Newtons Laws is essentially a statementof the Conservation of Momentum , with the second equation given here clarifyingexactly what momentum is according to Newton. We will learn later that Conservationof Momentum can be derived from Translation Symmetry of the Laws of Physics ,that is that the Laws of Physics are the same everywhere in the universe.

We can include the denition of Force and add a few more equations which add nonew physics.

X ~ F =

d~ pdt Dene Force.

~ v = d~ xdt

= ~ x Dene non-relativistic Velocity.

X ~ F = md2~ xdt2

2nd order equation in x for constant m.

X ~ F = m~ x 2nd order equation in x for constant m.

Depending on the Physics problem, we may integrate these rst order di ff erentialequations to get the solution or if the force depends on the position (or velocity), wemay prefer to solve a second order di ff erential equation in the position variableand its time derivatives.


12/305

12

Equations of Newtons Laws

~ p = m~ x

~ F = ~ p = m~ x

X ~ p = const.

1.2 Cartesian Coordinates

We may solve Newtonian Mechanics problems in any inertial coordinates system .That means we may choose any origin that is not accelerating. We can orient thecoordinate axes in any directions we want, but the directions must remain xed intime. We can transform to a frame moving with a constant velocity and the laws of Physics will be the same.

For non-relativistic problems, objects move in three space dimensions as a func-tion of time . Often at least one space dimension can be eliminated from the problemwith the proper choice of the coordinate system. The simplest physical coordinate sys-tem is a 3D Cartesian Coordinates after Descartes. These Cartesian coordinates


13/305

13

have three orthogonal axes along which the distance from the origin is measured.

~ x =3

Xi =1 x i ei xi are the 3 coordinates.ei ej = ij ei are the orthogonal unit vectors.

ij =1 i = j0 i 6= j

Kronecker delta.

x i = ~ x ei Coordinates measured parallel to axis.

We order these axes and dene our system to be right-handed by convention.

e1 e2 = e3 Right handed.

We use vectors to describe position and velocity of objects. Inner and outer productsof vectors can be written.

~ a ~ b =3

Xi=1 a i bi Dot product of two vectors.~ a ~ b = Xijk a i bj ijk Cross product of two vectors.

Here the Levi-Civita tensor or more descriptively the totally antisymmetric ten-sor has a value of 1 for a cyclic permutation of 123, a value of -1 for an anti-cyclicpermutation, and is zero by its nature if any index is repeated.

ijk = jik = ikj = kji Antisymmetric under interchange of indices.123 = 231 = 312 = 1 Cyclic permutations.

113 = 0 Zero for any repeated index.


14/305

14

See this section on Vectors.

The vector derivative ~ can be written in our vector notation.

~

=3

Xi=1d

dx iei The gradient operator.

2 = ~

~ =

3

Xi=1d2

dx2iDel squared.

Cartesian Coordinates

~ x =

3

Xi=1 x i eiei ej = ij

x i = ~ x ei

~ a ~ b = Xijk a i bj ijk~ =

3

Xi=1d

dx iei

We will also study and use Cylindrical Coordinates and Spherical Coordinates in thiscourse.

1.3 Physical Forces

The gravitational and electromagnetic forces are long range (inverse square)and therefore relevant to Classical Mechanics . The other two forces, the strongforce and the weak force, are e ff ectively very short range and thus only relevant inQuantum Mechanics. So really, there are just two classical physical forces, however,we can make springs and ropes that use Electromagnetism to apply a force. We canidealize these in many simple problems, even though its all really Electricity.

The formal study of EM forces is very important to Physics, but is thetopic of another course . But you can be sure that the Magnetic force does notviolate conservation of energy or momentum. It just makes us consider the energy


15/305

15

and momentum in the EM eld correctly. We will mainly deal with gravity and theidealized Electric force .

Example: Motion in a Constant Gravitational Field

Motion in Constant Gravity

~ v(t) = ~ v0 + ~ gt

~ x(t) = ~ x0 + ~ v0 t + ~ gt2

2

1.4 Motion with Friction

While friction is realistic for many systems, it often just obscures the real physicalprincipals, so most of this course will be done without friction . The problem of an object moving in a constant gravitational eld near the earth, will in most cases beinuenced by friction, so this is our chance to discuss Drag Force for motion througha uid like air or water.

At low velocities and for small objects , the ow of air around a falling objecttends to be Laminar which implies no turbulence. In this limit, the Stokes regime ,the Drag Force is given by:

F Stokesd = 3 d p~ vwhere is the viscosity of the uid (air), d p is the diameter of the object, and v is thevelocity. Of course this force is in the opposite direction of the velocity (although thisdepends on shape).

For larger velocities and larger objects , we enter the Newton regime whichis for turbulent ow. In this regime, the drag force is proportional to velocitysquared , since at higher velocity we hit more air molecules per second and give themhigher velocities too.

F Newtond = 12

Av 2C D v

Here is the density of the air, A is the cross sectional area of the object, and C D

isa drag coe ffi cient that depends on shape, velocity...


16/305

16

For the G7 shape above, the drag coe fficient C D is plotted below as a function of velocity. The at region below the speed of sound is the Newton regime where the forceis proportional to velocity squared as in the formula. C D increases rapidly then peaksclose to the speed of sound as the air in front of the projectile compresses increasingthe pressure applied. For supersonic velocities, the coe fficient decreases again.

For artillery, naval guns, and ries, the projectiles come out at supersonic velocities andslow down. For objects that we drop or throw, velocities are well below the speed of sound but terminal velocities are probably in the Newton regime. For small droplets,the Stokes regime may be relevant. Usually we use the dimensionless ReynoldsNumber to decide whether we are in the Laminar ow regime or the regime of turbu-lence,

Re = d pv

with the division between 1 and 10.


17/305

17

Linear and Quadratic Drag Forces

F Stokesd = 3 d p~ v

F Newtond =

1

2Av 2C D v

Example: Motion in Constant g with Linear Drag

Projectile Motion with Linear Drag

vx (t) = vx 0e gt

v t

x(t) = x0 + vx 0vtg

1 egtv t

vz (t) = vt + ( vz 0 + vt )egtv t

z(t) = z0 vt t + ( vz 0 + vt )vtg

1 egtv t

Example: Motion with Constant g and Drag Force proportional to Velocity SquaredFor a falling object we can integrate the equation of motion to get the velocity andintegrate a second time to get the position.

Falling Object with Quadratic Drag

v(t) = vt tanh gtvt

z(t) = z0 v2t

g ln cosh

gtvt

Similarly for an upward moving object we get the velocity, the distance as a functionof time, and the distance at the time the velocity goes to zero.


18/305

18

Upward Moving Object with Quadratic Drag

v(t) = vt tan arctanv0vt

gtvt

z(t) = v2t

g ln

cos arctan v0v t gtv tcos arctan v0v t

zmax = v2t

g ln sec arctan

v0vt

Example: RocketsFor a rocket out in space, the velocity depends on how much fuel has been expelledand not on the rate at which it was burned. The velocity calculated depends on thevelocity with which the fuel is expelled u and the remaining mass of the rocket m(t).

Rocket with No External Force

v(m) = v0 + u lnm0

m

Example: Rockets with Constant GravityTo overcome gravity, we cant just add more fuel. We have to burn it fast at least inthe rst stage.

Vertical Rocket in Constant ~ g

v(m) = v0 g

(m0 m) + u lnm0m

Example: The Falcon Heavy Rocket

1.5 Examples

1.5.1 Example: Motion in a Constant Gravitational Field


19/305

19

The acceleration of Gravity is independent of mass . In General Relativity wewill learn that the e ff ect of gravity and acceleration are indistinguishable. If we callthe acceleration due to gravity ~ g(~ x), any object will be accelerated at this rate. Nearthe earth, we can approximate the gravitational acceleration by a constant g. This isthe simplest case where we can integrate Newtons equations of motion. (In this class,

we expect you to do this integration, not just apply the results.)d~ vdt

= ~ v = ~ g(~ x) Acceleration due to Gravity.

~ v = ~ g Constant g for x


20/305

20

1.5.2 Example: Motion with Constant g and Drag Force proportional toVelocity

For a drag force ~ F d = D~ v and an object falling in a gravitational acceleration ~ g,we can compute the terminal velocity without needing to integrate the equations of motion. Write Newtons equation for object falling ( z direction).

v = Dm

v g = 0Consider a falling object. At terminal velocity, v = 0 , and we have

v = Dm

v g = 0 No acceleration at terminal velocity.

v = mgD Terminal v is down, negative.

v = vt = mgD

Dene vt to be positive.

vt mg

D Use positive vt .

We will use the constant vt computed from the drag force, in our solution ratherthan D.

Terminal Velocity with Linear Drag

vt = mg

D

Now we put the velocity on the left hand side of the the equation and timeon the right and integrate both sides up to time t. We put the velocity at t = 0 into


21/305

21

the limits of integration.

vt mg

D From above.

v = Dm

v g Equation of motion.vz =

gvt

vz g = gvzvt

+ 1 Equation of motion using vt .

dvzdt

= gvt

(vz + vt ) Rewrite equation.

v

v0

dvzvz + vt

= t

0

gvt

dt Integrate equation of motion.

ln v + vtv0 + vt =

gvt t Do the integral.

v(t) = vt + ( v0 + vt )egtv t Solve for vz .

dzdt

= vt + ( v0 + vt )egtv t Prepare to integrate again to get z(t).

z

z0

dz =t

0

vt + ( v0 + vt )egtv t dt Set up integral to get z.

z(t) = z0 vt t + ( v0 + vt ) vtg 1 egtv t Integrate to get z(t).

The initial conditions are represented by the two constants of integration v0 and z0 .We can check the answer for t = 0 and also for v0 = vt to see if we made algebraerrors.

v(t) = vt + ( v0 + vt )egtv t

z(t) = z0 vt t + ( v0 + vt )vtg

1 egtv t

For a projectile problem , the object will also have some initial velocity in the hor-

izontal direction. In the case of drag force proportional to velocity, we can work thisproblem separately as a function of t to, for example, compute the range of a projectile


22/305

22

with drag.

~ F d = D~ v Drag force.F xd = Dvx x-component of drag.vx =

g

vtvx Equation of motion in the x direction.

ln vvx 0

= gvt

t Integrate equation.

vx (t) = vx 0e gt

v t Solve for v(t).

x(t) = x0 + vx 0vtg

1 egtv t Integrate to get x(t).

Projectile Motion with Linear Drag

vx (t) = vx 0e gt

v t

x(t) = x0 + vx 0vtg

1 egtv t

z(t) = z0 vt t + ( v0 + vt ) vtg 1 egtv t

To nd the range of a projectile we need to nd t solving the z(t) = 0 equationnumerically (since it is transcendental), then calculate x(t).

1.5.3 Example: Motion with Constant g and Drag Force proportional toVelocity Squared

In this case, the z and x equations are coupled by the drag term .

~ F d = Cv2 v Drag force proportional to v2 .F xd = Cvx |v| Drag in x depends on total |v|v.

So for simplicity we will only consider the case of an object with zero horizontalvelocity .

We also have a second problem with the sign of the drag force . It clearly opposesthe motion but we dont have a good way to handle upward and downward moving


23/305

23

objects with the same equation. So we will have to do separate solutions for upwardand downward motion .

First lets calculate the terminal velocity for a falling object.

v = +C m v

2

g = 0 No acceleration at terminal velocity.v2t =

mgC

Solve for terminal velocity.

vt = r mgC Use positive terminal velocity in our equations.The solution for a falling object with drag proportional to v2 , approaches vt as the

exponentials become small with increasing time.

Terminal Velocity with Quadratic Drag

vt = r mgC Now we will try the case for a downward velocity . This means the drag force is inthe positive direction.

Appendix: Complex Exponentials.

Appendix: Review of Hyperbolic Functions.


24/305

24

v = C m

v2 g = gv2t

v2 g = gv2t

(v2 v2t ) Eq. for downward motion using vt .vt

dvv2

v2t

= gvt

dt Eq. of motion with v on LHS.

v = vt tanh substitution for v downdv = vt

cosh2 d for substitution

vt vtv2t cosh2 (tanh 2 1) d = gvt

dt do substitution

1(sinh 2 cosh2 ) d = gvt

dt algebra

d = gvt dt algebra!

0

d =t

0

gvt

dt v = 0 at t = 0

= gt

vtdo integrals

tanh 1 vvt

= gtvt

plug in

v(t) = vt tanh gtvt

plug in

We can integrate again to get the vertical position z.

dz = vtt

0

tanh gtvt

dt integrate to get . z

tanh( ax )dx = 1a ln (cosh( ax )) integral of tanhz(t) = z0 vt

vtg

lncosh gtv t

cosh(0) evaluate.

z(t) = z0 v2t

g ln cosh

gtvt

algebra


25/305

25

Falling Object with Quadratic Drag

v(t) = vt tanh gtvt

z(t) = z0 v2t

g ln cosh gtvt

We can check a few conditions to catch errors. We get v(t = 0) = 0 and v(t ) =vt . We get z(t = 0) = z0 .Now write the equation of motion for an upward moving object so the drag force is

down.

v = C m

v2 g = gv2t

v2 g = gv2t

(v2 + v2t ) Eq. for upward motion using vt .

vtdv

v2t + v2 = g

vtdt Get v on LHS.

vt

v

v0

dvv2t + v2

= gtvt

Set up integral with init. v.

dvv2t + v2 = 1vt

arctanvvt

subs. v = vt tan

arctanvvt arctan

v0vt

= gtvt

Do the integral.


26/305

26

arctanvvt

= arctanv0vt

gtvt

Get v on LHS.

vvt

= tan arctanv0vt

gtvt

Take tan of both sides.


gtvt Solve for v(t)

tv=0 = vtg

arctanv0vt

time to reach max height.

z

0

dz = vt

t

0

tan arctanv0vt

gtvt

dt integrate tangent

z(t) = v2t

gln cos arctan

v0

vt gt

vt

t

0evaluate

z(t) = v2t

g ln


z(t)

zmax = v2t

g ln

cos arctan v0v t arctan v0v tcos arctan v0v t

max height

zmax = v2t

g ln sec arctan v0

vt max height

Upward Motion with Quadratic Damping


gtvt

z(t) = v2t

g ln


zmax = v2t

g ln sec arctan

v0vt

This seems to pass all the sanity checks one can make.


27/305

27

1.6 Example: Rockets

The equation of motion for a rocket is di ff erent than for a normal object with a forceapplied because the mass of the rocket decreases as fuel is ejected . So both themass and the velocity of the rocket are changing with time. Our model of a rocket willassume that the fuel is ejected at a constant velocity in the frame wherethe rocket is at rest .

The total momentum is conserved . Consider the ejection of a small amount of mass dm with fuel velocity u. If we eject dm of fuel, the rocket mass change is dm.First lets compute without gravity, like we are accelerating in empty space.

pi = pf Conservation of Momentum.mv = ( m

dm)(v + dv) + dm(v

u) Eject fuel dm.

mv = mv dmv + mdv dmdv + dmv udm momentum of rocket + ejecta.0 = + mdv udm udm is the Thrust of the rocket.

dv = + um

dm Solve for dv.

dv = um

dm rocket dmrocket = dm.v

v0

dv = um

m 0

1m

dm integrate.

v(m) = v0 + u lnm0m

Solve for velocity.

The logarithmic gain in velocity makes it hard to get to really high velocities.Multistage rockets help but its not as good as using all the mass of the rocket as fuel.So the best way to get to high velocity it to have a small payload mass . Also clearlya high fuel ejection velocity is important. The best would be u = c with the

momentum of light being E

c .

Rocket with No External Force

v(m) = v0 + u lnm0m

For a rocket out in space, the velocity depends on how much fuel has been expelledand not on the rate at which it was burned.

1.6.1 Example: Rockets in Constant Gravity


28/305

28

Launching a rocket from earth to get into low earth orbit is close to working in aconstant gravity eld. Gravity changes the problem. We are familiar with the largeThrust needed to overcome gravity and get a rocket to take o ff . So here we will needto have a burn rate for the fuel that is more than su fficient to overcome gravity if arocket is to leave the earths atmosphere. Lets call the burn rate of fuel = dmdt .

The radius of the earth is about 6500 km and an orbit 200 km above the earth ispossible, so g is nearly constant. The problem is to get up to a high enough velocity toget into a stable orbit. There is a big gain by launching from near the equatorwhere we start with a high tangential velocity due to the rotation of the earth. Onehas to orbit in the same direction the earth is turning. Orbits in the other direction(retrograde) are much harder to achieve and decay faster. Often spy satellites are putinto polar orbits which are also costly.

We just need to add the acceleration of gravity to the equations above, and to dealwith dv = gdt versus dv = um dm by using the burn rate.mgdt = pf pi innitesimal change in p.

mgdt + mv = ( m dm)(v + dv) + dm(v u) calculate pf .mgdt + mv = mv dmv + mdv dmdv + dmv udm algebra.

mgdt = + mdv udm u is Thrust of rocket.

dv = gdt + um dm Solve for dv.dv =

g

dm rocket um

dm rocket use burn rate.

dv =g

um

dm rocket RHS all dm.

u > mg Thrust must beat gravity.v

v0dv =

m

m 0g

u

mdm integrate.

v = v0 g

(m0 m) + u lnm0m

solve for v.

Rocket in Constant Gravity

v(m) = v0 g (m0 m) + u ln

m0m

To overcome gravity, we cant just add more fuel. We have to burn it fast at least inthe rst stage.


29/305

29

1.6.2 Example: The Falcon Heavy Rocket

The announcement of the Falcon Heavy in early April, 2011 was a potential game-changer in the space launch industry. The Falcon Heavy is slated to launch twicethe payload of the Shuttle at about one-fteenth the cost of a Shuttle launch anapproximate 95% reduction in launch costs compared with the Shuttle!

http://www.spacex.com/falcon-heavy

http://www.nss.org/articles/falconheavy.html

The SpaceX Falcon Heavy Rocket scheduled for launch in 2015, will be the worldsmost powerful rocket. (The Saturn V of the U.S. Apollo program was more powerful,many years ago.) It has been designed with cost e fficiency in mind, largely using onetype of engine everywhere and basically using kerosene as fuel. Also, the boosters andcenter core are expected to return to the launch site, landing using their thrusters onextendable legs. Thus all the rst stage engines will be reusable.
http://www.spacex.com/falcon-heavyhttp://www.nss.org/articles/falconheavy.htmlhttp://www.nss.org/articles/falconheavy.htmlhttp://www.spacex.com/falcon-heavy


30/305

30

Its somewhat di fficult to get all the parameters of the Falcon Heavy rocket system,but I compiled some information from the web so that we can make a model.


31/305

31

Falcon Heavy Rocket DataFunction Orbital launch vehicle Or Lunar launch vehicleManufacturer SpaceXCountry of origin United StatesCost per launch (2014) $85M for up to 6,400 kg to GTO

Height 68.4 mDiameter 3.66 mMass 1,462,836 kgStages 2+Payload to LEO 53,000 kgPayload to GTO 21,200 kgStatus In DevelopmentLaunch sites KSC LC-39AVandenberg SLC-4E

Total launches 0Successes 0Failures 0First ight 2015 (projected)Boosters (Stage 0)No. boosters 2Engines 9 Merlin 1DThrust 5,880 kN (sl)Total thrust 17,615 kN (sea-level thrust of boosters plus core)Specic impulse Sea level: 282 secVacuum: 311 secBurn time UnknownFuel LOX/RP-1First stageEngines 9 Merlin 1DThrust 5,880 kN (sl)Specic impulse Sea level: 282 secVacuum: 311 secBurn timeFuel LOX/RP-1Second stageEngines 1 Merlin 1D VacuumThrust 801 kNSpecic impulse Vacuum: 342 secBurn time 375 secondsFuel LOX/RP-1


32/305


33/305

33

Falcon Heavy Information

FIRST STAGE: THREE ROCKET CORES Falcon Heavys rst stage consistsof three cores. All three cores operate together at lifto ff . About T+2:45 minutesinto ight, the center core throttles down while the two side cores continue at fullthrust until their fuel is nearly spent. At that point, pneumatic separators releasethe side cores, which plummet into the ocean, and the center core throttles up.

CENTER CORE For payloads heavier than 100,000 pounds, Falcon Heavy uses across-feed system to run fuel from the side cores to the center core, leaving thecenter core almost fully fueled after the side boosters separate. Whats left is theequivalent of a complete Falcon 9 rocket already high in space.

FUEL TANKS A liquid-oxygen tank at the top of each core feeds the engines througha center tube; the lower portion of the tank contains rocket-grade kerosene. The

propellants are turbo-pumped into each Merlin engines injector, where they aremixed and fed into the combustion chamber.

SECOND STAGE Powered by a single Merlin 1D engine modied to operatein the vacuum of space, the second stage delivers the nal push that gets thepayload into orbit. The engine can shut down and reignite as needed, enablingFalcon Heavy to deliver multiple payloads to di ff erent orbits.

FAIRING Falcon Heavy can carry either a Dragon capsuleSpaceXs free-ying space-craft, currently used to resupply the International Space Stationor up to 117,000pounds of payload (think multiple military and commercial satellites) enclosedin a shell 45 feet long and 17 feet in diameter. The fairing consists of two clam-shell-style halves made of an aluminum honeycomb core and carbon-ber facesheets. When the second stage nears the desired orbit, pneumatic pushers splitthe halves apart, exposing the payload.

MERLIN 1D ENGINE A single Merlin 1D generates 147,000 pounds of thrust atsea level, burning rocket-grade kerosene and liquid oxygen fed by a turbo-pumpinto the combustion chamber. Falcon Heavys liquid propellant has an advantageover solid fuel: Liquid-fueled engines can stop and restart in ight, whereas solid-fuel engines burn until they are spent. Through proprietary adjustments thatSpaceX wont disclose, engineers recently lightened the engine to increase itsefficiency, making it the most e fficient rocket booster engine ever built.


34/305

34

Falcon Heavy Launch Timing

T3:00:00 Falcon Heavy is ready on the launchpad at Cape Canaveral. Engineers timeliftoff to achieve the optimal ight path and desired orbit.

T0:10:30 The countdown begins. All actions from here forward are pre-programmed,

although Mission Control can abort the mission at any time.T0:02:30 The launch director issues the nal launch command.

T0:00:40 Propellant tanks are pressurized.

T0:00:03 First-stage engines ignite.

0:00:00 The on-board rocket computer commands the launch mount to release. Lifto ff .

T+0:01:25 The rocket reaches maximum aerodynamic pressure; mechanical stresspeaks.

T+0:02:45 The rocket has now burned enough fuel (thus decreasing its mass) thatthe center core engines can throttle down.

T+0:03:00 The side cores separate and fall into the ocean, while the center coresnine Merlin engines continue to burn for approximately 30 seconds.

T+0:03:30 The second stage separates from the remaining rst-stage core. The

second-stage engine ignites and continues toward orbit.T+1020 MINUTES When the rocket nears the desired orbit, the two halves of the

clam-shell fairing open and fall away. When in position, the payload separatesfrom the second stage. Both the fairing and second stage eventually fall back toEarth.

We can get a little more data by looking at the Falcon-9 Rocket which is in use. Therst stage of the Falcon-9 is essentially the same as one of the boosters or as the centercore of the Falcon Heavy.


35/305

35

Falcon 9 First Stage is Like Falcon heavy Core and BoostersType Falcon 9 v1.1 Stage 1Length 42mDiameter 3.66mInert Mass 18,000kg

Propellant Mass 385,000kgFuel Rocket Propellant 1Oxidizer Liquid OxygenLOX Tank MonocoqueRP-1 Tank Stringer and Ring FrameMaterial Aluminum-LithiumGuidance From 2nd StageTank Pressurization Heated HeliumPropulsion 9 x Merlin 1D

Engine Arrangement OctawebEngine Type Gas GeneratorPropellant Feed TurbopumpMerlin 1 D Thrust Sea Level: 654kN - Vac: 716kNEngine Diameter 1.25mEngine Dry Weight 450 to 490kgBurn Time 180sSpecic Impulse 282s (SL) 311s (Vac)Chamber Pressure 9.7MPa (1,410psi)Expansion Ratio 16Throttle Capability 70% to 100%Restart Capability YesIgnition TEA-TEBAttitude Control Gimbaled Engines (pitch, yaw, roll)Shutdown Commanded ShutdownStage Separation Pneumatically actuated mechanical colletsNotes Soft Splashdown Landing as a precursor

to full reusability of the First Stage


36/305

36

Falcon 9 Stage 2 is like Falcon Heavy Stage 2Type Falcon 9 v1.1 Stage 2Length 15mDiameter 3.66mInert Mass 4,900kg

Propellant Mass 90,000kgFuel Rocket Propellant 1Oxidizer Liquid OxygenLOX Tank MonocoqueRP-1 Tank MonocoqueMaterial Aluminum-LithiumGuidance InertialTank Pressurization Heated HeliumPropulsion 1 x Merlin 1D Vac

Engine Type Gas GeneratorPropellant Feed TurbopumpThrust 801kNEngine Diameter 1.25mEngine Dry Weight 450 to 490kgBurn Time 372sSpecic Impulse 340s (Est: 345s)Chamber Pressure 9.7MPa (1,410psi)Expansion Ratio > 117Throttle Capability YesRestart Capability YesIgnition TEA-TEB, RedundantPitch, Yaw Control Gimbaled EngineRoll Control Reaction Control SystemShutdown Commanded ShutdownReaction Control S. Cold-Gas Nitrogen ThrustersPayload Fairing Composite FairingDiameter 5.2mLength 13.1mWeight 1,750kg

First, lets try to account for the mass of the rocket plus payload.


37/305

37

Falcon Heavy Component Mass EstimatesComponent Mass Total Mass Scaled MassDry Mass of Boosters 18000 36000 43000Fuel Mass of Boosters 385000 770000 829000Dry Mass of Center Core 18000 18000 19000

Fuel Mass of Center Core 385000 385000 415000Dry Mass of Second Stage 4900 4900 5000Fuel Mass of Second Stage 90000 90000 97000Mass of Fairing 1750 1750 1900Mass of Payload 53000 53000 53000Sum of Masses 1,358,650 1,462,900Total Mass 1,462,836

We are miss the total mass by 7.5%. Its clear from the pictures and statements that

the booster are bigger on the Heavy compared to the Falcon 9 rst stage and there issomewhat more structural support needed. So lets just scale everything up to makethe total come out about right.

The other things we need to calculate the capability of this rocket is the thrust andthe burn time . Lets assume that the time integral of the Thrust is proportional tothe fuel burned. This is not exactly but OK. The total Thrust is 180(17615)+30(6000)kN s, while the total fuel is 1244000 kg, giving 0.3713 kg per (kN s).

Falcon Heavy Key Data EstimatesStage Thrust Burn Time Initial Mass Mass of Fuel Mass of Stage Dropped0 17615 kN 180 s 1462836 1177166 430001 6000 kN 30 242670 66834 190000+1 17615 190.2 1462836 1244000 620002 801 kN 372 s 156876 97000 0

This should be enough data to fairly compare a pure 2 stage system to the currentversion that can drop the boosters and still continue to thrust with the center core fora while, to a more extreme system where the boosters drop with the center core enginestill lled with fuel.

https://www.youtube.com/watch?v=ZwwS4YOTbbw

JULY 22, 2014SPACEX SOFT LANDS FALCON 9 ROCKET FIRST STAGEFollowing last weeks successful launch of six ORBCOMM satellites, the Falcon 9 rock-ets rst stage reentered Earths atmosphere and soft landed in the Atlantic Ocean.

http://www.spacex.com/news/2014/07/22/spacex-soft-lands-falcon-9-rocket-firs

Interviews with Elon Musk. https://www.youtube.com/watch?v=P79E0-3LeW8

https://www.youtube.com/watch?v=g92rP1Mi_oQ
https://www.youtube.com/watch?v=ZwwS4YOTbbwhttp://www.spacex.com/news/2014/07/22/spacex-soft-lands-falcon-9-rocket-first-stagehttps://www.youtube.com/watch?v=P79E0-3LeW8https://www.youtube.com/watch?v=g92rP1Mi_oQhttps://www.youtube.com/watch?v=g92rP1Mi_oQhttps://www.youtube.com/watch?v=P79E0-3LeW8http://www.spacex.com/news/2014/07/22/spacex-soft-lands-falcon-9-rocket-first-stagehttps://www.youtube.com/watch?v=ZwwS4YOTbbw


38/305

38

1.7 Homework Problems

1. If a projectile is red from the origin (on earth), with an initial velocity v0 , andat an angle above the horizontal, how much time does it take to cross a linestarting at the origin an at an angle < above the horizontal?

2. A projectile is red (on a at plain on earth) with a velocity v0 such that itpasses through two points both a height h above the the ground. The gun hasbeen adjusted for maximum range. Find the separation distance between the twopoints.

3. A mass is shot vertically upward (on earth), with an initial velocity v0 . Assumethat there is a drag force proportional to the velocity squared F d = kmv 2 . Interms of v0 and vt the terminal velocity, what is the velocity of the particle whenit returns to the point from which it was red?

4. What is the maximum angle above the horizontal with which a projectile can bered so that its distance from its point of ring (on earth) is always increasing?

5. A potato of mass 0.5 kg is shot out of coke-bottle-CO 2 cannon at an angle of 45 degrees. It goes a distance of 266 meters for a measured initial velocity of 70meters per second. If the drag force is ~ F = D~ v, what is the value of D?

6. Find the equation of motion for a rocket accelerating along the direction of an ex-ternal force F ext . Assume the exhaust velocity is u and that it burns kilogramsper second of fuel. Specialize to the case of a constant gravitational accelerationg. Calculate the speed of the Falcon Heavy Rocket after one minute given apayload of 20000 kg.

7. A rocket of initial mass m0 accelerates from rest in empty space. At rst itsmomentum increases but eventually it begins to decrease. What is the mass of the rocket when the momentum rst begins to decrease?

8. A rocket of initial mass m0 ejects mass at a rate = dmdt at a velocity u. The

rocket is subject to a drag force F d = bv. Calculate the velocity of the rocketin terms of m, the mass remaining.9. Compare the nal speeds of a single stage rocket with 60% of its mass in fuel to a

two stage rocket with 30% of its mass in fuel for each stage, and 10% of its massejected with the rst stage, so that only 30% of its initial mass remains when allthe fuel is burned.

10. Compare the nal velocities of the Falcon Heavy rocket launched vertically inconstant g, for three staging scenarios. a) With the boosters plus core operationas a single stage with all three components ring at full throttle; b) With thecurrent design having the core operate for 30 seconds after the boosters drop o ff ;and c) With a modied system where the three rockets re at full throttle forsome time but then the boosters drop o ff leaving the center core with 30% of theinitial fuel that was in stages 0 and 1.


39/305

39

1.8 Sample Test Problems

1. (10 points) A projectile of mass m is shot into the air with and initial velocityv0 at 30 from the horizontal. Assuming that the forces on the particle are~ F = mgy k~ v:a) Write down the equation of motion for the horizontal ( x) direction. (2)

mx = kx x + km

x = 0

b) Write down the equation of motion for the vertical ( y) direction. (3)

my = ky mg y + km

y + g = 0

c) Determine the maximum height. (5)The basic di ff erential equation is:

dvdt

+ km

v + g = 0

The quickest way to nd the max height is to nd h(v).

v dvdt = km v + g dydtv dv

km v + g

= dy0

v0 / 2

v dvkm v + g

= h

0

dy

mk

v0 / 2

0

v dvv + mgk

= h

mk

v0 / 2

0

1 mgk

v + mgkdv = h

mv02k

m2gk2

hln v +

mgk

i

v0 / 2

0= h

mv02k m2gk2 ln

v02 +

mgk

mgk

= h

h = mv0

2k m2g

k2 ln 1 +

kv02mg


40/305

40

A second way to solve this problem is nd the time at max height, then nd y atthat time.

dvdt

= km

v g

mk

v

vy 0dv

v + mgk=

t

0 dtmk

ln v + mgkvy0 + mgg != t

v + mg

k = vy0 +

mgk

ekt/m

v =

mg

k + vy0 +

mg

kekt/m = 0

ekt max /m = mgkvy 0 + mg

tmax = mk

ln mg

kvy0 + mg=

mk

ln 1 + kvy0

mg

y =t

0

mgk

+ vy0 + mg

kekt/m dt

y = mgk t

mk vy0 +

mgk ekt/m 1

h = mgk

tmax + mvy0

k

h = mgk

mk

ln 1 + kvy 0

mg+

mvy0k

h = mv0

2k m2g

k2 ln 1 +

kv02mg

2. (5 points) A rocket is launched vertically from the surface of the earth. Thepayload of a rocket is 1000 kg. The rocket engines expel fuel at 2500 m/s andare capable of expelling 20,000 kg of fuel per second. How much fuel is needed toreach earths escape velocity of about 11,000 m/s? (Neglect wind resistance andgravity for this calculation.) Now show that gravity would be neglegible.The basic equation to use is v = u ln m 0m , where m0 is the mass of the payloadplus fuel, m is the mass of the payload, and u is the fuel velocity. We will use


41/305

41

m f as the mass of the fuel.

v = u lnm p + m f

m p (1)

evu =

m p + m f

m pm p e

vu 1 = m f (1)

m f = 1000 e110002500 1 = 80 , 459 (2)

So the mass of the fuel is 80,450 kg which takes about 4 seconds to burn. Theeff ect of gravity is about 40 m/s on the velocity which is a very small fraction.(1)

3. (10 points) A rocket of total mass m0 is shot vertically from the surface of theearth with sucient thrust of overcome gravity imediatly. The rocket carries fuelof mass mf which it burns at a constant rate of = dmdt . It reaches a nalvelocity of v1 when the fuel runs out. We can neglect frictional forces during theacceleration of the rocket.a) Write down the (di ff erential) equation of motion.b) Find the velocity at which the fuel leaves the rocket.c) The rocket reaches a maximum height of h. It falls to the ground under the

inuence of gravity and air resistance. Assume that the air resistance for therocket is given by F air = v2 and is of course opposite to the direction of motion.What is the velocity as the rocket strikes the ground?

Since I dont have a large number of examples that I have put on tests for 110A, I listbelow the Homework Problems that I think could be test problems. The ones thatare excluded could be too time consuming, too dependant on detailed data, or could bea prove that problem which is not usually good for a test. There are many reasons I

might exclude a problem and such an exclsion should not be taken to mean that thattopic will not be covered. In general, the homework topics will be fair game, however,some homework problems are too long for a 50 minute test.

In this chapter, I think HW problems 1, 2, 4, shortened 5, 6, 7, and 8 couldbe test problems.


42/305

42

2 Review of Energy and Conservative Forces

In Newtons time, the equations of motion together with Newtons Gravity were a tri-umph. Physicists (or Natural Philosophers) could understand the orbits of the Plan-

ets and motion on earth with the same equations. The equations of motion werederived from experiments and basically told us that momentum is conserved .Later (19Th century) it was also shown from experiment, that Energy is Conservedand Kinetic Energy was dened along with Potential Energy . Of course Heat wasimportant to these experiments making the physics more complicated.

In relativity we will nd that Energy and Momentum are part of one vector in 4-dimensions. (Like the time and the space coordinate are part of a 4-vector.) Wewill also later show that translation symmetry in time and space gives us Energy and

Momentum conservation, but for now, we will take the non-relativistic result as we didNewtons Laws, from experiment. Energy conservation allows us to solve manyproblems quite simply .

2.1 Kinetic Energy

The Kinetic Energy of an object, in the non-relativistic limit, is given by

T = 12

mv 2 .

We may compute the rate of change in kinetic energy with time when a force isapplied to an object using Newtons laws.

dT dt

= ddt

12

mv 2 = md~ vdt

~ v = ~ F d~ xdt

Change in KE.

T 2

T 1

dT =B

A

~ F d~ x Integrate.

T 2 T 1 =B

A

~ F d~ x Change in KE = Work done.

The change in Kinetic energy is the line integral of the Force , or the Work .

2.2 Conservative Forces and Potential Energy


43/305

43

T 2 T 1 =B

A

~ F d~ x

In general this line integral will depend on the path taken to get from point A to pointB , but for a particular class of forces, the line integral will be independent of path and thus the change in kinetic energy will only depend on the endpoints .An example of a conservative force with which you are familiar is shown below. Fora constant gravitational acceleration , the potential energy only depends on theheight, and the line integral will prove to be independent of the path taken.

So for a Conservative Force , the work is independent of path from one point toanother. This will be true if the Force can be written as the gradient of a scalarfunction .

~ F = ~ U F is gradient of -U.~ B

~ A

d~ x ~ F = ~ B

~ A

dU = (U ( ~ B ) U ( ~ A)) Compute line integral.

U B U A = B

A

~ F d~ x Change in U independent of path.

T B T A =B

A

~ F d~ x Change in KE from above is opposite.


44/305

44

Not all possible forces satisfy this condition, that ~ F = ~ U . For example, a force~ F = ~ F = k1yx k2xy will not give an integral that is independent of path. Letsintegrate from (0,0) to (2,3) by two di ff erent simple paths.~ F = k1yx k2xy Example of non-conservati

(2 ,0)

(0 ,0)

~ F xdx +(2 ,3)

(2 ,0)

~ F ydy = k1(0)2 k2(2)(3) = 6k2 Path 1

(0 ,3)

(0 ,0)

~ F ydy +

(2 ,3)

(0 ,3)

~ F xdx = k2(0)3 k1(3)(2) = 6k1 Path 2 is di ff erent

The line integral depends on the path, unless k1 = k2 for this force.

Stokes Theorem , one of the basic theorems of calculus, relates the surface integralof the curl of a function to the line integral of the same function around the boundaryof the surface.

S

d~ x ~ F = S

d~ S ~ ~ F

A conservative force has zero curl and thus the line integral around any closed pathwill be zero.

~ ~ F = 0 Functions with zero curl.

S

d~ x ~ F = 0 Line integral around closed path is 0.

From this its easy to see that the line integral from point A to point B is independentof path and thus the force can be written as the gradient of a scalar function which wewill call the Potential Energy Function . Basically, if there were nonzero curl, wetravel around a little loop and gain or lose energy.


45/305

45

For our example, ~ ~ F = F

y x F x y z = ( k2 k1)z, which will be zero if k1 = k2 . Sowe can see that Energy conservation is not automatic for an arbitrary force. What hasbeen found in experiment, is that the forces that exist in nature do not violateconservation of energy .

2.3 Total Energy Conservation

For Conservative Forces like the Electric force and the Gravitational force, TotalEnergy is conserved . The magnetic force is perpendicular to motion so it does nowork. Clearly we will need other tools for the magnetic force (the vector potential).The weak and strong interactions also have been measured to conserve energy. Wealso know that in relativity , mass also contributes to the energy balance equations(E = mc2) and we know that in quantum mechanics, energy conservation canbe violated for short times but that after a long time the total energy is conserved.Even with all these caveats, Total Energy is Conserved in everything we have been

able to measure. (Physicists tend to believe that even in the Big Bang, some kind of energy conservation is still valid, but we are far from measuring this with any accuracy.)

Many problems can be most easily solved by using Energy Conservation ,rather than by integrating the equations of motion. We may also derive simple equa-tions of motion from Energy Consecration.

Of course problems which include friction do not conserve the sum of potential andkinetic energy. With friction, kinetic energy is converted into Heat but total

energy is still conserved. Friction makes solutions based on energy conservation lessuseful.


46/305

46

2.3.1 One Dimensional Problems

The utility of solving a problem based on energy conservation is most easily demon-strated in one dimension ( d = 1). Many 3D problems can be reduced to a1D problem using other conserved quantities and a good choice of coordinates,allowing similar solutions to those problems.

E = T + U = 12

mx2 + U (x) Write total E.

x(x) = r 2m (E U (x)) Solve for v(x).r m2

x

x 0

dx

p E U (x)

= t 0 Integrate to get x.

r m2x

x 0

dx

p E U (x)= t algebra.

Once the integral, which obviously depends on the form of the potential energy func-tion, is done, the equation can be inverted to solve for x(t). There are generallytwo solutions to this equation shown by the sign. Solutions can be found forany initial x

0 and energy E iff E

U (x

0). Of course the energy could also be

computed given the initial position x0 and velocity v0 . So for any initial (x0 , v0),the subsequent motion, (x(t), v(t)) , can be computed , by doing the integral.

2.3.2 Phase Space

Since the initial conditions of a problem can be written as ( x0 , v0), and the state of the system at a later time would be written as ( x(t), v(t)), it is useful to dene apair of variables to give the state of the system. Since the position and velocityare di ff erent variables and have di ff erent units , this is not like a normalmultidimensional space . We call this 2D space Phase Space .

Time Evolutions in Phase Space

ddt

xv =

vF (x )

m

Given some initial condition, the particle will follow some path through phasespace . The collection of the paths through phase space is called the phase portraitof the system.


47/305

47

Often the trajectories will have Turning Points where the total energy is equal tothe potential energy and thus the velocity goes to zero. The phase curves may beconned between two turning points and oscillate back and forth retracing thesame path .

If the particle is not conned between two turning points and has enough energy toescape to innity , the curve will not be closed.

It is possible to generalize the behavior of a Phase Curve in the vicinity of a FixedPoint , that is a point at which the derivative of the potential energy is zero


48/305

48

and the velocity is zero .

U 0(xfp ) = dU dx

= 0 Zero force at xed point.

vfp = 0 Zero velocity at xed point.

x = xfp + x Dene x.U (x) = U (xfp ) + U 0(xfp ) x +

12

U 00(xfp )( x)2 + ... Taylor series.

k = U 00(xfp ) Like HO if k > 0.

U (x) U (xfp ) + 12

k( x)2 Near the xed point.

E = T + U = 12

mv 2 + 12

k( x)2 Ellipse in phase space ( k > 0).

E = T + U = 12mv 2 +

12k( x)2 phase space hyperbola ( k < 0).

So near a xed point, the trajectories are ellipses for local minima and hyper-bolae near local maxima . Classically, a system could sit right at the xed point fora local minimum, or take an innite amount of time to reach a local maximum if theenergy is just right.

We can also write a linear phase space equation .

F (x) = dU dx k x Force law.

ddt

xv =

0 1km 0

xv Linear phase space equation.

Six dimensional phase space is used to follow the state of a particle in 3D. We willlater prove a theorem about phase space density.

2.3.3 Example: Phase Portrait for Harmonic Oscillator

A harmonic oscillator has a minimum in the potential at x = 0 and it is not anapproximation that E = T + U = 12 mv

2 + 12 k( x)2 , with k > 0. So the phase portrait

will be ellipses. For energy E , there will be turning points at x = q 2E k . The higherthe energy, the bigger the ellipse.


49/305

49

2.4 Conservation of Momentum and Angular Momentum

We know that Newtons laws are essentially a statement of the conservation of momentum . Consider two particles. If particle 1 exerts a force ~ F on particle 2, thenby Newtons third law, particle 2 exerts an equal and opposite force ~ F on particle 1.The momentum of each particle may be changed but the sum of the momentumwill be conserved . Its easy to see by summing over all pairs that this will also be truefor a system of particles . If an external force is applied to a system of particles, thenthe rate of change of the momentum will be equal to that external force, but, whateverapplies the external force will have its momentum changed. Nevertheless sometimeswe wish to consider the existence of external forces in Mechanics and we will have:

External Force Changes Momentum

~ p = ~ F ext

Similarly with angular momentum, we will nd that the change in angular momen-tum of a particle or a system of particles is given by the external torque .

External Torque

~ L = ~ r ~ p~ L = ~ ext

~ ext = ~ r ~ F ext


50/305

50

Consider the pair of particles again with a Central Force between them. A Centralforce acts along a line between the two particles (or a line toward the Centerof Mass ), which is in the same direction. Let us compute the change in angularmomentum due to the central forces between a pair of particles.

~ L =

ddt (~ r ~ p) =

d~ rdt ~ p + ~ r

d~ pdt Di

ff erentiate L.

d~ rdt ~ p = 0 Vectors parallel.

~ r d~ pdt

= ~ r ~ F = 0 Central force.

So for a central force:

L unchanged by Central Force

~ L = 0

Therefore it takes an external torque to change the angular momentum of a system of particles and we can be assured that the external agent will feel an equal and oppositetorque.

2.4.1 Example: Phase Portrait for Gravity

Since angular momentum is conserved in a central force problem, we can make an earlyanalysis of motion in a gravitational eld. Consider a satellite orbiting the earth. Theorbit will be in a plane and lets choose our (spherical) coordinate system so that itis the = 0 plane. The motion is then in the radial coordinate r , and the azimuthalcoordinate . This is then like polar coordinates in 2D. The kinetic energy is:

T = m2

r 2 + ( r )2

and the potential energy is:

U = GMm

rAngular momentum is conserved.

~ L = ~ r

~ p = r (mr )z = mr 2 z

Writing out the total energy with a xed angular momentum we nd that the coor-dinate is removed and we have a 1D problem in r .

E = mr 2

2 +

L2

2mr 2 GMm

r


51/305

51

There is an attractive potential that goes like 1r and a repulsive (pseudo)potential thatgoes like 1r 2 . The potentials all go to zero at innity.

The top graph shows the potential including the pseudo-potential. There is a minimumcorresponding to circular orbits r = 0. Near the minimum, the phase diagram isapproximately elliptical. Further away, the phase diagram takes on the shapes shown.For E > 0, the satellite escapes to innity as there is no turning point in the diagram.The satellite can never go to r = 0 due to the angular momentum.

2.5 Homework Problems1. A superball of mass M and a marble of mass m are dropped from a height h

(on earth) with the marble just on top of the ball (but not in contact with theball). Assume that the superball is perfectly elastic so that no energy is lost ina collision. The superball hits the oor and rebounds colliding with the marble,with both objects moving directly upward. Neglecting the sizes of the spheres,how high do the superball and the marble go?

2. A particle of mass m is under the inuence of a force F =

kx + kx

3

2 where k

and are positive constants. Determine the potential energy U (x) and discussthe motion. What happens if E = k 2 / 4?

3. What is the work done by the force ~ F = x2 x +2 xyy in going from the point (0 , 0)to (1 , 1) along each of the following 3 paths? a) Go along x to the point (1 , 0),


52/305

52

then along y to (1, 1). b) On the path y = x2 . c) On the path parametrized by twith x = t3 and y = t2 .

4. Which of the following forces are conservative? a) ~ F = k(xx + 2yy + 3zz) b)~ F = k(yx + xy + 0 z) c) ~ F = k(yx + xy + 0 z) Find the potential for theconservative forces.

5. A mass moves in a circular orbit in an attractive central force with the potentialU = kr n . Show that for this case, the Virial theorem T = n U 2 is obeyed. Whatdoes this mean for the gravitational potential?

6. A particle of mass m moves in the one-dimensional potential U (x) = U 0a 4 (x2a2)2 .(a) Sketch U (x). Identify the location(s) of any local minima and/or maxima,

and be sure that your sketch shows the proper behavior as x . (b) Sketcha representative set of phase curves. Be sure to sketch any separatrices whichexist, and identify their energies. Also sketch all the phase curves for motionswith total energy E = U 02 and for E = 2U 0 . (c) What is the time derivative of

the phase space coordinate xv written in terms of that coordinate. (d) Derive

an expression for the angular frequency of the motion when the system exhibitssmall oscillations about a potential minimum.

2.6 Sample Test Problems

Since I dont have a large number of examples that I have put on tests for 110A, I listbelow the Homework Problems that I think could be test problems. The ones thatare excluded could be too time consuming, too dependant on detailed data, or could bea prove that problem which is not usually good for a test. There are many reasons Imight exclude a problem and such an exclsion should not be taken to mean that thattopic will not be covered. In general, the homework topics will be fair game, however,some homework problems are too long for a 50 minute test.

In this chapter, I think HW problems 1, 2, 4 could be test problems.


53/305

53

3 Oscillations

As we have seen from the Taylor expansion of the Potential Energy near aminimum , the potential U (x) = 12 kx

2 is a common problem, at least for small oscil-

lations . This is the Harmonic Oscillator Potential and it is the potential for theHookes Law Force , F = kx . = 2 Angular Frequency times Period . = 2 is the frequency.

The harmonic oscillator is extremely important in Mechanics as well as Quantum Me-chanics and Quantum Field Theory because the solution has one pure frequency of oscillation .

3.1 Simple Harmonic Oscillator

U (x) = 1

2kx 2 Potential Energy -

F = dU dx

= kx gives Hookes Law force.mx = kx ma = F .

0 = + r km Dene osc. (ang.) freq.x = 20 x eq. in terms of 0 .

x(t) = A sin(0 t

) sine or cosine soln.

x = 0A cos(0 t ) diff erentiate to get v.x = 20 A sin(0 t ) = 20 x again to get a.

We have thus veried that the sine solution satises the equation of motion.

Simple Harmonic Oscillator

x = 20 xx(t) = A sin( 0 t )


54/305

54

The solution has two constants that must be set by initial conditions , theAmplitude A and the phase . Solutions to second order equations will have twoconstants of integration. We could equally well have chosen a cosine solution but thisis included in the sine solution above with the arbitrary phase, however another wayto write the solution would be x(t) = A sin(0 t) + B cos(0 t). Using the trig identities,

Trig. Sum and Di ff erence Formulas

sin( u v) = sin u cos v cosu sin v

cos(u v) = cos u cos vsin u sin v

Applying the rst of these relations to our solution we get another way to write thegeneral solution.

2 Ways to Write the General Solution

C sin(0 t ) = C sin(0 t)cos C cos(0 t)sin = A sin(0 t) + B cos(0 t)

Here A and B are the constants that must be set to match initial conditions. Wecan see that the original sine solution can be written this way with A = C cos andB = C sin .If the initial conditions are that the oscillator is at x = 0 with velocity v0 at t = 0, thenv00 sin(0 t) will be the simplest way to write the solution. If the initial conditions arethat the oscillator is at rest with x = A at t = 0 then A cos(0 t) will be the simplest.

3.2 Energy of the Oscillator

The kinetic energy can be easily calculated from the solution by di ff erentiating to getthe velocity. The potential energy is even easier.

x = ddt

(A sin( 0 t )) = 0A cos(0 t ) compute v.T =

12

mx2 = 12

m20 A2 cos2(0 t ) compute T .

U = 12

mx 2 = 12

k sin2(0 t ) compute U .E = T + U =

12

A2(m20 cos2(0 t ) + k sin2(0 t )) =

12

kA2 sum is total E .


55/305

55

The total energy is conserved . At x = A all the energy is potential and at x = 0all the energy is kinetic. The energy is proportional to the Amplitude squared.

3.3 Phase Curves of the Simple Harmonic Oscillator

As shown earlier, a Phase Curve is a plot of x versus x that maps the movement of a system through phase space. In the case of the harmonic oscillator its fairly easy tocompute the curves analytically.

x(t) = A sin( 0 t ) position.x = 0A cos(0 t ) and velocity.

This is very much like the usual parametrization of a circle in trigonometry, exceptthere is an extra factor of 0 and the units are di ff erent on the two axes. So this isntreally the equation for a circle but for an ellipse with the axes parallel to thecoordinate axes .

We could do another simple calculation.

x = 20 x Equation of motion.dxdt

dtdx

= 20 x/ x Divide eq. by v.dxdx

= 20 x/ x cancel dt.xdx =

20 xdx algebra.

x2 = 20 x2 + C integrate eq.x2 + 20 x

2 = C The equation of an ellipse.


56/305

56

This shows that the phase curve explicitly as the equation for an ellipse .

3.4 Damped Oscillations

Sometimes we wish to make oscillators that keep oscillating for a long time withoutmuch energy loss. Sometimes we want to make oscillations go away as quickly aspossible. Of course most real oscillators have some energy loss due to frictionand we need to understand it to optimize our oscillators or to keep oscillations fromdoing damage.

We will consider linear damping .

F d =

bx

So we will have a new di ff erential equation to solve .

mx + bx + kx = 0

We will solve this by substituting in an exponential solution , which could be areal or a complex exponential, then understanding what the solutions mean.See Appendix: Complex Exponentials.

x + b

m x +

k

mx = 0 Eq. of motion.

b2m

def. of .

0 r km def. of 0 .x + 2 x + 20 x = 0 The eq. of motion.

x = Ae rt Trial soln, r complex.

(r 2 + 2 r + 20)Aert = 0 plug in.

r 2 + 2 r + 20 = 0 solve this (characteristic) eq.

r = 2 p 4 2 4202 = q 2 20 2 solutions.There are two roots to this equation which means there are two solutions . Anylinear combination of these two solutions will be a solution so we need to

Documents

Physics Mechanics 110A notes UCSD