Classical Mechanics Notes

copyright 2012, George Gollin

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copyright 2012, George Gollin 2

Physics 325

Classical Mechanics I

George Gollin

University of Illinois at Urbana-Champaign

Fall 2012

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copyright 2012, George Gollin

Classical Mechanics I

Physics 325

Welcome!

Here is an introduction and syllabus for the course.

Lectures: Tuesday, Thursday, 11:00 12:20 in Loomis 151.

Discussion sections: Monday, at various times, in Loomis 236.

Office hours: To be arranged.

Weekly problem sets are to be submitted in class immediately

before the beginning of the Thursday lecture. Ill return them, in

class the following Thursday. We wont be using the yellow

homework boxes that are on the second floor of Loomis.

Text: Classical Dynamics of Particles and Systems, 5th edition,

Stephen T. Thornton, Jerry B. Marion. The fourth edition is fine

too, and MUCH less expensive. The third edition has a chapter on

chaos and nonlinear dynamics, and you can always borrow a later

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Physics 325, fall 2012 Introduction University of Illinois


edition when its time to take a look at that material next semester

if youve found an even earlier edition. (The fourth and third

editions can be purchased through online merchants.)

My lecture notes are available at Notes-n-Quotes, 502 E. John St.

in Champaign. You should buy your own copy and bring it to

class.

Take note:

Well use calculus all the time. No calculators or computers are to be used as computational

aids, or as portals to access reference sources unless I

explicitly give the OK for a particular problem. You are not

to use them when working problem sets, problem session

exercises, or exams.

There is to be no laptop, cellphone, iPhone, iPad usage during class. I do not object to snacks and beverages.

Ill hand out a short credit/no credit written exercise in most classes. They are intended to give me some feedback about

what is (and isnt) making sense to you. I expect theyll take

about five minutes to complete.

I encourage you to work together on problem sets, but please make sure everyone in your group participates in generating

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the solutions and understands how they were obtained. You

are not to submit solutions that you have found on the web, in

books of worked problems, in archival copies of earlier

offerings of the course, or any other source of that nature.

That would be cheating: submitting work for a grade that is

not your own.

There will be two (evening) midterms, no time limit, and one final exam. Exams count for approximately 75% of your

course grade.

Late homework will be penalized 50%. Homework more than one week late will not be graded. I will penalize you for

failing to submit homeworks, approximately a half-letter

grade for every two missed assignments. Homework counts

for approximately 20% of your course grade.

Attendance at lectures is mandatory. You are to arrive on time and stay until the end of class. If you are ill, have the

emergency dean send me the relevant documentation. If you

need to miss class for a job interview, athletic competition,

family event, or other legitimate reason let me know ahead of

time, and in writing. I will penalize you for unexcused

absences, probably a half-letter grade for every two missed

classes.

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Come to the Monday problem sessions: your grade will include points earned from your participation in these.

I will use email, sent to your UIUC email address, as an important means of communication. Please be sure to check

this at least once per day. I dont think the physics

departments bulk mail system is currently able to send

messages to anything other than your illinois.edu address, so

I wont be able to get class-wide announcements to you

through other services such as gmail.

All of the material I distribute is copyrighted. That means you are not to redistribute it without my permission. From

time to time I find that my courses material is offered for sale

on the web. That makes me cross, and generally triggers a

phone call to the Office of University Legal Counsel.

About academic integrity and plagiarism: you are responsible for knowing what the Student Code requires of you. Much of

it is common senseyou are to conduct yourself in an honest

manner, not represent the work of others as your own, and

provide full attribution for sources you use in your written

work. Cheating and plagiarism are unacceptable; lifting even

a significant portion of a single sentence from an existing

work is an act of plagiarism. See

http://admin.illinois.edu/policy/code/.

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Course staff

Lecturer: George Gollin, Professor of Physics. Loomis 437d

Teaching assistants: ---, Graduate student. ---, Graduate student. Graders: ---, Graduate student.

---, Graduate student. ---, Graduate student.

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Lecture Notes Contents

1. Newtonian Mechanics, mostly in one dimension Newtons Laws for motion in an inertial reference frame...........1-1 A comment about gravitational vs. inertial mass........................1-3 Invariance of Newtons Laws under Galilean transformations ...1-5 Solving the equation of motion of a particle................................1-8

1. Force is zero ........................................................................1-9 2. Force is constant................................................................1-10 3. Force is a function of time only ........................................1-14 4. Force is a function of position only...................................1-17 5. Force is only a function of velocity...................................1-24 6. Force is a function of everything.......................................1-30

Nonrelativistic rocket.................................................................1-32 2. Dynamics in three dimensions Conservative forces .....................................................................2-1

A reminder about the gradient operator ..................................2-1 Line integral around a loop as a test for a conservative force .2-5 Form of the equations in three dimensions .............................2-8 Interesting special cases ........................................................2-10 Examples of forces derivable from potentials .......................2-15 Zero curl as a test for a conservative force............................2-20 Two examples in three dimensions .......................................2-28

3. Systems of particles Definition of the center of mass...................................................3-1 Motion of the center of mass .......................................................3-2 Total angular momentum and torque...........................................3-4 Reduced mass coordinates for two-particle systems ...................3-5

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4. Oscillations An overview ................................................................................4-2 Force vs. displacement and period vs. amplitude ........................4-4

Nature of the motion close to an equilibrium point.................4-4 Simple pendulum, two ways: energy method; force method ..4-8 Behavior of the period...........................................................4-12 Successive approximation approach to the nonlinear problem ....

......................................................................................... 4-13 Next order solution................................................................4-21 Size of the effects induced by nonlinearities.........................4-26

A reminder about initial conditions ...........................................4-28 Phase diagrams ..........................................................................4-29 Inhomogeneous elastic deformations: introduction to tensors...4-32

Strain (displacement) vs. stress for identical springs ............4-35 Strain vs. stress for an inhomogeneous set of springs ...........4-36 Transformation properties of the elastic stress tensor under

rotations ............................................................................4-40 Rank-n tensor transformation properties...............................4-47 The outer product of two vectors is a tensor .........................4-49

Using complex exponentials to solve the oscillator equation ....4-51 Damped oscillations...................................................................4-55

Case 1: underdamped oscillations .........................................4-57 Oscillator Q (Quality factor) .................................................4-59 Case 2: overdamped oscillator ..............................................4-63 Case 3: critically damped oscillator ......................................4-64

Electrical analogues ...................................................................4-69 Driven oscillations .....................................................................4-73

Particular solution to the inhomogeneous equation...............4-76 Characteristics of xp...............................................................4-85

Non-sinusoidal driving forces....................................................4-96 Fourier series ...........................................................................4-100

Orthogonality relationships .................................................4-102 Square wave ........................................................................4-108 Periodic impulses of short duration.....................................4-113

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Impulse response and an introduction to Greens functions ....4-119 Response of an oscillator to a single, short impulse............4-121 Response of an oscillator to a non-periodic driving force...4-126 Greens functions ................................................................4-128 Greens function for the Coulomb potential........................4-130 Satisfying the initial conditions...........................................4-132 Some Greens function examples........................................4-134

5. Motion in rotating frames of reference A qualitative explanation for the existence of Coriolis forces.....5-2 Pendulums and hurricanes .........................................................5-13 Quantitative description of motion in rotating frames...............5-16 Velocities and accelerations ......................................................5-27 Effective force ...........................................................................5-32 Angular acceleration term..........................................................5-34 Centrifugal term.........................................................................5-34 Coriolis term..............................................................................5-36 The earths oblateness................................................................5-38 Coriolis effect on a hockey puck ...............................................5-39 Coriolis effect on a free-falling object .......................................5-41 Foucault pendulum analysis ......................................................5-44 6. An introduction to fluid dynamics Utility of conservation laws in fluid dynamics ............................6-2 Partial derivatives and convective derivatives.............................6-4 Conservation of mass...................................................................6-8 Current density and the mass in a macroscopic volume ............6-15 Density changes in a comoving frame ...................................6-16 (Ir)rotational flow ......................................................................6-18 Equations of motion for an ideal fluid .......................................6-22 Shear stress in a Newtonian fluid...........................................6-27 Conservation laws......................................................................6-33 Energy conservation ..................................................................6-36

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Stress tensor...............................................................................6-43 Including viscosity; the Navier-Stokes equations......................6-47 Water flowing through a long, cylindrical pipe .........................6-48 Summary....................................................................................6-53 7. Lagrangians and the calculus of variations Introduction .................................................................................7-2 Definition of the Lagrangian .......................................................7-4 The connection between the Lagrangian and Newtons laws......7-5 Using the Lagrangian...................................................................7-8

Sliding mass, sliding wedge ....................................................7-8 Nasty pendulum ....................................................................7-14

Definition of the (classical) action.............................................7-20 Classical action for a free-falling mass......................................7-21 Hamiltons principle ..................................................................7-23 Demonstration of the equivalence of Hamiltons principle and the Euler-Lagrange equations .............................................7-24

Setting up a calculus of variations approach to the problem .................................................................................7-25 Minimizing the action through proper choice of trajectory ...............................................................................7-29 The Euler-Lagrange equations and other minimization

problems ...........................................................................7-35 A different minimization problem .............................................7-36

Brachistochrone.....................................................................7-37 A few comments ........................................................................7-49

Generalized coordinates ........................................................7-49 Feynmans use of the principle of least action ......................7-51

More Lagrangian examples .......................................................7-53 Double Atwoods machine, done two ways..........................7-53 Brief recap.............................................................................7-61 Spherical pendulum...............................................................7-63 Conical pendulum .................................................................7-70 Small perturbations to the conical pendulum........................7-71

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A comment on generalized momenta ........................................7-78 The Hamiltonian as a Legendre transform of the Lagrangian ...7-81 Lagrangian and Hamiltonian for a particle in a magnetic field .7-90 Poisson brackets ........................................................................7-93 Math review 1. series approximations .............................................................m-2

Taylors Theorem...................................................................m-2 Binomial approximation.........................................................m-2 sin(x), for x in radians and x close to zero .............................m-4 cos(x), for x in radians and x close to zero.............................m-5

2. some geometry........................................................................m-6 right triangles .........................................................................m-6 triangles in general .................................................................m-7 circles .....................................................................................m-8

3. exponentials ............................................................................m-9 4. differential equations ............................................................m-10 5. Vectors..................................................................................m-13

Scalar product.......................................................................m-13 Cross product .......................................................................m-14 Differentiation of vectors .....................................................m-15 Integration of vectors ...........................................................m-16 Line integrals........................................................................m-17

6. Vector transformation properties ..........................................m-18 Under rotations.....................................................................m-18 Matrix multiplication and rotation matrices.........................m-20 How unit vectors transform under rotations.........................m-26 Successive rotations in three dimensions .............................m-27 Kronecker delta symbol .......................................................m-30 Levi-Civita symbol...............................................................m-31 Tensors .................................................................................m-32

7. Non-Cartesian coordinate systems........................................m-35 cylindrical coordinates .........................................................m-36 spherical coordinates ............................................................m-37

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8. partial derivatives..................................................................m-38 using partial derivatives to approximate changes in functions

........................................................................................m-40 gradient operator ..................................................................m-41 divergence operator ..............................................................m-44 curl operator .........................................................................m-45

9. quadratic formula..................................................................m-46 Physics 325 course calendar Week Day Date Time Room Class activity Topic

1 Tue 8/28 11 am LLP 151 Lecture 1 Newtonian Mechanics, one dimension1 Thu 8/30 11 am LLP 151 Lecture 2 Newtonian Mechanics, one dimension2 Mon 9/3 Labor Day 2 Tue 9/4 11 am LLP 151 Lecture 3 Newtonian Mechanics, one dimension2 Thu 9/6 12 am in class HW 1 due 2 Thu 9/6 11 am LLP 151 Lecture 4 Dynamics in three dimensions 3 Mon 9/10 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 2 material 3 Tue 9/11 11 am LLP 151 Lecture 5 Dynamics in three dimensions 3 Thu 9/13 11 am in class HW 2 due 3 Thu 9/13 11 am LLP 151 Lecture 6 Dynamics in three dimensions 4 Mon 9/17 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 3 material 4 Tue 9/18 11 am LLP 151 Lecture 7 Systems of particles 4 Thu 9/20 11 am in class HW 3 due 4 Thu 9/20 11 am LLP 151 Lecture 8 Oscillations 5 Mon 9/24 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 4 material 5 Tue 9/25 11 am LLP 151 Lecture 9 Oscillations 5 Thu 9/27 11 am in class HW 4 due 5 Thu 9/27 11 am LLP 151 Lecture 10 Oscillations 6 Mon 10/1 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 5 material 6 Tue 10/2 11 am LLP 151 Lecture 11 Driven oscillations 6 Thu 10/4 11 am in class HW 5 due 6 Thu 10/4 11 am LLP 151 Lecture 12 Driven oscillations 7 Mon 10/8 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 6 material 7 Tue 10/9 11 am LLP 151 Lecture 13 Driven oscillations 7 Thu 10/11 11 am in class HW 6 due 7 Thu 10/11 11 am LLP 151 Lecture 14 Impulse response 7 Thu 10/11 7 pm TBA Exam 1 Material from weeks 1 - 6 8 Mon 10/15 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 7 material 8 Tue 10/16 11 am LLP 151 Lecture 15 Impulse response 8 Thu 10/18 11 am in class HW 7 due 8 Thu 10/18 11 am LLP 151 Lecture 16 Impulse response

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9 Mon 10/22 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 8 material 9 Tue 10/23 11 am LLP 151 Lecture 17 Motion in rotating frames 9 Thu 10/25 11 am in class HW 8 due 9 Thu 10/25 11 am LLP 151 Lecture 18 Motion in rotating frames 10 Mon 10/29 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 9 material 10 Tue 10/30 11 am LLP 151 Lecture 19 Fluid dynamics 10 Thu 11/1 11 am in class HW 9 due 10 Thu 11/1 11 am LLP 151 Lecture 20 Fluid dynamics 11 Mon 11/5 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 10 material 11 Tue 11/6 11 am LLP 151 Lecture 21 Fluid dynamics 11 Thu 11/8 11 am in class HW 10 due 11 Thu 11/8 11 am LLP 151 Lecture 22 Fluid dynamics 12 Mon 11/12 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 11 material 12 Tue 11/13 11 am LLP 151 Lecture 23 Lagrangians and calculus of variations12 Thu 11/15 11 am in class HW 11 due 12 Thu 11/15 11 am LLP 151 Lecture 24 Lagrangians and calculus of variations12 Thu 11/15 7 pm TBA Exam 2 Material from weeks 7 - 12 Thanksgiving

13 Mon 11/26 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 12 material 13 Tue 11/27 11 am LLP 151 Lecture 25 Lagrangians and calculus of variations13 Thu 11/29 11 am in class HW 12 due 13 Thu 11/29 11 am LLP 151 Lecture 26 Lagrangians and calculus of variations14 Mon 12/3 3, 4, 5, 6, 7, 8 pm LLP 236 Problem session week 13 material 14 Tue 12/4 11 am LLP 151 Lecture 27 Lagrangians and calculus of variations14 Thu 12/6 11 am in class HW 13 due 14 Thu 12/6 11 am LLP 151 Lecture 28 Hamiltonian dynamics 15 Mon 12/10 3, 4, 5, 6, 7, 8 pm LP 236 Problem session week 14 material 15 Tue 12/11 11 am in class HW 14 due 15 Tue 12/11 11 am LLP 151 Lecture 29 Hamiltonian dynamics Tue 12/18 8 am TBA Final exam Everything

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Physics 325, fall 2012 Newtonian mechanics University of Illinois

Newtonian Mechanics, mostly in one dimension

Newtonian Mechanics, mostly in one dimension ........................1-1 Newtons Laws for motion in an inertial reference frame........1-1 A comment about gravitational vs. inertial mass.....................1-3 Invariance of Newtons Laws under Galilean transformations 1-5 Solving the equation of motion of a particle.............................1-8

1. Force is zero.......................................................................1-9 2. Force is constant ..............................................................1-10 3. Force is a function of time only .......................................1-14 4. Force is a function of position only .................................1-17 5. Force is only a function of velocity .................................1-24 6. Force is a function of everything .....................................1-30

Nonrelativistic rocket..............................................................1-32

Newtons Laws for motion in an inertial reference frame

1st law: No force uniform motion (constant vector velocity) 2nd law: Applied force = induced momentum change per unit time

3rd law: Bodies exert equal-and-opposite forces on each other.

Lets assume vG c so we can define p m= GG .

1st law 0 when 0dp Fdt = =G G

2nd law dp dvF mdt dt= = =G

maGG G .

copyright 2012 George Gollin PHY_325_lec_notes_01.doc

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The 1st and 2nd laws are related: the 1st follows from the 2nd.

The 3rd law comes from the first two.

Heres how: consider an isolated system (with no external forces

being applied).

1pG

2pG

1 2

1pG

2pG

1 2

Total momentum is 1 2P p p= +G G G .

Since no external force acts on the system, 0dPdt

=G

.

Therefore, ( )1 2 0d p pdt+ =

G G so 1 2dp dp

dt dt=

G G.

12 on

dp Fdt

=G G

1 and 2 1 2ondp Fdt

=G G

2 1 1on onF F 2 = G G


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This sounds odd, in a way: the gravitational force exerted on you

by the earth is equal, but opposite to the force exerted by you on

the earth.

Note that 2

2

dv d xF ma m mdt dt

= = =G GG G only works when m is constant.

If m changes (for example, a rocket) then we have to use

F dp d=G G t .

This differential equation lets us figure out how xG changes with time when we know the nature of F

G.

A comment about gravitational vs. inertial mass.

(Saywhut?)

Inertial mass is what determines how brisk/sluggish the response is

to a given force. Big mass small acceleration.

For example: charge Q, mass m, in an electric field EG

:


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F Q=G GE so F QEam m

= =G GG .

A peculiar thing about gravity:

F m :g=G G the charge seen by a gravitational field is the same thing as the objects mass (which determines how briskly the

object accelerates in response to a given force).

As best as can be measured, the gravitational mass (the charge

seen by gravity) exactly equals the inertial mass.

Coincidence? Noits a hint about gravity and the interplay

between mass, space, and time. This led Einstein to guess that the

effects of uniform acceleration and a static gravitational field were

equivalent: you couldnt tell which was which without looking out

the window of your lab.


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accelerationacceleration

acceleration

acceleration

gravity gravity gravity gravityacceleration

acceleration

acceleration

acceleration

accelerationaccelerationaccelerationaccelerationaccelerationacceleration

accelerationaccelerationacceleration

accelerationaccelerationacceleration

gravity gravity gravity gravitygravity gravity gravity gravity

Left: accelerating elevator being traversed by a light pulse.

Right: stationary elevator in uniform gravitational field.

Light bends in gravitational fields. Black holes, general

relativity

Invariance of Newtons Laws under Galilean transformations


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O

x

y

O

x

y

O

x

y O

x

y

O

x

y O

x

y

The primed frame has velocity 0vG seen from unprimed frame. Say

origin of (as viewed from O) at t = 0. Also, 0 is at O G

, .

x

, x x y& & y z z &

An event (useful even in non-relativistic situations) at , x tG as seen by observers in O will occur at ( )0 0x x x v t = G G G G and t t . =

Two events, ( )1 1, x tG and ( )2 2, x tG , will have

2 1 x x x = G G G , ( ) ( )2 0 0 2 1 0 0 1x x x v t x x v t = G G G G G G G 2 1t t t = , 2 1 2 1t t t t t = = = t


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So that 0 ,x x v t = G G G or 0x x vtt = G G G .

A velocity v seen from O will relate to vG G seen from O like this:

0v v v = G G G

A change in velocity will correspond to 2 v v G G 1vG

( ) ( )2 1 2 0 1 0v v v v v v v = = = G G G G G G vG .

Therefore

v vtt =

G G

so that 2 2

2 2 md x md xdt dt

=G G

Newtons Laws are left undisturbed by a Galilean transformation:

F ma ma= =G G G .


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We see the response to a gravitational pull obeying F ma=G G whether were at rest with respect to the earths surface or in

uniform motion.

We can choose the particular inertial frame that suits our

convenience (e.g., center-of-mass) without loss of generality.

Solving the equation of motion of a particle

Physics input:

1. (a differential equation) F ma=G G2. identify all the forces acting on a particle and model their

dependences on , , , etc.x v tG G

Lets categorize some of the kinds of forces we might encounter.

1. 0F = (easy: 0F = means vG G G is constant) 2. (example: falling body, close to the earths

surface, neglecting air resistance.)

constantF =G


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3. FG

a function of time only (uniform through all space)

4. FG

a function of position only (constant in time)

5. FG

a function of vG only (e.g., air resistance) 6. F

G a function of everything

Lets look at each case in turn.

1. Force is zero

Zero force means is constant since there is no acceleration.

Therefore v t .

vG

( )0v( ) 0v= G G G

Find ( )x tG by integrating: dxvdt dt dx x Cdt

= = = + G GG G G

0

where C is a constant of integration. G

Because we have 0v v=G G

0 0 0vdt v dt v t C v t x= = + = + GG G G G G

since ( ) 00tx t x= =G G .


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If youre troubled by the integration of a vector in vdt G , keep in mind that this is just shorthand for three terms: x y zv y v zv v x= + +G so that

x y z x yvdt v xdt v ydt v zdt xv t yv t z v t= + + = + + G z

when are constant. , , v v vzx y

Dealing with the integration constant CG

is simple:

( ) 0x t v t= + GG G C so ( ) 00x C x= GG G .

2. Force is constant

An example of this is a free-falling body close to the earths

surface if we can neglect air resistance.

Solve for the motion by integrating twice:


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2

2 so d dx dxd x d dxF m m Fdt m dt m d

dt dt dt dt dt dt = = = =

G GG GG G

or

1dxFt m Cdt

= +G GG

[Equation 1]

since the force is constant. Do some algebra:

( ) ( )1 1 so that dx dxFt C m Ft C dt m dt mdxdt dt = = = G GG GG G G

or

2

1 22Ft C t mx C = +G G GG . [Equation 2]

Recall that the force is constant so that a F m= GG . Define 1 1C C G G

m and 2 2C C G G

m

G

to rewrite Equations 1 and 2 this

way:

Equation 1 ; Equation 2 1at C v+ =GG 2

1 212

at C t C x + + =G GG G Therefore


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( )1 00C v v = G G G x and ( )2 00C x = G G G so that ( ) 2 0 012x t at v t x= + +

G G G G

y

.

Heres the classic constant-force problem: projectile motion.

We have 0 0 0cos sinv v x v = +G and would like to solve for where the projectile lands.

xy

xy

( ) 2 0 012x t at v t x= + +G G G G

Break this up into three equations:

( ) ( )( ) ( )( )

0 0

2 20 0

0

cos1sin ( 9.8 m/sec )2

(motion is in the plane)

x t x v t

y t y v t gt g

z t z xy

= += + + =

The projectile lands when 0y y= at a time . 0t >


1-12

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For convenience: place the origin at the projectiles starting point

so that . 0 0 0 0x y z= = =

We have

( ) ( ) 20 1sin 2y t v t gt= + so solve for the positive time that yields y = 0.

Two ways come to mind: either use the quadratic formula or just

factor the polynomial. (Its easier to factor the polynomial in this

case.)

( ) 20 1sin 02y v t gt= + = 01sin 02

t v gt + =

Solutions: both and 0t = 0 0sin 2 0 (so 2 sin )v gt t v g + = = will satisfy the equation. Its the t > 0 solution we want: recall that

. 9.8g

After time 02 sint v g= , the projectile has traveled this distance in x:


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20 0

02 sin 2 cos sin

cosv vx v g g

= = .

I want you to be capable of using various trig identities to simplify

expressions. An example: (12cos sin sin 2 ) = (see the Math review section of the notes).

Proving this: use cos , sin2 2

i i ie e e ei

i

+ = =

to write

2 2

2 2

1 1cos sin2 2 4

1 1 sin 22 2 2

i i i i i

i i

e e e e e ei i

e ei

i

+ + = = = =

So the projectile range is 2

0 sin 2vg

. Note the maximum at 45.

3. Force is a function of time only

If possible, integrate twice: d dxF mdt dt=

GG

so we have


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F d dxdt dt vm dt dt

= = G

dxdt

=G GG

and then

( )F dxdt dt dt dx x tm dt

= = = G G G G .

Since is a function of time only, the integral on the left can be

attacked, even if we need to do it numerically, via computer.

FG

An example: exponentially waning force:

( ) 0 tF t ma e x=G

We have: m22

d x mdt

= 0 ta e and also 2 22 2

d y d zm mdt dt

0= = .

Useful notation: 22 ,

dx d xx xdt dt so we can write

( ) 0 d tx a edt = .

Multiply by dt and integrate: ( ) 0 td x a e dt= or


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0 0tx c a e = where c0 is a constant of integration. [Equation 1]

We have 0 0dx tc a edt

= which can be integrated again:

( )0 0 tdx c a e dt = so that

( ) 21 0 0 tx t c c t a e = + + [Equation 2]

DONT jump to conclusions about the integration constants:

( ) ( )1 00 , 0c x c v .

From Equation 1: ( ) 00 0x c a= so 0 0c v a0= +

From Equation 2: ( ) 210 0x c a= + so 21 0 0 c x a=

As a result,

( ) ( ) /0 0 0 tx t v a a e = + and

( ) ( ) ( )2 20 0 0 0 0 t /x t x a v a t a e = + + + .


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4. Force is a function of position only

** For the time being, lets restrict ourselves to problems in one

dimension **

We have ( )22d xm mx Fdt = = x .

Theres a trick we can use to integrate this out:

( )mx F x= ( )mxx F x x= .

Integrate over time: ( )mxxdt F x xdt= . dxxdt dt dxdt

= = so ( ) ( )mxx dt F x x dt F x dx= = .

Also: note that 212

d mx mxxdt =

As a result, 2 21 12 2

dmxx dt mx dt mx Cdt = = +

Finally, we have ( ) 212

F x dx mx C= + . copyright 2012 George Gollin PHY_325_lec_notes_01.doc

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This is the work-energy theorem.

Define the kinetic energy 212T mv so that the change in kinetic energy is the same as the work done on the

particle.

( )0T T F x dx =

The work-energy theorem is is only guaranteed to work when F is

a function of position. Velocityor time-dependentforces are

more complicated. For velocity-dependent forces, work is

expended in other ways (heating something up) in addition to

changing the kinetic energy of the object.

I have no idea why the letter T is used for kinetic energy!

Lets define a special function V(x) (sometimes well call it U(x)

instead) so that

( ) ( )dV x F xdx

= .

Then

( ) ( ) ( ) ( ) ( ) ( )B BA A

dV xF x dx dx V B V A T B T A

dx= = = +


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34


We have a conservation rule in the making: T V = going from point A to point B so

0T V + = .

Define the total energy E = T+V as the sum of the kinetic and

potential energies. In this case,

0 E A B = so E is constant.

This gives us an equation in which the time doesnt appear

explicitly:

T V E+ = or ( )212

mv V x E+ = .

Therefore

( ) ( )2 E V xv xm =

and we can solve for velocity as a function of position.

Finishing the job:


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35


( ) dxv xdt

= so ( )dxdt

v x= and

( )( ) 122 E V xdxdt dx

v x m

= =

so that ( ) 122 E V x

t dm

= + x C

x

.

Crank out the integral, then invert to write x as a function of time.

Heres an example: (a spring). F k=

Step 1: define V(x) so that ( ) ( )dV xF xdx

= .

Then: ( ) ( )F x dx kxdx dV x= = . As a result, ( ) ( )21 0

2V x kx V= + (V(0) is a constant of integration).

If we decide to place the zero of potential energy at x = 0 then we

have

( ) 212

V x kx= .


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36


Step 2: determine E from the initial conditions on v and x since E is

constant:

2 2 20 0

1 1 1 12 2 2 2

E mv kx mv k= + = + 2x

Step 3: solve for v: ( )2

212

22E kx

E kxv xm m

= = .

For convenience, define , 2k m A E k . We have: ( ) 2 v x A x= 2 after a bit of algebra.

As a result, 2dx 2A xdt

= so 2 2

dxdtA x= .

I expect you to be able to deal with integrals of this sort. Use the

trig substitution sinx A = so that ( )1sin /x A = (sin-1 is the arcsine function):

2 2

dxdtA x= 2

cos coscos1 sin

A d ddt CA

= = = +


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37

37


since 21 sin cos = and ( )sin cosd A A d = .

As a result, ( )1 const. sin /t x + = A .

Step 4: solve for x as a function of time.

( ) ( )1sin const. sin sin /t x + = = A x A so that ( ) ( )sinx A tt = +

where Ive renamed the constant to be .

Both A and are constants of integration and must be determined from initial conditions on two of: x(0), v(0), E.

Simple harmonic motion!!

Why is SHM so important? Heres why. Say V(x) (potential

energy) looks like so:


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38


x

V(x)

x

V(x)

Local (stable) equilibrium points have 2

20 and 0dV d Vdx dx

= >

since the force felt by the particle is dVF dx= and theres no force

at an equilibrium point.

Taylor expand F(x) around an equilibrium point at x = x0:

( ) ( ) ( ) ( )0 00 00 1! n nndF d FF x F x x x xx x x xndx dxx= + + + += =

Keep in mind that ( )0 0F x = since were at an equilibrium point. Remember too that

0x x

dFdx =

is a constant since weve plugged in

0x x= .

If ( we have )0 1x x ( ) ( )0F x k x x where 0x x

dFkdx =

.


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39


The force is a restoring force, pointing opposite to the

displacement from x0thus the negative sign in front of k.

Most systems act like simple harmonic oscillators when displaced

a small distance from equilibrium.

5. Force is only a function of velocity

( )dvmx m F vdt

= = so that ( )m dv dt

F v= .

Perhaps we can do the integral, then invert ( )t v to find . ( )v t

We wont get anywhere trying to define a potential energy function

since that had assumed that ( ) ( )dV xF xdx

= and F here depends on velocity.

An example: linear damping force F m v= ( is a positive constant so the force opposes the velocity, slowing the object).

m dv mdt = v so dv dtv = .


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40


Integrate: or lndv dt v t Cv

= = + .

Use both sides as exponents: ln v t C Ce e e e t + = = or C tv e e =

Initial condition: so ( ) 00v t v= = 0Ce v= : 0 tv v e = .

Integrate again :

0tdxv v

dte = = so or 0 tdx v e dt= 0 tvx e C = + .

Initial condition: ( ) 00x x= so 0 00 0 or v vx C C x = + = +

which yields ( ) ( )0 01 tvx t e = + x .

Note that ( ) 0 0limt vx t = + x which is finite.

The object doesnt get all that far away from us; the bigger the

damping, the closer it stays.


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41


What about cubic damping? The damping force is stronger for

large v than in the case of linear damping, but falls off more

quickly for small v. Lets see what happens

Lets take a look at 312F mc= v (c is constant, not intended to be the speed of light.)

m 12dv mdt = 3cv so 32

dv cdtv

= .

Integrate to find 221 1 or ct C vv c

= + =t C+ .

At t = 0, 201vC

= so 201Cv

= . Therefore,

2

2 0 02 2 20 0 0

1 so that 1 1 1

v vv vct v v ct v ct

= = =+ + + .

Integrate again, since v dx dt= :


1-26

42

42


( ) 12 20 0 0 02 20 0

or 11 1

v vdx dx x dt v v ct dtdt v ct v ct

= = = =+ + +

This is an easy integral. Let 20 1u v ct + . Note that so we can write

20du v cdt

( ) 1 12 2 20 0 0 20

1 12 2

0 0

1

1 2

dux v v ct dt v uv c

u du u Cv c v c

+

= + =

= = +

or 200

2 1x v ct Cv c

= + + .

Since ( ) 00x x= we can solve for C: 0

0

2x Cv c

= + so 00

2C xv c

= . As a result, we have

20 0

0 0

2 21x v ct xv c v c

= + + .

Note that as x t .


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43

43


Another example: falling body with linear air resistance. Lets do

this in one dimension: the object falls straight down.

dvm m vdt

= + mg (g is negative if up is positive)

Integrate: dv dtv g = + v and define u g so du dv=

in order to write

1 du dtu = or 1 ln const.u t = + or ( )ln Cg v t = +

so that C tg v e e = .

At t = 0 so we can write C0g v e = ( )0 tg v g v e = or

( )0 0t

tg g v e g gv v

e

= = + .

Note that gv regardless of the initial velocity as . t

Integrate once more to get ( )x t (+x is up):


1-28

44

44


0tdx g gv v e

dt

= = + so 0

tg gdx v e dt = +

Note that nearly everything in the integrand is constant. We find

01 tg gtx v e

= + + C .

Use the fact that ( ) 00x x= to finish this off: 0 0 0 0

1 1 or g gx v C C x v = + = + so that

0 01 1tg gtx v e x v 0

g

= + + + .

After some algebra, we find

( )0 01 1 tg gx x v e t = + + .

It is instructive to look at x for large and small times.

Large t: 0 01 g gx x v t

+ + (note the constant speed)


1-29

45

45


Small t: expand te around 0t = :

( ) ( )2 31 ...2! 3!

t t te t = + + so ( ) ( )

2 3

1 ...2! 3!

t t te t = +

( ) ( ) ( )

( ) ( )

2 3

0 00

2 30 0 0 0

1 ...2! 3!

1 1 ...2 6

t

t tg gx t x v t

x v t g v t g v t

t

+ + + = + + +

Note the linear increase in x for small times.

6. Force is a function of everything

In general, this is a mess that needs to be addressed with a

computer. Some special cases do lend themselves to analytic

techniques, however.

For example, if is a product of functions: FG


1-30

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46


a) ( ) ( ) (, F v t f v g t= )( ) ( ) dvf v g t m

dt= so

( ) ( )mdvg t dtf v

= and perhaps you can do the integrals.

b) ( ) ( ) dvF f v h x mdt

= =

Use a trick: dv dv dx dvvdt dx dt dx

= = .

Then ( ) ( ) dv dvm mv f v hdt dx

= = x so that ( ) ( )mvdv h x dxf v

= .

Try to solve for v as a function of x then use v(x) = dx/dt to write

( )dx dt

v x= and crank it out as best as you can.

c) ( ) ( )F h x g t= get thee to a workstation!!

The general idea in all of this: Newtons laws and their

consequences (for example, conservation of momentum) will


1-31

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47


allow us to write differential equations we can solve, even though

we might be forced to resort to numerical (computer) techniques.

Nonrelativistic rocket

Lets work another example, also in one dimension, that lends

itself to analytic (non-computer) solution. Its a non-relativistic

rocket problem, neglecting gravity.

The rocket burns fuel, shooting it out the back as exhaust. The

exhaust speed with respect to the rocket is u; the rockets speed

with respect to an inertial frame is ( )v t . The tricky part is that the rockets mass decreases as it burns fuel.

How we deal with this: consider the momentum of the rocket and

fuel that it carries at time t, then equate this to the momentum of

the (slightly lighter, slightly faster) rocket at t dt+ added to the momentum of the exhaust gases. A little algebra will let us

construct a differential equation from the pieces of our momentum

conservation equation.

Momentum at time t carried by the rocket and its onboard fuel:


1-32

48

48


( ) ( ) ( )rocketp t m t v t=G G where m(t) is the combined mass of the rocket and the fuel it

contains at time t. Lets do our calculations in the inertial frame in

which we see the rocket moving at velocity v at time t.

At time t + dt the rocket is slightly lighter and moving slightly

faster. It carries momentum

( ) ( ) ( ) ( ) ( )rocketp t dt m t dt v t dt m t dm v t dv+ = + + = + + G G G G (Note that dm is negative.) The change in the rockets momentum

is

( ) ( ) ( ) ( )( ) ( )( ) ( )

rocketp m t dm v t dv m t v

m t dv v t dm dmdv

m t dv v t dm

= + + = + + +

G G GG G GG G

tG

since the product of the two infinitesimals dmdvG can be neglected.

The change in the momentum of exhaust gas flying through space

is just the product of its mass and its velocity:

( ) ( )exhaustp v t u d = + G G G mG

since the exhaust gas has mass dm and travels with velocity

. (Recall that u is with respect to the rocket.) ( )v t u+G G


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49

49


Because the rocket + fuel comprises a closed system, the net

momentum in the inertial frame we are using must remain

constant: so 0rocket exhaustp p + =G G

( ) ( ) ( ) (( ) 0

rocket exhaustp p m t dv v t dm v t u dm

m t dv udm

+ = + + + = =

G G G G G GG G

)

or

( ) ( ) so that udmm t dv udm dvm t

= =GG G G .

Solve for the velocity by integrating from the start of engine burn

(rocket mass is mi) to the end of engine burn (rocket mass is mf):

( ) ( ) so lnf f

f

ii i

m m v vm

f imm m v v

udm dv u m v v vm t

= =

= == =

G G G G G G

Recall that ( )ln ln lna b a b = so that ( ) ( )ln lnf i iv u m m u m m = = G G G f .

( is in the same direction as u G f iv vG G .)

You do better by increasing the exhaust velocity for a given

quantity of fuel than you do by increasing the fuel carrying

capacity of your rocket.


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50


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Physics 325, fall 2012 Dynamics in three dimensions University of Illinois

Dynamics in three dimensions

Dynamics in three dimensions.....................................................2-1 Conservative forces ..................................................................2-1

A reminder about the gradient operator .................................2-1 Line integral around a closed loop as a test for a conservative force .......................................................................................2-5 Form of the equations in three dimensions ............................2-8 Interesting special cases.......................................................2-10 Examples of forces derivable from potentials .....................2-15 Zero curl as a test for a conservative force ..........................2-20 Two examples in three dimensions......................................2-28

Conservative forces

A reminder about the gradient operator

Recall the definition of the gradient operator in Cartesian

coordinates in three dimensions (see the Math Review, if

necessary):

dx y zx dy z

+ + G

.

With this definition, we can calculate the change in a scalar

function when we shift from ( , ,u x y z ) to x x + GG G (we are

copyright 2012 George Gollin PHY_325_lec_notes_02.doc 2-1

53

53


assuming that G is very small):

( ) ( )

( )

x y

u u x u x

u u ux y z

u ux y z

z

ux y z

u

= + + +

= + + =

GG G

G

G G

This is what led us to this particular definition for the gradient: we

can use it to calculate for any scalar function u. u

A note of caution: in non-Cartesian coordinates, G isnt so simple.

For example, you might expect G to be r zr z

+ + in

cylindrical coordinates. That would be WRONG!

Heres why. Keep in mind that we want to use G this way: ( ) ( ) ( )u u r u r u = + = G G GG G .

Imagine that G points in the direction, as drawn below: copyright 2012 George Gollin PHY_325_lec_notes_02.doc 2-2

54

54


x

z

y

r

rx = r cosy = r sin = r^

x

z

y

r

rrrx = r cosy = r sin = r^

x = r cosy = r sin = r^

Notice that the length of G depends on r since, for a particular value of the angle , the length of G increases as r increases.

Allowing for G to point in an arbitrary direction, we have ( ) r zr r z = + +G

rather than the incorrect expression r zr z + + .

Since ( ) ( ) ( )u r u r u + = G G GG GG

is, in effect, forcing us into a

particular definition of , we are required to have

u r z

r r z

u = + +

G


55

55


so that dotting ( )uG with G to calculate ( )u G G will actually give us u rather than something else.

Heres a summary of in different coordinate systems. uG

Cartesian coordinates x, y, z : u x y z ux y z

= + + G

Cylindrical coordinates r, , z : u r zr r z

u

= + + G

z

a

x

z

y

r

x = r cosy = r sin

z

a

x

z

y

r

z

aa

x

z

y

r

x = r cosy = r sin


56

56


Spherical coordinates r, , :

sin

du rr r r

u = + +

G

sin cosx r = sin siny r =

cosz r =

a

x

z

yr

sin cosx r = sin siny r =

cosz r =

aa

x

z

yr

Line integral around a closed loop as a test for a conservative

force

Imagine we map out ( )U rG along a path from A to B like so:


57

57


A

B

dsGA

B

dsG

Along each short segment of path dsG along the route, U will change by ( )dU U ds= G G . If we add up (actually integrate: keep in mind that an integral is really just a big sum) the changes

in U over each of the small path segment, well get the net change

in U between A & B:

( ) ( ) ( ) ( )B BB iA A A

U B U A dU dU U ds U B U A = = = G G .

If the net force acting on our 1-particle system can be set equal to

the gradient of a scalar function of position so that F = UG G then

we can write ( ) ( ) B BA A

U B U A U ds F ds = = G GG GB

A

F ds

G

so that the work

done by the force G equals the reduction in potential

energy, and the gain in kinetic energy.


58

58



U

Note that the path taken doesnt enter into the result: as long as

we automatically find F = G GB

A

F ds U = G G .

All we need to know to calculate U is the location of the starting and ending points A and B of the path to calculate.

Forces that have this property are called conservative forces: if A

and B are the same point (true for any round trip), no net energy is

expended during the trip.

Not all conceivable forces are conservative. Friction is an obvious

example. But also: force fields like the ones sketched below are

non-conservative since you can find loops about which there is a

net energy gain or loss when you go around once.

andand

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59


Form of the equations in three dimensions

F ma=G G is shorthand for 3 equations:

( ), , , , , ,xmx F x y z x y z t= ( ), , , , , ,ymy F x y z x y z t= ( ), , , , , ,zmz F x y z x y z t=

If we can write

( ) ( ) ( ) ( ) , , , , , , , ,x y zF x x t x F x x t yF y y t zF z z t= + +G G G

Then the equations of motion for the , , x y z directions are

decoupled and becomes three (often simpler) equations: F ma=G G

( )( )( )

, ,

, ,

, ,

x

y

z

mx F x x t

my F y y t

mz F z z t

===

OFTEN we deal with forces that do not separate in


60

60


Cartesian coordinates. A familiar example: 1/r2 central force

(gravity, for example).

In spherical coordinates, 1 22 Gm mF

rr=G (easy: it separates).

In Cartesian coordinates, 2 2 2r x y z2= + +

and

[ ] sin cos sin cosr x y z = + +

x

z

y

r^

x

z

y

r^r^

so everything gets messy. Plugging in yields

[ ]1 22 2 2 cos sin sin sin cosGm mF x yx y z z = ++ +

G +

and since


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61


2 2 2 2

2 2 2 2 2 2sin , cos , sinx y x y

2x y z x y x y += =+ + + +=

we can conclude that

( ) ( ) ( )1 2 1 2 1 23 2 3 2 3 22 2 2 2 2 2 2 2 2 x y zGm m x Gm m y Gm m zF F F

x y z x y z x y z

= = =+ + + + + +

Our equations relating force to the particles coordinates do not

separate. Looks nasty!

Interesting special cases

In general, there isnt an analytic solution to the equations since

they can be quite complicated: the component of force in any

direction might be a function of many variables ( , , , , , ,x y z x y z t ).

But there are a number of interesting special cases that yield

simpler equations of motion which can (sometimes) be solved.

Here are some examples:


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62


1. Say sG is a constant vector (for example, x ). If FG is always perpendicular to sG , then the component of momentum parallel to sG will be constant.

Heres the proof:

( )d p sdpF s sdt dt

= = G GGG G G since sG is constant and can be brought

inside the derivative.

Because is perpendicular to FG

sG well always have so

that it will also be the case that

0F s =G G( ) 0d p s F sdt

= =G G G G : the value of

never changes since its time derivative is zero. p sG G

The length of the component of pG along a direction parallel to sG is s p sps s

=G GG G G

G . In order for p sG G to remain unchanged, the

component of parallel to pG sG must remain constant.

2. Another special case: define the torque exerted by a force to be

r F = GG G where r is measured with respect to a fixed point in an Gcopyright 2012 George Gollin PHY_325_lec_notes_02.doc 2-11

63

63


inertial reference frame. Define angular momentum to be

. Then, if L r p= G G G 0, L = GGLG

is constant and the object moves in a

plane perpendicular to . (Note that when FG

is a central force so

that , well automatically have r FFG G& r 0 =GG .)

Heres the proof:


( ) ( )d d d dpL r p rdt dt = = p r dt + dtGG G G G G G

0=dr p v mvdt

= G G G G .

Therefore,

dL dpr r Fdt dt

= = G G GG G .

So if 0 =G , then is constant. Also, since LG ( )p r p G G G , must be in a plane perpendicular to

pG

LG

.

You might not be comfortable with this part of the first step:

( )d d dpr p r rdt dt dt

= + pGG G G GG . If so, consider writing out the

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64


expression in terms of components and grinding it out:

( ) ( ) ( ) y x z y x zr p xp yp z yp zp x zp xp y = + + G G so

( ) ( ) ( ) ( )

y x z y x z

y xy x

zz y

x zx z

d dr p xp yp z yp zp x zp xp ydt dt

dp dpdx dyp x p y zdt dt dt dt

dpdy dz dpyp y p z xdt dt dt dt

dp dpdz dxp z p x ydt dt dt dt

= + + = + + +

+ + + +

G G

+

so that


65

65


( )

[ ]

y xx y y x

zy z z y

x zz x x z

x y y x y z z y

z x x z

y x

dp dpd r p v p x v p y zdt dt dt

dp dpyv p y v p z xdt dt

dp dpv p z v p x ydt dt

v p v p z v p v p x

v p v p y

dp dpx ydt dt

= + + + + + + +

= + + + +

G G

+

z

x z

dp dpyz y z xdt dt

dp dpz x ydt dt

+ +

or

( )d dp drr p v p r p rdt dt dt dt

= + = + dpG G GG G G G G G G .

Whew.

3. Last special case well discuss, no proof to be offered

If is only a function of position, and if FG B

A

F ds G G is independent of the path taken between A and B, then one can always define a


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66


potential such that ( )U rG ( ) ( )U r F r =G GG G

B B

A A

dT

. (We discussed this

earlier, but for one dimensional systems.) The great utility of U is

that the sum of potential and kinetic energies is constant in such a

system.

Recall the work-energy theorem: F ds= G G where we have defined kinetic energy in the usual way as 2 2T m= v

G

U

. Since

we can write ( ) )U r r (F=G GG

( )

B B

A A

T

U B

=

T =

( )T d F ds

U A

= = =

B B

A A

U ds dU

U

= = G GG G

Since we can define the total energy E T U + to find that E is conserved in systems of this sort: regardless of the choice

of A, B, and the path connecting them it is always true that

E T U+ U = + 0U = = . The total energy E is constant everywhere along the path linking A & B.

Examples of forces derivable from potentials


67

67


Here are some examples of forces derivable from potentials.

1. Uniform gravitational field:


zF mg= G .

Note that so ( ) mgz mgz =G const.U mgz= + is a solution.

We could have written the gradient in Cartesian coordinates and

integrated, if we wanted to brute-force it:

F mgz U x y z Ux y z

= = = + + G G

so

0U Ux y

= = and U can only be a function of z.

As a result, U dUz d

= z and we can integrate:

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68


dUF mgz zdz

= = G or dUmgdz

= so that mgdz dU= .

As before, we find that const.U mgz= +

Since the sum of kinetic and potential energies is constant, we have

21 constant2

mv mgz E+ = = while the object falls.

2. Inverse square law: 2krF

r=G

x

z

y

r

x

z

y

rr

Note that it is most natural to use spherical coordinates, since

always points inwards, towards the origin. Lets figure out what FG

( , ,U r ) is. (Clearly U is independent of , .)


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69


( )( )32 3 2 2 2 2

x x y y z zkr kr kr r x y z

+ + = = + +

G

What function u has 3krr

G as its gradient? Lets take a look: use the form of in spherical coordinates (see the Math Review).

( ) ( ) , , , ,sin

U r U r F rr r d r

= + + =

G G .

Since always points towards the origin and doesnt have a FG

or component, 0U U = = . As a result, U is only a function

of r. Therefore,

( )2

U r krU r

r r = =

G

Since U only depends on r, U dUr dr

= and we can integrate:

( )2

dU r krr

dr r = so ( ) 2dU r kdr r= or 2

kdU drr

= .


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70


We find

( ) constantkU rr

= + .

3. A last example: the Yukawa force, important in

understanding nuclear potentials. The force is a central force: it

depends only on r, not on or .

02

0

1r rk rF er r

r = +

G.

Note that the force becomes the same as the Coulomb force when

. For the strong nuclear force, (a fermi!) 0r r 130 10 cmr

The force is radial so

02

0

1r rU k rF U r er r r

r = = = +

G G.

Integrate this, just like you would for a total derivative:


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71


0 02 2

0 0

1 1r r r rU dU k r k re U er dr r r r r

dr = = + = +

We can take all sorts of brute-force approaches to this integral, but

take note of the following useful fact:

0 0 0 02 2

0 0

1r r r r r r r rd k k k k re e e edr r r rr r r

= + = +

We can use this to rewrite the integrand so that

0 02

0

1r r r r r rk r d k kU e dr e dr er r dr r r

= + = = 0 .

For nuclei, k ~ 1 GeV/fermi ~ 1.6 105 Joules/meter.

Zero curl as a test for a conservative force

In the examples above we derived position-dependent potentials

from conservative forces. The fact that we could find potentials

that depend only on position guaranteed the conservative

nature of the forces.

( )U xG


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72


The path independence of the potential energy is equivalent to the

fact that the net change in potential around any closed loop is zero:

path independence means that both staying put and also going on a

round trip will result in since the starting/ending points are

the same.

0U =


U

Theres a way to investigate whether or not a force is conservative

without having to integrate it to solve F = G G for the potential U.

Lets work with a microscopic version of the no-work-around-a-

closed-loop idea to develop another test for a conservative force.

Imagine that a force field acts on a particle as we move it

counterclockwise around a tiny rectangular loop, as shown in the

diagram.

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z

y

x

Rectangular loop is centered at (x, y)

y

x1

2

3

4

x-x/2 x+x/2

y-y/2

y+y/2

z

y

x

Rectangular loop is centered at (x, y)

yy

xx11

22

33

44

x-x/2 x+x/2

y-y/2

y+y/2

The midpoints of the four sides of the loop are at positions

( , 2, 0)x y y , ( 2, , 0)x x y+ , ( , 2, 0)x y y + , and ( 2, , 0)x x y .

The work done by the external force as we push the object around

the loop is . loop

F ds G Gv

If the loop is small enough we can approximate the integral: the

force that acts as we move the mass along each side can be well

represented as constant, with the value it assumes at the midpoint

of each side.

The force acting on our particle as it moves along side 1 is


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approximately 1 ( , 2)F F x y y= G G

, the value halfway between the

ends of side 1.

The forces acting as the particle moves along the other sides are

approximately 2 ( 2, )F F x x y= + G G

, 3 ( , 2)F F x y y= + G G

, and

4 ( 2,F F x x y= G G

) .

The line integral around the loop can be approximated as a

sum over the four sides:

F ds G Gv4

1i i

iloop

F ds F s=

G GG Gv with 1 s x x = G , 2 s y y = G , 3 s x x = G , and 4 s y y = G .

We have:


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4

1 ( , 2) ( 2, )

( , 2) ( 2, )

( , 2) ( 2, )( , 2) ( 2, )

( , 2) ( , 2)

( 2, ) (

i ii

x y

x y

x x

y y

F s F x y y x x F x x y y y

F x y y x x F x x y y y

F x y y x F x x y yF x y y x F x x y y

F x y y F x y y x yy

F x x y F

= = + +

+

= + + +

+ = + +

G G GGG G

2, )x x y

x yx

where I have grouped the terms and written the force in terms of its

x, y components: x yF F x F= +G

y ( and x yF x F F y F = =G G

).

Recall the definition of a partial derivative:

( ) ( ) ( )( ) ( )

, , ,

0, 2 ,

0

x x x

x x

F x y F x y y F x ylimy yy

F x y y F x y ylim yy

+ +

= 2

We can use this to rewrite


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4

1

yxi i

i

FFF ds F s x y x ydy dx=

= + G GG Gv .

If is conservative, FG

0F dx = G Gv and well have 0y x

F F x ydx dy

= .

Define the curl of a vector function FG

to be

y x z y x zF F F z F F x F Fx y y z z x = + +

G Gy .

With this we can write

( ) 0F ds F z x y = G G GGv .

Our conclusion is that conservative forces have zero curl.

Note that theres a compact way to write the definition of the curl

using notation described in the Math review. Recall the definition

of the Levi-Civita symbol ijk :


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( )( )

1 if is an even permutation of 123 123, 312, or 231

if is an odd permutation of 123 213, 132, or 321

if any two (or more) of , , are equal

10

ijk

ijkijki j k

+

=

Its not hard to see that

33 3

1 1 1 , ,

also written as ji k

ijk i j ijk i ji j k i j k

a b a b k a b k == == = =

= GG

and that

, ,

jijk

i j k i

FF k

x =

G G

where 1x x , 2x y , and 3x z .

Note that we can express F ds G Gv as a sum of line integrals around subloops:


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z

y

x

y

x 1 2

3 4

Along each small interior side we always have the edges of two

small loops touching, with opposite directions of travel. Since FG

is

the same, but sG in the opposite direction along interior loops adjacent edges, the contributions to a grand sum will cancel. As a

result,

small loopsbig loop loop

i

i

F ds F ds = G GG Gv v .

Lets define an area patch dAG

with an associated direction using

the right-hand rule:


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z

y

x

dy

dx

dA dxdy z=G

z

y

x

dy

dx

z

y

x

dydy

dxdx

dA dxdy z=G

By doing this we can rewrite the sum over small loops as an

integral:

( ) ( )tiny loop

F ds F zdxdy F dA = GG G G G GGv so

( )big loop

Ftiny loops i loop area

F ds F ds dA = = GG G GG Gv v G

This is Stokes Theorem, named for the Irish mathematical

physicist George Gabriel Stokes (1819-1903); it is true for any

vector field.

Two examples in three dimensions

1. Projectile in a uniform gravitational field with linear air

resistance.


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vLinear air resistance: F mgz m= G G .

Fortunately, this equation separates:

x xF m v m x = = y yF m v m y = =

( )z zF m v mg m g z = = +

As a result,

( ), , yx zx y zdvdv dvv v gdt dt dt v = = = + .

Easy to solve:

lnx xx

dv dt v t cv

= = + so cxv e e

t= .

At t = 0, 0x xv v= so 0c xe v= . The y solution works the same way: 0y yv v

te = .

Now for z: let zu v g + . Note that zdu dt dv dt= .

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We have ( )z zdv dug vdt dt u = + = so c tu e e = .

At t = 0 0zu v g = + so ( )0 tzu v g e = + .

As a result,

( )0 tz zv u g v g e g = = + .

To solve for ( )x tG , integrate and use ( ) 00x x=G G . The final result is

( ) 0 0 1tgt gz ex t x z v

= + +

G G G

Is this sensible? Lets check.

In the limit 0 ,x yv v

(no air resistance), this should look like a freely

falling body ( ) constant, etc. etc.

Use ( )212!

t te t = + +" to rewrite


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( )2 21 2 2

t te t t t

= + " .

As a result,

( ) ( )220 0 01 22 tx t x v t gt z v = + + G G G G "

Yep, looks ok. In the limit 0 we have ( ) 20 0 1 2x t x v t gt + zG G G .

2. Motion of a charged particle in a constant magnetic field.

The equations dont separate, but theres a handy trick youll find

useful in other circumstances.

Lets say and B B=G z 0E =G .

The Lorentz force is 1 2F k qE k qv B= + G G GG . The specific values for

depend on the units you choose for charge, length, etc. In

the commonly use SI system of units (Tesla, meters, Coulombs,

Joules, and so forth)

1 and k 2k

1 2=1 and 1k k = .


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We have

( )2 2 2 y xF k qv B k qv zB k qB v x v y= = = G GG G .

The equation F ma=G G becomes

2 or x yk qB k qBa v x

m m= = 2 y [Equation 1]

2 or yy xk qB k qBa v

m m= = 2 x [Equation 2]

0za =

Looks nasty: the equations are coupled, since the x component of

acceleration depends on the y component of velocity!

Heres the trick: define , u x iy w x iy + so that ,

2 2u w u wx y

i+ = =

Then add Equation 1 to i times Equation 2:

( ) ( ) ( )2

2 2

2

y

y

k qBx ym k qB k qBx i ix y i

m mk qBi i x

m

= + + = + = +=

x iy


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or

2k qBu im

= u .

Now subtract i times Equation 2 from Equation 1 and do a small

amount of algebra to find

2k qBw i wm

=

Integrate the u equation once: 2 2 so ik qB ik qBdu duu u

dt m u m= = = dt

or

( ) 2ln ik qBu tm

= + c .

Similarly, integrate the equation: w2 2 so ik qB ik qBdw dww w

dt m w m= = = dt

or

( ) 2ln ik qBw tm

= + c .


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A

To save effort better, Im going to be clever about redefining the

integration constant : c lnc i + . I will require A to be a real number.

Since

( ) 2 2ln lnik qB ik qBu t c t im m

= + = + A

we find (after using the left and right sides of the last equation as

exponents) 22 2ln ln

k qBik qB ik qB i tt i A t i A mm mu e e e Ae + + = = = .

Recall that cos sinie i = so that

22 2cos sin

k qBi tm k qB k qBu Ae A t i t

m m

+ = = + +

and that u x . iy= +

The real part of u is x while the imaginary part of u is . As a result,

y

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2cos k qBx A tm

= + and 2sin k qBy A tm

= + .

Integrating once more gives

2

2

sin k qBAmx tk qB m

= +

and

2

2

cos k qBAmy tk qB m

= + .

Note that 2

2 2

2

Amx yk qB

+ = independent of time.

Therefore, 2

2 2

2

Amx yk qB

+ =

or

2 220 0

k qBA x ym

= +


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2-36

so we can write

2 22 20 0

2

2 2 20 0

sin

sin

k qB k qBmx x y tm k qB m

k qBx y tm

= + = + +

+

and, similarly,

2 2 20 0 cos

k qBy x y tm

= + + .

We can solve for the remaining unknown : 0 0 sin cos tanx y = =

so

( )1 0 0tan x y = .

Note that the motion (projected onto the x-y plane) is a circle. If we

allow for the a non-zero (but constant) velocity in the z direction

then the particle will move in a helix.

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Physics 325, fall 2012 Systems of particles University of Illinois

Systems of particles

Systems of particles .....................................................................3-1 Definition of the center of mass................................................3-1 Motion of the center of mass ....................................................3-2 Total angular momentum and torque........................................3-4 Reduced mass coordinates for two-particle systems ................3-5

So far weve only been dealing with a single particle moving under

the influence of an externally applied force. A system of particles

exhibits more complicated behavior: external forces can act on the

system, and forces between particles in the system can also affect

things.

Definition of the center of mass

There are some useful definitions we can employ to simplify the

description of whats going on.

Define the position of the center-of-mass of the system to be

1R m rM

G G where M m

and the sum runs over all particles in the system.


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Lets define fG

to be the force exerted on particle by particle . We know f f =

G G from Newtons laws.

Motion of the center of mass

Useful fact: if external forces , , , ...F F F G G G

act on particles , , , the response of the center-of-mass position will be the same as if there were a single particle of mass i

iM m= at the point RG

being acted on by a force iFF G G . The particles dont even have to be bound together to make this true.

Proof: the th particle feels a net force which is a sum of the inter-particle forces f

G acting on it added to the external-to-the-system

force FG

. As a result,

m r F f

= + GGG . We could have included = in the sum, since a particle does not exert a force on itself: 0f f = =

G G.

Summing over all particles:


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90


,m r F f F f

= + = + G GG GG .

The double-sum on the right is zero because of Newtons law

concerning equal-and-opposite forces. Lets list terms explicitly to

see this:

12 13 14,

21 23 24

31 32 34

+

+

f f f f

f f f

f f f

= + + +

+ + ++ + +

G G G G G G G G G G

Note that , etc. etc. For each team 12 21 13 310, 0f f f f+ = + =G G G G

ijfG

in

the sum theres a corresponding term jifG

that appears and cancels

the ijfG

term: . 0ij jif f+ =G G

As a result, we can rewrite the equation ,

m r F f

= + GGG as . m r F F

= G GG

Also, since 2

2

dm r m rdt

= G G and 1R m rM

G G we can copyright 2012 George Gollin PHY_325_lec_notes_03.doc 3-3

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rewrite once more:

m r F MR F

= = G G GG . (QED!)

This works for things like exploding projectiles, where the external

force acting on individual pieces can vary significantly from piece

to piece. It works even if some pieces go on to orbit while others

fa

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Classical Mechanics Notes