PHYS.303-CLASSICAL MECHANICS II - LECTURE NOTES"PPT"

Cartesian Coordinates (x, y, z)

z

y

x

z

x

y

),,( zyxP

z

y

x

Cartesian Coordinates (x,y,z)

x

y

z z

y

x

A

xa

ya

za

xxaA

yyaA

zzaA

zzyyxxzyx aAaAaAAorAAAA

),,(

are unit vectors along the x-, y-, and z-directions zyx aandaa

,,

Circular Cylindrical Coordinates (ρ,

, z)

z

20

0

zzz aAaAaAAorAAAA

),,(

Relationships between Cartesian and

Cylindrical Systems

zz

x

y

x

y

yx

,)(tanor tan

,

11

22

zzyx ,sin,cos

Relationships between Cartesian

and Cylindrical Systems

x

y

ya

a

a

xa

zz

y

x

aa

aaa

aaa

cossin

sincos

zz

yx

yx

aa

aaa

aaa

cossin

sincos

Spherical Coordinates (r, , )

20

0

0

r

aAaAaAAorAAAA rrr

),,(

Relationships between Cylindrical

and Spherical Systems

z

zr

1

22

tan

cos

sin

rz

r

Relationships between Cylindrical


ra

a

a

za

aaa

aa

aaa

rz

r

sincos

cossin

aa

aaa

aaa

z

zr

sincos

cossin



x

y

x

y

z

yx

zyxr

11

221

222

tanor tan

tan

cos

sinsin

cossin

rz

ry

rx



ya

a

xa

a

ra

a

a

za

aaa

aaaa

aaaa

rz

ry

rx

sincos

cossincossinsin

sincoscoscossin

yx

zyx

zyxr

aaa

aaaa

aaaa

cossin

sinsincoscoscos

cossinsincossin

Differential length and volume

dzdddv

Cylindrical coordinates Cartesian coordinates

zyx adzadyadxld

dzdydxdv

zadzadadld

Differential length and volume

Spherical coordinates

adradradrld r

sin

dddrrdv sin2

Exercise 1-1

Given point P(-2, 6, 3) and vector

a) Express P and A in cylindrical and spherical coordinates.

b) Evaluate A at P in the Cartesian, cylindrical, and spherical systems.

yx azxayA

Chapter 2

Some Methods in the

Calculus of Variations

1- Statement of the problem

The basic problem of the calculus of variations is to determine

the function y(x) such that the integral J is an extremum

2

1

),(),(

x

x

dxxxyxyfJ

The functional J depends on the function y(x).

The limits of integration are fixed.

The function y(x) is to be varied until an extreme value of J is found.

xxyx,Y

Y(0, x) = y(x) is the function that yields an extremum for J

(x) has a continuous first derivative and

0)()( 21 xx

1- Statement of the problem (Cont.)

2

1

),,(),,()(

x

x

dxxxYxYfJ

If functions of the type Y(α,x)

are considered, the integral J

becomes a functional of the

parameter α

The condition that the integral has a stationary value

(i.e., that an extremum result) for all functions (x) is:

00

d

dJ

This is only a necessary condition; it is not sufficient.

2- Euler’s Equation

2

1

),,(),,()(

x

x

dxxxYxYfJ 2

1

x

x

dxd

df

d

dJ

dx

x

fYd

Y

fdY

Y

fdf

dx

x

Yd

YYddx

x

Yd

YdY

,

dxx

fdx

x

Yd

Y

Y

fdx

x

Yd

Y

Y

fdf

Y

Y

fY

Y

f

d

df

xdx

dYx

Y

; xxyx,Y

dxdx

d

Y

fx

Y

f

d

dJx

x

2

1

2- Euler’s Equation (cont.)

dxdx

d

Y

fx

x

2

1

can be integrated by part

dxdx

d

Y

fx

Y

f

d

dJx

x

2

1

dx

dvv

Y

f

dx

du

Y

fu

,

dxvuvudxvu

dxxY

f

dx

dx

Y

fdx

dx

d

Y

fx

x

x

x

x

x

2

1

2

1

2

1

00 21

2

1

xxbecausex

Y

fx

x

2- Euler’s Equation (cont.)

dxxY

f

dx

d

Y

fdxx

Y

f

dx

dx

Y

f

d

dJx

x

x

x

2

1

2

1

002

10

dxxy

f

dx

d

y

f

d

dJx

x

Because (x) is an arbitrary function (subject to the conditions

already stated), then

0

y

f

dx

d

y

fEuler’s equation

3- The “Second Form” of the Euler

Equation

A second equation may be derived from Euler’s equation that is

convenient for functions that do not explicitly depend on x: ( f/ x) = 0

dxx

fyd

y

fdy

y

fxyydfdf

,,

x

f

y

fy

y

fy

dx

dx

x

f

dx

yd

y

f

dx

dy

y

f

dx

df

x

f

y

fy

dx

df

y

fy

y

f

dx

dy

y

fy

y

fy

dx

dAlso

3- The “Second Form” of the Euler Equation (Cont.)

y

f

dx

dy

x

f

y

fy

dx

df

y

fy

dx

d

y

f

y

f

dx

dy

x

f

dx

df

y

fy

dx

d

0

y

f

y

f

dx

din view of the Euler equation. Therefore,

0

x

f

dx

df

y

fy

dx

d0

y

fyf

dx

d

x

f

0for constant

x

f

y

fyf

4- Functions with Several

Dependent Variables The Euler equation derived in the preceding section is the solution

of the variational problem in which it was desired to find the single

function y(x) such that the integral of the functional f was an

extremum. The case more commonly encountered in mechanics is

that in which f is a functional of several dependent variables:

xxyxyxyxyff ...,),(),(),(),( 2211

nixxyxyff ii ...,,2,1;),(),( Or simply

xxyxY iii ,

dxxy

f

dx

d

y

f

d

dJi

x

x i ii

2

1

The development proceeds

analogously as in section 2,

resulting in

4- Functions with Several Dependent Variables (Cont.)

002

10

dxxy

f

dx

d

y

f

d

dJi

x

x i ii

niy

f

dx

d

y

f

ii

...,,2,10

5- Euler’s Equations when

Auxiliary Conditions are

Imposed

Consider the case in which , and the auxiliary equation

or equation that is also called equation of constraint x,y,yff ii

0x,yg i

Example: x,z,z,y,yff 0x,z,ygand

0

dx

x

gdZ

Z

gdY

Y

gdg

0

dx

x

gdx

x

Zd

Z

Z

gdx

x

Yd

Y

Y

gdg

0

dx

x

g

x

Z

Z

g

x

Y

Y

gd

Z

Z

gY

Y

gdg

5- Euler’s Equations when Auxiliary Conditions are Imposed

(Cont.)

0

dx

x

g

x

Z

Z

g

x

Y

Y

gd

Z

Z

gY

Y

gdg

0

Z

Z

gY

Y

g

xxyx,Y 1 xxzx,Z 2

xY

1

x

Z2

021

x

Z

gx

Y

gZ

Z

gY

Y

g

Zg

Yg

x

x

1

2


(Cont.)

From section 4:

2

1

x

x

dxZ

Z

f

dx

d

Z

fY

Y

f

dx

d

Y

f

d

dJ

2

1

21

x

x

dxxZ

f

dx

d

Z

fx

Y

f

dx

d

Y

f

d

dJ

2

1

1

1

2

x

x

dxxx

x

Z

f

dx

d

Z

f

Y

f

dx

d

Y

f

d

dJ

2

1

1

x

x

dxxZg

Yg

Z

f

dx

d

Z

f

Y

f

dx

d

Y

f

d

dJ

002

1

1

0

x

x

dxxzg

yg

z

f

dx

d

z

f

y

f

dx

d

y

f

d

dJ


(Cont.)

0

zg

yg

z

f

dx

d

z

f

y

f

dx

d

y

f

zg

yg

z

f

dx

d

z

f

y

f

dx

d

y

f

11

z

g

z

f

dx

d

z

f

y

g

y

f

dx

d

y

f

The left-hand side of this equation involves only derivatives of f and g

with respect to y and y’ , and the right-hand sides involves only

derivatives with respect to z and z’ . Because y and z are both functions

of x, the two sides of the above equation may be set equal to a function

of x, which we write as : -λ(x).


(Cont.)

xz

g

z

f

dx

d

z

f

y

g

y

f

dx

d

y

f

11

z

gx

z

f

dx

d

z

f

y

gx

y

f

dx

d

y

f

For the general case of n dependent variables, yi, and k constraints,

gk(yi, x):

n...,,,iandm...,,,kx,yg

y

gx

y

f

dx

d

y

f

ik

k i

kk

ii

21210

0

6- Variational Notation

dd

dJJandd

d

dJJ

0

0

dx)x,Y,Y(fJ

x

x

2

1

xxyx,Y xxyxY ,

dx)x,Y,Y(fJ

x

x

2

1

dx)x,Y,Y(fJ

x

x

2

1

The symbol δ was used in the early days of the development of the

calculus of variations to indicate what we have called differentiation

with respect to the parameter α. It is just like the symbol d in a

differential except that it warns you that α and not x is the

differentiation variable.

6- Variational Notation (Cont.)

??x,Y,Yf dxx

fYd

Y

fdY

Y

fdf

dx

x

Yd

YYd,dx

x

Yd

YdY

dxx

fdx

x

Yd

Y

Y

fdx

x

Yd

Y

Y

fdf

dd

dx

x

fd

d

dx

x

Yd

Y

Y

fd

d

dx

x

Yd

Y

Y

fd

d

df

d

Y

Y

fd

Y

Y

ffd

d

df

dY

dd

YdY,d

Yd

d

dYY


YY

fY

Y

ff

Ydx

d

dx

dYY

dxY

dx

d

Y

fY

Y

fJ

x

x

2

1

Integrate by parts dxYdx

d

Y

fx

x

2

1

Ydx

dvYv,

Y

f

dx

du

Y

fu

vdxuuvdxvu

dxYY

f

dx

dY

Y

fdxY

dx

d

Y

fx

x

x

x

x

x

2

1

2

1

2

1


00 21

2

1

xx,dxdY

YYY

fx

x

dxYY

f

dx

ddxY

dx

d

Y

fx

x

x

x

2

1

2

1

dxYY

f

dx

d

Y

fJ

x

x

2

1

2

1

0

x

x

dxyy

f

dx

d

y

fJ

dxdd

dYYy

0

0


ddxxy

f

dx

d

y

fJ

x

x

2

1

0

equation Eulery

f

dx

d

y

fJ 000

Chapter 3

Hamilton’s Principle

Lagrangian and Hamiltonian

Dynamics

Dr. Abdelaziz Sabik

Physics Department – College of Sciences

Al-Imam Muhammad Ibn Saud Islamic University

1- Introduction

A particle’s motion in an inertial reference frame is correctly

described by the Newtonian second law equation:

amF

where F is the total force acting on the body.

In particular complicated problems, this equation can become

difficult to manipulate.

An alternate method of dealing with complicated problems is

contained in Hamilton’s Principle, and the equations of motion

resulting from the application of this principle are called

Lagrange’s equations.

2- Hamilton’s Principle

Of all the possible paths along which a dynamical system may move

from one point to another within a specified time interval, the

actual path followed is that which minimizes the time integral of

the difference between the kinetic and potential energies.

In terms of the calculus of variations,

Hamilton’s Principle becomes

2

1

0

t

t

dtUT

extremuman bemust i.e.2

1

t

t

dtUT

(1)

1t 2tt

)(tx

Path

مسار الجسم يشتق بإيجاد المسلك الذي يقلل الفعل


In fixed rectangular coordinates (x1,x2,x3), the kinetic energy, T, of a

particle is a function of :

ixTT

and if the particle moves in a conservative force field, the potential

energy is function of xi:

ix

ixUU

We define the difference of these quantities to be

ii xxLUTL , (2) where L is Lagrange function or

Lagrangian for the particle. Equation (1) becomes

2

1

0

t

t

ii dtx,xL (3)


The function L may be identified with the function f of the variational

integral (see Chapter 2) dxxxyxyf

x

xii

2

1

),(),(

If we make the transformations:

iiii

ii

ii

xxLxxyxyf

txxy

txxy

tx

,,,

The Euler’s equations (Chapter 2) corresponding to Equation (3):

3210 ,,i,x

L

dt

d

x

L

ii

(4)

These are the Lagrange

equations of motion for the

particle.


Example 1: Plane Pendulum

Problem: Obtain the Lagrange equations of motion for the plane

pendulum of mass “m”.

b

m0U

b

y

cosb

ld

abdld

Solution: Write the expressions of T

and U to determine L.

abadt

db

dt

ldv

222 bv

222

2

1

2

1mbmvT

)cos( bbmgmgyU



)cos1(2

1 22 mgbmbUTL

sinmgbL

2mb

L

2mb

L

dt

d

0

L

dt

dLApplying Lagrange equation leads to:

Calculate derivatives of L by treating as if it were a rectangular

coordinate:

0sin 2 mbmgb

0sin b

g

b

m

T

gm

x

y

xa

ya



Problem: Do we obtain the same equation if we use Newtonian

mechanics?

Solution: Applying Newton’s second law:

amgmT

yxyxy ayaxmamgaTaT

sincos

)2(cos

)1(sin

ymmgT

xmT

sincos

cos

sin

2

bbx

bx

bx

cossin

sin

cos

2

bby

by

by


sincossin)1( 2 mbmbT

)3(sin

cos2

mbmbT

cossin

sin

coscos)3(&)2( 2

22 mbmbmgmbmb

sin

sin

cos2

mbmgmb

0sinsincos 22 mbmgmb

0sin b

g which is identical with the equation of motion

obtained using Lagrangian mechanics.


Remarks regarding Example 1 (Lagrangian mechanics)

1- Example 1 has been solved by calculating the kinetic and

potential energies in terms of rather than x and then applying a set of

operations designed for use with rectangular rather than angular

coordinates.

2- No-where in calculations did there any statement regarding force.

3- Hamilton’s Principle allows us to calculate the equations of motion

of a body completely without recourse to Newtonian theory.

3- Generalized Coordinates

• Consider mechanical systems consisting of a collection of n

discrete point particles.

• We need n position vectors, i.e. 3n quantities must be specified

to describe the positions of all the particles.

• If there are m constraint equations that limit the motion of particle

by for instance relating some of coordinates, then the number of

independent coordinates is limited to 3n-m.

• One then describes the system as having s = 3n-m degrees of

freedom.


• If s=3n-m coordinates are required to describe a system, it is NOT

necessary these s coordinates be rectangular or curvilinear coordinates.

• One can choose any combination of independent coordinates as long

as they completely specify the system.

• These coordinates need not even have the dimension of length

(e.g. θ in our previous example).

• We use the term generalized coordinates to describe any set of

coordinates that completely specify the state of a system.

• Generalized coordinates will be noted: q1, q2, …, or simply as the qj.


• In some cases, it may be useful to use generalized coordinates

whose number exceeds the number of degrees of freedom, and

to explicitly take into account the constraint relations through the

use of the Lagrange Undetermined multipliers. Such would be the

case, for example, if we desired to calculate the forces of constraint.

• The choice of a set of generalized coordinates is obviously not

unique. We choose the set that gives the simplest equations of

motion.

• In addition to the generalized coordinates, we may define a set of

quantities consisting of the time derivatives of or

simply . We call generalized velocities. jq...,,,: 21 qqq j

jq


Transformations from Cartesian coordinates to generalized coordinates

sj

i

n

tqxtqqqxx jisii

,...,2,1

)components (Cartesian3,2,1

)particlespoint discreten (,...,2,1

,,,...,, ,21,,

tqqxx jjii ,,,,

Transformations from generalized coordinates to Cartesian coordinates

txxqqtxqq iijjijj ,,,, ,,,

There are m = 3n-s equations of constraint of the form

mktxg ik ,...,2,1,0,,

4- Lagrange’s Equations of Motion in

Generalized Coordinates

tqUtqqTL

xUxTL

jjj

ii

,,,

,,

tqqLtqqqqqqLL jjss ,,;...,,,;...,,, 2121

Hamilton’s Principle becomes: 02

1

dtt,q,qL

t

t

jj

We get Lagrange’s equations as

mnsjq

L

dt

d

q

L

jj

3...,,2,1,0

4- Lagrange’s Equations of Motion

in Generalized Coordinates

The conditions of validity of Lagrange’s equations:

1-The forces acting on the system (apart from any forces of

constraint) must be derivable from a potential (or several potentials).

2-The equations of constraint must be relations that connect the

coordinates of the particles and may be functions of the time. That is,

we must have constraint relations of the form

mktqg jk ,...,2,1,0,


Generalized Coordinates – Example 2

Example 2: The double pendulum

Problem: Consider the case of

two particles of mass m1 and m2

each attached at the end of a

massless rod of length b1 and b2,

respectively. The second rod is

also attached to the first particle.

Derive the equations of motion

for the two particles by using

Lagrangian mechanics.

1

2

1b

2b1m

2m

1x2x

1y

2y

x

y



1

2

1b

2b1m

2m

1x2x

1y

2y

x

ySolution: We choose 1 and 2 as the generalized

coordinates.

Starting with Cartesian coordinates:

111

111

cos

sin

by

bx

22112

22112

coscos

sinsin

bby

bbx

1111

1111

sin

cos

by

bx

2221112

2221112

sinsin

coscos

bby

bbx



We write an expression for the kinetic and potential energies for the

system:

2

2

2

22

2

1

2

112

1yxmyxmT

0)yat 0 U take(we2211 gymgymU

2

22

2

2

222121212

1

22

1

2

122

22

2

2

222121212

1

22

1

2

121

22

1

2

111

22

1

2

11

sin2

1sinsin

sin2

1cos

2

1coscos

cos2

1sin

2

1cos

2

1

bmbbm

bmbmbbm

bmbmbmT



2121212

2

2

2

22

2121212

2

1

2

12

2

1

2

11

sinsin2

1

coscos2

1

2

1

bbmbm

bbmbmbmT

2121212

2

2

2

22

2

1

2

12

2

1

2

11 cos2

1

2

1

2

1 bbmbmbmbmT

222112111 coscoscos gbmgbmgbmU

2221121 coscos)( gbmgbmmU UTL

2221121

2121212

2

2

2

22

2

1

2

12

2

1

2

11

coscos)(

cos2

1

2

1

2

1

gbmgbmm

bbmbmbmbmL



2221121

2121212

2

2

2

22

2

1

2

12

2

1

2

11

coscos)(

cos2

1

2

1

2

1

gbmgbmm

bbmbmbmbmL

11212121212

1

sinsin

gbmmbbmL

2122121

2

121

2

11

1

cos

bbmbmbmL

212122122121

2

121

1

sincos

bbmbmmL

dt

d



0sinsincos

sinsin

21

2

22121212212

1

2

12111212121212

bbm

bmmgbmmbbm

We obtain from Lagrange’s equation with respect to 1:

0sincos

sin

21

2

2212212

1

2

1211121

bbm

bmmgbmm

0sincossin 21

2

2212

1

2

21

21

1

1

b

b

mm

m

b

g

Or



We have:

2221121

2121212

2

2

2

22

2

1

2

12

2

1

2

11

coscos)(

cos2

1

2

1

2

1

gbmgbmm

bbmbmbmbmL

2222121212

2

sinsin

gbmbbmL

2112122

2

22

2

cos

bbmbmL

212112112122

2

22

2

sincos

bbmbmL

dt

d



We obtain from Lagrange’s equation with respect to 2:

0sincos

sinsin

21211211212

2

2

222222121212

bbm

bmgbmbbm

0sincossin 21

2

12112122

2

22222 bbmbmgbm

0sincossin 21

2

1211

2

12

2

2 b

b

b

g

Or



Problem:

A particle of mass m is constrained to

move on the inside surface of a smooth

cone of half-angle α. The particle is

subject to a gravitational force.

a) Determine a set of generalized

coordinates and determine the constraints.

b) Find Lagrange’s equations of motion

x

y

z

z

m g



Solution: Since the problem possesses cylindrical symmetry, we

choose , , and z as the generalized coordinates.

The equation of constraint:

So there are only two degrees of freedom for the

system. We choose , and as independent

generalized coordinates.

cotz

z

m

zadzadadld

zadadadld

cot

zadt

da

dt

da

dt

d

dt

ldv

cot zaaav

cot

222222 cot v



222222 cot v

222222 cot2

1

2

1

2

1

2

1 mmmTmvT

The kinetic energy is

The potential energy (if we choose U = 0 at z = 0) is

cotmgmgzU

so the Lagrangian is

cotcot2

1

2

1

2

1 22222 mgmmmL

UTL



cotcot2

1

2

1

2

1 22222 mgmmmL

The Lagrange equation for is 0

L

dt

dL

cot2 mgmL

2cot

mmL

22 cot1cot

mmm

L

dt

d

2sin

mL

dt

d

0sinsin

cos0

2

2

mmgm

L

dt

dL

Multiplying each term of the latest equation by , we find m

2sin



0cossinsin22 g

which is the equation of motion for the coordinate .


L

dt

dL

,0

L

2m

L

constant is then ,0 Since

LL

axis)-z about the momentumangular (theconstant 2 m



Problem:

The point of support of a simple pendulum

of length b moves on a massless rim of

radius a rotating with constant angular

velocity .

a) Obtain the expression for the Cartesian

Components of the velocity and acceleration

of the mass m.

b) Obtain the angular acceleration for the angle .

a

t

b

x

y

y

x

1x2x

1y

2y

g

m

Solution: The Cartesian components, x and y, of mass m are

sin)cos(21 btaxxx sin and),cos( 21 bxtax

cos and ),sin( 21 bytay cos)sin(21 btayyy



cos)sin( and ,sin)cos( btaybtax

The velocities are cos)sin( btaxdt

dx

sin)cos( btaydt

dy

Taking the time derivative once again gives the acceleration:

)sincos()cos( 22 btax

)cossin()sin( 22 btay

b) The infinitesimal displacement of m in Cartesian coordinates is

yx adyadxld

The velocity of m: yxyx ayaxa

dt

dya

dt

dx

dt

ldv



The kinetic energy is 222

2

1

2

1yxmTmvT

22 ]cos)sin([ btax

2222222 coscos)sin(2)(sin btabtax

22 ]sin)cos([ btay

2222222 sinsin)cos(2)(cos btabtay

222222

222222

sin2

1sin)cos()(cos

2

1

cos2

1cos)sin()(sin

2

1

mbtmabtma

mbtmabtmaT



]sin[cos2

1]cos)sin(sin)[cos(

)](cos)([sin2

1

2222

2222

mbttmab

ttmaT

2222

2

1]cos)sin(sin)[cos(

2

1 mbttmabmaT

2222

2

1)sin(

2

1 mbtmabmaT

The potential energy is )0at 0 (where yUmgyU

cos)sin( mgbtmgaU



The Lagrangian is

cos)sin(

2

1)sin(

2

1 2222

mgbtmga

mbtmabmaUTL

0 is for equation Lagrange The

L

dt

dL

sin)cos( mgbtmabL

2)sin( mbtmabL

2)cos()( mbtmabL

dt

d



0)cos()(

sin)cos(

2

mbtmab

mgbtmab

0)cos(

)cos(sin)cos(

22

mbtmab

tmabmgbtmab

0)cos(sin 22 mbtmabmgb

2

22 1)0)cos(sin(

mbmbtmabmgb

0)cos(sin 2 tb

a

b

g

)cos(sin 2 tb

a

b

g

Lagrange’s Equations of Motion in


Problem

A simple pendulum of length b and bob

with mass m is attached to a massless

support moving vertically upward with

constant acceleration a.

Determine the equations of motion.

Solution:

b

m

0Ux

x

y

a

1y

y

g

sinbx )2

1 where( 2

1 aty 2

2

1cos atby

We choose a fixed Cartesian coordinate system

with y = 0 and ẏ = 0 at t = 0. The position and

velocity of m become:

cos bx atby sin




The infinitesimal displacement of m is yx adyadxld

The velocity of m is yxyx ayaxadt

dya

dt

dx

dt

ldv

222

2

1

2

1yxmmvT

2222 )sin(cos2

1atbbmT

22222222 sin2sincos2

1tatabbbmT

222222 sin2)sin(cos2

1tatabbmT

2222

2

1sin

2

1tmatmabmbT



The potential energy is 0)yat 0 U(where mgyU

2

2

1cos mgatmgbU

The Lagrangian is UTL

22222

2

1cos

2

1sin

2

1mgatmgbtmatmabmbL


L

dt

dL

sincos mgbtmabL

sin2 tmabmbL

cossin2 mabtmabmb

L

dt

d



0cossinsincos 2 mabtmabmbmgbtmab

0sinsin 2 mabmbmgb

2

2 1)0sinsin(

mbmabmbmgb

0sinsin b

a

b

g

5- Lagrange’s Equations with Undetermined

Multipliers

Chapter 2 – Section 5

n...,,,iandm...,,,kx,yg

y

gx

y

f

dx

d

y

f

ik

k i

kk

ii

21210

0

For the general case of n dependent variables, yi, and k constraints,

gk(yi, x):

),(),(

)()(

tqgxyg

tx

qy

tx

qy

Lf

jkik

kk

ji

ji

Transformations:


Multipliers

We can rewrite the Lagrange equations as

k j

kk

jj q

gt

q

L

dt

d

q

L0

k j

kkj

q

gQ

the undetermined multipliers, , are closely related to the

Forces of constraint.

tk

generalized forces of constraint


Multipliers – Example 6

ra

x

y

x

y

0U

Problem:

A particle of mass m starts at rest on

top of a smooth fixed hemisphere of

radius a. Find the force of constraint,

and determine the angle at which the

particle leaves the henisphere.

Solution:

We choose the generalized coordinates to be r and . The constraint

equation is 0),,(1 artrg

ardadrld r

:m ofnt displaceme malinfinitesi The

araradt

dra

dt

dr

dt

ldv rr

:m of velocity The



2222

rrvararv r

2222

2

1

2

1

2

1 mrrmTmvT

y

rmgUmgyU

cos


The potential energy is

where U = 0 at y = 0

The Lagrangian is cos2

1

2

1 222 mgrmrrmUTL

The Lagrange equation with undetermined multipliers for r is

00 11

1

r

g

r

L

dt

d

r

L

r

g

r

L

dt

d

r

L m

k

kk



cos2 mgmrr

L

rmr

L

rm

r

L

dt

d

11

r

g

)1(0cos0 1

211

rmmgmr

r

g

r

L

dt

d

r

L

The Lagrange equation with undetermined multipliers for is

00 11

1

gL

dt

dLgL

dt

dL m

k

kk

cos2

1

2

1 222 mgrmrrmL

sinmgrL

2mr

L

22 mrrmr

L

dt

d

0),,( 11

gartrg



0)(constant isr rrar

Next, we apply the constraint to equations of motion (Eq. (1) & (2)):

)3(cos0cos 2

11

2 mgmamgma

Then these equations of motion become

002sin 1

2 mrrmrmgr

)2(02sin 2 mrrmrmgr

)4(sin0sin 2 a

gmamga

To find the force of constraint , we must determine . For this,

we integrate Equation (4). 1

2



,sina

g ,sin

a

g

dt

d

,sin

a

g

dt

d

d

d

sin

a

g

d

d

da

gd sin d

a

gd sin

constant)cos(2

1 2 a

g 0 and ,0 0,at t

a

g

a

g constantconstant0

a

g

a

g cos

2

1 2

a

g

a

g 2cos

22 cos2

1 mgma



cos2

cos2

1 mga

g

a

gma

cos2cos21 mgmgmg mgmg 2cos31

)2cos3(1 mg

Which is the force of constraint. The particle falls off the hemisphere

at angle 0 when λ1 = 0.

02cos30 01

2,483

2cos

3

2cos 1

00

6- A Theorem Concerning the Kinetic Energy

If we consider the fixed rectangular coordinates, the kinetic energy

of one particle will be expressed as:

We now wish to consider in more detail the dependence of T on

the generalized coordinates and velocities.

s...,,,j,t,qxx jii 21 dtt

xdq

q

xdx i

j

s

j j

ii

1

dt

dt

t

x

dt

dq

q

x

dt

dx ijs

j j

ii

1

s

j

ij

j

ii

t

xq

q

xx

1

3

1

2

2

1

i

ixmT


The following step is to find . For this let’s take, for

example, s = 2.

2

ix

2

1j

ij

j

ii

t

xq

q

xx

t

xq

q

xq

q

xx iii

i

2

2

1

1

t

xq

q

xq

q

x

t

xq

q

xq

q

xx iiiiii

i 2

2

1

1

2

2

1

1

2

t

x

t

xq

q

x

t

xq

q

x

t

x

t

xq

q

xq

q

xq

q

xq

q

xq

q

x

t

xq

q

xq

q

xq

q

xq

q

xq

q

xx

iiiiii

iiiiii

iiiiiii

2

2

1

1

2

2

2

2

2

2

1

1

2

2

1

1

2

2

1

1

1

1

1

1

2


t

x

t

xq

q

x

t

xq

q

x

t

x

t

xq

q

xq

q

xq

q

xq

q

xq

q

x

t

xq

q

xq

q

xq

q

xq

q

xq

q

xx

iiiiii

iiiiii

iiiiiii

2

2

1

1

2

2

2

2

2

2

1

1

2

2

1

1

2

2

1

1

1

1

1

1

2

2

2

2

1

1

2

2

22

22

12

12

1

1

21

21

11

11

2

t

xq

t

x

q

xq

t

x

q

x

qt

x

q

xqq

q

x

q

xqq

q

x

q

x

qt

x

q

xqq

q

x

q

xqq

q

x

q

xx

iiiii

iiiiii

iiiiiii


2

2

2

1

1

2

2

22

22

12

12

1

1

21

21

11

11

2

t

xq

t

x

q

xq

t

x

q

x

qt

x

q

xqq

q

x

q

xqq

q

x

q

x

qt

x

q

xqq

q

x

q

xqq

q

x

q

xx

iiiii

iiiiii

iiiiiii

22

1

2

1

2

1

2 2

j k j

ij

i

j

ikj

k

i

j

ii

t

xq

t

x

q

xqq

q

x

q

xx

2

11 1

2 2

t

xq

t

x

q

xqq

q

x

q

xx i

ji

s

j j

ikj

k

is

j

s

k j

ii

Therefore, in general:

6- A Theorem Concerning the Kinetic Energy 2

11 1

2 2

t

xq

t

x

q

xqq

q

x

q

xx i

ji

s

j j

ikj

k

is

j

s

k j

ii

23

11

3

11 1

3

1 2

1

2

1

t

xmq

t

x

q

xmqq

q

x

q

xmT i

i

ji

s

j j

i

i

kj

k

is

j

s

k j

i

i

cqbqqaT j

s

j

jkj

s

j

s

k

jk

11 1

A particularly important case occurs when the system is scleronomic,

so that the time does not appear explicitly in the equations of

transformation; then the partial time derivatives vanish:

000

c,b,

t

xj

i

23

1

3

1

3

1 2

1,,

2

1

i

i

i

i

j

ij

k

i

j

i

i

jkt

xmc

t

x

q

xmb

q

x

q

xmawhere


kj

s

j

s

k

jk qqaT

1 1

s

k

s

j

jjlklk

l

qaqaq

T

1 1

general,In

Under these conditions, the kinetic energy becomes

Now, we want to determine an expression for

s

l l

lq

Tq

1

Let’s take s = 2, so: 2222122121121111 qqaqqaqqaqqaT

222121112

2

2 qaqaqaq

T

222112222121

2

qaqaqaqaq

T

2

1

2

1

22

2 k j

jjkk qaqaq

T

2221122221212

2211112121111

2

2

1

1

qaqaqaqaq

qaqaqaqaqq

Tq

q

Tq


2221122221212

2211112121111

2

2

1

1

qaqaqaqaq

qaqaqaqaqq

Tq

q

Tq

2222211222221221

1221111121121111

2

2

1

1

qqaqqaqqaqqa

qqaqqaqqaqqaq

Tq

q

Tq

2222122121121111

2

2

1

1 2222 qqaqqaqqaqqaq

Tq

q

Tq

2

1

2

1

2

1

2l j k

kjjk

l

l qqaq

Tq

Tqqaq

Tq

s

l

s

j

s

k

kjjk

l

l 221 1 1

7- Conservation of Energy

j j

j

j

j

j

s

j

s

j

j

j

j

j

jj

dt

dt

t

L

dt

qd

q

L

dt

dq

q

L

dt

dL

dtt

Lqd

q

Ldq

q

LdL

tqqLL

1 1

becomes lagrangian theof derivative total then the,,,

But Lagrange’s equations are: jj q

L

dt

d

q

L

t

Lq

q

Lq

q

L

dt

dL

j j

j

j

j

j

t

Lq

dt

d

q

L

q

L

dt

dq

dt

dL

j j

j

jj

j


t

L

q

Lq

dt

d

dt

dL

t

Lq

dt

d

q

L

q

L

dt

dq

dt

dL

s

j j

j

j j

j

jj

j

1

t

L

q

Lq

dt

d

dt

dL

j j

j

or

t

L

q

LqL

dt

d s

j j

j

1

s

j j

jq

LqLH

1

We introduce a new function H, the quantity in the parentheses,

called the Hamiltonian of the system:

t

L

dt

dH

so that


If the Lagrangian is explicitly independent on time (i.e. )

then the total time derivative of H is zero and H becomes a constant.

0

t

L

constant0 then ,0 If1

s

j j

jq

LqLH

dt

dH

t

L


If the potential energy U does not depend explicitly on the velocities:

ixUU

If we exclude the possibility of an explicit time dependence in the

transformation equations:

)( jii qxx

Therefore: 0)(

j

jq

UqUU

jjj q

T

q

UT

q

L

H

q

LqL

j j

j

HTUTHq

TqUT

j j

j

2

HEUT where E is the total energy


The Hamiltonian H is equal to the total energy E only if the

following conditions are met:

1-The equations of the transformation connecting the rectangular

and generalized coordinates must independent of the time (the

system is scleronomic.)

2-The potential energy must be velocity independent.

8- Canonical Equations of Motion –

Hamiltonian Dynamics

If the potential energy of a system is velocity independent, then the

linear momentum components in rectangular coordinates are given by

ii

ix

L

x

Tp

By analogy, we extend this result to the case in which the

Lagrangian is expressed in generalized coordinates and define the

generalized momenta according to

j

jq

Lp



Lagrange’s equations are j

jjj

pdt

d

q

L

q

L

dt

d

q

L

The Lagrange equations of motion are then expressed by j

jq

Lp

j

jj j

j pq

LH

q

LqL

and ,We have:

LqpH jj Consequently, the Hamiltonian becomes:

We may solve Equation for the generalized velocities, ,

and express them as j

jq

Lp

jq

),,( tpqqq kkjj



tqqLLtpqHH kkkk ,,,,,

k

k

k

k

k

dtt

Hdp

p

Hdq

q

HdH

Thus, in equation , we may make a change of

variables from the set to the set and express

the Hamiltonian as

LqpH jj

),,( tqq jj ),,( tpq jj

k

kkkkkk tqqLqptpqH ,,,,

The total differential of H is therefore



k

k

k

k

k

kkkk dtt

Lqd

q

Ldq

q

LqdpdpqdH

we can also write

k

kkkkkkkk dtt

LqdpdqpqdpdpqdH

According to equation

k

kkkkkk tqqLqptpqH ,,,,



We have:

k

k

k

k

k

dtt

Hdp

p

Hdq

q

HdH

k

kkkk dtt

LdqpdpqdH

k

k

k

k

q

Hp

p

Hq

Hamilton’s equations of motion

or

Canonical equations of motion

t

H

t

L

The description of motion by these equations is termed

Hamiltonian dynamics



k

k

k

k

k

k

k

k

k q

Hp

p

Hqdt

t

Hdp

p

Hdq

q

HdH

and,

We have

t

H

dt

dH

then

k

kkkk dtt

HdpqdqpdH )(

dt

dt

t

H

dt

dpq

dt

dqp

dt

dH

k

kk

kk

t

Hpqqp

dt

dH

k

kkkk



1- If H does not explicitly contain the time ( ), then the

Hamiltonian is a conserved quantity (dH/dt = 0).

2- We have seen previously that the Hamiltonian equals the total

energy E=T+U if the potential energy is velocity independent and

the transformation equations do not explicitly contain the time.

3- If the potential energy is velocity independent and the

transformation equations do not explicitly contain the time, and if

, then H = E = constant.

0 tH

0 tH


Hamiltonian Dynamics – Example 7

Problem:

Use The Hamiltonian method to find the

equations of motion of a particle of mass

m constrained to move on the surface of

a cylinder defined by x2+y2=R2. The

particle is subject to a force directed

toward the origin and proportional to

the distance of the particle from the

origin: rkF

Solution: zadzadadld

0 dR zadzaRdld

zadt

dza

dt

dR

dt

ldv

zazaRv

2222 zRv

z

R

R

z

x

yF

g

m



)(2

1

2

1 2222 zRmmvT 222

2

1

2

1zmmRT

scoordinate sphericalin

sin

ld

rr adrardadrarkldFU

The potential corresponding to the force is F

2

2

1rkdrrkU )(

2

1

2

1 22222

222

zRkzyxkUyx

r

)(2

1

2

1

2

1 22222 zRkzmmRUTL

The generalized coordinates are and z, and the generalized momenta

are p and pz:



2

2

mR

pmR

Lp

m

pzzm

z

Lp z

z

22

2

2222

2

1

2

1

2

1

2

1

m

pm

mR

pmRTzmmRT z

m

p

mR

pT z

22

2

2

2

Because the potential energy is velocity independent and the

equations of transformation between rectangular and cylindrical

coordinates do not explicitly contain the time, then

UTppzH z ,,,

222

2

2

2

1

2

1

22kzkR

m

p

mR

pH z



The equations of motion are therefore found from the canonical

equations: (1)constant0

p

Hp

)2(kzz

Hpz

)3(2

2

mRpmR

p

p

H

)4(zz

z

pzmm

p

p

Hz

constant2 mRpEquations (1) & (3) give

The angular momentum about the z-axis is thus a constant of the

motion (the z-axis is the symmetry axis of the problem.)



Combining Equations (2) & (4), we find

kzzmpzm z

0 kzzm 0 zm

kz

m

kzz 2

0

2

0 where0

tAz 0sin

The motion in the z direction is therefore simple harmonic.



x

yx

y

b

0U

g

m

Problem:

Use the Hamiltonian method to find the

equations of motion for a spherical pendulum

of mass m and length b.

Solution:

adrardadrld r

sin

0 drbr

adbabdld

sin

a

dt

dba

dt

db

dt

ldv

sin ababv

sin

222222 sin bbv



222222 sin2

1

2

1

2

1 mbmbTmvT

y

bmgUmgyU

)cos( cosmgbU

cossin2

1

2

1 22222 mgbmbmbUTL

The generalized momenta are then

2

2

mb

pmb

Lp

22

22

sinsin

mb

pmb

Lp



Because the potential energy is velocity independent and the

equations of transformation between rectangular and spherical

coordinates do not explicitly contain the time, then

UTppH ,,,

U

T

mgbmb

pmb

mb

pmbH

cossinsin2

1

2

1 2

2

22

2

2

2

2

cossin22 22

2

2

2

mgbmb

p

mb

pH

The equations of motion are therefore found from the canonical

equations:



sinsin

cossin2

2 42

2

mgbmb

pHp

sinsin

cos32

2

mgbmb

pp

(1)constant0

p

Hp

2mb

p

p

H

)2(

sin22

mb

p

p

H

Combining Equations (1) & (2), we find constant)(sin22 mb

The angular momentum about the y-axis is thus a constant of the

motion (the y-axis is the symmetry axis of the problem.)



constant dt

dr

Problem:

Consider a simple plane pendulum consisting

of a mass m attached to a string of length r0.

After the pendulum is set into motion, the

length of the string is shortened at a constant rate

The suspension point remains fixed. Compute the

Lagrangian and Hamiltonian functions. Compare

The Hamiltonian and the total energy, and

discuss the conservation of total energy for the system.

x

yx

y

0U

g

m

r



Solution: The generalized coordinates are r and .

,constant dtdrdt

dr dtdr and

constant tr 00 constantconstant)0( rr

0rtr

sinrx cossin rrx cossin rx

cosry sincos rry sincos ry

22222 )sincos()cossin( rryxv

22222

222222

sincossin2cos

coscossin2sin

rr

rrv



2222 rv

2222

2

1

2

1

2

1 mrmmvT

cos)cos( mgrrmgmgyU

y

cos2

1

2

1 222 mgrmrmUTL

The kinetic energy, T, and the potential energy, U, of the system are

expressed as

so that the Lagrangian is

cos)()(2

1

2

10

22

0

2 rtmgrtmmL



LL

Lq

LqH

j j

j

2

0

2

0)(

)(rtm

prtmp

L

cos)(

)()(

2

1

2

1

)(

0

2

2

0

2

0

2

2

0

rtmg

rtm

prtmmp

rtm

pH

cos)(2

1

)(20

2

2

0

2

rtmgmrtm

pH

The Hamiltonian, H, is



cos)()(2

1

2

10

22

0

2 rtmgrtmmUT

The total energy, E=T+U, is

cos)()(

)(2

1

2

10

2

2

0

2

0

2 rtmgrtm

prtmmUT

cos)()(22

102

0

22 rtmg

rtm

pmUT

dt

dE

dt

dHmEH 2

000

dt

dE

dt

dH

t

HThe total energy is not

conserved.

Chapter 4

Central-Force Motion

Dr. Abdelaziz Sabik

Physics Department – College of Science


1- Reduced Mass

Consider the motion of a system consisting of two bodies affected

by a force directed along the line connecting the centers of the two

bodies (i.e., a central force).

21 )()( rdrFrdrFU

1m

2m

r

)(21 rF

)(12 rF

O

1r

2r

r

r

1r

2r

1rd

2rd

1mofPath

2mofPath)()()( 2112 rFrFrF

111 rrrd

222 rrrd

21 rrr

21 rrr

221112 )()( rdrFrdrFU

)()( 21 rdrdrFU

)()( 221121 rrrrrdrd

rdrrrrrrrdrd

)()( 212121

1- Reduced Mass

)()( rUrdrFU

radr

dUrUrF

)()(

1m

2m

1r

2r

R

CM

O

r

1m

2m

1r

2r

CM

Or

0

R

Mass ofCenter CM

)(a )(b

Figure. Two methods to describe the position of two particles.

(a) From an arbitrary coordinate system origin, and

(b) from the center of mass.

The force points radially and depends only on r. )(rF

1- Reduced Mass

The relative vector:

1r

2r

and are the position vectors for, respectively, m1 and m2.

21 rrr

The center of mass vector: 21

2211

mm

rmrmR

Translational motion of the system is uninteresting, so we use 0

R

00 2211

rmrmR

0)( 1211

rrmrm

0)( 2121

rmrmm

rmm

mr

21

21

0)( 2221

rmrrm

0)( 2211

rmmrm

rmm

mr

21

12

1- Reduced Mass

Lagrangian: )(2

1

2

1 2

22

2

11 rUrmrmL

Substitute in the Lagrangian and : 1r

2r

)(2

1

2

1 22

21

12

22

21

21 rUr

mm

mmr

mm

mmL

)(1

2

11

2

1 22

21

21

2

22

21

21

1

rUrmm

mm

mr

mm

mm

mL

)(11

2

1

21

22

21

21 rUmm

rmm

mmL

)(2

1

21

212

2

21

21 rUmm

mmr

mm

mmL

1- Reduced Mass

)(2

1 2

21

21 rUrmm

mmL

21

21

mm

mm

Reduced mass

)(2

1 2

rUrL

We reduced the problem of the motion of two bodies to an equivalent

one-body problem in which we must determine only the motion of a

“particle” of mass in the central field described by the potential

function U(r).

x

y

z

r

2- Conservation Theorems

Conservation of angular momentum

We define the angular momentum as prL

The vector is orthogonal to both and , by nature of the cross

product.

L

r

p

prdt

d

dt

Ld

dt

pdrp

dt

rd

dt

Ld

0

Frvvdt

Ld

0

Fr because, is proportional to and the cross product of

two parallel vectors is zero.

F

r

constant prL



If a particle is subject to a central force only, then its angular

momentum is conserved, and its motion takes place in a plane.

r

x

y

We have only a two-dimensional

problem, and the Lagrangian may

then be conveniently expressed in

a plane polar coordinates

rUrrL 222

2

1

)(2

1 2

rUrL

constant0

p

Lp

constant2

r

Lp



constant2 r

positive.or negative becan :Note

:by Denote p

Conservation of total energy

Discussion is restricted to conservative systems (i.e. E=constant)

constant UTE

)(2

1)(

2

1 2222 rUrrErUvE

)(2

1

2

12

22 rU

rrE

2

:Rememberr

3- Equations of Motion

)(2

1

2

12

22 rU

rrE

Assume U(r) is specified, and solve for ṙ :

2

2

2

2

1

2

1)( r

rrUE

22

22 )(

2

rrUEr

22

22

rUE

dt

drr

This equation can be solved for dt and integrated to yield the solution

t = t(r). An inversion of this result then gives the equation r = r(t).


We are interested in the equation of the path in terms of r and .

Starting with:

drr

drdr

dt

dt

dd

Substitute:

2r

22

22

rUEr

and

Integrating, we have equation of the orbit = (r):

)17(constant

22

)(

2

2

2

rUE

drr

r


We can also attack the problem using Lagrange’s equation for the

coordinate r:

0

r

L

dt

d

r

L

)(

2

1 222 rUrrL

r

rUr

r

L

)(2 rr

L

dt

dr

r

L

and ,

0)(

0 2

r

r

rUr

r

L

dt

d

r

L

)18()( 2 rr

r

rU


Making a change of variable: r

u1

r

rd

dt

dt

dr

rd

dr

rd

du

d

dr

dr

du

d

du222

111

2r

We have

rrr

rr

r

rd

du

2

222

11

Next, we write

r

dt

d

d

dtr

d

d

d

du

d

d

d

ud

2

2


rrd

ud

12

2

rrr

r

d

ud

r

2

2

2

2

2

2

2

2

2

22

2 1

d

ud

rr

)19(2

22

2

2

ad

udur

32

22

42

22

2

1

1

rr

rr

)19(3

2

22 bur

Substituting Equations (19-a) and (19-b) into Equation (18), we obtain

the transformed equation of motion:


3

2

2

2

22

2

21u

d

udu

uF

uF

uu

d

udu

d

udu

uF

111222

2

2

22

2

which we may also write as

)21(11

2

2

2

2

rFr

rrd

d

This form of the equation of motion is useful if we wish to find the

force law that gives a particular known orbit r = r().

3- Equations of Motion – Example 1

Problem:

Find the force law for a central field that allows a particle to move

in a spiral orbit given by r = k 2, where k is a constant.

Solution: We use Equation (21) to determine the force law F(r).

First, we determine

3422

2211

kk

k

kd

d

rd

d

462

2

32

2 1632211

kk

k

kd

d

rd

d

d

d

rd

d

2

2

42

2422 1

,,r

k

k

r

k

rkr


22

2

2

2 661

r

k

r

k

krd

d

Then

rFr

rrd

d2

2

2

2 11

From Equation (21): , we now determine F(r)

)(16

2

2

2rF

r

rr

k

rr

k

rrF

16)(

22

2

34

2 16)(

rr

krF


Problem:

Determine r(t) and (t) for the problem in Example 1.

Solution:

422

2

r

kdt

dr

dtk

ddtk

d2

4

42

Rearrangement of the above equation gives

Ctk

dtk

d2

5

2

4

5

1

where C is an integration constant


51

25

5)(

Ct

kt

or

51

2

5)(

Ct

kt

We solve for r(t): 52

2

2 5)(

Ct

kktrkr

52

5)(

Ct

ktr

The integration constant and angular momentum needed for

equations r(t) and (t) are determined from the initial conditions.

C


Problem:

What is the total energy of the orbit of the previous two examples?

Solution: The total energy is

We need ṙ and U(r).

)(2

1

2

12

22 rU

rrE

3

2

4

2

34

2 616)(

r

dr

r

drkdr

rr

krU

drrFrU )()(

.2

1

3

16)(

2

2

3

2

constrr

krU

2

2

3

2 1

2

12)(

rr

krU

where we have let U() = 0


2r

dr

dr

dt

dr

dr

d

dt

d

2rdr

dr

and

d

dr

rr

2

k

rk

d

drkr

and,22 rk

k

rk

d

dr22 :so

rkr

r 22

32

22 14

r

kr

and

We have the total energy of the orbit is )(2

1

2

12

22 rU

rrE

02

2

2

4

2

12

2

3

2

2

2

32

2

rr

k

rr

kE

E = 0 if U(r = ) = 0

4- Centrifugal Energy and the Effective

Potential

In the preceding expressions for ṙ, (r), and so forth, a common

term is the radical 2

2

2)(

rrUE

2

2

2)(

rrU c

If we interpret the quantity as a “potential energy”,

then the “force” that must be associated with Uc is

2

3

2

r

rr

UF c

c

Fc: centrifugal force (traditionally called)

Uc: centrifugal potential energy of the particle.


Potential

2

2

2)()(

rrUrV

effective potential energy

For the case of inverse-square-law central-force motion:

2)(

r

krF (gravitational attraction (F(r) < 0)), from which

constant)()(r

kdrrFrU

0constant0)( UAssume

The effective potential function for gravitational attraction:

2

2

2)(

rr

krV


Potential

r

1E

2E

3E

2

2

1r

4r3r2r

1r

2

2

2)(

rr

krV

)(

2

1 2 rVrE

)(02

1 2 rVEr

Figure: The effective potential, V(r), for gravitation attraction.


Potential

• If E ≥ 0 (e.g. E = E1): The particle moves toward the force center

(located at r = 0) from infinitely far away until it “strikes” the

potential barrier V(r1) at the turning point r = r1 and is reflected

back toward infinitely large r. the motion is unbounded r1 ≤ r <.

• If Vmin < E < 0 (e.g. E = E2): the values r2 and r4 are the turning

points, or the apsidal distances, of the orbit. The motion is bounded

r2 ≤ r ≤ r4.

• If E = Vmin (e.g. E = E3): r = r3, and then ṙ = 0 for all values of

the time; hence the motion is circular.

5- Planetary Motion – Kepler’s Problem

r

krU

r

krF )()(

2If

From Equation (17) we obtain:

constant

22

2

2

2

rr

kE

drrr

The integral can be evaluated if the variable is changed to r

u1

drr

du

2

1

constant

2

2 22

2

ukuE

dur


constant22 2

22

uu

kE

dur

Using the standard form of the integral

constant4

2sin

12

1

2

acb

bax

acbxax

xd

obtain we,2

and ,2

;1For 22

Eck

ba

2

2

2

21

82

22

sin)1(

1constant

Ek

ku


2

2

2

2

82

22

constant)sin(

Ek

k

r

As r goes to zero, the total energy E goes to infinity, and if we define

the origin of so that the minimum value of r is at = 0, then

2constant1)constant0sin(

cos)2

sin()constantsin(

2

22

2

2

2

2

2

2

2

8

21

2

12

2

2

82

22

cos

E

k

k

rk

k

Ek

k

r


2

2

2

21

11

cos

k

E

rk

2

2

2

21

11

cos

k

E

rk

Let us now define the following constants:

k

2 2

221

k

E

and

11

cos

rthen

cos1r

This is the equation of a conic section with one

focus at the origin. Conic sections are formed by

the intersection of a plane with a cone.

: eccentricity

2: latus rectum of the orbit


CircleVE0

Elipse0V10

Parabola0E1

Hyperbola0,1

min

min

E

E


axissemimajor 21 2 E

ka

For planetary motion, the orbits are ellipses.

axissemiminor 21 2 E

b

FF

a

a

b

minr

maxr

Figure 6. The geometry of

elliptic orbits is shown in

terms of parameters , ,

a, and b. F and F’ are the

foci.

11min ar

11max ar


Areal Velocity

d

)( 2tr

)( 1tr

rd

o

)()( 12 trtrrd

dtvrd

rdtrdA

)(2

1 dtvtrdA

)(2

1

)(2

1)(

2

1 arararvtr

dt

dArr

constant2

2

1

2

2

dt

dA

r

rdt

dA

Thus, the areal velocity is constant in time


Period for Elliptic Motion (τ)

dAdtdt

dA

2

2

A

dAdt00

2

A

2

The area of an ellipse is given by abA

EE

kab

22

22

)46(22

322223 EkEk



)47(2

2;

22

333

k

a

a

kE

a

kE

E

ka

From Equations (46) and (47), we obtain:

3

222 2

2

k

ak

3

22 4

ak

For planetary motion, we identify k = Gm1m2

2

21

2 r

mmG

r

krF

2211 /10674.6 kgmNG

G: gravitational constant


3

21

23

21

21

212

2

)(

44

ammG

amGm

mm

mm


32

2 4a

k

21

3

2

22 if

4mma

mG

Kepler’s problem

21

3

2

22 condition put thet didn' He .

4 Kepler,For mma

mG

The Kepler’s statement is correct only if the mass m1 of a planet

can be neglected with respect to the mass m2 of the Sun.


1- Planets move in elliptical orbits about the Sun with the Sun at

one focus.

2- The area per unit time swept out by a radius vector from the Sun

to planet is constant (dA/dt is constant).

3- The square of a planet’s period is proportional to the cube of the

semimajor axis of the planet’s orbit (τ2 proportional to a3).

Kepler’s Laws

Chapter 5

Motion in a Noninertial

Reference Frame

Dr. Abdelaziz Sabik

Physics Department – College of Science


1- Reference Frame

hkm /70 hkm /80

observer Stationary

A""Car B""Car

- To a passenger in car “A”, the speed of car “B” is 10km/h.

- To the stationary observer, the speed of car “B” is 80km/h.

Question: Which observer is correct ?

Answer: They both areǃ

This simple example demonstrates that the velocity of an object

depends on the frame of reference in which it is measured.

1- Reference Frame

Conclusion:

For the laws of motion to have meaning, the motion of bodies must

be measured relative to some reference frame.

Inertial Reference Frame

A reference frame is called an inertial frame if Newton’s laws are

valid in that frame.

For example, if a body subject to no force moves with constant

velocity in a certain coordinate system, that system is, by definition,

an inertial frame.

Consequence:

An inertial reference frame is one that is not accelerating.

1- Reference Frame

Fictitious Forces

A fictitious force, also called a phantom force, pseudo force,

d’Alambert force or inertial force, is an apparent force that acts on

all masses in a non-inertial reference frame.

The fictitious force, Ff , does not arise from any physical interaction

but rather from the acceleration a of the non-inertial reference frame

itself.

Example of fictitious forces in linear motion

0on Accelerati a

xy

x

y

observer

Inertial

observer

lNoninertia

A small sphere of mass m is hung

by a cord from the ceiling of a

car that is accelerating to the right.

1- Reference Frame

gm

T

According to the inertial observer at rest

- The forces on the sphere are the force T exerted

by the cord and the force of gravity.

- The acceleration of the sphere is the same as that

of the car.

- The acceleration is provided by the horizontal

component of T.

- The vertical component of T balances the force

of gravity.

)1(0cos

sin

mgTF

maTFamgmTF

y

x

1- Reference Frame

According to the noninertial observer riding in the car

-The cord still makes an angle with the vertical.

- The sphere is at rest and so its acceleration is zero.

- Introduction of a fictitious force to balance the horizontal

component of T.

- The net force on the sphere is zero.

gm

T

fF

)2(0cos

0sin0

mgTF

FTFamFgmTF

y

fx

lnoninertiaf

1- Reference Frame

Remarks concerning the example:

1- If we recognize that Ff = mainertial = ma, then (1) and (2) are

equivalent; therefore, the noninertial observer obtains the same

mathematical results as the inertial observer does. However, the

physical interpretation of the deflection of the cord differs in the

two reference frames.

2- The introduction of the fictitious force permitted us to apply

Newton’s second law to the sphere in noninertial frame.

3- In noninertial frame, the effective force is defined as

lnoninertiaamFgmTF feff

2- Rotating Coordinate Systems

Consider two systems of coordinate axes:

- "fixed" or inertial axes (O׳,x׳,y׳,z׳)

- "rotating" axes (O,x,y,z) that may be in motion with respect to the

inertial system.

r

PP

r

d

d

d

d

rd

rdr

x

y

x

y

x

x

y

y

OO OO

zz page ofout

Figure 1. two systems of coordinate axes. System (O,x,y,z) rotates

by an angle d with respect to (O׳,x׳,y׳,z׳). P is fixed in (O,x,y,z).


drrdfixed

In vectorial form: rdrdfixed

rotation of axis

ousinstantane theis where

d

rdt

d

dt

rd

fixed

dt

d

:rotation theoflocity angular ve The

systemcoordinate rotatingin fixed for Prdt

rd

fixed

where the designation “fixed“ indicates that the

quantity dr is measured in the fixed coordinate

system.


:system coordinate

fixed in the of change of rate time then thesystem coordinate

rotating therespect to with 0 velocity a has Ppoint If

r

dt

rd

rotating

)3(rdt

rd

dt

rd

rotatingfixed

The validity of Equation (1) is not limited to a displacement vector.

In fact, for any arbitrary vector , we have

)4(Qdt

Qd

dt

Qd

rotatingfixed

Q


:for (4)Equation Applying

Q

Angular acceleration is the same in both the fixed and rotating

systems.

rotatingrotatingfixed dt

d

dt

d

dt

d

0

)5()()( rotatingfixed


R

r

r

Ox

y

z

O

xy

z

P

)( axes fixed the

torelative O ofposition vector :

.)( axes fixed the

torelative P ofposition vector :

.)( axes rotating the

torelative P ofposition vector :

zyxO

R

zyxO

r

Oxyz

r

Consider now and the system (O,x,y,z) rotates arbitrary

relative to (O’x’y’z’). OO

fixedfixedfixed dt

rd

dt

Rd

dt

rdrRr


rdt

rd

dt

Rd

dt

rd

rotatingfixedfixed

fixed

ffdt

rdrv

fixed

fdt

RdRV

We define

velocity of point P relative to the fixed axes

linear velocity of the moving origin, O, relative

to the fixed axes.

rotating

rrdt

rdrv

velocity of point P relative to the rotating axes

)6(rvVv rf

therefore


x

y

z

O

xa

ya

za

tx

ty

tzxxa

yya

zza

zzyyxx aaa ˆˆˆ

axes rotating theoflocity angular ve theis

axes moving theof

rotation the todue velocity theis r

3- Centrifugal and Coriolis Forces

Newton’s equation, , is valid only in an inertial frame

of reference. amF

The sum of the forces acting on the particle as measured in the fixed

inertial system:

(7)

fixed

f

fdt

vdmamF

)8(fixedfixed

r

fixedfixed

f

dt

rdr

dt

vd

dt

Vd

dt

vd

From Equation (6):

)9(

fixed

fdt

VdR

rrr

rotating

r

fixed

r vavdt

vd

dt

vd

)10(


rotating

rr

dt

vda

where is the acceleration of the particle in the

rotating coordinate system (O,x,y,z).

rdt

rd

dt

rd

rotatingfixed

)11(rvdt

rdr

fixed

Combining Equations (6) – (11), we obtain

)12(2 rrff vmrmrmamRmamF


To an observer in the rotating coordinate system, the effective force

on a particle is defined as

)13(reff amF

)14(2 rfeff vmrmrmRmFF

system fixed the torelative system

coordinate moving theofon acceleratiangular the todue :

system. fixed the torelative system

coordinate moving theofon accelerati onal translati the todue :

system. inertial fixed

in the measured as particle on the acting forces theof sum the:

rm

Rm

F

f


.force Coriolis thecalled is termThis system.

coordinate rotating in the particle theofmotion todue :2

below). figure (seerotation ofcenter

thefrom outward directed is and force lcentrifuga :)(

rvm

rm

r

r

O

)( rm


Centrifugal and Coriolis forces are not forces in the usual sense of the

word. They are fictitious forces introduced by our desire to write

reff amF

We thus have: termslnoninertia rmRmamF ffeff

rv

rvcoF

coF

m

m

path Deflected

Figure. Overhead view of a horizontal

turntable that is rotating counterclockwise

relative to an inertial frame. The turntable’s

angular velocity points up out of the page.

as seen by observers on the turntable, the

two objects sliding on the table are subject

to Coriolis forces. Irrespective of the masses’

positions and velocities, the Coriolis force

always tends to deflect the velocity to the right.

4- Motion Relative to the Earth

The Earth is not really an inertial frame for the following reasons:

1- its orbital motion around the Sun experiences a centripetal

acceleration of about 4.4×10-3m/s2 directed toward the Sun.

2- its rotational motion about its own axis once every 24 hours, a

point on the equator experiences an additional acceleration of

33.710-3m/s2 directed toward the center of the Earth.

3- the motion of the solar system with respect to the local galaxy…etc


The motion of Earth with respect to an inertial reference frame is

dominated by Earth’s rotation about its own axis. The effects of

the other motions are small by comparison.

Therefore let’s explicitly take the Earth’s surface as the rotating

frame and the center of Earth as the fixed frame.


-The origin of the fixed

frame of reference at the

center of Earth (point O׳

in Figure).

-The origin of the rotating

frame of reference at latitude

λ and longitude (point O in Figure).

zz aggaRR

gR

ˆ and ,ˆ

: and toparallel z-Axis -

00

0

-Axis-x tangent to a great circle passing

through the North and South poles.

xzy aaay ˆˆˆsuch that is -Axis-

xa

za

x

y

z

N

S

EW

O

O

za ˆ

xa ˆ

ya ˆ

a

a

zaRR ˆ

0g

ya

zagg ˆ00

za ˆ

)(upz

)(easty

)(southx


)14(2 rfeff vmrmrmRmFF

We can describe the motion of a moving object close to the surface

of Earth by applying Equation (14):

In the fixed inertial frame: )15(0gmSF

Earth. toattraction nalgravitatio :gm

ngravitatio other than forces external theof sum :where

0

S

2

020 /8.9,ˆ smgaR

MGg z

E

)6378 (Earth spherical a assuming radius sEarth' the:

)109736.5(Earth of mass :

)/.10674.6(constant nalgravitatio :G

24

2211

kmRR

kgM

kgmN

E


Rdt

Rd

dt

RdR

rf

f

0ˆ

r

zdt

RdaRR

RR f

and

therefore

ffff

fdt

RdR

dt

d

dt

Rd

dt

RdR

)(

)/103.7(

in timeconstant y practicall is locity angular ve sEarth' of valueThe

5 srad

rneglect weSo)16(0

fdt

d

)17(Rdt

RdR

f

f


From Equations (14) to (17), we obtain:

reff vmrRmgmSF

20

force lcentrifuga theis rRmFCf

r

r

O

)( rm

z

rRr

Figure. Centrifugal force directed

outward from the center of rotation.


g effective theasEarth of surface the

on measure) (and experience what weis 0

rRgg

The direction of g at a given point

is in general slightly different from

the true vertical. The situation is

represented schematically in figure

with considerable exaggeration.

0g

g

)( Rr

force Coriolis theis 2 rCo vmF

In simpler terms: )18(Coeff FgmSF

4- Motion Relative to the Earth – Example 1

)ˆsinˆ(coscosˆcos)(

)ˆˆ(cos)(

)ˆ)2

(sin(ˆ)(

)ˆˆ(ˆ )(

)ˆˆ(ˆ)(

22

2

2

2

xz

yz

yz

zzz

zzz

aaRaRR

aaRR

aaRR

aaaRR

aRaaR

Problem:

Show that the small angular deviation δ of a plumb line from the

true vertical (i.e., toward the center of Earth) at a point on Earth’s

surface at a latitude λ is

Solution:

22

0

2

cos

cossin

Rg

R

Rgg

0

xa

zaza ˆ

a

ya


xzz aRaRagg ˆsincosˆcosˆ 222

0

xz aRaRgg ˆsincosˆcos 222

0

22

0

2

cos

cossintan

Rg

R

g

g

z

x

tan small very tan

22

0

2

cos

cossin

Rg

R

therefore

g

x

z

zg

xg


Problem: (Refer to Example 1)

What is the value of the maximum deviation (max)?

Solution:

)cos(2

2sin22

0

2

Rg

R

222

0

222

0

2

)cos(

))sin(cos2(2sin)cos(2cos2

2

Rg

RRgR

d

d

222

0

222

0

2

)cos(

)sincos2(2sin)cos(2cos2

2

Rg

RRgR

d

d

0sincos2sin2cos)cos(0 222

0

RRg

d

d


0sincossincos2)sin)(coscos( 22222

0 RRg

0sincos2sincoscossincos 222222422

0

2

0 RRRgg

0sincoscossincos 222422

0

2

0 RRgg

2

0

2222

0

2 sin)sincos(cos gRRg

2

0

2

0

2 sin)(cos gRg

45;tantan0

2

01

0

2

02

g

Rg

g

Rg

rad 002.0)45cos(2

90sin isdeviation maximum the

rad/s103.7 and m/s8.9 ,km6378 Using

22

0

2

max

52

0

Rg

R

gR


Problem:

Find the horizontal deflection from the plumb line caused by the

Coriolis force acting on a particle falling freely in Earth’s gravitational

field from a height h above Earth’s surface.

Solution: reff vmgmSF

2

freely) falling (particle0

S

reff amF

rrrr vgavmgmam

22

From the above equations, we obtain

)ˆsinˆcos(ˆzxz aaa


zyx

aaa

v

zyx

r

sin0cos

ˆˆˆ

zyxr ayaxzayv ˆ)cos(ˆ)sincos(ˆ)sin(

zzyxr aggazayaxa ˆandˆˆˆ

cos

)sincos(

sin

20

0

y

xz

y

gz

y

x

cos2

).()sincos(2

sin2

ygz

aEqxzy

yx

Let’s solve these using the

method of successive

approximations.


First write the equations for ω = 0. This is the zeroth-order

approximation.

This gives the following equations:

ion)approximatorder zeroth (,0,0 gzyx

After integration, these lead to: ).(,0,0 bEqgtzyx

After an other integration, we get the solutions:

2

2

1,0,0 gthzyx

where we have used initial conditions appropriate to dropping the

object from rest from high h.


cos2

).()sincos(2

sin2

ygz

aEqxzy

yx

).(,0,0 bEqgtzyx

We have:

We then plug Equations (b) into our original equations (a) to get the

first-order approximation:

ion)approximatorder -(first,cos2,0 gzgtyx

Integrating twice, we get: 23

2

1,cos

3

1,0 gthzgtyx

The only equation that has changed is the y equation, which shows

that there will be an eastward deviation (“y” has an easterly direction

as already mentioned in this section).


We can repeat this process again to get a second-order approximation

and so on, but the first order is good enough for our purposes which

to show the effect of Coriolis force on the free falling particle.

The time, T, for which the particle reaches the bottom is:

g

hTgThz

20

2

1 2

The horizontal deflection is then:

cos8

3

13

3

g

hgy cos

8

3

1or

3

g

hy

An object dropped from a height of h=100m at latitude =45 is

deflected approximately 1.55cm (neglecting the effect of air resistance)

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PHYS.303-CLASSICAL MECHANICS II - LECTURE NOTES"PPT"