Chapter VII Hamilton’s Principle-

Lagrangian & Hamiltonian Mechanics

.

Recommended problems: 7-1, 7-2, 7-3, 7-4, 7-6, 7-7, 7-10, 7-12, 7-13, 7-14,

7-15, 7-25, 7-26, 7-27, 7-29, 7-33, 7-34, 7-37, 7-39,

Hamilton’s Principle

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

point to another within a specific time interval (consistent with any constraint),

the actual path followed is that which minimizes the time interval of the

difference between the kinetic and potential energies, i.e.,

Defining the difference of T-U to be the Lagrangian as

)1.7(02

1

dtUTt

t

Eq.(7.1) now reads

)2.7(;, iiii xUxTtxxL

This means that the integral of T-U must be an extremum.

)3.7(0;,2

1

t

tii dttxxL

Comparing Eq.(7.3) and Eq.(6.3) with

We get the Euler’s equations as xyxytxLf ,

EquationLagrangex

L

dt

d

x

L)4.7(0

Example Find the equation of motion of the 1-Dimensional harmonic oscillator.

Solution The Lagrangian of the system is

2212

21 kxxmUTL

xm

x

Lkx

x

L

&

0xmdt

dkx 0xmkx 0 x

m

kx

Example Find the equation of motion of the simple pendulum shown.

Solution With respect to the top, the potential

energy of the ball is

m

L

Lcos

cosmgLU

And for the kinetic energy we have

22212

21 mLmvT

cos2221 mgLmLL

2&sin mLL

mgLL

Using Lagrange equation we get

0sin 2 mLdt

dmgL 0sin 2 mLmgL

0sin L

g

Generalized Coordinates

Generalized coordinates are any set of independent coordinates qi (not

connected by any equations of constraint) that completely specifies the state of

a system. The required number of generalized coordinates is equal to the

system’s number of degrees of freedom.

A single particle free to move in 3-dimensional space requires 3-coordinates to

specify configuration and hence has a 3-degrees of freedom. n-free particles

would require 3n coordinates and so on. .

For each constraint equation the number of generalized coordinated is

decreased by one coordinate. This means that if there are m equations of

constraints, then 3n-m coordinates are independent, and the system is said to

posses s=3n-m degrees of freedom. The equations of constraints must be

expressible of the form

)5.6(,2,1,2,10, mknitxf ik

Constraints that can be expressed in the form of Eq.(6.5) is called holonomic

constraint, otherwise it is nonholonomic.

Since the Lagrangian is a scalar function it is invariant to coordinates

transformation. In terms of the generalized coordinates, the Lagrange’s

equations can be written as

EquationsLagrangesjq

L

dt

d

q

L

jj

)6.6(,2,10

There are s of these equations, and together with the m equations of

constraints and the initial conditions, they completely describe the motion of the

system.

It is important to realize that the validity of Lagrange’s equations requires the

following 2-conditions:

1- The forces acting on the system (apart from any constraint forces) must

be conservative.

1- The Equations of constraints must be in the form of Eq.(6.5), i.e., the

constraint must be holonomic..

Example Find a suitable set of generalized coordinates for a point particle

moving on the surface of a hemisphere of radius R whose center is at the

origin.

Solution The particle is constrained to move on the surface, so we have

022222222 RzyxRzyx

yqxq 21 ,

If we choose the Cartesian coordinates our generalized coordinates will be

222 yxRz

Example Use the (x,y) coordinates system as shown in

the figure to find the K.E. T, P.E. U, and the Lagrangian L

for a simple pendulum moving in the x-y plane. Determine

the transformation equations from the (x,y) coordinates to

the coordinates . Find the equations of motion.

Solution Using the rectangular coordinates we have

2221 yxmT mgyU

mgyyxmUTL 2221

To transformation x&y into the coordinates we have

cos&sin lylx sin&cos lylx

cos222122

21 mglmlmgyyxmL

To find the equation of motion we apply Lagrange’s equation (Eq.7.4). We have

2sin ml

Lmgl

L 0sin

2

dt

mldmgl

0sin l

g

Example Consider the problem of a

projectile motion under gravity in 2-

dimensions. Find the equations of

motion in both Cartesian and polar

coordinates. x

y

vo

Solution Using the Cartesian coordinates

we have, taking U=0 at y=0,

2221 yxmT mgyU

mgyyxmUTL 2221

Here we have 2-generalized coordinates, x & y. So Lagrange’e equations give

0&0

y

L

dt

d

y

L

x

L

dt

d

x

L

For the x-coordinate we have 00 xmdt

d 0x

0ymdt

dmg gy

In polar coordinates, the generalized coordinates are r & . Now we have

sin22221 mgrUrrmT

sin22221 mgrrrmL

For the r-coordinate we have

0sin0sin 22 rgrrmdt

dmgmr

For the -coordinate we have

02cos0cos 22 rrrgrmrdt

dmgr

0

r

L

dt

d

r

L

0

L

dt

dL

Lagrange’e equations give for the 2-generalized coordinates, r & gives

Example A particle of mass m is constrained to move

on the inside surface of a smooth cone of half-angle

. Determine the generalized coordinates and the

constraints. Find the equation of motions.

Solution here we use the cylindrical coordinates r, , & z. The equation of constraint is

cotrz

So we have 2-degrees of freedom and the generalized coordinates are r & .

Now

22222122222

212222

21 csccot rrmrrrmzrrmT

cotmgrmgzU cotcsc 222221 mgrrrmL

For the r-coordinate we have

0

r

L

dt

d

r

L

0csccot0csccot 2222 rgrrmdt

dmgmr

For the -coordinate we have

0

L

dt

dL

constant00 22 mrmrdt

d

constant2 mrILBut

So we recover the conservation of angular momentum about the axis of

symmetry of the system.

Example The point of support of a simple

pendulum of length b moves on a massless rim

of radius a rotating with constant angular

velocity . Obtain the equation of motion of m.

Solution Taking the origin of the coordinates

be at the center of the rotating rim we get

sincos btax cossin btay

cossin btax sincos btay

The K.E & P.E are now

tbabamyxmT sin222222122

21

cossin btamgmgyU

cossinsin2222221 btamgtbabamL

The only generalized coordinate is . SO

sincos mgbtmbaL

tbabmL

sin2

tmbambL

dt

dcos2

Applying the Lagrange’e equation we get

sincoscos2 mgbtmbatmbamb

sincos22 mgbtmbamb

sincos2

b

gt

b

a

Note that for =0 we get

0sin b

g

Which is the equation of motion of the simple pendulum.

Example Find the frequency of a simple

pendulum placed in a rail-road car that has

a constant acceleration a in the x-direction.

Solution We choose a fixed coordinate

system with x=0, and v=vo at t=0. The

position of the mass m is

221sin attvlx o cosly

atvlx o cos sin ly

The K.E & P.E are now

2212

2122

21 sincos lmlatvmyxmT o

cosmglmgyU

cossincos2

212

21 mgllmlatvmL o

The only generalized coordinate is again . SO

sincossincossin 22 mglmllatvml

Lo

m

l

a

m

l

sinsin mglatvmlL

o

22 sincoscos

mllatvmlL

o

Applying the Lagrange’s equation

0

L

dt

dL

)(cossin HWasprovel

a

l

g

If the mass doesn’t oscillate with respect to the train, then 0

lll

a

l

g cossin0

g

al tan

Because the oscillations are small we can assume smallveryisl

lll

a

l

gcossin

Expand the sine and cosine functions we get

sinsincoscossincoscossin lllll

a

l

g

Using Taylor expansion for sin and cos to first order of we get

lllll

a

l

g sincoscossin

llll agagl

sincoscossin1

The first bracket is already zero leaving ll agl

sincos

2222cos&sintan

ga

g

ga

a

g

aAs l

2222

22ag

lag

agl

022

l

ag

We have SHM with frequency l

ag 222

Note that when the car is at rest (a=0) we get lg2

Example A bead slides along a smooth wire bent in the

shape of a parabola z=cr2. The bead rotates in a circle of

radius R when the wire is rotating about its vertical

symmetry axis with angular velocity . Find the value of c.

Solution Choosing the cylindrical coordinates we get

222221 rzrmT mgzU

The equation of constraint is rcrzcrz 22

222222221 4 mgcrrrrcrmL mgcrrrrcm

r

L24 222

222222 16822

14 rrcrrcr

r

L

dt

drrcrm

r

L

02441 22222 gcrrcrrcr

This equation can be simplified if r=R=constant. In such a case we have

02 2gc

gc

2

2

Example Consider the double pulley

system shown. Use the coordinates

indicated to determine the equations of

motion.

Solution If l1 & l2 be the lengths of the

rope hanging freely from each pulley. The

distances x & y are measured from the

center of the pulleys. So we have

ylxlxyxlxxx 213121 &&

yxxyxxxx 321 &&

23212

2212

121

2332

12222

12112

1

xymxymxm

xmxmxmT

ylxlgmxylgmgxmUUUU

213121

321

ylxlgmxylgmgxm

xymxymxmL

213121

232

1222

1212

1

Here we have 2-generalized coordinates, x & y. For the x-coordinate we have

0

x

L

dt

d

x

L

gxmgxmgmx

L321

xymxymxmx

L

321

gmmmyxmyxmxm 321321

For the y-coordinate we have

0

y

L

dt

d

y

L

gymgymy

L32

xymxym

y

L

32

gmmyxmyxm 3232

Lagrange’s Equations with Undetermined Multipliers

As we mentioned before, Holonomic constraint is the constraint that can be

expressed in the form of Eq.(7.5). Such a constraint enables us to find an

algebraic relations between coordinates. Any constraints that can be expressed

in terms of the velocities, i.e., in the form

Such constraints is nonholonomic unless the equations can be integrated to

yield relations among the coordinates. It is called semiholonomic.

)7.6(,2,1,2,10,, mknitxxf iik

Consider a nonholonomic constraint in the form of

)8.6(,2,10 niBxAi

ii

But if Ai & B have the forms

)9.6(,,& txfft

fB

x

fA

ii

Then Eq.(6.8) becomes

)10.6(0

dt

df

t

f

dt

dx

x

f

i

i

i

0constant, txf So the constraint is actually holonomic.

From Eq.(10) we conclude that any constraint expressible in the form

)11.6(0

dt

t

fdq

q

f

ii

i

Are equivalent to the constraint given by Eq.(6.5). Referring to the previous

chapter (Eq. 6.23) we can write

)12.6(0

ikk

ii q

ft

q

L

dt

d

q

L

)13.6(ik

kiq

ftQ

With k(t) are called the Lagrange undetermined multipliers. It its related to the

force of constraint. The generalized forces of constraint are given by

Example Reconsider the case of the disk

rolling down an inclined plane. Find the

equation of motion, the force of constrained,

and the angular acceleration.

Solution The K.E. & P.E of the disk are

22412

212

212

21 MRyMIyMT

sinylMgU

Where l is the length of the inclined plane. The Lagrangian is therefore

sin22412

21 ylMgMRyML

The equation of constraint is 0, Ryyf

Although the system has only one degree of freedom, we may continue to

consider both y & as generalized coordinates and use the method of

Lagrange undetermined multipliers. Eq.(6.12) now reads

0&0

fL

dt

dL

y

f

y

L

dt

d

y

L

1&&sin

y

fyM

y

LMg

y

L

0sin yMMg

R

fMR

LL

&

2

1&0 2

0

2

1 2 RMR

These 2-equations together with the equation of constraint is sufficient to solve

for the 3-unkowns. Now from the equation of constraint we have

RyRy

3

sin&

3

sin2,

3

sin2

Mg

R

ggy

To find the generalized forces we use Eq.(6.13) to get

3

sin

Mg

y

fQy

3

sin

MgRR

fQ

These 2-genralized forces are, respectively, the force of friction and the torque

required to keep the disk rolling without slipping.

Note that the problem can be solved using Eq.(6.6) by eliminating one of the

coordinates and keeping only one-generalized coordinates.

Example A particle of mass m starts at rest

on top of a smooth fixed hemisphere of

radius a. Find the force of constraints, and

determine the angle at which the particle

leaves the surface.

Solution The Lagrangian of the particle is

cos22221 mgrrrmUTL

And the equation of constraint is 0, aryf

The generalized coordinates are r & and Eq.(6.12) now reads

0&0

fL

dt

dL

r

f

r

L

dt

d

r

L

1&&cos2

r

frm

r

Lmgmr

r

L

0cos2 rmmgmr

0&&sin 2

fmr

Lmgr

L

02sin 2 mrrmrmgr

0&,0 rrarBut

0cos2 mgma 0sin 2 mamga sina

g

To find we have

sin

a

g

d

d

dt

d

d

d

dt

d

00

sin da

gd

a

g

a

g

cos

2

2

2cos32cos2coscos 2 mgmgmgmgmamg

The particle falls off the surface when =0 02cos3 omg

o1 2.483

2cos

o

Note that at the top of the hemisphere we have mg

Conservation Theorems Revisited

The Lagrangian that describes a closed system does not depend on time

explicitly, i.e.,

0

t

L)14.6(j

j jj

j j

qq

Lq

q

L

dt

dL

Using Eq.6.6, Eq.6.14 becomes

Conservation of Energy

j

j jj jj q

q

L

q

L

dt

dq

dt

dL

0

j jj

q

Lq

dt

d

dt

dL

)15.6(0or

j jj

q

LqL

dt

d

)16.6(constant Hq

LqL

j jj

If the potential energy doesn’t depend on velocities nor on time, explicitly, i.e.

U=U(qj)

jjj q

T

q

UT

q

L

Eq.6.16 becomes

)17.6(Hq

TqUT

j jj

T

q

Tq

j jj 2But

HTUT 2 )18.6(HEUT

This tells that the total energy is conserved

Conservation of Linear Momentum

Let the Lagrangian be given as , The change in L caused by

infinitesimal displacement is

ii xxLL ,

i

iiexr ˆ

)19.6(0

ii

iii

i

xx

Lx

x

LL

0But ix

0

ii

i

xx

LL )20.6(0

ix

L

From Lagrang’s Equation we conclude that

0

ix

L

dt

d

)21.6(constant

ix

L

constant

ii x

T

x

UT

)22.6(But constant2

21

iij

jii

pxmxmx

T

x

T

The linear momentum of a closed system is constant in time.

Conservation of Angular Momentum

If a system is rotated about an axis by an

infinitesimal angle , the radius r changes to r +r

r r+r

r

From the figure it clear that

sinBut rr

).(r 236o rr

).(similarly 246rr

)25.6(0

ii

iii

i

xx

Lx

x

LL

The change in L caused by infinitesimal rotation is

i

i

px

L

haveweEq.(6.21)fromBut

i

i

px

LsLagran haveweEq.'from

Eq.(6.25) now becomes

)26.6(0 i

iii

ii xpxpL

).(or 2760 rprp

Using Eqs.(6.23 & 6.245) eq.(6.27), becomes

).(or 2860 rprp

We can use the formula of triple scalar product to write

0 prpr

0prpr

)29.6(0 prdt

d

Because is arbitrary we must have )30.6(constant pr

The angular momentum of a closed system is constant in time.

The Hamiltonian & Hamilton’s Equations:

For simple dynamical systems the potential energy is a function of coordinates

alone and the kinetic energy is a quadratic function of velocities, i.e.,

)31.6(, iii qUqqTL

)32.6(iiiii

pqmq

T

q

L

With pi is called the generalized momenta. Now from Lagrange’s equation we

have

00

dt

dp

q

L

q

L

dt

d

q

L i

jjj )33.6(

ji

q

Lp

Now defining the Hamiltonian function as

)34.6(LpqHi

ii

With L is the Lagrangian given by Eq.(6.31)

The generalized force

Now from Eq.(6.32) we have

)35.6(2i i

ii i

ii

ii Tq

Tq

q

Lqpq

)36.6(2 UTUTTLpqHi

ii

That is, the Hamiltonian function represents the total energy of the system. Now

)37.6(

ii

ii

iiiii q

q

Lq

q

LpqqpH

But from Eq.(6.15) the 1ast and the 3rd term in the brackets cancel. Using Eq.

(6.33) for the 4th term, Eq.(6.37) now reads

)38.6( i

iiii qppqH

ii pqHHSince , )39.6(

ii

ii

i

qq

Hp

p

HH

Now comparing Eq.(6.21) with Eq.(6.22) we get

motionofequationssHamilton

pq

H

qp

H

ii

ii ')40.6(

Example Use the Hamiltonian method to find the

equations of motion of a particle of mass m m

constrained to move on the surface of a cylinder

of radius R. The particle is subjected to a gorce

directed toward the origin and proportional to the

distance from the origin: F=-kr

Solution The speed of the particle in cylindrical

coordinates is, from Eq.1.47

22221 zRmT

zr ezeerv ˆˆˆ

22212

21& zRkkrU

To find the Hamiltonian we have to find the generalized momenta. From

Eq.(6.32) we have

zm

z

LpmR

Lp z

&2

2221

2

2

2

22,,, zRk

m

p

mR

pUTppH z

Applying Hamilton’s equations, Eq.(6.40) we get

m

p

p

Hz

mR

p

p

H z

z

&

2

kzz

Hp

Hp z

&0

From the above 4-equations we can conclude that consatnt2 mRp

That is, the angular momentum about the z-axis is constant because there is no

torque about the z-axis. Also we have

0 kzzmkzpz

That is, the motion along the z-axis is SHM with frequency mk

Example Use the Hamiltonian

method to find the equations of

motion for a spherical pendulum

of mass m and length b.

Solution The K.E. & P.E of the particle are

cos&

sin 2222221

mgbU

bbmT

To find the Hamiltonian we have to find the generalized momenta. From

Eq.(6.32) we have

222 sin&

mbL

pmbL

p

0&sinsin22

2

Hpmgb

mb

pHp

This means that the momentum about the symmetric axis is constant.

cossin22

,,,22

2

2

2

mgbmb

p

mb

pUTppH

Applying Hamilton’s equations, Eq.(6.40) we get

222 sin

&mb

p

p

H

mb

p

p

H