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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.(.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 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 sin 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 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
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.,
)14.6(, iii qVqqTL
)15.6(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 )16.6(
ji
q
Lp
Now defining the Hamiltonian function as
)17.6(LpqHi
ii
With L is the Lagrangian given by Eq.(6.14)
Now from Eq.(6.145) we have
)18.6(2i i
ii i
ii
ii Tq
Tq
q
Lqpq
)19.6(2 VTVTTLpqHi
ii
That is, the Hamiltonian function represents the total energy of the system. Now
)20.6(
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.
(6.16) for the 4th term, Eq.(6.20) now reads
)21.6( i
iiii qppqH
ii pqHHSince , )22.6(
ii
ii
i
Hp
p
HH
Now comparing Eq.(6.21) with Eq.(6.22) we get
motionofequationssHamilton
pq
H
qp
H
ii
ii ')23.6(
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 mgbUbbmT
To find the Hamiltonian we have to find the generalized momenta. From
Eq.(6.15) we have
222 sin&
mbL
pmbL
p
cossin22
,,,22
2
2
2
mgbmb
p
mb
pVTppH
Applying Hamilton’s equations, Eq.(6.23) we get
222 sin
&mb
p
p
H
mb
p
p
H
0&sinsin22
2
Hpmgb
mb
pHp
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