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 Introduction.
Syllabus
 Lagrange’s Equation.
 Small Oscillations.
 Hamilton’s equations of motion.
 Canonical transformations and
Generating functions.
 Hamilton-Jacobi Equation.
Note : Attendance is important because the theory
and questions will be explained in the class.
References
1. Dynamics by A S Ramsey
2. Motion of a particle Rigid body and a Particle
S L Loni
3. Advanced Mechanics – Schaum Series
4. Any book related with Mechanics
System of Particles
With respect to a system of 3-dimensional coordinates, we
need 3n number of independent variables to describe the
position of a system consisting of n number of particles.
If there are k
then we need
variables.
number of constraints (restrictions ),
only 3n-k number of independent
We define Degrees of freedom of the system as 3n-k.
By looking at the system we can define these
independent variables and they are called generalized
coordinates and they are denoted by q1, q2,…, qn .
Constraints
3constraints can be classified as
Holonomic constraints.
2) Non-Holonomic constraints.
A Holonomic constraint can be expressed as a function
of the space coordinates and time.
1)
E.g. (1) A rigid body. Length between any two points is
constant. So a Holonomic constraint.
E.g. (2)
a
Consider a fluid inside a spherical vessel of radius
. If
the distance between the centre of the vessel and a fluid
particle is r, then we have r  a.
This is a non-Holonomic constraints.
So Degrees of freedom of a system means the
number of independent variables required to describe the
system.
After observing the system, one can define the
generalized coordinates.
Simple pendulum

Since system has one particle, need 3
independent variables.
Constraints :
• Particle moves in a plane.
• Length of the string is constant.
Therefore need 3-2=1 independent variable.
i.e. it has one degree of freedom.
Define the generalized coordinate as the angle
between downward vertical and the string.
Lagrange’s Equations
Result I
Suppose r i is the position vector of the ith particle and qj is the jth
generalized coordinate.
 r i  r i .

q j q j
Proof : Since
r i  r i q1 , q2,...................... , qn , t 

N
d ri 
j 1
r i
r i
dq j 
dt
q j
t
Eq. 1
Therefore
d ri
ri 
dt
1  N ri
ri 
 
dq j 
dt 
dt  j 1 q j
t 



N

j 1
Hence
 r i  r i .

q j q j
r i
r i
q j 
q j
t
Result II
d   r i   r i

.
dt  q j  q j
Proof :

N
Since
r i 
m 1
r i

q m  r i
qm
t

 r i  

q j q j 


N

m 1

N
m 1

r i
r i 
q m 
qm
t 

  r i 
  r i 

 q m 


q j  qm 
q j  t 

N

m 1
   r i 
   r i 
q m  


qm  q j 
t  q j 
d   r i 
 
dt  q j 
Hence the result is proved.
We have
 r i  r i

q j q j
Result 1
d   r i   r i



dt  q j  q j
Result 2
n
Kinetic energy of the system is
T

i 1

N
T



q j q j
i 1

N

i 1
T


q j
1

r i  ri 
mi
2 q j

N
i 1
1
mi r i  r i
2
 r i
mi r 
.
q j
1
mi r i  r i
2
Also

N
T


q j q j
i 1

N

i 1

1

r i  ri 
mi
2 q j

N
i 1

 r i
mi r i 
q j

N
i 1
1
mi r i  r i
2
r i
mi r i 
q j
By
Result 1
Hence
T

q j

T

q j

n
i 1
n
i 1
 r i
mi r i 
q j
Result 3
r i
mi r i 
q j
Result 4
Suppose the system of particles changed slightly without changing
the time t then Eq. 1 becomes

N
d ri 
j 1
N
Work done
dw 
r i
dq j
q j

 
F i .d r i
j 1
N

r i
q j
dq j
r i
Fi 
q j
dq j
Fi 
i 1
N

N
j 1

i 1
N
j 1
Here F i is the external force on the
ith particle.

N

j 1
N


j 1
N
i 1
r i
Fi 
q j
Q j dq j
dq j

N
Here Q j 
i 1
r i
Fi 
q j
is called generalized
force associated with
generalized
coordinate q j .
Also w  wq1 , q2 ,................q N  gives us

N
dw 
j 1
w
dq j
q j

N

j 1
N
w
dq j 
q j

N
Q j dq j
j 1

 w


 dq j  o


Q
j
 q

j


j 1
Since dq j are all independent above equation yields
w
Qj 
Eq. 2
q j
Applying Newton’s 2nd Law F i  mi ri to the ith particle
we have
r i
r i
 Fi 
 mi ri 
q j
q j
d
r i
 mi r i 
dt q j
 d  r i 
  r i 
d
  r i  

 mi   r i 

 q 
dt

q
dt
j
 j 
 
 d  r i 
  r i 
  r i  

 mi   r i 

 q 
dt

q
j
 j 
 

N
 Qj 
in1


i 1
d   r i   ri

dt  q j  q j
By
r i
Fi 
q j
 d  r i 
  r i 
d
  r i  

mi   r i 

 q 
dt

q
dt
j
 j 
 
Result 2
d

dt

n
i 1
r i
mi r i 

q j

n
i 1
When Result 3
and
equation, it reads as
d   r i 
mi r i  
dt  q j 
Result 4
d  T  T
Qj  


dt  q j  q j
This is called
are applied in above
T

q j
T

q j

n
i 1
n

i 1
Lagrange’s equation of motion.
 r i
mi r i 
q j
r i
mi r i 
q j
Special case
Suppose all the forces are conservative. i.e. there
exists a scalar function V  V q1 , q2 ,..............q N , t 
called potential function.

V
By the definition of the potential function
 0.
q j
Definition
Lagrangian or Lagrange’s Function L of the system is
defined as the difference of Kinetic energy and the
Potential energy and denoted by L.
i.e. L = T - V
Here dw  F . d r
 V. d r
 dV
w
V


q j
q j
Hence by
Eq. 2
w we get
Qj 
q j
V

 Qj
q j
d  T  T
 

dt  q j  q j
d  T  T V
 


0

dt  q j  q j q j
d  T V   T V 
 



0


dt  q j q j   q j q j 
V
 0.
Since
q j
 
d  
T  V   T  V   0
 
dt  q j
 q j
d  L  L


0
This is the Lagrange’s equation
dt  q j  q j
for a conservative system
E.g. ( Question No. 1 of Exercises )
A particle of mass m moves in a conservative force
field. Find the Lagrangian and equations of motions in
cylindrical polar coordinates.
Z
P-particle
O

X
Y
Let OQ= r , P
Q
  x, y, z 
Then r  OP
 r cos  i  sin  j  z k 
 r  r cos  r sin  i  r sin  r cos   j  z k
 r.r  r 2  r 2 2  z 2
1
 L  mr 2  r 2 2  z 2   V r , , z 
2


1
 L  m r 2  r 2 2  z 2  V r , , z 
2
Therefore
L
2
L

mr

 mr

r
L
V
L
V
2

 mr 

r
r


L
 mz
z
L V

z z
V
2

Hence equations of motions are mr  mr 
 o.
r
d 2  V
m r 
 o.
dt

V
mz 
 o.
z
 
E.g. ( Question No. 2 of Exercises )
Suppose that the particle, in the previous example moves
in the Oxy plane and V=V(r) only. If at time t=0 the
particle on the Ox axis of distance a and the velocity of
the particle is vo in the direction of the positive Oy axis.
Find the velocity of the particle.
Solution :
 
d 2  V
V
r 
 o.
The equations mr  mr 
 o. m
dt

r
V
mz 
 o. become
z
d 2
dV
2

m r    o
mr  mr 
 o.
dt
dr
 r 2  C
2
 C  r   r r
2
 C  avo  r   avo
2
 
2
 r  
2
2 2
a vo
 r 2 
2 2
a vo
3
r
.
So the equation
dV
mr  mr av  
 o. becomes
dr
d
2 2 1
mr  ma vo 3  V r   o
d
Further r  r
dr
r
dt
dr dr

dr dt
dr
 r
dr
3
2
dr 2 2 1
1d
Hence r  a vo 3  
V r .
dr
m dr
r


1d


2 2 1
r dr   a vo 3 
V r dr
r m dr


1 2
1 2 2 1 1
r   a vo 2  V r   C
2
2
m
r
1 2 2 1 1
o   a vo 2  V a   C
2
a m
1 2 1
 C  vo  V a 
2
m
 v 
2
a 2 vo2

1
1
2
2
 2 V r   vo  V a 
2
m
m
r

2
v
2
2
2
2
o
i .e. v  2 r  a  V a   V r .
m
r
Now we have velocity as a function of the distance
from the origin. When the velocity potential function is
given we can get the velocity using this equation..