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Mid Semester Examination - December 2010.
MAT 303 1.0 Classical Mechanics
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Sample Solutions.
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1.
A particle A of mass 3m is connected to a fixed point O by a light inextensible string of
length a, and the particle B of mass m is connected to the particle A by another light
inextensible sting of same length. The particles swing in a vertical plane such that both
strings taut. At time t, OA and AB make small angles  and  with the downward
vertical, respectively. Show that the kinetic energy and the potential energy of the system
can be expressed as
1
1
T  m a 2 4  2  2    2 and V  mga 4  2   2 .
2
2




Find the Lagrangian of the system and the Lagrange’s equations.
Determine the normal coordinates.
Solution
v  A, O
 2  a 2  2
v B , O
 2  a  cos   a  cos  2  a  sin   a  sin  2
 a 2  2  2a 2   cos      a 2  2 .
When  and  are small,
v B , O
 2  a 2  2  2a 2    a 2  2 . 5


1
1
3m a 2  2  m a 2  2  2a 2    a 2  2 .
2
2
1
 m a 2 4  2  2    2 . 5
2
T 


5
5
V  3mg a cos   mg a cos   a cos  
1
  mg a 4  2   2
5
2
1
1
5
 L  m a 2 4  2  2    2  mg a 4  2   2
2
2
L
L
 ma2 4     and
 mag 4   ma4      mg 4   0


5
5
Equation (1)
4 a  a   4 g  0










L
L
 ma2    and
 mga   ma     m g   0



5
5
a  a   g   0
Equation (2)
5
Equation (1) + k . Equation (2)
a4  k   a1  k    4 g   k g   0
Now, choose k such that
5
Equation (3)
4  k 1 k

 k2  4  k  2 .
4
k
5
5
When  k  2 ; Equation (3) becomes 6 a  3a   4 g   2 g   0 .


2g
2      0 . so, 2    is a normal coordinate.
 2   
3a
5
When  k  2 ; Equation (3) becomes 2a  a   4 g   2 g   0 .


2g
2      0 . so, 2    is a normal coordinate.
 2   
a
2.
5
Define a canonical transformation.
Show that
 2k q 
1 2
 is a canonical transformation.
P
p  k q 2 and Q  tan 1 
4k
 p 
Assuming that the generating function T of the form T  qi , Qi , t  satisfies that
T
T
pi 
and Pi  
, find a generating function T for the above transformation.
qi
Qi
Solution
canonical transformation : A transformation that leaves the Hamilton’s equations invariant
is called a canonical transformation.
 2k q  
 1 2
 
5
 
p dq  P dQ  p dq  
p  k q 2  d  tan1 
 4k
 
 p 



2k
 1 2
2
 p dq  
p  k q 
 4k
   p 2
 1   2 k q 

 


 p dq  q dp  


2


p




 2k  p dq  q dp  
1 2
 p dq 
p  4k 2 q 2 
2
2 2 
4k
 p  4 k q

1
 p dq   p dq  q dp 
2
5
1
 pq 
  p dq  q dp   d 
 , an exact differential.
2
 2 
so, given transformation is a canonical transformation.
5


5
5
Given transformation can be expressed as
p  2 k q cot Q and P 
5
1
2k q cot Q 2  k q 2
4k
 k q 2 cot 2 Q  k q 2
 k q 2 cos ec 2Q
pi 
T
qi
Pi  
 2kq cot Q 
5
T
 T  kq2 cot Q  f Q ,t 
q


T

 k q 2 cos ec 2Q  
kq2 cot Q  f Q ,t 
Qi
Q
 k q 2 cos ec 2Q  kq2 cos ec 2 Q 
 k q 2 cos ec 2Q  kq2 cos ec 2 Q 

f Q ,t 
Q
5
5
5


f Q ,t  
f Q ,t   0  f Q ,t   f t 
Q
Q
5
T  kq2 cot Q  f t 
5
State the Hamilton’s equations for a dynamical system.
3.
A particle of mass m moves in an Oxy plane under the influence of a potential function


V  2 k 2 m x 2  y 2 , where k is a positive constant. Obtain Hamilton’s equations for
the particle.
Initially, the particle is at the point  0 , 2k
 and moving in the positive Ox direction with
speed 4 k 2 . Solve the Hamilton’s equations and determine the path of the particle.
5
Hamilton’s equations for a dynamical system are
H
H
  p i
 qi and
5
qi
pi
L



1
m x 2  y 2  2 k 2 m x 2  y 2
2

5
Generalized momenta are given by,
px 
L
L
 mx and p y 
 my
x
y
5
5
5
H



Hamilton’s equations
5

x 
5
H
  p x  4 k 2 mx  p x
x
5
1
H
 y  p y  y
m
p y
5
5
H
H
  p i becomes
 qi and
qi
pi
H
1
 x  p x  x
p x
m


1
px 2  p y 2  2 k 2 m x2  y 2 .
2m
H
  p y  4k 2 my  p y
p y
5
1
1
p x  4 k 2 x and y  p y  4 k 2 y .
m
m
5
5
5
5
x  A1 cos 2kt  B1 sin 2kt and y  A2 cos 2kt  B2 sin 2kt .
Initial conditions are x  0 , y  2 k , x  4 k 2 and y  0 .
x  0  A1  0
x  4 k 2
y  0  B2  0
y  2k

 A2  2k
 B1  2k
x  2k sin 2kt and y  2 k cos 2 kt .
 x 2  y 2  4k 2 .
Hence, path is a circle.
5
5
5
5
With polar coordinates (difficult)
L


1
m r 2  r 2 2  2 k 2 m r 2
2
5
5
Generalized momenta are given by,
pr 
L
L
 mr and p    mr 2
r

5
5
H
1
1
pr 2 
p 2  2 k 2 m r 2 .
2
2m
2mr
Hamilton’s equations
5
5
H
H
  p i becomes
 qi and
qi
pi
H
1
 r  pr  r
p r
m
1
H
p  
  
2 
p
mr
5
H
1
  p r  
p 2  4 k 2 mr   p r
3
r
mr
5
H
  p  0  p
p

p  c0 constant.
5
5
 
Hence,  mr 2  p  co  mr r  p  co


5
Initially, r  2k and r  4 k 2  m2k  4 k 2  co  co  8 mk 3
Further
 mr 
1
1
pr  r and 
p 2  4 k 2 mr   p r
3 
m
mr
1
mr 3
p 2  4 k 2 mr  mr
 r

5


2
dr
1

 8 mk 3  4 k 2 mr
dr mr 3
dr 64 6

k  4k 2 r
3
dr r
1 2
32
r   k 6  2k 2 r 2  c1
2
r2
Using initial conditions, 0  
32
2k 
2
Solve this and show that r  2k .
5
5
k 6  2k 2 2k 2  c1  c1  0 .
1
32
64
 r 2   k 6  2k 2 r 2  r 2   k 6  4 k 2 r 2
2
r2
r2
Hence, path is a circle.
5
5
5
5