TUTORIAL 2 (MPHA021) MEMO 2021
1. The generalised coordinates for the system are  and  . The positions and velocities of the
pendulum are given by
r  a sin   a cos
and
r  a sin
v  a cos  a sin
and
v  a sin 
Kinetic energy is therefore



1
m a 2 2 cos2   sin 2   a 2 sin 2  2
2
1
 m a 2 2  a 2 sin 2  2
2
1
 ma 2  2  sin 2  2
2
The Potential energy is given by
U   mga cos
T




The Lagrange’s equation for  is given by
d  T  T
U





dt    

Therefore
T
T
 ma 2
 ma 2 2 sin  cos


Hence,
d
ma 2  ma 2 2 sin  cos   mga sin 
dt


U
 mga sin 


The Lagrange’s equation for  is given by
d  T  T
U

  
dt    

Therefore
T
 ma 2 sin 2 

T
0

U
0

Hence,
d
ma 2 sin 2   0
dt


(20)
2. The position and velocity of the pendulum are given by
r  R sin   R cos
v  R cos   R sin   R sin 
Kinetic energy is therefore
1
m  R 2 2  cos 2   sin 2    R 2 sin 2  
2
1
 m  R 2 2  R 2 sin 2  
2
1
 mR 2  2   sin 2  
2
The Potential energy is given by
U   mgR cos 
T
Therefore the Lagrangian is
L  T U
1
 mR 2  2   sin 2    mgR cos 
2
Hence, the generalised momenta are:
L
p
 mR 2


Solving for  and  , we get
p
mR 2
Therefore, Hamiltonian is given by
H  p  L

2
 p  1
 p  1
 p
 mR 2 
 mR 2 2 sin 2   mgR cos 
2 
2 
mR
2
mR



 2
p
1

 mR 2 2 sin 2   mgR cos 
2
2mR 2
Thus, Hamilton’s equations are

p
H
p

p mR 2
and
H
 mR 2 2 sin  cos   mgR sin 

(15)
TOTAL [35]