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Homework 9: # 8.19, 8.24, 8.25
Michael Good
Nov 2, 2004
8.19
The point of suspension of a simple pendulum of length
l
and mass
m
is con
strained to move on a parabola
z
=
ax
2
in the vertical plane. Derive a Hamilto
nian governing the motion of the pendulum and its point of suspension. Obtain
the Hamilton’s equations of motion.
Answer:
Let
x
0
=
x
+
l
sin
θ
z
0
=
ax
2

l
cos
θ
Then
T
=
1
2
m
( ˙
x
0
2
+ ˙
z
0
2
)
U
=
mgz
0
Solving in terms of generalized coordinates,
x
and
θ
, our Lagrangian is
L
=
T

U
=
1
2
m
( ˙
x
2
+2 ˙
xl
cos
θ
˙
θ
+4
a
2
x
2
˙
x
2
+4
ax
˙
xl
˙
θ
sin
θ
+
l
2
˙
θ
2
)

mg
(
ax
2

l
cos
θ
)
Using
L
=
L
0
+
1
2
˜
˙
qT
˙
q
where ˙
q
and
T
are matrices. We can see
˙
q
=
±
˙
x
˙
θ
²
T
=
±
m
(1 + 4
a
2
x
2
)
ml
(cos
θ
+ 2
ax
sin
θ
)
ml
(cos
θ
+ 2
ax
sin
θ
)
ml
2
²
with
1
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View Full Document L
0
=

mg
(
ax
2

l
cos
θ
)
The Hamilitonian is
H
=
1
2
˜
pT

1
p

L
0
Inverting
T
by
±
a
b
c
d
²

1
=
1
ad

bc
±
d

b

c
a
²
with the algebra,
1
ad

bc
=
1
m
2
l
2
(1 + 4
ax
2
)

m
2
l
2
(cos
θ
+ 2
ax
sin
θ
)
2
this is
=
1
m
2
l
2
(sin
2
θ
+ 4
ax
2

4
ax
cos
θ
sin
θ

4
a
2
x
2
sin
2
θ
)
=
1
m
2
l
2
(sin
2
θ

4
ax
sin
θ
cos
θ
+ 4
a
2
x
2
cos
2
θ
)
which I’ll introduce, for simplicity’s sake, Y.
=
1
m
2
l
2
(sin
θ

2
ax
cos
θ
)
2
≡
1
m
2
l
2
Y
So now we have
T

1
=
1
m
2
l
2
Y
±
ml
2

ml
(cos
θ
+ 2
ax
sin
θ
)

ml
(cos
θ
+ 2
ax
sin
θ
)
m
(1 + 4
a
2
x
2
)
²
T

1
=
1
mY
±
1

(cos
θ
+ 2
ax
sin
θ
)
/l

(cos
θ
+ 2
ax
sin
θ
)
/l
(1 + 4
a
2
x
2
)
/l
2
²
I want to introduce a new friend, lets call him
J
J
≡
(cos
θ
+ 2
ax
sin
θ
)
Y
≡
(sin
θ

2
ax
cos
θ
)
2
So,
T

1
=
1
mY
±
1

J/l

J/l
(1 + 4
a
2
x
2
)
/l
2
²
Proceed to derive the Hamiltonian,
H
=
1
2
˜
pT

1
p

L
0
2
we can go step by step,
T

1
p
=
1
mY
±
1

J/l

J/l
(1 + 4
a
2
x
2
/l
2
²±
p
x
p
θ
²
=
1
mY
±
p
x

(
J/l
)
p
θ
(

J/l
)
p
x
+ (1 + 4
a
2
x
2
/l
2
)
p
θ
²
and
˜
pT

1
p
=
1
mY
(
p
2
x

J
l
p
θ
p
x

J
l
p
θ
p
x
+
1 + 4
a
2
x
2
l
2
p
2
θ
)
the full Hamiltonian is
H
=
1
2
mY
(
p
2
x

2
J
l
p
θ
p
x
+
1 + 4
a
2
x
2
l
2
p
2
θ
) +
mg
(
ax
2

l
cos
θ
)
plugging in my
Y
and
J
H
=
1
2
m
(sin
θ

2
ax
cos
θ
)
2
(
p
2
x

2
cos
θ
+ 2
ax
sin
θ
l
p
θ
p
x
+
1 + 4
a
2
x
2
l
2
p
2
θ
)+
mg
(
ax
2

l
cos
θ
)
Now to ﬁnd the equations of motion. They are
˙
x
=
∂H
∂p
x
˙
θ
=
∂H
∂p
θ
˙
p
x
=

∂H
∂x
˙
p
θ
=

∂H
∂θ
The ﬁrst two are easy, especially with my substitutions.
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This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.
 Spring '12
 atkin
 Mass, Work

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