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Homework 1: # 1.21, 2.7, 2.12
Michael Good
Sept 3, 2004
1.21. Two mass points of mass
m
1
and
m
2
are connected by a string passing
through a hole in a smooth table so that
m
1
rests on the table surface and
m
2
hangs suspended. Assuming
m
2
moves only in a vertical line, what are the
generalized coordinates for the system? Write the Lagrange equations for the
system and, if possible, discuss the physical signiﬁcance any of them might have.
Reduce the problem to a single secondorder diﬀerential equation and obtain a
ﬁrst integral of the equation. What is its physical signiﬁcance? (Consider the
motion only until
m
1
reaches the hole.)
Answer:
The generalized coordinates for the system are
θ
, the angle
m
1
moves round
on the table, and
r
the length of the string from the hole to
m
1
. The whole
motion of the system can be described by just these coordinates. To write the
Lagrangian, we will want the kinetic and potential energies.
T
=
1
2
m
2
˙
r
2
+
1
2
m
1
(˙
r
2
+
r
2
˙
θ
2
)
V
=

m
2
g
(
R

r
)
The kinetic energy is just the addition of both masses, while V is obtained
so that
V
=

mgR
when
r
= 0 and so that
V
= 0 when
r
=
R
.
L
=
T

V
=
1
2
(
m
2
+
m
1
)˙
r
2
+
1
2
m
1
r
2
˙
θ
2
+
m
2
g
(
R

r
)
To ﬁnd the Lagrangian equations or equations of motion, solve for each
component:
∂L
∂θ
= 0
∂L
∂
˙
θ
=
m
1
r
2
˙
θ
d
dt
∂L
∂
˙
θ
=
d
dt
(
m
1
r
2
˙
θ
) =
m
1
r
2
¨
θ
+ 2
m
1
r
˙
r
˙
θ
1
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View Full Document Thus
m
1
r
(
r
¨
θ
+ 2
˙
θ
˙
r
) = 0
and
∂L
∂r
=

m
2
g
+
m
1
r
˙
θ
2
∂L
∂
˙
r
= (
m
2
+
m
1
)˙
r
d
dt
∂L
∂
˙
r
= (
m
2
+
m
1
)¨
r
Thus
m
2
g

m
1
r
˙
θ
2
+ (
m
2
+
m
1
)¨
r
= 0
Therefore our equations of motion are:
d
dt
(
m
1
r
2
˙
θ
) =
m
1
r
(
r
¨
θ
+ 2
˙
θ
˙
r
) = 0
m
2
g

m
1
r
˙
θ
2
+ (
m
2
+
m
1
)¨
r
= 0
See that
m
1
r
2
˙
θ
is constant, because
d
dt
(
m
1
r
2
˙
θ
) = 0. It is angular momentum.
Now the Lagrangian can be put in terms of angular momentum and will lend the
problem to interpretation. We have
˙
θ
=
l/m
1
r
2
, where
l
is angular momentum.
The equation of motion
m
2
g

m
1
r
˙
θ
2
+ (
m
2
+
m
1
)¨
r
= 0
becomes
(
m
1
+
m
2
)¨
r

l
2
m
1
r
3
+
m
2
g
= 0
The problem has been reduced to a single nonlinear secondorder diﬀerential
equation. The next step is a nice one to notice. If you take the ﬁrst integral
you get
1
2
(
m
1
+
m
2
)˙
r
2
+
l
2
2
m
1
r
2
+
m
2
gr
+
C
= 0
To see this, check by assuming that
C
=

m
2
gR
:
d
dt
(
1
2
(
m
1
+
m
2
)˙
r
2
+
l
2
2
m
1
r
2

m
2
g
(
R

r
)) = (
m
1
+
m
2
)˙
r
¨
r

l
2
m
1
r
3
˙
r
+
m
2
g
˙
r
= 0
(
m
1
+
m
2
)¨
r

l
2
m
1
r
3
+
m
2
g
= 0
Because this term is
T
plus
V
, this is the total energy, and because its time
derivative is constant, energy is conserved.
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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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