ME201 Advanced Dynamics (Fall 2007)
HW1 Solutions (
100 pts
)
1.
20pts
Write Newton’s equations of motion for two particles of masses
m
1
and
m
2
interacting via
Newton’s law of gravitation. How many degrees of freedom does the system have?
Without any further information the number of degrees of freedom this system has is 12. Each mass
has three degrees of freedom:
{
x, y, z
} ∈
R
3
and rotational degrees of freedom:
{
σ
x
, σ
y
, σ
z
} ∈
R
3
making six degrees of freedom for each mass, two masses, no constraints resulting in 12 degrees of
freedom.
Since this is an engineering class, we usually don’t think about rotating masses (which
rotate about themselves) and so we can exclude these resulting in a system with 6 degrees of freedom
total. Now lets look at the equations of motion. Assuming no spin, and no external forces, we begin
with the free body diagram below.
The force on each mass is

ˆ
F
2
=
ˆ
F
1
=
Gm
1
m
2
r
3
ˆ
r
, ˆ
r
= ˆ
r
2

ˆ
r
1
Figure 1: Free Body Diagram, Problem 1
where hat indicates a three vector and the
r
in the denominator is a magnitude (not a vector). From
Newton’s Laws we have:
m
1
¨
ˆ
r
1
=
ˆ
F
1
=
Gm
1
m
2
r
3
ˆ
r
m
2
¨
ˆ
r
2
=
ˆ
F
2
=

Gm
1
m
2
r
3
ˆ
r
These equations can be simplified slightly, while we can see that each of the two equations of motion
have three degrees of freedom resulting in a six degree of freedom system. As the question is a bit
open ended, one may perform a coordinate transformation to obtain a slightly different view of the
dynamics. Often relative distances are interesting in this type of problem. In order to retain
all
of
the information we can express the system by the difference in their position and by their center of
mass coordinates. In the coordinate that describes the differences we have
¨
ˆ
r
=
¨
ˆ
r
2

¨
ˆ
r
1
. We can just
plug in the equations above to get:
¨
ˆ
r
=

G
(
m
1
+
m
2
)
r
3
ˆ
r
(1)
ME201 Advanced Dynamics
1
HW1 Solutions
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This is one of two equations of motion and it has three degrees of freedom. The second coordinate is
the center of mass for this system which is ˆ
r
g
=
m
1
ˆ
r
1
+
m
2
ˆ
r
2
(
m
1
+
m
2
)
, and considering constant masses we have
¨
ˆ
r
g
=
m
1
¨
ˆ
r
1
+
m
2
¨
ˆ
r
2
(
m
1
+
m
2
)
, we can simply plug in the equations above to get:
¨
ˆ
r
g
=
1
m
1
+
m
2
Gm
1
m
2
r
3
ˆ
r

Gm
1
m
2
r
3
ˆ
r
¶
= 0
From this you may feel that the second three degrees of freedom have disappeared. This is partly
true only the case because their are no external forces and the internal forces act colinearly. Imagine
the case where external forces are indeed zero but initial conditions are such that the two mass are
tumbling through space while interacting colinearly, we need this coordinate to fully describe the
dynamics. So we are back to our original six degrees of freedom.
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