6Systems of particles

IA Dynamics and Relativity

6.1 Motion of the center of mass
Sometimes, we are interested in the collection of particles as a whole. For
example, if we treat a cat as a collection of particles, we are more interested in
how the cat as a whole falls, instead of tracking the individual particles of the
cat.
Hence, we define some aggregate quantities of the system such as the total
mass and investigate how these quantities relate.
Definition (Total mass). The total mass of the system is M =
P
m
i
.
Definition (Center of mass). The center of mass is located at
R =
1
M
N
X
i=1
m
i
r
i
.
This is the mass-weighted average position.
Definition (Total linear momentum). The total linear momentum is
P =
N
X
i=1
p
i
=
N
X
i=1
m
i
˙
r
i
= M
˙
R.
Note that this is equivalent to the momentum of a single particle of mass
M
at
the center of mass.
Definition (Total external force). The total external force is
F =
N
X
i=1
F
ext
i
.
We can now obtain the equation of motion of the center of mass:
Proposition.
M
¨
R = F.
Proof.
M
¨
R =
˙
P
=
N
X
i=1
˙
p
i
=
N
X
i=1
F
ext
i
+
N
X
i=1
N
X
j=1
F
ij
= F +
1
2
X
i
X
j
(F
ij
+ F
ji
)
= F
This means that if we don’t care about the internal structure, we can treat
the system as a point particle of mass
M
at the center of mass
R
. This is why
Newton’s Laws apply to macroscopic objects even though they are not individual
particles.
Law
(Conservation of momentum)
.
If there is no external force, i.e.
F
=
0
, then
˙
P = 0. So the total momentum is conserved.
If there is no external force, then the center of mass moves uniformly in a
straight line. In this case, we can pick a particularly nice frame of reference,
known as the center of mass frame.
Definition
(Center of mass frame)
.
The center of mass frame is an inertial
frame in which R = 0 for all time.
Doing calculations in the center of mass frame is usually much more convenient
than using other frames,
After doing linear motion, we can now look at angular motion.
Definition
(Total angular momentum)
.
The total angular momentum of the
L =
X
i
r
i
× p
i
.
How does the total angular momentum change with time? Here we have to
assume a stronger version of Newton’s Third law, saying that
F
ij
= F
ji
and is parallel to (r
i
r
j
).
This is true, at least, for gravitational and electrostatic forces.
Then we have
˙
L =
X
i
r
i
×
˙
p
i
+
˙
r
i
× p
i
=
X
i
r
i
×
F
ext
i
+
X
j
F
ij
+ m(
˙
r
i
×
˙
r
i
)
=
X
i
r
i
× F
ext
i
+
X
i
X
j
r
i
× F
ij
=
X
i
G
ext
i
+
1
2
X
i
X
j
(r
i
× F
ij
+ r
j
× F
ji
)
= G +
1
2
X
i
X
j
(r
i
r
j
) × F
ij
= G,
where
Definition (Total external torque). The total external torque is
G =
X
i
r
i
× F
ext
i
.
So the total angular momentum is conserved if
G
=
0
, ie the total external
torque vanishes.