Let's take a closer look at the center off mass.
Let's take a situation whereby this is my laboratory, and
this is the origin of my laboratory--
I have here lots of particles which
interact with each other.
They could be a cluster of all stars
somewhere out in the universe.
They interact with each other--
they could attract each other, they could repel each other
with electric charge, they could collide,
and they could explode.
Anything is allowed.
However, if the net external force--
if summing all the external forces on those together, the
net external force is 0, then the center of mass, which may
have a velocity in this direction, the v center of
mass relative to my laboratory, then the total
momentum can always be written as the total mass of this
entire cloud of particles times the velocity of the
center of mass.
In the absence of external force, this is conserved, and
that means that dP dt equals 0.
The velocity of the center of mass would be dr center of
mass dt: that's the definition of velocity.
In the absence of an external force-- a net external force--
the momentum of the center of mass is conserved and cannot
change, regardless of how these particles interact with
each other.
If there is no external force, because dP dt is the net
external force, then we notice that since F external equals
the total mass of all the particles times a of the
center of mass.
Notice that there is some remarkable
property going on here.
The property that you see here--
you take the derivative of this equation, and you got dP
dt, which is F, and you get dP dt, which is a.
The remarkable property is that if there is an external
force, that the center off mass behaves as if it were a
point particle, which has the total mass of all the little
particles together, and acted upon by the sum of the
external forces that act together on all these
particles together--
I have to add them all up, and that is not so obvious at all.
I would refer you to page 57 of your book.
Let us return now to a case where we don't have many
little particles, but that we have one big particle.
Since the center of mass acts as a point, and all the mass
is in that point, it's clear that when I hang an object-- a
crazy object--
on a string, that the center of mass, which acts like a
point, will always fall exactly below the vertical
line of the string.
If I have an object like this, and the center of mass may be
somewhere here, I could hang it like this, and then the
center of mass--
if I dampen out all the oscillations--
must lie exactly on the vertical, somewhere on this
vertical line.
I don't know where, but it's somewhere on
this vertical line.
I now suspend it from here, and the center of mass must
again be somewhere along the vertical line, and so where
the two intersect--
which is somewhere here--
there is the center of mass.
It's a nice way to determine the center of mass of some
crazy object-- this is not so crazy, but
I can make it crazy.
Suppose I take my bracelet off, and I put my bracelet
here on one side--
now it's a crazy object.
It's clear that the center of mass is shifting in this
direction, and notice this vertical line now
goes through here.
Somewhere here is the center of mass, and if I now suspend
it from here, the center of mass must also
fall below this line.
Where the two intersect, that is where the
center of mass is.
I leave you with the determination of centers of
mass, and you'll get lots of opportunities in
this problem set.
The center of mass is very special, indeed:
look at this equation.
Total--
you've seen it before--
times r of the center of mass.
Suppose you are not in the laboratory frame, but you are
in a frame of reference which moves with the center of mass.
Quite possible, because it's the same velocity of the
center of mass.
There are no external forces, and therefore, the center of
mass will continue with the same
velocity forever and ever.
Now you move in a frame of reference with
the center of mass.
That means that dr dt as seen from your center of mass frame
must be 0, because the center of mass is not
moving in your frame.
That means that this sum of m of i times dr i dt is also 0,
but this is also the sum of m of i times v of i, and that is
also the sum of each individual momenta of all
these little individual particles--
if I return to particles--
that interact with each other, and that must be 0.
If I move in the center of mass, the frame of reference,
then before and after the collisions--
so no external forces--
the momentum is always 0.
This is very characteristic, and this can come in very
handy in solving certain problems.