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We'd like to examine the
motion of two particles.
2
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Here's particle 1.
3
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And here's particle 2.
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And each particle
can have some motion.
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In between them is
the center of mass.
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And what we'd like
to do is figure out
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how to describe the motion
of each of these particles
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with respect to
the center of mass.
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So let's choose some
coordinate system.
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Origin down here.
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Here's our particle r1.
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Here's our particle r2.
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And here is our center of mass.
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Now in the reference frame
of the center of mass,
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we have position
vector r1 prime.
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And we have position
vector r2 prime.
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And what we'd like to
do is find expressions
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for r2 prime and
r2 prime in terms
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of the positions of r1 and r2.
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Now we call that the position of
the center of mass for this two
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body problem is given by
m2 m1 r1 plus m2 r2 divided
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by the total mass.
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And here we use
the vector triangle
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that r1 prime equals r1
minus center of mass.
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Now remember this vector
is equal to this vector
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minus that vector.
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Sometimes people like
to say the vector
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r1 is equal to r plus r prime.
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And that's how we get
that relationship.
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Now we can use our
result here that r1 prime
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is r1 minus m1 r1 plus m2
r2 divided by m1 plus m2.
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And when we combine
terms-- let's just do this
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so you can see it-- r1
minus m1 r1 plus m2 r2
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divided by the total mass.
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We now have the m1
r1 terms cancel.
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And we have a common m2 over
m1 plus m2 times r1 minus r2.
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Now r1 minus r2 is a vector
that goes from-- here is r1.
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Here's r2.
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So the vector r1 minus r2
is the relative position
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of vector 1 with respect to 2.
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And let's give that
a special name.
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We'll call that r1, 2, the
relative position vector.
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So we have m2 over m1
plus m2 r1, 2 is r1 prime.
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Now you can easily see
that if you interchange
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the indices 1 and 2, the
only thing that changes here
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is a sign.
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And if we interchange 1 and
2-- and this is an exercise
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that you can do--
then r2 prime is minus
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m1-- I'm interchanging
the indices.
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The minus sign came from
the interchange of 1 and 2.
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And so we get m1 over m1 plus
m2 with the minus sign r1, 2.
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Now what is the
significance of this result?
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If you know the
position of r1 and r2,
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you know the relative velocity.
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If you have information
about this relative position,
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if you know the relative
position vector,
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then you can separately get the
locations of the two objects
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in the center of mass frame.
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Now this quantity in
here will appear often.
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And I'd like to introduce
a new quantity called
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the reduced mass.
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And that reduced mass,
mu, is the product
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of m1 m2 over m1 plus m2.
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It's a simple exercise to see
that 1 over mu is 1 over m1
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plus 1 over m2.
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And we'll encounter
that a little bit later.
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Then I can write both
of these vectors--
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and this is our conclusion--
that r1 prime is the reduced
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mass.
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Notice we have an m2 here, so
we have to divide by m1 times
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the vector r1, 2.
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And r2 prime is minus
the reduced mass.
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Again, we now have
to divide by m2.
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And you see this nice symmetry
of m1 and 1, 2 and 2, r1, 2.
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And that's our key result.