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Let's try to find the center
of mass of a uniform object
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like a uniform rod.
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And let's assume
this rod is length L,
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and we want to find
the center of mass.
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Now, before I begin
this calculation,
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you can probably
already guess that it's
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going to be exactly in the
middle, and we'll verify that,
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but let's first define what
we mean by our center of mass
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for discrete particles.
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Recall that this
was a sum over all
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the particles in the system.
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So we'll take a label
J goes from 1 to N,
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and it was the mass of that
jth particle times the position
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vector that jth particle
with respect to some origin,
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and we're dividing
that by j equals
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from 1 to N of the total
mass in the system.
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Now, how do we
translate this equation
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for a continuous system?
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And let me just again show
that we had chosen an origin.
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Here was our jth particle
of mass mj and rj.
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So what we want to do
is draw the analogy,
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and here's how it works-- that
for each discrete particle,
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we're going to look at that
as some mass element delta mj.
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Our vector rj will go to a
vector for this mass element.
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I'll just write it delta m.
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And our sum from
j goes from 1 to N
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is actually going to go to
an integral over the body.
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So let's see how that looks.
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So first, we'll do it
with the total mass,
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m-- here we're summing
over j-- from 1 to N of mj.
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That goes to the
integral over the body.
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Now, the delta m,
when we take limits,
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because that's when an integral
goes, we'll write that as dm.
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So that becomes a
limit over the body.
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And likewise, our
sum j goes from 1
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to N of mj rj goes to an
integral over the body of dm
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vector r going to that element.
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So we can say in the
limit that this becomes
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r going to that element.
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Now, that means that our
continuous expression
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for the continuous
object is an integral
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over the body of dm r to
that element dm divided
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by an integral over
the body of dm.