WEBVTT
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I'd like to now calculate
the angular momentum of a two
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particle system.
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So here's one particle.
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Call that 1.
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And here's the other particle 2.
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And particle 1 has a momentum
in the downward direction.
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And particle 2 has a momentum
in the upper direction.
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Now, in this case, I want to
make some of the momentum--
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so this momentum
of the system is 0.
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And now I want to calculate
the angular momentum
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about two different points.
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So let's just draw an axis here.
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And let's choose one point A.
And we'll call this distance r,
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and this distance r.
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And let's choose
another point B,
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and we'll call that distance d.
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Now, in order to calculate
angular momentum of the system,
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I need to draw my vectors.
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Let's start with
the angular momentum
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about the point A. That's going
to be a vector from A to object
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1, cross the momentum of 1, plus
the vector from A to object 2,
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cross the momentum of 2.
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Now what's crucial
here, is at this point,
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things are a little
bit of a abstract,
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but we want to draw these
vectors, rA1 and rA2.
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So we're now constructing our
angular momentum diagrams.
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So there's rA1, and there rA2.
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but again, in order
to calculate this,
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we could do it geometrically
with right hand rules,
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but I now want to
introduce unit vectors,
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so I can do vector
decomposition at every point.
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So I'm going to choose a
unit vector, i-hat, j-hat,
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and k-hat.
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And now I can decompose
all of these vectors
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in terms of my unit vectors.
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This is the angular
momentum of the system.
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So now it's not hard at
all to write these out.
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rA1 is minus r in
the i-hat direction.
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Cross P1 is down.
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That's minus P1 in
the j-hat direction,
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where P1 is the magnitude .
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And rA2 is in the
plus i-hat direction.
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So that's r i-hat.
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And I'm crossing
that with P2, which
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I'll write right now as
P2 in the j-hat direction.
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Now remember, P2 has the
same magnitude as P1,
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but they're pointing
in opposite directions.
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So when I take
the cross project,
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i-hat cross j-hat is k-hat
minus sign minus sign.
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So I get rP1 k-hat.
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And now over here,
P2 is equal to P1.
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i-hat cross j-hat is k-hat.
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And I get another rP1 k-hat.
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And so I have 2rP1 k-hat.
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And that's the angular momentum
of the system about point A.
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Now I'd like to calculate the
angular momentum about B So
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let's draw the
same diagram, 1, 2.
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Here is P2.
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Here's P1.
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Here is B. And now
I'll draw my vectors.
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This is rB2 and this is rB1.
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This distance was 2r.
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This distance was d.
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I'm going to use the
same unit vectors.
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And so I get l for the
system about B. Again,
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I'll just write everything out.
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P1 plus rB2 cross P2.
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So I'm going to write
P2 as, in this case,
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it's equal to minus P1.
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And magnitudes P2 equals P1.
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In magnitude directions
are opposite.
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So rB1 is equal 2r plus D
i-hat, and it's pointing
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in the minus i-hat direction.
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Cross P1 is in the
minus j-hat direction.
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rB2 is also in the
minus i-hat direction,
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so that's minus D i-hat.
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And P2 is in the plus j
direction, so that is plus--
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the magnitudes are the same--
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P1 j-hat.
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We don't need the
plus sign there.
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And now I take the
cross products.
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i-hat cross j-hat is k-hat.
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A minus sign minus sign.
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That makes plus.
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So I have 2r plus D P1 k-hat.
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Now notice here I have i-hat
cross j-hat, which is k-hat,
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but there's a minus sign.
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So I have minus D P1 k-hat.
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Here I have plus D P1
k-hat minus D1 P k-hat.
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And so I get 2rP1 k-hat.
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And I have the same result.
L system A equals L system B.
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Now that's not a
coincidence in this problem.
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And the reason is
that whenever--
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so we can say it this way.
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Whenever the momentum
of the system is 0,
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then LA is independent of
the choice of the point A.
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So coming back to our
example, no matter where
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I picked our points A and B--
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any point I could
pick, anywhere I want--
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I could pick a
point C up there--
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I make this cross
product calculation.
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I would get exactly
the same answer, 2rP1k.