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Now, we've been discussing
steady uniform precession,

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which is the simplest possible
case of a phenomenon that

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can be much more complicated.

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As an example and the
case we've been discussing

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so far where we release a
gyroscope from rest when it's

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horizontal, very
careful measurements

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would show that the
initial motion isn't

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just steady precession but
has some additional motion

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superimposed on it.

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The spin axis sort
of bounces or nods

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up and down as it precesses.

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However, friction
at the pivot point

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causes the amplitude
of the nodding

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to rapidly decay until it
settles into steady precession.

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This nodding motion
is called nutation.

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In our example, it
decays so quickly

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that we don't even notice it.

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But if you had a perfectly
frictionless pivot point,

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it would be a more
noticeable effect.

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And it would persist
for much longer.

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Now, there's an
interesting point

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about our horizontal gyroscope
that I'd like to point out.

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So here is my rod of length d.

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This is my point s.

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Here's my wheel that's rotating
with some angular velocity.

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And again, there's
the r hat direction.

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That's the direction that little
omega vector's pointing in.

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That's the k hat direction.

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And that's the
theta hat direction.

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Now, before I release it,
before I release the gyroscope

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from rest, the angular
momentum vector

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points in the r hat direction,
as drawn at this instant.

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When I release it
from rest, there

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is a torque acting
in the theta hat

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direction that causes
the angular momentum

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vector to rotate.

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And we've seen this
in some detail.

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But there's something
else to keep

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in mind, which is that now
when the system is precessing,

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that means that the center
of mass of the wheel

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is orbiting around the
z-axis, the vertical axis,

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through this pivot point.

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So you can think of that as a
point mass with the full mass

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of the gyroscope moving in
a circular path around point

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s with a radius d.

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That implies that there must be
a component of angular momentum

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pointing in the k hat
direction, pointing along

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the z-axis, corresponding
to the translational motion

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of the center of mass, the
circular translational motion

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of the center of mass, of the
gyroscope around the z-axis.

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But where did that k
component come from?

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There was no initial z component
of the angular momentum.

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We said that initially the
angular momentum just pointed

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in the r hat direction.

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And there's no torque
in the k hat direction.

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The only torque is in
the theta hat direction.

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So how do we end up with some
angular momentum pointing

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in the k hat direction?

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This seems to violate
the conservation

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of angular momentum.

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The solution to this puzzle is
that, in fact, the gyroscope

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doesn't remain precisely
horizontal when I release it.

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Instead, it dips by
a very small angle.

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So that's the horizontal.

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After I release
it, the gyroscope's

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actually dipped downward
by a small amount.

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I've exaggerated it
considerably in this drawing.

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We'll call that
angle delta theta.

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That small dip gives the
small negative component

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of the spin angular
momentum that's

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able to balance
the plus z angular

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momentum due to the
motion center of mass.

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Now, because the spin
angular momentum vector

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has such a large magnitude--

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that's my spin angular
momentum vector--

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only a small angle
is necessary in order

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to get enough of a
negative Lz component

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to balance out the
Lz corresponding

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to the center of mass motion.

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OK?

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But what we see is that
although gyroscopes

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seem like a remarkable
system, they're not magic.

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And, in fact, angular
momentum is conserved.

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The larger little
omega is, the faster

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the spin angular velocity is.

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And, therefore, the
larger the vector L

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is, the smaller that angle
is and the less of a dip

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that you get.

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So now we can write the
exact angular momentum

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for the gyroscope.

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This is my distance d.

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This is my pivot point s.

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My spin angular velocity
looks like that.

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Here is the r hat direction, the
k hat direction, and the theta

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hat direction into the screen.

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And we're orbiting--
the precession is around

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the z-axis with an angular
speed capital omega.

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Now, recall that
the total angular

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momentum with respect to point
s can be written in two parts.

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There's the angular momentum
due to the translational motion

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of the center of mass.

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And we know the
center of mass is just

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orbiting around the z-axis.

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So I'll call that, in fact,
the orbital angular momentum.

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So this is due to
the translation

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of the center of mass
with respect to point s.

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And then the second term is
due to the rotational angular

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momentum, or spin
angular momentum,

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relative to the center of mass.

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So this is due to rotation
about the center of mass.

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So again, the general angular
momentum of a rigid body

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is equal to the center of
mass translational angular

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momentum plus the angular
momentum due to rotation,

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a pure rotation, about
the center of mass.

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So let's write each
of these terms.

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The orbital angular
momentum, that

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is the translational angular
momentum of the center of mass,

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is just due to the motion
of the center of mass

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of the gyroscope, which is
at radius d with respect

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to point s, moving in a circle
with angular speed capital

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omega.

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So that's equal to the
mass times the center

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of mass velocity times
the radius of the circle.

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And that angular momentum
is in the k hat direction.

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But the center of
mass velocity is

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just equal to-- since
it's a circular motion--

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is just equal to the
radius of the circle,

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d, times the angular speed
of the circular motion

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of the center of mass,
which is capital omega.

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So I can write this
as m capital omega d

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squared in the k hat direction.

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So that's the angular momentum
due to the translational motion

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of the center of
mass around point s,

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what I'm calling the
orbital angular momentum.

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Now, the spin angular
momentum, we've

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been talking about
the rapid spin

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of the wheel around its axis.

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So that's given by
the moment of inertia

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about that axis times the
spin angular velocity.

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And that's pointing in
the r hat direction.

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But there's a subtlety here.

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It turns out that is
not the only rotation

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about the center of mass that
this wheel is undergoing.

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And actually because of that,
I'm going to call this I1.

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And let me just
draw a picture here.

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So for my disk rotating
around this axis,

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the relevant moment of
inertia is what I'll call I1.

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And for a disk, we know that
would be 1/2 m r squared.

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The other rotation
is a subtle one.

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Notice in this
drawing, suppose I

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were to draw a dot on the
outside face of the wheel.

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When this wheel precessed
around 180 degrees

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to the other side,
that dot on the outside

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would be facing in the
minus r hat direction now

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or would be pointing to the
left rather than to the right.

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And what that means is
that this disk has actually

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rotated about its diameter.

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So this disk has rotated
around a diameter like this.

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And that rotation is at the
slower angular speed capital

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omega.

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And it takes one full orbit for
it to rotate entirely around.

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If that rotation
weren't happening--

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this wouldn't make
sense physically the way

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I have this set up.

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But as an object, if that
rotation weren't happening,

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what that would mean is that
this face with a dot on it

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would always be pointing to the
right as the disk moved around.

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If it had an independent pivot
point right at the center,

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it would be physically
possible for it to do that.

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That's not what's
happening in this case.

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In this case, this face is
always pointing outward,

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radially away from
the pivot point.

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And that results in a
rotation around this diameter.

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Now, it turns out that moment
of inertia, which I'll call I2,

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happens to be half the moment
of inertia for this axis.

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So in this case, it's
one 1/4 m r squared.

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So that is another
kind of rotation

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that's happening about
the center of mass.

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And so there's an additional
angular momentum term

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arising from that rotation.

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And that's equal to
I2 times the angular

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velocity of that rotation,
which is capital omega.

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And because the axis
there is the z-axis,

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this is pointing in
the k hat direction.

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So the total angular
momentum of the gyroscope

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is I1 times omega in the r
hat direction plus I2 times

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capital omega in
the k hat direction

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plus m capital omega d squared
in the k hat direction.

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The first term, the
r hat component,

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is the only part
that's rotating.

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So this is a rotating vector.

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The k hat terms are constant.

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This is the exact expression
for the angular momentum

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of a gyroscope.

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And now we see the gyroscopic
approximation more precisely

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is saying that this
rotating term dominates

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over the other two terms.

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So it's actually that this
term is very large compared

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to either of these two
terms, which, as we can see,

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is roughly equivalent to saying
that little omega is very

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large compared to big omega.