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DEEPTO CHAKRABARTY:
Now that we've

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seen how to describe
a rotating vector,

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we can use that to analyze
the motion of our gyroscope.

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So again, I'll draw a side
view of my pivot, my rod.

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Here's the wheel.

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I'll call this point
S. That's a distance d.

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And we'll assume that
the angular velocity is

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in that direction, so that
the spin angle velocity

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vector is pointing outwards
in the plus r hat direction.

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That's the k hat direction and
theta hat is into the screen.

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Now, again, the weight
is acting downwards

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at the center of
mass of the wheel.

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There's a normal
force acting upwards.

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So the spin angular momentum
with respect to point S

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is just equal to the
moment of inertia

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of the disk about
its center of mass

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times the angular
speed of the spin,

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and that's directed in
the plus r hat direction.

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And this is the angular
momentum with respect

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to point S. Now, the torque
with respect to point S, again,

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it's R cross F
relative to point S.

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That's going to be Mgd in
the plus theta hat direction.

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And for a fast omega, if omega
is a large angular speed,

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and therefore if
the angular momentum

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vector is a large
vector, then the torque,

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which acts perpendicular
to the angular momentum,

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will cause the angular
momentum vector to rotate.

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And so as a rotating
vector, we can

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write that the
magnitude of the time

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derivative of that
rotating vector

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is equal to the angular
velocity of the rotation

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times the length of that vector.

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And this is Ls.

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And so the length of that vector
is just I times little omega,

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and then we multiply
that by capital omega.

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Now, this quantity here,
dL dT, is the torque.

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So that's the magnitude
of the torque.

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But the torque we said is
equal to Mgd in the theta hat

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direction.

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And so the torque,
Mgd, is equal to I

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little omega times big omega.

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So I can solve for big omega.

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And big omega, which is the
angular speed of rotation

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of the angular
momentum vector, is

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Mgd divided by moment of
inertia, I, times little omega.

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Now recall, little omega
is the angular velocity,

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the angular speed of the spin
of the disk or the wheel.

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Capital omega is
the angular speed

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of the rotation of the
angular momentum vector.

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It basically tells
us the speed at which

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the center of mass of the wheel
orbits around the vertical axis

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through the pivot.

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We call capital omega the
precessional angular velocity.

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So the precessional angular
velocity is capital omega.

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And notice that the
faster little omega

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is, the bigger little omega
is, the slower the precession

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angular velocity is.

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Now, this expression
tells us what

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the magnitude of
capital omega is,

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what the magnitude of that
precessional angular velocity

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is, but it doesn't tell us which
way the system is processing,

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whether say, viewed
from the top,

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the motion is clockwise
or counterclockwise,

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or equivalently, which way
the vector, capital omega,

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is pointing.

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We know it must point
along the vertical axis,

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but does it point upwards,
in the plus K hat direction

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or downwards in the
minus K hat direction.

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To see that, we need to look at
which way the angular momentum

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vector is rotating.

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And there are two possibilities.

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So let's again go to a
top view of our gyroscope.

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So suppose here is
the pivot point,

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and here is my gyroscope.

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And the way I'm drawing things
is that r hat is this way,

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is the top view, so
theta hat is that way,

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and k hat is out of the screen.

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So in the example
that I did earlier,

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the angular momentum was
pointing in the plus r hat

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direction.

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So that's L sub s
pointing that way.

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If the torque is pointing
in the theta hat direction,

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then that's going to act to
rotate the angular momentum

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vector this way.

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And so the sense of
rotation will be like that.

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And so in that case,
looking down on the system,

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we would see a
counterclockwise rotation.

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And that's equivalent to omega
vector pointing in the plus k

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hat direction.

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Alternatively, suppose
the wheel were spinning

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in the other direction.

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OK, so the sense of rotation
of this wheel around the axle,

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we're in the opposite direction.

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In that case, even when
the wheel was on this side

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of the pivot --again,
this is a top view--

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the angular momentum
vector would be pointing

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in the opposite direction.

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So now this is my
angular momentum vector.

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It's pointing in the
opposite direction

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that the axle is pointing.

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Still along a line, but in
the minus R hat direction.

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But the torque would still be
in the theta hat direction.

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And so my new angular
momentum vector

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would rotate this
way, which would

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be equivalent to the axle
rotating in this direction.

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And that's equivalent to a
rotation in the opposite sense,

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clockwise as viewed
downward from the top,

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or in other words, with capital
omega hat vector pointing

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in the minus K hat direction.

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So the direction of
precession depends

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upon which way the
wheel is spinning,

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which way the spin angular
momentum vector is actually

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pointing.

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Now, I'd just like to point out
that there's an approximation

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that we've been making here.

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I've alluded to it,
but I want to make

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it very specific right now.

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We have been assuming that the
spin angular momentum vector

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L is large enough that
the torque vector, which

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is perpendicular, provides
only a small angular

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impulse that rotates L
without changing it in length.

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Now, as L rotates,
the direction of r hat

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and the direction of
theta rotate with it,

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and the torque is always
in the theta hat direction.

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So after any small delta T,
when the angular momentum

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vector rotates, the
instantaneous torque

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at that next instant is still
perpendicular to the rotating

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momentum vector.

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And so we're making
an approximation

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that L is large enough
that we can consider

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the instantaneous
angular impulse

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as a small perpendicular
perturbation.

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That's equivalent, it
turns out, to saying

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that, so this is the
approximation that we're

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making, that we've been
making so far, in this vector,

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is that little omega, the
spin angular velocity,

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is much, much
larger than capital

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omega, the precession
angular velocity.

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This is called the
gyroscopic approximation.

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We'll see a more precise
statement of it later on.

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But it's basically equivalent
to saying that the spin angular

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momentum vector is so large
that we can consider the angular

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impulse due to the
torque as causing

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a pure rotation of the
vector without any change

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in its length.

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It's important to keep
in mind that a vector can

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include both a
rotating component

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and a constant component.

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And in that case, it's important
to identify what the rotating

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component is in order to use
the analysis that we presented

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earlier.

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So for example, let's consider
the case of a tilted gyroscope,

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instead of one that's horizontal
and parallel to the ground.

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So here is my pivot point.

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That's the vertical.

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And instead of a
horizontal gyroscope,

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now I'll draw my gyroscope
at some angle like this.

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Here's my wheel.

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I'll call that angle
with the vertical phi.

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This distance is still d.

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And I'll use the usual
coordinate system,

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that r hat is
pointing out this way,

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k hat is pointing
vertically, and theta hat

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is pointing into the screen.

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Now, in this case, the gyroscope
will still process around

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the vertical axis
to the pivot point,

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and the angle phi
will remain constant.

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Now, the angular momentum
vector due to spin points now

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not in the r hat direction,
but again, outward along

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the axis of rotation.

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And that angular
momentum vector can

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be decomposed into two
components, an r hat component

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and a k hat component.

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So in other words, my
angular momentum vector

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can be written as the sum
of a vector pointing along

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Z axis or the k hat direction
plus a vector pointing

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in the radial direction.

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I can also write that as L sub
z times k hat plus L sub r times

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r hat.

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Now, notice as this
gyroscope precesses

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around, the z hat
component is constant,

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but the r hat component
rotates around.

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OK, so the r hat component,
the vector along the r hat

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direction, is a purely
rotating vector,

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whereas the z component
is a constant vector.

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So L sub z is constant
and L sub r is rotating.

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And so now, the
magnitude of dL dT,

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which is equal to the magnitude
of dLz dT plus dLr dT,

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well dLz dT, since that's a
constant vector, is just 0.

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So the time derivative just
involves the r component.

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And so that's equal to the
angular speed of rotation times

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the length of the rotating
vector, which is just the r

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component, not the z component.

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So this is multiplied
by L sub r.

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Now, in this particular
case, the r component

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of the angular momentum vector
is L, the angular momentum

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vector, times the sine of phi.

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So in this particular case, this
would be omega times the full L

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times the sine of phi.

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So in the case
that we have here,

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the angular momentum vector
consists of a constant part

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and a rotating part.

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And the magnitude of
its time derivative

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is equal to the angular
velocity rotation

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times the length of the
rotating part of the vector.