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DEEPTO CHAKRABARTY: A pivoted
rod held horizontal, parallel

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to the ground, and
released from rest

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will simply fall, rotating
about the pivot point.

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In particular, suppose
we have a mass attached

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to one end of a pivoted rod.

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So here is my pivot.

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Here's my pivoted rod, which
we'll assume is massless.

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And I have-- and
it has a length d,

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and I attach a
mass m to one end.

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If I let this go from
rest, it will simply

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fall, rotating about the pivot
point, which I'll call s.

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And it's easy to
understand that in terms

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of the action of gravity
which is acting downward

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and the resulting
torque about point s.

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If I replace this point mass
with a wheel of the same mass,

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so this disk is a
wheel with, let's say,

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a radius r and the same mass
m attached to a pivoted rod.

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The rod is massless
and has length d.

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If I hold this horizontal,
parallel to the ground,

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and release it from
rest, it will still just

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fall to the ground.

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Not a surprising result.

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What is surprising is if
I then spin up the wheel,

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so if I have the
wheel rotating about

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its axle with some large
angular velocity, little omega,

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and then I hold it horizontal,
parallel to the ground,

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and release it from rest, then
remarkably this wheel plus axle

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will not fall.

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It will remain horizontal,
parallel to the ground.

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But the center of
mass of the wheel

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will execute a small circular
orbit about the vertical axis

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through the pivot point.

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This remarkable and
very non-intuitive

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motion is called
precession and the system

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that's undergoing precession
is called a gyroscope.

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Let's see if we can
understand this behavior

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in terms of the angular
momentum of the system.

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So what I've drawn here is
a side view of the system.

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So let's define
some coordinates.

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So we have r-hat in the
radial outward direction,

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k-hat in the direction of the
z-axis, the vertical axis.

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And then we'll define theta-hat
pointing into the screen.

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OK, so that's a side view.

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I'd like to now draw a top view.

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So let's say we're looking
down along the z-axis

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from the top on the system.

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So now here is my
pivot point, and here

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is a top view of my wheel.

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Again, this is a distance d.

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Now this is still
the r-hat direction.

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This is the theta-hat direction.

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And the k-hat direction. is
pointing out of the screen.

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Let's draw the forces
acting on our diagram here.

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So the weight is acting at the
center of mass of the wheel.

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That's mg downward.

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And there's a normal
force acting upwards

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at the pivot point.

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The torque is just
given by r cross f,

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and so relative to
point s at the distance

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d times the weight mg.

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So the torque is mgd.

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And by doing r cross f
with the right hand rule,

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we see that it's directed in
the plus theta-hat direction.

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Now again, let's
suppose that I'm

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holding the wheel horizontal
and release it from rest.

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If the wheel is not spinning--

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so this is not rotating,
this is a stationary wheel

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that I'm holding horizontal,
and I release it from rest,

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then its initial
angular momentum,

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with respect to point s, is 0.

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Over a short time
interval, delta t,

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the torque which is acting
in the theta-hat direction,

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will cause an angular impulse or
change in the angular momentum.

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That change in angular
momentum, delta l vector,

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is equal to the torque
times that short time

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interval, delta t.

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And it will act in the
theta-hat direction.

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The change in
angular momentum will

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be in the theta-hat direction
because that's the direction

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that the torque is in.

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So for this case--

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and by the way, this view here,
I should have labeled this.

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This is a top view.

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So this is the side view.

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This is the top view
of the same system.

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So the torque is mgd in
the theta-hat direction.

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So in the side view, that's
going into the screen.

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And in the top view, that's
going pointing upward.

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That's the theta-hat direction.

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So in the case where
the wheel is not

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spinning, if we think
about the top view,

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the initial angular
momentum is 0.

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Right?

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So I'll just draw that as a dot.

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So that's l initial equals 0.

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And then I add a small torque,
or small angular impulse

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due to the torque,
delta l, which

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is in the theta-hat direction.

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So that's pointing in
the theta-hat direction.

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So that's my delta l.

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And when I sum those
together, I start out with 0.

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I add a small delta l.

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So my final angular momentum,
a time delta t later,

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is just equal to my delta l.

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So this is l final.

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And that's pointing in
the theta-hat direction.

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So a torque in the
theta-hat direction,

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into the screen
for the side view,

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is consistent with the
wheel falling down.

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So with theta-hat
pointing into the screen,

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the wheel will
basically fall this way.

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It's rotating about point s.

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And that's consistent
with the torque pointing

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in the theta-hat direction.

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Now as the wheel
falls, the torque

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continues to point in
the theta-hat direction.

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And so the angular
acceleration will increase.

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Now what happens if this wheel
is not stationary, but instead

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is spinning rapidly?

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In that case, the
torque remains the same.

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It's still mgd in the
theta-hat direction.

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But now the initial
angular momentum is not 0.

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Rather it is a very large vector
pointing along the spin axis.

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Let's choose the
sense of rotation

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such that l points in
the plus r-hat direction.

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That's actually the
way I've drawn it here.

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So in that case, the
angular momentum vector

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initially points in the
plus r-hat direction.

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So then what happens over
a short time, delta t?

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So now let's consider the case
where the wheel is spinning.

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So now my initial
angular momentum

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is a large vector pointing
in the r-hat direction.

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That's l initial.

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I'm adding a small
perpendicular vector, delta l,

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in the theta-hat direction.

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And so the sum of
those two things,

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if this is the original r-hat
direction, what's happened

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is that my new vector is at
a small angle with respect

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to the original r-hat direction.

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I'll call that
angle delta theta.

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So what I've done is I've
rotated my initial angular

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momentum vector by a small angle
without changing its length.

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Notice the two very
different situations.

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In one case, I start out
with 0 angular momentum.

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And all the angular
momentum I end up with

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comes from the angular
impulse due to the torque.

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In the second case, where
the wheel is spinning,

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I start out with a very large
initial angular momentum.

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I then add a small
angular impulse,

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small compared to my
initial angular momentum,

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in the perpendicular direction.

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And that causes not a change
in the length of the vector,

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but a change in its direction.

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Which means that the angular
momentum vector rotates.

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That's why the
system precesses when

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the wheel is rotating rapidly.

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Before we can understand
precession more carefully,

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it'll be useful to review
the mathematics of rotating

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vectors.

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So first, suppose we have
a vector that I'll call r1.

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And I'm going to add a
vector delta r to this.

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And I'm going to
have the condition

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that the length of
delta r is much,

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much smaller than the length
of my original vector r1,

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and that delta r is
perpendicular to R1.

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So here's my delta r.

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And so if I add
these two vectors,

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let's say r2 is the
sum of r1 plus delta r,

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so that's my vector r2.

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And I'm going to call
this angle delta theta.

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And now few things to note.

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First of all, since this
is a right triangle,

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notice that the length
r2 is equal to r1 divided

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by the cosine of delta theta.

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But if that angle is small, if
delta theta is a small angle,

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then cosine theta is
well approximated by 1,

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and so this is just
my original length r1.

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And since r1 is equal to r2,
I'm just going to call that--

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call them both r.

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And this is for
small delta theta.

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So that tells us that
the vector just rotates.

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When I have a large vector,
and I add a small perpendicular

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vector, the result is to
rotate the original vector

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without changing its length.

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As long as the angle is
small or, equivalently, as

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long as delta r is very small
compared to my original vector

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length, I'll get
a pure rotation.

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In addition, again
looking at this triangle,

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notice that delta r is
equal to [AUDIO OUT], which

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I can describe as r, times
the sine of delta theta.

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And again, if the
angle is small,

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if delta theta's a
small angle, then this

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is well approximated by
r times the angle delta

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theta in radians.

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And again, this is
for small delta theta.

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So now if I divide
this last equation

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by delta t, small time interval
in which this rotation is

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happening, then I write this
as delta r vector divided

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by delta t.

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And I take the
magnitude of that.

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That is equal to r delta
theta divided by delta t.

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And if I go to the limit
as delta t gets small,

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as delta t approaches
0, then I can write this

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as the derivative of the
vector r, the time derivative,

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I should say the
magnitude of the time

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derivative of vector r.

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And that's equal to the length
r times the time derivative

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of the angle d theta dt.

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00:13:04,350 --> 00:13:07,500
But d theta dt is just
the angular velocity.

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So I could write that
as r times capital omega

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for the angular velocity.

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And this is just
our familiar result

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that for circular motion, for
the rotation of a position

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vector, the velocity
is equal to the radius

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00:13:27,330 --> 00:13:34,044
of the circle times the angular
velocity of the rotation.

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00:13:34,044 --> 00:13:35,460
Another way of
thinking of that is

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that the magnitude of the
rate of change of the position

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00:13:39,780 --> 00:13:42,030
vector, which we
call the velocity,

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00:13:42,030 --> 00:13:44,520
is equal to the length
of the rotating vector

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00:13:44,520 --> 00:13:49,680
r times the angular
velocity of the rotation.

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00:13:49,680 --> 00:13:52,740
But there's nothing special
about the position vector

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that I used in order
to do this analysis.

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So if instead of a position
vector I considered any vector,

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00:14:02,970 --> 00:14:11,780
so let's say this is my
rotational motion of a vector

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00:14:11,780 --> 00:14:17,000
that I'll call A. So
that's A at time t.

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00:14:17,000 --> 00:14:25,790
And then at some later time,
that's A of t plus delta t.

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00:14:25,790 --> 00:14:33,260
This vector is my delta A. I'll
call this angle delta theta.

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00:14:33,260 --> 00:14:37,070
And so the vector A is
rotating in that direction

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with an angular
velocity capital omega.

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00:14:39,920 --> 00:14:45,560
Now in the example I just did,
my vector A is actually r of t.

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00:14:45,560 --> 00:14:47,390
But so everywhere
where I have an r here,

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00:14:47,390 --> 00:14:51,860
I can just write an A. This
is now just an arbitrary

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vector in space that is rotating
at an angular velocity capital

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omega.

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And what we see, using
the same analysis,

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we would find that the magnitude
of the time rate of change

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00:15:06,830 --> 00:15:10,670
of the vector, of
the rotating vector,

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00:15:10,670 --> 00:15:13,760
is equal to the length of
the rotating vector, which

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00:15:13,760 --> 00:15:20,450
is A, times d theta dt.

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00:15:20,450 --> 00:15:22,700
Or in other words, the
length of the rotating

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00:15:22,700 --> 00:15:26,750
vector times the angular
velocity of the rotation.

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00:15:26,750 --> 00:15:28,500
That's true for any vector.

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00:15:28,500 --> 00:15:29,000
OK?

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This is a general
result. v equals

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r omega is just a special
case of this general rule

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00:15:36,920 --> 00:15:40,010
where, in that case, my rotating
vector is a position vector.

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00:15:40,010 --> 00:15:43,280
But this is true for any vector
that's rotating in space.

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00:15:43,280 --> 00:15:46,610
In particular for a
rotating angular momentum--

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00:15:52,430 --> 00:15:58,440
for a rotating angular
momentum vector,

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00:15:58,440 --> 00:16:04,410
I have that the magnitude of the
time derivative of the rotating

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00:16:04,410 --> 00:16:09,111
angular momentum vector is
just equal to the length

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00:16:09,111 --> 00:16:11,610
of the angular momentum vector,
the magnitude of the angular

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00:16:11,610 --> 00:16:16,570
momentum, times the angular
velocity of rotation.

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00:16:16,570 --> 00:16:19,176
Now in addition, in the
particular case of a rotating

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00:16:19,176 --> 00:16:21,550
angular momentum vector or of
any angular momentum vector

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00:16:21,550 --> 00:16:27,160
rather, we know that the time
derivative of the angular

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00:16:27,160 --> 00:16:31,920
momentum vector is also
equal to the torque vector.

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00:16:31,920 --> 00:16:33,930
So I can set these
two things equal.

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00:16:33,930 --> 00:16:36,510
In the case of a rotating
angular momentum vector,

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00:16:36,510 --> 00:16:39,000
the magnitude of
the torque is given

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00:16:39,000 --> 00:16:43,260
by the magnitude of the
rotating angular momentum

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00:16:43,260 --> 00:16:46,600
vector times the angular
speed of rotation.

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00:16:46,600 --> 00:16:49,680
And that's just using
the general behavior

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00:16:49,680 --> 00:16:52,430
of a rotating vector in space.