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Let's consider the motion of
a car on a circular track,

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and the track is frictionless.

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And it's also banked.

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So this is the overhead
view of our circular track.

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It has radius, r.

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And here's our car moving
at a constant velocity.

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Now, from the side view when
we want to look at that bank

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turn-- let's draw a side view.

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So here's our side view, and the
car is moving with a velocity

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into the plane of the figure.

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Now this surface
here is frictionless.

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And what we'd like
to do is find out

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what speed the car can move
such that it doesn't slide

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up or down the inclined plane.

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So how should we analyze that?

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Well our approach will be to
apply Newton's second laws.

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Now what's very
important to realize

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is this is circular motion.

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And for circular motion
we know that the car

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is accelerating towards
the center of the circle.

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Now from the side view, towards
the center of the circle

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is in this direction.

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So the car is accelerating
radially inward.

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And that will guide how we
choose our coordinate system.

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And so we can then write
our free body force diagram.

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So let's begin
with the analysis.

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So we don't need to see
the overhead view anymore.

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So I'll just remove that, and
then we can start drawing.

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This is what we can refer to
as our acceleration diagram.

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And now let's draw
the force diagram

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on the car as our
choice of system.

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So here's our angle, phi.

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Because the
acceleration was inward

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we're going to choose a
radially outward coordinate

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and a vertical
coordinate, K hat up.

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Notice that this is
different than just a mass

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on a fixed incline
plane where we used unit

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vectors up and down
the inclined plane.

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The reason we choose
our unit vectors

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like that-- to
emphasize it again,

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is we already know this
is constrain motion.

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It's circular motion.

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Now what is the free body--
what are the forces on the car?

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Well there is the normal
force, the plane on the car,

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and the gravitational force.

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Now here-- whenever you're
doing problems like this

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remember that the
trig is crucial to get

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these angles right.

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So that's phi, and that's phi,
and that's our free body force

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diagrams.

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And now we can write
down Newton's second law.

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So we'll start out with
our usual approach,

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and we have two directions
that we have to consider.

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So in the radial direction
there is an inward component

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of the normal force, like that.

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And that's opposite the
angle, so it's pointing

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opposite our direction.

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So we have minus n sine phi.

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The gravitational force is
only in the negative K hat

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direction.

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And we know that the
acceleration is inward,

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and so there's a minus sign.

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We have the mass, and the
constraint for circular motion

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is that that's phi
squared over r.

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Where r was the
radius of that circle,

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this can be thought of
as the central point.

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Now for the k hat
direction, we have

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a component of the normal
force that's pointing up.

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I'll just draw that.

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That's adjacent to the angle,
so we have plus and cosine phi.

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And we have the gravitational
force downward, minus mg.

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And as far as the
vertical direction

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goes, because the car
is going in a circle,

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there is no acceleration up or
down in the vertical direction.

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Again, that's a constraint
in this problem.

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That's equal to 0.

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So in this problem, this
is the side that we know,

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and we're trying to figure
out up to the speed, v. Now,

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how do we analyze this problem?

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Well you can see that if
I write my two equations,

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this n sine phi equals
mv squared over r.

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And cosine phi equals mg.

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We have two equations.

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We have two unknowns, v and n.

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Many times people
just solve for n

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and try to find the equation--
and then substitute in,

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but you're also allowed
to divide two equations,

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and that's much easier.

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The masses cancel and we get the
relationship, that tan phi is v

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squared over rg.

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And so we have our result that
the speed that the car can

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travel on a frictionless
inclined plane

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and maintain uniform
circular motion

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is exactly the square
root of rg tan phi.

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And that's how we
analyze the motion

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of this car on a banked turn.

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What we would now
like to think about

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is what would happen if
you're traveling faster

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or slower than this speed.

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So suppose we have the
prime bigger than the speed.

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Now, what that means is
that the car is going faster

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and the new equilibrium-- if you
asked what would the radius be

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such that traveling
at v primed the car

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undergoes circular
motion, the prime

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would be equal to
r prime g tan phi.

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And so in order to
go with this speed

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you have to go at
a greater radius.

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Now what does that mean?

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Well, that means that if
the car is traveling at v,

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so it's in this circular motion,
and now the driver increases

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the speed to v
prime, the car will

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start to slide up the
inclined plane-- remember,

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it's frictionless--
until it reaches

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a-- as it starts slide
up the inclined plane

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it will get to this
new radius, r prime,

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but because a car will have
a little inertia it will

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overshoot that
speed, that radius,

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and then it will start to come
back down the inclined plane,

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and it will oscillate
about that point.

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It won't be sinusoidal
oscillations,

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but they'll be a
periodic oscillation

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about this new radius, r prime.

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The same thing, too, if we have
the double prime less than d,

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then the double prime is equal
to r double prime G tan phi.

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Now remember, this double
prime is not two derivatives.

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I'm just using
that as a notation

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to indicate different speeds.

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So if the car is going
along at speed, v, and slows

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down, what would happen
is the new equilibrium

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radius is smaller so the car
slides down the inclined plane

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until it gets to r double prime.

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It turns out that it will
overshoot that a little bit,

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and then start to move up.

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And, again, it will oscillate
around this new equilibrium

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length.

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So on a frictionless
inclined plane

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if you go faster then this
speed the car slides up.

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If you go slower than this
speed, the car slides down.