WEBVTT
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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.