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- [Narrator] Imagine that in an effort

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to make bowling more exciting,

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bowling alleys put a big loop-the-loop

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in the middle of the lane,
so you had to bowl the ball

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really fast to get the
ball up and around the loop

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and then only afterward, it
would go hit the bowling pins

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kinda like mini golf bowling
or something like that.

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Well if you were gonna build this,

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you'd have to know at the top of the loop,

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this structure's gonna have to withstand

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a certain minimum amount of force.

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You might wanna know how strong
do you have to make this.

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You can't have this thing breaking

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because it can't withstand
the force of the bowling ball.

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So let's ask ourselves that question.

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How much force is this loop
structure gonna have to be able

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to exert while this bowling
ball is going around in a circle

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and let's pick this point
at the top to analyze.

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We'll put some numbers in here.

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Let's say the ball was going
eight meters per second

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at the top of the loop.

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That's pretty darn fast
so someone really hurled

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this thing through here.

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Now let's say the loop
has a radius of two meters

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and the bowling ball has
a mass of four kilograms,

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which is around eight or nine pounds.

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Now that we have these numbers,
we can ask the question:

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How much normal force is there gonna be

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between the loop and the ball?

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So in other words, what is
the size of that normal force,

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the force between the two surfaces?

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This is what we'd have to
know in order to figure out

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if our structure is
strong enough to contain

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this bowling ball as it
goes around in a circle.

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And it's also a classic
centripetal force problem,

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so let's do this.

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What do we do first?

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We should always draw a force diagram.

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If we're looking for a force,
you draw a force diagram.

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So what are the forces on this ball?

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You're gonna have a force
of gravity downward,

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and the magnitude of the
force of gravity is always

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given by M times G, where
G represents the magnitude

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of the acceleration due to gravity.

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And we're gonna have a
normal force as well.

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Now which way does this
normal force point?

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A common misconception,
people wanna say that

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that normal force points up because

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in a lot of other situations,
the normal force points up.

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If you're just standing
on the ground over here,

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the normal force on you is upward

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because it keeps you from
falling through the ground,

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but that's not what this loop
structure's doing up here.

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The loop structure isn't keeping you up.

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The loop structure's keeping
you from flying out of the loop

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and that means this normal force is gonna

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have to point downward.

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So this is weird for a lot
of people to think about,

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but because the surface
is above this ball,

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the surface pushes down.

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Surfaces can only push.

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If the surface is below you,
the surface has to push up.

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If the surface was to the side of you,

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the surface would have to push right.

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And if the surface was
to the right of you,

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the surface would have to push left.

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Normal forces in other words, always push.

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So the force on the ball from the track

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is gonna be downward but vice versa.

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The force on the track from
the ball is gonna be upward.

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So if this ball were
going a little too fast

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and this were made out of wood,

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you might see this thing splinter

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because there's too much force pushing

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on the track this way.

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But if we're analyzing the
ball, the force on the ball

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from the track is downward.

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And after you draw a force diagram,

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the next step is usually,
if you wanna find a force,

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to use Newton's Second Law.

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And to keep the calculation simple,

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we typically use Newton's Second
Law for a single dimension

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at at time, i.e. vertical,
horizontal, centripetal.

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And that's what we're
gonna use in this case

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because the normal
force is pointing toward

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the center of the circular
path and the normal force

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is the force we wanna find,

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we're gonna use Newton's Second Law

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for the centripetal direction and remember

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centripetal is just a fancy word

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for pointing toward the
center of the circle.

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So, let's do it.

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Let's write down that the
centripetal acceleration

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should equal the net centripetal force

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divided by the mass that's
going in the circle.

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So if we choose this, we know that

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the centripetal acceleration
can always be re-written

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as the speed squared divided by the radius

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of the circular path that
the object is taking,

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and this should equal
the net centripetal force

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divided by the mass of the
object that's going in the circle

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and you gotta remember how
we deal with signs here

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because we put a positive sign over here

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because we have a positive sign

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for our centripetal acceleration

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and our centripetal
acceleration points toward

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the center of the circle always.

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Then in toward the center
of the circle is going to be

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our positive direction,

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and that means for these forces,

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we're gonna plug in forces toward

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the center of the circle as positive.

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So let's do that.

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This is the part where most
of the problem is happening.

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You gotta be careful here.

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I'm just gonna plug in.

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What are the centripetal forces?

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To figure that out, we just
look at our force diagram.

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What forces do we have in our diagram.

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We've got the normal force
and the force of gravity.

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Let's start with gravity.

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Is the gravitational force
going to be a centripetal force.

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First of all, that's the
question you have to ask.

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Does it even get included in here at all?

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And to figure that out you ask:

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Does it point centripetally?

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I.e. does it point toward
the center of the circle?

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And it does so we're gonna
include the force of gravity

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moreover because it points
toward the center of the circle

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as opposed to radially away
from the center of the circle.

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We're gonna include this as
a positive centripetal force.

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Similarly, for the normal
force, it also points

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toward the center of the circle,

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so we include it in this calculation

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and it as well will be a
positive centripetal force.

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And now we can solve for the normal force.

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If I solve algebraically,
I can multiply both sides

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by the mass and then I'd
subtract MG from both sides.

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And that would give me the
mass times V squared over R

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minus the magnitude of
the force of gravity,

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which if we plug in numbers,
gives us four kilograms

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times eight meters per second squared,

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you can't forget the square,

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divided by a two meter
radius minus the magnitude

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of the force of gravity which
is four kilograms times G

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which if you multiply that
out gives you 88.8 newtons.

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This is how much downward
force is exerted on the ball

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from the track but from
Newton's Third Law,

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we know that that is also
how much force the ball

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exerts upward on the track.

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So whatever you make this loop out of,

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it better be able to
withstand 88.8 newtons

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if people are gonna be rolling
this ball around the loop

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with eight meters per second.

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Now let me ask you this.

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What if the ball makes
it over to here, right?

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So the ball rolls around and
now it's over at this point.

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Now how much normal force
is there at this point?

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Is it gonna be greater than, less than,

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or equal to 88.8 newtons.

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Well to figure it out, we
should draw a force diagram.

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So there's gonna be a force of gravity.

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Again, it's gonna point straight down,

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and again, it's gonna be equal to

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at least the magnitude
of it will be equal to

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the mass times the magnitude
of acceleration due to gravity.

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And then we also have a normal force,

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but this time, the normal
force does not push down.

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Remember, surfaces push outward

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and if this surface is
to the left of the ball,

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the surface pushes to the right.

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This time our normal
force points to the right.

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And let's assume this a well oiled track

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so there's really no
friction to worry about.

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In that case, these would
again be the only two forces.

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So what about the answer to our question.

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Will this normal force
now be bigger, less than,

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or equal to what the normal
force was at the top.

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Well I'm gonna argue it's gotta be bigger,

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and I'm gonna argue it's
gonna have to be much bigger

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because when you plug in over here,

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into the centripetal forces,

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you only plug in forces
that point radially.

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That is to say centripetally,

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either into the circle,
which would be positive,

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or radially out of the circle,
which would be negative.

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If they neither point into
nor out of the circle,

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you don't include them in
this calculation at all

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because they aren't
pointing in the direction

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of the centripetal acceleration.

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In other words, they're not causing

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the centripetal acceleration.

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So for this case over
here, gravity is no longer

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a centripetal force because
the force of gravity

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no longer points toward
the center of the circle.

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This force of gravity is
tangential to the circle.

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It's neither pointing into nor out of,

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which means it doesn't factor into

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the centripetal motion at all.

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It merely tries to speed
the ball up at this point.

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It does not change the ball's direction,

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which means it doesn't
contribute to making this ball

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go in a circle, so we don't
include it in this calculation.

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So when we solved for the normal force,

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we'd multiply both sides by M,

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we would not have an MG anymore.

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So we wouldn't be subtracting this term

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and that's gonna make
our normal force bigger.

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Moreover, the speed of
this ball's gonna increase

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compared to what it was up here.

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So as the ball falls
down, gravity's going to

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speed this ball up and now
that it's speed is larger,

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and we're not subtracting
anything from it,

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The normal force will be
much greater at this point

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compared to what it was
at the top of the loop.

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So recapping, when you wanna solve

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the centripetal force problem,

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always draw your force diagram first.

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If you choose to analyze the forces

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in the centripetal
direction, in other words,

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for the direction in toward
the center of the circle,

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make sure you only plug
in forces that are into,

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radially into the circle or
radially out of the circle.

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If they're radially into the
circle, you make them positive.

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If they were radially out of the circle,

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you would make them negative.

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And if they neither point radially inward,

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toward the center of the circle

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or radially outward, away
from the center of the circle,

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you just do not include
those forces at all

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when using this centripetal direction.