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- [Instructor] People find
centripetal force problems

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much more challenging than
regular force problems,

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so we should go over at
least a few more examples,

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and while we're doing
them, we'll point out

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some common misconceptions
that people make along the way.

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So, let's start with this example,

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and it's a classic.

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Let's say you started with a yo-yo

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and you whirled it around vertically,

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and I think this is called
the around the world,

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if you want to look it up on YouTube,

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looks pretty slick.

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They whirl it around, goes
high, and then it goes low.

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So this is a vertical circle,

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not a horizontal circle.

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We're not rotating this ball
around on a horizontal surface.

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This ball is actually
getting higher in the air

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and then lower in the air.

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But for our purposes, we just need to know

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that it's a mass tied to a string.

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Let's say the mass of the yo-yo

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is about 0.25 kilograms,

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and let's say the length of the string

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is about 0.5 meters.

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And let's say this ball is going about

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four meters per second when
it's at the top of its motion.

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And something you might want to know

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if you're a yo-yo manufacturer

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is how much tension should
this rope be able to support,

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how strong does your string need to be.

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So let's figure out for this example,

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what is the tension in the string

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when this yo-yo is at its maximum height

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going four meters per second?

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And if it's a force you want to find,

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the first step always is to
draw a quality force diagram.

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So let's do that here.

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Let's ask what forces are on this yo-yo.

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Well, if we're near Earth

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and we're assuming we're going to be

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near the surface of the
Earth playing with our yo-yo,

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there's gonna be a force of gravity

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and that force of gravity is
gonna point straight downward.

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So the magnitude of that force of gravity

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is gonna be m times g,
where g is positive 9.8.

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g represents the magnitude

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of the acceleration due to gravity,

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and this expression here represents

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the magnitude of the force of gravity.

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But there's another force.

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The string is tied to the mass,

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so this string can pull on the mass.

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Strings pull, they exert
a force of tension.

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Which way does that tension go?

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A lot of people want to draw
that tension going upward,

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and that's not good.

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Ropes can't push.

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If you don't believe me, go get a rope,

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try to push on something.

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You'll realize, oh yeah, it can't push,

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but it can pull.

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So that's what this rope's gonna do.

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This rope's gonna pull.

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How much?

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I don't know.

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That's what we're gonna try to find out.

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This is gonna be the force
of tension right here,

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and we'll label it with a capital T.

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We could have used F with the sub T.

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There's different ways
to label the tension,

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but no matter how you label it,

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that tension points in towards
the center of the circle

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'cause this rope is pulling on the mass.

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So, after you draw a force diagram,

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if you want to find a force,

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typically, you're just gonna
use Newton's second law.

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And we're gonna use this formula as always

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in one dimension at a time

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so vertically,
horizontally, centripetally,

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one dimension at a time

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to make the calculations
as simple as possible.

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And since we have a
centripetal motion problem,

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we have an object going in a circle,

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and we want to find one of those forces

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that are directed into the circle,

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we're gonna use Newton's second law

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for the centripetal direction.

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So we'll use centripetal acceleration here

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and net force centripetally here.

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So in other words, we're gonna write down

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that the centripetal acceleration

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is gonna be equal to the
net centripetal force

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exerted on the mass that's
going around in that circle.

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So because we chose the
centripetal direction,

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we're gonna be able to replace

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the centripetal acceleration
with the formula

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for centripetal acceleration.

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The centripetal acceleration's
always equivalent

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to v squared over r,

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the speed of the object squared

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divided by the radius of the circle

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that the object is traveling in.

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So we set that equal to
the net centripetal force

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over the mass,

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and the trickiest part here,

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the part where the failure's
probably gonna happen

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is trying to figure
out what do you plug in

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for the centripetal force.

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And now we gotta decide

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what is acting as our centripetal force

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and plug those in here
with the correct signs.

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So, let's just see what
forces we have on our object.

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There's a force of tension
and a force of gravity.

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So, when you go try to figure
out what to plug in here,

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people start thinking, they
start looking all over.

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No, you drew your force diagram.

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Look right there.

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Our force diagram holds
all the information

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about all the forces that we've got

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as long as we drew it well.

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And we did draw it well.

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We included all the forces,

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so we'll just go one by one.

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Should we include, should we even include

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the force of gravity in this
centripetal force calculation?

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We should because we're
gonna include all forces

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that point centripetally and remember,

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the word centripetal is just a fancy word

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for pointing toward the
center of the circle.

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And this force of
gravity does point toward

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the center of the circle.

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So we're gonna include this
force in our centripetal force.

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It's contributing, in other words,

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to the centripetal force.

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It's one of the forces
that is causing this ball

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to go in a circle,

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so we include it in this formula.

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So mg is the magnitude
of the force of gravity.

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We have to decide, do we
include that as a positive

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or a negative?

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Many people want to
include it as a negative

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'cause it points down,

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but when we're dealing with
the centripetal direction,

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it's inward that's gonna be positive,

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not necessarily up
that's gonna be positive.

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If up happens to point in,

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then we'd consider it positive.

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So if we were down here,

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up is positive.

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But up here, down is positive

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'cause it points in toward
the center of the circle,

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and if we're over here,

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diagonally up and left would be positive

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because any force that would be pointing

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toward the center of the circle

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is gonna be included as a positive sign

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and there's a reason for that.

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The reason we're including

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toward the center of
the circle as positive

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is because we chose to write

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our centripetal acceleration as positive.

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And since we know the
centripetal acceleration

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points toward the center of the circle,

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if we make the centripetal
acceleration positive,

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we've committed to toward
the center of the circle

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as being positive as well.

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In other words, we could have decided that

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out of the circle's positive,

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but if we did that,

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this centripetal acceleration
that points inward

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would have had to be included
with a negative sign over here

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and that's just weird.

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Nobody does that.

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So we choose into the circle as positive.

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That makes our centripetal
acceleration positive,

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but it also makes every
force that points inward

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positive as well.

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And long story short,
this force of gravity

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is gonna be counted as a
positive centripetal force

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since it points inward toward
the center of the circle.

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And we keep going.

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We've got another force here.

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We've got a force of tension.

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Do we include it in here?

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Yes we do because it points
toward the center of the circle.

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And do we include it as
a positive or a negative?

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Since it also points toward
the center of the circle,

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we're gonna include it as a
positive centripetal force.

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It is also one of the forces

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that causes this ball to
move around in a circle.

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In other words, the combined force of both

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gravity and the force of tension

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are making up the net
centripetal force in this case.

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So now, if we want to
solve for the tension,

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we just do our algebra.

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We'll multiply both sides by the mass.

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Then, we'll subtract mg from both sides,

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and if we do that, we'll
end up with the tension

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equals m v squared over r

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minus mg, which if we plug in numbers,

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we'd get that the tension in the rope

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is 5.55 Newtons.

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So this is to be expected.

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We subtracted the force of
gravity, the magnitude of it

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from this net centripetal force,

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so this term here represents

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the total amount of
centripetal force we need

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in order to cause this
yo-yo to go in a circle.

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But the amount of tension we need

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is that amount minus the force of gravity,

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and the reason is, the force
of gravity and tension together

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are both acting as the centripetal force.

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So, neither one of them have to add up

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to the total centripetal force.

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It's just both together
that have to add up

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to the centripetal force,

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and because of that, the
tension does not have to be

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as large as it might have been.

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However, if we consider the case

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where the yo-yo rotates down
to the bottom of its path,

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down here, once the yo-yo
rotates down to this point,

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our force diagram's gonna look different.

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The force of gravity
still points downward,

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the force of gravity's
always gonna be straight down

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and the magnitude is always
gonna be given by m times g.

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But this time, the tension points up

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because the string is
always pulling on the mass.

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Ropes can only pull.

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Ropes can never push,

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so this rope is still pulling the mass,

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the yo-yo toward the center of the circle.

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So now, when we plug in over here,

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one of these forces is gonna be negative.

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Before they were both positive

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and they were both positive

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because both forces pointed
toward the center of the circle.

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Now, only one force is pointing toward

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the center of the circle,

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and we can see that that's tension.

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Tension's pointing toward
the center of the circle.

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Gravity's pointing away from the center,

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radially away from the center.

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That means tension still
remains a positive force,

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but the force of gravity
now, for this case down here,

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would have to be considered
a negative centripetal force

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since it's directed away from
the center of the circle.

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So, if we were to calculate the tension

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at the bottom of the path,

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the left hand side would
still be v squared over r

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'cause that's still the
centripetal acceleration.

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The mass on the bottom would still be m

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'cause that's the mass of
the yo-yo going in a circle.

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But instead of T plus
mg, we'd have T minus mg

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since gravity's pointing radially out

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of the center of the circle.

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And if we solve this expression

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for the tension in the string,

258
00:08:43,561 --> 00:08:45,355
we'd get that the tension equals,

259
00:08:45,355 --> 00:08:47,114
we'd have to multiply both sides by m,

260
00:08:47,114 --> 00:08:49,089
and then add mg to both sides.

261
00:08:49,089 --> 00:08:50,686
And we'd get that the
tension's gonna equal

262
00:08:50,686 --> 00:08:53,626
m v squared over r, plus m g.

263
00:08:53,626 --> 00:08:57,247
This time, we add m g to
this m v squared over r term,

264
00:08:57,247 --> 00:08:59,228
whereas over here, we had to subtract it.

265
00:08:59,228 --> 00:09:00,983
And that should make sense conceptually

266
00:09:00,983 --> 00:09:04,086
since before, up here,
both tension and gravity

267
00:09:04,086 --> 00:09:06,141
were working together to add up

268
00:09:06,141 --> 00:09:08,009
to the total centripetal force,

269
00:09:08,009 --> 00:09:10,077
so neither one had to be as big

270
00:09:10,077 --> 00:09:11,451
as they might have been otherwise.

271
00:09:11,451 --> 00:09:15,249
But down here, not only is
gravity not helping the tension,

272
00:09:15,249 --> 00:09:17,487
gravity's hurting the centripetal cause

273
00:09:17,487 --> 00:09:20,358
by pulling this mass out of
the center of the circle,

274
00:09:20,358 --> 00:09:21,841
so the poor tension in this case

275
00:09:21,841 --> 00:09:25,044
not only has to equal the
net centripetal force,

276
00:09:25,044 --> 00:09:27,858
it has to add up to more than
the net centripetal force

277
00:09:27,858 --> 00:09:30,305
just to cancel off this negative effect

278
00:09:30,305 --> 00:09:31,724
from the force of gravity.

279
00:09:31,724 --> 00:09:32,977
And if we plugged in numbers,

280
00:09:32,977 --> 00:09:35,076
we'd see that the tension
would end up being bigger.

281
00:09:35,076 --> 00:09:37,478
We'd actually get the
same exact term here,

282
00:09:37,478 --> 00:09:40,440
except that instead of
subtracting gravity,

283
00:09:40,440 --> 00:09:41,999
we have to add gravity

284
00:09:41,999 --> 00:09:44,294
to this net centripetal force expression.

285
00:09:44,294 --> 00:09:48,485
And we'd get that the tension
would be 10.45 Newtons.

286
00:09:48,485 --> 00:09:51,130
So recapping, when solving
centripetal force problems,

287
00:09:51,130 --> 00:09:53,517
we typically write the v squared over r

288
00:09:53,517 --> 00:09:56,397
on the left hand side as
a positive acceleration,

289
00:09:56,397 --> 00:09:58,351
and by doing that, we've selected in

290
00:09:58,351 --> 00:10:00,370
toward the center of
the circle as positive

291
00:10:00,370 --> 00:10:01,826
since that's the direction

292
00:10:01,826 --> 00:10:03,602
that centripetal acceleration points,

293
00:10:03,602 --> 00:10:06,005
which means that all
forces that are directed

294
00:10:06,005 --> 00:10:07,604
in toward the center of the circle

295
00:10:07,604 --> 00:10:09,129
also have to be positive.

296
00:10:09,129 --> 00:10:10,688
And you have to be
careful because that means

297
00:10:10,688 --> 00:10:14,414
downward forces can count as
a positive centripetal force

298
00:10:14,414 --> 00:10:16,707
as long as down corresponds to

299
00:10:16,707 --> 00:10:18,178
toward the center of the circle.

300
00:10:18,178 --> 00:10:20,116
And just because a force was positive

301
00:10:20,116 --> 00:10:21,644
during one portion of the trip,

302
00:10:21,644 --> 00:10:23,664
like gravity was at
the top of this motion,

303
00:10:23,664 --> 00:10:26,328
that does not necessarily
mean that that same force

304
00:10:26,328 --> 00:00:00,000
is gonna be positive at some
other point during the motion.