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- [Narrator] I want to show you how to do

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a slightly more sophisticated

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centripetal force problem,

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and this one's a classic.

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This is the one where there's a mass,

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tied to a string,

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and that string is secured to the ceiling,

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and the mass has been
given an initial velocity,

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so that it swings around
in a horizontal circle.

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So, this mass is going to
maintain a constant height,

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it's not moving up or down,

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but it revolves in a horizontal circle,

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so if you were to view
this thing from above,

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from below, it looks something like this.

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You'd see the ball tracing
out a perfect circle.

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If you were looking at
this from right below.

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And the question I want
to ask is: two things.

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What is the tension in the rope?

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And what has to be the speed of the mass?

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And these are given variables.

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We know the mass is three kilograms.

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We know the length of
the rope is two meters.

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And this rope is making an
angle of thirty degrees,

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with respect to this
vertical line right here.

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Now this problem's a
classic for a good reason.

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If you don't have a clean
conceptual understanding

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of what we really mean by the terms

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when dealing with centripetal forces.

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And if you don't have a problem solving

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strategy to use to tackle problems,

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this problem will expose you.

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So that's why we should go over this,

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to make sure you have
a clean understanding

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of exactly what we mean
by all the different terms

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and to show you there's
a strategy you could use

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to solve any centripetal force problem.

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And that strategy is this.

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First, draw a quality force diagram

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for the object or objects in your problem.

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So, let's do that first.

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The forces that are acting
on this three kilogram sphere

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are the force of gravity.

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You'll have a force of
gravity straight down.

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That's gonna be m times g.

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So the way we find the force of gravity

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is with the formula mass times 9.8.

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And the only other object
that's touching this mass

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is the rope.

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So the only other force on this mass

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is the force of tension.

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So I'm gonna to label that force

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with a capital T.

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This'll be the total force of tension

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from the rope.

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And, really, the only other step

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in this problem solving strategy,

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there's really only two steps,

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use Newton's second law.

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But only use it for one
direction at a time.

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And we typically only have two directions.

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I mean, we've got this vertical direction

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or we've got this horizontal direction.

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We've got a 50-50 shot.

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I mean, we've gotta pick one of the two.

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It's not like you can
pick the wrong direction.

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If you pick the wrong direction,

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it'll just mean that you can't solve

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because there'll be too many variables.

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But what you write down in that process

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is probably gonna be useful
later in the problem anyway,

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so don't erase it.

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If you pick a direction
that you can't solve,

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because you have too many variables,

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just go to the other direction,
and pick that direction.

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I know what direction to pick,

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cause I've done this one before.

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But if you don't know,

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the worst thing you do is freeze up.

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You gotta try something.

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If you try the wrong direction,

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no big deal,

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there's only one other direction to pick.

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So I'm gonna choose to
analyze these forces

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in the vertical direction.

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So I'm gonna say that the acceleration

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in the vertical direction,

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this y direction,

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is equal to the net force

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in the y direction divided by the mass.

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And now I simply ask myself,

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what is the acceleration
in the vertical direction?

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It's not 9.8.

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A lot of people want to say negative 9.8.

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But that's only if this
ball were free-falling.

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And this ball is not free-falling.

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In fact, this ball is not even changing

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it's vertical height.

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It's remaining at the
same constant height,

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and that means not only is
there no vertical velocity,

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there's no vertical acceleration.

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Since the ball's not
even moving vertically.

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So, on the left hand side here,

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this is great,

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we can put a zero.

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Zero's are wonderful.

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They make calculations easier.

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So this is going to equal the net force

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in the vertical direction.

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So one force is gonna to be

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the force of gravity.

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So I'll put that in here.

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You'll have m times g.

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That force is definitely a vertical force.

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But you gotta be careful,

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if we're gonna consider
upward is positive,

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this mg is a negative force.

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So in terms of y-directed forces,

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downward, typically,

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the convention is that you
choose that to be negative.

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So I'm gonna say that mg is negative.

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And now we ask ourselves,

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are there any other vertical forces?

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Well, we just look at our force diagram.

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So our force diagram will tell us

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if there's any other vertical forces.

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We look over here,

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the only other force
is this tension force.

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And part of this tension force

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is vertical.

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We can't put in the entire tension

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into this formula.

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We could only do that if this tension

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was directed vertically upward,

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but it's not.

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Part of it's vertically upward,

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and part of it's horizontal.

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So part of this tension force

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is directed this way.

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I'm gonna call that the x component

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of the tension.

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And then part of this tension
is directed vertically.

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I'll call that the y
component of the tension.

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So we can only plug in this y component

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of tension into this formula over here,

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since we're plugging in y-directed

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forces into this equation.

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So I could just write plus Ty.

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But I don't wanna do that.

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I wanna solve for what T is,

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not what Ty is.

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So I wanna write this Ty in terms of T.

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And I can do that.

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Look at the triangle this is making.

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This tension and this Ty

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is gonna make an angle right here.

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And that angle has to be the same angle

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that the rope makes
with this vertical line.

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So the tension is not the rope.

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The tension is a force
exerted by the rope.

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And the tension lies along the
same direction as the rope.

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But the tension is not the rope.

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This is a really common misconception.

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Sometimes people think,

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oh, the tension is two meters, right?

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No.

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That's not even a force.

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That's the length of the rope.

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And, yes, the length of the rope

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lies along the same direction
as the tension force,

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but the tension is different

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from the length of the rope.

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However, this angle between the tension

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and this vertical component of the tension

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is the same as the angle between this rope

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and the vertical direction.

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So that means we can label this

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as thirty degrees right here.

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And now I can figure out

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what is the vertical
component of the tension

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in terms of T and in terms of theta?

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I got a right triangle.

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Here's a right angle.

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This side that I wanna find out,

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Ty, is adjacent to this thirty degrees.

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Since it's adjacent,
we're gonna use cosine.

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In other words,

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we're gonna say that cosine of theta

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equals adjacent over hypotenuse.

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People get confused sometimes.

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They're like, wait,

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I thought vertical was
always the opposite side.

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It'd be the opposite side
if we knew this angle,

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But we don't know that angle.

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We know this vertical angle.

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And that means that this vertical side

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is adjacent to that angle.

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Since we know the angle,

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we can write the cosine of thirty degrees

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is gonna equal the adjacent side.

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Now the adjacent side is Ty.

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And the hypotenuse is gonna be T.

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People get freaked out here.

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They're like, I've got too many variables.

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But that's okay.

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We can label this total hypotenuse is T.

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Even though we don't
know it, it's alright.

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We're gonna manipulate symbols

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and we're gonna solve for Ty.

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If I multiply both sides by T, I get Ty.

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The vertical component of the tension

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is equal to the total tension

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times cosine of thirty degrees.

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This is the vertical
component of the tension.

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And this is the force that I can plug into

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my vertical net force.

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So I'm gonna add,

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cause this points upward,

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T, the total tension,

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times cosine of thirty.

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And then I have to divide by my mass.

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So I can solve this for T now.

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If I multiply both sides by m.

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M times zero is gonna be zero.

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I'll move the mg over

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and then I divide by cosine of thirty.

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To get that the tension in the rope

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is gonna equal mg,

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the force of gravity divided by

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cosine of thirty.

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And if we plug in numbers

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we'll get that T equals,

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the mass was 3 kilograms,

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g is always 9.8.

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Divided by cosine of thirty.

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Gives us a tension of about 33.9 Newtons.

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I'll just say that that's 34 Newtons.

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So that's the tension in the rope.

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That's the first thing we wanted to find.

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We just found it over here.

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That is the tension in the rope.

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So let's do the next part.

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Let's try to find the speed.

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People get a little concerned now.

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They're like, what do I do?

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Don't deviate from the plan.

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We drew our force diagram.

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We used Newton's second law

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for one of the directions.

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You still got work to do,

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then do Newton's second
law for another direction.

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We're just gonna do this
for the x direction.

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So we'll do the x direction.

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I'll see that the acceleration
in the x direction

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equals the net force in the x direction

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divided by the mass.

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And I'll ask myself the same question.

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Is there any acceleration
in the x direction?

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There wasn't any acceleration
in the y direction.

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You might think there's
not any in the x direction.

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But there is.

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This mass is going in a circle.

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That means there's gonna be centripetal

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acceleration in this direction.

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So this horizontal
direction is, essentially,

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just the centripetal direction.

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So to make that more clear,

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I'm just gonna put ac and Fc.

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And whenever you have
centripetal acceleration,

269
00:07:58,948 --> 00:08:01,280
we can replace that with v squared.

270
00:08:01,280 --> 00:08:04,298
The speed squared divided by the radius.

271
00:08:04,298 --> 00:08:06,481
And that's gonna equal
the net centripetal force

272
00:08:06,481 --> 00:08:07,589
over the mass.

273
00:08:07,589 --> 00:08:10,180
What force is acting in
the centripetal direction?

274
00:08:10,180 --> 00:08:11,664
Well, you can figure that out.

275
00:08:11,664 --> 00:08:14,194
It's just the force that was
acting in the x direction.

276
00:08:14,194 --> 00:08:16,059
Cause this is the x direction.

277
00:08:16,059 --> 00:08:17,893
The x direction is the direction

278
00:08:17,893 --> 00:08:19,312
that happens to be pointing

279
00:08:19,312 --> 00:08:21,036
toward the center of the circle.

280
00:08:21,036 --> 00:08:23,034
That's why the x direction here

281
00:08:23,034 --> 00:08:24,358
is just the centripetal direction.

282
00:08:24,358 --> 00:08:26,682
So to figure out which
forces are centripetal,

283
00:08:26,682 --> 00:08:28,228
I just look at my force diagram.

284
00:08:28,228 --> 00:08:29,725
I drew this for a reason.

285
00:08:29,725 --> 00:08:31,259
I drew this so when I look at it,

286
00:08:31,259 --> 00:08:33,294
I can figure out what forces are vertical,

287
00:08:33,294 --> 00:08:34,265
to put in over here,

288
00:08:34,265 --> 00:08:36,033
and what forces are centripetal,

289
00:08:36,034 --> 00:08:37,138
i.e. horizontal,

290
00:08:37,138 --> 00:08:38,272
to put in over here.

291
00:08:38,272 --> 00:08:39,647
The only force that's horizontal

292
00:08:39,647 --> 00:08:41,884
is the horizontal
component of the tension.

293
00:08:41,884 --> 00:08:43,732
That's this Tx.

294
00:08:43,732 --> 00:08:46,910
But, again, instead of
just plugging in Tx,

295
00:08:46,910 --> 00:08:49,161
we'll plug in what this component is.

296
00:08:49,161 --> 00:08:51,804
in terms of the angle
and the total tension.

297
00:08:51,804 --> 00:08:52,798
Just like we did over here.

298
00:08:52,798 --> 00:08:54,795
Ty was T cosine thirty.

299
00:08:54,795 --> 00:08:56,628
So it might not be that big of a surprise

300
00:08:56,628 --> 00:08:59,219
that Tx, the horizontal component,

301
00:08:59,219 --> 00:09:02,140
is just gonna be T sine thirty.

302
00:09:02,140 --> 00:09:03,213
If you don't believe that,

303
00:09:03,213 --> 00:09:04,303
you can prove it to yourself.

304
00:09:04,303 --> 00:09:05,136
Think about it.

305
00:09:05,136 --> 00:09:07,866
This Tx component is the
opposite to this angle.

306
00:09:07,866 --> 00:09:09,594
And for opposite we use sine.

307
00:09:09,594 --> 00:09:12,578
So we can say sine of thirty would be

308
00:09:12,578 --> 00:09:15,184
the x component, which
is the opposite side,

309
00:09:15,184 --> 00:09:17,436
over the hypotenuse, which is T.

310
00:09:17,436 --> 00:09:18,807
And if you solve this for Tx,

311
00:09:18,807 --> 00:09:20,549
you multiple both sides by T,

312
00:09:20,549 --> 00:09:23,354
you, indeed, get that Tx is just T,

313
00:09:23,354 --> 00:09:26,228
the total tension, times sine thirty.

314
00:09:26,228 --> 00:09:27,998
So we can plug that back in over here.

315
00:09:27,998 --> 00:09:29,270
We know that the only component

316
00:09:29,270 --> 00:09:31,094
that's acting as a centripetal force,

317
00:09:31,094 --> 00:09:33,680
i.e. that's pointing towards
the center of the circle,

318
00:09:33,680 --> 00:09:34,786
is this x component,

319
00:09:34,786 --> 00:09:37,979
which we just found is
T sine thirty degrees.

320
00:09:37,979 --> 00:09:40,224
Now you see why picking
this direction first

321
00:09:40,224 --> 00:09:42,118
wouldn't have allowed us to solve,

322
00:09:42,118 --> 00:09:43,830
cause we wouldn't have known the speed, v,

323
00:09:43,830 --> 00:09:45,745
and we wouldn't have known the tension T.

324
00:09:45,745 --> 00:09:47,895
Only because we chose
the y direction first,

325
00:09:47,895 --> 00:09:49,211
we were able to find the tension.

326
00:09:49,211 --> 00:09:50,665
And now that we know this tension,

327
00:09:50,665 --> 00:09:51,812
being 34 Newtons,

328
00:09:51,812 --> 00:09:53,623
we can plug that in over here

329
00:09:53,623 --> 00:09:55,065
and solve for our speed.

330
00:09:55,065 --> 00:09:57,485
But there's a really common
mistake that people make here.

331
00:09:57,485 --> 00:10:00,967
People really wanna
plug in r as two meters,

332
00:10:00,967 --> 00:10:01,800
cause they're like,

333
00:10:01,800 --> 00:10:03,678
hey, you gave me two meters over here,

334
00:10:03,678 --> 00:10:04,729
I'm gonna use it.

335
00:10:04,729 --> 00:10:06,027
That's an r, right?

336
00:10:06,027 --> 00:10:06,860
Isn't that r?

337
00:10:06,860 --> 00:10:07,871
No, that is not r.

338
00:10:07,871 --> 00:10:10,769
R here, always, in the centripetal formula

339
00:10:10,769 --> 00:10:12,657
for the acceleration, this r represents

340
00:10:12,657 --> 00:10:14,988
the radius of the circle that the object

341
00:10:14,988 --> 00:10:15,945
is moving in.

342
00:10:15,945 --> 00:10:17,639
And this object doesn't move in a circle

343
00:10:17,639 --> 00:10:18,805
with a radius of two.

344
00:10:18,805 --> 00:10:20,965
The radius of the circle this object's

345
00:10:20,965 --> 00:10:22,429
traveling in looks like this.

346
00:10:22,429 --> 00:10:24,486
That's the radius of the circle.

347
00:10:24,486 --> 00:10:25,756
And that's not two meters.

348
00:10:25,756 --> 00:10:26,686
How do we find that?

349
00:10:26,686 --> 00:10:28,309
We'll again use trigonometry.

350
00:10:28,309 --> 00:10:30,473
We're just gonna say that
we've got a right triangle.

351
00:10:30,473 --> 00:10:32,561
This time, though, we're
gonna make a right triangle

352
00:10:32,561 --> 00:10:33,661
out of the length,

353
00:10:33,661 --> 00:10:35,118
not of the force.

354
00:10:35,118 --> 00:10:35,979
Not of this tension.

355
00:10:35,979 --> 00:10:38,337
We're making a right
triangle out of the length.

356
00:10:38,337 --> 00:10:40,709
But, again, we know that
side's a right angle.

357
00:10:40,709 --> 00:10:42,366
We know that this side's thirty degrees.

358
00:10:42,366 --> 00:10:45,209
So we say that this radius
is the opposite side

359
00:10:45,209 --> 00:10:47,109
of that thirty degrees.

360
00:10:47,109 --> 00:10:48,542
So we're gonna use sine theta.

361
00:10:48,542 --> 00:10:50,279
We're gonna say that the sine of theta,

362
00:10:50,279 --> 00:10:51,112
which is thirty,

363
00:10:51,112 --> 00:10:52,551
equals the opposite side,

364
00:10:52,551 --> 00:10:53,418
and that's r,

365
00:10:53,418 --> 00:10:57,001
divided by the total
length of the string, L.

366
00:10:57,001 --> 00:10:58,879
And, so if I solve this for the radius,

367
00:10:58,879 --> 00:11:01,077
I get the radius of the circle

368
00:11:01,077 --> 00:11:02,439
that this ball is traveling in

369
00:11:02,439 --> 00:11:05,088
would be L times sine of thirty,

370
00:11:05,088 --> 00:11:07,253
where L would be this two meters.

371
00:11:07,253 --> 00:11:09,106
So we'll plug that back in over here.

372
00:11:09,106 --> 00:11:11,109
We'll say that v squared divided by

373
00:11:11,109 --> 00:11:13,578
r, which is L sine thirty,

374
00:11:13,578 --> 00:11:16,981
is gonna equal T sine
30 divided by the mass.

375
00:11:16,981 --> 00:11:17,855
And now we can solve,

376
00:11:17,855 --> 00:11:19,942
we can multiple both
sides by L sine thirty.

377
00:11:19,942 --> 00:11:22,353
And we get that v squared is gonna equal

378
00:11:22,353 --> 00:11:23,822
T sine thirty

379
00:11:23,822 --> 00:11:25,588
times L sine thirty

380
00:11:25,588 --> 00:11:27,603
divided by the mass of the sphere.

381
00:11:27,603 --> 00:11:29,650
And since I want to find v not v squared,

382
00:11:29,650 --> 00:11:31,360
I'll take a square root of both sides,

383
00:11:31,360 --> 00:11:33,063
and when I take a square
root of both sides,

384
00:11:33,063 --> 00:11:35,621
I end up with v is the square root

385
00:11:35,621 --> 00:11:38,488
of t sine thirty, L sine
thirty over the mass.

386
00:11:38,488 --> 00:11:39,565
So if we plug in our numbers,

387
00:11:39,565 --> 00:11:42,530
we get that v is the square root of T,

388
00:11:42,530 --> 00:11:43,728
which is 34 Newtons,

389
00:11:43,728 --> 00:11:46,187
times sine of 30, times L,

390
00:11:46,187 --> 00:11:48,800
and this L is referring
to this total length

391
00:11:48,800 --> 00:11:49,843
which is two meters,

392
00:11:49,843 --> 00:11:51,181
times sine of thirty.

393
00:11:51,181 --> 00:11:54,503
All divided by the mass,
which was three kilograms.

394
00:11:54,503 --> 00:11:58,093
Which, if you solve, gives
you a speed of about 2.38.

395
00:11:58,093 --> 00:12:01,020
So I'll just say 2.4 meters per second.

396
00:12:01,020 --> 00:12:03,635
Which is the speed that
we were trying to find.

397
00:12:03,635 --> 00:12:05,236
So, recapping, when you're solving

398
00:12:05,236 --> 00:12:07,191
a sophisticated centripetal force problem,

399
00:12:07,191 --> 00:12:09,974
be sure to draw a quality force diagram.

400
00:12:09,974 --> 00:12:12,941
Then use Newton's second
law for a single direction.

401
00:12:12,941 --> 00:12:15,461
And only plug forces in that direction

402
00:12:15,461 --> 00:12:16,688
into the net force.

403
00:12:16,688 --> 00:12:18,431
If the direction you choose

404
00:12:18,431 --> 00:12:20,962
happens to lie along the
centripetal direction,

405
00:12:20,962 --> 00:12:23,287
i.e. it points toward
the center of the circle,

406
00:12:23,287 --> 00:12:25,356
then you can use v squared over r,

407
00:12:25,356 --> 00:12:26,922
for your centripetal acceleration,

408
00:12:26,922 --> 00:12:28,524
But, again, only plug in forces

409
00:12:28,524 --> 00:12:30,218
that lie along that direction

410
00:12:30,218 --> 00:12:31,911
for the centripetal force.

411
00:12:31,911 --> 00:12:34,311
And make sure that you understand
when we say the radius,

412
00:12:34,311 --> 00:12:36,657
we're talking about the
radius of the circle

413
00:12:36,657 --> 00:00:00,000
that the object is traveling in.