Yo-yo in vertical circle example | Centripetal force and gravitation | Physics | Khan Academy

- [Instructor] People find
centripetal force problems

much more challenging than
regular force problems,

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

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

some common misconceptions
that people make along the way.

So, let's start with this example,

and it's a classic.

Let's say you started with a yo-yo

and you whirled it around vertically,

and I think this is called
the around the world,

if you want to look it up on YouTube,

looks pretty slick.

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

So this is a vertical circle,

not a horizontal circle.

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

This ball is actually
getting higher in the air

and then lower in the air.

But for our purposes, we just need to know

that it's a mass tied to a string.

Let's say the mass of the yo-yo

is about 0.25 kilograms,

and let's say the length of the string

is about 0.5 meters.

And let's say this ball is going about

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

And something you might want to know

if you're a yo-yo manufacturer

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

how strong does your string need to be.

So let's figure out for this example,

what is the tension in the string

when this yo-yo is at its maximum height

going four meters per second?

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

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

So let's do that here.

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

Well, if we're near Earth

and we're assuming we're going to be

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

there's gonna be a force of gravity

and that force of gravity is
gonna point straight downward.

So the magnitude of that force of gravity

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

g represents the magnitude

of the acceleration due to gravity,

and this expression here represents

the magnitude of the force of gravity.

But there's another force.

The string is tied to the mass,

so this string can pull on the mass.

Strings pull, they exert
a force of tension.

Which way does that tension go?

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

and that's not good.

Ropes can't push.

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

try to push on something.

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

but it can pull.

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

This rope's gonna pull.

How much?

I don't know.

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

This is gonna be the force
of tension right here,

and we'll label it with a capital T.

We could have used F with the sub T.

There's different ways
to label the tension,

but no matter how you label it,

that tension points in towards
the center of the circle

'cause this rope is pulling on the mass.

So, after you draw a force diagram,

if you want to find a force,

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

And we're gonna use this formula as always

in one dimension at a time

so vertically,
horizontally, centripetally,

one dimension at a time

to make the calculations
as simple as possible.

And since we have a
centripetal motion problem,

we have an object going in a circle,

and we want to find one of those forces

that are directed into the circle,

we're gonna use Newton's second law

for the centripetal direction.

So we'll use centripetal acceleration here

and net force centripetally here.

So in other words, we're gonna write down

that the centripetal acceleration

is gonna be equal to the
net centripetal force

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

So because we chose the
centripetal direction,

we're gonna be able to replace

the centripetal acceleration
with the formula

for centripetal acceleration.

The centripetal acceleration's
always equivalent

to v squared over r,

the speed of the object squared

divided by the radius of the circle

that the object is traveling in.

So we set that equal to
the net centripetal force

over the mass,

and the trickiest part here,

the part where the failure's
probably gonna happen

is trying to figure
out what do you plug in

for the centripetal force.

And now we gotta decide

what is acting as our centripetal force

and plug those in here
with the correct signs.

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

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

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

people start thinking, they
start looking all over.

No, you drew your force diagram.

Look right there.

Our force diagram holds
all the information

about all the forces that we've got

as long as we drew it well.

And we did draw it well.

We included all the forces,

so we'll just go one by one.

Should we include, should we even include

the force of gravity in this
centripetal force calculation?

We should because we're
gonna include all forces

that point centripetally and remember,

the word centripetal is just a fancy word

for pointing toward the
center of the circle.

And this force of
gravity does point toward

the center of the circle.

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

It's contributing, in other words,

to the centripetal force.

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

to go in a circle,

so we include it in this formula.

So mg is the magnitude
of the force of gravity.

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

or a negative?

Many people want to
include it as a negative

'cause it points down,

but when we're dealing with
the centripetal direction,

it's inward that's gonna be positive,

not necessarily up
that's gonna be positive.

If up happens to point in,

then we'd consider it positive.

So if we were down here,

up is positive.

But up here, down is positive

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

and if we're over here,

diagonally up and left would be positive

because any force that would be pointing

toward the center of the circle

is gonna be included as a positive sign

and there's a reason for that.

The reason we're including

toward the center of
the circle as positive

is because we chose to write

our centripetal acceleration as positive.

And since we know the
centripetal acceleration

points toward the center of the circle,

if we make the centripetal
acceleration positive,

we've committed to toward
the center of the circle

as being positive as well.

In other words, we could have decided that

out of the circle's positive,

but if we did that,

this centripetal acceleration
that points inward

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

and that's just weird.

Nobody does that.

So we choose into the circle as positive.

That makes our centripetal
acceleration positive,

but it also makes every
force that points inward

positive as well.

And long story short,
this force of gravity

is gonna be counted as a
positive centripetal force

since it points inward toward
the center of the circle.

And we keep going.

We've got another force here.

We've got a force of tension.

Do we include it in here?

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

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

Since it also points toward
the center of the circle,

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

It is also one of the forces

that causes this ball to
move around in a circle.

In other words, the combined force of both

gravity and the force of tension

are making up the net
centripetal force in this case.

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

we just do our algebra.

We'll multiply both sides by the mass.

Then, we'll subtract mg from both sides,

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

equals m v squared over r

minus mg, which if we plug in numbers,

we'd get that the tension in the rope

is 5.55 Newtons.

So this is to be expected.

We subtracted the force of
gravity, the magnitude of it

from this net centripetal force,

so this term here represents

the total amount of
centripetal force we need

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

But the amount of tension we need

is that amount minus the force of gravity,

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

are both acting as the centripetal force.

So, neither one of them have to add up

to the total centripetal force.

It's just both together
that have to add up

to the centripetal force,

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

as large as it might have been.

However, if we consider the case

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

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

our force diagram's gonna look different.

The force of gravity
still points downward,

the force of gravity's
always gonna be straight down

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

But this time, the tension points up

because the string is
always pulling on the mass.

Ropes can only pull.

Ropes can never push,

so this rope is still pulling the mass,

the yo-yo toward the center of the circle.

So now, when we plug in over here,

one of these forces is gonna be negative.

Before they were both positive

and they were both positive

because both forces pointed
toward the center of the circle.

Now, only one force is pointing toward

the center of the circle,

and we can see that that's tension.

Tension's pointing toward
the center of the circle.

Gravity's pointing away from the center,

radially away from the center.

That means tension still
remains a positive force,

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

would have to be considered
a negative centripetal force

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

So, if we were to calculate the tension

at the bottom of the path,

the left hand side would
still be v squared over r

'cause that's still the
centripetal acceleration.

The mass on the bottom would still be m

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

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

since gravity's pointing radially out

of the center of the circle.

And if we solve this expression

for the tension in the string,

we'd get that the tension equals,

we'd have to multiply both sides by m,

and then add mg to both sides.

And we'd get that the
tension's gonna equal

m v squared over r, plus m g.

This time, we add m g to
this m v squared over r term,

whereas over here, we had to subtract it.

And that should make sense conceptually

since before, up here,
both tension and gravity

were working together to add up

to the total centripetal force,

so neither one had to be as big

as they might have been otherwise.

But down here, not only is
gravity not helping the tension,

gravity's hurting the centripetal cause

by pulling this mass out of
the center of the circle,

so the poor tension in this case

not only has to equal the
net centripetal force,

it has to add up to more than
the net centripetal force

just to cancel off this negative effect

from the force of gravity.

And if we plugged in numbers,

we'd see that the tension
would end up being bigger.

We'd actually get the
same exact term here,

except that instead of
subtracting gravity,

we have to add gravity

to this net centripetal force expression.

And we'd get that the tension
would be 10.45 Newtons.

So recapping, when solving
centripetal force problems,

we typically write the v squared over r

on the left hand side as
a positive acceleration,

and by doing that, we've selected in

toward the center of
the circle as positive

since that's the direction

that centripetal acceleration points,

which means that all
forces that are directed

in toward the center of the circle

also have to be positive.

And you have to be
careful because that means

downward forces can count as
a positive centripetal force

as long as down corresponds to

toward the center of the circle.

And just because a force was positive

during one portion of the trip,

like gravity was at
the top of this motion,

that does not necessarily
mean that that same force

is gonna be positive at some
other point during the motion.