Bowling ball in vertical loop | Centripetal force and gravitation | Physics | Khan Academy

- [Narrator] Imagine that in an effort

to make bowling more exciting,

bowling alleys put a big loop-the-loop

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

really fast to get the
ball up and around the loop

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

kinda like mini golf bowling
or something like that.

Well if you were gonna build this,

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

this structure's gonna have to withstand

a certain minimum amount of force.

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

You can't have this thing breaking

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

So let's ask ourselves that question.

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

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

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

We'll put some numbers in here.

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

at the top of the loop.

That's pretty darn fast
so someone really hurled

this thing through here.

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

and the bowling ball has
a mass of four kilograms,

which is around eight or nine pounds.

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

How much normal force is there gonna be

between the loop and the ball?

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

the force between the two surfaces?

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

if our structure is
strong enough to contain

this bowling ball as it
goes around in a circle.

And it's also a classic
centripetal force problem,

so let's do this.

What do we do first?

We should always draw a force diagram.

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

So what are the forces on this ball?

You're gonna have a force
of gravity downward,

and the magnitude of the
force of gravity is always

given by M times G, where
G represents the magnitude

of the acceleration due to gravity.

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

Now which way does this
normal force point?

A common misconception,
people wanna say that

that normal force points up because

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

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

the normal force on you is upward

because it keeps you from
falling through the ground,

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

The loop structure isn't keeping you up.

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

and that means this normal force is gonna

have to point downward.

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

but because the surface
is above this ball,

the surface pushes down.

Surfaces can only push.

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

If the surface was to the side of you,

the surface would have to push right.

And if the surface was
to the right of you,

the surface would have to push left.

Normal forces in other words, always push.

So the force on the ball from the track

is gonna be downward but vice versa.

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

So if this ball were
going a little too fast

and this were made out of wood,

you might see this thing splinter

because there's too much force pushing

on the track this way.

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

from the track is downward.

And after you draw a force diagram,

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

to use Newton's Second Law.

And to keep the calculation simple,

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

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

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

because the normal
force is pointing toward

the center of the circular
path and the normal force

is the force we wanna find,

we're gonna use Newton's Second Law

for the centripetal direction and remember

centripetal is just a fancy word

for pointing toward the
center of the circle.

So, let's do it.

Let's write down that the
centripetal acceleration

should equal the net centripetal force

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

So if we choose this, we know that

the centripetal acceleration
can always be re-written

as the speed squared divided by the radius

of the circular path that
the object is taking,

and this should equal
the net centripetal force

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

and you gotta remember how
we deal with signs here

because we put a positive sign over here

because we have a positive sign

for our centripetal acceleration

and our centripetal
acceleration points toward

the center of the circle always.

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

our positive direction,

and that means for these forces,

we're gonna plug in forces toward

the center of the circle as positive.

So let's do that.

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

You gotta be careful here.

I'm just gonna plug in.

What are the centripetal forces?

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

What forces do we have in our diagram.

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

Let's start with gravity.

Is the gravitational force
going to be a centripetal force.

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

Does it even get included in here at all?

And to figure that out you ask:

Does it point centripetally?

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

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

moreover because it points
toward the center of the circle

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

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

Similarly, for the normal
force, it also points

toward the center of the circle,

so we include it in this calculation

and it as well will be a
positive centripetal force.

And now we can solve for the normal force.

If I solve algebraically,
I can multiply both sides

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

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

minus the magnitude of
the force of gravity,

which if we plug in numbers,
gives us four kilograms

times eight meters per second squared,

you can't forget the square,

divided by a two meter
radius minus the magnitude

of the force of gravity which
is four kilograms times G

which if you multiply that
out gives you 88.8 newtons.

This is how much downward
force is exerted on the ball

from the track but from
Newton's Third Law,

we know that that is also
how much force the ball

exerts upward on the track.

So whatever you make this loop out of,

it better be able to
withstand 88.8 newtons

if people are gonna be rolling
this ball around the loop

with eight meters per second.

Now let me ask you this.

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

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

Now how much normal force
is there at this point?

Is it gonna be greater than, less than,

or equal to 88.8 newtons.

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

So there's gonna be a force of gravity.

Again, it's gonna point straight down,

and again, it's gonna be equal to

at least the magnitude
of it will be equal to

the mass times the magnitude
of acceleration due to gravity.

And then we also have a normal force,

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

Remember, surfaces push outward

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

the surface pushes to the right.

This time our normal
force points to the right.

And let's assume this a well oiled track

so there's really no
friction to worry about.

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

So what about the answer to our question.

Will this normal force
now be bigger, less than,

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

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

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

because when you plug in over here,

into the centripetal forces,

you only plug in forces
that point radially.

That is to say centripetally,

either into the circle,
which would be positive,

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

If they neither point into
nor out of the circle,

you don't include them in
this calculation at all

because they aren't
pointing in the direction

of the centripetal acceleration.

In other words, they're not causing

the centripetal acceleration.

So for this case over
here, gravity is no longer

a centripetal force because
the force of gravity

no longer points toward
the center of the circle.

This force of gravity is
tangential to the circle.

It's neither pointing into nor out of,

which means it doesn't factor into

the centripetal motion at all.

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

It does not change the ball's direction,

which means it doesn't
contribute to making this ball

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

So when we solved for the normal force,

we'd multiply both sides by M,

we would not have an MG anymore.

So we wouldn't be subtracting this term

and that's gonna make
our normal force bigger.

Moreover, the speed of
this ball's gonna increase

compared to what it was up here.

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

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

and we're not subtracting
anything from it,

The normal force will be
much greater at this point

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

So recapping, when you wanna solve

the centripetal force problem,

always draw your force diagram first.

If you choose to analyze the forces

in the centripetal
direction, in other words,

for the direction in toward
the center of the circle,

make sure you only plug
in forces that are into,

radially into the circle or
radially out of the circle.

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

If they were radially out of the circle,

you would make them negative.

And if they neither point radially inward,

toward the center of the circle

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

you just do not include
those forces at all

when using this centripetal direction.