- [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.