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# Yo-yo in vertical circle example

In this video David explains how to find the tension in a string that is whirling a yo-yo in a vertical circle. Created by David SantoPietro.

## Want to join the conversation?

- please, how do we get (v^2)/r=a ? thanks(2 votes)
- Earlier in this playlist there's a video Sal made showing how to derive v^2/r without calculus. https://www.khanacademy.org/science/physics/two-dimensional-motion/centripetal-acceleration-tutoria/v/visual-understanding-of-centripetal-acceleration-formula

There's also a calculus one that's really good if you know calculus.(27 votes)

- (4:29) I don't understand why we include gravitational force as a centripetal forces because it is not always pointed at the centre of the circle. when the yoyo is at its lowest point, fg is pointed downwards.(7 votes)
- You are right. We don't always include gravitational force as contributing to the centripetal force. The example at4:29is only about the case where the yoyo is at the highest point where gravity is pointed downwards and centripetally, so we include it.

When the yoyo is at its lowest point, gravitational force points downwards, which is opposite to the centripetal direction. So we have to subtract it from the tension to get the centripetal force.

When the yoyo is at the leftmost or rightmost point, we don't include gravitational force because the gravitational force at that point is pointing tangentially to the circle, which is unrelated to the centripetal direction.(9 votes)

- What if the yo-yo is on the left and right points of the circle (let's say it's turning counter-clockwise)? How would I include gravity (acceleration) in the circular velocity?(4 votes)
- (at7:29) Since the force of gravity is always pointing downward (or pointing in vertical direction), it has no horizontal component if you're considering the points at 3 o'clock and 9 o'clock. Therefore, you don't have to include the force of gravity. If you're considering other points on the left or right, you need to include the horizontal component of the force of gravity. You only include the horizontal component of a force when solving these problems because only forces with some horizontal direction can affect the circular motion of the object. Hope that helps!(11 votes)

- David gave examples in vertical circles. There is also a video on horizontal circle. But what if the yo-yo is swinging at another angle to the horizontal? For example, if the yo-yo has a mas 5g, the string length is 0.5m and gravity is 9.8m/s^2, and it is swung at an angle of 36 degrees to the horizontal (as opposed to 90 degrees when the yo-yo is swinging vertically) and accelerating at 7.9m/s in the direction it is moving, how do I figure out the tension?

Thanks,

Gor(6 votes)- I think you should be able to figure out the tension force if you draw your force diagram, with the horizontal and vertical components of the tension force in their respective directions.

You should also specify the direction in which the Yo-yo is moving. Since you have said it is 'accelerating', I'm sure the Yo-yo is not climbing up the circle which it is tracing in space.

--HIH--

Ramana(2 votes)

- If the yo-yo is at the bottom, why would mg still be included if Fc includes any forces that are pointing toward the center of the circle?(3 votes)
- Gravity doesn't go away, so at the bottom the force that is providing the centripetal acceleration needs to overcome it.(4 votes)

- Why do we not include normal force to calculate net forces?(3 votes)
- Normal force is between two surfaces in contact. Given that there are no surfaces in contact here, normal force does not exist.(3 votes)

- Wouldn't the velocity at the bottom be greater than at the top due to conservation of energy? the ball is gaining kinetic energy while losing potential energy all the while still being pulled by the rope. This would change the calculation(4 votes)
- What if the Yo-yo string were pointing in a direction other than directly up or down, for example a slanting direction or directly horizontal to the left or right. What would the sign of Fg and Tension be in that case?(3 votes)
- Fg is always down and T is always in the direction of the string.(3 votes)

- What would you do if the yoyo wasn't directly at the top of the arc or the bottom of the arc? What would you plug in for gravity?(4 votes)
- whichever component acts towards the center(0 votes)

- Why is there no normal force opposing gravity?(0 votes)
- Think about what a normal force is. Where would it come from in this case?(6 votes)

## Video transcript

- [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.