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Centripetal force problem solving

In this video David gives some problem solving strategies for centripetal force problems and explains many common misconceptions people have about centripetal forces. Created by David SantoPietro.

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  • leaf grey style avatar for user uchechinwasike
    At , how does less Normal force cause you to be airborne? Shouldn't it be the other way round because the centripetal force is more therefore you'd be pulled more towards the centre of the circle?
    (88 votes)
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    • blobby green style avatar for user davidsantopietro
      Good question. Less normal force does indeed mean that the net centripetal force increases. However, since the speed is increasing you need more centripetal force for the bike to go in the same radius circle. But the smallest the normal force can be is zero, so after that point, the centripetal force is maxed out at mg and any increase in speed will not be able to have a corresponding increase in centripetal force, so the bike will no longer be able to travel in a circle of that same radius (which means it loses contact).
      (73 votes)
  • male robot hal style avatar for user RandomDad
    How many kind of forces are there in physics?
    (5 votes)
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    • blobby green style avatar for user davidsantopietro
      There are only four fundamental forces: strong nuclear force, weak nuclear force, electromagnetic force, gravitational force (note: it could be considered to be three forces since electromagnetism and the weak nuclear force can be considered two sides of the same more encompassing electro-weak force). But in macroscopic scenarios, we call the various types of "contact forces" which are usually just specific manifestations of the electromagnetic force by different names: Tension, Normal force, elastic/spring force, friction, etc. are all mostly electromagnetic in nature (since bonds between atoms/molecules are due mainly to electromagnetism) but it is useful to classify them by different names.
      (39 votes)
  • aqualine seedling style avatar for user Victoria
    At , David says that the force of tension is pointing towards the centre. But in previous videos, David said that the force of tension is only a PULL, not a PUSH. But if the force of tension is pointing towards the centre, wouldn't that be contradictory?
    (7 votes)
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  • leafers ultimate style avatar for user rezatahmid
    If david said that the centrifugal force doesnt exist then why doesnt the ball plummet into the center if nothing is pushing it outwards?
    (7 votes)
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    • aqualine ultimate style avatar for user Jude
      The ball plummets to the center because of the centripetal force if no other forces are there. However, if you add a force perpindicular to the centripetal force pulling inwards, the two forces create a circular movement of the ball.
      (4 votes)
  • blobby green style avatar for user Dibaloke Chanda
    What I understand about centrifugal force is following -
    From an inertial frame of reference we do not need to count the centrifugal force.But from a
    non-inertial reference frame (in this case a rotating reference frame) we need to consider the centrifugal force.For example if I was on a merry-go-round and I observe and calculate acting forces on the merry-go-round, I have to consider centrifugal force(along with other real forces).But if I was standing next to the merry-go-round and calculate the acting forces on the merry -go -round ,I do not need to consider centrifugal force.I only need to consider real forces (centripetal force, tension force etc).

    But centripetal force( observed from a inertial reference frame) and centrifugal force (observed from a non-inertial rotating reference frame) can be expressed through the same equation which is mv^2/r .

    Am I right?
    (3 votes)
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    • starky tree style avatar for user Caleb
      Centrifugal force is what we call a "pseudo-force" - it doesn't actually exist, but many people believe it does. This misconception stems probably from how certain objects that are not part of (or attached to) the moving body experience an outwards pull.

      For example, say you are driving in the passenger seat of a car. While the car is driving straight, you experience no abnormalities. However, when the car turns left or right, you lean in the opposite direction. Is this centrifugal force? The answer is no. What's happening is your body (which is not attached to the car) tries to follow Newton's 1st Law while the car is in circular motion - you are trying to keep moving straight but the car is impeding your efforts, dragging you to the side. There is no outwards "centrifugal" force involved - the car is pulling your body out of its straight path and taking your body (against your will) along the curve.

      A situation that disproves centrifugal force would be a catapult. If centrifugal force DOES exist, then the payload from, let's say, a trebuchet* would first fly and curve higher into the sky, thus violating the law of gravity, a some time before beginning to fall. However, since this is not the case and the payload fired from any catapult travels a parabolic path that can be predicted given the initial velocity and angle of release, we must assume that centrifugal force DOES NOT exist.

      Hope this helps.



      *see http://www.real-world-physics-problems.com/trebuchet-physics.html
      (11 votes)
  • spunky sam blue style avatar for user bilalquetta457
    Why do we feel slightly weightless when passing over the hill?If the normal force is less than weight then we should feel more weighted.Can we say that our weight adjusts itself equal to the Normal force?
    (8 votes)
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    • blobby green style avatar for user liudav.hel
      Feeling weightless is feeling the gravitational force, hence why the normal force was less than the gravitational. When we feel more weighed that's due to the combination of the forces. The gravitational force accelerates you down, but the normal force counteracts it. We feel the pressure the normal force causes. This is why, when the normal force decreases, we feel less weighed down, because of the decrease in pressure.
      (4 votes)
  • blobby green style avatar for user Giovanni Numpaque
    After minute 12, when you're explaining the bike over the hill problem, what would be the centripetal force if you're only halfway up the hill? Would it still be the force of gravity? Would you break the force of gravity up into different vectors, choosing an angle from a perpendicular tangent, and proceed with trig? Thanks
    (3 votes)
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    • aqualine ultimate style avatar for user vinnv226
      Yes. You would break gravity into two components: one pointing in the centripetal direction, and one pointing perpendicular to this direction (tangent to the hill). The component in the centripetal direction is your centripetal force.
      (5 votes)
  • duskpin ultimate style avatar for user Nishant Mishra
    The convention we chose makes it feel like centripetal force is helping normal force to counteract gravity. Does that mean that the centripetal force points out from the hill, in the same direction as the normal force? It sounds weird that the "centripetal" force would do that.
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      The centripetal force points toward the center of the circle. That's why it's called centripetal. Gravity IS the centripetal force at the top of the hill. Gravity pulls down, normal force pushes up. The normal force is less than the gravitational force because the hill is curving downward away from the bike, and the bike is accelerating downward. Downward is toward the center of the circle, at the moment the bike is atop the hill.
      (5 votes)
  • leaf orange style avatar for user Orangus
    I don't understand that thing that:

    If the puts the gravity as the centripetal force, it means that he deals with the single instant when the bike is on the top of the hill(because only on the top it equals to gravity, on other leverages it should be a part of gravity, because gravity points downwards, not inside the circle)

    But, how can then the magnitude of the normal force be NOT equal to the magnitude of the gravitational force, I think it does for a single instant, the instant he is dealing with, it certainly is lower than the gravity when the bike is on lower leverage(because it's not perpendicular to the surface), and it will get lower as the bike would move to that direction, but at that single instant, it MUST be equal to gravity, at the moment when the force of gravity is perpendicular to the surface.
    (4 votes)
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  • piceratops tree style avatar for user Andres Saavedra
    I'm trying to understand why does the normal force decrease when the bicycle goes over the hill. Until know I have reached the conclusion that there is a part of the force of gravity which is dedicated to creating the centripetal acceleration and this part of the force of gravity is the one "missing" from the normal force, Is my reasoning correct?
    (2 votes)
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Video transcript

- [Instructor] There are unfortunately quite a few common misconceptions that many people have when they deal with centripetal force problems, so in this video, we're gonna go over some examples to give you some problem solving strategies that you can use as well as going over a lot of the common misconceptions that people have when they deal with these centripetal motion problems. So, to start with, imagine this example, let's say a string is causing a ball to rotate in a circle. And to make it simple, let's say this ball is tracing out a perfect circle, and let's say it's sitting on a perfectly frictionless table so this would be the bird's eye view. This is the view from above. What it would look like from the side would be something like this. You'd have the ball tied to the rope and then you nail some sort of stake in the middle of the table. You tie the rope to the stake, and then you give the ball a push. And the ball's gonna take this circular path on the table when we view it from the side. But when we view it from above, you see this path traced out. So this is a bird's eye view that you would see if you were looking down from above the table, and this would be the side view. So let me ask you this question. What force is causing this ball to go in a circle? Now, a lot of people want to answer that question with the centripetal force. They'd say that it's the centripetal force that points inward that causes this ball to go in a circle, and that's not wrong. It's the truth, but it's not the whole truth. And the reason is that when we say centripetal force, all we really mean is a force that's directed toward the center of the circle. So saying the force that causes this ball to go in a circle is the centripetal force is a little unsatisfying. It'd be like answering the question, what force balances the force of gravity while the ball's on the table with the answer, the upward force. I mean, yeah, we knew it had to be an upward force, but that really doesn't tell us what force it is. Similarly, just saying the centripetal force just tells us what direction the force points. It doesn't really tell us what type of force this is, so to answer this question over here in a better way, if someone asked you what force counteracts gravity that keeps the ball from falling through the table, instead of saying upward force, it'd be better to just say that's the normal force. And we can do better over here as well. Instead of just saying the centripetal force, we could say what kind of force this is. It's gotta be one of the forces that we already know about. I mean, it's gotta be either the force of friction or normal force or tension or the force of gravity. The centripetal force isn't a new type of force. It's just one of the forces we already know that happens to be pointing toward the center of the circle. And that's important because this is our first, big common misconception. People think the centripetal force is a new kind of force, but it's not. It's just one of the forces we already know that happen to be pointing toward the center of the circle and that happen to be causing an object to move in a circle. So in this case over here, what force is it? Well, there's a rope tied to this mass, and that rope's gonna pull on it. And when a rope pulls, we call that the force of tension, so I'm gonna call this the tension. So that's a little better. Now we know what kind of force is acting as the centripetal force. Now, be careful out there. Sometimes, people want to do this, they're like, oh yeah, there's a force of tension, and there's also a centripetal force. But that's just crazy because this tension is the centripetal force. I wouldn't draw it twice anymore than I'd come over here and say, yeah, there's a normal force, there's also upward force. The upward force is the normal force. I wouldn't draw it again. Similarly, over here, I'm not gonna draw the centripetal force twice. The tension was the centripetal force. I mean, it's possible you could have two forces inward. Maybe there's two ropes and you had a second tension over here pulling inward, but you'd better be able to identify what force it is before you draw it. Don't just call it F centripetal, so you might be like, yeah, yeah, I get it. The centripetal force is just an extra title we give to a force that happens to point toward the center of the circle, but how would I ever solve a problem like this? What strategy do I use? I've got forces that are up, that are down, that are in. So let me show you how to solve some problems and some things to keep in mind. So let me add some numbers in here. So let's say I told you this. Let's say the mass of the ball was two kilograms, the rope's length was 0.5 meters, and the ball is traveling around the circle at a constant speed of five meters per second. So what kind of question might you be asked if given a problem like this? A possible question would be, well, what's the force of tension in the rope? And so, now's a good time for me to let you in on a little secret. The secret to solving centripetal force problems is that you solve them the same way you solve any force problem. In other words, first, you draw a quality force diagram. And then you use Newton's second law for one of the directions at a time. And if the direction you chose to analyze Newton's second law for didn't get you to where you needed to be, just do it again. Use Newton's second law again for another direction, and that'll get you to where you need to be. So in other words, let's draw a quality force diagram. We've got forces, but they're kind of all over here. This side view's gonna better illustrate all the forces involved. So we've already got the normal force upward and the force of gravity downward. Now, I'm gonna draw this tension pointing inward, that's the force that's acting as the centripetal force. Now, we're gonna use Newton's second law for one of the directions. Which direction should we pick? Well, which force do we want to find? We want to find this force of tension, so even though I could if I wanted to use Newton's second law for this vertical direction, the tension doesn't even point that way, so I'm not gonna bother with that direction first. I'm gonna see if I can get by doing this in one step, so I'm gonna use this horizontal direction and that's gonna be the centripetal direction, i.e., into the circle. And when we're dealing with the centripetal force, we're gonna be dealing with the centripetal acceleration, so over here, when I use a and set that equal to the net force over mass, if I'm gonna use the centripetal force, I'm gonna have to use the centripetal acceleration. In other words, I'm gonna only plug forces that go into, radially into the circle here, and I'm gonna have the radial centripetal acceleration right here. And we know the formula for centripetal acceleration, that's v squared over r, so I'm gonna plug v squared over r into the left hand side. That's the thing that's new. When we used Newton's second law for just regular forces, we just left it as a over here, but now, when you're using this law for the particular direction that is the centripetal direction, you're gonna replace a with v squared over r and then I set it equal to the net force in the centripetal direction over the mass. So what am I gonna plug in up here? What forces do I put up here? I mean, I've got normal force, tension, gravity. A common misconception is that people try to put them all into here. People put the gravitational force, the normal force, the tension, why not? But remember, if we've selected the centripetal direction, centripetal just means pointing toward the center of the circle, so I'm only going to plug in forces that are directed in toward the center of the circle, and that's not the normal force or the gravitational force. These forces do not point inward toward the center of the circle. The only force in this case that points toward the center of the circle is the tension force, and like we already said, that is the centripetal force. So over here, I'd have v squared over r, and that would equal the only force acting as the centripetal force is the tension. Now, should that be positive or negative? Well, we're gonna treat inward as positive, so any forces that point inward are gonna be positive. Is it possible for a centripetal force to be negative? It is. If there was some force that pointed outward, if for some reason there was another string pulling on the ball outward, we would include that force in this calculation, and we would include it with a negative sign, so forces that are directed out of the circle, we're gonna count as negative and forces that are directed into the circle, we're gonna count as positive in here. And if they're not directed into or out, we're not gonna include them in this calculation at all. Now, you might object. You might say, wait a minute. There is a force out of the circle. This ball wants to go out of the circle. There should be a force this way. This is often referred to as the centrifugal force, and that doesn't really exist. So when people say that there's an outward force trying to direct this ball out of the circle, they're usually referring to this centrifugal force, but this doesn't exist. It turns out this is not a real thing if you're using a good reference frame. There is no natural outward force for something going in a circle. You might object. You might be like, wait a minute. If I let go of this ball, it flies out of the circle. Won't it go flying off this way? And no, it won't. If you let go of the string right now, for some reason the string broke, at this moment this ball would not veer off that way. There's no force pushing it to the right. The ball, if the string broke, would just follow Newton's first law. It says it would just travel in a straight line with constant velocity, and it would roll off the table. So the reason you have to pull on the rope to get the ball to go in a circle is not because there's an outward force but because this ball wants to maintain its velocity. It has inertia, it wants to keep moving in a straight line, but you have to keep pulling on it to keep changing the direction of this velocity. So even though many people think there's an outward centrifugal force that's just naturally occurring on an object going in a circle, there is not. So finally, we can come back over to here. I can put my mass here. I can finally solve for my force of tension. If I do this, I'll multiply both sides by mass, and I just get that the force of tension is mass times the speed squared over the radius of the circle, and if I plug in my values, the mass was two, the speed was five, and you can't forget to square it. You divide by the radius which was 0.5, and you get that the force of tension had to be 100 Newtons. So in this case, the force of tension, which is the centripetal force, is equal to 100 Newtons. Now, some of you might be thinking, hey, this was way too much work for what ended up being a really simple problem. Why did we have to go through all the trouble of stating all of this problem solving strategy? And I agree. This one was easy, but other problems won't be easy. And if you don't have some sort of problem solving framework to fall back on, you'll be shooting blind and that's a lonely, lonely place to be. So let's use this same procedure, but let's look at a new problem. Let's say, you have this. Let's say you were riding your bike over a circular hill. So this gray line represents the pavement, and it starts off flat. But then the pavement veers upward and it creates this concrete hill that you ride over and then down and you ride over to this side. And all this purple circle is representing is the fact that if you were to continue this crest of the hill around into a circle, it would form this shape, so that gives us a way to define what the radius is of this top part of the hill. So, let's put some numbers in here. Let's say the radius of this hill was eight meters. Let's say the mass of you and your bike together are about 100 kilograms. And let's say you're riding over this hill at six meters per second. And let's say I asked you, what's the size of the normal force exerted on you and your bike as you ride over the crest of this hill at six meters per second? Now, let me show you what you can't do because most people would try to do this. They really want to say that the normal force is just gonna be equal to the force of gravity. Therefore, since the force of gravity is mg, the normal force should just be mg, but that can't be right. If the forces on an object are balanced and they cancel, the object is just gonna maintain its velocity, size, and direction, so this object, since it's going to the right, this bike would just continue going to the right and it would just hover straight off this hill. That'd be awesome, but that doesn't happen. This bike moves downward. It accelerates downward after this moment since it rides down the hill, so the downward force has got to be bigger. The force of gravity's gonna be bigger than the normal force 'cause if it wasn't, this bike would just hover off into space. So how do you solve this problem? We use the same strategy we used before. We're gonna draw a force diagram, but we already did that. We're gonna use Newton's second law for one of the directions, and the direction we're gonna pick is the vertical direction. Now, is that vertical direction the centripetal direction? Yeah, it is because look at into the circle is downward. Because this bike is at the crest of the hill, down corresponds to pointing toward the center of the circle and upward corresponds to pointing away, radially away from the center of the circle. So, since I'm dealing with the centripetal direction, we plug in the formula for the centripetal acceleration, and the part where you have to be most careful is what you plug into the centripetal forces. Remember that into the circle counts as positive and out of the circle counts as negative. So both of these forces, normal and gravity, are gonna be included, but only one of them are gonna be included with a positive sign. Think about which one. Can you figure out which force would be included in here with a positive sign? If you said the force of gravity, you're right, which is weird. Usually, we treat the force of gravity as negative because it points down, but for centripetal forces, what we care about is into or out of the circle. So, I'm gonna treat gravity as a positive centripetal force. Gravity is the force pointing toward the center of the circle, and the normal force in this case is gonna be a negative centripetal force since it's directed out of the center of the circle. And then, we divide by our mass. And so, if we solve this for the normal force, if you do some algebra, we'll multiply both sides by m, we move over the F N and then move the m v squared to the other side and what we end up getting is that mg minus m times v squared over r is equal to the normal force, which if we plug in numbers, gives us 100 kilograms times 9.8 minus 100 kilograms times the speed squared, that's gonna be six meters per second squared, divided by the radius of the circle we're traveling in which is eight meters, and you end up getting 530 Newtons. So the normal force on you and your bike as you ride over this hill is 530 Newtons. That is not equal to your weight. This is less than your weight. The force of gravity on you is gonna be m times g, that would be about 980 Newtons. So you experience less normal force, and this is natural. This is what happens when you ride over a hill fast. You feel slightly weightless as you go over that hill. If you've ever gone with a car a little too fast over a hill, you feel that whoa in your stomach, and you're like, hey, that was cool. That was the weightlessness you felt for a moment. If you go too fast, if you go too fast, this normal force will become zero. You'll subtract so much m v squared over r here, the normal force becomes zero. When that happens, you do become airborne, so be careful driving over those hills. If you drive too fast, you'll become airborne since your normal force is gonna become zero. So, recapping, when you solve centripetal force problems, be sure to draw a quality force diagram. Then use Newton's second law for one of the directions at a time. If you use the centripetal direction, the direction pointed radially into the circle, you can say that the acceleration in that direction is v squared over r, but be sure to only plug in forces that are directed radially, that is to say, forces that are pointed into or out of the circle. If they point into the circle, they're gonna be positive forces, and if they point out of the circle, they're gonna be negative forces.