Vertical Centripetal Forces - Video Tutorials & Practice Problems

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Vertical Centripetal Forces

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Hey guys, so now you understand the basics of uniform circular motion and centripetal forces, occasionally going to have problems where you have objects that are traveling not in a horizontal plane, but rather in a vertical plane. A classic example of this is a roller coaster that goes around in a loop de loop. But I'm gonna show you this video is we're actually going to solve these problems the exact same way in the same exact equations. There's just a couple of things that are different. For example, we have some other forces to consider, like MG. What also happens? What's so special about vertical centripetal forces is that the speed actually isn't going to be constant because throughout the motion, the MG, the weight forces constantly pulling you down, it's changing your speed. So even though the speed is not going to be constant in vertical centripetal forces were still going to use the same equations like F equals M A and v square bar. So let's go ahead and get to our example here. So we have a roller coaster cards and it's going around a vertical rupe and we have a radius of 10. So we have the radius of this loop. This r equals 10 right here. So we have the speed at the bottom of the loop and the top, we're told the speed at the bottom, I'm just going to call this V bottom or VB is 10. I'm sorry, this is the 30 it's gonna be 30. And then at the top of the loop we are speed. I'm gonna just going to label this, we're going this way at the top of the loop, we're actually going, I'm going to label this as V. T. We're going at 20 m per second. We have the speeds of both the bottom of the top. We wanted to impart a is we want to figure out what is this centripetal acceleration and the normal force when you're at the bottom of the loop. So basically what is um a C. What is the centripetal when you're at the bottom? So how do we solve this? Well like any centripetal forces or centripetal acceleration problem, we're just going to use this equation here. V squared over R. It actually doesn't depend or it doesn't matter that we're traveling in a vertical loop because only just depends on V. And R. That's all there is to it. So this A C at the bottom is going to be V bottom squared divided by our, so we don't need to get into our F. Equals M. A. We don't have to do anything like that because this is just centripetal acceleration. So we can calculate this. This is going to be 30 squared divided by the radius which is 10 If you go out and work this out what you're gonna get is 90 m per second squared. So that's the answer. So that's the a centripetal. Now the second part of the problem of part A is that we actually have to calculate the normal force on the seat. That's that's basically acting from the seat of the roller coaster card on you while you're inside of it. So there's going to be a normal force that's going to be in at the bottom. So to figure out a force now we're actually gonna have to figure out our F. Equals M. A. So we're gonna get into our centripetal forces here. So in order to figure out the normal force, I'm gonna have to write out my F. Equals M. A. So this f centripetal equals M. A. C. Now I've already calculated with this A. C. Is remember that there is usually an extra step. We have to rewrite our A. C. As V. Square bar. But I actually don't have to do that here because I've already calculated with a C. Is just the 90 that we calculate before. So what I have to do is I have to expand out my centripetal forces here. So one of the forces that are acting on you when you're at the bottom of the loop. Well it turns out I want to move this over. There's actually two. So we have an MG that points downwards. This is RMG here. Um Let's see. Do we have any tension forces? No, there's no tensions. Um but there is a contact force, there's no friction, anything like that. But there's a contact force. It's basically the force that is on you that's acting from the seat of the cards. When you're sitting down, the force that keeps you up is the normal force that keeps you from crashing through the floor like this. So you have these two forces normal and MG. Now because these two forces point in opposite directions are different directions, we're going to have to do that we haven't considered before is going to have to establish a direction of positive. Unfortunately, there are some rules, very simple ones to determine the signs of your centripetal forces. Here's the way it works. Any forces that point towards the center of your circle are going to be positive. We know that as you travel around the circle, you're always going to be accelerating towards the circle. So any forces that point in that direction are going to be positive. Like our normal force here, any forces that point away from the center are going to be negative like this MG here And lastly, any forces that point perpendicular, meaning 90° to the direction of the center. So for example, if I had like a friction force that was going in this direction, right? Just making it up, then that would actually be zero because that doesn't contribute anything to the centripetal acceleration. So that's really the rules. So now that we have that rules those rules, we're gonna go ahead and expanded on some of all forces. So our normal force is going to be positive and then R. M. G. Is going to be negative because it points away so those are through to forces and this is going to equal M times A C. Alright, so we want to figure out is the normal force that's at the bottom. So this is gonna be NB so N. B. And you're gonna move this over to the other side is going to equal M. A. C. Plus we'll be going to write this out. This is gonna be M. A. C. Plus MG. Which really if you can group together the ems is just gonna be M. Times A. C. Plus G. So if you work this out, your normal force is just gonna be the mass which is 70 right? Where our mass is 70 kg is given to us here at the at the top. Uh The a centripetal that we just calculated was 90. That's what we calculate over here. Plus R. G. Which is going to be 9.8. If you go and work this out, what you're gonna get, you're gonna get about 7000 newtons. All right, so that's how you calculate the centripetal forces. We're gonna do the exact same thing now. Except the only thing that's different is that now we're gonna be at the top of the loop and our speed is going to be 20. All right. So, we're going to calculate this very quickly. A centripetal here, at the top is going to be v at the top divided by our same equation. Except now we just have a different velocity that we're considering. VT is 20. So, we're gonna calculate this this is gonna be 20 squared divided by our and which are are is just equal to 10. And you're gonna get a centripetal acceleration of 40 m per second squared. So, what happens is is here you're a centripetal is 90. But here, when you're at the top you're a centripetal is 40 and it's because you're traveling slower, your speed has gone down because you've gone up this hill. So, again, speed is not constant, but we can still calculate our centripetal acceleration. Alright? So let's take a look at the normal force. Now, normal force at T. We're gonna get that by using F equals M. A. So f centripetal equals M A C. We actually already calculated this. What are the forces? Well here at the top you actually have to forces you have an MG that points down and the normal force from the seat also points down on you as you're going around the circle. The normal force is going to point towards the circle. So what happens is that both your N and your MG are actually both going to be positive in this case. So this equals M A C. So when you move this over to the other side which are going to get I'm just gonna move this over here is N. Equals and you're going to have M times a c minus G. Notice the difference. So here, at the bottom of the loop it was a C plus G. Because one of your forces with negative and here you actually have to subtract it. So when we calculate this is going to be at the top and this is gonna be M. T. equals 70 times. Um this is gonna be your a centripetal, which is 40 -9.8. You've got to work this out. What you're gonna get is about 2100 newtons. So, we can see here that at the bottom of the loop, the normal force that's acting on you is much much larger because basically have a ton of speed and your normal force actually has to counteract the your your weights and also accelerated towards the circle like that. So the normal force we knew is much stronger. All right, so let's take this one guys, let me know if you have any questions

2

Problem

Problem

Suppose a 1,800-kg car passes over a bump in a roadway that follows the arc of a circle of radius 20m. What force does the road exert on the car as the car moves over the top of the bump if the car moves at a constant 9 m/s?

A

10350 N

B

24930 N

C

17640 N

D

16830 N

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