Vertical Centripetal Forces - Online Tutor, Practice Problems & Exam Prep

Vertical Centripetal Forces

Hey, guys. So now that you understand the basics of uniform circular motion and centripetal forces, occasionally, you're 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. What I'm going to show you in this video is that we're actually going to solve these problems the exact same way using 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 also special about vertical centripetal forces is that the speed actually isn't going to be constant because throughout the motion, the \(mg\), the weight force, is constantly pulling you down. It's changing your speed. So even though the speed is not going to be constant in vertical centripetal forces, we're still going to use the same equations like \(f = ma\) and \(\frac{v^2}{r}\). So let's go ahead and get to our example here. We have a roller coaster kart, and it's going to run a vertical loop and we have a radius of 10. So we have the radius of this loop, this \(r = 10\) 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 \(v_{\text{bottom}}\) or \(v_b\), is 10. Oh, I'm sorry. This is 30. This is going to be 30. And then at the top of the loop, our speed, I'm 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 meters per second. So we have the speeds at both the bottom and the top. What we want to do in part a is we want to figure out what is the centripetal acceleration and the normal force when you're at the bottom of the loop. So basically, what is, \(a_c\)? 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, \(\frac{v^2}{r}\). It actually doesn't depend or it doesn't matter that we're traveling in a vertical loop because it 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 \(v_{\text{bottom}}^2\) divided by \(r\). So we don't need to get into our \(f = ma\). 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 \(\frac{30^2}{10}\). If you go ahead and work this out, what you're going to get is 90 meters per second squared. So that's the answer. So that's the centripetal. Now the second part of this problem, part a, is that we actually have to calculate the normal force on the seat that's basically acting from the seat of the roller coaster cart on you while you're inside of it. So there's going to be a normal force, it's going to be \(n_{\text{bottom}}\). So to figure out a force now, we're actually going to have to figure out our \(f = ma\). So we're going to get into our centripetal forces here. So in order to figure out the normal force, I'm going to have to write out my \(f = ma\). So this is \(f_{\text{centripetal}} = m a_c\). Now I've already calculated what this \(a_c\) is. Remember that there's usually an extra step where we have to rewrite our \(a_c\) as \(\frac{v^2}{r}\), but I actually don't have to do that here because I've already calculated what this \(a_c\) is, which is the 90 that we calculated before. So what I have to do is I have to expand out my centripetal forces here. So what are the forces that are acting on you when you're at the bottom of the loop? Well, it turns out I'm going to move this over. There's actually 2. So we have an \(mg\) that points downwards. This is our \(mg\) here. Let's see. Do we have any tension forces? No. There's no tensions, but there is a contact force. There's no friction or anything like that, but there is a contact force. It's basically the force that is on you that's acting from the seat of the cart. When you're sitting down, the force that keeps you up is the normal force that's what 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 or different directions, we're going to have to do that we haven't considered before is you're 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 degrees 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 that up, then that would actually be 0 because that doesn't contribute anything to the centripetal acceleration. So that's really the rules. So now that we have those rules, those rules, we can go ahead and expand those sum of all forces. So our normal force is going to be positive, and then our \(mg\) is going to be negative because it points away. So those are the two forces, and this is going to equal \(m \cdot a_c\). Alright. So what we want to figure out is the normal force that's at the bottom, so this is going to be \(n_b\). So \(n_b\), and you're going to move this over to the other side, is going to equal \(m \cdot a_c + m \cdot g\), which really if you can group together the \(m\)'s, it's just going to be \(m (a_c + g)\). So if you work this out your normal force is just going to be the mass which is 70, right, where our mass is 70 kilograms. It's given to us here at the top. The centripetal that we just calculated was 90. That's we calculated over here, plus our \(g\) which is going to be 9.8. If you go ahead and work this out, what you're going to get is you're going to get about 7,000 newtons. Alright. So that's how you calculate the centripetal forces. We're going to do the exact same thing now, except the only thing that's different is that now we're going to be at the top of the loop and our speed is going to be 20. Alright. So we're going to calculate this very quickly. Centripetal here at the top is going to be \(v_{\text{top}}\) divided by \(r\). Same equation, except now we just have a different velocity that we're considering. Considering. \(v_t\) is 20. So we're going to calculate this. This is going to be \(\frac{20^2}{r}\), and which \(r\) is just equal to 10, and you're going to get a centripetal acceleration of 40 meters per second squared. So what happens is here your \(a_{\text{centripetal}}\) is 90, but here when you're at the top, your \(a_{\text{centripetal}}\) 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's 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 going to get that by using \(f = ma\). So \(f_{\text{centripetal}} = m a_c\). We actually already calculated this. What are the forces? Well, here at the top, you actually have 2 forces. You have an \(mg\) that points down, and the normal force from the c also points down on you. As you're going around in the circle, the normal force is going to point towards the circle. So what happens is that both your \(n\) and \(mg\) 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, what you're going to get, I'm just going to move this over here, is \(m\) equals and you're going to have \(m(a_c - g)\). So notice the difference. So here at the bottom of the loop, it was \(a_c + g\) because one of your forces was negative, and here you actually have to subtract it. So when we calculate this, this is going to be \(n_t\), and this is going to be \(n_{\text{top}}\) equals 70 times, this is going to be your \(a_{\text{centripetal}}\) which is 40 minus 9.8. If you go ahead and work this out, what you're going to get is about 21100 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 you have a ton of speed and your normal force actually has to counteract your weight and also accelerate towards the circle like that. So the normal force on you is much stronger. Alright. So that's it for this one, guys. Let me know if you have any questions.

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