10. Conservation of Energy

Rollercoaster Problems

# Rollercoaster Problems

Patrick Ford

499

1

3

Was this helpful?

Bookmarked

Hey guys. So occasionally in energy conservation, you'll run across these roller coaster type problems. We have a car that's traveling in a roller coaster around these curved or even circular paths, the good news is we know how to solve curved and circular path problems. When we took a look at centripetal force is the tricky part of these problems is figuring out whether target variables asking you for so you can figure out which equation you're gonna use. I'm gonna show you how to determine that in this video. We're really just gonna see that we're just gonna use either ethical or energy conservation depending on the two different types of variables that were asked for. Let's go ahead and check out our problem here. So we have a roller coaster carts without seat belts that is traveling around a loop de loop of radius five. So this is our equals five here. So we have two parts of this problem and I want to sort of lay them both out so we can figure out what equation we're going to use to solve both of them in part A. We're gonna figure out the minimum speed that we need at point B. So that the passengers would not fall. We've seen this type of problem before in centripetal forces in the vertical direction. What we're looking for here, remember is that we're looking for a speed here, at the top of the loop. So we're gonna look for vB so that the passengers don't fall out of their seats. So we're looking for here is we're looking for V. B. Now, before I start that, I want to point out what part B is asking this for. And the second part of the problem, we're figuring out the height of the cart at point A. So that we can reach this point here with the speed that we find. So we're looking for here is the heights of this hill that we're starting off from. And so that brings me to an important point here. An important conceptual points. These problems are actually asking you to do two different things in the first part. We're just looking at one point, we're looking at what is the speed at B. So that the passengers don't fall. So we're looking at really just analyze one point here, what's the velocity at the top of this loop And in circular path problems. When the target variable refers to one points, you're going to use F equals M. A. This is gonna be no different than our centripetal forces problems. When you have a target variable there first to two points like we have in part B. Then we're actually gonna use energy conservation in part B. We're looking for the heights at A. So that we can reach B. With the speed that we find, so notice how they're referring to two points here, we're gonna have to sort of compare and contrast. We're gonna have to set up an initial and final and set up an energy conservation equation. Alright. So let's go ahead and get to it. Let's take a look at part A. So in part of what we're gonna do is we're gonna solve this by using F. Equals M. A. So this is F. C. Equals M. A. C. So we're gonna have to do the sum of all forces. And you might be wondering where this is, where does this velocity come into play? Well remember that your your centripetal acceleration equation which is right here has a V inside of it, that's what it really comes from. So you're gonna expand this out and you're gonna write this as M. V. B squared over R. So this is where this variable comes in. So we have to figure out what are the forces that are acting on your body or on these passengers. Um So let's go ahead and do that. So here at point B you have to force is you're gonna have an MG. That always points down and there's another force that's acting on you because you're constantly traveling in a circle. So what happens is the passengers, they want to continue flying off in this direction here at this tangential velocity. But the thing that keeps them going in a circle is actually the force from the seat of the road coaster carts. So there is a normal force that is keeping them going in a circle like that. So these are both positive forces because they point towards the center of the circle. So when I expand out my terms here, I'm gonna get a MG and a normal force. They're both positive is equal to M. V. B squared over R. Alright, so I want to figure out what this VB is. But let's take a look at my variables. I don't have the mass, I know what the G. Is, I don't have the normal force and I know what the radius is. So I've got these two unknowns in this problem, the mass and the normal force. And I'm gonna have to sulfur both of them if I want to figure out what this velocity is. So let's see, can we cancel out the masses? Well, I can't yet because remember, I can only cancel the masses if the MSR in every single term in this equation. But I actually have an end here that doesn't have an M. Inside of it. So I can't cancel that out. So can I sulfur what this normal forces And actually we can and we've seen this type of problem. Remember what this phrase is telling you when the passengers don't fall out of their seats, is that the normal force is equal to zero. So basically, in order to figure this problem out, you have to figure out what would happen if the passengers actually did fall out of their seats. It's kind of counterintuitive. But if they did fall out of there then basically there will be no more normal force because there's there but would come off of the seats there would be no surface push anymore. So you kinda have to figure out the minimum speed. So that that doesn't happen right? So this normal force actually goes away and now what we can do is now because this normal force goes away we can cancel out the masses and therefore we've kind of gotten rid of both of our unknowns here. So we end up with is we end up with G. Equals V. B squared over R. So what I'm gonna do is I'm gonna move the R. To the other side and then take the square roots so we're gonna get VB is equal to the square root of G. R. Or the square root of 9.8 times the radius of five. What you're gonna get is exactly seven m per second. So this is the minimum speed. VB equals seven. If you were traveling at 7.1 then you could have been going a little bit slower and that's not the minimum speed. If you're going at 6.999 then that's actually too slow. And you would actually come off of the seats and you would start to fall. Right so seven is the minimum speed where that doesn't happen. Let's take a look at B. Now and be we said we're gonna set up an energy conservation equation because we're looking at what happens from A. To B. So we're gonna set up an energy conservation equation. This is gonna be K. Initial. You initial work done by non conservative equals K. Final plus you. Final. So let's go ahead and eliminate and expand all the terms. We're really just gonna stick to the steps now. Right. We have an energy conservation equation, we're gonna eliminate and expand the terms. So do we have any initial kinetic energy? Remember, that's the kinetic energy here at the top of the hill. Well, maybe because, you know, we're sort of rolling off this car, we're gonna have to go down at some some speed like this. So maybe do we have some potential energy? Yes, we do because we have some heights here. So we have both of these. Um Do we have any work done by non conservative forces? There's no work done by you and there's also no friction in the problem. So, no, we don't. Do we have any kinetic energy here at point B. We just figured out that the speed is gonna be seven. So there's definitely gonna be some kinetic energy. Do we also have some heights here at B. Well, yes, because here at the top of the loop we have some this is going to be Y. B. So we definitely have some height above the ground because we're considering the ground to be are zero points. So we actually have four of these terms here. Let's go ahead and write out our terms. So I've got my one half, this is going to be M V A squared plus my potential energy which is going to be M G Y. A. This equals one half Mv B squared plus MG times Y B. So as we can see, the masses are actually gonna cancel out because they're in every single term of our problem here. So, let's take a look. Um I don't have the speed at A. I'm looking for the height at a. This is actually my target variable right here. Uh G. I can figure out, I have V. B. And I also have what why actually, I don't know what it is. I don't know what the height of this loop is. So, I've got these two unknowns in this problem, right? I've got actually three unknowns. I'm gonna have to figure out either. I'm actually gonna have to figure out both via and yB. So, let's take a look at this V. A. Here. Can we figure that out? Well, this is gonna be the speed that the cart has initially so that it can sort of start rolling down and then finally reach this loop here. Well, what you have to realize about this problem is that when they're asking you for the minimum height needed, we actually have to assume that the velocity is equal to zero. So basically the idea here is that if you had any the speed at the top of this hill than actually your heights could have been lower because you have some additional kinetic energy. So we're asking for the minimum heights were also assuming that the initial speed is equal to zero. So that's kind of how we also have to realize this problem. What that means is that there is no kinetic energy, initial. And so the whole entire term goes away. Alright, so basically let me go ahead and start solving out some writing out some numbers here. So this whole term goes away and I've got let's see 9.8 times my Y. A. Equals and then I've got one half and then VB remember VB is actually just the number that we calculated in part A. That's the seven m per second, That's seven squared plus. And then we've got 9.8. And then what about the height of this loop? Or what about the height here at point B. Well, the only variable that I know that has to do with a distance. And this problem is actually the radius of the loop itself, we're told that the radius is equal to five. So the That means that this distance here is five. And this distance to the top of the loop is also five. So the height at the top of the loop is actually going to be 10, which is two times the radius here. And that's actually generally gonna be true for all your problems. The height of the loop is actually just going to be twice the radius, so h loop is just equal to two are in general. Alright. So basically this is actually an equal 10 here as we just figured this out. All right, so now what happens is I'm gonna get Y. A. Is equal to this is gonna be 9.8 Y. A. When you sort of soul everything over here, you're gonna get 100 and 22.5. When you divide you're gonna get that Y. A. Is equal to 12.5 m and that is your final answer. Now, this should make some sense. You have a height that you calculated is equal to 12.5. This makes sense because you're gonna have to start a little bit higher than the top of the loop, which is at Y equals 10 in order for you to roll down the loop and then still have some speed when you get here to the top of the loop. Alright, so it makes sense. We've got a number that's slightly above 10, we got 12.5 and that's it for this one guys. So let me know if you have any questions

Related Videos

Showing 1 of 8 videos