Rollercoaster Problems - Online Tutor, Practice Problems & Exam Prep

In roller coaster problems, understanding centripetal force and energy conservation is crucial. To find the minimum speed at the top of a loop (point B), use the equation for centripetal force, where the normal force is zero when passengers are about to fall. The minimum speed, $\sqrt{g*r}$, is calculated as 7 m/s. For height at point A, apply energy conservation, leading to a height of 12.5 m, ensuring the cart has enough potential energy to reach point B with the required speed.

Rollercoaster Problems

Hey, guys. So occasionally in energy conservation, you run across these roller coaster type problems. We have a car that's traveling on 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 looked at centripetal forces. The tricky part of these problems is figuring out what the target variable is asking for, so you can figure out which equation you're going to use. I'm going to show you how to determine that in this video. We're really just going to see that we're going to use either f equals ma or energy conservation depending on the two different types of variables that we're asked for. Let's go ahead and check out our problem here. So we have roller coaster karts without seat belts that are traveling around a loop de loop of radius 5. So this is r equals 5 here. We have two parts to 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 going to 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 going to look for vb, so that the passengers don't fall out of their seats. Now before I start that, I want to point out what part b is asking us for. In 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 height of this hill that we're starting off from. And so that brings me to an important point here an important conceptual point. 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 the speed at b so that the passengers don't fall. So we're looking at really just analyzing 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 point, you're going to use f equals ma. This is going to be no different than our centripetal forces problems. When you have a target variable that refers to two points like we have in part b, then we're actually going to use energy conservation. In part b, we're looking for the height 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 going to have to sort of compare and contrast, we're going to 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 a, what we're going to do is we're going to solve this by using f equals ma. So this is fc = mac. So we're going to have to do the sum of all forces and you might be wondering where does this velocity come into play. Well, remember that your centripetal acceleration equation, which is right here, has a v inside of it. That's where it really comes from. So remember you're going to expand this out and you're going to write this as mvb2 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. So let's go ahead and do that. So here at point B, you have two forces, you're going to 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 roller coaster karts. 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 going to get mg and a normal force, these are both positive, is equal to mvb2 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 going to have to solve for 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 m's are in every single term in this equation. But I actually have an n here that doesn't have an m inside of it, so I can't cancel that out. So can I solve for what this normal force is? 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 0. 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 their seats, then basically there would be no more normal force because their butt would come off of the seats, there would be no surface push anymore. So you kind of 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 = vb2 over r. So what I'm going to do is I'm going to move the r to the other side and take the square roots. So we're going to get vb = gr or the square root of 9.8 × 5, what you're going to get is exactly 7 meters per second. So this is the minimum speed, vb = 7. If you were traveling at 7.0001, 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 7 is the minimum speed where that doesn't happen. Let's take a look at part b now. In part b, we said we're going to set up an energy conservation equation because we're looking at what happens from a to b. So we're going to set up an energy conservation equation. This is going to be kinitial, uinitial, worked on by non-conservative, equals kfinal plus ufinal. So let's go ahead and eliminate and expand all the terms. We're really just going to stick to the steps now. Right? We have an energy conservation equation. We're going to 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 going to have to go, you know, down at some speed like this, so maybe. Do we have some potential energy? Yes, we do because we have some height here. So we have both of these. 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 going to be 7, so there's definitely going to be some kinetic energy. Do we also have some height here at b? Well, yes, because here at the top of the loop, we have some height. This is going to be yb. So we definitely have some height above the ground because we're considering the ground to be our zero point. 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 mva2, plus my potential energy, which is going to be mgya. This equals, one half mvb2 plus mgyb. So as we can see, the masses are actually going to cancel out because they're in every single term of our problem here. So let's take a look. I don't have the speed at a. I'm looking for the height at a. This is actually my target variable right here. G I could figure out, I have vb and I also have what yb actually, I don't know what yb 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 actually got 3 unknowns. I'm going to have to figure out either I'm actually going to have to figure out both va and yb. So let's take a look at this va here. Can we figure that out? Well, this is going to be the speed that the cart has initially so that it can sort of start rolling down and then, finally reach this loop here. But 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 0. So basically the idea here is if you had any speed at the top of this hill, then actually your height could have been lower because you have some additional kinetic energy. So we're asking for the minimum height, we're also assuming that the initial speed is equal to 0. So that's kind of how we also have to realize in 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, some writing out some numbers here. So this whole term goes away and I've got, let's see, 9.8 × my equals and then I've got 1 half and then vb, remember, vb is actually just the number that we calculated in part a. That's the 7 meters per second. That's 7 squared plus, and then we've got 9.8s, 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 in this problem is actually the radius of the loop itself. We're told that the radius is equal to 5, So that means that this distance here is 5, and this distance to the top of the loop is also 5. So the height at the top of the loop is actually going to be 10, which is 2 times the radius here, and that's actually generally going to be true for all your problems. The height of the loop is actually just going to be twice the radius. So hloop = 2 r in general. 등록된 패션 칼럼니스트 송은영입니다. Alright. So basically, this is actually going to equal 10 here as we just figured this out. Alright. So now what happens is I'm going to get ya = this is going to be 9.8 to ya. When you sort of solve everything over here, you're going to get a 122.5. When you divide, you're going to get that you is equal to

More Rollercoaster Problems

Hey, guys. Let's go ahead and work this one out together. So we have a big loop de loop of radius r that this cart is trying to go around, and we're just dealing with letters here. All we know is that this loop de loop is of radius r. Now what happens is we're trying to figure out the minimum speed that we need at the very bottom of the loop. So, we're trying to figure out how if we're going around this loop like this, what's the minimum speed that we need so that we reach the top, but just barely. So, what happens here? Are we using just one point or are we comparing 2 points? We're trying to figure out the speed we need at the bottom, or I'm going to call this my initial here, so this is my v initial, so that I can reach the top, which is going to be my final. That's 2 points. So, I'm going to use energy conservation. So 2 points just means I use energy. It's going to be energy conservation here. Once I draw my diagram, I'm just going to go ahead and start writing my energy conservation equation. So this is k initial plus u initial plus work done by non-conservative equals k final plus u final. We have to do this because we're traveling in a curved or circular path. So, let's go ahead and expand out the terms and eliminate. We're looking for some kinetic energy because we're looking for what's the minimum speed, that's going to be v initial and we're again we're just working with letters. Is there any potential energy initial? Well, if you consider the bottom of the loop the place where you have y equals 0, then there is no gravitational potential energy. There's also no work done by non-conservative forces because you're just sitting there watching the cart and there's also no friction. What about k final? Well, that's actually the tricky part about this problem. We're trying to figure out the minimum speed so that we just barely reach the top and the other important part is that the cart has to stay locked to the tracks. So, what does that mean? It means that if the cart weren't locked to the tracks, if the cart were going very, very slow at the top of the loop, it would just basically just fall off the tracks. If the cart remains locked, then that means what happens is it can actually go very, very slow as it reaches the top and it won't have to necessarily fall. So, what does this just barely reach the top actually mean? What it means here is that we're trying to figure out the minimum speed so that the speed here at the top is actually equal to 0. So, that's the important sort of conceptual point they need to realize. So, this kinetic final here actually has to go away, and the reason for that is we're looking for the minimum speed here. So, we're looking for some sort of limit or boundary condition. What happens is if this minimum speed that we calculate were any less than what we calculated, then that means that this that this cart wouldn't actually reach the top, right? If we were traveling any less than this minimum speed, we would get all the way up here, but then we would basically fall back down like this. If this, if the speed here at the top weren't equal to 0, let's say the speed here at the top were equal to 5, then whatever speed that we calculated down here actually could have been less than what we calculated. Right? So, that's why we have to set this equal to 0, so we just barely reach the top. So now what happens is we're going to set our 12mvinitial² equal to and that's the gravitational potential energy. This is going to be mgyfinal. So we've risen some height like this. This is going to be my y final. So, I can go ahead and cancel out the masses in my expression, and again, I would just want to solve or figure out an expression for this v initial. So, I'm going to move the one-half to the other side and I get vinitial2 is equal to 2g. Now I can figure out an expression for y final. Remember, I'm not going to use y final because I'm given the radius is r. If you take a look here, what happens is you've actually risen sort of twice the radius. There's just the diameter. So your y final is actually equal to 2r. So, that's the that's the expression I'm going to substitute in here. So now all you have to do is just take the square roots. So, we're going to take the square root and this really just becomes 4gr, and that is your expression. One last thing that you could have done, but you totally could have left this like this, is just pull the 4 outside of the square root, and you would have simplified this. It would have become 2gr. Either one of these is fine and you give full credit for these answers. Alright. So that's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is the minimum speed required at the top of a roller coaster loop to prevent passengers from falling out?

The minimum speed required at the top of a roller coaster loop can be determined using the centripetal force equation. At the top of the loop, the normal force is zero when passengers are about to fall. The equation is:

$\sqrt{(g\times r)}$

where $g$ is the acceleration due to gravity (9.8 m/s²) and $r$ is the radius of the loop. For a loop with a radius of 5 meters, the minimum speed is:

$\sqrt{(9.8\times 5)}=7 m/s$

How do you calculate the height needed at the start of a roller coaster to ensure it reaches the top of a loop with a specific speed?

To calculate the height needed at the start of a roller coaster to ensure it reaches the top of a loop with a specific speed, use energy conservation principles. The equation is:

$K_i+U_i=K_f+U_f$

where $K_i$ and $U_i$ are the initial kinetic and potential energies, and $K_f$ and $U_f$ are the final kinetic and potential energies. Assuming the initial speed is zero, the equation simplifies to:

$mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f$

Solving for $h_i$ gives the height needed. For a loop with a radius of 5 meters and a final speed of 7 m/s, the height is approximately 12.5 meters.

What role does centripetal force play in roller coaster problems?

Centripetal force is crucial in roller coaster problems, especially when dealing with curved or circular paths. It is the force that keeps the roller coaster car moving in a circular path and is directed towards the center of the circle. The centripetal force equation is:

$F=\frac{m{v}^2}{r}$

where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of the path. Understanding centripetal force helps in determining the minimum speed required at the top of a loop to prevent passengers from falling out and in analyzing forces acting on the car at different points of the track.

How do you apply energy conservation in roller coaster problems?

Energy conservation in roller coaster problems involves equating the total mechanical energy at different points of the track. The principle states that the sum of kinetic energy (K) and potential energy (U) remains constant if no non-conservative forces (like friction) do work. The equation is:

$K_i+U_i=K_f+U_f$

where $K_i$ and $U_i$ are the initial kinetic and potential energies, and $K_f$ and $U_f$ are the final kinetic and potential energies. This approach helps in determining the height needed at the start to ensure the car reaches a specific point with the required speed.

What equations are used to solve roller coaster problems involving circular paths?

To solve roller coaster problems involving circular paths, two main equations are used: the centripetal force equation and the energy conservation equation. The centripetal force equation is:

$F=\frac{m{v}^2}{r}$

where $m$ is the mass, $v$ is the velocity, and $r$ is the radius. The energy conservation equation is:

$K_i+U_i=K_f+U_f$

where $K_i$ and $U_i$ are the initial kinetic and potential energies, and $K_f$ and $U_f$ are the final kinetic and potential energies. These equations help determine speeds and heights at various points on the track.