1

concept

## Banked Curve

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Hey, guys. So now that we've talked about flat curve problems in this video, I want to show you how to solve the other kind of curve problem, which is called the Banked curve. You've definitely experienced this in everyday life you've ever driven along the highway. You'll notice that the turns on a highway or the exits aren't completely flat. They're actually banked or inclined. And that's basically to help you turn in a circle like this. All right, And so if you've ever watched a NASCAR race, you'll notice that the ends of the track you have these really steeply inclined turns that all the cars are going around to help them turn right. That's basically what banks curve is. All right. So the idea here is that, unlike flat curve problems, flat curves, you have objects going in horizontal circles because of static friction for banked curves, you're actually going around in a circle because you're on a frictionless incline. So the idea here is that if you're in a car and you're on that incline, the thing that's actually helping you turn is not actually friction. It's basically just the incline itself as you're going around in a circle. And so the idea is you're gonna accelerates towards the center of the circle like this. This is gonna be your A centripetal, but there's gonna be no friction on these inclines, okay? And so what happens here is now that we kind of understand the problem, we're actually gonna skip these two points. We're gonna start the example, we'll go back to them in just a second here. So the idea here is we have an 800 kg race. Cars. This is RM equals 800. All right, so we have the incline, which is angled at 37 degrees. That's our data. This is basically this data right here, and we have the radius of the curve. So basically, we have this turn like this and the radius of this curve, this is R R. Is equal to 200. So we want to do in This problem is we want to figure out the exact speed. So we're trying to figure out some kind of velocity so that the car does not slide up or down the incline. Remember that these cars are sorry. These inclines are friction list. There's no friction. So What happens here is there's a special speed that you have to go so that you're not sliding up or down if you're going too slow. Like, for instance, if you stop sliding and there's no friction, you're gonna go down the incline. And if you're going too fast, you're actually gonna shoot up the incline, eventually fly off the ramp like this. So we're trying to figure out that sweet spot, right? That that perfect speed that's gonna keep you at the middle of the incline. Okay, So just like with flat curve problems, the way that we solve this velocity, we're just going to take a look at our equations. We know we have some equations that relate the velocity to period and frequency, but we don't have any of those variables in this problem here. So instead, what we do is we can relate this velocity to this centripetal acceleration. The V squared over R. And if we're trying to figure out this centripetal acceleration is v squared over r, the way that we do that is we look at F equals M A and the centripetal direction. That's what we have to do. So the first step really is. We're just gonna draw the free body diagram, right? That's what we always do with problems. So let's check this out for the car that's on the incline. We have the weight force that acts straight down, that's mg. And then there's no other forces like applied forces, tension or friction. The only thing that it's basically keeping this on the incline is the normal force, which acts perpendicular to the surface. All right, so those are two forces. That's our free body diagram. Now, the way that we normally solve this problem these problems before is our acceleration pointed either down or up the incline. So we tilted our coordinate system to basically line up our X direction with the parallel to the slope. But now what happens is remember, we're accelerating in a circle and it's purely horizontally. So in these kind of problems, because the centripetal direction is horizontal, we're actually going to use an until tid coordinate system. We're not gonna tilt our coordinate system where we're basically just gonna use our normal X and y axis. So we're just gonna go back to normal X and y axis here. This is like the one case of inclined plane problems or you're not gonna tell that coordinate system. Okay, so what happens here is now you actually don't have to decompose RMG because it points straight down and that's along our access, the one that the force that does need to be decomposed is the normal force. So this gets broken up into N X and then N y, and we can see here is that the NX is actually the only force that points in the direction of the centripetal acceleration. Right? Your NX is pointing to the right. So in these problems here, without friction, objects accelerate on banked curves not because of the weight or friction force, but actually because of the normal force. Specifically the x component of that normal force. This nx. So we write on our F equals m A and start writing out our forces. There's only one. It's really just our necks and because it points along the direction, it's positive, right? So now what we do is we write em a C and we just replace this with v squared over R. So we really just have to come up with an expression for this velocity here. It's very similar to how we dealt with flat curves. All right, so we need to come up with an expression for this NX because we're never given that in problems. So let's go ahead and decompose, right? We know the NX and why you're going to have to get broken up using sign and co sign. So which one is it? Well, what happens here is that in these problems, this angle theta is actually the same as this angle over here, which is kind of bad, because that angle is with respect to the Y axis. And that's usually not the one that we want we wanted to with respect to the X axis. Unfortunately, there's no really good way around this. So what happens here is that we just basically have to use the flipped trig function So NX is actually gonna be Ensign Data and N y is going to be n cosign theta. It's the same exact principle as in our normal incline plane problems. This mg excellent with sign and why I went with co sign. It's the same thing here. So, really, this annex becomes Ensign data and you have m V squared over R. All right, so now again, what we need to do as we look at this normal force were usually never get given that in problems and the way we solve normal force and inclined planes is we look at the Y axis, right? We look at the other axis that we haven't yet looked at, so we're gonna look at some of all forces in the Y axis. This equals mass times acceleration. So we're really just looking at these forces over here. Now, remember that the car is not accelerating vertically. It's only accelerating in the centripetal direction like this. So what happens is all of our forces have to cancel. So really R N y and r m g have to be equal to each other. And remember, the n y is just n cosign theta. So remember we came over here because we wanted an expression for N, and now we have it. So now we have this end is equal to mg, divided by cosign theta. And I'm just gonna plug this right back into this equation here. So really, what we end up with is we end up with MG over CoSine of theta times. The sine of theta equals M v squared over r were super close. All we have to do is just some tidying up. So there's two important things that happened here. Um, one thing you'll notice is that the masses will cancel from both sides. And the other thing is that you end up with a sign over a co sign. So on the left side, you end up with G times the tangent of theta, and this equals V squared over R. And the last thing we do is we move the r to the other side with G r times, the tangent of theta. So if you take a look at this equation here, this equation was very similar to the equation that we ended up with flat curbs, and that's why I've written it that way. So this is basically going to be your shortcut equation. Whenever you're given multiple choice problems and you have banked curves, this is if you all, as long as you memorize this equation, you'll be able to solve the questions in no time. All right, All right. So the last thing we have to do is just take the square root and plug in all of our numbers. So you've got square up to 9.8. We've got the radius, which is 200 then we've got the tangent of 37 degrees and we end up with is a velocity of 38.4 m per second. That's our answer. If you go any slower or faster than this, you're gonna start going up or down the ramp. So the last point I want to make here is basically just a summary of all of our equation that we've seen for flat curve and banked curve problems for flat curve. Remember you on a flat surface and the thing that kept you going around with static friction. So you have static friction, but you have no incline. And so therefore, your velocity equation has to include that variable, which is mu static. So this is G R. Times, mu static, right? Basically just includes the variable that you have in those problems. And for banked curves, it was the opposite for banked curves. You have no friction, but you do have some incline, so your velocity equation has to include that variable So this is gonna be gr times the tangent of theta. All right, So these are your two equations for flattened banked curve problems. They're gonna be super helpful for you when you have multiple choice problems. If you have anything where you have to solve kind of like I did here, you're gonna have to go through this whole process, unfortunately. All right, so that's it for this one, guys.

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Problem

A bobsled turn banked at 78° is taken at 24 m/s. Assume it is ideally banked and there is no friction between the ice and the bobsled. Calculate the centripetal acceleration of the bobsled.

A

1100 m/s

^{2}B

2.08 m/s

^{2}C

46.1 m/s

^{2}D

1.92 m/s

^{2}Additional resources for Banked Curves

PRACTICE PROBLEMS AND ACTIVITIES (1)