Guys, 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. If you've ever driven along a 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. Alright? And so, if you've ever watched a NASCAR race, you'll notice at 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 a banked curve is. Alright? So the idea here is that unlike flat curve problems, 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 going to accelerate towards the center of the circle like this. This is going to be your centripetal, but there's going to be no friction on these inclines. Okay? And so what happens here is, now that we kind of understand the problem, we're actually going to skip these two points, and we're going to start the example. We'll go back to them in just a second here. So the idea here is we have an m=800 kilogram race car. This is our m=800. Alright. So we have the incline, which is angled at θ=37 degrees. That's our theta. This is basically this theta 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 our r=200. So what we want to do in this problem is we want to figure out the exact speed so that the car does not slide up or down the incline. Remember, These inclines are frictionless. There is 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 slowly, for instance, if you start sliding and there's no friction, you're going to go down the incline. And if you're going too fast, you're going to shoot up the incline and eventually fly off the ramp like this. So we're trying to figure out that sweet spot; that perfect speed is going to 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 to 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 the centripetal acceleration, v2∕r. And if we're trying to figure out this centripetal acceleration, v2∕r, the way that we do that is we look at F=ma in the centripetal direction. So that's what we have to do. So the first step really is we're just going to 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 are no other forces like applied forces, tensions, or friction. The only thing that's basically keeping this on the incline is the normal force, which acts perpendicular to the surface. Alright? So those are our two forces. That's our free body diagram. Now the way that we would normally solve these problems 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 horizontal. So in these kinds of problems, because the centripetal direction is horizontal, we're actually going to use an untilted coordinate system. We're not going to tilt our coordinate system. We're basically just going to use our normal x and y-axis. So we're just going to go back to the normal x and y-axis here. This is like the one case of inclined plane problems where you're not going to tilt that coordinate system. Okay? So what happens here is now we actually don't have to decompose our mg because it points straight down and that's along our axis. 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 what we can see here is that the n_x is the only force that points in the direction of the centripetal acceleration. Right? Your n_x 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 n_x. So when we write out our F=ma and start writing out our forces, there's only one. It's really just our n_x, and because it points along the direction, it's positive. Right? So now what we do is we write ma_c and we just replace this with v2∕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. Alright? So we need to come up with an expression for this n_x because we're never given that in problems. So let's go ahead and decompose. Right? We know that n_x and n_y are going to have to be broken up using sine and cosine. 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 want it 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 n_x is going to be nsinθ and n_y is going to be ncosθ. It's the same exact principle as in our normal inclined plane problems. This mg_x went with sine, and y went with cosine. It's the same thing here. So really, this n_x becomes nsinθ and you have mv2∕r. Alright? So now, again, what we need to do is we look at this normal force. We're usually never given that in problems. And the way we solve the normal force in 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 going to look at some wall 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 this centripetal direction like this. So what happens is all of our forces have to cancel. So really our n_y and our mg have to be equal to each other. And remember that n_y is just ncosθ. So remember we came over here because we wanted an expression for n, and now we have it. So now we have this n is equal to mgdivided by cosθ and I'm just going to plug this right back into this equation here. So really, what we end up with is we end up with mg divided by cosθ times the sine of θ equals mv2∕r. We're super close. All we have to do is do some tidying up. So there are two important things that happen here. 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 sine over a cosine. So on the left side, you end up with gtanθ, and this equals v2∕r. And the last thing we do is we move the r to the other side, you know, with grtanθ. So if you take a look at this equation here, this equation is very similar to the equation that we end up with for flat curves, and that's why I've written it that way. So this is going to be your shortcut equation whenever you're given multiple-choice problems and you have banked curves. As long as you memorize this equation, you'll be able to solve the questions in no time. Alright? Alright. So the last thing we have to do is just take the square root and plug in all of our numbers. So we've got the square root of 9.8. We've got the radius, which is 200, and then we've got the tangent of 37 degrees. And what we end up with is a velocity of 38.4m/s. That's our answer. If you go any slower or faster than this, you're going to 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 equations that we've seen for flat curve and b...

8. Centripetal Forces & Gravitation

Banked Curves

8. Centripetal Forces & Gravitation

# Banked Curves - Online Tutor, Practice Problems & Exam Prep

1

concept

### Banked Curve

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#### Video transcript

2

Problem

ProblemA 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

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PRACTICE PROBLEMS AND ACTIVITIES (9)

- A 1125-kg car and a 2250-kg pickup truck approach a curve on a highway that has a radius of 225 m. (a) At what...
- A car drives at a constant speed around a banked circular track with a diameter of 145 m. The motion of the ca...
- (II) How large must the coefficient of static friction be between the tires and the road if a car is to round ...
- An airplane feels a lift force L perpendicular to its wings. In level flight, the lift force points straight u...
- A banked curve of radius R in a new highway is designed so that a car traveling at speed v₀ can negotiate the ...
- (II) A 1250-kg car rounds a curve of radius 68 m banked at an angle of 14°. If the car is traveling at 85 km/h...
- An airplane traveling at 510 km/h needs to reverse its course. The pilot decides to accomplish this by banking...
- A car drives at a constant speed around a banked circular track with a diameter of 145 m. The motion of the ca...
- Consider a train that rounds a curve with a radius of 530 m at a speed of 160 km/h (approximately 100 mi/h ).(...