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Banked turn

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A banked turn is the term used to describe a car riding along a circle with inclined edges. The angle at which a turn is banked refers to the angle of incline of the given path. The benefit of such a structure is that there are forces other than that of friction to keep the car on its designated path. Banked turns also have applications to aviation.

1 Flat surfaces
2 Frictionless banked turns
3 Banked turns with friction
4 Aviation
5 See also
6 Sources

Flat surfaces

To begin our analysis of banked turns we shall first consider the forces acting on a car traveling on a flat circle. In this situation, the force that is keeping the car on its path is that of friction. The goal of our analysis will be to determine a minimum/maximum speed a car can travel to remain on its path. An important note is which way friction points in our analysis. When considering maximum velocity, friction points towards the center of the circle, as the car is being "pushed" outside and friction opposes the direction of motion. For minimum velocity the opposite is true. For the purpose of this example we will only analyze maximum velocity. Because friction in this case acts as the centripetal force, or the force "pulling" the car into the circle, we have the following equation: <math>{mv^2\over r} = \mu mg</math> The left side of the equation is the equation for centripetal force while the right side is friction, or the coefficient of static friction multiplied by the car's normal force. Solving this equation for velocity we get: <math>v = {\sqrt{r\mu g}}</math>

Frictionless banked turns

As opposed to a car riding along a flat circle, inclined edges add an additional force that keeps the car in its path and prevents it from being "dragged into" or "pushed out of" the circle. This force is the horizontal component of the car's normal force. In the absence of friction, the normal force is the only one acting on the car in the direction of the center of the circle. Therefore, we can set centripetal force equal to the horizontal component of the normal force: <math>N\sin \theta ={mv^2\over r}</math> Because there is no motion in the vertical direction, the sum of all vertical forces acting on the system must be zero. Therefore we can set the vertical component of the car's normal force equal to its weight: <math>N\cos \theta = mg</math> Solving the above equation for the normal force and substituting this value into our previous equation, we get: <math>{mv^2\over r}= {mg\tan \theta}</math> Solving for velocity we have: <math>v= {\sqrt{rg\tan \theta}}</math> This provides the velocity that in the absence of friction and with a given angle of incline and radius of curvature, will ensure that the car will remain in its designated path.

Banked turns with friction

When considering the effects of friction on the system, once again we need to note which way the friction force is pointing. When calculating a maximum velocity for our automobile, friction will point down the incline and towards the center of the circle. Therefore we must add the horizontal component of friction to that of the normal force. The sum of these two forces is our new centripetal force: <math>{mv^2\over r}= \mu N\cos \theta +N\sin \theta </math> Once again, there is no motion in the vertical direction, allowing us to set all opposing vertical forces equal to one another. These forces include the vertical component of the normal force pointing upwards and both the car's weight and vertical component of friction pointing downwards: <math>N\cos \theta =\mu N\sin \theta +mg</math> By solving the above equation for mass and substituting this value into our previous equation we get: <math>{v^2\left(N\cos \theta -\mu N\sin \theta \right)\over rg}= \mu N\cos \theta +N\sin \theta </math> Solving for v we get: <math>v= {\sqrt{rg\left(\sin \theta +\mu \cos \theta \right)\over \cos \theta -\mu \sin \theta }}</math> This equation provides the maximum velocity for the automobile with the given angle of incline, coefficient of static friction and radius of curvature. By a similar analysis of mimimum velocity, the following equation is rendered: <math>v= {\sqrt{rg\left(\sin \theta -\mu \cos \theta \right)\over \cos \theta +\mu \sin \theta }}</math> The difference in the latter analysis comes when considering the direction of friction for the mimimum velocity of the automobile (towards the outside of the circle). Consequently opposite operations are performed when inserting friction into equations for centripetal force and vertical forces.

Aviation

The physics of banked turns has applications in aviation as well as automobiles. When completing a turn in the air, a plane is forced to turn at a banked angle. First we shall consider the forces acting on a plane at all times. Its weight points downwards, and counteracting its weight is lift. When the lift force is larger than weight, the plane can achieve flight. Lift is caused when there is an interaction between a solid body and a fluid. The difference in the two medium's velocity generates the upward force. A thrust force points forward in the direction of motion and opposing this is a drag force. Designers of aircraft must consider how to reduce the drag force through wing design etc. For a visual, go to: Forces visual A plane's turning motion can be described in three ways. 'Pitch' describes how much the front of the plane points downwards or upwards. 'Yaw' describes a plane's motion through a horizontal circle, as if it were rotating around a vertical axis through its center. Finally, 'roll' describes a plane rotating around a horizontal axis running from its front to its back. Roll is therefore intrinsically linked with banking. In fact, a plane's banking angle is the complement to its roll angle. When a plane banks at an angle, its lift force can be broken into horizontal and vertical components. Because the horizontal component points towards the center of the circle around which it is turning, this acts as the centripetal force. Thus we can establish the following relationship: <math>{mv^2\over r}= L\sin \theta </math> where L is the lift force and theta is the banking angle. Solving this equation for velocity we get: <math>v= {\sqrt{rg\tan \theta }}</math> Thus the velocity of the aircraft is dependent on the radius of curvature and the angle at which it is banked.