Banked With No Friction
At the Indianapolis Motor Speedway there are 4 turns that have the tracked banked at nine and twelve degrees. Compared to a flat track, inclined edges add an additional force that keeps the cars on their path. This force prevents the cars from being pulled towards the center or pushed out.
Why It Matters
 When a car is driving on a flat surface in a straight line, the following 4 forces are acting on it:
 the weight of the car
 the normal force
 the force applied by the tires
 the frictional force

 These 4 forces keep the car moving in a straight line. There are no forces that are perpendicular to the other two forces that would allow the car to turn in a circular path.
 If a car is on a banked turn, centripetal force is an additional force on the car. Usually the normal force just has a vertical component, but in the case of a banked curve it has both a vertical and a horizontal component. In the case where friction exists, it must be added to the horizontal normal component since the frictional force points towards the center of the arc.
 Watch what happens when friction does not exist:
http://www.youtube.com/watch?v=ZwqLLpNuAQU
What Do You Think?
Using the information provided above, answer the following questions.
 The equation for the maximum velocity of a car on a banked curve is given as: \begin{align*}v=\sqrt{\frac{rg(\tan \theta  \mu_{s})}{1+\mu_{s} \tan \theta}}\end{align*}. What does this equation simplify to when there is no banked corner?
 Why does the normal force have a horizontal component when the car is on a banked curve?
 What does the equation for the maximum velocity of a car on a banked curve simplify to when there is no friction?