Gravity and Motion
Physics (Year 12)
A banked track is a track which is on an incline. They have an advantage over horizontal tracks because objects travelling on a banked track do not depend on the sideways frictional force needed to travel in circular motion.
Imagine a car going around a circular, horizontal track in uniform circular motion. Looking at the vertical axis, the force of gravity pointing down is balanced by the normal force pointing up. Looking at the horizontal axis, the only horizontal force is the sideways force of friction which the road exerts on the car’s tyres and is pointing towards the centre of the circle. Hence, this is the net force (i.e. centripetal force) and is what causes the car to accelerate towards the centre of the circle.
Thus, if the track was frictionless (eg. ice), the car would not be able to move in circular motion and instead move in a straight line.
This is where banked tracks have an advantage. Since the track is banked, the force of gravity and the normal force are unbalanced. The normal force has a larger magnitude and is pointing in a different direction to the force of gravity. When you add up the 2 vectors, we end up with a vector pointing towards the centre of the circle. This vector is the net force (i.e. centripetal force) and is simply the horizontal component of the normal force and is what causes the car to accelerate towards the centre of the circle. Therefore, a banked track would allow a car to travel in circular motion even if the path is frictionless.
Using the triangle formed by the vector addition of the force of gravity and the normal force, we can deduce that:
This is the equation that is commonly used for banked track questions and it can only be used in scenarios where the force of friction is not present. This is because if the force of friction is present, the triangle we used to derive the equation becomes invalid. The reason to that is there will be 2 vectors pointing towards the centre of the circle; the horizontal component of the frictional force and the horizontal component of the normal force. Hence the equation is not useful in this scenario.
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