# Motion of a car on a banked road

1. I read that the maximum possible speed of a car on a banked road is greater than that on a flat road. I know the formula for finding velocity in each case. But I am unable to figure out the reason by comparing the equations.

2. I read about the ideal speed when the frictional force is zero. I cannot understand how can the speed determine the frictional force. Frictional force should only depend on the nature of surfaces in contact. How is it possible that at a certain speed the frictional force is zero?

3. Also I cannot understand that if the velocity is greater than the ideal velocity, then what will be the change in the motion of the car?

• Can you tell us where you read this? Point 2. does not sound right. – Void Apr 8 '19 at 9:11
• I have read it in many books that there is a velocity given by v=[R.g.tan( theta)]^1/2 where frictional force is not needed to provide the necessary centripetal force. – Ashok Sharma Apr 8 '19 at 9:20
• @AshokSharma If we're neglecting friction, then we're not strictly talking about a car on a banked road (which can use the static friction exerted on the road by the wheels to accelerate even around flat, non-banked curves). We're talking instead about a wet bar of soap sliding around a banked bathtub; the intuition from that scenario is far more helpful here (for example, the soap cannot turn on non-banked curves at all due to the lack of friction, and the normal force is the only thing that allows it to change direction). – probably_someone Apr 8 '19 at 9:54
• Perhaps it's best to refer to physics.stackexchange.com/questions/106405/… and then decide if you still have questions left. – Dlamini Apr 9 '19 at 1:17

1. The higher the speed at which a car takes a bend, the greater the centripetal force that is required to keep the car following the curve of the bend. On a flat bend the normal force from the road surface is $$mg$$ and the whole centripetal force is supplied by the friction between the car's tires and the road surface, which is usually assumed to be proportional to the normal force. On a banked bend a component of the normal force acts towards the centre of the curve, and so provides part of the centripetal force.