If a car is moving on a banked frictionless road one of the component of normal reaction force acts as the centripetal force required for the turn. But normal reaction force is a reaction force. Technically seeing, the the component of normal reaction force should be the result of the centripetal force required for the turn. And turning needs some friction. So how can the road be frictionless? Please explain.
[Physics] Banking of road
centripetal-forceforcesfree-body-diagramnewtonian-mechanicsvectors
Related Solutions
(1) is correct, but (2) and (3) are not. Consider a car turning counterclockwise around a track that is banked inwards. There are three forces acting on the vehicle: the weight of the car $\vec{W}$, the normal force $\vec{N}$, and the force of friction $\vec{f}$. Breaking these up into Cartesian coordinates gives:
$$\vec{W} = -mg\hat{y} \\ \vec{N} = N \left(\cos\phi \hat{y} - \sin\phi \hat{x}\right) \\ \vec{f} = f \left(-\sin\phi \hat{y} - \cos\phi \hat{x}\right)$$
Getting the direction correct for the friction is a bit subtle. We know from the $\phi \rightarrow 0$ limit that the friction has to point roughly in the direction of the turn, and also be perpendicular to the normal force, so for these coordinates, it must point down and to the left.
Since the car is moving in a circle in the horizontal plane, the forces in the $\hat{y}$ direction must cancel. Setting these equal to each other and considering the limiting case $f = \mu_s N$ gives an equation that can be solved for $N$. Then you can sum up the horizontal forces an insert this value to get the centripitel force.
There are some good reasons why you should not take a sharp turn at high speeds.
1) On a flat road, the force of static friction is what provides the centripetal force to accelerate you through a curve. Unfortunately, there is a maximum value for static friction that depends greatly on the mass of the vehicle. The heavier it is, the more static friction you can use to get you through a turn. The centripetal force required to get you through the turn is proportional to the inverse of the radius of the turn and the velocity of the vehicle squared. This means that sharp turns at high speeds require very large centripetal forces to complete. If you are in a light vehicle, the force required to make a high-speed turn may be more than static friction can provide. The result, in this case, is that the wheels will slip and slide outward from the turn and you'll quickly find out how your vehicle handles off-road (if nothing worse). Even a heavy vehicle shouldn't be used for this because the force required scales proportional to the mass of the vehicle as does the maximum static friction, which means if the light vehicle would skid, adding weight to it changes nothing.
2) The static friction available to you also depends on the contact area of the wheels (and the material, etc). Having more wheels means more static friction. Freight trucks usually have many large wheels to provide extra contact area and a higher maximum static friction. However, the mass distribution can play an important role. The mass distribution of the vehicle determines where the center of mass is. If you take a sharp turn at high speeds with enough friction to get through the turn but your center of mass is not very close to the ground, then you could have problems. The friction forces that make you turn act on the bottom of the wheels. If your center of mass is above this, that makes these forces produce a net torque on the vehicle. The higher the center of mass, the more torque these forces apply to the vehicle. Since the vehicle is free to rotate (or pivot) around the point where the outside wheels meet the road, once the torque from friction overcomes the torque due to gravity on the inside half of the vehicle, it's going to rotate. The result is the vehicle rolls onto its side (and possibly keeps rolling). And remember, the centripetal force is proportional to the square of the velocity and the inverse of the radius of the turn. So the faster it is and sharper the turn, the easier it is to roll the vehicle.
Best Answer
Roads are banked so that we get a component of normal reaction to assist in the turn if friction is not sufficient. For example on very sharp curves on road, in absence of banking, the tires can skid and the results can be deadly. Now if friction is absent, the only force which can help us in turning is the normal reaction. But normal reaction is perpendicular to the road and we want a component of it in the radial direction. So we bank the roads to provide us with the necessary centripetal acceleration.
And normal reaction is a reaction by surface of the road to the weight of the car as well as the centripetal acceleration needed to keep the car in circular motion.