Type of friction on circular road

Question

Why is the friction static while driving a car (at constant speed) on a horizontal circular road? Please explain in details.

Answer 1

Suppose you have a mass connected to the end of a string. You swing the mass in a circular horizontal path at constant speed. You feel a force pulling on your hand. Per Newton's third law that force is equal and opposite to the force that the string is applying to the mass. That force is called the centripetal (center seeking) force acting on the mass and is responsible for the circular motion of the mass. It causes the mass to have a constant acceleration towards the center. If the string was cut, the mass would fly off in a straight line, per Newton's first law.

Now let the mass be the car moving at constant speed in a circular path on a level road. There is no string connecting the car to the center of the circular path. So what force is responsible for keeping the car in the circular path? What is the source of the necessary centripetal force? It is the force of friction between the tires and the road.

I needed where is the car trying to move radially outwards since friction opposes the relative motion,please help me with this

The car is not trying to move radially outward. It tries to move in a straight line. Before the car attempts to turn it is moving in a straight line at constant speed, i.e., constant velocity. It will continue to move in a straight line unless acted upon by a net force. Per Newton's first law: an object at rest stays at rest and an object in motion stays in motion with the same speed and same direction unless acted on by a net force.

If there were no friction between the tires and the road, turning the steering will will not result in the car turning (the tires will slide on the road). In order for the car to be able to turn some force is needed directed radially inward. The only available force is the static friction force directed radially inward.

The centripetal force is not opposing any radial outward force. The radially outward force the appears to occur in the non-inertial frame of the vehicle is a pseudo or fictitious force called the centrifugal force. In the inertial frame of the road the only force acting on the car making it turn is the centripetal force, the static friction force between the tires and the road.

Hope this helps.

Answer 2

The car takes the corner because its wheels are turned (by the steering mechanism) at an angle to its direction of motion. There is still rolling resistance that is trying to slow the wheels down, but this is overcome by the car's engine which exerts a rotational torque on the wheels via the gearbox and transmission (just as it does when the car is going in a straight line). But the inertia of the car is now trying to move the wheels sideways as well, and this gives rise to an opposing sideways friction force (parallel to the axis between the wheels) which is much greater in magnitude than the rolling resistance. It is this sideways friction that provides the centripetal force that changes the car's direction of motion. Of course, if the car's speed is too high, the sideways friction will not be sufficient to turn the car and it will skid off the road.

Answer 3

I completely agree with the wonderful answer of @Bob D sir.

However I would like to add a few things to help your understanding even better.

First of all,what a lot of people seem to believe is that friction always opposes motion.This is not true.Friction may also help in motion. Eg We can walk only because friction pushes us forward when we push our feet back against the ground.

Friction is helping the motion here in circular motion,too.

Secondly ,as @Bob D sir pointed out, we need a centripetal force here and the only possible source is the ground reaction to the tyres. This ground reaction is by definition called as friction force. Notice that I am laying stress on 'by definition'.

P.S. The horizontal component of ground reaction is called friction and the vertical component is called normal force.