# Frictional Force in Circular Motion

1. Sep 27, 2014

### rahil27

Hi Guys,

I am trying to understand that how does the frictional force on a car moving in a circle is directed towards the centre? I do understand that kinetic friction acts tangentially to the motion and opposite to the velocity vector but why does static friction act towards the centre? I did read some forums where it was explained that frictional force provides the centripetal force and therefore it is directed towards the centre but that does not explain how does it act towards the centre and not in any other direction? Any help would be greatly appreciated

2. Sep 27, 2014

### Staff: Mentor

In static friction, there is no relative movement between the surfaces. Within the contact patch of a car tire, the rubber is not slipping relative to the road surface (unless the car is in a skid). In cornering, the road exerts a frictional force on the rubber passing through the contact patch. This force is directed inward toward the center of the turn, and provides the centripetal force needed to accelerate the car radially. The force is transmitted from the contact patch to the rest of the car by the entire tire deforming. As long as the frictional force on the contact patch divided by the normal force (which is on the order of the tire pressure times the contact patch area) does not exceed the coefficient of static friction between the rubber and the road, the rubber will not slip relative to the road, and the car will not skid.

Chet

3. Sep 27, 2014

### TumblingDice

Welcome to PF, rahil27!

That sounds right. The friction is providing an opposite force to counter the centrifugal force from a stationary observers perspective.

Well, that would contradict the uniform circular motion. If the friction was applying a force in other directions, there would have to be another force countering it to maintain the circular motion. (Like rockets aimed sideways on the car...)

4. Sep 27, 2014

### rahil27

That does make some sort of sense on how the frictional force is transmitted from the road to the tire via the stationary contact patch but I still don't understand why is it towards the centre of the circle?

5. Sep 27, 2014

### dean barry

A body continues in a straight line unless acted upon by an external force.

6. Sep 27, 2014

### Staff: Mentor

Are you familiar with the somewhat easier to visualize question of what happens when you tie a weight to a string and swing it in a circle? The weight wants to move in a straight line tangent to the circle; the string pulls it off that straight line onto the circular path. What direction is the force of the string acting? Clearly, towards the center of the circle because that's the direction the string is lined up on.

The turning car is the same general idea except that the force comes from the friction between tires and ground - if the car is moving in a perfect circle at a constant speed the force on it has to be directed towards the center of the circle.

7. Sep 27, 2014

### rahil27

But how does the contact between the tire and the ground create a frictional force which is always towards the centre of the circle and not any other direction?

8. Sep 27, 2014

### TumblingDice

That is the direction it's resisting against. If you were in the car you'd feel the centrifugal force acting on you towards the outside of the circle. The tires are experiencing that same force in the same direction. Inside the car you have the door to stop you, as it applys an equal force against you (towards the center of the circle). The tires have the road to "grab onto", and the friction is in the same - equal and opposite to centrifugal, towards the center of the circle.

9. Sep 27, 2014

### Staff: Mentor

Well, if the tires were rolling directly forward, there would be no force to make the car go in a circle. So it must have something to do with the fact that, when you turn the steering wheel, the front tires reorient and roll in a direction different from the direction of the center of mass of the car. Can you figure out how this works?

Chet

10. Sep 27, 2014

### A.T.

It's not always exactly towards the center, if the wheels are used for propulsion/braking. Even the rolling resistance of freewheeling wheels requires a tangential component.

It's mostly towards the center on freewheeling wheels, because of the way the wheels are aligned. Any component in a different direction just spins the wheels without much resistance. Only perpendicular to the track the wheels can have a significant resistance.

11. Sep 27, 2014

### TumblingDice

That's a good point, A.T. And good for rahil for beinging persistent. The car couldn't keep going in a circle without being propelled. That friction will only apply to the tire(s) the drive train is engaged to.

12. Sep 27, 2014

### rahil27

So it seems like when you turn the steering wheel, the tire(s) point in the tangential direction while the centre of mass will be in a direction somewhere between the tangential and the normal. Is that correct?

13. Sep 27, 2014

### rahil27

This does make sense from the reference frame of the observer inside the car but what about the observer in an inertial frame? Instead of centrifugal force force will inertia be acting away from the circle and friction will try to oppose the normal component of the inertia?

14. Sep 27, 2014

### Staff: Mentor

Not exactly. The center of mass travels in the tangential direction, while the tires point in a direction somewhere between the tangential and the inward normal.

Chet

15. Sep 27, 2014

### rahil27

So what I understand now is that when turning, the tires point in a tangential direction to the circular motion while the centre of mass or inertia will be in a direction somewhere between tangential and normal. The tangential direction of inertia will help the car move in the circle while the normal component will acts in a direction radially outwards which will be opposed by the frictional force between the tires and the ground and therefore it will act towards the centre. Am I in the right direction?

16. Sep 27, 2014

### rahil27

But if the tires point in a direction somewhere between the tangential and the inward normal and the frictional force will acts perpendicular to tires, so then wouldn't the centre of circular motion of tires and car be different?

17. Sep 27, 2014

### TumblingDice

The names of the forces and what they are attributed to are different between an IRF on the road and the non-IRF of the car. Specifically, the non-IRF will introduce ficticious forces (terminology).
Per wiki:
As opposed to the non-IRF:

18. Sep 27, 2014

### Staff: Mentor

Yes, a little bit. Think of riding a bicycle, with the front tire pointing in a different direction than the body of the bike. But, everything straightens out when you complete the turn, and reorient the car wheels in line with the car.

Chet

19. Sep 27, 2014

### rahil27

So the frictional force acts towards the centre of the circle formed by the wheels and compared to this circle the centre of mass will be in a direction between tangential component(tangent to the tire) and outward normal

20. Sep 27, 2014

### Staff: Mentor

Yes. This sounds correct.
If you are interested in this kind of thing, there are many other aspects of the geometry of the steering/suspension system that you may find fascinating: caster, camber, toe in. A good book that explains all this with simple diagrams is Stokel, Auto Mechanics Fundamentals.

Chet

21. Sep 27, 2014

### TumblingDice

@Chestermiller - can you help me to understand the part I underlined? My understanding was that Uniform Circular Motion equations explain the COM movement, but I thought that just meant how the forces are explained to act on the mass (based on its COM). IOW, the COM never changes in the non-inertial frame and only moves in a circle from an inertial observer's perspective in the rest frame of the circle. The underlined part of the quote make it sound as though the COM is being described as between a parallel to the circle tangent and a perpendicular to the circle tangent, and all my brain sees is a 45 degree line between tangent and perpendicular, not a location of COM. Can you untangle my thought process and understanding? Thank you!

22. Sep 27, 2014

### Staff: Mentor

The path that the contact patch of the front tire follows is different from the path that the center of mass of the car follows. The car COM lags behind the contact patch motion, and they only realign after the turn has been completed. The force of ground on the front tire contact patch produces a moment on the car with respect to its center of mass, and causes the car to rotate. After a 90 degree turn, the entire car has rotated 90 degrees.

Chet

23. Sep 27, 2014

### TumblingDice

O.K... hmmm So, does this mean the frictional force is slightly askew from normal centripital for the COM (?), and the COM information in the quote was meant to confirm the OP's understanding of this?

24. Sep 28, 2014

### Staff: Mentor

Sorry, I don't understand this question.

Chet

25. Sep 28, 2014

### Staff: Mentor

Not in the hypothetical ideal car with a rigid suspension and precisely neutral suspension alignments, travelling in a perfect circle with no air resistance or rolling resistance in undriven wheels - the vector sum of the frictional forces through all four tires must be centripetal and pass directly through the COM so that there is no rotational moment.

Real cars are waaayy more complicated: the slightest change transfers kinetic energy between the potential energy in the flexed tire sidewalls and suspension; the suspension geometry changes; the loading on each tire and hence the frictional force at that tire changes. Anyone who has ever driven a '70s-vintage 911 Porsche (the same effect can be seen, less strikingly, in many mid-engine RWD cars of the era) can attest to the way that you can steer the rear end with with just the throttle pedal.