Frictional Force in Circular Motion

In summary: The force of gravity applies a force on the weight that is directed towards the center of the circle, and that is why objects tend to move in a circle around the center of the Earth.
  • #1
rahil27
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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
 
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  • #2
rahil27 said:
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
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
Welcome to PF, rahil27!

rahil27 said:
I did read some forums where it was explained that frictional force provides the centripetal force and therefore it is directed towards the centre
That sounds right. The friction is providing an opposite force to counter the centrifugal force from a stationary observers perspective.

but that does not explain how does it act towards the centre and not in any other direction?
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
Chestermiller said:
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
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
A body continues in a straight line unless acted upon by an external force.
 
  • #6
rahil27 said:
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?
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
Nugatory said:
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.
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
rahil27 said:
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?
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.
 
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  • #9
rahil27 said:
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?
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
 
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  • #10
rahil27 said:
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?
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.
 
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  • #11
A.T. said:
It's not always exactly towards the center, if the wheels are used for propulsion/braking.
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
Chestermiller said:
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
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
TumblingDice said:
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.

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
rahil27 said:
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?
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
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
Chestermiller said:
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

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
rahil27 said:
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?
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:
From the viewpoint of an inertial reference frame stationary with respect to the road, the car is accelerating toward the center of the circle. This acceleration is necessary because the direction of the velocity is changing, despite a constant speed. This inward acceleration is called centripetal acceleration and requires a centripetal force to maintain the circular motion. This force is exerted by the ground upon the wheels, in this case thanks to the friction between the wheels and the road.

As opposed to the non-IRF:
From the viewpoint of a rotating frame, moving with the car, there is a fictitious centrifugal force that tends to push the car toward the outside of the road (and to push the occupants toward the outside of the car). The centrifugal force balances the friction between wheels and road, making the car stationary in this non-inertial frame.
 
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  • #18
rahil27 said:
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?
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
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
rahil27 said:
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
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
 
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  • #21
rahil27 said:
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

@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
TumblingDice said:
@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!
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
Chestermiller said:
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...
O.K... hmmm o_O 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
TumblingDice said:
O.K... hmmm o_O 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?
Sorry, I don't understand this question.

Chet
 
  • #25
TumblingDice said:
O.K... hmmm o_O So, does this mean the frictional force is slightly askew from normal centripetal for the COM (?), and the COM information in the quote was meant to confirm the OP's understanding of this?

Not in the hypothetical ideal car with a rigid suspension and precisely neutral suspension alignments, traveling 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.
 
  • #26
Nugatory said:
Not in the hypothetical ideal car with a rigid suspension and precisely neutral suspension alignments, traveling in a perfect circle with no air resistance or rolling resistance in undriven wheels...
Thank you! I like to separate the difference between correct answers on undergrad physics exams, from in-depth analysis that would appear in someone's thesis. :p
 
  • #27
Chestermiller said:
Sorry, I don't understand this question.
I'll give it another go. Does this mean that the friction force is not precisely in the direction of the circle's center? You're indicating it is askew (not parallel) with the vector of the centripital force acting on the COM.

EDIT: Or the friction force direction is different because the tires are simply reacting to their own location and COM on their own circle rather than the COM of the system?
 
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  • #28
TumblingDice said:
I'll give it another go. Does this mean that the friction force is not precisely in the direction of the circle's center? You're indicating it is askew (not parallel) with the vector of the centripital force acting on the COM.

EDIT: Or the friction force direction is different because the tires are simply reacting to their own location and COM on their own circle rather than the COM of the system?
When you turn the tire, it now points in a different direction than the direction that the rest of the car is moving. This causes a sideways frictional component of force on the tire patch. This sideways component of force has a component normal to the direction of the rest of the car, and is transmitted to the rest of the car through the suspension. This provides the centripetal force to get the car body to turn.

Chet
 
  • #29
I think most of the posts above are a bit off topic. The second one by Tumbling Dice is on topic. I think it is misleading to get into circular motion, centripetal forces and all that. The issue is in the mechanics of wheels.

A wheel is generally idealized as light, compared to a car, and rigid, and with point contact, and with bearings with negligible friction. Thus the force on the tire from the road can have a component towards the center of the wheel, a component along the axle, and only a small component (related to
the relative inertia of the wheel relative to the car) in the car direction. Ignoring the support force (towards the axle) the only possible significant friction force is parallel to the axle.

This is all consistent of the job we all expect wheels to perform: they allow easy motion in the direction of travel and the keep the car on the road
(only moving in the direction the car is oriented).

All of what I wrote above doesn't depend on discussion of circular motion, of centripetal forces or of the laws of sliding or static friction.

And all of it is a little be wrong because: wheels are not quite rigid, bearings have a little friction, and the sheel-road contact is not at a point.
Some of this is discussed, at least a bit, in my book (pdf here: google ruina text, page 175).
 
  • #30
Hi Andy. Welcome to Physics Forums!

I looked over your reply and also the section in your book that you referred to. It all seems to refer strictly to straight-ahead motion. But the OP was asking about cornering forces and kinematics. Other posts that have appeared in Physics Forums have addressed frictional forces in straight-ahead motion. However, I like the discussion in your book quite a bit. In fact, the entire book looks great, and I'm bookmarking it. Very nicely done.

Chet
 

1. What is frictional force in circular motion?

Frictional force in circular motion is the force that acts between two surfaces in contact, in a direction opposite to the motion. It is caused by the roughness of the surfaces and the interlocking of their microscopic irregularities.

2. How does frictional force affect circular motion?

Frictional force can slow down or stop circular motion by opposing the motion and causing energy loss. It can also change the direction of the motion by providing a centripetal force towards the center of the circle.

3. What factors affect the magnitude of frictional force in circular motion?

The magnitude of frictional force in circular motion depends on the coefficient of friction between the two surfaces, the normal force, and the speed of the object. Additionally, the roughness and material of the surfaces can also affect the frictional force.

4. How can frictional force be reduced in circular motion?

Frictional force can be reduced by using smoother surfaces, reducing the normal force, and decreasing the speed of the object. Lubricants can also be used to reduce frictional force by creating a layer between the surfaces.

5. Can frictional force be beneficial in circular motion?

Yes, frictional force can be beneficial in circular motion by providing the necessary centripetal force to keep an object in circular motion. It can also be used in braking systems to slow down or stop circular motion in a controlled manner.

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