# Circular motion of a particle around a track -- what provides the centripital acceleration?

Vigorous
Suppose a particle is moving around a circular track of radius R at speed v. To bend around a circle some agency has to exert an acceleration towards the center of the circle. I analyze the forces acting on the particle, its weight and the normal force and there is no acceleration in the vertical direction. Why is it the friction of the circular track that provides the centripetal acceleration?

static friction- is whatever force required to keep the object from moving and is always directed opposite to the velocity of the moving particle.
Accordingly, for a particle in circular motion, friction would be directed opposite velocity vector (tangential). I also thought about it in another way, the particle has an inertia to continue its linear motion and not follow a circular path, so to maintain a constant speed, friction acts in the radial direction to keep the velocity magnitude constant along the circle at all times and by this reasoning the speed is constant. But why should friction act this way and not act as I talked earlier opposite to the velocity vector?

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## Answers and Replies

Mentor
static friction- is whatever force required to keep the object from moving and is always directed opposite to the velocity of the moving particle.
Neither of those statements is necessarily true. It can be in any direction on the surface depending on the way they interact, and can exist for a moving object. Consider a car. The tires (usually) interact with the road via static friction, enabling the car to accelerate, decelerate or turn.
Accordingly, for a particle in circular motion...
If static friction applied per your previous statement, you wouldn't be able to extrapolate to motion. E.G., if the object is sliding, then it is dynamic friction that acts on it.
I also thought about it in another way, the particle has an inertia to continue its linear motion and not follow a circular path, so to maintain a constant speed, friction acts in the radial direction to keep the velocity magnitude constant along the circle at all times and by this reasoning the speed is constant.
That doesn't make any sense/contradicts Newton's laws.

Either way, you haven't actually described a real object traveling in a circle, so it isn't really possible to explain how the centripetal force is applied or whether static and/or dynamic friction apply. Please describe a real object traveling in a circle, and we'll analyze the forces.

Vigorous
Neither of those statements is necessarily true. It can be in any direction on the surface depending on the way they interact, and can exist for a moving object. Consider a car. The tires (usually) interact with the road via static friction, enabling the car to accelerate, decelerate or turn.
I extrapolated my stated from an 1 D motion example where one is trying to push an object but is getting nothing in return because friction is exactly balancing my push. So I cannot describe friction unless the mechanism within the object is specified, but why and how does friction cause a car to turn?

That doesn't make any sense/contradicts Newton's laws.
Why?

Mentor
I extrapolated my stated from an 1 D motion example where one is trying to push an object but is getting nothing in return because friction is exactly balancing my push.
In that case you are correct -- but you didn't describe that case in your opening post.
So I cannot describe friction unless the mechanism within the object is specified, but why and how does friction cause a car to turn?
The tires turn/align to establish a circular path and static friction with the road surface pushes the tires/car toward the center of the turn.
Why?
I also thought about it in another way, the particle has an inertia to continue its linear motion and not follow a circular path...
In constant velocity motion, inertia is irrelevant. Objects in constant speed motion stay in constant speed motion per Newton's 1st law.
...so to maintain a constant speed, friction acts in the radial direction to keep the velocity magnitude constant along the circle at all times and by this reasoning the speed is constant.
That is at best oddly put. The reason for the constant speed is no force in the direction of motion, so I guess you are saying force acting in the radial direction isn't acting in the direction of motion so it doesn't affect the speed.

Lnewqban
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static friction- is whatever force required to keep the object from moving and is always directed opposite to the velocity of the moving particle.
It is not. Kinetic friction but not static is always opposite to the velocity. The direction and magnitude of static friction adjusts themselves to provide the observed acceleration. The magnitude does so only up to a limit. For example, when you start walking from standing at rest, the acceleration is forward. The normal force and gravity are vertical, so the only force that can provide the horizontal forward acceleration is a horizontal forward force of static friction. In the case of the turning car, stating friction is towards the center of the circle, what is necessary to provide the observed acceleration. If the road is frictionlessly slick, there would be no observed acceleration and the car would move in a straight line according to Newton's first law.

Physics guy, hutchphd, russ_watters and 1 other person
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... Why is it the friction of the circular track that provides the centripetal acceleration?
...
It is not friction what forces the change of direction, but the geometry of the trajectory followed by the surface and the reactive forces that it induces.
Same for a car negotiating a curve, or a rollercoaster, or a motorcycle inside a spherical cage.

russ_watters
It is not friction what forces the change of direction, but the geometry of the trajectory followed by the surface and the reactive forces that it induces.
And one of these forces can be friction.

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Kinetic friction resists relative motion between the surfaces in contact and acts opposite to the direction of that relative motion.

[Barring unusual situations such as meshing diagonal grooves]

Static friction takes on whatever value it needs in whatever direction it needs to avoid relative motion between the surfaces in contact, so long as the magnitude is within the static limit.

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A.T. and hutchphd
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Both kinetic friction resists relative motion between the surfaces in contact and acts opposite to the direction of that relative motion.

[Barring unusual situations such as meshing diagonal grooves]

Static friction takes on whatever value it needs in whatever direction it needs to avoid relative motion between the surfaces in contact, so long as the magnitude is within the static limit.
I don't see that. When I start from rest and run to the right, my motion relative to the floor is to the right. The accelerating force of static friction is also to the right. When I plant my feet on the floor and slide across it until I stop, my velocity relative to the floor is still to the right but the force of kinetic friction is to the left.

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I don't see that. When I start from rest and run to the right, my motion relative to the floor is to the right. The accelerating force of static friction is also to the right. When I plant my feet on the floor and slide across it until I stop, my velocity relative to the floor is still to the right but the force of kinetic friction is to the left.
I think you caught me mid-edit. I distinguished static and kinetic after that.

It is the motion at the contact patch that counts. Not the jiggly motion of all that stuff above the contact patch.

The floor pushes your foot forward because your body is pushing your foot back. Newton's second law says that the required force from floor on foot to keep foot stationary relative to the floor is then forward. Static friction.

Alternately, if the floor is not as sticky as you had expected, the floor pushes your foot forward because your body is pushing it back, causing a relative rearward slipping motion of foot across floor and a corresponding forward push of floor on foot. Kinetic friction. Sadly inadequate to prevent a nose-floor interaction.

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I think you caught me mid-edit. I distinguished static and kinetic after that.
I probably did.

Static friction takes on whatever value it needs in whatever direction it needs to avoid relative motion between the surfaces in contact, so long as the magnitude is within the static limit.
This works for me, but I find it easier to advise "when static friction is the net force, it must be in the direction of the acceleration."

This works for me, but I find it easier to advise "when static friction is the net force, it must be in the direction of the acceleration."
That's true for any force, so it doesn't tell us anything specific about static friction.

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This works for me, but I find it easier to advise "when static friction is the net force, it must be in the direction of the acceleration."
If you have tie downs on the load in the bed of your pickup truck then that rule does not tell you much about the direction of the force of static friction.

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If you have tie downs on the load in the bed of your pickup truck then that rule does not tell you much about the direction of the force of static friction.
As long as the truck and the load move as one, it tells me that the net force of static friction on the load is equal to the mass of the load multiplied by the acceleration of the truck and points in the direction of the truck's acceleration. For example, if the truck is moving in reverse while the brake is applied, it tells me that the acceleration of the load is in the forward direction.

To explain it your way, I would have to say something like: If static friction were not there, when the truck is backing while braking, the load will tend to move backward per Newton's first. Therefore static friction must be in the forward direction to prevent it from doing that. However, if the relative motion is prevented from occurring, this means that it is zero, i.e. we are looking at the system in the non-inertial frame of the load in which case an extra step is required to argue that the net force on the load in the inertial (lab) frame is ma. Sure.

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As long as the truck and the load move as one, it tells me that the net force of static friction on the load is equal to the mass of the load multiplied by the acceleration of the truck and points in the direction of the truck's acceleration.
No. It does not. The tie downs exert force as well.

As long as the truck and the load move as one, it tells me that the net force of static friction on the load is equal to the mass of the load multiplied by the acceleration of the truck and points in the direction of the truck's acceleration.
You are assuming that static friction is the net force (along a certain axis), which is a trivial special case that doesn't require much explaining.

The general case is that static friction is one of several forces acting. To find static friction here you have to apply Newton's Laws and solve for the force that produces a motion with no relative motion at the contact patches. This is the same as finding the normal contact force for a general case.

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No. It does not. The tie downs exert force as well.
You are assuming that static friction is the net force (along a certain axis), which is a trivial special case that doesn't require much explaining.
Yes, if static friction is not the only force contributing to the acceleration, then of course the other forces need to be taken into account. In that case one knows that static friction must be parallel to the surface but the direction may not be obvious. One has to draw the FBD, assume a direction for the static friction, write Newton's 2nd and solve for the magnitude fs. If the magnitude turns out to be negative, the arrow representing fs in the FBD must be flipped. That's standard stuff and I think we all agree. It's just that we have different preferred methods for finding the direction of static friction.

Here is a problem that I used to assign when I taught intro mechanics. A yo-yo is pulled by two strings exerting forces F1 and F2 as shown in the figure. The yo-yo is rolling without slipping at constant speed to the right. Find the direction and magnitude of the force of static friction given that F1 = 30 N.

10 N to the right.

It is not clear to me that the direction of static friction in this problem can be deduced a priori without at least doing some back of the envelope scribbling. Going the FBD route is an application of the principle that the force of static friction is "whatever is necessary to provide the observed acceleration but up to a certain point". It never fails and is easy to explain. That's my preference.

Physics guy
Suppose a particle is moving around a circular track of radius R at speed v. To bend around a circle some agency has to exert an acceleration towards the center of the circle. I analyze the forces acting on the particle, its weight and the normal force and there is no acceleration in the vertical direction. Why is it the friction of the circular track that provides the centripetal acceleration?

static friction- is whatever force required to keep the object from moving and is always directed opposite to the velocity of the moving particle.
Accordingly, for a particle in circular motion, friction would be directed opposite velocity vector (tangential). I also thought about it in another way, the particle has an inertia to continue its linear motion and not follow a circular path, so to maintain a constant speed, friction acts in the radial direction to keep the velocity magnitude constant along the circle at all times and by this reasoning the speed is constant. But why should friction act this way and not act as I talked earlier opposite to the velocity vector?
A common question that confused even me when I first learned about centripetal acceleration while moving in a circular track. The answer to this question is simple because you seemed to be confused with why friction does not act in the direction of tangential velocity vector in circular motion along a circular track. The answer is yes, it does act in the direction of tangential velocity but we are not concerned about this kinetic friction because this friction will be in the direction opposite to the velocity vector and does not contribute to the centripetal acceleration required to keep the object in circular motion. This can understood by taking the parallel component of kinetic friction along the radius of circle which will be zero. So as you have mentioned we can intuitively think of static friction (not kinetic friction which acts opposite to the direction of tangential velocity) as a force which prevents the object from going in a linear path and 'forcing' it to move in a circular path.

...we are not concerned about this kinetic friction because this friction will be in the direction opposite to the velocity vector ...
If there is no slippage there is no kinetic friction (just rolling resistance). So even less reason to be concerned with kinetic friction.

Physics guy
If there is no slippage there is no kinetic friction (just rolling resistance). So even less reason to be concerned with kinetic friction.
But that is only true in an ideal case. In reality there will always be some amount of friction while moving in circular track since pure rolling can't be expected.

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But that is only true in an ideal case. In reality there will always be some amount of friction while moving in circular track since pure rolling can't be expected.
Lots of friction if centripetal force is to be provided. Little frictional loss of energy if slipping is minimal.

Going the FBD route is an application of the principle that the force of static friction is "whatever is necessary to provide the observed acceleration but up to a certain point".
And if the motion is not known, you have to calculate the motion assuming no slippage and the required static friction for it, and check if it can be achieved.

Lots of friction if centripetal force is to be provided. Little frictional loss of energy if slipping is minimal.
For completeness: Kinetic friction can provide a part of the centripetal force if the vehicle is sliding/drifting through a turn.

The key here is, as noted earlier by @jbriggs444, that kinetic friction opposes the relative motion of the surfaces in contact, which can be different (even opposite) from the motion of the vehicle relative to the ground. During hard acceleration with slipping wheels kinetic friction propels the vehicle forward.

jbriggs444
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And if the motion is not known, you have to calculate the motion assuming no slippage and the required static friction for it, and check if it can be achieved.
Yes, but only if both μs and μ[SUBk[/SUB] are mentioned explicitly in the statement of the problem. One needs the former for the reasons you mentioned. If there is no slippage all is good; if there is slippage, then one needs μ[SUBk[/SUB] to find the acceleration. The assumption is that if μs is not mentioned in problems involving static friction, there is no slippage anywhere.

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The key here is, as noted earlier by @jbriggs444, that kinetic friction opposes the relative motion of the surfaces in contact, which can be different (even opposite) from the motion of the vehicle relative to the ground. During hard acceleration with slipping wheels kinetic friction propels the vehicle forward.
Yes, the opposition to relative motion is key. It can be carried over to EM induction and eddy current where terms like "magnetic friction" and "magnetic braking" are used. An eye-popping demonstration is a rolling ceramic disk magnet on an aluminum incline. It will slow down relative to a non-magnetic disk as expected. However, if you angle it towards the edge before you release it, it will not fall off but will twist away from the edge and stay on the incline following an S-shaped path from one edge to the other all the way down.

Vigorous
The tires turn/align to establish a circular path and static friction with the road surface pushes the tires/car toward the center of the turn.
When I turn the wheels, the plane in which the wheels are rotating is displaced by some angle equal to part of the circumference of the circular track traversed). Friction opposes the relative motion of the two planes. I can't grasp how from this friction will act radially inwards?

Homework Helper
When I turn the wheels, the plane in which the wheels are rotating is displaced by some angle equal to part of the circumference of the circular track traversed). Friction opposes the relative motion of the two planes. I can't grasp how from this friction will act radially inwards?
The friction resists the relative motion of the mating surfaces. Those surfaces are physical, not virtual. The plane in which the wheels are rotating is virtual, not physical.

russ_watters
Vigorous
I don't see that. When I start from rest and run to the right, my motion relative to the floor is to the right. The accelerating force of static friction is also to the right. When I plant my feet on the floor and slide across it until I stop, my velocity relative to the floor is still to the right but the force of kinetic friction is to the left.
Picture a runner in the right direction, let the runner plant his left foot on the floor, he is exerting a force down and backward on the floor, the left leg stays in contact with the floor for a few milliseconds, and the right leg is in mid air, after those milliseconds, the force exerted exceeds the maximum limit of static friction and slippage occurs. By Newtons third law, the ground exerts an equal and opposite force on the runner but do we identify this force as static or kinetic friction

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Friction opposes the relative motion of the two planes.
Wrong. It opposes the relative motion of the contact surfaces. This has been explained several times already.

Looking only at the plane of the wheel ignores the rotation of the wheel. It would only make sense if the wheel was locked.

Picture a runner in the right direction, let the runner plant his left foot on the floor, he is exerting a force down and backward on the floor, the left leg stays in contact with the floor for a few milliseconds, and the right leg is in mid air, after those milliseconds, the force exerted exceeds the maximum limit of static friction and slippage occurs. By Newtons third law, the ground exerts an equal and opposite force on the runner but do we identify this force as static or kinetic friction
Before slippage occurs, static friction. After slippage occurs, kinetic friction.

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Nugatory and russ_watters
I can't grasp how from this friction will act radially inwards?
What static friction opposes is the relative motion at contact that would occur if there was no friction (marked red in the diagram).

How the particle moving around a circular track described in post #1 ended up having wheels and a steering mechanism?

See post #3:
but why and how does friction cause a car to turn?

Lnewqban
Physics guy
Picture a runner in the right direction, let the runner plant his left foot on the floor, he is exerting a force down and backward on the floor, the left leg stays in contact with the floor for a few milliseconds, and the right leg is in mid air, after those milliseconds, the force exerted exceeds the maximum limit of static friction and slippage occurs. By Newtons third law, the ground exerts an equal and opposite force on the runner but do we identify this force as static or kinetic friction
You seem to be confused that friction is a reaction force to the force you apply to move any object. In fact it is not. It is a force that occurs due to electromagnetic interaction between the 2 surfaces in relative motion. In fact if you were to take an object to outer space and push it, it is not friction that is reacting to your push (because action reaction pair always act on different bodies) it the push of the object on you. You can see that you will be pushed backwards when you try to push the object forward in space. This can be recreated by standing on a roller skate and trying to push the wall, you seem to go backward even though you push the wall forward, this is the reaction force that the wall exerts on you when you exert a force on the wall.

Vigorous
Vigorous
As I turn the wheels to the left the wheels are pushing to the right against the floor. Friction allows the floor to push back against the wheels allowing them to turn left. If there was no friction then the object would fly off the circular trajectory and continue in the direction of the tangential velocity.