Car turning and centripetal force

[aspodkfpo]

Homework Statement

Conceptual qualms.

Relevant Equations

No equations.

1. When a car turns there is a centripetal force towards the centre. This centripetal force is labelled as a static frictional force. I don't understand where this static frictional force arises from. Friction is meant to oppose motion, but I don't see the motion that is parallel to the friction in this case. The wheels of a car turn, but if the right side turns outwards, the left side turns inwards and that motion cancels out.

2. Also, I was wondering what would happen if the car's velocity increases in magnitude as it turns.

3. In what case, would the centripetal force become kinetic friction? And if it was kinetic friction, how would the path deviate from being circular?

 

[haruspex]

Friction is meant to oppose motion

Friction opposes the tendency to relative motion of the two surfaces.
For kinetic friction it acts oppositely to the actual relative motion.
For static friction it is the minimum magnitude force to prevent relative motion.

Without friction, what would happen as the car corners?
In what direction would be the minimum magnitude force to prevent that?

 

[Cutter Ketch]

1. Yes, I can see where it isn’t obvious. Forget about turning for a second. When the car is traveling straight the force of friction acts at the bottom of the tire in the direction directly opposed to the motion, that is to say backwards. However, that does not stop the car. Due to the bearings and the axle the tires are free to spin in that direction. Now take the same case: the car traveling straight, and turn the wheels slightly. The force of friction still acts in the same direction: toward the rear of the car. But this time the wheels are not free to rotate in that direction. Instead you must break the force into components. The largest component is in the plane of the wheel and continues to turn the wheel as before. It’s that smaller component which is perpendicular to the wheel which turns the car. Also, in rotating the car, the front wheels drive the rear wheels into a similar situation where the instantaneous direction of motion is not aligned with the plane of rotation of the tire and an inward component of force results.

 

[Cutter Ketch]

2. Without slipping, the angle of the tires defines the radius of the turning circle. If you accelerate, you will continue to turn along the same circle. This means the centripetal force must increase as $\frac {v^2} r$. “Static” friction is a force of constraint. It provides whatever force is necessary to prevent slipping. However, there is a maximum force that static friction can provide. If you accelerate beyond the maximum force of static friction the tire will skid. Once it starts sliding the frictional force stays constant at the force of kinetic friction.

 

[haruspex]

what would happen if the car's velocity increases in magnitude as it turns.

If the car is accelerating in the direction of motion at the same time then the net acceleration is at some angle to the velocity. The static frictional force will be in the opposite direction.
But if it starts to skid then, at first at least, the wheels will still be rotating at the appropriate rate for the car's forward motion, so the relative velocity of the surfaces will be normal to the velocity and the kinetic friction will act in the centripetal direction.
If the driver brakes or presses the accelerator then that will change the relative motion of the surfaces.

In what case, would the centripetal force become kinetic friction?

Kinetic friction is an applied force, whereas centripetal force is that component of the net force which is orthogonal to the velocity, so one cannot 'become' the other.
They do become equal, however, when the frictional force is purely orthogonal to the velocity and no other applied force has such a component. This will happen when cornering without skidding if there is no other lateral force, such as a crosswind.

The wheels of a car turn, but if the right side turns outwards, the left side turns inwards and that motion cancels out.

Couldn't follow that. Is this for a single wheel? What do you mean by a side of a tyre turning inwards?

 

[Tom.G]

https://g.co/kgs/5zMTGD

also:
https://www.google.com/search?&q=define+centripetal+force

 

[haruspex]

https://g.co/kgs/5zMTGD

also:
https://www.google.com/search?&q=define+centripetal+force

Several of those erroneously restrict it to circular paths.
Wikipedia is, as usual, reliable: https://en.m.wikipedia.org/wiki/Centripetal_force

 

[Cutter Ketch]

I don’t believe his question is: how does centripetal force cause circular motion? I believe his question is: how does friction which seems like it should only act parallel to the motion generate a force perpendicular to the motion?

 

[aspodkfpo]

If the car is accelerating in the direction of motion at the same time then the net acceleration is at some angle to the velocity. The static frictional force will be in the opposite direction.
But if it starts to skid then, at first at least, the wheels will still be rotating at the appropriate rate for the car's forward motion, so the relative velocity of the surfaces will be normal to the velocity and the kinetic friction will act in the centripetal direction.
If the driver brakes or presses the accelerator then that will change the relative motion of the surfaces.

Kinetic friction is an applied force, whereas centripetal force is that component of the net force which is orthogonal to the velocity, so one cannot 'become' the other.
They do become equal, however, when the frictional force is purely orthogonal to the velocity and no other applied force has such a component. This will happen when cornering without skidding if there is no other lateral force, such as a crosswind.

Couldn't follow that. Is this for a single wheel? What do you mean by a side of a tyre turning inwards?

I don’t believe his question is: how does centripetal force cause circular motion? I believe his question is: how does friction which seems like it should only act parallel to the motion generate a force perpendicular to the motion?

If we look at a diagram as such, https://physics.stackexchange.com/q...centripetal-force-during-the-turning-of-a-car, it proposes that the wheels push back and static friction (which is the force opposing changes to relative motion) causes the car to go inwards. However, I have qualms about this, as it seems like the backside of a wheel pushes outwards, while the frontside of that same wheel pushes inwards, generating a net sum of static friction=0?

Rather, I feel that if considering the car as a whole system which pushes outwards, generates centripetal force inwards makes more sense?

 

[Cutter Ketch]

However, I have qualms about this, as it seems like the backside of a wheel pushes outwards, while the frontside of that same wheel pushes inwards, generating a net sum of static friction=0?

No, I think you should think of it as one contact patch producing one force of friction in one direction. The friction isn’t putting a vertical torque on the tire (although the linear force of friction plus the angular momentum of the wheel will result in gyroscopic torque). The one linear force of friction at the contact patch is then broken into two components: one parallel to the wheel where the wheel is free to rotate and one perpendicular to the wheel which pushes the car like a rigid body.

 

[Cutter Ketch]

Rather, I feel that if considering the car as a whole system which pushes outwards, generates centripetal force inwards makes more sense?

This is very true. My argument about the front tires is just the beginning of the answer. That force doesn’t act at the center of gravity, so it is also a torque. It isn’t directed perfectly perpendicular to the motion either. If I used the same argument at the rear tires it would seem friction is pushing the rear outward, but you know it’s not. That’s because the torque from the front is trying very hard to throw the rear out and friction will no longer be just directly backwards from the motion.

Arguing that way causes a mess and gets confusing. It is much easier to start with the car moving in a circle and say what the forces must be. However that approach didn’t answer your question, so I started with the front tires and motion in a straight line. However, once you are satisfied how the wheel and axle can generate a nearly perpendicular turning force, leave that behind and go right to the car moving in a circle and the sum of forces and torques.

 

[jbriggs444]

Arguing that way causes a mess and gets confusing. It is much easier to start with the car moving in a circle and say what the forces must be

So true for so many things. Bicycle chains, stones on strings, trains on rails, books on tables. We know that there are constraints in place and we deduce what forces must exist to produce motion that is consistent with those constraints. [Nobody bothered to point this out in so many words when I was in school]

 

[aspodkfpo]

So true for so many things. Bicycle chains, stones on strings, trains on rails, books on tables. We know that there are constraints in place and we deduce what forces must exist to produce motion that is consistent with those constraints. [Nobody bothered to point this out in so many words when I was in school]

How would you deduce that there is static friction? And that it acts towards the centre?

 

[Cutter Ketch]

How would you deduce that there is static friction? And that it acts towards the centre?

Ignoring wind resistance the only external forces acting on the car are gravity, the normal force from the ground, and friction at the tires. The only one of those that acts parallel to the ground is friction. (assuming flat ground, no banking). The car IS moving in a circle. Therefore there must be a centripetal force, and that force can only be provided by friction. Therefore the force of friction will be equal to the required centripetal force.

That approach isn’t very helpful regarding your question of how friction comes to produce a centripetal force. But once you are satisfied that it can, the above approach is the simplest way to calculate.

 

[aspodkfpo]

Ignoring wind resistance the only external forces acting on the car are gravity, the normal force from the ground, and friction at the tires. The only one of those that acts parallel to the ground is friction. (assuming flat ground, no banking). The car IS moving in a circle. Therefore there must be a centripetal force, and that force can only be provided by friction. Therefore the force of friction will be equal to the required centripetal force.

That approach isn’t very helpful regarding your question of how friction comes to produce a centripetal force. But once you are satisfied that it can, the above approach is the simplest way to calculate.

Also, would it be true that at very point around a circle, that the car experiences a traction force forwards and a friction force backwards that is equal in magnitude and thus maintain that velocity? Since I assume that the wheels will always be spinning.

 

[jbriggs444]

Also, would it be true that at very point around a circle, that the car experiences a traction force forwards and a friction force backwards that is equal in magnitude and thus maintain that velocity? Since I assume that the wheels will always be spinning.

Friction and traction are the same thing.

If we assume all wheel drive and a good differential then all of the tires are providing their fair share of just enough forward force to balance the rearward wind resistance. There is no rearward frictional force of road on tire at all.

 

[Cutter Ketch]

Friction and traction are the same thing.

If we assume all wheel drive and a good differential then all of the tires are providing their fair share of just enough forward force to balance the rearward wind resistance. There is no rearward frictional force of road on tire at all.

Yes, that’s true for the drive wheels, but non-drive wheels will resist turning some. There will be rolling resistance from the deformation of the tires and reality of bearings not being frictionless, etc, so the force of friction on the non-drive wheels is rearward. If you throw it into neutral all wheels will have rearward friction and the car will slow down. At low speed that will be greater than wind resistance.

 

[jack action]

Let's start with the basics.

The friction force does not exist on its own. It always reacts to an opposite force. Friction doesn't appear out of thin air. And you don't need relative motion, only an opposing force.

Let's imagine we have a $100\ kg$ block on a surface with static friction coefficient of 0.7 and a kinetic fiction coefficient of 0.5.

The static friction force is: $100 \times 9.81 \times 0.7 = 687 N$. The kinetic friction force is: $100 \times 9.81 \times 0.5 = 491 N$. What do those forces represent?

The calculated static friction force represents the maximum friction force before the block will begin to move. But you can push the block with a force of $200 N$ and the opposing friction force will also be $200 N$ and the block will not move, because it is less than $687 N$.

If you reach $687 N$, then the friction bonds will partially break and the block will begin to move and the friction will lower to the kinetic friction force of $491 N$.

How is it different with a wheel? A wheel is succession of block rotating around an axis. When a force is applied sideways (parallel to the axis), the same mechanics apply: Until the force is not high enough to break the maximum static friction, the wheel doesn't move. Higher, and the wheel starts to slide sideways.

If the force is perpendicular to the wheel axis, then there is a moment created, since the friction force is at a distance $r$ apart (i.e. the wheel radius). If the force is less than the maximum static friction force, then the block doesn't slide and the wheel pivot about its axis (because of the moment created). The 'pivoting' creates a forward motion of the axis (since the block doesn't move) and removes the block from the ground and introduces the next block, where the cycle is repeated.

But if the perpendicular force is greater than the maximum static force, the block slides, so the wheel axis doesn't move. With the moment still present, the wheel is actually spinning freely.

 

[Lnewqban]

1. When a car turns there is a centripetal force towards the centre. This centripetal force is labelled as a static frictional force. I don't understand where this static frictional force arises from. Friction is meant to oppose motion, but I don't see the motion that is parallel to the friction in this case. The wheels of a car turn, but if the right side turns outwards, the left side turns inwards and that motion cancels out.

1) The order of the facts is not in the order you explain.
With minimum or zero coefficient of friction (like when driving on ice), you can turn the front wheels all you want, but the car follows the first law of Newton and continues moving at constant speed on a straight trajectory.
The driver has to force the huge mass of a car to accelerate, decelerate or deviate from a straight trajectory.
The only tool he/she has to achieve any of the above is the static friction that exists between the constantly changing contact patches of the tires and the asphalt of the road.

To accelerate: More torque on the wheels means higher rearward tangential force and longitudinal friction force.
To decelerate: Resisting to rotation torque on the wheels means higher forward tangential force and longitudinal friction force.

To deviate from straight trajectory: The rotational axis of the front wheels converge in a point (located along the axis of the rear wheels) means lateral force appears on the contact patches (induced by change of direction of the acceleration vectors (linear and angular) or second law of Newton) which is resisted by lateral friction force of equal magnitude and opposite direction (following the third law of Newton).

As the front wheels rotate, successive sections of the perimeter of the tire make contact with the surface of the road.
During the instant that each section of rubber is supporting its weight of the car, the nature of the friction is static, since rubber is not sliding respect to the asphalt.

As rubber and tire's carcass deform under that lateral force, the following section of the perimeter of the tire that makes contact with the surface the next instant lands a little off the ideal trajectory (towards the outside of the curve).
That may create the illusion that friction is less than static (because the tire seems to be crabbing out), but that is not the case, the phenomenon is called lateral slip, which is the angle between the direction it is moving and the direction it is pointing.

Please, see:
https://en.wikipedia.org/wiki/Slip_angle

2. Also, I was wondering what would happen if the car's velocity increases in magnitude as it turns.

The magnitude of that lateral force depends on the square of the forward velocity of the car and on the inverse of the radius of the turn or curvilinear trajectory.

3. In what case, would the centripetal force become kinetic friction? And if it was kinetic friction, how would the path deviate from being circular?

In extreme cases when the magnitude of the lateral force becomes too high, because excessive forward velocity of the car and/or insufficient radius of the turn or curvilinear trajectory, it can overcome the available forces of static friction of tires/asphalt.
The tires then start sliding sideways and the magnitude of the lateral friction that resist that slide is less that before the slide, because the coefficient of kinetic friction is always a little smaller than the static coefficient.