What happens when a car turns?

[Stephen Bulking]

A car moving at constant speed is in uniform circular motion, thus having centripetal acceleration of ##a=\frac{v^2}{R}##. The force associated with this acceleration is known to be friction. But friction, in nature, appears as an opposition to the relative motion between two surfaces whether it be static or kinetic friction (Here, the car is neither "sliding" away from the center of the imaginary circle it is rounding nor is it sling towards it, hence the friction of our interest is static friction). Since friction, the centripetal force is directed towards the center of the circle, is there any force acting on the car in the opposite direction? Apparently, no... So is it inertia that friction is fighting against? If so, which is the direction of motion that friction is opposing? Is it the original straight-line motion of the car? Why is it directed to the center of the circle in the first place? Just because it needs to cause radial acceleration?

 

[DrStupid]

Since friction, the centripetal force is directed towards the center of the circle, is there any force acting on the car in the opposite direction? Apparently, no...

Correct. But there is an opposite force acting on the ground.

So is it inertia that friction is fighting against? If so, which is the direction of motion that friction is opposing? Is it the original straight-line motion of the car?

Correct again.

Why is it directed to the center of the circle in the first place?

Because the wheel is turned off the straight-line direction.

Just because it needs to cause radial acceleration?

Just because the driver wants the car to drive in a circle (assuming he has the car under control).

 

[PeroK]

Summary:: When a car running at constant speed turns, rounding a flat curve or moves in uniform circular motion, there is a centripetal force holding it in the circular path, what force is this? And how does it work?

A car moving at constant speed is in uniform circular motion, thus having centripetal acceleration of ##a=\frac{v^2}{R}##. The force associated with this acceleration is known to be friction. But friction, in nature, appears as an opposition to the relative motion between two surfaces whether it be static or kinetic friction (Here, the car is neither "sliding" away from the center of the imaginary circle it is rounding nor is it sling towards it, hence the friction of our interest is static friction). Since friction, the centripetal force is directed towards the center of the circle, is there any force acting on the car in the opposite direction? Apparently, no... So is it inertia that friction is fighting against? If so, which is the direction of motion that friction is opposing? Is it the original straight-line motion of the car? Why is it directed to the center of the circle in the first place? Just because it needs to cause radial acceleration?

Those links show the kinematic situation, but don't attempt to explain how the position of the front wheels on the road results in a lateral force on the car. Those notes effectively assume that there is a net force at right angles to the direction of motion of the car - and that must result in uniform circular motion.

To explain circular motion this is all you need: constant force at right angles to the direction of motion. In this case caused (somehow!) by friction between the tyres and the road.

To explain why the position of the front wheels results in this force is another matter. Note that there is also a torque on the car. Not only is the centre of mass of the car executing circular motion, but the car is rotating about the centre of mass, with the same angular frequency. I.e. once the centre of mass has completed one circle the car has also rotated once about its centre of mass.

You might want to investigate this yourself, but the book you are using does not analyse precisely how the turning force is generated.

 

[jbriggs444]

Note that there is also a torque on the car.

To be clear, there is no need for torque causing rotation about a vertical axis as the car smoothly turns right or left. This turning action is uniform. No change in rotation rate or angular momentum about a vertical axis.

There is a torque needed against the tendency of the car to lean outward, however. The force of friction on the tires is not aligned with the center of mass of the car. Without a restoring torque, the car would rotate about a horizontal axis -- rolling over.

 

[PeroK]

In uniform circular motion there is no torque. There is rotation, yes. But there is no change in rotation and, thus, no need for torque.

The car definitely rotates as it performs a circular lap. At the halfway point the car is pointing in the opposite direction and has, therefore, done half a rotation about its centre of mass.

 

[jbriggs444]

The car definitely rotates as it performs a circular lap. At the halfway point the car is pointing in the opposite direction and has, therefore, done half a rotation about its centre of mass.

No torque is required for rotation.

 

[PeroK]

No torque is required for rotation.

Are you saying that the force is directed through the centre of mass of the car?

 

[Stephen Bulking]

Correct. But there is an opposite force acting on the ground.
Correct again.
Because the wheel is turned off the straight-line direction.
Just because the driver wants the car to drive in a circle (assuming he has the car under control).

Thank you for your quick response, it really helped confirm my thoughts

 

[miltos]

Summary::
Here, the car is neither "sliding" away from the center of the imaginary circle it is rounding nor is it sling towards it

The answer is to the term "tyre slip angle"
Have a look here https://en.wikipedia.org/wiki/Slip_angle and here https://suspensionsecrets.co.uk/tyre-slip-angle/

You could also find usefull the Ackermann steering geometry: https://en.wikipedia.org/wiki/Ackermann_steering_geometry

 

[Stephen Bulking]

Those links show the kinematic situation, but don't attempt to explain how the position of the front wheels on the road results in a lateral force on the car. Those notes effectively assume that there is a net force at right angles to the direction of motion of the car - and that must result in uniform circular motion.

To explain circular motion this is all you need: constant force at right angles to the direction of motion. In this case caused (somehow!) by friction between the tyres and the road.

To explain why the position of the front wheels results in this force is another matter. Note that there is also a torque on the car. Not only is the centre of mass of the car executing circular motion, but the car is rotating about the centre of mass, with the same angular frequency. I.e. once the centre of mass has completed one circle the car has also rotated once about its centre of mass.

You might want to investigate this yourself, but the book you are using does not analyse precisely how the turning force is generated.

Thank you very much, I will try not to be confused by the position of the front wheels, but for future reference, because I am really interested, do you have any suggestion for a book that would analyze such situation in details?

 

[PeroK]

Thank you very much, I will try not to be confused by the position of the front wheels, but for future reference, because I am really interested, do you have any suggestion for a book that would analyze such situation in details?

See post #9 above.

 

[jbriggs444]

Are you saying that the force is directed through the centre of mass of the car?

Let us assume for the sake of discussion that we are adopting a reference point that is level with the center of mass of the car and assessing angular momentum about that point.

I am saying that there is no change in angular momentum about this point. The force of friction from the road involves no torque about a vertical axis. So no additional torque about the vertical axis from steering inputs is required in steady state.

You are correct to point out that when using this reference point, the force of friction from the road involves a non-zero torque about a horizontal axis. This torque must be countered in order to keep angular momentum constant. An imbalance in vertical force from the tires together with downforce from gravity suffices to provide that countering torque.

 

[Stephen Bulking]

The answer is to the term "tyre slip angle"
Have a look here https://en.wikipedia.org/wiki/Slip_angle and here https://suspensionsecrets.co.uk/tyre-slip-angle/

You could also find usefull the Ackermann steering geometry: https://en.wikipedia.org/wiki/Ackermann_steering_geometry

I will certainly go through the links you have provided, thank you very much.

 

[PeroK]

Let us assume for the sake of discussion that we are adopting a reference point that is level with the center of mass of the car and assessing angular momentum about that point.

I am saying that there is no change in angular momentum about this point. The force of friction from the road involves no torque about a vertical axis. So no additional torque about the vertical axis from steering inputs is required in steady state.

You are correct to point out that when using this reference point, the force of friction from the road involves a non-zero torque about a horizontal axis. This torque must be countered in order to keep angular momentum constant. An imbalance in vertical force from the tires together with downforce from gravity suffices to provide that countering torque.

That's a good point. There's no torque in the steady state scenario.

 

[A.T.]

But friction, in nature, appears as an opposition to the relative motion between two surfaces whether it be static or kinetic friction

In static friction there is no relative motion between the contact surfaces, so how could static friction be opposed to it? Static friction is like a constraint force, that takes whatever direction/magnitude is needed (within the limits dictated by the normal force and static friction coefficient) to prevent the relative motion. So macroscopically static friction depends on all the other forces acting on the object. Microscopically it depends on the deformation of the materials in contact.

Why is it directed to the center of the circle in the first place?

Because the wheel can spin around it's axis, but not around other axes. So if it's not aligned with the direction of motion, it will deform such that mainly horizontal forces parallel to its axis are generated, but not horizontal forces perpendicular to its axis, which just make it spin instead of deforming it.

If the spinning of the wheel confuses you here, just consider the steerable blade of an iceboat: it allows motion in one direction, but resists motion perpendicular to that. So it's effectively the same type of constraint as a wheel.

 

[Dale]

Summary:: When a car running at constant speed turns, rounding a flat curve or moves in uniform circular motion, there is a centripetal force holding it in the circular path, what force is this? And how does it work?

But friction, in nature, appears as an opposition to the relative motion between two surfaces whether it be static or kinetic friction

In order for a wheel to move without slipping it must roll in the direction it is pointing*. If it moves transverse to the direction it is pointing then it must slip. Static friction acts to oppose that slipping by a lateral force that turns the direction of travel towards the direction the wheel is pointing.

*for a non-rigid wheel (eg a rubber tire) there can be a difference between the direction the wheel is pointing and the direction of travel even without slipping. This is allowed by deformations in the wheel material.

 

[hutchphd]

That's a good point. There's no torque in the steady state scenario.

But in order to initiate the turn (and of course to maintain dynamic stability once in the turn) requires lateral (axial) forces from the rear wheel(s) be considered.