What is the effect of centripetal force on velocity?

[rcgldr]

In the real world, cornering reduces speed if not compensated for with a bit of additional throttle. The lateral load on the tires increases the deformation at the contact patch, and due to the same principle as rolling resistance, additional energy is lost due to hysteresis during the increased deformation and recovery. The lost energy is converted into heating of the tires.

As an extreme example, Formula racing cars can take some high speed turns at full throttle, but lose speed due to the high (around 4) g cornering forces, even though the throttle is kept floored. With infrared on board cameras, you can see the tires heat up from braking or cornering loads. Note how quickly the tires cool when not under load, so the amount of energy being added to the tires during turns and being dissipated as seen on the straights is significant. The insides of the front tires are hotter due to the amount of camber (inwards lean) used.

 

[Kajal Sengupta]

I do not know whether I am breaking the rules of the forum but after so many posts it is becoming so weird that those who know about it also may get confused. A force can do a lot of things, one of which is changing direction. Around a bend the force exactly does that without changing the speed. If you know what is speed , velocity and acceleration then you will realize that it does not go against Newton's second law of motion.

 

[sophiecentaur]

those who know about it also may get confused.

Not confused about the basics. What is confusing is that the question shifts from theory to practice and back again and it is never clear what is actually being discussed.

 

[Rupert Young]

In the real world, cornering reduces speed if not compensated for with a bit of additional throttle. The lateral load on the tires increases the deformation at the contact patch, and due to the same principle as rolling resistance, additional energy is lost due to hysteresis during the increased deformation and recovery. The lost energy is converted into heating of the tires.

As an extreme example, Formula racing cars can take some high speed turns at full throttle, but lose speed due to the high (around 4) g cornering forces, even though the throttle is kept floored.

Interesting, this is what I am trying to get to. So what is the source of the lateral force, is it friction due an increase in pressure of the tyres on the ground? Does centrifugal force come into this?

How can we go about quantifying the loss of speed?

 

[rcgldr]

So what is the source of the lateral force.

What causes a car to turn is that the front tires and rear tires don't point in the same direction. If you were to extend the axis of the steered front wheels and straight rear wheels, the intersection of where those extended axis cross is the center of the circle that the car tries to turn. The tires flex, so the actual turn radius is a bit larger than that.

There is a Newton third pair law of friction forces, the tires exert an outwards force onto the pavement, the pavement exerts an inwards (centripetal) force onto the tires.

How can we go about quantifying the loss of speed?

This would be complicated. It's a combination of factors, cornering load, deformation of the tires due to a lateral load, and the hysteresis factor of the tire compound.

 

[votingmachine]

Interesting, this is what I am trying to get to. So what is the source of the lateral force, is it friction due an increase in pressure of the tyres on the ground? Does centrifugal force come into this?

How can we go about quantifying the loss of speed?

I think the problem I have with this is that you are asking for the non-ideal answer, without a real way to quantify. If you yank the wheel hard and start to slide sideways, that is a lot more friction than if you barely nudge it from straight. And really, a constant position of the accelerator (floored, or at a constant gas flow) is not the same as constant force ... there is a power curve for the engine and the change in speed of the car corresponds to a change in rpm of the engine, which means the horsepower of the engine output is different.

If you barely nudge the steering wheel from straight, to a new circular path of a really large radius, you probably won't see any change. If you turn hard, you start to use some of your force for the change in direction, and less for maintaining the speed. You break the force into two components, parallel to the straight ahead direction, and perpendicular to the straight ahead direction. Only the straight ahead component acts on "speed". The perpendicular component acts on direction.

So say you are applying 100 hp, and the speed is constant at 100 mph (that means the total resistance is also exactly 100 hp, or you would go faster). You turn the wheel to 45-degrees. You still apply 100 hp, but it has to be looked at as 70.7 hp forward, and 70.7 hp sideways (Pythagorus again: 70.7^2 + 70.7^2 = 100^2). That means you slow down, as the 100 hp was enough to equal the 100 mph forces of resistance (air resistance, machine friction, etc).

It is not easy to estimate the steady state speed of the car at 70.7 hp. If you measured the straight-line speed vs hp for the car, then you could calculate the speed reduction for any angle of turn. Assuming the angle of turn does not increase machine friction, which seems unlikely as an assumption for any large angles. And again, since the engine rpm changed with the speed, you have to make assumptions about the power output for the constant gas flow.

You are changing directions. That uses force from the engine. That leaves less force for speed. You slow down. How much is going to depend on a lot of things about the individual car.

 

[rcgldr]

If you turn hard, you start to use some of your force for the change in direction.

If it wasn't for hysteresis type losses due to deformation related to lateral loads, then lateral force would require no energy. Lateral forces don't perform any work.

 

[votingmachine]

If it wasn't for hysteresis type losses due to deformation related to lateral loads, then lateral force would require no energy. Lateral forces don't perform any work.

That doesn't make any sense to me. If a ball is lobbed to me and I whack it with a bat exactly sideways, I can hit it very far, with an exactly lateral force. That lateral force is clearly doing work. If I apply an exactly lateral force, I do nothing to the component of velocity from the lobbed force.

 

[sophiecentaur]

What causes a car to turn is that the front tires and rear tires don't point in the same direction.

That is very true for conventional cars but you can still achieve cornering with all your wheels being steered and even a unicycle will corner quite happily ( not for the dispraxic, though). It only requires a lateral force.
Some examples of four wheel steered cars.

 

[A.T.]

I can hit it very far, with an exactly lateral force.

If you increase the speed, the force you applied wasn't perpendicular to velocity all the time.

 

[votingmachine]

If you increase the speed, the force you applied wasn't perpendicular to velocity all the time.

Right. I was using an impulse, but really it can't be instantaneous.

Regardless, the lateral force changes the velocity (vector direction). I'm puzzled by how that force is not doing work. It is a force, applied across a distance ... the definition of work seems to apply.

 

[jbriggs444]

Regardless, the lateral force changes the velocity (vector direction).

It is not a lateral force.

Perpendicular to the initial velocity but not perpendicular to the final velocity. That's not lateral.

 

[A.T.]

It is a force, applied across a distance ...

Yeah, across not along. Hence work is zero.

 

[votingmachine]

It is not a lateral force.

Perpendicular to the initial velocity but not perpendicular to the final velocity. That's not lateral.

Perhaps you are talking about the lobbed ball and bat ... I see that error. The force has to go across time and will go in the direction of the hit. I agree that was a poorly thought out example.

But turning a car takes no work? The lateral force does no work? Somehow there is something I am not seeing.

If the car is front wheel drive, and the wheels are turned, all the force is now directed along the direction of the forward rotation of the wheels. The car was going straight, and the force was in that direction, exactly balancing the resistance. Now it is in a different direction. If it is at 45-degrees, it divides into a parallel to velocity and perpendicular to velocity component. And only the parallel matters for the forward speed. It seems to me that the work is still the 100 hp x distance. Dividing it into perpendicular and parallel should not lose work.

It seems that I am missing something in these comments. Again, I see that the lobbed ball story was a bad example.

 

[A.T.]

all the force is now directed along the direction of the forward rotation of the wheels.

Not true.

 

[olivermsun]

Perhaps you are talking about the lobbed ball and bat ... I see that error. The force has to go across time and will go in the direction of the hit. I agree that was a poorly thought out example.

But turning a car takes no work? The lateral force does no work? Somehow there is something I am not seeing.

If the lateral force is truly lateral, that is, across the path of the car rather than along, then it is not doing work.

If the car is front wheel drive, and the wheels are turned, all the force is now directed along the direction of the forward rotation of the wheels. The car was going straight, and the force was in that direction, exactly balancing the resistance.

The resistance obviously resists the forward motion of the car whether it is turning or not, and hence resistance makes the problem more complicated than just the problem of turning. In that case, then the answer is yes, the engine is doing work to maintain forward speed. However, I think you should first work out your questions about the idealized case (no friction/resistance), in which case you don't need to consider the forward force exerted by the engine.

 

[rcgldr]

What causes a car to turn is that the front tires and rear tires don't point in the same direction.

That is very true for conventional cars but you can still achieve cornering with all your wheels being steered ...

My statement was the front and rear tires don't point in the same direction, this could be front wheel steer, rear wheel steer, all wheel steer. My next statement mentioned front wheel steer, so to generalize it, if you were to extend the axis of the front wheels and rear wheels, the intersection of where those extended axis cross is the center of the circle that the car tries to turn. The tires flex, so the actual turn radius is a bit larger than that. This assumes an idealized case where Ackerman like steering setup is used on the steered wheels so that the extended axis of all wheels cross at the same point.

 

[sophiecentaur]

But turning a car takes no work? The lateral force does no work? Somehow there is something I am not seeing.

No in principle. The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there] That will be zero but that's the same old 'work done on' paradox that applies to many mechanics situations and which gets people going so much. The fact is that there are many components of forces times distances that are involved in a real case because there is motion in the direction of several of the forces involved in a real tyre. One of the reasons that steel wheels and rails are so successful is that the situation on bends is much nearer the ideal.
I could go so far as to say that the 'work done on' the car is irrelevant and not worth getting bothered about. The work that is done is on distorting the tyres and the road surface etc. etc..

 

[votingmachine]

No in principle. The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there] That will be zero but that's the same old 'work done on' paradox that applies to many mechanics situations and which gets people going so much. The fact is that there are many components of forces times distances that are involved in a real case because there is motion in the direction of several of the forces involved in a real tyre. One of the reasons that steel wheels and rails are so successful is that the situation on bends is much nearer the ideal.
I could go so far as to say that the 'work done on' the car is irrelevant and not worth getting bothered about. The work that is done is on distorting the tyres and the road surface etc. etc..

For some reason, my intuition leads me to completely different thinking with a train on rails, and a car. I see that error now.

The force (torque of the tires) seems directed by the tire's direction. And then the force has to be divided into the component parallel to the forward direction of the car, and the sideways direction of the car.

No one needs to correct that ... it was an explanation of my intuitive thinking, not a statement of the physics.

 

[jbriggs444]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

 

[votingmachine]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

The OP was specifically about a real case, not ideal. With the accelerator applying constant gas/torque, and the net resistance equal to that, and the speed constant.

EDIT: That leads to confusion as people have shifted between "real" and "ideal". I've been trying to stick with "real". With net resistance and net forward forces equal and constant forward speed. Then a turn of the steering wheel to a new, steady angle.

 

[sophiecentaur]

different thinking with a train on rails, and a car.

It brings to mind what Newton's Third Law tells us. Resistance is very low with rails and steel wheels and Work is Force times Distance. The force (at constant speed) in the direction of the rails cannot be very great - or the train would be accelerating - so the work can't be much. So you have large (lateral) force and no distance and small (tangential) force in the direction where there is distance.
Some modern trams use rubber wheels because, perhaps, low noise is more important than efficiency.

 

[rcgldr]

The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there]

Work done would be a radial force times change in radius. While the radius is changing, the path spirals inwards or outwards, a component of the radial force is in the direction of that path (so it's not a centripetal "only" force), so work is done. A centripetal force is always perpendicular to the path, so doesn't perform work.

 

[sophiecentaur]

Work done would be a radial force times change in radius. While the radius is changing, the path spirals inwards or outwards, a component of the radial force is in the direction of that path (so it's not a centripetal "only" force), so work is done. A centripetal force is always perpendicular to the path, so doesn't perform work.

Yes, of course. That makes perfect sense! Cheers

 

[sophiecentaur]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

I don't think that's right because the direction in which the 'footprint' points is not in the plane of the wheel. The process of compressing the front of the footprint and releasing it at the back involves hysteresis and the forces (torques) do not cancel. This will produce a torque which 'fights' against the applied torque on the steering wheel.

 

[A.T.]

I've been trying to stick with "real". With net resistance and net forward forces equal and constant forward speed. Then a turn of the steering wheel to a new, steady angle.

Addressed here:
https://www.physicsforums.com/threa...-force-on-velocity.885192/page-2#post-5570267

 

[jbriggs444]

I don't think that's right because the direction in which the 'footprint' points is not in the plane of the wheel. The process of compressing the front of the footprint and releasing it at the back involves hysteresis and the forces (torques) do not cancel. This will produce a torque which 'fights' against the applied torque on the steering wheel.

Be careful. The steering wheel does not apply any propulsive torque, that's the engine's job. The torque that the steering wheel fights is due to the rearward offset of the contact patch from the steering axis, i.e. caster.

 

[Rupert Young]

This would be complicated. It's a combination of factors, cornering load, deformation of the tires due to a lateral load, and the hysteresis factor of the tire compound.

Can we simplify things by considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects.

Would those nearer the centre (on a path of higher curvature) move at a slower speed?

 

[rcgldr]

This would be complicated. It's a combination of factors, cornering load, deformation of the tires due to a lateral load, and the hysteresis factor of the tire compound.

Can we simplify things by considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects. Would those nearer the centre (on a path of higher curvature) move at a slower speed?

The loss in speed is due to tires consuming energy (conversion into heat). The amount of energy consumed by tires is increased when cornering. In a scenario where there are no energy losses, then cornering would not reduce speed (this is a self fulfilling statement, no energy lost means that kinetic energy and the related speed don't change).

 

[A.T.]

considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects. Would those nearer the centre (on a path of higher curvature) move at a slower speed?

No, faster:
https://en.wikipedia.org/wiki/Orbital_speed#Mean_orbital_speed