What is the effect of centripetal force on velocity?

[Rupert Young]

If a car is being driven with a constant force (the angle of the pedal is kept constant) and it goes around a curve, will the velocity change? If so, is this due to centripetal force?

 

[nasu]

Of course, it will change. If it did not change then you will go in a straight line and not around a curve.

 

[Rupert Young]

Don't you mean that the velocity changes because you are going around a curve?
Could you clarify why it is changing, given that the forward force is the same? Why does going around a curve exert a counter force?

 

[jbriggs444]

Don't you mean that the velocity changes because you are going around a curve?
Could you clarify why it is changing, given that the forward force is the same? Why does going around a curve exert a counter force?

Ummm, because there is a sideways force?

 

[Rupert Young]

Ummm, because there is a sideways force?

What is the source of this sideways force?
People, it might be obvious to you, but it is not to me, which is why I am asking the question, so a more in depth answer would be much appreciated.

 

[CWatters]

Velocity has components speed and direction. Going around a corner means the direction is changing so the velocity is changing. The speed need not be changing.

Forum rules discourage us just giving people the answers because they learn more when they figure things out for themselves. Have a think about the forces that act on a car. Have you tried to push one sideways? Why would that be hard? Why do you have to slow down in wintery conditions?

 

[Rupert Young]

Velocity has components speed and direction. Going around a corner means the direction is changing so the velocity is changing. The speed need not be changing.

Ok, I thought velocity and speed were the same thing. So, if a car is being driven with a constant force and it goes around a curve, does that mean the speed also remains constant?

Forum rules discourage us just giving people the answers because they learn more when they figure things out for themselves. Have a think about the forces that act on a car. Have you tried to push one sideways? Why would that be hard?

Because it is heavy?

Why do you have to slow down in wintery conditions?

Because the road is slippery?

 

[jbriggs444]

Ok, I thought velocity and speed were the same thing. So, if a car is being driven with a constant force and it goes around a curve, does that mean the speed also remains constant?

The idea that "velocity" and "speed" are different is drilled into one's head constantly in first year physics. As others have said, "speed" is how fast you are going without regard to direction. By contrast, "velocity" takes direction into account. 50 miles per hour forward is a different "velocity" than 40 miles per hour forward and 30 miles per hour leftward even though both are exactly the same "speed".

In an ideal car -- perfect wheels with no rolling resistance and no sideways slip, and no air resistance, there would be no reason to put your foot on the accelerator at all. You could coast at constant speed forever. Since your car requires force to maintain a constant speed, we need to consider what makes your car less than ideal.

So.. how about you tell us? Why does your car slow down if you do not keep your foot on the gas?

 

[CWatters]

Because the road is slippery?

So what is the car relying on to turn a corner?

 

[A.T.]

Because the road is slippery?

What does this mean in terms of forces?

 

[Rupert Young]

The idea that "velocity" and "speed" are different is drilled into one's head constantly in first year physics.

Well, I'm not a physicist, and I'm hoping this forum is open to those physicsally challenged!

As others have said, "speed" is how fast you are going without regard to direction. By contrast, "velocity" takes direction into account. 50 miles per hour forward is a different "velocity" than 40 miles per hour forward and 30 miles per hour leftward even though both are exactly the same "speed".

So, if you add a leftward force (wind, say) of 30 mph to a car that was going 40 mph it would then be going 50 mph? (computed by Pythagoras I guess)

In an ideal car -- perfect wheels with no rolling resistance and no sideways slip, and no air resistance, there would be no reason to put your foot on the accelerator at all. You could coast at constant speed forever. Since your car requires force to maintain a constant speed, we need to consider what makes your car less than ideal.

So.. how about you tell us? Why does your car slow down if you do not keep your foot on the gas?

What does this mean in terms of forces?

I feel you want me to say friction. But is that a force? Air resistance? Both?

 

[Rupert Young]

So what is the car relying on to turn a corner?

Turning the wheel? Is that exerting a sideways force? Does that effect the speed?

 

[jbriggs444]

So, if you add a leftward force (wind, say) of 30 mph to a car that was going 40 mph it would then be going 50 mph? (computed by Pythagoras I guess)

Yes, that's Pythagoras in action.

But you you cannot directly equate a leftward force with a leftward addition to velocity. There is a relationship via quantities known as momentum or impulse. A leftward force applied for a specified time interval will produce an impulse which is the product of the two. The leftward impulse divided by the mass of the car will tell you how much its leftward velocity changes as a result.

 

[Rupert Young]

Yes, that's Pythagoras in action.

Just to clarify, are you answering yes to the question as I'd thought the answer would be no. Are you saying that if the forward speed of a car is 40 mph, which is then exposed to a leftward force (wind, say) of 30 mph, the car would then speed up (over time) to go at a forward speed of 50 mph?

I would have thought that the speed would still be 40 mph, but it would be pushed to the right to change direction, if the steering wheel was not turned.

If the steering wheel was turned, into the wind, the speed and direction of the car would remain the same. Yes?

 

[jbriggs444]

Just to clarify, are you answering yes to the question as I'd thought the answer would be no. Are you saying that if the forward speed of a car is 40 mph, which is then exposed to a leftward force (wind, say) of 30 mph, the car would then speed up (over time) to go at a forward speed of 50 mph?

No, as I said, the relationship between force and speed is not direct. The relationship between wind speed, then force, then car speed is even less direct. And if you do not turn the steering wheel, a lateral force will (ideally) have zero effect on the vehicle.

If the steering wheel was turned, into the wind, the speed and direction of the car would remain the same. Yes?

If you turn the steering wheel, the car turns. When you normalize the wheel, the car retains its current heading. The wind does not enter in.

Edit: If you want to imagine an airplane, then wind velocity adds as a vector to airplane velocity to get ground velocity, but that's not the subject matter here.

 

[A.T.]

I feel you want me to say friction. But is that a force?

Have you thought about looking it up yourself?
https://en.wikipedia.org/wiki/Friction

 

[Rupert Young]

No, as I said, the relationship between force and speed is not direct. The relationship between wind speed, then force, then car speed is even less direct.

Let's come back to speed later, but first can we clarify this point.

And if you do not turn the steering wheel, a lateral force will (ideally) have zero effect on the vehicle. If you turn the steering wheel, the car turns. When you normalize the wheel, the car retains its current heading. The wind does not enter in.

This doesn't seem right to me. If there is a lateral force on an object surely that object will move laterally. If it is wind then the car will be moved laterally (maybe only slightly), so it would be necessary to turn the steering wheel into the lateral force to maintain the heading. Generally we don't notice the effects of wind except when we go across a windy bridge and we have to make an abrupt correction with the steering wheel to maintain our heading.

Similarly, if a car is being driven along a straight road but one which slopes from left to right the force of gravity would draw the car to the right. However, by turning the steering wheel to the left the straight heading of the car is maintained. In essence we would be applying an upward force to compensate for the downward force of gravity. Do you not agree?

 

[Rupert Young]

Have you thought about looking it up yourself?
https://en.wikipedia.org/wiki/Friction

Ah yes, as I thought,

Friction is not itself a fundamental force.

​

 

[A.T.]

Ah yes, as I thought,

The fundamental force behind friction is electromagnetic.

 

[jbriggs444]

L
This doesn't seem right to me. If there is a lateral force on an object surely that object will move laterally. If it is wind then the car will be moved laterally (maybe only slightly), so it would be necessary to turn the steering wheel into the lateral force to maintain the heading. Generally we don't notice the effects of wind except when we go across a windy bridge and we have to make an abrupt correction with the steering wheel to maintain our heading.

Similarly, if a car is being driven along a straight road but one which slopes from left to right the force of gravity would draw the car to the right. However, by turning the steering wheel to the left the straight heading of the car is maintained. In essence we would be applying an upward force to compensate for the downward force of gravity. Do you not agree?

No, I do not agree. The rightward turning of the steering wheel creates a rightward angle of the front wheels on the car which tends to cause the car to progressively turn farther and farther right. This is desirable if it maintains the desired course while the vehicle "crabs" left under the influence of a leftward wind. However, if one maintains the steering in this position (ignoring caster effects on the steering mechanism) the car will continue to turn further right than is desired. Ideally you want a one-time rightward steering input to counter a continuing leftward force.

In addition, I used the word "ideally" in my earlier posting. Ringing in non-ideal effects is not appropriate.

 

[Rupert Young]

No, I do not agree. The rightward turning of the steering wheel creates a rightward angle of the front wheels on the car which tends to cause the car to progressively turn farther and farther right. This is desirable if it maintains the desired course while the vehicle "crabs" left under the influence of a leftward wind. However, if one maintains the steering in this position (ignoring caster effects on the steering mechanism) the car will continue to turn further right than is desired. Ideally you want a one-time rightward steering input to counter a continuing leftward force.

In addition, I used the word "ideally" in my earlier posting. Ringing in non-ideal effects is not appropriate.

I am not quite sure what you are disagreeing with as your first statement seems to agree and the second doesn't. Could you clarify? My query regards your statement that a leftward force has no effect on an object (ideally or otherwise), which doesn't seem right, or I am misunderstanding what you mean?

So, in this,

The rightward turning of the steering wheel creates a rightward angle of the front wheels on the car which tends to cause the car to progressively turn farther and farther right. This is desirable if it maintains the desired course while the vehicle "crabs" left under the influence of a leftward wind.

are you saying that it is necessary to turn the wheel to the right to compensate for a force coming from the right? If so, that makes sense and agrees with what I said.

In this,

However, if one maintains the steering in this position (ignoring caster effects on the steering mechanism) the car will continue to turn further right than is desired. Ideally you want a one-time rightward steering input to counter a continuing leftward force.

you appear to be saying that you need to turn the wheel to the right and than back again to "normal" to compensate, i.e. an impulse turn. Is that what you are saying? I don't think that could be correct because after the impulse (one-time rightward steering input) you would be back to square one with the lateral force once again moving the car laterally. I'd say you have to maintain a turned wheel, all the time the lateral force exists, in order to maintain the straight-ahead heading. In other words, although the wheel is turned the car does not turn, due to the opposing lateral force. Yes?

 

[olivermsun]

I think the confusion/disagreement arises because in one case what seems to be described is an impulsive force leading to a small permanent yaw which would continue to compensate for the slippage caused by the side force. In the second case what is being described is a continuing force exerted through the steering mechanism and tires which exerts the compensating force.

 

[A.T.]

I'd say you have to maintain a turned wheel, all the time the lateral force exists, in order to maintain the straight-ahead heading. In other words, although the wheel is turned the car does not turn, due to the opposing lateral force. Yes?

You don't necessarily need steering to compensate for a lateral force. If you have sufficient traction, the static friction at all the wheels can have a lateral component, which balances the lateral force.

 

[olivermsun]

You don't necessarily need steering to compensate for a lateral force. If you have sufficient traction, the static friction at all the wheels can have a lateral component, which balances the lateral force.

Perhaps in an idealized case, but in the real world case, the car/wheels/tires will probably want to move sideways at some rate that the driver will simply compensate for.

 

[sophiecentaur]

Instead of using a car on a road as your 'simple' model, there is a simpler one of a locomotive on a Perfect, frictionless) railway line. The force on the loco, when it follows a curve is not friction; it is a lateral force resulting from the fact that the track in front is angled slightly. to one side If the track is frictionless (a further simplification) then no energy can be lost so there will just be a change of direction and no change in speed. There are practical issues on a real track and one of them will be that the resistive forces from the track (rolling resistance) may actually increase when going round a curve, so a freewheeling loco would slow down - or a driven loco would need more fuel to maintain its speed. But that is a step too far if you want to get the basics first.
If you read and absorb the basics of speed, velocity and a few other quantities for moving bodies then there will be no conflict in your mind. The Science of basic stuff like this works on consistent rules and people can easily get confused when they are told something that follows those rules but their 'experience' is of situations involves other factors that the simple rules don't cover. So that's the next step . . . . . . I'm afraid that the dreaded Maths comes into this stuff very early on.

 

[A.T.]

Perhaps in an idealized case, but in the real world case, the car/wheels/tires will probably want to move sideways at some rate...

I'm assuming sufficient traction as stated. Static friction doesn't let the wheels move where they "want".

...that the driver will simply compensate for.

Even given slippage, you still don't need to "maintain a turned wheel to maintain the straight-ahead heading" as Rupert Young suggests. You can have the entire vehicle oriented slightly off it's movement direction, against the lateral force.

 

[olivermsun]

I'm assuming sufficient traction as stated. Static friction doesn't let the wheels move where they "want".

No, but compliance in the tires can allow the wheels/car to "walk" sideways even if the tread itself doesn't slip sideways on the pavement.

Even given slippage, you still don't need to "maintain a turned wheel to maintain the straight-ahead heading" as Rupert Young suggests. You can have the entire vehicle oriented slightly off it's movement direction, against the lateral force.

See my post #22.

 

[Rupert Young]

Even given slippage, you still don't need to "maintain a turned wheel to maintain the straight-ahead heading" as Rupert Young suggests. You can have the entire vehicle oriented slightly off it's movement direction, against the lateral force.

Ok, I think there is some agreement that it is necessary to turn the wheel to compensate for a lateral force, though some dispute about whether the car itself is redirected or just the wheels themselves are. Whichever is the case, it could be described that we turn the steering wheel until we perceive the heading of the car in the desired direction when driving.

But perhaps we could go back to the original question. It was speed I was interested in so I'll rephrase the question.

If a car is being driven with a constant force and it goes around a curve, will the speed change? If so, is this due to centripetal force?

I'd be interested help in how this could be quantified and modeled mathematically. Any help to get started on this would be appreciated.

 

[olivermsun]

sophiecentaur's post #25 gives a good suggestion for how to start with a simple model of a vehicle going around a curve.

 

[A.T.]

Ok, I think there is some agreement that it is necessary to turn the wheel to compensate for a lateral force, though some dispute about whether the car itself is redirected or just the wheels themselves are.

There are different options as post #22 describes.

I'd be interested help in how this could be quantified and modeled mathematically.

How ideal/realistic you want to model it? What exactly do you mean by "being driven with a constant force" ? What is the direction of that force?

 

[Rupert Young]

sophiecentaur's post #25 gives a good suggestion for how to start with a simple model of a vehicle going around a curve.

Well, I did read it but didn't see how to get to the next step. I know some basic stuff about bodies moving in a straight line (f=ma), but not about any effects due to curves. So, I might need an explanation as if I were a small child (which I might be).

 

[Rupert Young]

How ideal/realistic you want to model it? What exactly do you mean by "being driven with a constant force" ? What is the direction of that force?

The forces that would have the main effects, particular interest in centripetal force. As in the original question if the angle of the throttle pedal is kept constant, but basically a constant force exerted by the engine. The direction is forward, relative to the car.

 

[Kaura]

If I understand correctly then the magnitude of velocity can be the same when going around the curb
However the vector components of velocity will change
You can think of a car going around the curb as being similar to a ball on a string being twirled around or the Moon orbiting the Earth

 

[sophiecentaur]

Several of the last bunch of posts have introduced a number of extra concepts that are not really helping to explain what's going on in the most basic situation of a car (/railway loco) going round a circular bend. Please sort out the basics first.
A ball on a string, in space, orbiting round a hand at constant speed is the best model to start with. This will have a constant centripetal force. If the ball is changing speed or if their is any friction or motion under gravity then things get harder. (But not impossible)

 

[A.T.]

basically a constant force exerted by the engine. The direction is forward, relative to the car.

Whether the speed changes or not then depends on whether that engine force is greater or less than the restive forces (rolling resistance, aerodynamic drag). The centripetal force doesn't change the speed per definition, just the direction.

 

[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.