Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

In summary, the conversation discusses the application of physics in modeling the dynamics of a vehicle. The individual is unsure about whether to consider the number of wheels in calculating the friction force and if the total mass of the vehicle should be used or if a single wheel's force is sufficient. The conversation also mentions the relationship between angular and linear acceleration and the need to consider the angular inertia of all wheels. The concept of combining forces from different wheels is also brought up for simplification. The conversation ends with a discussion about the work of frictional forces in this scenario.
  • #1
Saalz
5
1
Homework Statement
Vehicle acceleration from wheel torque
Relevant Equations
Ftan = Torque / r
F = m * a
I was modeling the dynamics of a vehicle for a project, and started doubting about the way of applying physics in this particular case.

The thing is, I know the torque in the wheels from the torque the electric motor I designed do provide, multiplied by the gearbox ratio. I also know the radius of the wheels, and so, the tangential force in the surface of the tyre.

I'm under the case the wheels do not slip and therefore, the velocity at the circumference of the wheel its the same as the velocity of the car.

Thus, I do apply F = m * a, and isolate acceleration in that equation to calculate it as the ratio between the friction force on the road equal to the tangential force in the wheels, and the mass of the vehicle.

My question following that is, do wheel number matter? As the force is calculated from a single wheel, do I have to multiply it by the number of wheels the vehicle has as each one generates it's own friction force in the road (and, not even of the same magnitude sometimes, as in turns, the differential makes each wheel rotate at different speeds)?

Also, the mass used in the equation is the total mass of the vehicle and each wheel holds a percentage of the total mass of the vehicle. Does that even matter if I use the force in a single wheel?

Or I'm just overthinking and it's as simple and physically correct as using the friction force generated by one wheel and the total mass of the vehicle to know the linear acceleration it has? In that case, why?

Thanks for the replies.
 
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  • #2
Welcome!
You should also consider the angular inertia of all the wheels.
 
  • #3
Lnewqban said:
Welcome!
You should also consider the angular inertia of all the wheels.
What would be the procedure then?
 
  • #4
Saalz said:
What would be the procedure then?
If there is no-slip between the tire and the road there is a relation between the angular acceleration of the wheel and the linear acceleration of the vehicle that must be held.

Then you get a system of equations using the following laws (one for a driven wheel, one for a non-driven wheel, and one for the entire vehicle):

$$ \sum \tau = I \alpha $$

$$ \sum F = m a $$

The last equation contains elements of the other two which you will use to solve for ##a##.
 
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  • #5
Saalz said:
What would be the procedure then?
For all rotating parts of the car, the engine must use energy for increasing both, the linear velocity and the angular one.
The heavier and bigger the tires, the higher their moment of inertia and the slower they can be accelerated.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

:cool:
 
  • #6
Lnewqban said:
For all rotating parts of the car, the engine must use energy for increasing both, the linear velocity and the angular one.
The heavier and bigger the tires, the higher their moment of inertia and the slower they can be accelerated.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

:cool:
This made me realize that we have to include a friction force acting on the non-driven wheels too (to the OP think about its direction in comparison to the static friction acting on the driven wheels). Even though they are not driven, they too have angular acceleration, and the force of static friction acting on those wheels is responsible for it. I edited my post to show that.
 
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  • #7
If you look at the top view of your vehicle, you will have longitudinal and lateral forces acting on each wheel. All of these forces should be taken into account to evaluate not only the resulting motion of the vehicle but also the weight transfer on each wheel that will in turn influence the longitudinal and lateral forces on these wheels.

That being said, simplification can be made by combining forces from different wheels together - for example, wheels on the same axle when examining the side view of the vehicle. For simplicity and good approximations, combining all wheels into a single wheel might be acceptable - for example, assuming the rolling friction acts on a single wheel with the entire weight of the vehicle on it.

Don't forget that if a single source drives more than one wheel, its power is split between all of them. The ratio will depend on the mechanical design of the drivetrain.
 
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  • #8
Does static friction force do work in this case? It seems like the frictional forces must be doing work, but it often seems like this is "explained away" in some way I still don't fully understand.

If I'm doing a force balance to solve the OP's problem, there are frictional forces acting as external forces on the vehicle. The rear frictional forces assisting acceleration and the front tires frictional forces opposing it. These forces must be doing the work...they are the only external forces (in the direction of motion). But I so often hear about static frictional forces not doing work because the point of contact between the tire and the road is instantaneously at rest w.r.t the road. Am I still confused, because I feel a bit unsure of my understanding?[EDIT]
I believe I'm confusing the case where the wheel rolls without slipping at constant velocity. That must be the context in which static friction does no work. I feel like I've been over this a lot, and it still escapes me initially.

But then again...in both cases ( accelerating/not accelerating) the point of contact is instantaneously at rest w.r.t. the ground. So, I still am unsure what the actual resolution is! It all seems like a bit of a circular argument. It feels like whether or not the static frictional force does work or not is whether or not the body is accelerating, and whether or not the body is accelerating is whether or not the static frictional forces are doing work.
 
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  • #9
erobz said:
The rear frictional forces assisting acceleration and the front tires frictional forces opposing it.
To be clear, it is the rolling resistance that opposes the motion, not friction.

Also, there is not necessarily an acceleration.

erobz said:
But I so often hear about static frictional forces not doing work because the point of contact between the tire and the road is instantaneously at rest w.r.t the road.
Although the tire-road contact point is at rest, the axle is not. The geometry of the wheel transfers the friction force to the axle that is not at rest, thus doing work on the vehicle.

erobz said:
It feels like whether or not the static frictional force does work or not is whether or not the body is accelerating,
Acceleration is not a criterion to do work. Work is a force that is displaced.

erobz said:
and whether or not the body is accelerating is whether or not the static frictional forces are doing work.
Work is not a criterion to cause acceleration. Acceleration is present if there is a resultant force acting on a body. A body at rest can have an acceleration. Otherwise, its speed would never increase.
 
  • #10
1663966269220.png


jack action said:
To be clear, it is the rolling resistance that opposes the motion, not friction.

Also, there is not necessarily an acceleration.

Front wheels: There must be a torque that is causing the angular acceleration of the wheel. Which implies there is a force acting at the point of contact.

$$ \circlearrowright^+ \sum \tau = R f_f = I \alpha = I \frac{a}{R} $$

Rear Wheel:

$$ \circlearrowright^+ \sum \tau = T - R f_r = I \alpha = I \frac{a}{R} $$

Total Vehicle:

$$ \rightarrow^+\sum F = 2 \left( f_r - f_f \right) = Ma $$

jack action said:
Although the tire-road contact point is at rest, the axle is not. The geometry of the wheel transfers the friction force to the axle that is not at rest, thus doing work on the vehicle.

The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?

Is this an incorrect model, a failed idealization? I don't want to be leading the OP astray, because this is what I had in mind.
 
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  • #11
If a concentrated moment is acting on the free end of a rod, which is articulated or hinged at the opposite end, that rod will rotate about that hinge.
Now, think of many rods forming the rays of a circunference, as well as many hinges that instantaneously form (contact point or patch of asphalt and rubber) in rapid succesion moving forward.

The top end of each ray will happily rotate around its pivot, only that the road constrains that rotation.
The next ray-pivot pair appears, and so on.
Rather than a rotation of each open end, we obtain a forward translation of the many free ends, now named the driving axle (concentrated moment or driving torque) of the car.

Ch6Fig7C.jpg
 
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  • #12
erobz said:
Front wheels: There must be a torque that is causing the angular acceleration of the wheel. Which implies there is a force acting at the point of contact.
There is a torque. The torque is caused by the misalignment between the normal load and the normal reaction force at the tire-road contact. That misalignment is caused by the deformation of the wheel and/or ground. That rolling resistance torque can be defined as a perpendicular force acting at the tire-road contact patch when you assume both normal forces are aligned.

erobz said:
The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?
I'm not sure I understand. If you remove the wheel (no matter its size), the friction force transfers directly to the vehicle's axle, thus pushing the vehicle. Do a free body diagram (FBD) for the wheel and another one for the vehicle without the wheel. With the wheel FBD, the wheel center has the same velocity as the axle, thus the wheel also moves forward as it turns.
 
  • #13
jack action said:
There is a torque. The torque is caused by the misalignment between the normal load and the normal reaction force at the tire-road contact. That misalignment is caused by the deformation of the wheel and/or ground. That rolling resistance torque can be defined as a perpendicular force acting at the tire-road contact patch when you assume both normal forces are aligned.
I'm not talking about rolling resistance though. Does there have to be rolling resistance in the simplest physical model for the front wheels to turn? The static friction force is not capable of causing the rotation?
 
  • #14
Can someone just show the appropriate set of equations that describes the motion of the car in the diagram? I feel that conceptually it will be easier to examine a relevant chunk of the EOM.
 
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  • #15
Lnewqban said:
If a concentrated moment is acting on the free end of a rod, which is articulated or hinged at the opposite end, that rod will rotate about that hinge.
Now, think of many rods forming the rays of a circunference, as well as many hinges that instantaneously form (contact point or patch of asphalt and rubber) in rapid succesion moving forward.

The top end of each ray will happily rotate around its pivot, only that the road constrains that rotation.
The next ray-pivot pair appears, and so on.
Rather than a rotation of each open end, we obtain a forward translation of the many free ends, now named the driving axle (concentrated moment or driving torque) of the car.

View attachment 314520
Sorry, the massage is not being received here with regard to whether the force of static friction does work. The car is accelerating. There are only forces of friction acting externally on the car as a whole that I can see? If there was no friction the car cannot accelerate.
 
  • #16
erobz said:
There are only forces of friction acting externally on the car as a whole that I can see?
There need to be forces on the motor to keep it from counter-rotating (like a clown car). The car would not be very useful if, when you stepped on the gas, the motor rotated. It is the motor that does the work on the wheel
erobz said:
The axles in the model above would seem to be incapable of doing work by having the friction force transferred to them. They are non-dimensional in this idealization?
In your simplified model, there is no engine. In the simplest model just think of the engine as supplying "negative rolling friction" to the rear wheels. In fact it is supplying a torque which is countered by lessening weight on the front wheels. The work is done by the engine. Energy comes from liquified dinosaaurs. There is a net force on the car from the road and yet the road does no work.
 
  • #17
For the front wheel, the axle pushes on the wheel center. The friction force reaction must be equal to that axle force. Those two misaligned forces cause a torque.

That torque must be equal to another torque otherwise, the wheel would accelerate. (Assuming constant speed here.) What is that resistive torque? It is the one due to the misalignment of the two normal forces acting on the wheel I mentioned in my previous post.

So by pushing on the wheel center, you create a misalignment of the normal forces, and you have to push as hard as required by this resistive torque created, which depends only on the wheel/road deformation and the normal force acting on the wheel.

The force from the axle, pushing the wheel, moves so there is a displacement. Work is required to move a force.
 
  • #18
Can someone identify ##F_{net}## using work/energy?

$$ \begin{align} \int F_{net} dx &= \frac{1}{2} m v^2 + 4 \frac{1}{2} I \omega^2 \tag*{} \\ &= \frac{1}{2} m v^2 + \frac{1}{2 R^2} 4 I v^2 \tag*{} \\ &= \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \tag*{} \end{align} $$

If we differentiate both side with respect to ##x##:

$$ \begin{align} \end{align} $$

$$ \frac{d}{dx} \int F_{net} dx = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \implies F_{net} = \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) \right) 2 v \frac{dv}{dx} \tag*{} $$

$$ \implies F_{net} = \left( \frac{mR^2 + 4I}{R^2}\right) a \tag*{} $$

$$ \implies a = \frac{F_{net} R^2}{mR^2 + 4I} \tag*{} $$

So where is ##F_{net}## on the diagram?
 
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  • #19
hutchphd said:
There need to be forces on the motor to keep it from counter-rotating (like a clown car). The car would not be very useful if, when you stepped on the gas, the motor rotated. It is the motor that does the work on the wheel

In your simplified model, there is no engine. In the simplest model just think of the engine as supplying "negative rolling friction" to the rear wheels. In fact it is supplying a torque which is countered by lessening weight on the front wheels. The work is done by the engine. Energy comes from liquified dinosaaurs. There is a net force on the car from the road and yet the road does no work.
Isn't the engine in my model represented by the torque acting on the wheels?
 
  • #20
erobz said:
Sorry, the massage is not being received here with regard to whether the force of static friction does work. The car is accelerating. There are only forces of friction acting externally on the car as a whole that I can see? If there was no friction the car cannot accelerate.
That is correct, both extremes are zero static friction (wheels on ice) and maximum value of adherence.
The car does not move forward in the first case, even if the driving wheels where spinning, as you well know.
 
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  • #21
jack action said:
It is the one due to the misalignment of the two normal forces acting on the wheel I mentioned in my previous post.
Yes flexion of the tire requires work to be done. But this is not "work done by the friction force" in the usual sense of that term. This is a squirm issue. The simplest model is a train
 
  • #22
The train will not move without friction force.
The train will not move without gravity force.
The train will not move without engine.

Only the engine does work.
 
  • #23
hutchphd said:
The train will not move without friction force.
The train will not move without gravity force.
The train will not move without engine.

Only the engine does work.
But the engine is modeled as the applied torque ##T##. Are the equations in #10 incorrect? #18? What are the proper EOM?

I'm afraid if someone doesn't set the record straight soon with a reasonable set of equations, they are going to call me a "crank" and shut this kid's thread down for being a disruptor. :rolleyes:
 
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  • #24
Picking up with Work/Energy replacing the LHS by the work done by a constant torque ##2T## ( ## T## is applied to each wheel).

$$ \frac{d}{dx} \int 2T \, d \theta = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \frac{d}{dx}\frac{2T}{R} \int \, dx = \frac{d}{dx} \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) v^2 \right) \tag*{} $$

$$ \implies \frac{2T}{R} = \left( \frac{1}{2} \left( \frac{mR^2 + 4I}{R^2}\right) \right) 2 v \frac{dv}{dx} \tag*{} $$

$$ \implies \frac{2T}{R} = \left( \frac{mR^2 + 4I}{R^2}\right) a \tag*{} $$

$$ \implies a = \frac{2TR}{mR^2 + 4I} \tag*{} $$

This and the system of equations in #10 using Newtons Laws yield equivalent results.
 
  • #25
hutchphd said:
Yes flexion of the tire requires work to be done. But this is not "work done by the friction force" in the usual sense of that term. This is a squirm issue. The simplest model is a train
I agree. That's how I began with post #9. I'm merely trying to show that rolling resistance is NOT a friction force.
 
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  • #26
Not a crank. There is reason for confusion here. I have suggestions for how to think about this.

1) I am walking down the street. Is the road doing work on me? No. I am not a rigid body and it is the internal work of my muscles that does the work to deform me in such a way that the road friction pushes me forward
2) the automobil;e is similarly continuously deforming itself by forcing the rear wheels to rotate. It is not a rigid body and the road does no work. The work is done by the engine to deform the body (axles-wheels) to make friction propel me

I will not derive the equations because I am too damned lazy but the crux of the confusion is as stated. Maybe we should do a complete treatment
 
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  • #27
hutchphd said:
Not a crank. There is reason for confusion here. I have suggestions for how to think about this.

1) I am walking down the street. Is the road doing work on me? No. I am not a rigid body and it is the internal work of my muscles that does the work to deform me in such a way that the road friction pushes me forward
2) the automobil;e is similarly continuously deforming itself by forcing the rear wheels to rotate. It is not a rigid body and the road does no work. The work is done by the engine to deform the body (axles-wheels) to make friction propel me

I will not derive the equations because I am too damned lazy but the crux of the confusion is as stated. Maybe we should do a complete treatment
But I just got to the same result as #10 without considering friction at all via work/energy. That is usually worth something. That seems to be in alignment with what you are conveying here. However, considering the equations in #10 which appear to be equivalent, the net force on the car is the friction force. They seem to be different sides of the same coin.
 
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  • #28
erobz said:
But the engine is modeled as the applied torque ##T##. Are the equations in #10 incorrect? #28? What are the proper EOM?
I asked you to do the FBD for the wheels and the body separately, which you haven't done. So here we go:

Rear [driven] wheels:

$$m_w a = F_{f_r} - F_{a_r}$$
$$W_r + m_w g = F_{N_r}$$
$$I_w \frac{a}{R} = T_w - F_{f_r} R - F_{N_r} d$$

Body:

$$m_b a = F_{a_r} - F_{a_f} - F_R$$
$$W = W_r + W_f$$
$$W L_r = W_f L$$

Front wheels:

$$m_w a = F_{f_f} - F_{a_f}$$
$$W_f + m_w g = F_{N_f}$$
$$I_w \frac{a}{R} = - F_{f_f} R - F_{N_f} d$$

So you have a system of 9 equations with 9 unknowns, namely:

##a##: body acceleration;
##F_{f_r}, F_{f_f}##: rear & front friction forces;
##F_{a_r}, F_{a_f}##: rear & front axle forces;
##W_r, W_f##: rear & front weights;
##F_{N_r}, F_{N_r}##: rear & front normal forces;

And the known values:

##m_w, m_b##: wheel & body mass;
##I_w##: wheel inertia;
##R##: wheel radius;
##T_w##: wheel input torque;
##d##: misalignment distance for rolling resistance (actually equals the rolling resistance coefficient times the wheel radius, see ##F=\frac{Nb}{r}## on Wikipedia);
##F_R##: Body resistance (drag, etc);
##W##: body weight;
##L##: wheelbase;
##L_r##: CG to rear axle distance.

There might be some mistakes (maybe some + or - signs) and it could be more detailed, but that should be what you are looking for.
 
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  • #29
I appreciate you elaborating on that model, but I was after the minimum physical model that captures the basic idea. A simple physics model. Not the model an engineer would use. I had already provided the FBD for that.

I think you misunderstood what I am concerned about. If you neglect all that “real” stuff, you get the most basic result that can be used to talk about the fundamental physics that drives the model. That is where this stuff is confusing for me, and probably where the OP is confused as well. Fundamentally, there appears to be no difference between what you did and what I did. Which is good news! The difference just lies in level of the accounting. That’s not what I’m worried about though.
 
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  • #30
erobz said:
If you neglect all that “real” stuff, you get the most basic result that can be used to talk about the fundamental physics that drives the model.
If you want to neglect the "real" stuff, then neglect the wheels as well. Then you have a simple block with the friction forces acting on it (including the one for rolling resistance). You can also increase the vehicle's mass to compensate for the rotating components' inertia when accelerating (equivalent mass concept).
 
  • #31
jack action said:
If you want to neglect the "real" stuff, then neglect the wheels as well. Then you have a simple block with the friction forces acting on it (including the one for rolling resistance). You can also increase the vehicle's mass to compensate for the rotating components' inertia when accelerating (equivalent mass concept).
I feel like the wheels are a minimum model for which this concept of "work done by friction" can be looked at.
 
  • #32
If you don't want to include the wheels, you do the FBD for the vehicle as a whole, with the friction forces where the wheels are supposed to be;

If you want the wheels included, you do a FBD for the body and one FBD for each axle, and analyze them together, as I did in post #28.

If you want the powertrain included as well, you do a FBD for the body, one FBD for each axle, and one for each component of the powertrain.

Your choice.

But using wheels that move relative to the body while still considering them as part of the body won't work.
 
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  • #33
So, what you are saying the result of #10 being equivalent to #24 is a happy accident? I must consider some unknown force as pushing on the axles to solve the problem (namely F_a in your analysis)?
 
  • #34
erobz said:
Picking up with Work/Energy replacing the LHS by the work done by a constant torque ##2T## ( ## T## is applied to each wheel).
...

$$ \implies a = \frac{2TR}{mR^2 + 4I} \tag*{} $$

This and the system of equations in #10 using Newtons Laws yield equivalent results.
@erobz , what is in contradiction (in your mind) with this result?

 
  • #35
erobz said:
However, considering the equations in #10 which appear to be equivalent, the net force on the car is the friction force. They seem to be different sides of the same coin.
They are in fact exactly that and they are consistent. There is a net external force on the car provided by the difference in the rear ground-tire friction and the front.
erobz said:
So, what you are saying the result of #10 being equivalent to #24 is a happy accident? I must consider some unknown force as pushing on the axles to solve the problem (namely F_a in your analysis)?
There is a net external force on the car provided by the difference in the rear ground-tire friction and the front. This does not say that the ground does work on the car because the cantact point is static. The contact point is static because the car is not a rigid body and is continuously deformed by the engine (it takes energy to make the rear wheels turn. There is also some external "rotational friction" as expertly detailed by @jack action. But the motive energy does not come from an external force. It is internal energy from the long ago sun channeled through dead dinosaurs.
There is no unknown force required: this is not a rigid block.
This question has now been asked and answered. Finis.
 

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