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Consider the case of an accelerating motorcycle, the engine converts potential energy into mechanical energy which is used to generate a Newton third law pair of forces at the rear tire's contact patch, a backwards force exerted from the contact patch onto the roar, coexisting with a forwards force exerted by the road onto the contact patch. To simplify things, assume there are no losses in the process, no drag, no rolling resistance, etc., so that any decrease in PE (chemical potential energy of the fuel/battery) ends up as an increase in KE (kinetic energy).

The "abstract" object in this case is the contact patch of the rear tire. Although there is no relative motion between the surface of the tire and the road at the contact patch (static friction), the contact patch itself is moving at the same velocity as the motorcycle (ignoring load related deformations). This is easier to visualize if you use the center of area of the contact patch as the instantaneous position of the contact patch.

The road can't generate power, but the point of application of the force the road exerts onto the contact patch moves with the same velocity as the motorcycle. So power could be stated as the force exerted by the road times the velocity of the point of application of that force, the contact patch, which is the same as the velocity of the motorcycle (assuming a flat road).

The road also can't perform work, but the integral sum of force(versus position of contact patch) times distance the contact patch moves, could be used to calculate the "work" done that originated from the engine.

I did a rethink on this. Power = force exerted on motorcycle · velocity of motorcycle. The fact that the road is not moving doesn't affect the road's ability to exert a force on the moving motorcycle, because it applies the force to the contact patch of the tire, where the tread is not moving with respect to the road, but is moving with the negative of the motorcycles velocity with respect to the motorcycle. Due to the rolling motion and the torque from the engine, the motorcycle's tire and wheel transmit the force from the road onto the rear wheel axis with the same force from the road and at the velocity of the motorcycle. The road could be considered as part of the power transmission sequence that uses the engine's power to accelerate the motorcycle.

In this case the velocity of the contact patch is the same as the velocity of the motorcycle, but consider a drum being angularly accelerated by a spinning tire, in this case the contact patch is not moving, but the surface of the drum is. The drum could be replaced with a cable that loops between two spools, so that the acceleration of the cable at the contact point is linear. In this case, the contact patch is not moving, and power = force exerted on cable · velocity of cable.

The fact that the tire surface is not moving with respect to the road at the contact patch is the reason that a non-moving road can apply a force to a moving motorcycle.

So what I refer to as an abstract object is just a way of referring to something that moves at the same speed as the object the force is being applied to, and was my attempt to deal with rolling motion of the rear tire in the case of the motorcycle.

The point of application of force being at the contact patch does have consequences, such as a wheelie if the acceleration is sufficient.

From a strict physics viewpoint, the interface between tire and road convert angular power (torque times angular velocity) into linear power (force x linear velocity), so no net work is done. However, it is common practice to state what the rear wheel horsepower is for a motorcycle, and this can be calculated as force times speed. This can be done using a chassis dynamometer, but it is also possible to determine force through torque sensors (tranducers), allowing rear wheel horsepower to be determined in real time while riding, and some riders buy the equipment that includes torque sensors and data recording for their (racing) track bikes.

rcgldr
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  • Power is being applied to the tires, and the friction of the tires on the road applies that power to the car. Be careful which object you select when you use power equations. – David White Jan 27 '20 at 20:22
  • are you sure that the contact patch moves relative to the road? its velocity is zero at the bottom of the tire, and twice the speed at the top –  Jan 27 '20 at 20:22
  • What do you mean by " so that any decrease in PE (potential energy) ends up as an increase in KE (kinetic energy)". Is your motorcycle running down a hill or something? If it is running on a level surface there is no change in potential energy, so what's up? Are you talking about chemical potential energy of the fuel? – Bob D Jan 27 '20 at 20:32
  • @Wolphramjonny - This is different than a point on the outermost part of a tire, which moves in a cycloid pattern. I'm using the term "contact patch" as used by the tire dynamics people, the contact patch moves with the vehicle, and both the tread and the road "flow" through the contact patch. – rcgldr Jan 27 '20 at 20:32
  • @BobD - the engine is extracting potential energy from the fuel it consumes, or if it's an electric motorcycle, the engine is extracting potential energy from a battery. By no losses, I mean that PE + KE = constant. – rcgldr Jan 27 '20 at 20:33

2 Answers2

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The torque applied to the wheel by the engine causes causes a backward force on the road surface. Per Newton's third law the backward force on the road causes an equal force acting forward on the tire by the road, at the location you call the "contact patch". This forward acting force is due to static friction between the tire and the road. It will continue to equal the backward force as long as the maximum static friction force of $μ_{s}N$ is not exceeded, in which case the tire will slip. For the maximum static friction force $μ_s$ is the coefficient of static friction between the tire and the road and $N$ is the normal force acting on the drive wheel due to that portion of the motorcycles weight acting down on the drive wheel.

Assuming no drag, rolling resistance, or other external (to the car) forces are acting on the motorcycle, then the static friction force is the only external force acting on the motor cycle and is therefore directly responsible for propelling it forward. The road is, in fact, doing work on the car to propel it forward.

This fact is somewhat difficult to understand because the road is clearly not an energy source. The source is the engine that creates the backward force on the road which, in turn, creates the forward static friction force that is responsible for doing the work. So the energy comes from the power transmission system which transfers energy from the source (fuel in the engine) to the car by virtue of a series of interactions ultimately ending up as the static friction force of the road acting on the car.

Hope this helps.

Bob D
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  • I added a re-think section to my question. The road isn't moving, but it's able to exert a force onto a moving motorcycle due to the rolling motion (static friction) of the driven rear tire. – rcgldr Jan 27 '20 at 21:04
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    I think a good way to word it would be that the road is part of the power transmission system of the car, even though no power is actually provided by the road; just like no power is actually provided by the transmission elements in the car; they just help move (transmit) the power through the system to where we actually want it – JMac Jan 27 '20 at 21:05
  • @JMac I like the suggestion modified a bit. See my revision. – Bob D Jan 27 '20 at 21:33
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    This answer is not correct. The energy of the car is not increasing therefore no work is being done in the car. Work is a transfer of energy, no energy is transferred, therefore no work is done. – Dale Jan 28 '20 at 03:08
  • @Dale Just to be clear, is it your position that static friction can’t do work? – Bob D Jan 28 '20 at 07:32
  • No, that is not my position. Static friction does work in many cases. This is not one of them. P=F.v and here v=0. Static friction only does work when v!=0 – Dale Jan 28 '20 at 11:51
  • @Dale How can you say the energy of the motorcycle is not increasing when the OP says "consider the case of an accelerating motorcycle"? And what is $v!$? – Bob D Jan 28 '20 at 11:57
  • @Dale And check out the following: https://physics.stackexchange.com/questions/525337/how-does-a-car-gain-kinetic-energy/525342#525342 in particular look at JMac's answer which I believed was the correct one. – Bob D Jan 28 '20 at 11:59
  • The OP also says that any increase of KE is exactly offset by a decrease in PE. The fact that the energy does not change is explicitly stated in the problem. $v$ is the velocity of the material at the point where the force is being applied. Here, that is zero, not the velocity of the center of mass. – Dale Jan 28 '20 at 12:13
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    @Dale The PE referred to by the OP is the chemical potential energy of the engine fuel. It is not mechanical PE (e.g. Gravitational PE) hence the conservation of mechanical energy, PE+KE does not apply here. The motorcycle accelerates. It gains KE. Work is done. The only external force acting on the cycle in the forward direction causing the cycle to accelerate is the static friction force. It does the work. The energy ultimately comes from the fuel. I see nothing stated by the OP that the energy of the motorcycle doesn't change. Note: see the OP edit about rethinking the situation. – Bob D Jan 28 '20 at 12:53
  • I am not talking about conservation of mechanical energy, I am talking about conservation of total energy which does apply. The total energy is constant so no work is done. Comments are not for extended discussion. I encourage you to ask about this on a discussion oriented site like www.physicsforums.com – Dale Jan 28 '20 at 13:00
  • @Dale I’m sorry, maybe it’s me, but you are not making any sense to me. The OP says the motorcycle accelerates. That means its velocity and kinetic energy is increasing. The work –energy theorem states that the net work done on an object equals its change in kinetic energy. Therefore, net work is being done on the motorcycle. – Bob D Jan 28 '20 at 13:41
  • @Dale On the other hand, I would agree with you if the cycle were moving at constant velocity since that would mean the static friction force is being opposed by equal dissipative forces (air drag friction, rolling resistance, etc.), making the change in kinetic energy zero and net work zero. But the OP clearly states, “assume there are no losses in the process, no drag, no rolling resistance, etc.” So that doesn’t apply here. With that, I have said all I have to say. – Bob D Jan 28 '20 at 13:42
  • @BobD "The work –energy theorem states that the net work done on an object equals its change in kinetic energy. Therefore, net work is being done on the motorcycle" This is a misunderstanding of what "net work" means. The "net work" does not represent the work done by a specific force, even when that force is equal to the net force. Again, a full explanation requires a substantial discussion and cannot be done in comments. – Dale Jan 28 '20 at 17:06
  • @Dale Net work is done by the net force of all the external forces applied to the object. Once again, based on the OP statements the ONLY external force acting on the cycle is the static friction force. That makes it the net force. Dale, I'm finished here. – Bob D Jan 28 '20 at 17:43
  • But you cannot use the net work to determine the physical work done by any individual force. They are not the same even in the case where the net force equals an individual force. This is your key mistake in your answer. You are wrong to attribute the “net work” to work done by the individual force at the road. In other words, net work gives you the work that would be done if the net force were applied to the center of mass. It does not tell the work done by an individual force at its individual point of action. Again, I encourage you to go to a discussion oriented forum. – Dale Jan 28 '20 at 17:48
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Power is defined as: the rate at which work is done or the rate at which energy is transfered from one place to another or transformed from one type to another. https://physics.info/power/

“the rate at which work is done” is given by the formula $P=\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of application of the force. In the example of a lossless motorcycle the point of application of the force is the bottom of the tire, which has $\vec v= 0$.

However, it is exactly this point which is in question. Is the correct velocity for calculating the power equal to the velocity of the material at the contact patch or is it equal to the velocity of the contact patch? Therefore, to resolve this we will look at the other parts of the definition to see if one interpretation of $\vec v$ is more consistent with the remainder of the definition than the other.

“the rate at which energy is transfered from one place to another”. Due to the conservation of energy, if energy were transferred at the contact patch then the energy of the car would change. Since the energy of the car does not change it is clear that the rate at which energy is transferred across the contact patch is zero. So by this part of the definition the power is zero. This is consistent with $\vec v$ representing the velocity of the material at the contact patch, but inconsistent with $\vec v$ representing the velocity of the contact patch.

There do exist passive devices which transmit power from one location to another, such as shafts, ropes, gears, and levers. However, in all such devices there is one location on the device where positive $P$ is done and another where an (ideally) equal amount of negative $P$ is done. This not the case at the contact patch.

“or transformed from one type to another”. At a typical contact patch the only transformation of energy is from mechanical energy to thermal energy. By assumption that is zero in this case. In this problem the only transformation of energy is in the engine where energy is transformed from potential to mechanical. It therefore makes sense to speak of the power of the engine despite the constancy of the vehicle’s total energy. But at the contact patch any energy is mechanical and remains mechanical. So this portion of the definition is also consistent with $\vec v$ representing the velocity of the material at the contact patch, but inconsistent with $\vec v$ representing the velocity of the contact patch.

Therefore, both of the other parts of the definition of power indicate that the power provided by the contact patch is zero. This corresponds with the definition that $\vec v$ is the velocity of the material at the contact patch.

The reason this question tricks so many people is because the force from the road does change the motorcycle’s momentum. However, it is important to know that momentum and energy are distinct concepts. They are related but not the same. A force is the rate of change of momentum, not the rate of change of energy. Therefore it is possible for a force to change an object’s momentum without changing its energy. This is one example, though there are many other similar examples.

In the end the mechanical power transferred by a force $P=\vec F \cdot \vec v$ is always calculated using the velocity of the material where the force is applied, which is zero for the motorcycle example.

Dale
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  • Note that I updated my question with the added separated section to note the dilemma of the term "contact patch" since it refers to the interface between two objects, and the "contact patch" may or may not be moving. A "contact patch" only has a velocity if the tire it references also has a velocity, so it is situation specific. – rcgldr Jan 28 '20 at 07:13
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    By the same logic, is no work is done either $W=\vec F \cdot \vec s$ ? If no work is done, then what is responsible for the increase in the motorcyles KE as it accelerates (in a zero loss situation, PE + KE = constant, so a decrease in PE is matched by an increase in KE: ΔPE + ΔKE = 0)? – rcgldr Jan 28 '20 at 07:25
  • The velocity of the contact patch is not relevant, only the velocity of the material at the contact patch matters. Since $W=\int P \ dt$ if $P=0$ then $W=0$. For details see https://www.physicsforums.com/threads/is-there-any-work-done-by-static-friction-when-accelerating-a-car.983216/#post-6287428 – Dale Jan 28 '20 at 12:04
  • @Dale I like your argument, but find it interesting that the gain of KE of the car can surely be $calculated from$ $$\Delta E_k=\text{frictional force}\times \text{distance moved by contact patch},$$ in other words from what might be called the pseudo-work. – Philip Wood Jan 28 '20 at 14:39
  • @Dale And the equation I just offered can be justified because we are agreed that the frictional force adds momentum to the car, and hence kinetic energy, because $E_k=p^2 /2m$. Specifically, for motion in a straight line,$$dE_k=mvdv=m\frac{dv}{dt} vdt=Fdx.$$ And this is done without recourse to the $concept$ of work. – Philip Wood Jan 28 '20 at 15:04
  • @Phillip Wood you are correct. The important thing here is to recognize that a change in KE is not the same thing as work. Since there are multiple kinds of energy KE can change without work, as is the case here. – Dale Jan 28 '20 at 15:09
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    @Dale "KE can change without work" That's a rather bolder claim. I need to be convinced that we can't attribute the car's gain in KE to work done by the engine torque. – Philip Wood Jan 28 '20 at 15:39
  • @PhilipWood we cannot usually do "convincing" in comments on SE. I encourage you to visit on www.physicsforums.com where we can do a more in-depth discussion. But certainly the internal engine torque does provide work on different subsystems internally. That is not the issue. The issue is the road and the car as an overall system. The road does 0 work on the car. The force from the road changes the momentum, but not the energy of the car, therefore it does no work. No work means no energy is transferred into or out of the car. The KE changes without energy moving across the system boundary. – Dale Jan 28 '20 at 18:08
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    @Dale I would argue that the energy is reflected off the system boundary, which is a direct interaction with it. The road doesn't change the net energy in the car, but it directly facilitates the energy transfer from the rotation of the wheels to the linear kinetic energy of the car. – JMac Jan 28 '20 at 18:17
  • @JMac I have no problem calling it a "direct interaction" or "directly facilitates". It is just not "work" which must involve a transfer of energy and that transfer does not happen here. I am not sure about energy reflection, I would have to work out the math before I would argue either for or against that. – Dale Jan 28 '20 at 18:27
  • @Dale - I updated my question with a 3rd section to note that no work is done, but that the calculation of static friction force times velocity of motorcycle is a valid way to calculate the rear wheel horsepower of a motorcycle, and that there are instrumented recording kits that use torque sensors to calculate static friction force, and then use that force times velocity to calculate rear wheel horsepower in real time, to be recorded with other data into storage. These kits are used mostly by riders running on road courses. Similar kits are used during development of vehicles. – rcgldr Jan 30 '20 at 14:27
  • @rcgldr those devices are mounted on the axle. They measure power at the axle not at the ground, and they only measure power in the reference frame of the device which is different from the power in the reference frame of the earth. – Dale Jan 30 '20 at 16:41
  • @Dale - Those devices measure torque, power is is a calculation, and once calibrated, produce nearly identical numbers seen on a dynamometer. They can also include force calculations. There are power estimator calculators that calculate an estimated power based on 1/4 drag elapsed time or trap speed, and these would be earth based reference frames. The accuracy of the estimators can be pretty close if the parameters for a specific vehicle and environmental conditions are taken into account. – rcgldr Jan 30 '20 at 18:15
  • @rcgldr sure, but my point is that it is not relevant to my answer. Those devices do not indicate that my answer is incorrect. They are measuring something different than what we are discussing here. You clearly only want answers that agree with your preconceived view. It is wrong, but it is a common mistake, so you will find plenty of people making the same mistake and will be willing to tell you that you are right. That is what has happened here. The discussion on PF was much better, but you ignore all of that and come here and immediately accept whomever agrees with you. Your loss. – Dale Jan 30 '20 at 18:35
  • @Dale - I've already conceded that using "contact patch" was a bad idea, and that no net work is done by static friction. This is different that stating power = force · speed. As another example, Formula 1 race cars have both torque and load sensors. The load sensors are integrated into the suspension and can determine forces in all 3 dimensions, including the externally sourced forward force of the axles onto the suspension (the drive train is protected from linear loads). My only point here is the concept of driven wheel horsepower is valid and used by people in the industry. – rcgldr Jan 30 '20 at 21:55
  • @rcgldr oh, sorry, I thought you were going backwards again. Yes, P=F.v is valid, although due to the plentiful confusion it would be good to be explicit about which P, F, and v you mean. On an axle or a driveshaft there certainly is power. I am not sure how the power distributes through the wheel. I have been looking to see if anyone has images of stress distributions in a wheel or axle. – Dale Jan 30 '20 at 22:23