Suppose you are in a battery-powered car stopped on an ordinary level road. Then you start driving. The question is, physically speaking, what makes the car go?
There are two possible answers:
It turns out that both answers are 100% correct! If the engine isn’t working, the car won’t go. And if you don’t have suitable friction between the tires and the road, the car won’t go.
If we look more closely, we see that two different questions are being answered:
Some people are shocked to hear that the energy comes from inside the car while the momentum comes from outside the car. But it’s true.
Let’s calculate the energy transfer explicitly. Doing the calculation and interpreting the result is particularly simple for an observer in the center-of-mass frame. In this frame, both the car and the earth are initially at rest. Then the car gains momentum p and gains kinetic energy p^{2}/2m, where lower-case m is the mass of the car. Meanwhile, the earth receives momentum −p and gains kinetic energy p^{2}/2M, where upper-case M is the mass of the earth. All of this energy (p^{2}/2m + p^{2}/2M) is taken from the battery via the engine. Of this, an amount p^{2}/2M needs to be transferred across the car/road boundary. This term is really tiny, because M is so big. The important thing is the extraction of a certain amount of energy from the battery, by processes entirely internal to the car. The amount of energy that must be extracted and converted is p^{2}/2m + p^{2}/2M or approximately p^{2}/2m.
If you are not interested in details, you can skip this part.The calculation is somewhat trickier for an observer moving with velocity u relative to the center of mass. There will be an additional term in the energy equations. The car will gain an additional kinetic energy −u·p and the earth will gain an additional +u·p. This represents an additional transfer of energy across the car/earth boundary. This additional transfer could be positive, negative, or zero, depending on the direction of u relative to p. As you can see, this term is independent of the masses involved. For rather modest velocities u, this term is larger than the amount energy extracted from the battery, which is the same as before, namely p^{2}/2m + p^{2}/2M or approximately p^{2}/2m.
This additional, frame-dependent term has no physical significance. (Frame-dependent terms never do.) In any frame, the important energy is the energy extracted from the battery, namely p^{2}/2m + p^{2}/2M or approximately p^{2}/2m. The additional frame-dependent transfer does not require any additional energy from the battery, and does not require any additional force to be exerted at the car/road boundary.
The math for this is:
ΔKE_{(car)} =
(−m u − p)^{2} 2m −
(−m u)^{2} 2m =
m^{2} u^{2} −2m u·p + p^{2} 2m −
m^{2} u^{2} 2m =
−u·p +
p^{2} 2m ΔKE_{(earth)} =
(−M u + p)^{2} 2M −
(−M u)^{2} 2M =
M^{2} u^{2} + 2M u·p + p^{2} 2M −
M^{2} u^{2} 2M =
u·p +
p^{2} 2M ΔKE_{(total)} =
p^{2} 2m +
p^{2} 2M (1)
Point to remember: Energy is important. Momentum is important. When in doubt, ask yourself
These are two different questions, and might have two different answers. When a car accelerates, a goodly amount of momentum is necessarily transferred across the car/road boundary. In contrast, in the center-of-mass frame, virtually no energy is transferred across the car/road boundary. In any frame, center-of-mass or otherwise, the energy that matters is the energy extracted from the battery, which we can divide into an important car-related piece p^{2}/2m and an unimportant earth-related piece p^{2}/2M.
At this point, some people are still skeptical. They say the car was initially not moving, and later it was moving, and the "movingness" had to come from somewhere. The car can’t create motion from scratch. The car can’t create kinetic energy from scratch.
Well, let’s be careful. Terms like "movingness" and "motion" aren’t precise physics terms. If by "motion" you mean momentum, then yes, we’ve agreed that the car cannot make its own momentum. The momentum needs to be transferred across the road/car boundary. On the other hand, if by "motion" you mean kinetic energy, that’s a different story. Given a supply of momentum, the car most certainly can create its own kinetic energy, by processes internal to the car. The kinetic energy is not created from scratch, but it is converted from other forms:
There is no law that says kinetic energy is conserved. There is a law that says the total energy is conserved, but the various subcategories of energy are not separately conserved.
Some people are still not convinced. They say a baseball cannot acquire kinetic energy unless someone transfers energy to it by pushing on it, i.e. by doing work on it. There’s even a theorem about this, called the "Work / Kinetic Energy" theorem.
Well, what’s true for baseballs isn’t necessarily true for cars. If you look at what the Work/KE theorem actually says (reference 1 and reference 2), you will find that it applies only to point particles. It does not generally apply to objects that have internal structure. And it absolutely certainly does not apply to objects that have on-board energy conversion machinery!
If you’re desperate, you can contrive a scenario where the Work/KE theorem applies, such as the following:
That’s an amusing model system, but it is completely contrived. It has essentially nothing to do with how an ordinary car acquires its kinetic energy and momentum. The thrust of the car deforms the road a little bit, but not enough to transfer any significant energy to or from the car.
A much better model of what makes a car go is the following:
Of course, the rod will deform the wall – but not very much. We can think about the deformation, x, in terms of Hooke’s law, F = k x. The energy stored in this deformation is E = ½ k x^{2} or equivalently E = ½ F^{2} / k. Brick walls are proverbial for their large k values, so for any particular force F, the energy transferred to/from the wall is very tiny.
The deformation involves a force F and a distance x. We can contrast this with what’s going on inside the car: you are applying a force F during the entire acceleration phase, during which time quite a large length of rod passes through your hands – incomparably large compared to x. That should make it clear that energy conversion processes internal to the car are incomparably more significant than any energy transfer to/from the external world.
Because a real car has internal structure, the Work/KE theorem does not apply to the car as a whole. The car gains kinetic energy even though no work is being done on the car by any outside agent. If you want to apply the Work/KE theorem to an object with internal structure, you need to decompose the object into pointlike components, and apply the Work/KE theorem to each component separately. This may or may not be worth the trouble.
Energy is primary and fundamental. Momentum is primary and fundamental. Work is an entirely secondary notion. If work helps you understand what the energy is doing, fine. If not, forget about work.
Here’s a problem that is often discussed in physics classes. It is basically a simple problem, but it can be confusing if you analyze it the wrong way. Since it is a thinly-disguised version of the problem we’re discussing, we might as well deal with it now.
An ice skater is initially at rest, standing on the rink near a brick wall. Then she pushes against the wall and goes sliding away. Once again there are two different questions, and two different answers:
The wall does not deform significantly. The Work/KE theorem does not apply to the skater as a whole, because the skater has very significant internal structure.
My recommendation would be to not bother with the Work/KE theorem at all. You don’t need it in this case. If you want to figure out how fast the skater is going, you can get that from the momentum, as follows:
However, if you insist on pursuing the Work/KE angle, the rules are the same as always: you must consider the various subparts of the system and apply the theorem to each part separately. To give you glimpse of how this might be done, we can approximate the skater as a two-part system: Part "A" (arm) and part "B" (body and everything else).
The arm pushes against the wall, and the wall pushes back with a force F. No significant work is done at this boundary, because the displacement is so small. Remember, this is the proverbial brick wall. Meanwhile, the arm pushes against the body at the shoulder joint. This is the same force F, if we neglect smallish terms like the ratio of arm mass to everything else. We see that significant work is done by the arm on the body. The shoulder moves a very significant distance while subjected to the force F. Roughly speaking, this is where the kinetic energy of the body comes from. But since the arm is still part of the skater, we must conclude that this kinetic energy was produced internally. No significant work or energy of any kind crossed the wall/skater boundary.
Here’s another amusing example: water rockets. People often ask me why the rocket needs both water and compressed air. To a good approximation, the air provides the energy and the water provides the momentum.
For any given volume, given a limit on max operating pressure, you can store vastly more energy in pressurized air than in pressurized water. That’s because virtually all of the water goes in while the pressure is zero ... it just pours in as a liquid. Then it takes only a tiny bit more liquid to raise the pressure a great deal. In contrast, most of the air goes in after an appreciable pressure has been built up, so you’re doing a lot of work against this pressure.
Meanwhile, it doesn’t do much good to have a big supply of energy if you don’t have a good way to transfer momentum. Air doesn’t weigh very much, so just throwing the on-board air out the back would not transfer a useful amount of momentum. Water is a thousand times denser, so throwing water out the back works a lot better.
You can build your own water rockets from scratch. Details are available at various sites on the web. Sometimes you can buy a complete high-quality system at the toy store.
In any case, make sure you hydrostatically test your rockets. Choose a design pressure, and test to make sure the rockets can withstand the maximum design pressure plus a safety margin. Test them before the first flight, and re-test them whenever there is any suspicion they have been damaged or otherwise weakened. Never exceed the design pressure when there is air in the system – only during hydrostatic testing. See section 3.5.
In section 3.4 we noted that at any given pressure, you can’t store much energy in a tank full of pressurized water. That’s a disadvantage if you’re building water rockets, but in many other situations it can be used to great advantage.
Imagine you want to test a system of large tanks and pipes that are supposed to hold highly-pressurized gas. You want to make sure everything is strong enough, but you’d better not test it using pressurized gas. If/when the gas pressure exceeds the strength of the tank, the result might be a small leak ... or it might be a violent explosion.
The proper technique is called hydrostatic testing. Fill the system with a liquid (e.g. water or hydraulic oil) and get rid of any air pockets. Then pressurize the liquid to the desired working pressure, plus some safety margin. If something breaks, you will get a leak, not an explosion.
Earlier we said it was important to have friction between the tires and the road. Let’s look into that. It is useful to recognize several types of friction that must be considered separately:
If an object holds itself by static friction to something stationary, there is no energy transfer. The work is F · dx, where dx is zero.
If an object attaches itself by static friction to something moving, that’s a different story. Take for instance a cable-car attached to the moving cable by static friction. Energy is transferred via the cable to the car as it gets hauled up the hill. This energy transfer is non-dissipative.
If you apply a force to an object in a static-friction situation, for moderate forces it moves in proportion to the force, but only in a very small proportion. It’s like a spring with an enormous spring constant, x = F / k, where k is very large.
If the force F gets too large, the object will start sliding, and it is no longer a static-friction situation. Once it starts sliding, it will probably keep sliding, and x will grow without bound. The coefficient of static friction µ_{s} is a dimensionless number that tells us (very approximately) the maximum force that the object can resist without sliding. Therefore the frictional force F_{f} is subject to the inequality
|F_{f}| ≤ µ_{s} |F_{n}| (2) |
where F_{n} is the normal force holding the object to its mate. The coefficient of static friction depends on the materials involved and also on how they were prepared. (Ice will hold against Teflon if they’re both rough enough.)
Note that equation 2 won’t tell you how much force is developed in a typical static situation; it only tells you the maximum force that could be developed. You can’t use the coefficient µ_{s} to infer anything about the spring constant k. Also note that equation 2 is not a vector equation. It relates the magnitude of two vectors that are in completely different directions.
A good example of dynamic friction is a block sliding across a table. If you look at the microscopic details, this process is very complicated. It involves all sorts of asperities on the top of table and the bottom of the block. Let’s consider just one of these asperities. It is part of the table. It reaches up and makes contact with the block. This is locally and temporarily in static contact with a moving object. Energy is transferred, and stored in the springiness of the asperity. Very soon thereafter, the asperity breaks free and wiggles like crazy, dissipating energy in the form of sound, ultrasound, and heat. Then the process repeats itself. Meanwhile something similar is happening with thousands or millions of other asperities.
Assume a steady sliding motion, propelled by a steady force F. Energy is being dissipated. It is hard to describe this situation in terms of work = F · dx. We know the macroscopic force that the block exerts on the table, and the table exerts on the block. But what shall we use for dx? The table isn’t moving, so you might think dx is zero. Meanwhile the block is moving, so you might think dx is the entire distance the block travels. But neither of those is right. In order to understand dissipation via the Work/KE theorem, we would need to know the movement dx of the point(s) at which the force is applied and we definitely don’t know that. We know the macroscopic force and the macroscopic distance, but that doesn’t tell us anything useful about the forces and the motions of all those microscopic asperities. See reference 1.
A rolling object is an interesting intermediate case. It looks dynamic, but during natural roll it has more in common with static friction than dynamic friction.
Throughout this section, unless stated otherwise, we assume the object is undergoing natural roll i.e. with negligible slipping; more general situations will be discussed in section 4.4.
The tire as a whole is moving relative to the road, but at the one place that matters, there is no relative motion. This is the proverbial place where the rubber meets the road. The rubber is locally and temporarily in static contact at this point.
The following analogy may help you visualize the forces involved.
In the frame where the road is (essentially) at rest, a rolling contact can transfer momentum but (essentially) no energy.
It is possible to distinguish two kinds of quasi-static rolling friction, namely the kind perpendicular to the rolling motion, which is important during cornering, and the kind along the direction of motion, which is important for providing straight-line thrust.
The rolling motion of the tire is never exactly like static friction, hence the “quasi” in the name quasi-static rolling friction. There are several reasons for this.
For one thing, the tire has a flat spot of nonzero size where it touches the road. You can’t change that part of the tire from round to flat without bending the tread and stretching the sidewall. This bending and stretching leads to frictional dissipation, even when the tire is conforming the natural roll scenario as closely as possible. Indeed it’s hard to give a 100% sensible definition of natural roll. Things would be simple if we could say that the linear velocity v of the center of the axle is matched to the angular velocity ω in the “natural” way, namely v = ω r where r is the radius, but it’s not 100% clear what to use for the effective radius, because of the flat spot.
Things get even more non-ideal if we look at intermediate amounts of skid.
The amount of frictional force in this case is probably more-or-less what you would expect for dynamical friction, much less than what you would expect for static friction or quasi-static rolling.
Notice that you can’t figure out the energetics just by observing center-of-mass motion of the car. You can observe the distance dX the car travels, and you can observe the acceleration and infer the traction force (using F=ma). So there is an energy FdX involved. In the case of pure skidding you know this is all dissipated at the tire/road interface. In the case of pure rolling none of it is dissipated at the tire/road interface; it might be dissipated in the brake shoes, or it might not be dissipated at all if the car has regenerative braking. In the intermediate case, you have no idea how much dissipation there is unless you know Y, the amount of skid.
Note that “good” tires should have large values of both kinds of friction discussed in section 4.3, and small values of the dissipation discussed in this section. Without additional context, it is hard to know what is intended if somebody talks about “the” coefficient of rolling friction.
So far, in our discussion of what makes the car go, we have mainly covered what the energy is doing and not doing. Now let’s have a look at the momentum.
Up to this point, we have considered a situation where the car could transfer arbitrarily large amounts of momentum to the earth without difficulty, i.e. without transferring any significant energy in the process. It is amusing to look at what happens when this is no longer true.
Here’s a demo I learned from Scott Goelzer. Make a platform out of lightweight but stiff cardboard. Place it on rollers. Drinking straws make nice lightweight rollers – assuming they’re straight and have a nice round cross-section. Place a model car on the platform. Start the motor. The platform goes flying. The car doesn’t go anywhere (not until the platform is completely gone and the car falls onto the underlying table).
We can gain some insight by applying the p^{2}/2m rule again. In an unaccelerated frame, such as the center-of-mass frame, if the car gains momentum p the platform must be given momentum −p. The energy transferred to the platform represents practically all of the available energy, because the mass of the card is so small (compared to the car). This is quite a contrast to the car in direct contact with the road+earth system, where the energy transferred to the road was practically nothing, because the mass of the earth is so large (compared to the car).
Point to remember: Energy and momentum are very different things, but they are both important. If the car doesn’t have a supply of energy, it won’t go. If the car doesn’t have a reservoir of momentum (e.g. the earth) and a good coupling thereto, it won’t go.
The idea of momentum transfer is intimately connected with the idea of pseudowork, as discussed in reference 1.
We’ve been using a battery-powered car as our main example. This was chosen for a reason, to avoid various quibbles about the workings of a gasoline-powered car. But they are somewhat interesting quibbles, so let’s discuss them now.
A gasoline engine does not simply take in gasoline and put out kinetic energy. Here’s a somewhat more complete list of what’s needed:
In a typical car, the tank of gasoline is considered the reservoir of energy. It’s not super-clear what that means in detail, but clearly, in ordinary terrestrial applications, fuel is the hardest-to-obtain item on the foregoing list. It’s also clear that the reactants (octane+oxygen) have markedly higher energy than the reaction products (H_{2}0 and CO_{2}).
So we can say (somewhat sloppily) that there is energy stored in the gasoline, or we can say (somewhat more precisely) that there is energy stored in the gasoline + oxygen.
We can even go so far as to say that there is energy stored in the chemical bonds in the octane molecule (C_{8}H_{18}) and the oxygen molecule (O_{2}). But be careful! The way in which energy is stored in those bonds is slightly non-obvious. You should not think that energy is stored in those bonds the way helium is stored in a helium balloon. If you pop the balloon, the helium flies out. But if you break a chemical bond, the energy does not fly out – that’s diametrically the wrong picture.
A chemical bond containing a high level of energy is a very weak bond. If you added any more energy to it, the molecule would fall apart entirely. Conversely, a chemical bond containing a low level of energy is a very strong bond.
Combustion consists of breaking relatively weak bonds and replacing them with relatively strong bonds. It is the final bond-making step that liberates energy. Figure 1 is a picture of the energy levels of the chemicals. The reaction proceeds from left to right. When the chemicals fall from the initial state (high chemical energy) to the final state (low chemical energy), the excess energy is liberated and can be used for other purposes such as propulsion. It’s just like the way a heavy book liberates energy when it falls off a high shelf.
________ / ions + \ ____________/ radicals \ C8H18 + O2 \ \ \___________ CO2 + H2O
There is not much to be gained by arguing whether a particular bond has an energy that is "greater than zero" or "less than zero". In classical physics, you can choose the zero of energy to be anywhere you like. What matters is whether the reactant molecules have higher energy than the product molecules. Molecules with fewer and/or weaker bonds make good fuels.