It seems to me that it’s probably an issue of too much friction, not too much inertia. The friction on a non-moving object (static friction) is generally greater than the friction on a moving object (kinetic friction), so if you’re trying to start all the cars at once you’re fighting a greater force (the static friction of all of them) than if you’re starting them one at a time (where you’re fighting the static friction of one plus the kinetic friction of all the ones you’ve already started).
If we imagine a world of zero friction, then even a tiny force would be enough to get all the train cars moving, albeit with a very tiny acceleration (a = F/m). Sure, by applying the force F to just the first car you get a greater initial acceleration of that first car, but once you’ve got the whole train moving its speed should be the same as if you’d applied that same force to the entire train for the same amount of time (again, this is ignoring friction). Specifically its speed would be the integral of the applied force with respect to time (which gives the total change in momentum), divided by the total mass of the train.
No doubt the Perfect Master knows all this and is just dumbing it down…
Assume each loaded box car weighs 100,000 pounds. Assume further that there are 100 cars in the train. When the LAST car in the train begins to move, the coupling that holds the engine to the FIRST car has 10,000,000 pounds of stress on it. Once all the cars are in motion, the stress of course is greatly reduced.
That’s why trains start very slowly even if they have enough power available to go faster.
Your argument doesn’t really make sense to me, Fir na tine. The strength of the couplings might limit the amount of force the engine can apply without tearing the coupling off, but the question I’m addressing isn’t “Why limit the amount of force applied by the engine?”, it’s “Why have slack in the couplings so that you can start the train one car at a time instead of all at once?”
The fact that the full train weighs 10 million pounds doesn’t mean there’s “10 million pounds of stress on the first coupling.” The weight of the train is a force in the vertical direction, which is entirely counterbalanced by the normal force of the ground supporting the train. The external forces in the horizontal direction are only the force of the engine and the frictional force.
There’s an internal force between the first coupling and the rest of the train, but (again ignoring friction) this is basically equal in magnitude to the force exerted by the engine (actually a little less because the coupling itself has some mass, but this is negligible compared to the mass of even one freight car).
Also, it is worth noting that if there is kinetic friction being induced in a moving vehicle of any kind, it is not going to move very far at all. It might seem counter-intuitive, but you only have kinetic friction if wheels are slipping and the vehicle is not moving forward.
This is true of where the wheel makes contact with the ground (or I guess I should say “with the track” in this case), because a point on the wheel’s surface actually has zero instantaneous velocity at the point where it touches the ground. Here is a good illustration of this.
There will still be friction on the axle, though. But even for an equivalent frictional force the work done to slow the train would be much less, because it’s applied over a much shorter distance (due to the axle having a much smaller radius than the wheel).
I had never actually thought about it before, but tim314 is right. In a perfectly frictionless situation (or at least, a situation lacking all of the friction we don’t like, but still with all the friction we do like), it would take just as long for the prime mover to get the train up to speed by pulling all of the cars at once, or clanking them along one at a time.
I started to address this comment, then saw your followup. There are a couple different issues here.
First off, static friction is friction where the contact between two objects does not slip. Kinetic friction is slip along the surface of contact. Trains use steel wheels on steel tracks. The engines’ wheels actually have varying amounts of slip and stick, depending upon how much torque is applied to make them move. When the engine tries to start from rest, the wheels actually slip a bit as the engine tries to turn the wheels to move forward.
The wheels on the cars tend to operate more with static friction. They roll rather than slip along the track.
However, you do point out that the axles do slide in order to rotate. This is accommodated with bearings, which are designed to limit the friction.
What makes it hard for a train to get moving all at one time? The engine only gets so much friction between the steel wheels and the steel track. As it tries to apply load, the wheels quickly start spinning, thus only providing kinetic friction contact. There’s only so much torque that can be applied, so only so much force the engine can give. Multiple engines working in tandem gives more ability, but the limit still applies, just at a higher level.
If the engine (suppose just 1 for simplicity) tries to pull the whole train at one time, it is not fighting static friction of the wheels vs kinetic friction of the moving wheels - rolling wheels are also static friction. How much does the friction in the bearings vary?
Thinking about it, the only force resisting movement is friction. When the engine applies a pull, that pull is resisted by the internal friction of the parts through to the friction of the wheels with the track. If the engine tries to pull all cars at once, that friction combination is sufficient to prevent motion. But if the engine starts itself moving and achieves a bit of momentum, then when it stretches the first coupling, it only fights the resistance (i.e. friction) of one car. Hmmm.
I guess I agree that it is friction, but it is the internal friction of parts, not friction between wheels and track. Which you didn’t actually say, so yeah.
That all does make me wonder, even apart from friction at the axle, just looking at a wheel on a rail: is there a static rolling friction and a dynamic rolling friction, analogous to static and dynamic sliding friction? I don’t recall ever coming across that in any physics course.
If it’s rolling without sliding then the frictional force exerted on the wheels by the ground will be static friction, as mentioned above. Only when it starts sliding will there be kinetic friction between the wheel and ground.
In fact, it is the friction with the ground that causes the wheels to rotate*. The cars will start sliding exactly when the maximum static friction that can be exerted is not enough to keep the wheels turning at the necessary speed to keep up with the train’s forward motion.
This is true for all the cars except the engine. The engine is different because it’s applying its own power to turn the wheels, and then it is the friction of these wheels with the ground that actually propels the train forward – sort of the opposite of how it works for the rest of the cars.
(By “ground” of course I mean “track”. I tend to reason about a train by thinking of an automobile with a series of carts trailing behind it, neglecting the fact that it’s on rails. ;))
Yeah, I suppose my point about having to overcome the static friction of one car at a time is really only relevant to the internal friction, since the friction between the wheels and the track should always be static friction if there’s no slipping.
That said, I should point out that just because the friction exerted by the track on the wheel is always static friction, it doesn’t mean it’s constant. The maximum static friction that the surface can exert on a train car of a given weight is constant, but the actual friction exerted (assuming no slipping) will be whatever it needs to be to keep the wheels spinning at the necessary speed. The friction on the axle provides some torque that will need to be canceled by the torque exerted by the frictional force exerted by the track. So more friction on the axle means a greater frictional force must be exerted by the track as well.
That said, the frictional force exerted by the track on the edge of the wheel need only be a fraction of the frictional force exerted on the axle in order to counterbalance the torque, because the axle has a much smaller radius than the wheel. So I think probably it’s friction on the axle that matters more.
The torque exerted on the edge of the wheel does need to do a bit more than counterbalance the torque on the axle when the wheel is speeding up (as opposed to maintaining constant speed), but the difference will depend on the moment of inertia of the wheel and axle, so it scales with their mass, not the much larger mass of the whole train car.
Don’t trains ever have to stop on slight uphill grades? If so, there would be no slack in the couplings. Does that mean the full train can’t get started again from that spot? Do they have to back up a bit before going forward?
No, that’s not what I meant. A rolling cylinder will come to a stop. It’s not due to sliding friction, because it isn’t sliding. And the part touching the surface is stationary at the point where it’s touching, and moving vertically in the vicinity of that point, so static friction won’t come into play either.
Loss of energy would come from the surface and/or cylinder compressing slightly at the point of contact. This could easily differ with material independently of the sliding or static friction.
A WAG, but if the caboose (when they had them) or the cars at the end of the train were the ones applying the braking, rather than the engine, the front of the train could still be compressed, allowing the engine to start the train one car at a time to begin. There would have to be some coordination to release the brakes of the back cars at the right time.
Tim314 is correct, “rolling” is a process that involves static friction, because the point of contact is not slipping. That is what makes an object “roll”.
However, an object can spin about its axis and then come in contact with a surface. If the spin speed is different than the translation rate, then there will be slippage at the moment of contact until friction synchs the speeds. The friction at work in that situation is kinetic friction, until the speeds synch, at which point it becomes static friction.
A spinning object at a spinning faster than its translation rate when it makes contact would appear to be rolling and be experiencing kinetic friction. How long in undergoes kinetic friction will depend on the size/mass of the object, the difference between the speeds, and the surface materials.
This is actually a good point that feeds into my next observation.
Suppose you did have the train at rest but pulled tight, no slack. When the engine begins to try to drive, the wheels can only apply so much torque before they begin to slip, at which point the force the engine can apply drops a bit. Therefore, an engine has only so much force it can apply. Above that speed, the wheels slip along the track, and it doesn’t matter how fast they go, the friction force does not increase.
Meanwhile, the internal friction of the mechanisms is a responsive force. Friction resists in equivalence to the force applied until the applied force exceeds the amount the friction can hold, at which point the motion begins. So with the train pulled tight, all of the internal mechanism friction is reacting in unison to the force applied by the engine. Since the engine can only give so much pull and then it maxes out, how many cars does it take before the sum of the cars give enough friction that the engine cannot overcome the threshhold, so the train cannot move?
Now consider the train with slack. When the entire train including the engine is at rest, there is no momentum. As the engine begins to move, it builds up speed and thus momentum, at which point the coupler draws tight. The momentum of the engine then gets coupled to the momentum of the first car, at which point it provides the kick to get the first car moving. Now the engine keeps applying force, so both the engine and first car get more speed and more momentum, until the second coupler draws tight, and then the momentum is linked to the third car. Etc down the line.
This process is how momentum plays a role. If all the couplers were tight, there would be no momentum transfer, but the loose couplers allow momentum transfer to help overcome the internal friction of each car’s axles.
Correct?
I’m trying to figure out where “inertia” enters the picture. Inertia just means the train needs applied force to start moving, which is coming from the engine. Right?
Consider a cylinder that isn’t quite circular, but instead is a regular 1000-sided polygon in cross-section. We want to describe that as a circular cylinder with coefficients of friction. This is what normal friction is, in large part, where microscopically, surfaces aren’t really flat, and those peaks and valleys contribute to friction.
The cylinder is sitting on a flat surface, on one of its sides. If you tip the cylinder so one of its ends is higher, the usual friction quantities come into play. How far you tip it before it slides depends on the static friction.
Now tip the surface in the direction to roll the cylinder. If the static friction is large enough, it won’t move until you get to a certain angle, then it will start rolling. This is what will be described by static rolling friction.
The cylinder isn’t slipping, so it has nothing to do with sliding friction. It also doesn’t depend on static friction. The static friction could be arbitrarily large, and the cylinder will start rolling at the same angle. The sliding friction could also become arbitrarily large, and not affect when the cylinder starts rolling. On the other hand, a 2000-sided “cylinder” would start rolling at a smaller angle, and so would have a smaller static rolling friction.
So this is a property analogous to normal static friction, which is independent of normal static or dynamic friction. You wouldn’t really have an N-sided cylinder, of course, but there will still be imperfections in the surface contributing to this static rolling friction.
An old train guy told me they started that way because of inertia, a train doesn’t have enough traction to pull the entire train from a dead stop. They would start the train until they felt a jerk and the first car started to move, then slow DOWN, until the last car started to move, then they would speed up a little bit at a time. On longer trains they would speed up and slow down several time until the whole train was moving. He told me it was because if they didn’t slow down they would break the coupling on the cars at the end of the train.
He also told me that passenger trains, generally, were stopped while going slightly down hill so they didn’t have to pickup the cars one at a time. The hill would allow the train to move all the cars at one time.
