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Question about Conserved Quantity Transfer from Train Track to Train

  1. Jul 3, 2012 #1
    Hi there,

    I'm deeply ignorant about physics, but I'm curious about what conserved quantities (if any) are transferred from a section of railroad track to a train that's rolling along the track. I take it that in many situations if the track were absent, the gravitational attraction between the earth and the train would mire the train in the ground below or otherwise stop the train before it could reach locations that it would have reached if it had been rolling along the section of track. Does this mean that the track plays an important role in keeping the train on course by transferring linear momentum to the train and cancelling out the force of gravity? Also, from what I understand the surface on which a wheel is rolling exerts torque on the wheel. Does this mean that the track transfers angular momentum to the wheels of the train? Would the track transfer energy to the train?

    I very much hope that this question makes sense. I'd deeply appreciate any help you could give me with it.

    Thank you so much,
    Howard
     
  2. jcsd
  3. Jul 3, 2012 #2
    Hi Howard, you're not quite as ignorant of physics as you seem to think. I'll do my best to answer your questions.

    As for "conserved quantities" transferred between the track and the train, I can think of only one; The magnitude of vertical force applied to the train from the tracks always equals the train's weight, "cancelling out the force of gravity," as you say. This prevents the situation that you describe where the wheels of the train might sink into the ground where it's too soft.

    Another thing the track does for the train is to reduce the rolling friction. The less contact the train wheels make, the easier the train can move along. By providing a hard and narrow surface for the train to ride on, the tracks allow it to move much easier than if it were riding on something like rubber tires.

    As for transfer of momentum, the track always maintains 0 momentum (mostly) while the train's momentum will remain constant as it travels at a constant speed, thus there's no momentum transferred between the train and the track. This is good, since momentum transfer from the train to the track would mean the train is losing momentum (slowing down).

    The energy transfer is mostly from the train into the track, due to friction. You're probably familiar with this energy, as it's what causes the track to be hot after a train has passed.

    It's probably best not to think about torque applied to the wheels by the track, as torque in 3 dimensions gets a bit complicated, but it is fun to think about what's causing the torque and why it's designed that way. For an explanation of that, you might want to watch this short recording of Richard Feynman, about the question: "What keeps a train on the track?"
    http://youtu.be/y7h4OtFDnYE
     
    Last edited: Jul 4, 2012
  4. Jul 10, 2012 #3
    Thanks so much for the very helpful response, Nessdude 14. I'm just having trouble understanding: since the vertical force applied to the train from the track is needed to keep the train from digging into the ground, is it correct to say that the track transfers linear momentum to the train (and that this is responsible for the train's momentum remaining constant, as opposed to changing on account of its accellerating in the direction of the earth)?

    Also, because I don't think I understand energy transfer, I wasn't quite getting why there isn't energy transfer from the track to the train. If the track exerts a force on the train equal to the train's weight, why doesn't the track count as doing work on the train, and thus transfering energy to the train? I'm very sorry for my profound ignorance here; if I could be set straight about the concepts involved I would be most grateful.

    One other related question: is it correct to say that the earth transfers any conserved quantities to the train (linear momentum?) as the train moves down the track, since, absent the gravitational force exerted by the earth, the train would not move along the surface of the earth but along a path tangent to its surface?

    Thanks again, so very much!
    Howard
     
  5. Jul 10, 2012 #4

    jbriggs444

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    Work is defined as the product of force times distance in the direction of the force

    The train is moving horizontally. The force of the tracks on the train wheels is vertical. The relevant product is zero. No work is done.

    Things get trickier if you think about the situation when the train either applies its brakes or uses the engine to speed up.
     
  6. Jul 10, 2012 #5
    Thanks very much for this, jbriggs444. I guess I'm still a bit confused about how you're supposed to think about distance in the direction of the force. If, absent the presence of the track, the train WOULD have moved in the vertical direction (namely, downwards / towards the center of the earth), then why can't we count the train's movement in the direction of the surface of the earth as involving movement in the direction of the vertical force applied from the tracks to the train (upward / away from the center of the earth)?

    Thanks again!
    Howard
     
  7. Jul 10, 2012 #6

    jbriggs444

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    Would have, could have, might have, should have -- none of it counts.

    The train does not, in fact, move vertically, so the product is zero.
     
  8. Jul 10, 2012 #7
    Thanks, jbriggs444, that's SUPER helpful! I take it that this means that, since thing1 only counts as transferring momentum to thing2 if thing2's momentum is different AFTER interacting with thing1 (it's irrelevant what would have happened had the things not interacted), and the train's momentum remains constant, the track does *not* transfer linear momentum?

    Does the vertical force exerted by the track on the train involve *any* transfer of a conserved quantity from the track to the train? Is the only conserved-quantity transfer between the track and train the transfer of energy from the train to the track in the form of friction?

    Thanks so much again!
    Howard
     
  9. Jul 10, 2012 #8

    jbriggs444

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    "Work" is a transfer of energy and is computed as force times distance moved in the direction of the force.

    "Impulse" is a transfer of momentum and is computed as force times time.

    One could look at the situation and say that the force of the tracks on the train is transferring upward vertical momentum to the train.

    But at the same time, gravity is pulling the train downward and transferring downward vertical momentum to the train.

    Ordinarily one looks at net force. The net force here is zero. The upward supporting force of the rails on the train is equal to the downward pull of gravity. There is no net force and the train does not accelerate upwards or downwards. So rather than talk about a transfer of momentum, one simply notes the obvious -- there's no momentum change taking place.
     
  10. Jul 10, 2012 #9
    Thanks once more jbriggs444! I'm sorry for being silly and not understanding the relevant difference here between work and impulse. I'm sorry also if these questions about momentum transfer are strange; they are motivated in large part by my reading about certain philosophical theories of causation, which describe causal interactions in terms of the transfer of conserved quantities. I wanted to understand what those sorts of theories would say about the role of the tracks as a cause of the position of the train.
     
  11. Jul 11, 2012 #10

    jbriggs444

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    If your interest is causation then you might be better served by thinking about the concept of "equilibrium" and, in particular, "stable equilibrium".

    Why does the train travel in a straight line?

    Because any deviation from straight line travel along the tracks results in a restoring force that tends to move the train back onto that straight line path.

    If the train deviates to the left, the flange on the left hand wheels contacts the rails and is pushed to the right.

    If the train rides up on the flange, gravity pulls it back down.

    If the train rises off the tracks, gravity pulls it down.

    If the train sinks into the tracks, Hooke's law pushes it (very strongly) back up.
     
  12. Jul 11, 2012 #11
    As explained by jbriggs, no work is done by the tracks since the track only forces the train vertically, and the train does not move vertically. That's assuming the track is completely flat, however; If the train travels uphill, you could say the tracks are doing negative work (slowing the train down), and when the train travels downhill, the tracks will do positive work (speeding the train up). The actual cause of this work is gravity, however, and most physicists would say the work isn't done by the tracks, but by the force of gravity (although it's really a combination of the two). In order to understand work better, you might want to investigate vectors, as work is actually the dot product of the force and displacement vectors.

    Another way to understand the energy transfer is just by examining the energy present and thinking about how it changes. The tracks have very little energy (ignoring mass-energy equivalence), most of the energy they contain is due to thermal energy from the earth and sun. The train has quite a bit more energy since it's heavy and it's moving along the tracks, it has a lot of kinetic energy. After the train has moved along the tracks, the temperature of the tracks will rise (meaning it has gained thermal energy). This gain of energy is from friction between the wheels and the track, and this friction is also slowing the train down (removing its kinetic energy). Assuming we've covered all of the significant energy sources, using the law of conservation of energy, we can conclude that the energy gain in the tracks came from the train's kinetic energy.
     
  13. Jul 11, 2012 #12

    jbriggs444

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    If the train is coasting uphill, losing speed, the track still does no work on the train. The force of the track on the train is still at right angles to the motion of the train.

    Same if the train is coasting downhill gaining speed. The force of the tracks on the train is "normal" to the surface of the tracks.

    As I had indicated up-thread, things get tricker if you consider the brakes and engines. Those can result in the existence of a force from the tracks on the train in the direction of motion of the train.

    With respect to causation, we would typically not say that the tracks cause the train to slow down when the brakes are appllied. We would typically ascribe that cause to the engineer who applied the brakes. We might admit that the tracks play a role -- for instance because they are made of steel instead of ice.
     
  14. Jul 11, 2012 #13
    Hi jbriggs444,

    Thanks again - very helpful as usual. Most philosophers who work on causation think that all there really is are things that, as you say, "play a role", and that all of these things are equally real causes of an effect. When we single out one thing (e.g. the action of the engineer) as "the cause" we're really just reporting on a cause that we happen to find particularly striking or salient. Since the track does no work on the train, a conserved-quantity-transfer theory can't count its interaction with the train as causal on account of its transferring energy to the train. But if we can count the tracks as transferring momentum to the train, and without this momentum-transfer the train's position would have been different, I think that most versions of conserved-quantity-transfer theory can count the interaction between the track and the train (and thus if we like the track itself) among the causes of the train's position (which is in large part what I was hoping to confirm).
     
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