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Very Wierd Physics

  1. Jan 17, 2005 #1
    This is set in reality.

    You have a truck, set with high wheels(railroad wheels). Your driving on the railroadtracks going 50km/h. It's completely dry. You stop easily and quickly.

    The wierd thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

    How is this explained?
     
  2. jcsd
  3. Jan 17, 2005 #2
    are you saying it speed up when you hit the brake? that is not possible.... what if you keep hiting the brake... will your truck keep speeding up? so the truck can principally go from east coast to west coast without using any fuel..
     
  4. Jan 17, 2005 #3

    Integral

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    The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.
     
  5. Jan 17, 2005 #4
    vincentchan no you only gain 8-10km/h then it starts to lower.

    Integral thats amazing, at first I denied the possibility, then I thought it might be a friction thing.

    University of Toronto concluded they could not understand how it's possible.
     
  6. Jan 17, 2005 #5

    dextercioby

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    :rofl: Then University of Toronto should start reading...phyics... :uhh:

    Daniel.
     
  7. Jan 18, 2005 #6

    Bystander

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    The problem statement doesn't explicitly include any applied force --- it implies a constant 50km/h (maybe 200 N/m2 air resistance) balanced by engine and drive train, transitioning to zero driving force upon application of brakes (wheel lock-up is implied, or must be assumed), and the rotational kinetic energy of the wheels is conserved (some dissipation in the brakes stopping the rotation) to be added to that of the truck (I ain't gonna do the integration of the cycloid across a railroad wheel) as it hydroplanes along the rails.
     
  8. Jan 19, 2005 #7

    LURCH

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    Do you have a link or reference to this experiment?
     
  9. Jan 19, 2005 #8
    so what you are saying is that when the brakes hit the friction is reduced because of the wet tracks - ie the sliding friction is less than the rolling friction. it isn't that a force is added, but the frictional force is reduced so there is a net increase in force, hence the acceleration. is this the same sort of thing that happens on slippery snow where it is more advantageous to pump the brakes (hence, maintain some of the rolling wheel friction) rather than slamming the brakes and locking the wheels?
     
  10. Jan 19, 2005 #9
    physicsisphirst well i guess so. Very interesting stuff eh?
     
  11. Jan 19, 2005 #10

    GENIERE

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    I'm off to the local pub to make a few bets! If I were fair-minded I'd split the winnings with Integral, that's if I were fair-minded.
     
  12. Jan 19, 2005 #11
    but instead your absent-minded?
     
  13. Jan 19, 2005 #12
    Then why not just make a train that slides on a wet track instead of rolling wheels? :rofl:
     
  14. Jan 20, 2005 #13

    EL

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    This seems like a nice explanation.


    However this does not make sense to me:
    What force is applied more than friction when the wheels are locked???
     
    Last edited: Jan 20, 2005
  15. Jan 20, 2005 #14

    DaveC426913

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    Things to factor in:

    1] A speedometer does not measure actual vehicle speed. It measures the rotation of the tires. If the wheels happen to be spinning faster (for whatever reason and however briefly), the speedometer will show an increase in speed.

    2] The engine is still trying to drive the vehicle, even when the brakes are applied. In a small car, the engine quickly drops to a lower rev, but I'll bet those monster diesels don't slow to idle quite so easily.

    A combination of one or both these factors will likely yield a sensical answer.



    <trite> Everybody's looking at this as an idealized study in Newtonian physics. It's not ideal, it's a locomotive! This is what you get when you send a bunch of physicists out to do a mechanic's job! :smile: </trite>
     
    Last edited: Jan 20, 2005
  16. Jan 20, 2005 #15
    This only works if a constant force is applied to the truck even when the brakes are on. Else it would violate conservation of energy
     
  17. Jan 20, 2005 #16
    it's a good thing that u mentioned that, for some time i didn't understand what was going on.....
     
  18. Jan 20, 2005 #17
    hey! that's a good point! in fact, that is the point since the force that does exist is that of friction.

    even though we have less friction by the application of the brakes there is presumably no force being applied by the locomotive when the brakes are on. so what does that leave us with? it leaves us with the inertia of the locomotive going at say x km/hr and a frictional force that has just been reduced.

    so here is a similar set up:

    say you have a curling stone that you slide over a carpet that is on a rink. what will happen when the rock hits the edge of the carpet and slides onto the ice? will it speed up or not?
     
    Last edited: Jan 20, 2005
  19. Jan 20, 2005 #18
    but the static friction provides the force that makes the train go forwards, while the kinetic friction (when braking) is directed against the direction of the trains motion and wil only slow it down...

    (it is not the case that the non-braking train has to overcome static friction while the braking train has to overcome kinetic friction)
     
  20. Jan 20, 2005 #19

    Gokul43201

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    I cant see how the truck will speed up !! :confused: The mass of the wheels is negligible compared to that of the truck.
     
    Last edited: Jan 20, 2005
  21. Jan 20, 2005 #20
    It are railroad wheels, wich are pretty heavy I guess. Maybe their rotational energy is about enough to acomplish the acceleration. Let's see:

    [tex]T=\frac{1}{2}I\omega ^2=\frac{1}{2}(\frac{1}{2}MR^2) (\frac{v}{R}) ^2=\frac{1}{4}Mv^2[/tex]

    So if the mass of the truck is x times that of the total mass of the wheels this can by conservation of energy accomplish a change in velocity given by:

    [tex]\frac{1}{2}m(v+\Delta v)^2=\frac{1}{4}\frac{m}{x}v^2[/tex]
    [tex]\Delta v=v[\sqrt{1+\frac{1}{2x}}-1][/tex]

    So for v=50km/h this can only account for the observed increase in speed when the mass of the wheels is comparable to the mass of the truck. I guess this is not a very realistic scenario, and how would this energy be converted to translational energy?
     
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