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Several random questions

  1. Jun 24, 2006 #1
    As some of you may know, I've got a lot of questions all the time... and so recently I've come up with a list of those I have yet to be answered, some of which I was able to sort out through a bit of googling, and searching my past threads, but here's what I couldn't quite manage to figure out with ease, although I should have some explanation of my thinking for each one:


    -About diffraction, I know that as a wave reaches a corner, it will bed around it, the deviation from it's straight path is dependent on the wavelength and the width of the opening or something like that anyway... The waves will always bend around, it's just that when the wavelength is small compared to the opening, not too much of the wave will bend, much will simply retain it's forward direction. Now what I'm wondering is if the point source of a wave for instance was right along the wall, would the waves still diffract around? I'm guessing they would, considering the point source would, albeit only a little, be directing the waves towards the opening.


    -Part 2 of my first one, say the point source of the wave was right in the wall, for example, a hole was carved into it, with a speaker placed inside... this is kind of hard to describe... but would the wave still bend around the corner then? I don't think it would, seeing as the waves coming from the speaker would automatically form a semicircle along the edge of the wall, and thus not be travelling in the direction of the opening at all... I don't know... is my thinking right on this one?


    -I saw this in a cartoon... don''t laugh, I don't know if it's real or not, that's why I'm asking; what would *theoretically* happen to a person travelling at a speed greater than that of light in a circular fashion? And, I don't know if it matters any, but what if it wasn't in a circle, but rather straight up and down? And straight forward? I haven't a clue on this one, in the cartoon, the character essentially melted. :surprised


    -When a roller coaster goes down a hill, it does this obviously because of gravity, the force pulling it down... but what about when it goes up a hill, what allows this to happen? I think it's momentum related, could be wrong though... and yes, I did some research on momentum before hand, learned the basics of it, if it's even related...


    -Last, I'm pretty sure, just confirm this for me... from one of my other threads, I learned exactly why you slip on ice, the friction you experience is kinetic friction, which is smaller than static friction, so I'm guessing this is why it's also hard to stop, right? When you apply the brakes... the force is small because it's kinetic friction as opposed to static friction. Not sure if it's right, but it;s just my thinking... also, why is the friction kinetic instead, and why is kinetic friction smaller than static friction?

    Well, that's all I have for now... I know there's a lot to sift through... but please someone help soon. I don't expect anyone to tackle all of the questions at once either... unless of course they're really basic and simple answers... but whatever you prefer. I just hope that I can recieve some help soon, and of course, with all my questions, thanks in advance. :smile:
     
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  3. Jun 24, 2006 #2

    rbj

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    you need to split these up into their own separate threads.
     
  4. Jun 25, 2006 #3
    im gonig to try and answer these as best as possible, i have limited knowledge in this subject :P

    for the ice question, there is almost no friction between your foot and the ice, that is why you slip. static friction is the force that keeps something at rest when you push it to make it move. its static all the way UNTIL it starts to move. kinetic friction is the force that acts in the opposite direction when something is moving.

    the rollercoaster question...the chains move it up the hill, unless you are talking about after it moves down a hill then up a hill, its got to be momentum that carries it up the hill.

    thats all i can answer (PLEASE DONT BAN ME IF I AM WRONG) :biggrin:
     
  5. Jun 25, 2006 #4

    Hootenanny

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    I think I've answered the diffraction questions already on a previous thread.
    It is not momentum which is important here, it is the velocity and conservation of energy(you could consider momentum here but it is easier with energy). For the roller coaster to reach the top of the hill the kinetic energy at the bottom of the hill must be greater than the change in the potential energy (and any work done to to drag and friction). Do you follow?
    As discussed on our previous thread on friction the maximum force of friction is depedend on the normal reaction force and the coefficent of friction. Now, the coefficent between ice and rubber, well ice and anything infact is extreamly low, which means that if a force makes your move (eg someone pushing you), the friction force is so small there is no resistance to the force, so you (or you foot) changes momentum very quickly. This rapid change in position of your foot causes you to become unbalanced and fall over, while everyone else stands around laughing at you :biggrin: . As for the kinetic friction being less than the static friction coefficient, this is not entirely the case and can usually be attributed to experimental error due to contamination of the test surfaces. If the surfaces are cleaned thoroughly, then this difference almost always disappears.
     
    Last edited: Jun 25, 2006
  6. Jun 25, 2006 #5
    Ok, I think I'm getting this pretty well now... When the car starts moving, the friction is kinetic friction... and since kinetic friction is dependent on the coefficient of friction(and the normal force I believe), this is small here(the coefficient), since the surface is icy... And if the kinetic friction is small, then there isn't much force between the two, therefore it's hard to move properly because the forces are small. Therefore the force backwards(when brakes are applied) is also small, because kinetic friction is small... Ok, I think that's right... but let's see here, the applied force by the car to get moving responds to static friction, right? If it continues to apply this force throughout moving, why isn't there a large reaction force to keep it from slipping? I don't know, I think I'm missing something here...

    Also, I do follow the part with the energy hootenanny spoke of, we learned about that in class... about how you can't have hills higher than the first one or the ride won't have enough energy to make it up(and of course they must be smaller as well given the fact that some energy is lost due to friction/drag). but what about the momentum part mentioned here?

    Ok, I'm going to be gone for a few hours, and here's a couple more questions(I know it's a lot but I don't want to have 5+ threads all with my name on them).

    Also in a cartoon... :uhh: there was a string a certain height above the ground, and someone said that at that height it wouldn't make a shadow... I'm assuming this means the central maxima of the light wave would have been distributed wide enough to make it seem as though there was no shadow... right?

    And yet another comes(also somewhat biology related I suppose)... my physics teacher said(and I'm not doubting this or anything), that elderly people tend to get myopia(short sightedness) because gravity pulls their eyeballs down throughout their life, and thus they become elongated after a time... Now I'm curious as to why this doesn't happen sooner, I'm assuming it's due to the fact that there's vitreous humour in the eye ball, which slows down the acceleration of the top of the eye downwards. Someone tell me if I'm right? Also, the same principle I guess, say you have two blocks, and place a wooden board on top(supported at each end by one of the blocks), the middle isn't supported much. Basically, the middle would sag after a while right? Because gravity pulls it down and the only opposing force is the resistance from the air... wait a second, the air should also be applying a force equal to mg, right? Since you are being pulled into the air with a force of mg... Doesn't that mean that when an object falls, the air friction should be equal to the force of gravity? Since you are being pulled into the air with this force, the reaction force should be the same, then why is air friction always seen to be so small? Argh, this is becoming complicated, but I hope someone helps soon.

    I know there's a lot there, but please help.
     
  7. Jun 25, 2006 #6

    arildno

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    I'd rather say your physics teacher is pulling your leg.
    Some form of cell degradation due to aging is a far more likely reason.
     
  8. Jun 25, 2006 #7
    Didn't sound like he was joking... although I can understand your point as well, can't they both be responsible?
     
  9. Jun 25, 2006 #8

    arildno

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    Well, it boils down to an estimate of relative strengths:
    Your teacher says that the strain induced by the stress due to the influence of gravity is a significant factor here.
    It's up to him to back up this astonishing claim with some empirical data.

    We DO know, however, that the process of aging occasions a lot of cellular change, so why should suddenly gravitational stress displace such explanations in the case of myopia among the elderly??
     
  10. Jun 25, 2006 #9
    Well, I know about cellular damage and the like, maybe he was just... I don't know. But isn't it possible? Anyways, I think my other questions are more important than some random thing my physics teacher said... true or not.
     
  11. Jun 25, 2006 #10

    russ_watters

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    People get myopia because the muscles in their eyes that flex the lens get weak and the lens itself gets less pliable.
     
  12. Jun 25, 2006 #11
    Ok ok, it seems my teacher was incorrect, or at least not giving the full story... what about the other questions now?
     
  13. Jun 25, 2006 #12

    Hootenanny

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    I think part of the problem here is you are treating the car as a particle which is sliding on the ice, rather than wheels which are rolling on the ice.
    You could solve problems such as these by considering momentum, forces and acceleration, still however, momentum does not really need to be considered, it is rather the velocities you are working with.
    Or perhaps it was midday...?:wink:
     
  14. Jun 25, 2006 #13

    DaveC426913

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  15. Jun 25, 2006 #14

    Gokul43201

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    If you're talking about sound waves, then you'll see the bending even if the speaker is embedded in the wall. A trick that serves as a good way to figure how the wave will go is Huygens' Principle (look it up) of secondary wavelets emerging from every point in the wavefront.


    This is a cartoon and there is no science involved. If a thing is going faster than light, there is no way it can interact with us and transfer information about itself. (ie:we can know nothing about how it's behaving). If it did, this would result in all kinds of paradoxes and contradictions.


    No it's not. The gravity helps, but if the coaster has some speed to start off, it would keep moving for a while, even if there were no gravity.

    Thinking in terms of momenta is messy. The better way to think of this is in terms of energies. Either way, the important message is that there are certain quantities that are conserved, and understanding these conservation laws will explicate many of your quandaries.


    You often slip on ice because, when you step on it (and it's close to 0C or 32F) you melt a tiny bit of it (either by direct heat transfer and/or due to pressure melting) to make a film of water. This film of water acts as a low friction bearing that makes you slip. This, at least, was the conventional wisdom on the matter.

    Recent studies have shown however, that in cold ice, there's often not enough pressure or heat transfer to melt the ice. Why then, is it still slippery? The answer seems to be that the surface of ice consists of weakly bonded, nearly planar structures that are a few molecules thick. This "quasi-fluid" (look up that term) skin is what makes ice slippery (in a similar way to the slipperiness of the surface of graphite).
     
    Last edited: Jun 25, 2006
  16. Jun 25, 2006 #15
    The wheels are sliding, are they not? The car doesn't come into contact with the ice. I figured the car was constantly applying the backwards force to the ice, and thus should still recieve the appropriate reaction force, what am I missing here?

    Also about the sound waves questions... how is it that the waves would still bend around the corner, even if the speaker was embedded in the wall? From another thread, it was said, that in diffraction through an aperture, the waves would not diffract through any adjacent holes, simply because the waves are not travelling towards them in any way... of course if you think about Huygen's principle as stated, which I believe mentions that each point on the wave acts a point source... then wouldn't the waves always bend around? Here's a crude diagram to show my thoughts:

    (- wall
    . space)

    ---....---....---


    Say a wave passes through the left aperture, then as it bends around, when it reaches the second hole, won't it diffract around, considering huygen's principle? When the diffracted waves reach the second aperture, the points there act as a point source, and thus the waves would still go through the opening, right? Do I even have huygen's principle right, is this making sense to anyone?

    Also, there's not even a theory on what would happen to someone travelling at light speed? I know that it can't be measured really, but has there ever been even an idea of what may happen? If not, I'll leave this one alone.
     
  17. Jun 25, 2006 #16

    Gokul43201

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    You're saying that the wave should be able to pass downwards through one slit and go back up the next one? Well, it should, and there's an amplitude associated with it going back up. However, there's a larger amplitude associated with the part of the wave that comes directly down this second aperture (some of which, will then go up the first aperture). The net effect is that the wave travels downwards through both apertures (assuming I've understood your picture correctly - there's a source on top, and 2 apertures below?). It's a question of learning how to correctly calculate and add these amplitudes.

    You'll find more (on phasor addition) in any of the later editions of Resnick, Halliday & Krane (the chapter on diffraction and interference) or in Feynman Lectures (Vol 1).
     
    Last edited by a moderator: Jul 2, 2006
  18. Jun 26, 2006 #17
    What if there's no wave coming out down from the second aperture, then the net effect would be up through that one, right?
     
  19. Jun 27, 2006 #18
    I just thought up something else, say you have the following situation:


    ...|
    ...|
    ...|
    ...|

    Ignoring the dots, say that the line is barrier(without holes in it). Now say a wave a pporaches the barrier, does the wave automatically surround the barrier?

    .)
    .)..|
    .)..|
    .)..|
    .)..|
    .)

    Say the curved lines are a single wavefront, and they approach the barrier, as soon as the wave reaches the edges of the barrier, will they curve around, forming a complete 90 degree angle to the original direction? Sorry if this is vague or anything, but if no one understands I'll try and explain a little more clearly.
     
  20. Jun 27, 2006 #19
    Someone please help, I'll try to clarify if this is ambiguous at all.
     
  21. Jun 30, 2006 #20
    Please someone help. I also have a couple more after these ones.
     
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