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Idle thought experiment / question

  1. Apr 6, 2009 #1
    I've wondered before about a little toy problem, and I think I've solved it, but just to be sure, I'll let you guys have it and see what you get.

    Alright, so you have a table and a (possible infinite) stack of circular disks of radius R and of neglible thickness (each disk is identical, having uniform density and mass M).

    Say you take one disk, and put it so that a little bit is hanging off the side of the table. Clearly, if you move it so that more than half of it is off the table, it will fall. However, by moving it back just a little and adding a second disk, you can extend its "reach" a little further. By moving these back in the right proportions and adding a third disk, you can extend it a little further.

    The question is what is the maximum range of stacked disks? For the case when higher-up disks have to be strictly to the right of lower disks, I got that the maximum reach is simply R. I haven't solved it for the general case of arbitrary stackings of disks, but I suspect that if one is allowed to stack them arbitrarily, you can get arbitrarily far away from the table.

    However, I also wonder how the problem changes if you include things like friction? And what if you only have a fixed finite number of disks?

    Any thoughts would be greatly appreciated. I have another one involving springs if you guys like this one.
  2. jcsd
  3. Apr 6, 2009 #2


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    It seems to be that the net effect would be the same as if you simply clamped the first disk to the edge of the table with a vise of some sort. If so, you should be able to extend it as far as the friction allows a gripping force.
  4. Apr 6, 2009 #3
    "It seems to be that the net effect would be the same as if you simply clamped the first disk to the edge of the table with a vise of some sort."

    You mean for the general case where you can stack them arbitrarily? Because then I agree, you could put one hanging off half way (or more) and another one behind it, and then stack 1000 over the intersection to "clamp" it down. I'm not sure how frictional gripping forces play into this. Would you care to elaborate?

    But for the case where this isn't allowed, what do you get?
  5. Apr 6, 2009 #4

    Vanadium 50

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    Can you go infinitely far? You have to keep the center of gravity over the table, and each disk no more than half-way over the one underneath. A gradual stacking will always let you do this.
  6. Apr 6, 2009 #5
    This experiment was discussed in SCIENCE magazine for domino overhang theory about two months ago. It showed that stacking over intesections was better than the traditional sequential series 1/2 1/3 1/4 1/5 1/6 etc. Because the discs are circular and not rectangular, the exact sequences wll be slightly different. See "The Joys of Longer Hangovers" SCIENCE 13 Feb 200 pg. 875
  7. Apr 6, 2009 #6


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    Yes. Sorry; I misunderstood the question.
  8. Apr 6, 2009 #7
  9. Apr 6, 2009 #8
    See the bottom half of the attached page for the article on the big hangover.

    Attached Files:

  10. Apr 7, 2009 #9
    I heard that one a while ago somewhere. The overhang can be infinite--although it takes a lot of disks to get each incremental overhang distance.

    I suppose an easy way to think this through would be that you could stack a thousand disks and make a fairly solid support for having disks overhang. You could stack those thousand disks, slightly offset, on a million disks which have an even stronger support. Those you could stack, slightly offset, on a million for another almost identical shift over. Given an infinite supply of disks, you could go on forever.
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