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Does a ring contract faster than a disc?

  1. Dec 2, 2011 #1
    Household physics question: Before I left town for 3 weeks the lock on my apartment door was loose in its encasement. I had to hold it in place while turning the key or the inner disc would rotate uselessly inside the outer ring:

    http://scott-shepherd.com/share/forums/lock.jpg [Broken]

    When I came back after Thanksgiving the lock was fixed. So, assuming no opportunistic handyman came by, I think it must have tightened as a result of contraction in the cold weather. Does this make sense? Is it a demonstrable principle that a ring contracts faster than the circumference of a disc of the same material?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 2, 2011 #2
    Well, I am not 100% sure that a disk has more "freedom" to contract than a disk since it is hollow and no matter in the inside to oppose it (even though this matter is also contracting)...but another explanation to your problem is that the two pieces could very well be made of different materials even though they appear very similar.
     
  4. Dec 2, 2011 #3
    No. If the ring and the disk are made of the same material they will contract by the same percentage which is equal to their coefficient of expansion. The space between them will also shrink. It's possible that something in the lock is now rubbing against the ring and that's why it doesn't spin freely any more.

    http://en.wikipedia.org/wiki/Coefficient_of_expansion
     
  5. Dec 3, 2011 #4
    Yes the volume changes by the same coefficient (if the materials really are the same in both). I was wondering if there was something about the geometry of the ring that might make its inner circumference contract differently than the outer circumference of the disc.

    With the disc it seems safe to assume that if its volume shrinks by a proportion P, its radius shrinks by cuberoot(P) and so its circumference by 3√P2. (This presumes that the height and radius contract in equal proportion.)

    volume(disc) = hπr2
    P * volume = (cuberoot(P)h) * π * (cuberoot(P)r)2

    With the ring I'm not so sure. Probably the outer and inner radii contract equally, at the same proportion as the radius of the disc, but one could also effectively decrease the volume by increasing the inner radius and decreasing the outer. Not that this would answer my question, since it would make the lock looser, but it shows that there's more than one way to imagine the shrinkage of the ring...

    volume(ring) = hπ(r22-r12)

    A related question, probably not applicable to the lock: do solids expand liquidly? Like if a cylinder of an expanding solid material x was surrounded by an adamant material y that was not expanding, would the cylinder "squeeze out", expanding through its height to achieve the volume increase while its radius remained constant?
     
  6. Dec 3, 2011 #5
    There is a "bit" of "liquidly" as you call it.

    Depending on the strength ratio of the inner cylinder respect to the outer cylinder, the inner cylinder will make the outer one expand a bit, too...so, that the inner does not have to lengthen too much...

    ...but, yes, if a cylinder is being squeezed in the radial direction, it will grow a bit length-wise (in the axial direction)...look up Poisson's Ratio
     
  7. Dec 3, 2011 #6

    sophiecentaur

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    Couldn't you put the whole thing down to a bit of corrosion inside the lock?
     
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