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Relativity, space contraction and rotation

  1. Mar 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A bar (lo = 10mm) is moving along the x axis with speed v according to a referential R.

    Meanwhile a table, paralel to the bar is moving along the z axis, with speed u. There's a circular hole in the table with radius = 5mm.

    From R perspective the bar has l < 10mm and it fits the hole on the table. But, from the point of view of R', resting in relation to the bar, the hole is smaller than the length of the bar.

    A. Does it make sense to ask how the bar fits in the hole? Explain

    B. show that in R' the bar and the table are no longer paralel, and theres a angle A between them and tg(A) = uvg/c², g= sqrt(1/(1-v²/c²))

    C. show that the point C, in the middle of the whole approaches the point B, in the middle of the bar following the line z'=-u/gv x'



    2. Relevant equations



    3. The attempt at a solution
    Im really clueless here. Anyone care to explain this problem to me?
     
  2. jcsd
  3. Mar 28, 2008 #2

    tiny-tim

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    Hi imphat! :smile:

    Hint: leave A until you've done B and C.

    Can you do B? What have you tried? :smile:
     
  4. Mar 28, 2008 #3
    hi tim, thx for the fast reponse once again

    the only thing i did so far was to draw some drafts of what happens...

    i was thinking maybe rotating the axis on the R' referential, in a way that the movement of the hole is done along 1 single direction, lets say, the z' axis... but i really dont see how the table gets not to be parelel to the bar on R'
     
  5. Mar 28, 2008 #4

    tiny-tim

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    … avoid doom …

    Hi imphat! :smile:

    No no no …

    In relativity, trying to think up a logical answer usually spells doom.

    You have to apply the Lorentz equations!

    Just to start you off …

    Two corners of the table are at (t,a,0,ut) and (t,b,0,ut) in R (with a and b constant).

    So where are they in R´? :smile:
     
    Last edited: Mar 28, 2008
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