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Moment of inertia of Sierpinski carpet

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Hello
    sorry for my english, i know its bad.
    The problem i have is to count the moment of inertia of Sierpinski carpet around the point lying in middle of this figure. I would be very thanks to you for solving my problem :)


    2. Relevant equations
    i dont know what it mean in english..


    3. The attempt at a solution
    to use integrals, limes, other analysis methods
     
  2. jcsd
  3. Oct 14, 2009 #2
  4. Oct 14, 2009 #3
    what figure?
     
  5. Oct 14, 2009 #4
  6. Oct 14, 2009 #5
    If you are writing "Olimpiada Fizyczna", you shouldn't ask for solution there. If you are not I'll show you the answer in 2 days.
     
  7. Oct 15, 2009 #6
    what i am writing?
    can be in two days, not a problem...
     
  8. Oct 16, 2009 #7
    but what is a difference if you answer now or in 2 days? :bugeye:
     
  9. Oct 16, 2009 #8

    Borek

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    Staff: Mentor

    I suppose that's the deadline for submitting answers - rules prohibit pariticipants from discussing questions and solutions on public forums earlier.
     
  10. Oct 16, 2009 #9
    Borek is right.
    I don't know English good, so please forgive me mistakes.
    "Olimpiada Fizyczna" is some kind of competition in Poland.

    If we have big carpet with size 3r and bulk 8m (not 9, because middle of carpet is empty), x is a factor in moment eguation (I don't know how to name it)
    I- moment of small carpet (r,m)
    I=x*m*r^2
    Ic-moment of big carpet 8m,3r
    Ic=x*8m*(3r)^2 //I can't prove it, but i know taht it works (from simulation in Excel...)
    From Steiner: (http://en.wikipedia.org/wiki/Parallel_axis_theorem)
    Ic=4(I+mr^2)+4(I+m(r*2^0,5)^2) //r*2^0,5 is distance from the middle of big carpet to middle of small carpet in corner.
    Everything in 4, because we have 4 small carpets next to centre of empty carpet and 4 in corners
    first Ic=Ic sceond and for I in second Ic equation we substitute x*m*r^2
    and finally we get
    x=3/16
     
  11. Oct 16, 2009 #10

    Borek

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    Staff: Mentor

    Physics Olympiad, national level.

    So once you have explained that I am right you have posted solution two days before the deadline? Am I missing something?

    Edit: looks like the deadline was yesterday:

    http://www.kgof.edu.pl/

    "Termin wysyłania części I" means "deadline to submit 1st part".

    So far link to solutions is inactive, perhaps they will be pubslihed on Monday.
     
  12. Oct 16, 2009 #11
    interesting... it is a coincidence anyway :P
    thanks for solving this, it took a long time for me to solve it :)
     
  13. Oct 16, 2009 #12
    I wrote 2 days 2 days ago.
    Deadline was 15.10.2009 (I am sure), so I think, that i could post the solution.
    Am i right?
     
  14. Oct 16, 2009 #13

    Borek

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    Staff: Mentor

    Yep, seems everything is OK - hence my edit, I checked details after posting.

    Startujesz czy kibicujesz?
     
  15. Oct 16, 2009 #14
    Startuję
     
  16. Oct 16, 2009 #15

    Borek

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    Staff: Mentor

    Powodzenia :smile:

    Just before Mentors will ban us both - I have asked ununbium if s/he starts or just follows the competition. "Startuję" means "I take part", "powodzenia" means "good luck". That's probably enough Polish for October.
     
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