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Tricky circular motion question

  1. Nov 18, 2005 #1
    Ok here it goes:

    A mass m1 moves without any friction in a circle with radius r on a table. On to this mass a string is attached that goes through a hole in the table and is attached to another mass m2 under the table.

    http://img358.imageshack.us/img358/5636/circularmotion0ny.gif [Broken]

    There is no friction between the rope and the table. Derive a formula for the radius r in terms of m1, m2 and the time T for one complete revolution.

    What I did is the following:

    F= mtot * a, with a= ac= v^2/r
    Therefore F= mtot * (v^2/r)

    v= 2pi r/T. Substitution gives:

    F= mtot * ((2pi r/T)^2)/ r = (mtot 4pi^2 r^2)/ (r T^2) = (mtot 4pi^2 r)/ T^2

    so this gives r= (FT^2)/ (mtot 4 pi^2)...

    But how will I get rid of the F term in the formula? Or is there some other way to solve this problem?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 18, 2005 #2


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    Gold Member

    You know that Fc = T. In other words, the tension of the rope caused by m2, equals the centripal force of the sppining m1.
  4. Nov 18, 2005 #3
    Ah ok, so the formula for r will be r= (m2 g T^2)/ (m1 4 pi^2) right? I mean not r= (m2 g T^2)/ (m1+m2)(4 pi^2) (I just thought because it says r=(m2 g T^2)/ (mtot)(4 pi^2) that I need to write m1 + m2....)
  5. Nov 18, 2005 #4
  6. Nov 18, 2005 #5

    Doc Al

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    Staff: Mentor


    What you did earlier:
    is incorrect, since only m1 is centripetally accelerated.

    Don't try applying Newton's 2nd law to both masses together; that's too complicated. Instead, apply it to each mass separately.
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