1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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

    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook