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Angular momentum astronauts

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Two astronauts, each having a mass M, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speeds v. (Use M, d, and v as appropriate in your equations for each of the following questions.)

    By pulling on the rope, one of the astronauts shortens the distance between them to d/3.
    (c) What is the new angular momentum of the system?


    2. Relevant equations

    L = mvrsin[tex]\theta[/tex]

    L = I[tex]\omega[/tex]

    3. The attempt at a solution

    I calculated the angular momentum when the two astronauts were a distance d between each other and got the correct answer, L = Mvd

    For the new angular momentum, I thought I would do the same thing just replacing r in the first equation listed above:

    distance between astronaut and pivot point = (1/6)d

    L (per astronaut) = Mv(1/6)d
    2L = (1/3)Mvd

    The website I'm using is telling me it's wrong though... can anyone help?


    P.S. in the picture it shows the pivot point being the center of the rope connecting the astronauts.
     
  2. jcsd
  3. Feb 25, 2009 #2

    Delphi51

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    Homework Helper

    Angular momentum is conserved, so it is still mvd where v and d are from the original situation. It is also m*v2*d2, where the new v2 is larger than the old v and the new d2 is smaller than the old d.
     
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