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Conservation of momentum/energy

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/cEqXb24 [Broken]

    2. Relevant equations
    Ki = Kf
    Li = Lf

    3. The attempt at a solution
    So I tried to solve this using conservation of energy as well as conservation of momentum, but only conservation of momentum gave the correct answer. Why can't conservation of energy be used in this situation?
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 14, 2015 #2
    Here is the work:
    Both of these moments of inertia already include the body:
    Iinitial = 2.83
    Ifinal = 0.9625
    ωinitial = 0.40 rev/s

    Using conservation of momentum:
    Iinitialωinitial = Ifinalωfinal
    (2.83)(0.4) = (0.9625)ωfinal
    ωfinal = 1.2 rev/s

    Using conservation of energy:
    Wnon-conservative forces = 0 and no potential energy.
    Kinitial = Kfinal
    (1/2)Iinitialωinitial2 = (1/2)Ifinalωfinal2
    (2.83)(0.402) = (0.9625)(ωfinal2)
    ωfinal = 0.69 rev/s
  4. Mar 14, 2015 #3
    Also, if there were a non conservative force such as friction, momentum would not be conserved because it would be an external force.
  5. Mar 14, 2015 #4


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    Staff: Mentor

    The skater has to do work to bring the arms closer to the body, and this increases the kinetic energy.
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