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Kinetic Energy Problem - from GRE

  1. Nov 28, 2003 #1
    This is #32 on the GRE practice exam from www.gre.org.
    Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle, as shown above. The assembly is to be given an angular velocity w about an axis perpendicular to the triangle. For fixed w, the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to
    [A is located at the center of the triangle, B is located at a corner of the triangle]
    a. 3
    b. 2
    c. 1
    d. 1/2
    e. 1/3
    The answer is B, but I get 3 for an answer.
    This is an easy problem, but for some reason I don't know why I am not getting it. Can someone else work this out and tell me if you get the right answer, and how you arrive at it?
  2. jcsd
  3. Nov 28, 2003 #2

    Moments of Inertia of Point masses are MR^2 when you're rotating through the center of the triangle the Radius is L/(3^.5) and there are three masses when you're rotating about the corner the radius is L and there are two masses. So it works out something lize this...

    B=2*M*(L)^2 : A=3*M*[L/(3^.5)]^2

    which reduces to B=2ML^2 and A=ML^2
    this works because Omega is fixed and the only portion of the kinetic energy changing is the Rotational which is equivalent to a change in the net moment of inertia. I think that is right...

    A bored CSMPhysicist.
  4. Nov 28, 2003 #3
    I misread the question. I was treating the sticks as the particles instead of them being massless. I did the problem again, and got 2.
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