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Homework Help: Rotational inertia Problem

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Three identical balls, with masses of M, 2M, and 3M, are fastened to a massless rod of length
    L as shown. The rotational inertia about the left end of the rod is:

    http://img13.imageshack.us/img13/1424/1234sjy.jpg [Broken]

    2. Relevant equations

    I = MR^2

    3. The attempt at a solution

    I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

    I total = 3ML^2/2

    It says the answer is 3ML^2/4 though. Thanks for the help
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 3, 2009 #2

    I = mr^{2}

    I = (3M)0^{2} + 2M(\frac{L}{2})^{2}) + ML^{2}


    I = \frac{1}{2}ML^{2} + ML^{2}

    I = \frac{3}{2} ML^{2}

    I get the same as you.

    The definition of the moment of inertia is;

    SUM( miri2 ) So I'm preaty certian that we are following the correct method. So perhaps it is some arithmetical mistake we are both making.

    2/4 is 1/2 yes and 1/2mr2 + mr2 is mr2(1/2 + 1)
    (1/2 + 1) = 3/2. Nope. I think your answer book might be flawed.

    UNLESS you are meant to use the distribution of mass in a sphere for the first mass (3M). Mindyou, because they havn't given you a radius for the sphere I would assume not.

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