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Homework Help: The moment of inertia of a solid sphere

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid ball of mass M and radius is connected to a thin rod of mass m and length L as shown. What is the moment of inertia of this system about an axis perpendicular to the other end of the rod?

    Image: http://imageshack.us/photo/my-images/35/helpfy.jpg/

    2. Relevant equations

    Itotal = MD2 + Icm, where Icm = moment of inertia of the center of mass

    The moment of inertia of a solid sphere is given by: 2/5 * mr2
    The moment of inertia of a rod being rotated as shown is: 1/3 * mr2

    3. The attempt at a solution

    The center of mass of the sphere is a distance L away from the axis of rotation and has a moment of inertia of 2/5 * MR2. What I have a problem with is the rod. I know the rod's moment of inertia is given by 1/3 * mr2 (for this situation), and the distance from the center of mass of the rod to the axis of rotation is L/2.

    Here is my question: our lecture slides were made goofy and it gives both answer (A) and answer (C) as the correct answer. I think answer (A) is the correct answer because we are calculating Itot by adding the moment of inertia for the solid sphere and rod. With that said, I am still confused if answer (A) is the correct answer because the parallel axis theorem is given by Itotal = MD2 + Icm. Why is the MD2 term just ML2 (mass of sphere a distance L from the axis of rotation), and not the sphere PLUS the rod? I may have worded my questions a little weird, let me know if I am being confusing. Thanks (:
    Last edited: Aug 12, 2011
  2. jcsd
  3. Aug 12, 2011 #2


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    Homework Helper

    Yes A is correct, not C.

    For the rod about its own center, I = (1/12)ML2

    about its end (like where the axis is in the question) is I = (1/3)ML2

    Sphere about the axis in the question = (2/5)ML2+ML2

    (the rotating axis for the sphere must be moved to where it is in the diagram. The rotating axis for the rod is already the axis in the question)
  4. Aug 12, 2011 #3
    Okay I get it, thanks for clearing that up for me! I apply the parallel axis theorem to the sphere, and the moment of inertia for a rod with the axis of rotation at the end is (1/3)ML^2
  5. Aug 12, 2011 #4


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    Homework Helper

    Yep. In order to get the total inertia, all components must be 'brought to rotate about the same axis' if you get what I mean.
  6. Aug 12, 2011 #5
    Yep, makes sense. Thank you for your time!
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