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Moment of Inertia

  1. Jan 9, 2006 #1
    Hi. I have problem calculating the moment of inertia of a hollow cylinder if an extra mass in shape of smaller solid cylinder is embedded a distance x from the center of the larger cylinder.
    Firstly, I use the moment of inertia of a hollow cylinder, which is [tex]I_{larger cylinder}=m_2R_2^2[/tex]. I also know that the moment of inertia for the small solid cylinder is [tex]I=0.5 M_1r_1^2[/tex]. Using the parallel axis theorem, I can calculate the moment of inertia of the whole thing, which is [tex]I_{total}=m_2R_2^2 + 0.5 M_1R_1^2 + M_1x^2 [/tex]. Is that right?
    Secondly, how can I calculate the moment of inertia of the whole cylinder about P? Can i use the parallel axis theorem once again to 'move' the axis a distance R? So that the whole inertia will be [tex]I_{total}=m_2R_2^2 + 0.5 M_1R_1^2 + M_1x^2 + (m_2+M_1)R^2[/tex]?
    [​IMG]
     
  2. jcsd
  3. Jan 9, 2006 #2

    Doc Al

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    Staff: Mentor

    Right. This is the moment of inertia of the whole thing about the central axis.

    Careful! The parallel axis theorem tells you how to find the moment of inertia about any (parallel) axis if you know the moment of inertia about the center of mass. But the central axis is not the center of mass! Instead, use the parallel axis theorem for each piece separately and add them.
     
  4. Jan 12, 2006 #3
    I know understand it.
    Thank you very much....
     
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