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Moment of inertia

  1. Nov 16, 2004 #1
    i need to find the moment of inertia of this figure
    http://s93755476.onlinehome.us/stuff/knight.Figure.13.54.jpg [Broken]
    and express my answer in terms of M, L, m_1, and m_2.

    i tried using the parallel axis theorem using (1/12)ML^2 since it is a thin rod with the axis of rotation about the center but im not sure what to use for Md^2. in I = I_cm + Md^2.
    i was thinking maybe I = (1/12)ML^2 + (m_1)(1/2L)^2 + (m_2)(1/4L)^2. but it didn't work.

    any help is appreciated.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Nov 16, 2004 #2
    I believe you are on the right track. Use the regular formula to find the moment of inertia of the rod. Now you have to add to that the moments of inertia of the masses. Since you are not given radius values, I believe you are meant to assume these are point masses. You should have formulas for the moment of inertia of a point mass.
  4. Nov 16, 2004 #3

    Doc Al

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    Staff: Mentor

    No need to use the parallel axis theorem.
    That value of I is the rotational inertia of that object--that's the answer (if you make the reasonable assumption that the masses can be treated as point masses, not spheres). I don't know what you mean by "it didn't work".
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