Find Moment of Inertia of Figure w/ M, L, m_1 & m_2

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To find the moment of inertia of the given figure in terms of M, L, m_1, and m_2, the correct approach involves using the moment of inertia formula for the rod and adding the contributions from the point masses. The moment of inertia for the thin rod is (1/12)ML^2, and the point masses can be treated as point masses with their respective distances from the axis of rotation. The formula should be I = (1/12)ML^2 + (m_1)(1/2L)^2 + (m_2)(1/4L)^2. The parallel axis theorem is unnecessary in this case, as the masses can be considered point masses. This approach should yield the correct moment of inertia for the figure.
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i need to find the moment of inertia of this figure
http://s93755476.onlinehome.us/stuff/knight.Figure.13.54.jpg
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 I am 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.
 
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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.
 
quick said:
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 I am not sure what to use for Md^2. in I = I_cm + Md^2.
No need to use the parallel axis theorem.
i was thinking maybe I = (1/12)ML^2 + (m_1)(1/2L)^2 + (m_2)(1/4L)^2. but it didn't work.
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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