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Center of Mass/Moment of Inertia Question

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Masses M1 and M2 are separated by a distance L. The distance of the center of mass of the system at P from M1 as shown above would be:
    (A) (M1L)/(M2)
    (B) ((M2+M1)L)/M1
    (C) ((M2+M1)L)/M2
    (D) (M2L)/(M1+M2)
    (E) (M1L)/(M1+M2)

    The moment of inertia of the system about the center of mass at P would be:
    (A) (M1+M2)L^2
    (B) [(M1+M2)/(M1M2)]L^2
    (C) (M1M2L^2)/(M1+M2)
    (D) (M1L^2)/(M1+M2)
    (E) (M2L^2)/(M1+M2)

    All of the 1s and 2s should be subscripts, I'm just lazy.

    2. Relevant equations

    Xcom=(M1X1+M2X2)/(M1+M2)
    I=ML2?

    3. The attempt at a solution

    I got the first part, that ended up being simple. The answer was D. I don't understand how to get the moment of inertia though. I tried doing I=mL^2=>I=(M1+M2)(Choice D)^2, but it didn't work. Help!

    Kids were telling me it was C, but I don't know how to get that.
     

    Attached Files:

  2. jcsd
  3. Feb 22, 2012 #2

    ehild

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    The moment of inertia of a point mass M at a distance d from a point is Md2, with respect to that point. You have masses M1 and M2. What is their distances from the CM?

    ehild
     
  4. Feb 22, 2012 #3
    I have the distances to the CM from M1. I need to do the same thing for M2 then. But where do I go from there? Once I have these two distances, do I plug them both in for d separately and add?
     
  5. Feb 22, 2012 #4

    ehild

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    Yes, square the distances, multiply by the masses and add.

    ehild
     
  6. Feb 22, 2012 #5
    I got it. Thanks!
     
  7. Feb 22, 2012 #6

    ehild

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    You are welcome. :smile:

    ehild
     
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