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Moment of Inertia for Rotational Kinematics

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    A thin uniform rod of mass (M) and length (L) is bent at its center so that the two segments are now perpendicular to each other.

    a. Find its moment of inertia about an axis perpendicular to its plane and passing through the point where the two segments meet.


    2. Relevant equations
    I = (integral) r^2 dm



    I have a picture drawn (attached) but don't know how to proceed from here.

    Thank you for any help
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2011 #2
    Won't the inertia be twice that of a bar of length L/2 and mass M/2 about its end?
     
  4. Sep 9, 2011 #3
    I would think that the moment of inertia would be:

    I = (1/3)(m) [(L^2) / 4] for each rod...
    = (m*L^2) / 12 for each rod

    Therefore the total moment of inertia should be
    I = (m*L^2) / 6 //sum of the two rods


    But it said that answer was wrong...

    Thank you for the help
     
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