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Energy in rotational motion

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Small blocks with mass m, are clamped at the ends and at the center of a light rod of length L. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one third of the length from one end.
    Neglect the moment of inertia of the light rod.


    2. Relevant equations
    [tex]\sum{I} = m_{1}r_{i} + m_{2}r_{i}... [/tex]

    3. The attempt at a solution
    my answer would be
    1/3 at one end and 2/3 because the other farther
    I = m([1/3]L)^2 + m([2/3]L)^2
    I = 1/9(mL^2) + 4/9(mL^2)
    I = 5/9(mL^2)
     
  2. jcsd
  3. Oct 2, 2007 #2

    andrevdh

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    There is another block at the centre of the rod.
     
  4. Oct 2, 2007 #3
    so the block at the center has a distance of:
    since it is at the center
    1/2 - 1/3 = 1/6 is its distance from the end point

    I = 1/9(mL^2) + 4/9(mL^2) + 1/6(mL^2)
    I = 13/18mL^2
     
  5. Oct 2, 2007 #4

    learningphysics

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    you should have (1/6)^2
     
  6. Oct 2, 2007 #5
    oh yeah thanks LP
     
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