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Calculating Inertia

  1. Dec 5, 2007 #1
    Two particles, each with mass m, are fastened to each other and to a rotation axis by two rods, each with length L and mass M. The combination rotates around the rotation axis with angular velocity ω. Obtain an algebraic expression for the rotational inertia of the combination about the axis.

    I = m1r1^2 + m2r2^2
    I = ML^2 + M(2L)^2
    I = 5ML^2

    where am I going wrong? or am I not taking some factors into consideration?

    thanks
     
  2. jcsd
  3. Dec 5, 2007 #2
    anyone?
    this seems like it should be an easy problem but it really has me stumped

    thanks
     
  4. Dec 5, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You forgot to include the rotational inertia of the rods.
     
  5. Dec 5, 2007 #4
    so inertia of the particles is found with:
    I = m1r1^2 + m2r2^2
    I = mL^2 + m(2L)^2
    I = 5mL^2

    so inertia of the rods:
    I = (1/3)(2M)(2L)^2
    I = (8/3)ML^2

    total inertial = 5mL^2 + (8/3)ML^2

    thanks
     
  6. Dec 5, 2007 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me. (Assuming the rods and masses are arranged in a straight line perpendicular to the axis.)
     
  7. Dec 5, 2007 #6
    they are

    thanks again for your help
    much appreciated
     
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