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Inertia: blocks attached to rod of negligible mass

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass.
    Find the formula for I


    2. Relevant equations
    I=MR^2


    3. The attempt at a solution

    I=m(0.5 L)^2 + m(0.5 L)^2
    I=2(0.5L)^2 m
    I=0.5L^2 m

    am i right?
     
  2. jcsd
  3. Dec 9, 2008 #2

    tiny-tim

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    Hi cantgetno! :smile:
    mmm …

    i] moment of inertia isn't measured in metres

    ii] about which point is I being measured?

    iii] what about the mass in the middle? :wink:
     
  4. Dec 9, 2008 #3
    Re: Inertia

    oh its around the middle
    so that makes the mass in the middle mass x 0

    metres? m=mass
     
  5. Dec 9, 2008 #4

    tiny-tim

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    D'oh! m is mass … should have spotted that! :redface:

    In that case, everything is fine. :biggrin:
     
  6. Dec 9, 2008 #5
    Re: Inertia

    oh crap
    i put 1/2 l m^2 ... im an idiot

    thanks anyway
     
  7. Dec 9, 2008 #6

    tiny-tim

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    hee hee :biggrin:

    Useful tip: keep your m's in the same place :wink:

    for some reason, you shifted them from the left to the right …
    … which makes it much easier to make a mistake, and much less easy to spot a mistake. :cry:
     
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