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Moment of inertia of a door?

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform, thin, solid door has height 2.2 m, width .87m and mass 23kg. Find the moment of inertia for rotation on its hinges.

    2. Relevant equations

    I=∫x^2dm

    3. The attempt at a solution
    The way I set it up was that I made into strips of length H (height) and width dr. dm=λHdr, where λ is mass over the length, H. But when I do the integral ∫r^2λHdr from 0 to r, I end up with ⅓mr^3, but it should be ⅓mr^2. Is lambda supposed to m/rH? If so, can someone explain why please?
     
  2. jcsd
  3. Nov 28, 2014 #2

    Orodruin

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    You can check this, in order for dm to be a mass, ##\lambda## must have dimensions of mass/length^2. It is the mass density per unit area of door.
     
  4. Nov 28, 2014 #3
    Oh, well that makes sense, thank you. Sometimes I feel like it's the simplest things that I get stuck on.
     
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