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Moment of Inertia Door

  1. Apr 4, 2006 #1
    A uniform, thin, solid door has a height of 2.2 m, a width of 0.87 m, and a mass of 23 kg. Find its moment of inertia for rotation on its hinges.

    Are any of the data unnecessary?
    the width of the door is unnecessary
    the mass of the door is unnecessary
    no; all of the data is necessary
    the height of the door is unnecessary

    First off, the height of the door should be unnecessary since the distance in moment of inertia is perpendicular to the force being applied? Second I'm having problems finding what I equation to use for a door about the hinge?
     
  2. jcsd
  3. Apr 4, 2006 #2
    [tex]I = \int r^2 \rho dA[/tex]
    Here r is the perpendicular distance to the hinge, \rho is the (surface) density of the door, and dA is the area differential.
    The height of the door will come into the area differential.
     
  4. Apr 5, 2006 #3
    Are you sure, the correct answer said the height wasnt needed?
     
  5. Apr 5, 2006 #4

    Doc Al

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    Staff: Mentor

    The height is not needed. If you do the integral that Euclid gave, using the mass density [itex]\rho[/itex], the height will drop out of the answer.

    Are you supposed to solve this using calculus? If so, set up the integral.

    Or are just supposed to get the answer using known formulas for the rotational inertia of common shapes? If so, since height doesn't matter, what formula would apply?
     
  6. Apr 5, 2006 #5
    Its not supposed to use calculus...I wasn't sure which moment of inertia would apply..i was thinking maybe 1/3MR^2 but not sure
     
  7. Apr 5, 2006 #6

    Doc Al

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    That's the one. Since height doesn't matter, the moment of inertia of a door about an edge is the same as that of a rod about one end.
     
  8. Mar 15, 2011 #7
    The Formula to use is

    I= (mass*((width^2)+(thickness^2)))/12
     
  9. Mar 16, 2011 #8

    Doc Al

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    Staff: Mentor

    Eh... no. (And you're 5 years too late anyway!)
     
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