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Inertia of a rotating door

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A 20 kg solid door is 220 cm tall, 93 cm wide.

    2. Relevant equations
    What is the door's moment of inertia for rotation about a vertical axis inside the door, 14 cm from one edge?

    3. The attempt at a solution
    I know that if the pivot point is at the edge of the door, the equation is 1/3ML^2, but I don't know how to find the formula for this question...
     
  2. jcsd
  3. Nov 23, 2008 #2

    berkeman

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    Staff: Mentor

    Hint: Think of the door as two doors, joined at the axis 14cm in....
     
  4. Nov 23, 2008 #3

    djeitnstine

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    Gold Member

    Oops thinking of angular momentum
     
  5. Nov 23, 2008 #4
    Do you know how the equation for the pivot point of the edge of the door is calculated?
    It is done by using the moment of inertia of an axis thru the center of the door plus M * R^2 where R is the perpendicular distance between the two axes. You can confirm this equation by starting with the moment of inertia of a thin rectangular plane thru the center and adding the M * R^2 to get the moment of inertia of a thin rectangular plane on the edge.

    BTW: The equation you have for the pivot point at the edge of the door is wrong (rather, incomplete). You need to have another term for the length of the door.
     
  6. Nov 23, 2008 #5

    berkeman

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    Staff: Mentor

    Might be a typo. L is angular momentum, and I is the Moment of Inertia. Just follow the PF Library link that was automatically added to your term moment of intertia:

    https://www.physicsforums.com/library.php?do=view_item&itemid=31
     
  7. Nov 23, 2008 #6
    Or you could do that :P
     
  8. Nov 23, 2008 #7

    djeitnstine

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    Gold Member

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