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Rotating Door Question

  1. Jun 4, 2007 #1
    1. The pQuestion Details:
    A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 94 kg. A person pushes on the outer edge of one pane with a force of F = 77 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.

    There is a picture of your typical rotating set of four doors with the distance from the center to the end of a door being 1.2 m

    There are two things that might have tripped me up on this problem. The first is deciding which moment of inertia equation to use and the other is the fact that there are four panes.

    Based on my text, I believe the equation to use for inertia is teh solid plate one? If i rotate the picture of the solid plate in my text so that it stands upright...it appears to be the same as one of these doors.


    Force of torque will be 77N x 1.2m = 92.4

    Now, i may be wrong here also, but i reason that the overall acceleration will be i/4 that of a single door as there is four times more weight to push... therefore i decided to solve the angular acceleration for a single door and divide that number by 4.

    Ft = I ang. acc. (abrv. as x)

    92.4 = (1/12 mL^2)(x)

    92.4 = (1/12 (94)(1.2^2)x

    x = 8.1914

    I would then divide this by 4, and have the angular acceleration and the answer.

    Is this at all correct?? would really appreciate some guidance
  2. jcsd
  3. Jun 4, 2007 #2

    Doc Al

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    Staff: Mentor

    Everything looks good except your formula for the moment of inertia of a plate about one edge (which is the same as a thin rod about one end, by the way). Realize that you are not rotating about the center of the plates, but about an edge.
  4. Jun 4, 2007 #3
    Ah...i see what you are saying sir...it should be (1/3)mL^2

    just a matter of seeing things in the correct way

    thanks so much
  5. Jun 4, 2007 #4

    Doc Al

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    Staff: Mentor

    Exactly. Now you're good to go.
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