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The acceleration of falling backwards

  1. Jun 6, 2009 #1
    I was wondering how you could analyse such things as the acceleration of someone falling directly backwards. The head clearly moves faster than the feet but how can the acceleration of each be measured? What about the acceleration in the middle?

    I guess on a amore general level i am looking for an answer that is more simplified than analyzing a person's fall. assume that the person is a flat board on the earth's surface and their mass is evenly destributed throughout the board.
     
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  3. Jun 6, 2009 #2

    ZapperZ

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    This is actually a rather common problem in classical mechanics using the lagrangian/hamiltonian method. It is typically an upright cylinder that is tipping over, with some coefficient of friction between the bottom of the cylinder and the surface. In fact, another variation to this problem is the falling chimney.

    Zz.
     
  4. Jun 6, 2009 #3

    rcgldr

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    You could consider a rod hinged at one end and consider the period between rod nearly vertical, to rod horizontal.

    For a rod rotating about and end point the moment of inertia = mass x length2 / 3 and would be constant.

    The torque = 1/2 length x sin(angle_from_vertical) x weight

    The angular acceleration = torque / inertia

    Then you'd need to determine the relationship of angle versus time. If the rod is vertical, it doesn't fall, so you need to start with some small angle. The equation is complicated, using a pendulum as an example where moment of inertia = mass x length2, 3 times that of a rod pivoting at one end, but otherwise also a constant:

    http://en.wikipedia.org/wiki/Pendulum_(mathematics)

    http://en.wikipedia.org/wiki/Pendulum_(derivations)

    I use a spreadsheet to do a crude numerical integration of time, to make a graph of angle versus time starting at +179 degrees, where 180 degrees means vertical:

    http://jeffareid.net/misc/rod2.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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