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What will happen if a elevator containing a spring attached to it fall freely

  1. May 3, 2012 #1
    in elevator before cable break
    mg=kx
    in frame of elevator after cable break
    F(gravity)+F(pseudo)+F(spring)=0
    answering question is not tough but i am interested in other things that will happen with this system.

    Will the elevator move with a constant acceleration with respect to a person standing on the surface of earth.???? (If mass of block attached to spring and elevator are comparable.)


    I think elevator will not move with constant acceleration ; but the center of mass of spring mass plus elevator will move with constant acceleration. since acceleration of spring block is not g (acceleration due to gravity) so elevator will not move with constant acceleration...

    How will you define the motion of block after cable breaks (with respect the earth's frame)???
    I can't do this.:frown:
     
  2. jcsd
  3. May 3, 2012 #2

    gneill

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    Staff: Mentor

    It's an interesting question that you pose. Presumably the center of mass of the free-falling mass-spring system will be well behaved and accelerate downwards uniformly as you've stated. The motions of the individual component masses in the center-of-mass frame of reference should behave like the usual two masses connected by a spring and perform SHM about the center of mass with appropriate motion amplitudes that depend upon their relative masses; The spring begins with an initial tension. Superimpose one motion upon the other to obtain the net motion of the individual masses.

    As for writing and solving the equations of motion for such a system, I'd be tempted to convert the system to an analogous electrical circuit and apply the tools of circuit analysis (Nodal analysis using Laplace Transforms in particular). There's a well established methodology for converting such mechanical systems to electrical equivalents. See, for example, this link.

    The "circuit method" will allow you to determine expressions for the velocities of the individual masses (the node voltages will be the analogs of the mass velocities). Integrate to find position expressions.
     
  4. May 3, 2012 #3
    oh my god!!
    these electrical circuits destroys my heads electrical circuits(hahaa..)
    sorry gneil but it is around impossible for me to understand all those electrical circuits and their analogous mechanical circuits... however hanks for link... it adds up a new thing in my mind that there is also an electrical newton..
    I would like to understand in simple mechanics way.
     
  5. May 3, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you can always tell and EE- want to change every thing to a circuit problem!

    Initially, the mass applies force mg to the spring and so it is extended (F= kx) by x= mg/k. Once every thing is in free fall, the spring no longer feels that force (well, it does but so does the elevator frame it is attached to so, relative to the frame there is no gravitational acceleration) so the only motion is that due to the extended length, just as if you had laid the spring on a table so it felt no gravitational force, stretched it and let it go. That is, with x measured from the "natural position" of the spring, md^2x/dt^2= -kx with initial value x= mg/k.
     
  6. May 3, 2012 #5

    gneill

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    Staff: Mentor

    :smile: I can understand your position.

    For a purely mechanical analysis I would suggest:
    1. Accept the premise that when the elevator cable is cut the elevator car and the spring-mass become what is essentially an isolated system.
    2. The center of mass of that system will follow the usual trajectory that a falling mass does.

    Using this information it remains to solve the problem of the oscillations of two masses connected by a spring which are released from rest with a given initial tension in the spring. This sub-problem is to be solved in the center of mass frame of the two bodies. Energy and momentum conservation would seem to be applicable tools.
     
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