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Trampoline mechanics

  1. Jun 16, 2010 #1

    I'm just trying to understand the mechanics of a trampoline, as i think it will better understand my understanding of mechanics as a whole. This is my model:

    We have some plastic fabric which we will assume is inelastic. A circular trampoline where we have 30 identical springs which are in parallel? If we assume that the force. I have a few questions about how this will work.

    Obviously when someone jumps on it a force is applied to the trampoline fabric which passes it on to the springs. the springs are stretched and exibit an extension (x) which is proportional to the force applied. The inverse of the total spring constant of the system (K) would be equal to the inverse of the sum of all the individual springs (30k) so 1/K =1/30k.

    Secondly when a person lands on the trampoline they have a momentum of Mv and the trampoline has a momentum of 0, obviously this is an inelastic collision as kinetic energy is lost by air resistance, friction etc. But i don't understand then how this follows the conservation of momentum, surely when the person leaves the fabric they are going to be slower that before, so where does the rest of the momentum go. I can't see any recoil on my trampoline.

    If anyone can help me on this question and tell me where my thinking or assumptions have gone wrong i would be very appreciative.

  2. jcsd
  3. Jun 16, 2010 #2

    Doc Al

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    Staff: Mentor

    Forget trampolines for a moment. You're standing still on the ground and then suddenly jump up with some momentum. How is that possible?
  4. Jun 16, 2010 #3
    Doesn't the earth recoil very very slightly?
  5. Jun 16, 2010 #4


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    Solving trampoline exactly is extremely difficult. But you can make some approximations to make it relatively simple.

    The springs around the trampoline are pre-stretched some distance, compared to which the change is going to be small. So I would suggest treating it as a constant force that results in tension in the surface.

    Next assumption would be to say that deflections are small compared to diameter of trampoline. That is, the angles of the surface are always small. That's probably going to be the worst of the assumptions.

    Finally, assume that height of the perimeter is fixed.

    Putting it all together will give you a 2-dimensional wave equation with circular boundary. You will need Bessel Functions to solve that. You can find some discussion of it in this Wikipedia article. First, find homogeneous solution (nobody on the trampoline) and then figure out what adding a weight to it is going to do.

    The dynamics you get out of this should be very similar to real trampoline.

    Precisely. Now picture that you aren't standing on ground directly, but rather on something that rests on the ground. Then extrapolate to case of trampoline.
  6. Jun 16, 2010 #5
    Ok sounds a bit to complex for me then, but thanks anyway.
  7. Jun 16, 2010 #6

    Doc Al

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    Absolutely. The total momentum of you + earth remains the same. The same force that propels you upward, pushes the earth away.

    It's the same basic idea with the trampoline.
  8. Jun 16, 2010 #7
    You also have to remember that each time you bounce you put more energy into the system with your muscles. Adding your momentum to the muscular interaction gives increasing bounce height.
  9. Jun 16, 2010 #8


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    Side note - a competition trampoline has minimal air resistance. In the case of a "ozzie" (Australian) "bed", the surface resembles a net, and it's somewhat elastic.

    Back to the momentum issue, the center of mass of the earth, trampoline, and person remains the same, along with the linear momentum. Angular momentum is also conserved, any change in angular momentum the person experiences coexists with an equal and opposing change in angular momentum of trampoline and earth.
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