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Mass dropped with preloaded spring

  1. Oct 16, 2015 #1
    Hello everybody,
    I'm trying to figure out how to calculate xadd, the additional compression of a preloaded spring (initially compressed of xp) on which a mass is dropped from a height "h".
    I also wonder :
    - if the final compressed spring length is the same with or without preload ?
    - if the spring receive an additional compression xadd >0 whatever the drop height ? (even at h=0.001 ?)

    I have some remembering about my academic studies (Energy conservation laws applied to spring) but I feel quite uncomfortable since I did not practice for a long time. Thanks a lot for any kind of help.


  2. jcsd
  3. Oct 16, 2015 #2


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    It is not the same.
    Yes it does.

    Consider the energy of the cylinder, just before the frame hits the ground:
    E1 = m*g*(L-xp) + ½*m*v2
    Now consider the energy, E2 , when the cylinder has been brought to a halt ( last figure ).

    The difference in energies: ( E1 - E2 ) must have been absorbed by the spring, so:

    ( E1 - E2 ) = xadd Fspring(x) dx
    Last edited: Oct 16, 2015
  4. Oct 19, 2015 #3
    Hello Hesch !
    Thanks to your help I did it this way :
    E1 (just before hit) = m*g*(L-xp) + ½*m*v2 + ½*k*xp2
    with v = √(2*g*h)
    E2 (just before rebund phase) = m*g*(L -xp -xadd) + 0 + ½*k*(xp + xadd)2

    Then E1 = E2 led me to this factorisation form (xadd as variable) :
    ½*k*xadd2 + (k*xp - m*g)*xadd - m*g*h = 0

    but I actually changed my mind about the study result and decided to get xadd as an input and to find the corresponding height ("how high can I drop the cylinder to get exactly 100% of potential energy absorbed by the spring").

    ==> h = [½*k*xadd2 + (k*xp - m*g)*xadd] / (m*g)

    k: 10 000N/m
    m: 1kg
    g: 9.81 m/s2

    L: 0.010 m
    xp: 0.003 m
    xadd_max: 0.006 m (considering a minimal spring contiguous wire height of 0.001 m)

    calculation led to h_max = 0.0307m (max height avoiding potential energy causing impact on the ground)

    to confirm the results, here is what I got using a mechanism simulation software :
    (velocity as initial condition of the dynamic study (just before hit) : v = √(2*g*h) = 776.067 mm/s)

    Good to see that lowest reached position is 1mm according to minimal spring height used is previous calculation (contiguous wire)

    I hope the results are not accidentally good :nb).
    Thanks again Hesch ! :wink:
  5. Oct 19, 2015 #4


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    What ?? So you gave up finding an algebraic solution, and started playing with numbers instead ?

    Well, I calculated that

    E1 = m*g*(L-xp) + ½*m*2*g*h = m*g*(h+L-xp)
    E2 = m*g*(L-xp-xadd)

    E1-E2 = m*g*(h+xadd)

    Espring = k*(xp*xadd + ½*xadd2 )

    More calculations leads to:

    ½*xadd2 + (xp-1)*xadd - h*m*g/k = 0 ( solve xadd )

    I'm not sure if this is correct, but anyway it makes sense that the calculations ends up in a 2. order equation, giving two solutions:
    One when the box is dropped with the spring below the mass, the other when the box is dropped upside down. :smile:
  6. Oct 21, 2015 #5
    Hello Hesch,
    I need to re-do your calculations now ;)
    Thank you for your help and explaination !
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