Mass dropped with preloaded spring

1. Oct 16, 2015

carb

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. Oct 16, 2015

Hesch

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
3. Oct 19, 2015

carb

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

Spring
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 .
Thanks again Hesch !

4. Oct 19, 2015

Hesch

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)

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.

5. Oct 21, 2015

carb

Hello Hesch,
I need to re-do your calculations now ;)
Thank you for your help and explaination !