Mass dropped with preloaded spring

[carb]

Hello everybody,
I'm trying to figure out how to calculate x_add, the additional compression of a preloaded spring (initially compressed of x_p) 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 x_add >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.

 

[Hesch]

the final compressed spring length is the same with or without preload ?

It is not the same.

the spring receive an additional compression x_add >0 whatever the drop height ? (even at h=0.001 ?)

Yes it does.

Consider the energy of the cylinder, just before the frame hits the ground:
E₁ = m*g*(L-x_p) + 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 F_spring(x) dx

 

[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 !

 

[Hesch]

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").

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)

E_spring = 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.

 

[carb]

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