Impulse & force of falling object 1

[jacobd]

Some time ago I came across an equation for calculating impulse of a falling object, given its mass and how far it falls.

the equations were:

Velocity: v^2 = 2gH (I'm curious how this was derived)
Impulse: J = m(2gH)^1/2


What I was interested in was estimating the forces generated by a 'rigid' falling object colliding with another 'rigid' object. By dividing the impulse by the time that elapses during the collision the forces could be estimated.

I wanted to use this to estimate the forces generated by a 1200# weight, attached to a 'solid' object with a chain, falling a given distance until the slack in the chain is taken up. The worst case is what I'm after, so I'm making assumptions that some items are 'rigid' and 'solid' although I know they will flex and absorb some of the force, etc...

I'm trying to come up with a reasonable collision time. I know it is hard to predict without knowing the specifics, but should I be in the ballpark of 1/10 sec, 1/100 sec, maybe even less time?

An example I worked out:

1200# weight falls 2 in:

m = 37.3 slugs
F = [37.3(2 * [32.2 ft/sec^2] * .17 ft)^.5] / .1 sec = 4084#

this seems a lot more reasonable than 40,840# based on a shorter elapsed time of .01 sec. Any comments are appreciated.

JD

 

[jacobd]

That was a miscalculation on my part in the above example, I just realized I didn't take the square of my result before dividing.

The results I got would be 2020# at .01 secs.

This just doesn't sit right with me. Should I be adding these values to the initial 1200# to get a total load? I'm probably totally off base here with the whole thing...

 

[IRstuff]

Suggest you peruse your textbooks to find the derivation of v^2=2gh. Something to do with potential energy?

TTFN

 

[chicopee]

check this out for the derivation of a free falling object:
dV/dT=-g
=(dV/dS)(dS/dT)=(dV/dS)(V)

VdV=-gdS
therefore
(1/2)(V2^2-V1^2)=-g(S2-S1)
letting V2=V,V1=0,S2-S1=-H
V^2=(2gH)^.5
voila!!
In regard to your impulse, you forgot your momentum-- remember your Dynamics"Impulse and Momentum" all derived from Newton F=d(MV)/dT keeping the mass constant for your case.

 

[IRstuff]

An interesting derivation, but the standard physics derivation for a body falling from a resting position:

kinetic energy=potential energy
1/2mv^2=mgh
v^2=2gh

TTFN

 

[chicopee]

IR stuff- your and my methods show that there is more than one way to skin a cat.

 

[GregLocock]

Your big problem is that you need to decide a time in which to decelerate the mass. I strongly suggest that a solution based on a time of the order of 1/(1/(2*pi)*sqrt(k/m))

will give more sensible results than any other simple approach. k is the stiffness of the chain, in appropriate units.

Likely errors are of the order of a factor of 10 for this sort of calculation. Real systems are often softer than you'd expect.





Cheers

Greg Locock

 

[Speedy]

Have a look at this previous thread:

thread404-36240

As you can see, while predicting the velocity and energy of impact are straight forward, predicting the max force is a lot more complicated.

Cheers

 

[jacobd]

Speedy,

That was a good link. As i figured coming up with a time is one of the more difficult variables. Looking around the web I came across an example of a duck colliding with a passenger plane. They figured a 12,000# impact load from a 1# duck based on the velocity of the airplane being 600mi/hr (880 ft/sec).

I believe the method they used to estimate the collision time was something like the duck was approx. 1 ft long, so the time that occured to basically pancake the duck and completely change its velocity from whatever it was to the velocity of the airplane (going the opposite direction) was 1/880 sec.

obviously this produced a very high resulting force.

 

[brad]

Chicopee--
I think what IRstuff was alluding to was an error in your final line. After your derivation, the final solution should read:

V^2=(2gH) or
V=(2gH)^0.5

Which is consistent with IRstuff's equation. The rest of your derivation works, it just looks like you took the square root of the right side while not taking the left.

Brad

 

[IRstuff]

Thanks for the vote of confidence; I actually bleeped past it, assuming it to be a typo.

What was interesting was that it was done with derivatives, while the usual derivation of potential energy involves an integration of work performed against the gravitational field.

The duck is interesting. I've never come across that light a duck, but then, I mostly eat them as opposed to pancaking them. ;-)

TTFN

 

[davidlivingstone]

I favour a different approach. Guessing a stopping time is just guessing. Better to estimate the stiffness of the chain. This can be done by calculation, experiment or reference to maker’s data.

At the point of arrest of the load the energy absorbed by the chain is the same as the kinetic energy of the falling load, which in turn is the same as the potential energy it lost in falling to the point of arrest [assuming no significant losses in falling].

If the chain obeys Hooke's law [a reasonable assumption], the energy absorbed is:

U = (ds)/2 = mgh

Where d = deflection
s = stiffness of chain
m = mass of falling object
h = distance object has fallen

and the maximum force, F = ds

 

[aybee]

unless you know the time taken to stop or the distance the you cant work this out. the formula you have now is the kinetic energy equation or moving object what you need is Force= Mass x decellaration and unless you can work out the decelleration which needs time to stop or distance to stop then it cant be done..AYBEE

 

[chicopee]

You are correct bradh. When I took the square root on the right side, forgot to remove the square on the left side. I'll try to be more carefull w/ the basics.

 

[i278]

Let me say up front that I've never had need to use it, but wouldn't Castigliano's Theorem work here? You'd figure out applied energy using all the above kinematics. Set that equal to the strain energy of whatever is taking the load ( in this case a chain) and solve for deflection. Take deflection and work back to an equivalent static load. It's so easy when you don't remember all the details.

Anyone have any working experience with Castigliano's Theorem?

 

[Maui]

There is a relatively simple way to solve this problem without needing to know the collision time. Model the problem as a falling object (your 1200 lb weight) connected to a fixed, immovable object by a chain. When the weight falls 2 inches, the kinetic energy that it gained is equal to the gravitational potential energy that it lost. At this point we can say that

Kinetic Energy = mgH

where m is the mass of the object, g is the acceleration due to gravity, and H is the distance through which it fell. The object will come to a stop when the slack in the chain is gone, and the chain is loaded in tension. The chain will stretch some amount as the falling weight decelerates. Assume that the chain deforms elastically, and that no permanent deformation occurs. Obviously this is a simplification of the actual process, but it is a good approximation. In this model the kinetic energy of the falling object is converted to elastic strain energy in the chain, which is given by

Elastic Strain Energy = [0.5V(sigma)^2]/E

where V is the volume of the chain material, sigma is the elastic stress in the chain, and E is the modulus of elasticity. Setting this equal to the kinetic energy,

mgH = [0.5V(sigma)^2]/E

Re-arranging,

sigma = [2EmgH/V]^0.5

This will allow you to solve for the maximum elastic stress in the chain. But since the stress is the force per unit area,

Force = A*sigma

where A is the cross-sectional area of the chain. This should allow you to calculate the maximum force that the falling object will produce on the support chain. Note that the value of the stress you calculate above should be below the yield strength of the material from which the chain is constructed. If this is not the case, then the assumptions used to arrive at this result will be violated and the answer will be invalid. Let me know how you make out.


Maui

 

[brad]

Maui,
Your assumptions would be appropriate when applied to a purely-tensile member. Chain links do not behave in this fashion. Their stress-field is more complicated.
Brad

 

[Maui]

Brad, I realize that. You could view the way I modeled the cahin as consisting of one long chain link. The model is not meant to provide jacobd with an exact answer - but it will give him an upper limit on the force applied to the chain. And the answer that is derived should be a good approximation.


Maui

 

[Speedy]

Consider 2 scenarios,
A - an object landing on a trampoline and
B - an object landing on a cushioning air-bag

Assume the trampoline is perfect elastic and and the airbag has perfect cushioning, defined as constant deceleration force. For both to have the same max force, the deceleration distance of the airbag is exactly half that of the trampoline. (The rectangle area of load versus deflection is twice that of the triangle.)

This shows how much the material properties such as viscoelasticity effect the max force of impact.