Calculate velocity of a spring loaded object

[applico]

I am trying to calculate the velocity of a spring loaded object. So say I have a marble at rest and then I deflect and release a spring and the spring hits the marble I need to know the velocity that the marble will travel at. I do have a formula but the units I am a little unsure about. Also, I have 2 resources, one says to use the "deflected" dimension of the spring and the other says to use the "compressed lenth" of the spring. Big difference. Any insight would be appreciated.

-Bill

 

[IRstuff]

"1/2mv^2=1/2kx^2 assuming zero friction and a sliding (rather than rolling) marble" as well"

That was not from the OP.

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[IDS]

"1/2mv^2=1/2kx^2 assuming zero friction and a sliding (rather than rolling) marble" as well"

That was not from the OP.

True.

What do you deduce from that fact?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[PNachtwey]

What do you deduce from that fact?

I don't know except it is apparent that the OP and many on this forum don't understand the problem. This is not an easy problem. It is way outside the math skills of most unless it is dumbed down to simple STUDENT problem where a lot of parameters are assumed to be 0. I doubt there is a perfect solution but a very good iterative solution can be found.

I am still plugging away at this like a big jig saw puzzle. I know the last part will be using Runge-Kutta 4 to iteratively solve the equation. That part is easy. The problem is the spring which is the part that interests me. I haven't found a good solution except to divide the spring in the small length wise pieces so each piece is the the mass of the spring divided by the number of pieces. Obviously the answer gets better as I divide the spring up into more pieces but that requires addition differential equations. Someone must of worked out how a spring expands with no load before but I can't find it.

Is the marble in contact with the spring all the time or does it hit it like snooker cue?

I am assuming the spring is a coil type of spring because that is what I am interested in.



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[handleman]

There are reasons why this problem is always simplified to the student level and why you aren't finding a cookbook solution... Solving it has very, very little benefit. Heck, if you want to get super-accurate, why not go ahead and take into account the internal material damping of the spring wire as it deforms elastically? As you are finding, the math is pretty complex, but I think you'll find that all the stuff you are calculating add up to have about as much effect as the variation in spring rate across two different batches of raw wire. The mass of the spring is pretty negligible to the final velocity on any reasonably-sized spring. That is, until you get up into heavy springs at a really high oscillation rate, like the valve springs on a 10k rpm indy car engine. As an academic exercise, I applaud your efforts and hope you enjoy the math, but if your answer ends up more than 20% different from the energy balance method I would be surprised.

-handleman, CSWP (The new, easy test)

 

[btrueblood]

Ah, but handleman, don't you think it's fun to send the sparky's down the calculus trail to the complexity morass?

 

[drawoh]

PNachtwey,

I do not understand the problem because it has not been clearly explained. If the marble is pushed down on the spring and then released, the problem is easy to solve. If the spring is compressed separately and shot at the marble, we need to make assumptions about the mechanism. We need to account for friction. We need to account for the possibility that the spring will rotate as it hurtles, and not act like a spring when it contacts the marble.

--
JHG

 

[drawoh]

I think there is a quick and dirty analysis.

If you know the spring rate and the compression distance, you know the strain energy of the spring as you get ready to release it. If you assume 100% efficiency of your mechanism, you use the mass of the marble to work out velocity. Then all you have to do is guess at all the efficiencies.

F[sub]max[/sub] = k[δ]

E = F[sub]max[/sub][δ]/2 = k[δ][sup]2[/sup]/2

Also, E = mv[sup]2[/sup]/2 = (w/g)v[sup]2[/sup]/2

You need the spring rate[ ]k, the spring compression[ ][δ] and the marble mass[ ]m. Work out velocity[ ]v.

You are on your own, working out efficiency.

--
JHG

 

[handleman]

Umm, that's what I said in the first reply to this thread.


-handleman, CSWP (The new, easy test)

 

[rb1957]

what's the difference between ...
a) compressing the spring d1 with the marble in contact with it, and
b) compressing the spring d1, but the marble rests (on some catcher) ahead of the spring.

in both cases the spring energy is the same, you've done the same work on the spring, so you'd expect the results (on the marble)would be the same (ok, very similar 'cause the losses are slightly different). i guess the biggest difference is that if the marble's motion is wrt some fixed datum, then it's starting point will be different in the two scenarios.

Quando Omni Flunkus Moritati

 

[handleman]

As I mentioned before, any reasonably sized spring relative to a marble will have pretty negligible mass w.r.t. the marble. In rb's "b" scenario, most of the spring's kinetic energy will, upon impact, be dissipated in much the same way as it would if the marble were fixed. The marble will travel only slightly farther than it would if the spring were released exactly at the marble's position.

Energy transfer from a coil spring directly to a marble is going to be horribly inefficient. Add a "striker" to the end of the spring to try to increase this efficiency and suddenly you have an entirely different problem. However, in either case the contribution of the spring's mass to the problem is pretty negligible.

-handleman, CSWP (The new, easy test)

 

[Cockroach]

This is a first year physics problem, you do it in one of those lab seminars on springs. It is not complex, finding iterative solution sets because there is a worry about standing waves translating through the spring. For God's sake man, this was solvable in Newton's day without computers!

DrawOH got it. A few before that post too. RB1957 let buddy know how to get his spring constant. This problem is over.

Regards,
Cockroach

 

[rb1957]

"buddy" can look that up for himself

Quando Omni Flunkus Moritati

 

[PNachtwey]

Ah, but handleman, don't you think it's fun to send the sparky's down the calculus trail to the complexity morass?

I do a lot of modeling using systems of differential equations. I have had to learn how to do it because the hydraulic and mechanical can't.
I know the answer to how to model the spring now. Do you?



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[handleman]

So... Now that you know... How different was the result from the simple energy calc?

-handleman, CSWP (The new, easy test)

 

[PNachtwey]

So... Now that you know... How different was the result from the simple energy calc?

No different. It did take some calculus but I expected that.
Now there needs to be something for the internal resistance of the spring. A spring doesn't oscillate forever.
I don't see internal damping specifications anywhere.
There is actually a lot of information on springs on the web but one must dig for it. I found info on cantilever springs too.





Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[GregLocock]

In cases where the internal behaviour of the spring matters the intrinsic damping of the spring is very small compared with that added to the system by the contacts at the end of the spring, etc.

Typical damping factor for a pure steel homogenous structure in isolation is around 0.1%. Very easy to measure approximately using the logarithmic decrement method.

Cheers

Greg Locock


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[PNachtwey]

Typical damping factor for a pure steel homogenous structure in isolation is around 0.1%. Very easy to measure approximately using the logarithmic decrement method.

Damping factors are not expressed in percent.
What does 0.1% mean?



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[GregLocock]

% critical damping.



Cheers

Greg Locock


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[PNachtwey]

% critical damping.

So 0.1% is a damping factor of 0.001.
If so then a spring alone with a natural frequency of 100Hz would have a time constant of 1/(2*PI*100*0.001)=1.59 seconds. It takes about 5 time constants for an oscillation to decay with 1% so that would be almost 8 seconds. I don't see where spring oscillate for that long. The damping factor of most springs must be be much higher.



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[GregLocock]

Tested in a vacuum with true free free boundary conditions did you?



Cheers

Greg Locock


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