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

 

[handleman]

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

-handleman, CSWP (The new, easy test)

 

[applico]

That's what I have as well. Will the mass of the object be in kilograms? How about the spring.....is the "x" the deflected amount or the compressed length amount. meaning if I have a 1" lg spring and am deflecting it .25 would "x" be the .25 or the .75? THANK YOU!!!

-Bill, CSWP

 

[TheTick]

x is the amount of deflection from free length.

Units of mass are a matter for you to decide. g, kg, or lbm. Just be consistent. Just be sure to use mass and not weight.

In simple terms, the formula calculates the energy contained in the deflected spring and converts it to kinetic energy, assuming 100% of the spring energy is converted to the marble's kinetic energy.

 

[btrueblood]

And to add to Tick's post, realize that the spring is not massless, and thus as parts of the spring are accelerated to push the marble, the spring itself gains some kinetic energy. I.e., you will never get 100% conversion of spring energy to marble kinetic energy.

 

[Tmoose]

Hi Applico,

From your description I do not think you are compressing the spring, and releasing the spring while the marble is in contact with the spring from the beginning of its extension.

It sounds to me instead either:
1 - The spring is released from fully compressed, and spanks the marble before full extension while the base of the spring is still grounded.
2 - You compress the spring, then release it so it launches itself and strikes the marble.

 

[handleman]

Frankly, from your description I don't think you are an engineer. This is high-school physics.

-handleman, CSWP (The new, easy test)

 

[drawoh]

applico,

Your description of your problem is ambiguous. Are you using the marble to compress the spring? F=ma, therefore, a=F/m. Are you causing the spring to strike the marble at some velocity? You are working out kinetic energy, and energy absorbed in impact.

I agree with handleman.

--
JHG

 

[rb1957]

he could be playing marbles in the office ...

but it's a sort of interesting question ...
you compress the spring, then release it. ok, it extends with a force F = kx; but this force is accelerating the small mass of the spring. i think that the spring will over-extend, and then retract, etc ... approaching it's original position as an undamped single degree of freedom dynamic system response. that imples that there's a difference between over-compressing the spring (ie past the marble) and releasing it (to strike the marble with some velocity) as opposed to pushing the spring back with the marble (so it's not over compressed) ... i think ...

Quando Omni Flunkus Moritati

 

[applico]

handleman & drawoh

Although I do appreciate your input I never claimed to be anybody but someone with a question. No mud slinging required. I think you need to take it down a couple notches.

 

[TheTick]

Do lighten up, folks. The man is here for help.

Applico:
All due respect, your lack of knowledge is apparent. We will try to advise as best as possible. However, that may be difficult without a certain base level of knowledge that is usually assumed among engineering professionals.

 

[Cockroach]

Yeah, I agree with Applico and TheTick. Manners have been horrible in this forum for quite some time now.

Anyway, toy cannon scenario. Load a marble and compress the spring, trigger release. How far will the marble go? The range is dependent on angle of trajectory and muzzle velocity off the compressed spring. Total energy of the spring equals total energy imparted to the marble of known mass. Assume potential energy during the compression of the spring is negligible, so you get HandleMan's solution for velocity, v=sqrt(k/m) for spring constant k and marble mass m.

If the cannon is inclined B to the horizontal, then the marble will have a range of R = v^2 sin2B / g, g = acceleration due to gravity.

We're talking a toy cannon here, small mass spring with negligible properties for inertia. I'm trying not to over think the problem and give buddy an answer to be used for his example.

Regards,
Cockroach

 

[LittleInch]

I am curious that you use the term deflect rather than compress. It opens up the possibility that you have a spring which you are pulling sideways in an arc to then strike your marble, or whatever, at some unknown distance from the deflected position of the spring?? As opposed to compressing it axially along it's length.

I would go down the energy route to start with and then do some tests as it sounds quite complex with many variables.

A sketch or a drawing always goes down well in trying to explain your problem to others and can demonstrate that you are serious. It's very easy to upload by clicking on the link under the post box.

If I can mis interpret what you say then you can see how use of words without drawings can lead to errors in understanding.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[rb1957]

i thought it was like 'roach's answer, except that the spring was being compressed more than the marble's position, like launching a pin-ball (for those other old farts who can still remember)

Quando Omni Flunkus Moritati

 

[IRstuff]

The OP's scenario needs to take into account a semi-elastic collision between the ball and the spring, so a simple kinetic energy analysis would probably be optimistic

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[PNachtwey]

It doesn't look like you guys are getting anywhere with this and I too find the problem interesting when one takes into account the spring.

I looked on the internet and couldn't find the canned solution for handling the mass of the the spring.

I am going to solve this with my Mathcad. It is super simple if one assumes the mass is weightless but it isn't. It will just take a little calculus to find how the spring will accelerate. The rest will require some differential equations.
A spring also has internal resistance so it doesn't vibrate forever.

Also, if you simply compress the spring, like those in a ball point pen, then release it it will fly up for a distance. Also, the spring should vibrate while in the air or after pushing the marble.

I do have a question. Is the other side of the spring connected to something? I don't know if it will make any difference to the answer.



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[IDS]

It doesn't look like you guys are getting anywhere with this and I too find the problem interesting when one takes into account the spring.

If we are ignoring losses due to friction and inelastic behaviour of the materials it's quite simple.

The spring will stay in contact with the marble so long as it has some compression. At the point where the spring returns to its original length the centre of mass of the spring will have a velocity of V/2, where V is the velocity of the marble, so the total kinetic energy will be:

(m(spring).(V/2)^2 + m(marble).V^2)/2 = Kx^2/2 where K is the spring stiffness and x is the spring compression.

So V = x.(K/(m(spring)/4 + m(marble)))^.5

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[LittleInch]

The OP though says the spring "hits the marble". I'm with rb1957 on this, sounds like there is an impact of a moving spring on a stationary marble with all the consquential issues of re-bound, start distance of the marble away from the compressed position etc etc.

Sounds quite complex to me - I would go for some experiments my self as the position thing looks crucial to me plus the thing at the end of the spring - is it hard or soft? will the marble ricochet off the end of the spring or be carried along with it for a while as it acceperates??

Too many here seem to be assuming that the marble will be in contact with the spring at all times. Read the OP carefully = that's not what he said and he hasn't come back to correct / provide more details yet.

Simple sounding question, but in reality I think there are a lot of variables and quite complex.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[IDS]

The OP though says the spring "hits the marble".

Well he also says it is a "spring loaded" object, and that he has "1/2mv^2=1/2kx^2 assuming zero friction and a sliding (rather than rolling) marble" as well.

So why not first sort out how it works for the simple situation, before worrying about scenarios that may not be applicable, and which in any case would need more information?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[LittleInch]

How about it applico?? Tell us which supposition is correct.

Are you bending or compressing the spring ?

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

Don't think this post is going to go much further without some response / input from you.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way