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Acceleration from a spring

  1. Jul 21, 2008 #1
    I have a spring of known rate preloaded inside a device where it is contacting a free object. The spring is pulled back a known distance and then released. The spring continues until it reaches the stop which is where its initial preload is and then the object continues. I need to know the acceleration of that object so I can determine the force. I know the mass of the object.

    Here's some number to go with it to help.

    Spring free length = 2.300"
    Preloaded Height = .866"
    Spring Rate = 4.255 lbf/in
    I am compressing the spring .095" more to a height of .771" then letting go.
    The object has a mass of 4.9g

    Thanks for the help.
  2. jcsd
  3. Jul 22, 2008 #2
  4. Jul 22, 2008 #3


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    Staff Emeritus
    Science Advisor

    Why do you need to obtain the force? Is this homework?
  5. Jul 22, 2008 #4
    not homework

    SO I can compare it to other masses and springs with different length strokes acting on them to see what I need to achieve the desired force.
  6. Jul 22, 2008 #5


    Staff: Mentor

    The spring force is determined by Hookes law.
  7. Jul 22, 2008 #6
    Yes I undestand that, but its not a constant force so it is a force that is decreasing before it stops pushing on the object. Am I to use the highest force? or does it need to take into acount the entire stroke to get the correct acceleration?
  8. Jul 22, 2008 #7


    Staff: Mentor

    The force is not constant, therefore the acceleration is not constant either. There is no one correct value for the acceleration.
  9. Jul 22, 2008 #8
    yes, but I just want the acceleration when the spring stops and the object continues.
  10. Jul 22, 2008 #9


    Staff: Mentor

    When the spring stops and the object continues the acceleration is 0. Just before the spring stops the acceleration is given by Hookes law at the "preload" compression distance.

    Are you sure you want the acceleration at that point, or do you really want the velocity as the object leaves the spring?
  11. Jul 22, 2008 #10


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    Gold Member

    Easy, when the spring stops so does the acceleration.
  12. Jul 22, 2008 #11
    Well if I only have the velocity I cannot find the force.
  13. Jul 22, 2008 #12
    You have used both metric and imperial units in your description. For simplicity's sake it is better to stick to one system, preferably S.I. (i.e. metric) units.

    I am not sure that I understand your unfamiliar term: "spring rate", which takes the unit [tex]\[\frac{{{\rm{pounds \times feet}}}}{{{\rm{inches}}}}{\rm{ = }}\frac{{{\rm{work}}}}{{{\rm{distance}}}}\][/tex].) In the context of your question, I can only think of one possible definition of "spring rate": the work required to compress the spring by one unit of distance. (In S.I. units: joules per metre, J/m.)

    This is where the question becomes confusing. J/m is a meaningless unit for a typical spring because a typical spring is linearly elastic, i.e. it obeys Hooke's Law. And Hooke's Law, [tex] F = -kx$[/tex], states that the increase in force required to deform (i.e. stretch or compress) a spring by some small distance is not constant, but proportional to the displacement that the spring is from rest. (In other words: "The further you deform it, the harder it is to deform it more").

    This leads to the formula for the potential energy stored in a linearly elastic spring: [tex]\[U = {\textstyle{1 \over 2}}kx^2 \][/tex]. Notice that [tex]\[U \propto x^2 \][/tex]. The idea of a "spring rate" with unit J/m implies [tex]\[U \propto \left| x \right|\][/tex], and therefore that [tex]\[F\][/tex] is constant.

    And, just as a linearly elastic (or "Hookean", according to Wiki) spring obeys [tex] F = -kx$[/tex], ("extension is proportional to load"), your spring would obey a different rule: [tex] \[F = a\][/tex] where a is some constant. At equilibrium, the extension would either be zero (where load < a) or at a maximum (where load > a).

    If I understand your question correctly, you would like to know how the force (due to the spring) on your projectile object varies with time. If you can clarify what type of spring you are using -- Hookean, or as described above -- then we can help you with your question!

    - m.e.t.a.

    Edit: Perhaps it would be more accurate to define the force of the non-Hookean spring as: [tex]$F = - a\hat x$[/tex], i.e. "The force is equal to some constant, a, in the direction: [tex]\[ - \hat x\][/tex]".
    Last edited: Jul 22, 2008
  14. Jul 23, 2008 #13
    The spring rate is the k. It is the number multiplied by the deflection to give you the force at that given deflection. Which means at the preload of .866" the load is 6.101 lbs and at the release .771" the load is 6.506 lbs.

    When you use spring software and order springs from a manufacturer they use the term Spring Rate. Pretty standard in the industry.

    My question is how do I find the acceleration at the release since the load is decreasing? I need to calculate what stroke and preload I need on the spring to get me the amount of force I am wanting.
  15. Jul 23, 2008 #14
    That would mean that spring rate = spring constant, [tex]\[F = - kx\][/tex]. If the two are the same thing, that makes everything easier. My confusion was due to the unit of your spring rate/constant:

    Where possible I don't use inches, gallons, yards and so on; I work with S.I. units. So correct me if I was wrong in reading the above as "4.255 pound-feet per inch". (I don't know if "f" necessarily stands for "feet".) It just didn't make sense to me that the unit for a spring constant for a Hookean spring should be [tex]\[\frac{{energy}}{{displacement}}\][/tex] (J/m) instead of [tex]\[\frac{{force}}{{displacement}}\][/tex] (N/m). So I misunderstood something somewhere.

    Wait -- pounds, not pound-feet? Was the "f" in your first post a typo? If not, what does that "f" mean? This is going to drive me crazy. From here on, I'll assume the unit for your spring rate to be pounds per inch, not pound-feet per inch.

    Anyway, your question:

    "Release" could mean two things here. It could mean the moment that the ball starts to be accelerated (at h = 0.771"), or the moment that it ceases to continue to be accelerated (at h = 0.866"). I'll give both answers, in that order.

    A linearly elastic spring has displacement/deflection (I like your word best) proportional to its load. The load is a force, and force is linearly related to acceleration: [tex]\[F = ma\][/tex]. Assuming a massless spring, and also assuming that 100% of the energy released from the spring is converted to kinetic energy of the object, we have:

    If "release" is taken to mean the moment the object's acceleration starts:
    Deflection of spring at maximum compression: [tex]\[ - \left( {2.30 - 0.771} \right)\]
    [/tex] = -1.53" (taking "upwards" as positive)
    Force on the object at this instant: [tex]\[F = - kx = - (4.255 \times - 1.53)\][/tex] = 6.51 lb
    Conversion to S.I. units: 6.51 lb = 28.9 N
    Acceleration of the object due to this force:[tex]\[\frac{{28.9}}{{0.0049}}\][/tex] = 5,910 m/s2

    If "release" is taken to mean the moment the object's acceleration ends:
    Deflection of spring at moment of preload: [tex]\[ - \left( {2.30 - 0.866} \right)\]
    [/tex] = -1.43"
    Force on the object at this instant: [tex]\[F = - kx = - (4.255 \times - 1.43)\][/tex] = 6.10 lb
    Conversion to S.I. units: 6.10 lb = 27.1 N
    Acceleration of the object due to this force: [tex]\[\frac{{27.1}}{{0.0049}}\][/tex] = 5,540 m/s2
  16. Jul 24, 2008 #15
    I understand that calculation I was able to figure that out myself. The problem is that during testing we have found that the longer the spring is acting on the object the faster it is going when the spring stops acting on it. Using just one of the force numbers is not really how it acts. I have tried a much weaker spring with a longer deflection acting on the same mass object and got more force. So I need a way to account for stroke (the total amount of deflection acting on the object in this case .095")

    I consider release the moment when the spring starts to accelerate the object because I have a device that pulls the spring and object back the desired stroke and releases them. Thats how I know taking a straight force is not giving me the right answer I need something that accounts for stroke and the total time the spring is accelerating the object.

    Oh, I am not sure what the f is for its really just lbs but my spring software shows it as lbf/in. Other spring manufacturers only show it as lb/in.
  17. Jul 24, 2008 #16


    Staff: Mentor

    Are you still sure that you don't want the final velocity rather than the force?
  18. Jul 25, 2008 #17


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    Homework Helper

    The acceleration at the stop is the ratio of force / mass, and the force at the stop is always the same. A real world issue here is the total amount of mass involved, since the the total mass includes the mass of the spring itself. What's missing from your original equation is the mass of the spring. Given this information, then you should be able to calculate the acceleration of both the spring mass and object mass knowing the spring rate. I'm not sure how complicated the situation of the rate of spring expansion versus the springs mass versus spring constant gets.
  19. Jul 25, 2008 #18
    I could make assumptions that would get me close using the final velocity. How would that be found with the acceleration decreasing? Could I just average the initial and final forces to find an average acceleration and then find the time of stroke and use that to find the velocity? Or am I way off?
  20. Jul 25, 2008 #19


    Staff: Mentor

    The best way to do it is with conservation of energy. m v² = k (x1² - x2²)
  21. Jul 25, 2008 #20
    is that mass x v^2?
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