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I Rate of Spring Extension

  1. May 8, 2017 #1
    Hi,

    So I am currently working on a rather simple problem of a projectile being launched by a spring at a certain angle. Ignoring friction, from conservation of energy we know that the velocity of the launched projectile would be ##v = \sqrt{\frac{kx^2}{m}}## (with ##m## being the mass of the projectile, ##k## the spring constant and ##x## the initial displacement of the spring). Then I applied simple kinematics to get that $$\text{Range} = \frac{v\cos{\alpha}}{g}\left(\sqrt{(v\sin{\alpha})^2+2gY} + v\sin{\alpha}\right )$$ (##Y## is the initial height and ##\alpha## is the angle of the launch relaive to the horizontal).
    What I was slightly confused about was to what extent this model would be applicable to a real world scenario in terms of just the spring extension process. That is, if we still assume that the losses due to external friction are negligible can we be sure that the extending spring itself is going to reach the computed velocity ##v## even if there is no projectile mass launched (i.e. the only mass the spring has to accelerate is the mass of itself)?
    For instance, let's say that we have a spring with a very large ##k = 10^6 ~\text{N/m}## and we compress it by ##10 ~\text{cm}##. If we take the effective mass that the spring has to push to be about ##m = 0.1 ~\text{kg}## the final velocity should be about ##316~\text{m/s}## according to the theoretical model
    The question I'm asking here is whether, if we neglect external friction, can we be sure that the endpoint of the spring is going to approximately reach the computed theoretical velocity for a given mass ##m## and a given value of ##x##?
    Even though it strictly follows from conservation of energy that it would be the case, I'm not sure if that's what would actually happen in a real world scenario. Would the internal non-conservative work done in the system due to internal friction be highly significant and thus make the final velocity calculation invalid?
    I would appreciate any sort of feedback on this question :)
     
  2. jcsd
  3. May 8, 2017 #2

    scottdave

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    I don't think it is accelerating the entire mass of the spring at the same rate. Here is a video from Smarter Every Day, that you may find interesting. It talks about rubber bands and slingshots, but some of the same principals can apply. and
     
  4. May 8, 2017 #3
    Thank you for the response. Yes, I watched those videos back in the day.
    What I'm more confused about is how the increase of the kinetic energy of the whole spring is represented by the motion of a given point in the spring in the case when we consider the spring just extending by itself.
    I tried first using the fact that that for an arbitrarily shaped body the kinetic energy is: $$\iiint_\mathcal{B} \frac{1}{2}\rho v^2 \,dV$$ Using this relation (unless there are some other formulas we could use) we need to find the velocity as a function of space coordinates ##v(x,y,z)##, so that ultimately we can plug in the coordinates of the spring endpoint.
    For the sake of simplicity, I parametrically defined the shape of the spring as a helix:
    $$\boldsymbol r(t)=\left [\frac{t}{2\pi N}, \cos{t},\sin{t} \right ]$$ (where ##N## is the number of "twists" in the spring wiring per meter of the spring length)
    I'm kind of stuck here, because the curve parametrization should be generalized to include the thickness of the spring, so that the integral can be evaluated over an explicit 3-D space.
    But even then I'm not sure how I would go about finding the velocity distribution on the spring.
     
  5. May 8, 2017 #4
    The question you are asking seems to be "what is the effect on the dynamics of a spring response if, rather than assuming that the spring has no mass, it has a uniformly distributed mass along the helix in its undeformed state?" Is that correct?
     
  6. May 8, 2017 #5
    Yes, that would basically be my question and I'm seeking to find an analytical solution to the velocity field of the spring. Sorry if my initial statement was unclear.
     
  7. May 8, 2017 #6
    In that case, would you be willing to make the approximation the spring behaves mechanically as a sequence of small infinitesimal springs? In this case, the local mechanical behavior of the spring would be described by $$T(x)=kL_0\frac{d u}{d x}\tag{1}$$ where T is the tension and u is the axial displacement of a material element in the deformed configuration of the spring that was at location x in the undeformed configuration, k is the spring constant of a spring that is of length ##L_0##. So, if T were uniform along the length of the spring, for example, the spring would experience a homogeneous elongation, and, if we integrated between ##x = 0## and ##x = L_0##, we would obtain:
    $$T=kL_0\frac{u(L_0)}{L_0}$$where $$u(L_0)=(L-L_0)$$So, combining these two equations, we would have (for a uniform tension along the length):
    $$T=k(L-L_0)$$
    So, Eqn. 1 is predictive of the correct tension in the spring for a uniform deformation of the spring. And it will predict the local tension in the spring if the axial deformation of the spring is not uniform.

    Is this satisfactory to you so far?
     
  8. May 8, 2017 #7
    Thanks for the answer. This all certainly makes sense, but what I am aiming for is incorporating this result into the dynamics of the spring itself. So assuming that the mass and the geometry of the spring is not to be neglected, how could we find the magnitude of the velocity of a certain point on the spring (with position ##x'##) at a certain time ##t## (while the spring is extending)?
    Edit: Since in the problem we defined ##x## as the displacement of the spring (you expressed it as ##L-L_{0})## let's define the position of the point along the spring relative to its attachment point to be ##x'##.
     
  9. May 8, 2017 #8
    Well, suppose we let ##\rho## represent the mass per unit initial undeformed length of the spring. Then the mass of the section of the spring between initial axial locations x and ##x+\Delta x## is ##\rho \Delta x##. If we do a force balance on this section of the spring, we can write:
    $$\rho \Delta x\frac{\partial^2 u}{\partial t^2}=T(x+\Delta x)-T(x)$$where ##\partial^2u/\partial t^2## is the local axial acceleration.
    Taking the limit of this equation as ##\Delta x## approaches zero yields:
    $$\rho \frac{\partial^2 u}{\partial t^2}=\frac{\partial T}{\partial x}$$ If we substitute our equation for the local tension into this equation, we obtain:
    $$\frac{\partial^2 u}{\partial t^2}=\left(\frac{kL_0}{\rho}\right)\frac{\partial ^2u}{\partial x^2}$$
    This is the wave equation for the spring (in terms of the local displacement), where ##kL_0/\rho## is the square of the wave velocity. So the spring can experience axial wave-like behavior like a Slinky.

    So, for a spring with mass, all we need to do is solve this PDE, subject to appropriate boundary conditions, for the axial displacement u as a function of axial position and time.
     
  10. May 8, 2017 #9
    Thank you, this the bit I was missing. Completely forgot that the wave equation can be actually derived from the Hooke's law :)
    So the solution to the PDE would be obtained as Fourier series with Fourier coefficients computed from eigenfunction orthogonality, but I'm sort of confused about the physical interpretation of the function variables for the spring. For instance, if I want to find the velocity of the point 2 cm below the endpoint of the spring at the time when the spring is maximally elongated, how would I do so using the PDE solution?
     
  11. May 8, 2017 #10
    The velocity is du/dt. The solution you get is for u(x,t). I would recommend solving this equation numerically, unless the imposed transient loading is pretty simple. Don't forget possible solutions to the wave equation based on the d'Alembert form of the solution.
     
  12. May 8, 2017 #11
    Oh I see. And yes, in the end I would probably solve the PDE numerically.
    What I'm still confused about is applying the obtained solution to the problem. The way you derived the given equation, is ##u## the change in position between the initial location ##x## of a material element in the undeformed configuration, or does function ##u## describe the position of the material element relative to the attachment point of the spring (since the spring is attached to an inertially fixed point in the problem)?
     
  13. May 8, 2017 #12
    It is the change in position between the initial location in the undeformed configuration.
     
  14. May 9, 2017 #13
    Okay, thanks.
    So I have also attempted to numerically solve the PDE, but I'm not sure about the boundary conditions that should be imposed with this problem statement. For the initial conditions, assuming that the geometry of the whole spring is compressed linearly, at ##t=0## we can say that ##u(x,0)=u_{0}## (##u_{0}## is a constant) and that ##\frac{\partial u}{\partial t}(x,0)=0##. Now for the boundary condition at ##x=0##, since the spring is just attached to a fixed point ##u(0,t)=0##. But then what would be the other boundary condition (perhaps for the moving endpoint of the spring)?
     
  15. May 9, 2017 #14

    Nidum

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    When you release the spring you go from an end condition where there is a force applied to one where there is no force applied . The change is notionally instantaneous . You are familiar with use of step functions ?

    This analysis is very similar to that used for determining the output response of electronic circuits and control systems to step function inputs .

    Just for interest :

    There is an upper limit to the velocity that springs can expand at when released . This is set by the speed at which a stress wave can travel through the material of the spring .

    Damping effects in real springs can be significant but are very hard to quantify except by experiment .
     
    Last edited: May 9, 2017
  16. May 9, 2017 #15
    If you take the partial derivative of wave equation in u with respect to x, you get the same wave equation in ##\epsilon=\partial u/\partial x##, where ##\epsilon## is the strain. This also means that the same wave equation is likewise satisfied in terms of the tension T. This might be easier to work with than the equation in u. For example, if you suddenly release the far end of the spring, the tension T suddenly becomes zero at x = Lo. Such a boundary condition would be ideally handled analytically by a d'Alembert form of solution.
     
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