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Physicality of a force from Taylors CM

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Problem 2.14 from Taylor's text. The problem involves caluculating the equations of motion for the force of form shown below. Additionally it asks to calculate max distance from start time. That is all well and fine.

    My question is, where is this force seen in Nature? Is it seen in any common problems in physics? I would think that it probably has an application somewhere.

    $$F=-F_0 \exp{v/V}$$

    I think it is probably just a better approximation for drag forces. That is,
    $$F+F_0=-F_0 v/V +-F_0 v^2/V+-F_0v^3/V^3+....$$
    The linear and quadratic terms are included in the Exp function.
    Where $$F_0$$ is the force on the object at zero speed, like an ambient background or something..

    2. Relevant equations

    newtons second law

    3. The attempt at a solution

    Any comments will be appreciated.
     
  2. jcsd
  3. Jul 23, 2013 #2
    Also, sorry for not better stating the force equation. v is the speed of the particle, and V is a positive constant, F_0 is also a positive constant.
     
  4. Jul 24, 2013 #3
    I dont think it is necessarily a better approximation for drag. I mean I guess it depends what you are comparing it too? Certainly there are better models of drag (although, have you know it, the Navi-Stokes Millenium problem deals with this exact problem, and it hasn't been solved!). But on the other hand, it's a good approximation. In laminar regions drag acts linearly with velocity, and in turbulent regions it acts as velocity squared...as you probably know. But the point I'm making it the linear and squared flow are approximations, but to add velocity cubed etc. to approximate it doesn't work...especially since the taylor series expansion always has the same sign. So drag isn't converging into a region where it's dependent upon a combination of linear and squared drag.

    With that said...it still makes a great approximation for drag, which is probably why it's being used (are you in the drag section by any chance?) Usually functions that approximate things to second order or higher are great approximations, and this function does just that. This function also tends to be extremely easy to manipulate, which might be why it's being used. We can use an easier function to illustrate a point about drag...or whatever else we're approximating.

    In nature...hmm. I can't thing of anything that has that exact equation but there could be something. Drag is probably the closest thing. Then again, there's plenty of natural phenomenon that I'm not too familiar with =P
     
  5. Jul 24, 2013 #4
    Thanks for the response, spaderdabomb. I was just thinking a better approximation in certain regimes (i.e. certain Reynolds numbers) than the force, $$F(v)=-bv-cv^2$$. Though like you say, for most applications with this type of drag, it is probably accurate enough to look at terms up to v^2.

    "especially since the taylor series expansion always has the same sign"

    Thats why there is the negative constant thrown into the function before we Taylor expand. This way once we expand you get terms with coefficients like b, c and etc.. in the above force I wrote earlier. Plus you get some additional further accuracy perhaps.

    $$F(v)+F_0=-\frac{F_0}{V}v-\frac{F_0}{V^2}v^2-\frac{F_0}{V^3}v^3-...-...-..$$
    $$F(v)+F_0=(-bv-cv^2)-dv^3-...-...-..$$

    Perhaps what you are saying is that the cubic and higher terms are just unphysical as we never see those in Nature (I don't think so). I am thinking just strictly in a mathematical sense and being able to more accurately model a function, this time the drag force.

    Definitely the next step would be to subject the particle to the forces of the continuum model and deal with the Stokes. I am not quite delving into that with this problem haha.

    And yes, this problem is from the drag section of Taylor CM - Chap 2
     
  6. Jul 25, 2013 #5
    Well I was mostly referring to the exact equation when I said I couldn't think of anything in nature that models it. But...depends what you call nature. The discharge time of a capacitor? Well...that follow this equation multiplied by a few constants almost exactly. But does that count as nature? I kind of think it does if you ask me. If you want to boil things down to "nature" then really, what are we talking about? There's only 4 different fundamental interactions known between particles...strong interaction, weak interaction, electromagnetic interaction and gravitational interaction. None of those really have a cubed term in them, so to speak.
     
  7. Jul 25, 2013 #6
    Nature being everything we observe and phenomenon that arise from it. Certainly, the microscopic nature of a gas gives rise to its properties overall. Namely, the electrostatic forces attracting each of those molecules. These types of forces crop up, because for the scales that they are relevant they nearly completely capture the phenomenon we look to describe, in this case, air resistance. These forces are an effect that can be derived from the fundamentals. Or just inferred empirically. In this case, for this problem, I will not derive it from the fundamentals. Which if you wanted to would entail studying the Stokes flow around an arbitrarily shaped object, in a certain Reynolds number regime.

    I think that the cubic term and higher orders may give this phenomenological force more accuracy, when needed.
     
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