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Homework Help: Infinite time for an object to get K=0 conservative Force

  1. Nov 10, 2018 #1
    1. The problem statement, all variables and given/known data
    Think that we have only conservative force in this question and no nonconservative force like friction exists. Also## U ##means potential energy (eg Gravitational Potential Energy),##K## means Kinetic Energy (##1/2 mv^2##) and ##E = K + U = constant##

    We have a arbitrary Potential energy (U) in terms of position (x) diagram like the picture. The object is at first, in a position like ##x_0## and wants to get to point ##x_1##. At ##x_1## we have ##dU / dx = 0## and Also the initial Mechanical Energy (E = K + U) of the object at first is equal to the U at point ##x_1##. We want to prove that it takes infinite time -theoretically- for this object to get to the point ##x_1##. (The movement is 1 Dimensional)
    question physics 2.png
    2. Relevant Equations
    ##E = K + U = Constant##
    ##K = 1/2 m v^2##
    ##F = m a##
    ##F = -dU/dx##
    Taylor Series at ##x = x_1## for function ##f(x)##: ##f(x) = f(x_1) + f'(x) (x - x_1) + f''(x) (x - x_1)^2 / 2! + ...##
    All forces are conservative in this question.

    3. The attempt at a solution

    At first we know that the K at ##x_1## equals to zero. Intuitively, I can understand the at the end of the path, the acceleration goes to zero and also the velocity goes to zero (because ## K -> 0 ==> V -> 0##). But I don't know how to prove this. I know that F = -dU/dx.

    For the start, I get this from taylor series but I don't know what to do with it:

    ##U(x) = U(x1) + dU/dx (x - x1) + O(x^2)## -> ##U(x) = U(x1) + (-m a) (x - x1)
     
    Last edited: Nov 10, 2018
  2. jcsd
  3. Nov 10, 2018 #2

    PeroK

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    What would happen to a particle that started at ##x_1## with ##v = 0##?
     
  4. Nov 11, 2018 #3
    It will not move at all. because ##dU/dx = 0 -> F =0 -> a= 0 , v =0##. But I want to derive something that mathematically and not intuitively shows for particles started at ##x=x_0## with ##v>0##, the time that they will get to ##x = x_1## is going to infinite. (something like ## t = 1 / (x-x_1)##). Probably using Taylor Series.
     
  5. Nov 11, 2018 #4

    PeroK

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    To make things easier, you could let ##E = 0## and note that, for a local maximum, ##U'(x) = 0## and ##U''(x) \le 0##.
     
    Last edited: Nov 11, 2018
  6. Nov 11, 2018 #5
    I think ##U''(x) \leq 0## at local maximum. However, I don't know how to relate this equations to time. We don't know anything about how U is related to t so we can't easily integrate something like ##(U(x) - U(x_1) )dt^2 = - m (x - x_1)d^2 x##
     
  7. Nov 11, 2018 #6

    PeroK

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    Yes, of course. I've corrected that.

    What about using ##v^2 = \frac2m(E -U(x))##?
     
  8. Nov 11, 2018 #7
    Maybe using this We can get:

    ##E = U(x_1) => U(x) = E + 0 + U''(x_1)(x-x_1)^2 / 2 + ...##

    So ##v^2 = U''(x_1) (x-x_1)^2 / m)##. Now what to do with that? Maybe I can say that t at the last part approximately equals to ##t = (x-x_1) / v## so I will get something like ##t= m / \sqrt {(-U''(x_1))}## Which is not what we want.

    Even if I say something like ##1/2 a t^2 + v t = 0## (##v## of the last moments), we get t =0 or ##t = -2U''(x_1) (x-x_1)^2 / ( -m^2 U'(x))## and I don't get something reasonable from this.

    So what can I do from Here?
     
  9. Nov 11, 2018 #8

    PeroK

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    First, a mathematical observation.

    It's possible, although I haven't checked it out, that you could construct a pathological function ##U(x)## that breaks this rule. So, you may need ##U(x)## to be "well-behaved": i.e. a physically possible potential. One common assumption to make in these cases is that the derivatives of ##U## are uniformly bounded. I.e.

    ##\exists \ M > 0 \ ## such that ##\forall x, n \ |U^{(n)}(x)| < M##

    Alternatively, and less rigorously, you could note that, for small enough ##x-x_1##

    ##E - U(x) \approx E - U(x_1) - (x-x_1)U'(x_1) - \frac12 (x-x_1)^2U''(x_1)##

    And in this case, therefore, you have:

    ##E - U(x) \approx - \frac12 (x-x_1)^2U''(x_1) = k^2(x-x_1)^2 ##

    For some ##k## and ##x## close to ##x_1##. Let's ignore the case where ##U''(x_1) = 0##. You could do that as an additional piece of work if you want.

    Now assume that the particle gets close enough to ##x_1## in a finite time for the above to hold - some point ##x_2##, say. Again, you may require ##U## to be well-behaved for this assumption to hold (edit: this assumption is safe enough, although you might want to prove it!). Then we have:

    ##\Delta t = \int_{x_0}^{x_1}\frac{dx}{v} = \int_{x_0}^{x_2}\frac{dx}{v} + \int_{x_2}^{x_1}\frac{dx}{v}##

    This is the time to get from ##x_0## to ##x_1##. Can you finish things from there?
     
    Last edited: Nov 11, 2018
  10. Nov 11, 2018 #9
    I think the integral is something in form of ##C ln (x_1-x)## so when x goes to x_1, log goes to -infinity and because of this, the time is infinity. Is this reasoning correct?

    Also I must note that because the problem has a diagram like the one above, we can assume that the U(x) is well-behaved.
     
  11. Nov 11, 2018 #10

    PeroK

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    Yes, it hinges on the fact that ##\int_0^1 \frac{dx}{x^s}## diverges for ##s \ge 1##.
     
  12. Nov 11, 2018 #11

    TSny

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    Is it possible to construct a shape of the potential energy "hill" such that the time is finite rather than infinite?

    For convenience, choose the origin of the x-y coordinate system at the top of the hill. In this coordinate system, suppose the hill is described by

    ##U(x) = -|x|^{1.8}##. The first derivative of U exists at x = 0 with U'(x) = 0. But higher order derivatives do not exist at the origin. I don't know if this would be considered "pathological" according to @PeroK 's comments in post #8.

    The graph of U(x) vs. x looks like an OK hill.

    upload_2018-11-11_14-50-0.png


    If the particle starts at ##x_1## with just enough initial speed to reach ##x = 0## with zero speed, then I get that the time to travel from ##x_1## to ##x = 0## is

    t = ##10\sqrt{\frac{m}{2}} x_1^{0.1}##
     
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