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Problem with solving equation of motion

  1. Mar 16, 2013 #1
    Hello everybody, this is my first time posting here, so if I make any errors, correct me please. :)

    1. The problem statement, all variables and given/known data
    We have a body of mass 2 kg and it starts to move with constant velocity 5 m/s along a straight line. After 2 seconds a force that decreases with square of the distance starts to act on a body. Amount of the force at the moment it started to act is 10 N. After what time the speed of the body will be equal to half of its initial value? How far from the starting position body will be at that moment?

    I hope everything is clear since this is rough translation from my language. :)

    So, we have:

    [itex]m = 2~kg[/itex] - mass of a body
    [itex]v_0 = 5~m/s[/itex] - initial constant velocity
    [itex]{F_x}_0 = 10~N[/itex] - amount of force that starts to act at position ##x_0##
    [itex]{t_x}_0 = 2~s[/itex] - time that passes when body reaches position ##x_0##
    ##F \propto 1/x^2## - force that decreases with square of a distance
    ##x_0## - distance that body moves with constant velocity

    2. Relevant equations
    Ok, now we can set the equation of motion. Since we have time and velocity with simple ##x_0~=~v_0*{t_x}_0## (because first 2 seconds body moves with constant velocity) we can calculate distance ##x_0## which is 10 m.
    For force F we can write ##F = k/x^2##.
    Given the initial amount of force and distance ##x_0## that we calculated we get that ##k = 1000~Nm^2##

    Equation of motion would then be:

    ##\ddot{x}*m = -k/x^2##
    3. The attempt at a solution
    Now the fun part. This is second-order nonlinear ordinary differential equation, which complicates things a bit.
    So, basically I did this:

    ##\ddot{x} = -k/mx^2##

    Then I made substitution ##\dot{x} = p(x)##, here p is function of x which means that ##\ddot{x} = p\dot{p}##.

    That yields:

    ##p\frac{dp}{dx} = -a\frac{1}{x^2}## where ##a = k/m##

    I separate the variables and integrate which yields:

    ##p^2/2 = a/x + C1##

    Then I get constant C1 by plugging initial values:
    ##\frac{{v_0}^2}{2} - \frac{k}{mx_0} = C1 \Rightarrow C1 = -37.5##

    Now I can calculate ##x_1##, a position where body has half of its initial velocity, that is 2.5 m/s:

    ##\frac{{\dot{x_1}}^2}{2} = \frac{k}{mx_1} - 37.5## where ##\dot{x_1} = \frac{\dot{x_0}}{2}##

    which gives ##x_1 = 12.31~m## (note: ##x_1## is distance from ##x_0## because of initial conditions)

    The problem is, how to get time? If I go and integrate this DE second time, I would get ##x(t)## and solve it, but the integral I have to solve is pretty nasty and I don't know how to solve it. Is there maybe some other way or some kind of catch that I'm missing here?
     
    Last edited: Mar 16, 2013
  2. jcsd
  3. Mar 16, 2013 #2

    Simon Bridge

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    Welcome to PF;
    OK - the problem statement is incomplete. However, from the context and what follows I can fill in the gaps. The origin for the force is the same place that the object started from, and the force acts in the opposite direction to the initial velocity of the object. Is that correct?

    So you can end up with: $$mx^2\ddot{x}=-k \; : \; \dot{x}(2)=5\text{m/s}\; , \; x(2)=10\text{m}$$

    ... surely, if ##\dot{x}=p##, then ##\ddot{x}=\dot{p}## ?
     
    Last edited: Mar 16, 2013
  4. Mar 17, 2013 #3
    Thank you.

    Yes, sort of. The force starts to act after 2 seconds, first 2 seconds body moves with constant velocity. And force acts in opposite direction of the initial velocity on the same x-axis, yes.
    That's what I came up with also.
    Erm, not really. This is 2nd-order nonlinear DE we are dealing with and in this substitution p is function of x, which means when you derive x' you apply a chain rule.
    See (2) here.

    I understand this is mostly a math problem, but I'm not very good at physics so maybe I am missing something here which would allow me to get that time without messing with integrals. Because you will get something like this:

    ##\int \frac{1}{\sqrt{\frac{a}{x} -~b}}\,dx = \int\,dt##

    And first integral is not really a nice one :biggrin:.
     
  5. Mar 17, 2013 #4
    I don't know if you set up the integral correctly or not, but by factoring out a b you can do a nice trigonometric substitution for it. The math is a tad bit hairy, but if you are already doing differential equations, it shouldn't be too much of a problem.
     
  6. Mar 17, 2013 #5

    rude man

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    Just a thought, may not work:

    F(x)/m = a = -k/mx2 = dv/dt
    k obtained from F(10) = 10
    dv/dt = dv/dx dx/dt = v dv/dx
    v dv/dx = -k/mx2
    which can easily be solved by separation of variables, yielding v(x) using your given initial condition.

    Then x2 d2x/dt2 = -k/m
    = x2 d/dt dx/dt = x2 dv/dt = x2 dv/dx dx/dt
    Again use separation of variables to solve for v(t).
     
  7. Mar 17, 2013 #6

    Simon Bridge

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    Ah... and you've adjusted the original post to clarify that I see, well done.

    Yeh - wen you make a substitution like that you lose the time information.

    Looks like the first part of rude_man's post is pretty much what you did and the second half contains a way forward. You may not need the equation of x(t) (or t(x) as it may be) because you already have the v and you've found the x. Let us know how you got on.
     
  8. Mar 17, 2013 #7
    Actually the equation x''=-k/x^2 yields an x function that can't be written in terms of elementary functions.
     
  9. Mar 18, 2013 #8
    Hm, I wrote it in the first place, maybe you missed it at first ;)

    Well, I've managed to solve integral with a special substitution. A bit of a hairy math, but solvable.
    Yes, it can be.
     
  10. Mar 18, 2013 #9
    If you can solve for x in this type of equation in elementary functions, ln(f(x)) + g(x) = t let me know
     
  11. Mar 19, 2013 #10

    Simon Bridge

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    ... and the "last edited" time makes that ambiguous. It is possible though and I have no reason to argue ;)
    Great - what was your solution?
     
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