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Integrate Acceleration to Find Time?

  1. Mar 29, 2012 #1
    This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

    1. The problem statement, all variables and given/known data

    A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

    2. The attempt at a solution
    [tex]F = ma = \frac{k}{d}[/tex]
    where d is the distance from 0 and k is a proportionality constant. Therefore, [tex]a = \frac{d^2x}{dt^2} = \frac{k}{md}[/tex]
    I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

    Thanks,
    Trapezoid
     
  2. jcsd
  3. Mar 29, 2012 #2
    Well, imagine that the question asked you for an equation that gave the position of the particle as a function of time. E.g. [itex]s(t) = a_0 + a_1t + a_2t^2[/itex]. You know how to do this, right? Now, once you have the equation, just put the distance traveled on the LHS and solve for [itex]t[/itex].
     
  4. Mar 29, 2012 #3
    [tex]x\frac{d^2x}{dt^2} = \frac{k}{m}[/tex]
    d is x. Is that what you were forgetting?
     
  5. Mar 29, 2012 #4

    HallsofIvy

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    You say "d is the distance from 0 and k is a proportionality constant" but you don't say what "x" is! I think you mean x to be also the distance from 0 at time t so your differential equation is really
    [tex]\frac{d^2x}{dt^2}-= \frac{k}{mx}[/tex]

    You will have to solve the differential equation for x, then set it equal to 0 and solve for t.

    Now, that is not easy to integrate directly but you can use a technique called "quadrature".
    Let v= dx/dt so that [itex]d^2x/dt^2= dv/dt[/itex]. By the chain rule, [itex]dv/dt= (dv/dx)(dx/dt)= v dx/dt[/itex]. So your equation is
    [tex]v\frac{dv}{dx}= \frac{k}{mx}[/tex]
    a separable equation.

    [tex]v dv= \frac{k}{m}\frac{dx}{x}[/tex]
    Integrating,
    [tex](1/2)v^2= \frac{k}{m}ln(|x|)+ C[/tex]
    Since the object was released "distance L" (x= L) at rest (v= 0) we have
    [tex](1/2)0^2= \frac{k}{m}ln(L)+ C[/tex]
    so
    [tex]C= -\frac{k}{m}ln(L)[/tex]
    and the equation for v becomes
    [tex]\frac{1}{2}v^2= \frac{k}{m}ln(x)- \frac{k}{m}ln(L)= \frac{k}{m}ln\left(\frac{x}{L}\right)[/tex]
    (I am assuming that L is positive and as x changes from L to 0, it will be positive so we don't need the absolute value.)
    [tex]v= \frac{dx}{dt}= \sqrt{\frac{2k}{m}ln\left(\frac{x}{L}\right)}[/tex]

    Integrate that, solve for x as a function of t, set it equal to 0, and solve for t.
     
  6. Mar 29, 2012 #5
    Thanks for all the help everybody. I'll take a closer look to see if I can't find the answer from here.
     
  7. Mar 29, 2012 #6
    I thought it would just be
    [tex]x\frac{d^2x}{dt^2} = \frac{k}{m}[/tex]
    [tex]\int \int x dxdx = \frac{k}{m} \int \int dtdt[/tex]
    [tex]\frac{x^3}{6}=\frac{k}{m}\frac{t^2}{2}+c[/tex]
    [tex]x^3=3\frac{k}{m}t^2+c[/tex]
    Here is a nice junction to find c (assuming x = 0 is at the point and x = L is the initial condition, L > 0):
    [tex]L^3 = 0 + c \rightarrow c = L^3[/tex]
    then
    [tex]x^3=3\frac{k}{m}t^2+L^3[/tex]
    [tex]x(t)=\left ( 3\frac{k}{m}t^2+L^3 \right )^{1/3}[/tex]

    You then solve for the time for which x(t) = 0. Note, k must be < 0 for such a solution.
     
  8. Mar 29, 2012 #7
    Thanks HallsofIvy and RoshanBBQ,

    Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral [itex]\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}[/itex] which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

    Trapezoid
     
  9. Mar 29, 2012 #8
    It just depends on how you set up the problem.I'm not entirely comfortable with the answer I found, though. Consider when t -> infinity. You would expect x -> 0. My solution has x become the 3rd root of an increasingly negative number...
     
    Last edited: Mar 29, 2012
  10. Apr 5, 2012 #9
    Hi all,

    Sorry to bump but we got the solutions to the assignment so I thought I'd post some more guidance in case anybody else is having trouble with a similar question.

    The method outlined by HallsofIvy is correct. Without giving too much away, to solve the last integral, substitute [itex] u = \frac{L}{x} [/itex] and use the gamma function.
     
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