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Homework Help: Why does substitution F = ma in this problem not work?

  1. Oct 16, 2012 #1
    We have a gravitational force on Planet X [tex]F=mγy^2[/tex] and we want to know the particle's final velocity. I know how to get the right answer, but I am wondering how come this doesn't work.

    [tex]F = mγy^2[/tex]
    [tex]ma = mγy^2[/tex]
    [tex]∫a dt= γ∫y^2 dt[/tex]
    Integral from 0 to t, I take v_0 = 0 y_0=0
    [tex] v = γy^3/3[/tex]
  2. jcsd
  3. Oct 16, 2012 #2


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    You don't say what y is, but shouldn't a gravitational force be inversely proportional to some distance squared, not directly proportional?
  4. Oct 16, 2012 #3
    Yes it should. But that isn't the case here. I'll quote the book:

    "Near to the point where I am standing on the surface of Planet X, the graviational force on a mass m is a vertically down but has magnitude mγy^2, where γ is a constant and y is the mass's height above the horizontal ground...although most unusual [the gravitational force] is still conservative."
  5. Oct 16, 2012 #4


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    You're integrating the position over the time variable. Your integral looks like

    [itex]\int y(t)^2 dt[/itex]

    in reality and you don't know what the functional dependence of y(t) is. What you did was basically:

    [itex]\int y^2 dy[/itex]

    which is very different from the first integral.
  6. Oct 16, 2012 #5
    Oh, thanks! Your right, I have absolutely no idea how y changes as a function of time, it all makes sense now.
  7. Oct 16, 2012 #6


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    Here's a trick if [itex]a = \ddot{y}[/itex]:
    m\ddot{y} &= m\gamma{}y^2 \\
    \ddot{y} &= \gamma{}y^2 \\
    \frac{\mathrm{d}\dot{y}}{\mathrm{d}t} &= \gamma{}y^2 \\
    \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}t} &= \gamma{}y^2 \\
    \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\dot{y} &= \gamma{}y^2 \\
    \int_{\dot{y}_0}^{\dot{y}}\dot{y}\mathrm{d}\dot{y} &= \int_{y_0}^y\gamma{}y^2\mathrm{d}y \\
    \frac{\dot{y}^2 - \dot{y}_0^2}{2} &= \gamma{}\frac{y^3 - y_0^3}{3} \\
    \dot{y} &= \sqrt{\dot{y}_0^2 + \frac{2\gamma}{3}\left(y^3-y_0^3\right)}
  8. Oct 16, 2012 #7
    I like that trick! I recall a professor of mine showed it to me last year when I was trying to solve a problem. I know it is separation of variables and playing around with the math, but does this trick have a specific name?
  9. Oct 17, 2012 #8


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    The left part is basically the kinetic energy (and its time-derivative before the integration), so it is related to energy conservation.

    ##F=m\gamma y^2## gives a potential of
    ##V(y)=\frac{m\gamma}{3} y^3##

    The total kinetic energy ##\frac{m}{2}\dot{y}^2 + V(y)## is conserved, write it down, and you get the same result without integration.
  10. Oct 17, 2012 #9
    Thanks, that is how the textbook tells you to solve it as well, that is to use conservation of energy. But I just wanted to try something different and was wondering why it didn't work.

    I didn't realize that you can use ## for latex, thanks a lot!
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