1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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

    Pengwuino

    User Avatar
    Gold Member

    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

    jhae2.718

    User Avatar
    Gold Member

    Here's a trick if [itex]a = \ddot{y}[/itex]:
    [tex]
    \begin{align*}
    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)}
    \end{align*}
    [/tex]
     
  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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why does substitution F = ma in this problem not work?
  1. F=Ma Problem Help (Replies: 2)

  2. F= MA problem (new) (Replies: 3)

  3. F= MA problem (Replies: 1)

  4. Why F=ma (Replies: 2)

  5. F=ma zipline problem (Replies: 7)

Loading...