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SHowing that conservation of Energy is related to Newton's motion

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Given: [itex]\frac{1}{2}m (\dot x)^2 + mg x = E[/itex]

    Gravitational force is mg.

    We need to show that by solving this DE that we can confirm that the conservation of energy correctly describes one-dimensional motion (the motion in a uniform field). That is, that the same motion is obtained as predicted by Newton' equation of motion.

    3. The attempt at a solution

    My attempt was as follows:

    [itex]\frac{1}{2}m\dot x + mg x = E[/itex] so by moving things around a bit I can reduce this to

    [itex](\dot x)^2 = \frac{2(E - mgx)}{m}[/itex]

    and from there I can solve the DE:

    [itex]\dot x = \sqrt{\frac{2(E - mgx)}{m}} \Rightarrow \int dx =\int \sqrt{\frac{2(E - mgx)}{m}}dt[/itex]

    from there I can move some variables some more:

    [itex]\frac{\sqrt{m}}{\sqrt{E-mgx}} \int dx = \int \sqrt{2}dt \Rightarrow \sqrt{m} \int \frac{dx}{{\sqrt{E-mgx}}} = \int \sqrt{2}dt[/itex]

    trying a u substitution where [itex]u=\sqrt{E-mgx}[/itex] and [itex]du = -mgdx[/itex] I should have [itex]\frac{\sqrt{m}}{mg} \int \frac{du}{{\sqrt{u}}} = \int \sqrt{2}dt[/itex] which gets me to [itex]\frac{\sqrt{m}}{mg} \sqrt{u} + c = \sqrt{2}t[/itex]

    going back to what I substituted for u I have:
    [itex]\frac{\sqrt{m}}{mg} \sqrt{E-mgx} + c = \sqrt{2}t[/itex]

    and when I do the algebra I get (after moving the c over and squaring both sides):

    [itex]\frac{m}{m^2g^2}{E-mgx} = {2}t^2 - 2\sqrt{2}c+ c^2[/itex]
    [itex]\frac{E}{mg^2}-\frac{x}{mg} = {2}t^2 - 2\sqrt{2}c+ c^2[/itex]
    [itex]-\frac{x}{mg} =-\frac{E}{mg^2}+ {2}t^2 - 2\sqrt{2}ct+ c^2[/itex]
    [itex]x(t) =\frac{E}{g}- 2mgt^2 + 2\sqrt{2}mgct- c^2mg[/itex]

    I am unsure of the last step. I see something that looks like an equation of motion there -- and since [itex]gt = v = \dot x[/itex] I can see that term, and [itex]gt^2 = x[/itex]. But I am not quite clear on what to do with the energy term.

    I know, I know, don't just plug in formulas. But I presume that the prof gave us the instructions he did for a reason and it wasn't just to confuse us, though he has succeeded in that regard. :-)

    Anyhow, any assistance is appreciated.
     
  2. jcsd
  3. Oct 14, 2013 #2

    tiny-tim

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    Hi Emspak! :smile:

    hmm … you want …
    sooo you want an equation with x'' in it

    hint: what's the most obvious way of getting x'' out of the original equation? :wink:
     
  4. Oct 14, 2013 #3
    But.. I did that, I came up with a solution for the DE. It's x(t). mg = force, int he solution I got, and I suppose I could rewrite the solution as [itex]x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x = c^2mg[/tex]. But I am still not sure about this.
     
  5. Oct 14, 2013 #4
    But.. I did that, I came up with a solution for the DE. It's x(t). mg = force, int he solution I got, and I suppose I could rewrite the solution as [itex]x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x - c^2mg[/itex]. But I am still not sure about this.
     
  6. Oct 14, 2013 #5
    sorry for doubling up...
     
  7. Oct 14, 2013 #6

    tiny-tim

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    no, you started with x' and x, and you eliminated x' (by integrating)

    try getting to something with x'' x' and x :smile:
     
  8. Oct 14, 2013 #7
    is this what you are referring to? You're being a bit cryptic...

    [itex]x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x - c^2m\ddot x [/itex] which becomes

    [itex]x(t) = \frac{E}{g}- (2m-mc^2) \ddot x t^2+ 2\sqrt{2}\dot x[/itex]

    [itex]x(t) = \frac{E}{g}- m(2-c^2) \ddot x t^2+ 2\sqrt{2}\dot x[/itex]

    EDIT: looking at this I can see the energy term could be moved over and you'd end up with Energy = equation of motion. Is that what we are after here ?
     
  9. Oct 14, 2013 #8

    tiny-tim

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    d/dt (1/2 mx'2) = mx'x'' :wink:
     
  10. Oct 14, 2013 #9
    (sigh) that isn't helping much, there are no x'x'' terms in the equation as I re-wrote it. I am not at all clear on how that derivative is helpful, unless you are alluding to something that should happen in the E term.
     
  11. Oct 14, 2013 #10

    tiny-tim

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    try differentiating the whole original equation …
     
  12. Oct 14, 2013 #11
    How about this: when you talk about getting x out of the equation where in the derivation I did should that be happening? Did I do the whole bloody thing wrong or what? Or was I correct that there's a last step here? If I plug a KE equation in for E I would have [itex] \frac{m(gt)^2}{g} [/itex] there. But I suspect that is not what you are taking about doing.
     
  13. Oct 14, 2013 #12
    OK, at that point I end up with [itex] m \dot x \ddot x + mg \dot x = 0[/itex], yes?
     
  14. Oct 14, 2013 #13

    tiny-tim

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    yup! :smile:

    and now divide by x' :wink:
     
  15. Oct 14, 2013 #14
    OK, we can do that because it would show up on both sides, right? [itex]m \ddot x = -mg[/itex]. Now since g was the acceleration (earth's gravity) and we have [itex]\ddot x[/itex] in the same place we've just shown that the equation of motion works here, or at least that in a 1-D case in a field that it does, correct?
     
  16. Oct 14, 2013 #15

    tiny-tim

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    correct! :smile:

    (and if you divide by m, you get x'' = -g, exactly as newton would have said)
     
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