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Newton's equations

  1. Jan 30, 2006 #1
    LOng question but not that hard apparently...
    Consider a body of mass m falling vertically from rest near the earth's surface. if air resistance is ignored then Newton;s equation [tex] m \ddot{y} = -mg [/tex] (1.1) has the solution [tex] \tilde{y} (x) = -\frac{1}{2} gt^2 + y_{0} [/tex](1.2) where Yo is teh initial position of the body. If air resistnace is taken into account Newton's equation must be modified to [tex] m \ddot{y} = -mg - \beta \dot{y} [/tex] (1.3) where beta is a positive constant. Assuming taht air resistance can be considered to b a small pertubartion (i.e. [tex] \left|\beta \dot{y} \right| << \left| mg \right| [/tex]) we can approximate (1.3) by the equation [tex] m \ddot{y} = -mg - \beta \dot{\tilde{y}} [/tex](1.4) where y(t) (tilde) is the unperturbed solution (with air resistance ignored) given by 1.2

    Solve 1.4 and discuss how this perturbative solution which we call Yp(t) compares to Y(t) tilde and determine under waht conditions (i.e. for what values of t) the perturbative solution is valid.

    now [tex] \dot{\tilde{y}} (x) = -gt [/tex]
    so am i simply going to substitute this into 1.4 and solve for y[t]?
    doing that gets [tex] m \ddot{y} = -mg + \beta gt [/tex]
    and [tex] y(t) = \frac{gt^3 \beta}{6m} - \frac{gt^2}{2} + tC_{2} + C_{2} [/tex]
    is this the right way to go?
     
  2. jcsd
  3. Jan 31, 2006 #2
    Should the beta-ydot bit be positive since it opposes the motion of the mass. air resistance doesn't make it go faster.

    Solve 1.4 and discuss how this perturbative solution which we call Yp(t) compares to Y(t) tilde and determine under waht conditions (i.e. for what values of t) the perturbative solution is valid.

    To solve that equation (1.4), I would solve for [tex]\dot{y}[/tex] and then integrate to get a function for y. I don't think that the two models are compatible since one completely ignores air resistance, while the other takes it into account, so I might wrong, but that is what I would do.
     
  4. Jan 31, 2006 #3
    so then what i have doen is correct? Since y tilde is totally diffferent from y itself. However y does depend on t , and x from my understanding. SO i cna solve for Yp(t) which i have given in my first post. But y tilde (t) ... would that be using this equation?

    [tex] m \ddot{\tilde{y}} = -mg - \beta \dot{\tilde{y}} [/tex]
    so here i am exclusively solving for Y tilde(t) .
     
  5. Jan 31, 2006 #4

    HallsofIvy

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    Assuming that [itex]\beta[/itex] is small enough to be ignored, the "non-perturbative" solution, from my"= -mg is, of course, y= -(g/2)t2+ y0 and y'= -gt

    Now, replace y' on the right of the equation by -gt:
    [tex]my"= -mg- \beta(-gt)= -mg+ \beta gt[/tex]
    That should be easy to integrate.
     
    Last edited: Jan 31, 2006
  6. Jan 31, 2006 #5
    whats that latex you typed halls?

    isnt that what i did in my first post though?
     
  7. Feb 2, 2006 #6
    ok for the perturbed system im solving
    [tex] m \ddot{y} = -mg + \beta gt [/tex]
    integrate twice and i get
    [tex] Y_{p} (t) = \frac{ygt^3}{6m} - \frac{gt^2}{2} + C_{1} t + C_{2} [/tex]

    the second part of this question asks under waht conditions this solution is valid. There seem to be no resitrictions on the value of t. So this is valid for all t?
    ALso need to solve (1.3)
    taht is
    [tex] m \ddot{y} = -mg - \beta \dot{y} [/tex]
    integrate once

    for the unperturbed system i get
    [tex] m \ddot{y} = -mg - \beta \dot{y} [/tex]
    integrate once and i get
    [tex] \dot{y} = -gt - \frac{\beta y}{m} + C [/tex]
    seem to be having trouble iwth this differential equation... any ideas on how to solve this?
     
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