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Particle moving under gravity

  1. Nov 11, 2009 #1
    1. A particle moves vertically under gravity and a retarding force proportional to the square of its velocity. If v is upward or downward speed, shot that a = +/-g -kv^2, where k is a constant. If the particle is moving upwards, show that its position at time t is given by;

    z = z0 +(1/k)lncos[rootgk (t0-t)] where z0 and t0 are integration constants.




    2. Relevant equations



    3. I proved the a = +/-g -kv^2 but I spent almost an hour looking at the next part with no results. I integrated twice to get x(t) = -gt^2/2 - kv^2t^2/2 + z0 + t0

    I have no idea what to do after this, I mean I dunno how you bring the cos into the problem. This hint is also given; lncosx = -1/2 ln(1 + tan^2x)

    Cheers
     
  2. jcsd
  3. Nov 11, 2009 #2

    gabbagabbahey

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    How exactly are you integrating [itex]-kv^2[/itex] with respect to time, if [itex]v(t)[/itex] is some unknown function of time?:wink:

    Instead, realize that [itex]a(t)=\frac{dv}{dt}=-g-kv^2[/itex]...which gives you a separable 1st order ODE....surely you've come across those before?:wink:
     
  4. Nov 11, 2009 #3
    Oh yes sorry I thought for some reason you could just integrate it with respect to time. But I don't see how changing it into a first order would help with the problem.
     
  5. Nov 11, 2009 #4

    gabbagabbahey

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    Have you not studied 1st order separable ODEs in any of your calculus courses yet?
     
  6. Nov 11, 2009 #5
    Yes we did them last year but I don't see how doing that will bring tan into the question.
     
  7. Nov 11, 2009 #6

    gabbagabbahey

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    Solve the ODE for v(t) and then use the fact that [itex]v(t)=\frac{dz}{dt}[/itex] to solve your resulting equation for z(t)...
     
  8. Nov 11, 2009 #7
    Alright cheers, will try that.
     
  9. Nov 11, 2009 #8
    Did it but I knew that a(t) = dv/dt anyway
    Still got the same results when I integrated
     
  10. Nov 11, 2009 #9

    gabbagabbahey

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    If you show me your steps, I can point out your error(s)...
     
  11. Nov 11, 2009 #10
    a(t) = dv/dt = -g-kv^2
    integrate, v(t) = -gt -kv^2t + t0

    integrate again to achieve z(t):

    Z(t) = -gt^2 - kv^t^2 + z0
     
  12. Nov 11, 2009 #11

    gabbagabbahey

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    No, you are making the same mistake you did in your first attempt...

    [tex]\int kv^2dt=kv^2t[/tex]

    only if v(t) is a constant....which it clearly isn't.

    Open up your old calc text and review how to solve separable 1st order ODEs.
     
  13. Nov 12, 2009 #12
    Ah yes, sorry I totally forgot what to do, didn't realize. Here's my solution;

    dv/dt = -g -kv^2
    integral(dv/-g-kv^2) = integral(dt)
    integral[(-1/g+kv^2)(dv)] = integral(dt)
    -1/k*integral[v^2 + (root(g/k)^2)] = integral(dt)
    Used math's tables; integral(1/x^2 + a^2) = 1/a inversetan x/a

    Cancelled terms etc, v = dz/dt

    Got z;
    z = zo - root(g)/root(k) ln [cos(root(gk))(to-t)]

    However the answer should be;
    z = zo + (1/k) ln [cos(root(gk))(to-t)]

    Any ideas?
     
  14. Nov 12, 2009 #13

    gabbagabbahey

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    Looks like you might have forgotten the factor of [itex]1/a[/itex] from

    [tex]\int\frac{1}{v^2+a^2}dv=\frac{1}{a}\tan^{-1}\left(\frac{v}{a}\right)+C[/tex]

    ....but unless you post the rest of your steps I can't tell for sure.
     
  15. Nov 12, 2009 #14
    I didn't;
    when applying that inverse tan I got;

    -1/k[root(k)/root(g) inversetan(v*root(k)/root(g))] = T-To
     
  16. Nov 12, 2009 #15

    gabbagabbahey

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    Okay, and what did you get when you solved that equation for v(t)?
     
  17. Nov 12, 2009 #16
    v(t) = root(g)/root(k) * tan (root(gk)*T1)

    where T1 = T-T0
     
  18. Nov 12, 2009 #17

    gabbagabbahey

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    What happened to the negative sign you had before?
     
  19. Nov 12, 2009 #18
    I just changed it mutiplied across by -1 and got; -T + T0 on the right hand side which I called T1
     
  20. Nov 12, 2009 #19

    gabbagabbahey

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    Okay, so now you have

    [tex]v(t)=\frac{dz}{dt}=\sqrt{\frac{g}{k}}\tan\left(\sqrt{gk}(t_0-t)\right)[/tex]

    Now use the variable substitution [itex]u=\sqrt{gk}(t_0-t)[/itex] to integrate the RHS....
     
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