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Initial velocity required for a certain displacement

  1. Sep 16, 2013 #1
    Apologies, this is most likely basic to most of you, and there are probably much better ways to go at this than I have.

    1. The problem statement, all variables and given/known data

    An 8mm wide cone of a hypothetical substance which does not melt, with a mass of 2.7 grams, is shot straight up at a 90 degree angle. Only accounting for fluid drag and gravity, what are the initial velocities required for it to:
    a) reach a height of 55000 m
    b) be travelling at 50% of its initial velocity at that height
    c) travel there in 1 second

    2. Relevant equations
    As far as I know:
    [itex]F_D=1/2\rho v^2C_DA[/itex]
    [itex]F=MA[/itex]

    3. The attempt at a solution
    My attempt was to create an equation for displacement vs time:
    [itex]\delta =\rho C_DA/2M[/itex]
    [itex]v=v_i-(\delta v^2+g)t[/itex]
    After isolating v:
    [itex]v=\sqrt{\frac{v_i}{\delta t}-g\delta+(\frac{1}{2\delta t})^2}-\frac{1}{2 \delta t}[/itex]
    After integrating it, I got (ignore the vector stuff, I don't know why I put that there):
    XxbL0JQ.png
    Which is far as I can go without a mathematica to isolate d for me.

    Is there a specific method to go about this, to make it simpler? Or do I need to use a mathematica (don't have one atm, not really sure how to use them)? Or is this all completely wrong?
     
  2. jcsd
  3. Sep 17, 2013 #2

    BruceW

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    Unfortunately, this bit is not right. I'm guessing you had dv/dt then multiplied by t to get v. But this does not work, because v^2 is a function of t - not a constant. Some advice: Start with dv/dt, but don't integrate the whole thing with respect to t. hint: use separation of variables. p.s. this problem is a bit tricky, but you can get there in the end.
     
  4. Sep 17, 2013 #3
    Sorry, new to this. Probably should've mentioned that this is actually my older brother's homework, which I thought I'd try my hand at.

    Would this be right for the velocity part?
    [itex]ma=m\frac{dv}{dt}=mg+kv^2[/itex] (where [itex]k=\frac{1}{2}\rho C_D A[/itex])
    ∴[itex]\frac{dv}{dt}=g+\frac{kv^2}{m}[/itex]
    [itex]dt=\frac{dv}{g+ \frac {kv^2}{m} }[/itex]
    And then integrate both sides, and isolate v after?
     
    Last edited: Sep 17, 2013
  5. Sep 18, 2013 #4

    BruceW

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    that's it, that's the right way. nice work. The rest of the exercise is just doing a difficult integral and rearranging. Also, you've now got the force to be in the positive direction, so you have 'down is positive', and the initial velocity will be negative. There is no problem with doing it this way round, but you do need to remember which way you have done it.
     
  6. Sep 18, 2013 #5
    Okay, modifying my equation so that it's negative. Forgot about that.

    [itex]dt=\frac{dv}{-g- \frac {kv^2}{m} }[/itex]

    About the integration: am I supposed to get an indefinite integral, or integrate from 0 to the variable?
     
  7. Sep 18, 2013 #6

    BruceW

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    it is a definite integral. You can think of what the limits are. What is the velocity when t=0 and what is the final velocity that you are interested in?
     
  8. Sep 18, 2013 #7
    Okay. So would the first scenario be something like [itex]\int_{0}^{t}dt'=\frac{1}{g}{}\int_{v_i}^{0}\frac{dv'}{-1-c^2v'^2}[/itex] where [itex]c=\sqrt{\frac{k}{mg}}[/itex]?
     
  9. Sep 18, 2013 #8

    BruceW

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    yep, that's right.
     
  10. Sep 18, 2013 #9
    So:
    [itex]t=\frac{tan^{-1}(cv_i)}{cg}[/itex]
    [itex]tcg=tan^{-1}(cv_i)[/itex]
    [itex]v_i=\frac{tan(tcg)}{c}[/itex]
    Doesn't really feel right.
    Tried integrating from v_i to v:
    [itex]t=\frac{tan^{-1}(cv_i)}{cg}-\frac{tan^{-1}(cv)}{cg}[/itex]
    [itex]tcg=tan^{-1}\frac{cv_i-cv}{1+c^2v_iv}[/itex]
    [itex]\frac{cv_i-cv}{1+c^2v_iv}=tan(tcg)[/itex]
    [itex]v=\frac{cv_i-tan(tcg)}{c^2v_itan(tcg)+c}[/itex]
    Am I on the right track?
     
  11. Sep 19, 2013 #10

    BruceW

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    what method did you use to get to this
    ##t=\frac{tan^{-1}(cv_i)}{cg}##
    I don't think this is right. But your equation
    ##\int_{0}^{t}dt'=\frac{1}{g}{}\int_{v_i}^{0}\frac{dv'}{-1-c^2v'^2}##
    is right. It is a bit tricky to go from here. You can probably find it in a list of tricky integrals, if you are satisfied with doing it that way. Or, you can get the answer yourself, with a bit of guesswork/intuition.
     
  12. Sep 19, 2013 #11
    So, was the second integration also incorrect?
     
  13. Sep 19, 2013 #12

    BruceW

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    yes, the second one was also incorrect. you should be integrating the speed from vi to zero, so the limits was not the problem with the first integration.
     
  14. Sep 20, 2013 #13
    I tried looking around. I couldn't find the problem, but I must admit, I'm kinda confused.
     
  15. Sep 20, 2013 #14

    BruceW

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    you've done the clever bit already by getting
    ##\int_{0}^{t}dt'=\frac{1}{g}{}\int_{v_i}^{0}\frac{dv'}{-1-c^2v'^2}##
    The stuff you still need to do is find out what the function on the right-hand-side will integrate to. This is the tedious/maybe a bit boring bit. Have you had much practice at doing lots of different kinds of integrals? There are several different standard 'tricks' which (possibly) can make an integral into a form which you know how to integrate.
     
  16. Sep 20, 2013 #15
  17. Sep 21, 2013 #16

    BruceW

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    ah! yes, you're totally right. Sorry about that, I was thinking of another way in which it can be written. I forgot that it can be written in the arctan form.

    OK, your answer of
    ##v_i=\frac{tan(tcg)}{c}##
    Is correct. It does look a bit weird, but if you think about it for a while, it starts to make sense. (Or at least, sounds plausible).

    You've done question c), so now on to a) and b). They ask for distance. You have an equation for velocity. So what is your next step?
     
  18. Sep 22, 2013 #17
    Integrate t?
     
  19. Sep 22, 2013 #18

    BruceW

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    yep, that's the way. And you've already written down the speed as a function of time:
    ##v=\frac{cv_i-tan(tcg)}{c^2v_itan(tcg)+c}##
    So you are not far now. Nice work, you are good at manipulating those inverse tangents! The integral from here is not too difficult. look at how you might be able to relate the numerator and denominator.

    edit: I thought you wrote "integrate it", not "integrate t"... I'm not sure how integrating t would help.
     
  20. Sep 23, 2013 #19
    Forgot that [itex]\rho[/itex] is proportionate to displacement. So, I'm guessing my next step would be solving another differential equation, [itex]\frac{dD}{dt}= ...[/itex]?
     
  21. Sep 24, 2013 #20

    BruceW

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    the density of the fluid is proportional to the displacement of the object? Since it is shot straight up, I guess this is the same as saying that the fluid density is proportional to the height? So the fluid density increases with height?

    Anyway, yeah, I guess you will need to start again, since before you were (implicitly) assuming that the density was a constant.
     
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