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Need help with fluid mechanics problem set by lecturer :[

  1. Sep 25, 2009 #1
    Note: Have been taught nothing about fluid mechanics yet, this task was set to help develop self learning skills, yet with some research i've carried out i'm still at a brick wall so i hope you don't mind helping me!

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

    Explain the phenomenon that causes a noticeable difference in the freefall time of different sixed balls.

    Determine the value of engineering co-efficients

    Formulate a mathematical moel that takes in account gravity and this other phenomenon to predict the fall time for a spherical object from a height of 6m (I'm supposed to do an experiment next week to put my equations to test, for now just use the letters)

    2. What I've done + Relevant equations

    Obviously the phenomenon is drag.

    Researched about drag and found out that

    [itex]\mathbf{F}_d= -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v}[/itex]

    Where F_d: Force of drag, p=density of fluid( 1.1877 @298K/25C),
    v=speed of object
    a=reference area ( [itex]\frac{\pi.d^2}{4}[/itex] for a sphere),
    C_d=drag coefficient(0.47 for a smooth sphere),
    v^=is the http://www.mathhelpforum.com/wiki/Unit_vector" indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 25, 2009 #2

    djeitnstine

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    So what exactly is your problem or what you need help with? It would be nice if you stated the issue you're having.

    So far you're on the right track. This does not seem like an extremely difficult project at your level to perform.
     
  4. Sep 25, 2009 #3
    What do i do now to develop the mathematical model i'm just confused and lost now
     
  5. Sep 25, 2009 #4

    djeitnstine

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    Well in any physical problem such as this one, the first thing you want to do is draw a diagram of the sphere and check the forces acting on it. This is called a free body diagram.

    Ideally, there will only be two major forces acting on your sphere as it falls, can you identify them?
     
  6. Sep 25, 2009 #5
    Gravity pulling the sphere downwards, then upwards drag right? :s
     
  7. Sep 25, 2009 #6

    djeitnstine

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    Exactly. Now since you have one of the two required equations, you now need to find an equation for free fall for constant acceleration.
     
    Last edited: Sep 25, 2009
  8. Sep 25, 2009 #7
    so then using f=ma

    mass.gravity - drag = mass.acceleration due to gravity

    then putting in the drag eqn i get

    [itex]mg - -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v} = mg[/itex]

    this right?, then just substitute in all the values? doesnt seem right to me argh
     
  9. Sep 25, 2009 #8

    djeitnstine

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    That's night quite right. You're trying to get the time that it will take for the object to fall. The idea is to find some equation that relates acceleration and time due to gravity.

    You need to do a little research, I do believe that equation is somewhere in the forum (hint you can search the PF library).

    What you want to do is construct some equation of the form [tex] x = f(t) [/tex] (x: displacement) that is, displacement as a function of time.
     
  10. Sep 25, 2009 #9

    djeitnstine

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    The reason why I said to look for the forces at play here is so that you can get an idea of what type of equations you will need to predict its motion.
     
    Last edited: Sep 25, 2009
  11. Sep 25, 2009 #10
    [itex]


    & s && = \tfrac12(u+v)t
    [/itex]
    [itex]s= ut + \tfrac12 at^2

    [/itex]

    is it the top one i should be using?
     
  12. Sep 25, 2009 #11

    djeitnstine

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    Good. The bottom one is a better approximation to what you want.
     
  13. Sep 25, 2009 #12
    so then i put

    [itex]6 = (0)t + 0.5(9.8)t^2[/itex]
    [itex]6 = 4.9t^2[/itex]
    [itex]t = \sqrt\frac{6}{4.9}[/itex]
     
  14. Sep 25, 2009 #13

    djeitnstine

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    Looks good to me. 6m is not that high and this is a good first approximation of the time it will take to land.
     
  15. Sep 25, 2009 #14

    djeitnstine

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    Just as a pointer, this equation assumes no drag and constant acceleration.
     
  16. Sep 25, 2009 #15
    yeah i figured but i need one that takes into consideration the drag :\
     
  17. Sep 25, 2009 #16

    djeitnstine

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    Are you familiar with basic calculus? You can derive the equation yourself.
     
  18. Sep 25, 2009 #17
    Im good with calculus but i've never applied calculus to mechanics

    all i've heard is that vdot = dV/dt or something lol
     
  19. Sep 25, 2009 #18

    djeitnstine

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    Ok good. So remember the equation you placed. Lets assume down to be positive and up to be negative. The correct form of the equation you placed earlier would be

    [tex]mg-\frac{1}{2}\rho v^2 A C_d = ma [/tex] -don't worry about the vector for now we are in 1D so its not needed.

    now let some constant [tex]k = \frac{1}{2}\rho A C_d [/tex]

    Now, show me what the resulting equation looks like and how can we get an explicit equation in displacement (x or s)from this.
     
  20. Sep 25, 2009 #19
    1stly wasnt the equation a double minus as k= -0.5pv^2ACd and mg - Fd = ma => mg - -k = ma

    so mg + k = ma?

    and im confused by your last line, how do i get an explicit equn in displacemnt sorry im so terrible atm
     
  21. Sep 25, 2009 #20

    djeitnstine

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    The drag equation you showed assumed that [tex]F_d[/tex] acts opposite to velocity correct?

    Now in my statement I said down is positive and up is negative, if the ball is falling down what direction does drag act? and what is its sign according to my convention in the equation.

    (look at your free body diagram again)

    Also in the constant k I did not include the [tex]v^2[/tex] term look carefully and try again.

    How to get the equation as a function of displacement? isn't it true that [tex] a=\frac{dv}{dt} [/tex] and that [tex] v= \frac{ds}{dt}[/tex] ?
     
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