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Math Help - [Help Me!] Linear motion equations with drag resistance

  1. #1
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    [Help Me!] Linear motion equations with drag resistance

    I've been taught about the normal linear motion equations now i've been set this task

    1. Identify the phenomenon that causes objects to hit the ground at different times [Completed]
    2. Determine the value of engineering coefficients
    3. If possible attempt to develop a mathematical model that takes both gravity and this phenomenon into account to accurately predict the fall time for different spherical objects.
    Other information

    Acceleration: 9.81ms^-2
    Fall height of object(s): 6M
    Shape of object(s): Sphere


    What I've done

    For part 1: I deduced the phenomenon is air resistance/drag.

    For part 2: Researched about drag and found out that

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

    Where F_d: Force of drag, p=density of fluid(1.1877 @298K/25C),
    v=speed of object
    a=reference areaπ(piD^2/4 for a sphere),
    C_d=drag coefficient(0.47 for a smooth sphere),
    v^=is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

    For part 3

    Not sure for this but was thinking of using v=u+at rearranging for time, draw force diagram calculate the modulus of the force then put in values
    Last edited by Kevlar; September 25th 2009 at 10:01 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kevlar View Post
    I've been taught about the normal linear motion equations now i've been set this task

    1. Identify the phenomenon that causes objects to hit the ground at different times [Completed]
    2. Determine the value of engineering coefficients
    3. If possible attempt to develop a mathematical model that takes both gravity and this phenomenon into account to accurately predict the fall time for different spherical objects.
    Other information

    Acceleration: 9.81ms^-2
    Fall height of object(s): 6M
    Shape of object(s): Sphere


    What I've done

    For part 1: I deduced the phenomenon is air resistance/drag.

    For part 2: Researched about drag and found out that

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

    Where F_d: Force of drag, p=density of fluid(1.1877 @298K/25C),
    v=speed of object a=reference areaπ(piD2/4 for a sphere),
    C_d=drag coefficient(whats this for sphere?),
    v^=is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

    For part 3

    Not sure for this but was thinking of using v=u+at rearranging for time, draw force diagram calculate the modulus of the force then put in values
    You should find that for a body dropping vertically in a resisting medium at high Reynolds Number that

     <br />
\dot{v}=-g+kv^2<br />
    where:

    k=\frac{\rho\; C_d A}{2m}

    The simplification is possible because we are only interested in the situation where v is negative.

    CB
    Last edited by CaptainBlack; September 25th 2009 at 09:53 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    You should find that for a body dropping vertically in a resisting medium at high Reynolds Number that

     <br />
\dot{v}=-g+kv^2<br />
    where:

    k=\frac{\rho\; C_d A}{2m}

    The simplification is possible because we are only interested in the situation where v is negative.

    CB
    no idea what v dot or reynolds number is, did some research found vdot is acceleration i think, and didnt really understand anything about reynolds no. something about the viscosity of air.

    Yet i still have no idea what you're on about.

    Hate this problem based learning wish they taught us the stuff first then set the task >_> not the reverse argh
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Kevlar View Post
    no idea what v dot or reynolds number is, did some research found vdot is acceleration i think, and didnt really understand anything about reynolds no. something about the viscosity of air.

    Yet i still have no idea what you're on about.

    Hate this problem based learning wish they taught us the stuff first then set the task >_> not the reverse argh
    the dot denote the time derivative:

    \dot{v}=\frac{dv}{dt}.

    Google for Reynolds number. High Reynolds number is the condition needed for the drag to be proportional to v^2, at low Reynolds number drag is proportional to v.

    CB
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