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Thread: Trying to formulate then solve DE [1stOrder notsure what type]

  1. #1
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    Trying to formulate then solve DE [1stOrder notsure what type]

    This is about dropping a spherical object from 6m high and creating and solving a mathematical model that includes drag.

    What i've done so far is.

    Got that the force of drag is this

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

    Drawn a free body diagram showin gravity acting downwards[+ve] and drag opposing this[-ve]

    Then using F=ma

    $\displaystyle mg -{1 \over 2} \rho v^2 A C_d=ma$

    (ignored the vector as im working in 1D [V hat])

    Then i changed $\displaystyle -{1 \over 2} \rho v^2 A C_d$ by removing $\displaystyle v^2$ to create the constant $\displaystyle k=\frac{1}{2}\rho A C_d$

    To get

    $\displaystyle mg-k v^2=ma$

    then using $\displaystyle a=\frac{dv}{dt}$

    i get $\displaystyle mg-k v^2=m\frac{dv}{dt}$

    now i'm trying to rearrange this to get my differential eqn this is where i get stuck

    from the above im doing this

    $\displaystyle dt(mg-k v^2)=m{dv}$

    $\displaystyle dt=\frac{m}{mg-k v^2}dv$

    now what do i do?
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  2. #2
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    Quote Originally Posted by Kevlar View Post
    This is about dropping a spherical object from 6m high and creating and solving a mathematical model that includes drag.

    What i've done so far is.

    Got that the force of drag is this

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

    Drawn a free body diagram showin gravity acting downwards[+ve] and drag opposing this[-ve]

    Then using F=ma

    $\displaystyle mg -{1 \over 2} \rho v^2 A C_d=ma$

    (ignored the vector as im working in 1D [V hat])

    Then i changed $\displaystyle -{1 \over 2} \rho v^2 A C_d$ by removing $\displaystyle v^2$ to create the constant $\displaystyle k=\frac{1}{2}\rho A C_d$

    To get

    $\displaystyle mg-k v^2=ma$

    then using $\displaystyle a=\frac{dv}{dt}$

    i get $\displaystyle mg-k v^2=m\frac{dv}{dt}$

    now i'm trying to rearrange this to get my differential eqn this is where i get stuck

    from the above im doing this

    $\displaystyle dt(mg-k v^2)=m{dv}$

    $\displaystyle dt=\frac{m}{mg-k v^2}dv$

    now what do i do?
    Integrate both sides.
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  3. #3
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    I was thinking of doing

    $\displaystyle dt = (\frac{1}{g-k v^2}dv)$

    then changing the denominator to

    $\displaystyle dt = \frac{1}{\sqrt g-v \sqrt k)(\sqrt g-v \sqrt k}dv$

    $\displaystyle \int dt =\int \frac{1}{(\sqrt g-v \sqrt k)(\sqrt g+v \sqrt k)}dv $

    what partial fractions should i get on the right hand side before integrating

    $\displaystyle \int (\frac{1}{\sqrt g-v\sqrt k}+\frac{1}{\sqrt g+v\sqrt k})$
    this?, and if so what does it integrate to, or can someone show me how to integrate this

    would i get this?

    $\displaystyle t= ln(\sqrt g-v\sqrt k)+ln(\sqrt g+v\sqrt k)$
    Last edited by Kevlar; Sep 26th 2009 at 06:55 AM.
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  4. #4
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    Try expressing it in terms of $\displaystyle \tanh^{-1}$ since:

    $\displaystyle \int\frac{1}{1-u^2}du=\tanh^{-1} u$

    You can put:

    $\displaystyle \int_{v_0}^{v}\frac{m}{mg-kv^2}dv$

    into that form right?

    Or just leave it as logs but you need factors of $\displaystyle \sqrt{k}$ in your integration above since you're letting $\displaystyle u=a+bv$ and $\displaystyle du=bdv$. Whatever works for you.
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