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Math Help - 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

    \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

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

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

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

    To get

    mg-k v^2=ma

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

    i get 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

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

    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

    \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

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

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

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

    To get

    mg-k v^2=ma

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

    i get 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

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

    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

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

    then changing the denominator to

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

    \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

    \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?

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

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

    You can put:

    \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 \sqrt{k} in your integration above since you're letting u=a+bv and du=bdv. Whatever works for you.
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