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Dissipative Function of Air Drag

by fobos3
Tags: dissipative, drag, function
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fobos3
#1
Jul7-10, 06:54 PM
P: 35
We have [tex]\dfrac{d}{dt}\left(\dfrac{\partial \mathcal L}{\partial \dot{q}_j}\right)-\dfrac{\partial \mathcal L}{\partial q_j}=-\dfrac{\partial \mathcal F}{\partial \dot{q}_j}[/tex].

If the force of friction is [tex]F=-kv^2[/tex] how do you go about calculating [tex]\mathcal F[/tex]

My idea is as follows

[tex]F=-kv^2 \dfrac{\textbf{v}}{|\textbf{v}|}[/tex]

In 2D we have

[tex]F=-k\dfrac{\dot{x}^{2}+\dot{y}^{2}}{\sqrt{\dot{x}^{2}+\dot{y}^{2}}}\left(\ begin{array}{c}
\dot{x}\\
\dot{y}\end{array}\right)[/tex]

[tex]F=-k\left(\begin{array}{c}
\dot{x}\sqrt{\dot{x}^{2}+\dot{y}^{2}}\\
\dot{y}\sqrt{\dot{x}^{2}+\dot{y}^{2}}\end{array}\right)[/tex]

And the Lagrange equations become

[tex]\dfrac{d}{dt}\left(\dfrac{\partial\mathcal{L}}{\partial\dot{q}_{j}}\rig ht)-\dfrac{\partial\mathcal{L}}{\partial q_{j}}=-k\dot{q}_{j}\sqrt{\sum_{i=1}^{n}\dot{q}_{i}^{2}}[/tex]

Is this correct?
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