- #1
fobos3
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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}}\right)-\dfrac{\partial\mathcal{L}}{\partial q_{j}}=-k\dot{q}_{j}\sqrt{\sum_{i=1}^{n}\dot{q}_{i}^{2}}[/tex]
Is this correct?
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}}\right)-\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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