# Dissipative Function of Air Drag

1. Jul 7, 2010

### fobos3

We have $$\dfrac{d}{dt}\left(\dfrac{\partial \mathcal L}{\partial \dot{q}_j}\right)-\dfrac{\partial \mathcal L}{\partial q_j}=-\dfrac{\partial \mathcal F}{\partial \dot{q}_j}$$.

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

My idea is as follows

$$F=-kv^2 \dfrac{\textbf{v}}{|\textbf{v}|}$$

In 2D we have

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

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

And the Lagrange equations become

$$\dfrac{d}{dt}\left(\dfrac{\partial\mathcal{L}}{\partial\dot{q}_{j}}\right)-\dfrac{\partial\mathcal{L}}{\partial q_{j}}=-k\dot{q}_{j}\sqrt{\sum_{i=1}^{n}\dot{q}_{i}^{2}}$$

Is this correct?

Last edited: Jul 8, 2010