Dissipative Function of Air Drag

[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?

 

[eljose79]

Yes, your approach is correct. You have correctly applied the formula for calculating friction force in terms of velocity, and your Lagrange equations are also correct. However, it is important to note that this approach assumes that the friction force is the only external force acting on the system. If there are other forces present, they should also be included in the Lagrange equations.