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I am a bit confused on how this works. For instance if I wanted to look at an object moving in 2 dimensions only subject to gravity (and assuming that the potential is just mgy), I get that my Lagrangian is:

$$\frac12m(\dot x^2+\dot y^2)-mgy$$

I can easily solve the Euler-Lagrange equations, to see that $\ddot y=-g$ but is there a way I could do this with generalized forces. I feel that this is an awful example. I vaguely remember my professor using generalized forces to find friction force. Any help with understanding generalized forces would be appreciated.

The reason I want to know is that I am trying to model a curveball (baseball) and treating the lift and drag as generalized forces then using the E-L equations to solve for the motion.

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1 Answers1

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I) In case of a point particle with mass $m$ (and no moment of inertia), the best one can do seems to be to model the friction/drag via a Rayleigh dissipation function ${\cal F}(v^2)$ with a friction/drag force

$$\tag{1} {\bf F}_f~:=~-\frac{\partial {\cal F}(v^2)}{\partial {\bf v}} ~=~-2{\cal F}^{\prime}(v^2){\bf v},$$

i.e. the Lagrange equations read

$$\tag{2} \frac{d}{dt} \left(\frac{\partial L}{\partial {\bf v}}\right)-\frac{\partial L}{\partial {\bf r}}~=~{\bf F}_f, $$

with Lagrangian

$$\tag{3} L~=~\frac{1}{2}mv^2-mgz.$$

II) One may prove using the methods of this and this Phys.SE answers, and this mathoverflow answer, that the friction/drag force (1) does not have a velocity-dependent potential $U({\bf r},{\bf v})$.

III) In case of a rotating ball with non-zero moment of inertia and Magnus effect, the situation becomes more complex (and model dependent). We leave it for others to investigate this in detail.

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  • So I could essentially use (2), but for each $q_i$ I would have different forces of friction, based on my angular velocity and initial velocity. Because the drag goes as $\vec v$ and lift goes as $\vec{\omega} \times \vec v$ – yankeefan11 Dec 05 '13 at 01:55