Virtual Work

So We’ve talked about mdc equations before in math monkey. They are usually in the form  m\: \ddot{x} + d\: \dot{x} + cx\: = 0\:

m Being the Mass, d the dissapation (dampening) and c the spring constants.

I won’t go into how one builds this equation up, there are many and I covered in math monkey pretty in depth on how to do it with the basic math pendel. One of these days I may get into a more indepth analysis of maybe a car or a real double pendle. Anyways, a part that wasn’t explained at all was the “Virtual Work” of the system, or namely the Q\: part of the Lagrange equation

\mathrm{Q}=\frac{d}{dt}\frac{\delta\mathrm{L}}{\delta\mathrm{\dot{\phi}}}-\frac{\delta\mathrm{L}}{\delta\mathrm{\phi}}

So whats Q? Well according to wiki:

Virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement will be different for different displacements.

Or whatever that means. In mathemagics, it means this:

Q\:= \displaystyle\sum_{i=1}^{n}\mathrm{M_i(t)}\frac{\delta\phi}{\delta\mathrm{x}}+\mathrm{F_i(t)}\frac{\delta\mathrm{r}}{\delta\mathrm{x}}+\mathrm{F_{di}(t)}\frac{\delta\mathrm{r_d}}{\delta\mathrm{x}}+\mathrm{R_i(t)}\frac{\delta\mathrm{r}}{\delta\mathrm{x}}

Namely the Sum of the Moment, external force, friction and dampening times their directional vector components first derivative. Or Well thats what it looks like when I read it anyways. I still haven’t quite got the hang/understanding of it. But I’ve never claimed to know what I’m doing. Just that I look like I know what I’m doing 😉

Well to calculate all this out theres quite a bit of work to do. I recently had to do such a thing in a system I had to describe and while digging through my notes I happened to fall on a trick I seemed to write to myself in a exam cheatsheet. Normally, you would calculate the vectoral force (for dampening in this example) and multiple it with its directional derivative like so:

\mathrm{F_{di}(t)}\frac{\delta\mathrm{r_d}}{\delta\mathrm{x}}

F could look like this (or rather, does):

\mathrm{\underline{F}_{d}(t)} = -d \dot{x} \vec{e_x}

and r would look like maybe this:

\delta\mathrm{r} = x\vec{e_x} +y \vec{e_y}

So you can imagine there are a bunch of steps to get this done, each partial div. blah blah multiplying, blah blah.

Well good news everyone!

I guess it’s not a surprise cause I already said I found a cheat. Well here it is:

\mathrm{\underline{F}_{d}(t)} = -d \Delta\Delta\dot{x} = -d \Delta\dot{x}^2

So if you’ve already kinematically described your system, You can take the derivatives you already know, square them, add them together and multiply by -d and you have your answer immediately.

Yay! I’ve tested this conjecture exactly once with 100% success rate. So that’s good math in my books. I wish I could credit where I found this from…but since it’s on my cheat sheet. I guess I’ll have to claim it as my own.

\mathrm{e}^{\sqrt{2}}

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