## Wednesday, August 15, 2012

### Derivation of the Work/Energy Principle

In my previous post on methods for particle dynamics, I stated that, by integrating Newton’s Second Law over a distance, we could make use of ‘Work/Energy’ methods. This is true as far as it goes, but is certainly not the whole truth. It’s important to realize what’s actually happening, lest we fall into a trap of some sort.

Newton’s Second Law, is, of course, a vector equation:

So I have to be a little more specific when I say 'Integrate over a distance.' Start tossing around vectors willy-nilly, and you'll end up with results that don't make any sense. Anyway, there are two main questions regarding what needs to happen here:

1) In order to integrate a function properly, there should be some 'infinitesimal' (very small) quantity involved. (i.e.,  'dx' would be the tiny change in x that we always see in textbooks). What quantity should we multiply each side of the equation by in order to integrate?

2) Which vector operation should be used here, dot or cross product?

Answering the first leads to the second. We want to evaluate the total force exerted over some distance, right? For a particle with position vector r, its change of position from moment to moment will be dr. (dr/dt is the velocity of that particle, without the 'dt' dr is just a very small change of position). If we add up the force exerted over every infinitesimal distance, we'll get what we want.

Now, to properly extract a result from the vectors F and dr, we have to choose a proper operation. The cross product would produce...a perpendicular vector. Moreover, if F was exerted entirely in the direction of dr, then our vector would have no magnitude--in other words, this is not the proper operation to use here. The dot product it is!

So we're travelling along a path from point 1 to point 2, integrating along the way. Note that, since the dot product has been invoked, there will be no direction information left here. Energy methods are purely scalar-based, and this is both a blessing and a curse.

At this point, we're halfway there: The left-hand integral is defined as work. It has units of [N*m] and is better thought of as 'useful work.' More on that later.

To proceed on the right-hand side, though, requires some trickery. How can we possibly integrate a in terms of dr? By finding an element they share in common: v, the velocity vector.

Acceleration, of course, is the time derivative of velocity:

So we can perform that substitution into the integral above. What about dr? Since velocity is the time integral of position:

Putting everything together, this is what it looks like:

$\int_{1}^{2}\overrightarrow{F}\cdot d\overrightarrow{r} = \int_{1}^{2}m\left (\frac{ d\overrightarrow{v}}{dt} \right )\cdot \overrightarrow{v}dt$

The 'dt' terms cancel, and the a vector is gone! Work is independent of both time and acceleration! At this point, the right-side integral is easy to evaluate:

$\int_{1}^{2}\overrightarrow{F}\cdot d\overrightarrow{r} = \int_{1}^{2}m\left\overrightarrow{v} \right \cdot \overrightarrow{dv} = \frac{mv_{2}^2}{2} - \frac{mv_1^2}{2}$

(I sort of skipped a step, but it's trivial: v dot dv is simply v*dv, since they are in the same direction. Good to remember that we are no longer dealing with vector quantities, though).

Those final terms are, of course, kinetic energy. (We know this thanks to good friend Willem Jacob Gravesande, among others).

So that wasn't too bad, as derivations go. A few closing thoughts on work and energy:

- These are line integrals we're talking about here. (I'm new with LaTex, so I couldn't deduce how to properly indicate a line integral). Technically, this is important; practically, it's not. v dot dv is an exact differential, and the line integral of an exact differential behaves like a normal integral--it's path independent

- Speaking of path independence, this is just one more quantity that Work and Energy are independent of. The others: time and acceleration.

- Direction of travel does matter, but only in a broad sense: If the force F is applied in the direction of travel, it does positive work, increasing your speed. If F is applied against your direction of travel, you slow down. Makes sense.

- Beware of traps such as: 'How much work does it take to hold a 45 lb weight above your head for 30 seconds?' You are obviously applying 45 lbs of force to hold the object up, which is probably sapping a lot of energy from your muscles, but no work is done from a mechanics standpoint. Furthermore, constraint forces--those applied not in the direction of travel--do no work. Unless the constraint moves. But that's another story.

The online LaTex editor I'm using can be found here. Also, reference for this section came, once more, from Engineering Mechanics: Dynamics; Plesha/Costanzo/Grey, 1st edition.