# How should I REALLY have done this continuous moment formula derivation?

One of my more recent hobbies is making "swords" out of PVC conduit, foam, and duct tape. As all the weight is forward of the cross guards, I needed to put some amount of some substance inside the hilt to counter-balance it. Curious as to how much counter-weight I needed, I decided to see if I could develop a continuous version of the typical moment formula (moment = force * distance, or $M = F*D$) as there was no way the counter-weight would be close enough to a point mass to use the usual formula. Here's how I set about it...

$m$ = mass
$l$ = length
$d$ = fraction of whole

For example, if I have a rod where $m = 80 kg$ and $l = 16 ft$ and cut it into 8 pieces, $d = .125$ and the mass of each piece will be $m*d = 10kg$ and the length of each piece will be $l*d=2ft$.

Thus, treating each piece as a point mass centered at the point closest to the pivot, adding up the moments gives:

$\sum F*D$

$=\displaystyle \lim_{d \to 0} \; \sum_{k = 0}^{\frac{1}{d}} mld^2k$

$=\displaystyle \lim_{d \to 0} \;\; mld^2\sum_{k = 0}^{\frac{1}{d}}k$

$=\displaystyle \lim_{d \to 0} \;\; mld^2 \left( \frac{ \left( \frac{1}{d} \right) \left( \frac{1}{d}+1 \right) }{2} \right)$

$=\displaystyle \lim_{d \to 0} \; \frac{ml}{2} + \frac{mld}{2}$

$=\displaystyle \frac{ml}{2}$

The problem I've noticed with this formula is that it means that the further half of an object has the exact same moment as the nearer half (a rod twice as long has twice the moment, not four times as it should be). This is clearly wrong, but I don't know where I made an invalid assumption. Can someone else see it?

By the way, I had a brain flash while writing this and realized the simplest (and correct) way to derive it is as follows:

$\displaystyle \int_{a}^{b} (F*D)\;dD = \frac{F}{2}(b^2-a^2)$

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Are you trying to establish reaction moments, or Mass-moment-of-inertia? It is not clear what the goal of the counterweight is. Do you want to establish where the center of gravity is going to be, or do you want to control the reactions from an impact (look at axis of percussion) ? – ja72 May 8 '11 at 0:10
You may want to look up the "parallel axis theorem" and be done with it. – dmckee May 8 '11 at 0:13
@ja72: Mass-moment-of-inertia. Without a counterweight, such a foam sword would be too top-heavy and hard to control. @dmckee: I don't think that's quite what I'm going for. – El'endia Starman May 8 '11 at 0:18
It tells you how to find the moment of inertia of a body around an axis offset from it's CoM if you already know the moment of inertia for rotations through the CoM. – dmckee May 8 '11 at 0:26
@dmckee: Ah, I see. However, that wouldn't have been useful to me as I literally put the sword's cross guards on the foot of a bed and rested a textbook on the handle to determine the approximate amount of moment produced by the foam and duct tape along the blade... :P – El'endia Starman May 8 '11 at 0:32

I'm not sure I understand what you're really trying to calculate, or what you're doing to try to calculate it. What are $F$ and $D$, for instance? And what is $\Delta x$? That variable is normally used for an increment length, but you seem to be using it as an index that labels the various pieces of the rod.

From what I can gather, your example calculation seems to be calculating the first moment of mass of the rod around one end. The first mass moment in one dimension can be defined as

$$I_x(c) = \int (x - c)\ \mathrm{d}m$$

where $x$ is a coordinate that measures along the length of the rod, and $c$ is the position of the end in this coordinate system.

If you use this formula for the half of the rod that is closest to the chosen end, you get

$$I_x(0) = \int_0^{l/2} x\frac{m}{l}\mathrm{d}x = \frac{ml}{8}$$

and for the other half,

$$I_x(0) = \int_{l/2}^{l} x\frac{m}{l}\mathrm{d}x = \frac{3ml}{8}$$

So the moment of the further half is indeed more than that of the closer half. In your calculation, I don't see any place where you distinguish between the two halves of the rod; all you're doing is calculating the moment of the whole thing. I'm not sure how you're concluding that your formula implies that both halves have the same moment. It doesn't.

If you're looking into minimizing the resistance of your swords to swinging, what you should be looking at is the second moment of mass, also known as the moment of inertia.

$$I_{xx}(c) = \int (x - c)^2\ \mathrm{d}m$$

This is the quantity that measures the resistance of an object to rotation around the point $c$. So to make the sword as easy to swing as possible, you want the moment of inertia around the hilt to be as small as possible. (At least, I'm guessing so, but I don't know anything about wielding a sword specifically)

It turns out (I'll omit the mathematical detail, unless you're interested) that the way to do that involves distributing the weight so that the center of mass is inside the hilt of the sword. To put it another way, the moment of inertia around a point $c$ for a given mass and length is smallest when the center of mass is at the point $c$. Mathematically, the way to figure out the center of mass is to divide the first moment of mass by the total mass,

$$x_\text{CM} = \frac{\int x\ \mathrm{d}m}{\int\mathrm{d}m}$$

Of course, in practice, you wouldn't bother with all this mass. Just move the counterweight such that the sword is balanced around the point where you're going to hold it.

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I edited my post in response to a couple observations you made. Also, decreasing the moment of mass about a pivot at the cross guards DOES decrease the moment of inertia, so I only care about the moment of mass. What I'm really going after is why the methodology I used gives an erroneous answer. – El'endia Starman May 8 '11 at 1:53
Dear @El'endia, as far as I can say, David has answered your question. More explicitly, it is not true that your own formula gets the same contribution from both halfs. The first half gives $\sum_{k=0}^{1/2d} k$ in the sum while the second half gives $\sum_{k=1+1/2d}^{1/d} k$ in the sum. You may check that these two sums are not equal and indeed, they give $1/4$ and $3/4$ of the total result, just like David wrote. – Luboš Motl May 8 '11 at 4:44
@Lubos: Hmmm...point noted. – El'endia Starman May 8 '11 at 5:06