Radius of Gyration

Radius of gyration. Symbol: $k$. Let’s figure out what this guy actually is.

First thing — the radius of gyration is tied to moment of inertia, $I$. You remember $mr^2$ for each particle, right? That’s the bread and butter.

OK but now we want to look at $I$ from a slightly different angle.

So what is $I$, really? Check this out — can we maybe describe it as… the sum of the squared distances $r_i$ from the axis of rotation, each one multiplied by a “weight” $m_i$??? @?@?

(That line was in the book. Once it clicked for me, I was like — yeah, that’s actually a pretty clean way to put it.)

Oh and — real quick — why did we square $r_i$ before multiplying by the weight $m_i$?

Because depending on where you plant the origin, $r_i$ can come out positive or negative, right? Squaring kills the sign. We squared them and summed so that flipping doesn’t mess us up.

It’s just about capturing “how far apart things are” — distance only, no direction.

So to recap what we’ve got so far: take each $m_i$, multiply by $r_i^2$, and throw them all into one big sum.

$$\sum m_i r_i^2$$

That’s $I$. Cool.

Now — what happens if we take $I$ and divide by the total mass $m$?

Can we read that as… canceling out the weights $m_i$? Like, stripping them off?

$$\frac{\sum m_i r_i^2}{m}$$

And then — we slap a square root over the whole thing. What’s that move doing?

Isn’t it basically undoing the “exaggeration” from earlier? You know — the one we made by squaring all the $r_i$ and summing them up?

What we’ve built here is precisely the mean of the squared distances. (Kinda reminds me of the geometric mean, weirdly.)

And that thing — that mean of squared distances — is written $k$, and in English it’s called the radius of gyration.

$$\sqrt{\frac{\sum m_i r_i^2}{m}} = k = \sqrt{\frac{I}{m}}$$

Common sense check: if the mean distance is large, it’d be hard to rotate, right????

(Or… wait, is it hard to stop the rotation? Hmm.)

Ahhh — either way, the physical feel you get from $k$ is basically the same as the feel you get from $I$.

Honestly, $k$ might even capture it better — because at the root of $I$, the whole idea is this averaging process, and $k$ is exactly that average laid bare.

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Originally written in Korean on my Naver blog (2015-01). Translated to English for gdpark.blog.