# Moment of Inertia of Infinite small rods! [closed]

Say we have a symmetrical and homogenous rod with mass $$M$$ and length $$L$$ and a rotating axis in the middle of the rod, dividing it into two halves. Its moment of inert is $$I_{cm}=\frac{M{L^2}}{12}$$ Now we cut rod in half, the halves in halves, etc and continue that infinite times, while keeping them in the shape of the original rod. The question is what is the moment of inertia of all the rods combined. My answer is $$I=\frac{2M{L^2}}{3}$$ and I want to check if I am right. By doing the process about for a few cuts, I concluded that for n cuts, the small rods will have $$I_n=\frac {M{L^2}}{12}\frac 1{4^{n-1}}$$ with the axis as described above. Then I used the Parallel axis theorem for the first few cases for one side of the half original rod. When increasing the rods I used it for the two rods that were formed by a cut, then using it again in the same way to find the new axis of the four rods that were formed by a cut and so on. Then I arrive at the expression :$$I_{n_2}=\frac {M{L^2}}{12}\frac 1{4^{n-1}} + \frac {M{L^2}}{12}(\sum_{n=1}^n \frac1{4^n})$$ The geometric series converges $$\space$$ $$\sum_{n=1}^\infty \frac1{4^n}=\frac13$$ $$\space$$ and $$\space$$ $$\lim_{n\to \infty}{\frac 1{4^{n-1}}}\to0$$ $$\space$$ so $$\space$$ $$I_{n_{\infty}}=\frac{M{L^2}}{3}$$. So, in the end, we have $$I=\frac{2M{L^2}}{3}$$ (I am really sorry if my explanation is complicated and hard to follow. Ask me if you have a question pertaining to the explanation. If you understood, can you check the answer?)

## closed as off-topic by Aaron Stevens, ZeroTheHero, Buzz, Kyle Kanos, Jon CusterJan 9 at 14:09

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• I could not follow the argument ,but why would the moment of inertia change if the pieces you cut remain in the same place? – Wolphram jonny Jan 7 at 0:43
• It wouldn’t, so the calculation has a mistake. – G. Smith Jan 7 at 0:55

I'm not sure if it is helpful to offer an answer, since the question is based on a mistake in your algebra (as has been pointed out in the comments) rather than a physics concept that would be useful to others. However, because your approach resembles (loosely) the "decimation" procedure used in renormalization group calculations, and one approach used for self-similar structures, I'll make a few comments.

Consider the first step in your halving process. For each half, $$L\rightarrow\frac{1}{2}L$$, $$M\rightarrow\frac{1}{2}M$$, so $$I\rightarrow\frac{1}{8}I_0$$ where $$I_0=ML^2/12$$. But there are two such rods, so the contribution to the moment of inertia of the combined system, $$I_1$$, is $$\frac{1}{4}I_0$$. For the parallel axis theorem part, the shift of axis is $$\frac{1}{4}L$$, each mass is $$\frac{1}{2}M$$, so each rod contributes $$\frac{1}{32}ML^2$$. But there are two rods, so the contribution to $$I_1$$ is $$\frac{1}{16}ML^2=\frac{3}{4}I_0$$. So the net result is $$I_1=\frac{1}{4}I_0+\frac{3}{4}I_0=I_0$$. Not surprisingly, the moment of inertia is the same when calculated these two different ways.

You can think of this as a recursion relation between successive levels of a decimation process. Your aim seems to be to iterate this process to deal with subsequent halvings. Somewhere in the algebra behind the phrase "Then I arrive at the expression ...." you have gone wrong. I don't think that it is "on-topic" for this site to start discussing where you went wrong (maybe the previous paragraph will help you spot your error). However, in case you are interested, here is a pointer to the literature on renormalization group calculations; this is also covered in textbooks such as Introduction to modern statistical mechanics by D Chandler, and Statistical mechanics of phase transitions by J Yeomans.

Even if that area of physics is outside your current interests, you should be aware that the aim of calculations like this is often not to obtain an infinite series from the recursion relation (which, as you have found, can be tricky to formulate) but instead to find a fixed point of the transformation. In your case, I think the closest analogy is that the equation $$I_1=I_0$$ is already the fixed point, so there's not much value in taking things further.

However, your approach may have some value when one wants to calculate the moment of inertia of self similar structures (fractals). See for example this answer to a question about the Koch snowflake, and these lecture notes which include the example of a self-similar one-dimensional object. Those notes also show how to use the parallel axis theorem to deduce $$I=\frac{1}{12}ML^2$$ for your rod, without doing any integrals.

An easy way to find out what the MMOI of a complex body is, to calculate the angular momentum for a unit rotation.

Your approach is the slice the rod up in $$Δx = \tfrac{L}{n}$$ increments and addup each contribution. If there are $$n$$ slices, then $$i=1\ldots n$$ and $$x_i = L (\tfrac{i}{n}- \tfrac{n+1}{2n})$$ is the location of each slice relative to the center. Also each slice has mass $$m_i = \tfrac{M}{n}$$. Now rotate the rod, and add up all the angular momenta. Each slice moves with speed $$v_i = x_i \omega$$.

$$H = \sum_{i=1}^n x_i (m_i v_i) = \sum_{i=1}^n (m_i x_i^2) \omega = I_{\rm cm} \omega$$

so the MMOI is

$$I_{\rm cm} = M L^2 \sum_{i=1}^n \frac{(2i-n-1)^2}{4n^3}= M L^2 \left( \frac{n^2-1}{12 n} \right) = \frac{M}{12} L^2 \left(1 - \frac{1}{n^2} \right)$$

So obviously when $$n \rightarrow \infty$$ the MMOI is as expected $$I_{\rm cm} = \frac{M}{12} L^2$$.

But why isn't the MMOI correct it for finite slices? Well, each slice is rotating as well as translating and we have not accounted for this. Each slice has length $$Δx$$ and the angular momentum contribution is $$I_i \omega = (\frac{m_i}{12} Δx^2) \omega$$.

$$H = \sum_{i=1}^n \left( x_i (m_i v_i) + I_i \omega \right) = \sum_{i=1}^n \left( m_i x_i^2 + \frac{m_i}{12} Δx^2 \right) \omega$$

Now the MMOI is evaluated as

$$I_{\rm cm} = M L^2 \sum_{i=1}^n \left( \frac{(2i-n-1)^2}{4n^3}+\frac{1}{12n^3} \right) = M L^2 \left( \frac{1}{4n^3} \frac{n^3}{3} \right) = \frac{M}{12} L^2 \; \checkmark$$