Addition of moments of inertia

[dyn]

Hi
To calculate the moment of inertia of a large sphere , I can calculate I for a smaller sphere and then add to it , I for the spherical shell that added to the smaller sphere makes the larger sphere.
Does this general process apply to all shapes ? If so , does this process have a name ie. is it a specific theorem ? If so , what is it called ?
Thanks

 

[Vanadium 50]

Parallel axis theorem.

 

[dyn]

I don't think it is The Parallel Axis theorem because the axis is not changing. I have heard that "moments of inertia add". This seems to fit the problem I referred to but what exactly does it mean ? Is it a specific theorem ?

 

[Vanadium 50]

<sigh>

 

[Negatratoron]

Can you draw a picture of what you're asking?

Generally the moment of inertia of a body about an axis is $\int r^2 dm$ where $r$ is the distance to the rotational axis and $dm$ measures the mass of the object. This is equal to $\iiint_V r^2 \rho(x,y,z) dx dy dz$ where $V$ is the region of space the object occupies, $r$ is a function of $x, y, z$ giving the distance to the rotational axis, and $\rho(x,y,z)$ is the mass density at the point $(x,y,z)$.

If you simply overlay two objects of mass $m_1$ and $m_2$ on top of each other, the integral becomes $\int r^2 d(m_1 + m_2) = \int r^2 dm_1 + \int r^2 d m_2$ or equivalently $\iiint_V r^2 (\rho_1(x,y,z) + \rho_2(x,y,z)) dx dy dz = \iiint_V r^2 \rho_1(x,y,z) dx dy dz + \iiint_V r^2 \rho_2(x,y,z) dx dy dz$ so it is simply the sum of their moments of inertia.

 

[dyn]

The example is of a large sphere, If I remove a smaller sphere from its centre I am left with the outer hollow shell. To caculate I for the outer shell I can calculate I for the larger and smaller spheres and subtract the smaller from the larger. This seems to apply the principle mentioned in #5 above. Is it just called "moments of inertia add" ? Does it apply to all shapes as long as they are measured relative to a common axis

 

[Lnewqban]

These may help:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Please, see the cases for cylinders and spheres.

 

[Ibix]

It's just from the definition of moment of inertia, as noted in #5. If you've got mass in some volume $V$ then the moment of inertia about some axis is $\int_V\rho r^2dV$, where $r$ is the distance to the axis. But an integral is just the limiting case of a sum. You are perfectly at liberty to do the sum in two stages and write it as $\int_{V_1}\rho r^2dV+\int_{V_2}\rho r^2dV$, as long as the two volumes $V_1$ and $V_2$ combine to make up the original volume $V$.

Or you could see it as a consequence of angular momentum conservation. Define the moment of inertia as $I=L/\omega$, where $L$ is the angular momentum and $\omega$ is the angular velocity. Imagine the two parts of your whole are almost, but not quite, touching and are rotating with the same angular velocity about the same axis. The total angular momentum is $L=I_1\omega+I_2\omega$. Now the two bodies are connected by a massless thread. The angular momentum can't change due to its conservation. $\omega$ can't change just because of this tiny thread. So the combined moment of inertia needs to be $I=I_1+I_2$.

 

[sophiecentaur]

I have heard that "moments of inertia add".

That's hardly surprising when you realize that the MI is defined as the Integral of elemental MIs. It will get more complicated when the CMs are not on the same axis but it still works consistently because of the definition.
It's easy to get confused unless you start from the beginning and follow the 'rules'. Trying to impress intuition, initially, on an explanation is doing things in the wrong order. They gave you the Maths as a useful tool and not as something to argue against.

 

[etotheipi]

Parallel axis theorem.

Please excuse my ignorance; how does the parallel axis theorem apply here? I thought the result follows due to the linearity of the integral operator, allowing the decomposition of $\iiint_V \rho r^2 dV$, as others have noted.

 

[Vanadium 50]

Look at the derivation (it's a theorem after all) and you can see the recipe for adding new contributing elements no matter where they are.

 

[etotheipi]

Look at the derivation (it's a theorem after all) and you can see the recipe for adding new contributing elements no matter where they are.

I still don't follow. All the parallel axis theorem tells me is that I can find the MoI of the same distribution of mass about some other axis if I know the MoI about a parallel axis through the CoM, by means of an additive constant.

 

[Vanadium 50]

Please look at derivations of it. They all show, by example, how to add up the contributing elements, no matter where they are. It seems better to point at existing derivations than to copy it all in a post.

If you understand the theorem, you understand the answer to the OP's question. Of course if you just look at the name and say "This isn't what I want so I won't look at it" you won't get anything out of it.

 

[wrobel]

I am not sure whether I understand a question but if you have a body $\mathcal B$ and its inertia tensor about a point $A$ is $I$ and you have another body $\mathcal B'$ with the inertia tensor about the same point $I'$ then the inertia tensor of $\mathcal B\cup\mathcal B'$ is $I+I'$

 

[etotheipi]

If you understand the theorem, you understand the answer to the OP's question. Of course if you just look at the name and say "This isn't what I want so I won't look at it" you won't get anything out of it.

I am quite aware of various different derivations of the PAT, I just don't see how it relates directly to the OP. You have a concentric sphere and shell through which I can position a single axis and subsequently perform the integration (or apply the known results), either together, or by adding the individual results; this choice follows from being able to split the region of integration.

The PAT just transforms the integral for a given distribution of mass, giving you an extra term. This is IMO distinct to asking "why can I then split up the region of integration into two parts, and then add them up again?"

 

[Vanadium 50]

Well, then there is no point in repeating what I said, is there?

 

[etotheipi]

Well, then there is no point in repeating what I said, is there?

It was just out of interest at your first comment, it's always a good idea to try and learn as much as possible from people more knowledgeable, like yourself! In this case I am still not sure I understand what you had in mind, but I'll think on it!

 

[dyn]

I would say that the explanation for my OP is "moments of inertia add". See #8. I just think that it's a shame that this principle is not a specific theorem

 

[etotheipi]

I would say that the explanation for my OP is "moments of inertia add". See #8. I just think that it's a shame that this principle is not a specific theorem

You could say it comes under the umbrella of the principle of superposition.

 

[jbriggs444]

$$\int_a^bf(x)+g(x)\ dx = \int_a^bf(x)\ dx + \int_a^bg(x)\ dx$$

 

[robphy]

I would say that the explanation for my OP is "moments of inertia add".

As written, it's incomplete since you need to specify the condition you mentioned that the axes have to be the same.
More completely, I think it's better to write

"moments of inertia with respect to the same axes add".​

Without the "with respect to the same axes", it's not correct (although it would be for "mass").
As similar comment would be applied to computing first moments in a center of mass calculation.

... or electric fields due to some configurations of charges.

You could say it comes under the umbrella of the principle of superposition.

I would agree with this description. (I use that when I teach.)

 

[sophiecentaur]

This has gone on too long but we are pretty much there. The MI is always relative to a specified axis. It cannot be assumed to be about the CM. So , using the term MI properly, MIs add. It's hardly worth losing sleep or keyboard strokes over this imo.

 

[hutchphd]

You could say it comes under the umbrella of the principle of superposition.

It is a special case of the parallel axis theorem (when the axes are parallel and also coincident ). It requires no separate discussion.....Really!

 

[etotheipi]

It is a special case of the parallel axis theorem (when the axes are parallel and also coincident ). It requires no separate discussion.....Really!

Now I feel very stupid . I still can't see how the principle of superposition follows from the parallel axis theorem. Here's how I would prove the superposition in this case; we have two overlayed density distributions $\rho_1(\vec{r})$ and $\rho_2(\vec{r})$. Then $$I_{tot} = \iiint_V (\rho_1(\vec{r}) + \rho_2(\vec{r}))r^2 dV = \iiint_V \rho_1(\vec{r}) r^2 dV + \iiint_V \rho_2(\vec{r})r^2 dV = I_1 + I_2$$
As for the parallel axis theorem, we suppose the vector from the CM to the new axis is $\vec{r}'$, and that $\vec{R}$ is measured from the CM whilst $\vec{r}$ is measured from the new axis, such that $\vec{r} = \vec{R} - \vec{r}'$: $$I_p = \iiint_V \rho \vec{r}^2 dV = \iiint_V \rho (\vec{R} - \vec{r}')^2 dV = \iiint_V \rho \vec{R}^2dV -2\vec{r}' \cdot \iiint_V \rho \vec{R} dV + \left( \iiint_V \rho dV \right) \vec{r}'^2 $$ The second term is zero via. the definition of the CM, so we end up with: $$I_p = I_{CM} + md^2$$ where $|\vec{r}'| = d$. But how does the latter derivation relate to the principle of superposition?

 

[robphy]

It is a special case of the parallel axis theorem (when the axes are parallel and also coincident ). It requires no separate discussion.....Really!

There is no addition of [new] mass associated with the "parallel axis theorem",
unlike the "adding of moments of inertia" by the addition of mass, as described in the OP.

 

[dyn]

I also cannot see any relation to the Parallel Axis theorem. The principle of superposition or moments of inertia about a common axis add seems to cover my OP

 

[vanhees71]

I still don't follow. All the parallel axis theorem tells me is that I can find the MoI of the same distribution of mass about some other axis if I know the MoI about a parallel axis through the CoM, by means of an additive constant.

The parallel-axis theorem is about how the moment of inertia changes when the axis of rotation is changed to another axis parallel to it.

What you were asking is the moment of inertia of a body around one fixed axis consisting of different parts. For this situation you simply don't need the parallel-axis theorem.

 

[sophiecentaur]

For this situation you simply don't need the parallel-axis theorem.

...except in as far as the total MI of two objects would involve choosing (and stating) a common axis of rotation and shifting the MIs to that common axis. That's only a practical note but pretty important if you happen to know the individual MIs (which will be about their individual CMs).

 

[etotheipi]

...except in as far as the total MI of two objects would involve choosing (and stating) a common axis of rotation and shifting the MIs to that common axis. That's only a practical note but pretty important if you happen to know the individual MIs (which will be about their individual CMs).

Right, but for this situation the axis passes through the CM of both objects already .

 

[hutchphd]

There is no addition of [new] mass associated with the "parallel axis theorem",
unlike the "adding of moments of inertia" by the addition of mass, as described in the OP.

In practice my use of the theorem always involves construction of a new object from lesser simpler pieces. If you enjoy a hundred special rules then you are wired different from me.

The construction of any object from two pieces is covered by the parallel axis theorem even if done sequentially in time. And the derivation of the parallel axis theorem must contain the case where the axes of the two objects coincide. I prefer one theorem not three.

 

[etotheipi]

The construction of any object from two pieces is covered by the parallel axis theorem even if done sequentially in time. And the derivation of the parallel axis theorem must contain the case where the axes of the two objects coincide. I prefer one theorem not three.

In the case where the axes coincide, the parallel axis theorem reduces to $I_1 = I_2$; i.e. the MoI of the same distribution about the same axis is, of course, the same. However I don't see how this helps.

As @robphy notes, the parallel axis theorem doesn't cover the addition of the MoI's of different components about some axis to obtain the MoI of the whole configuration about that axis. That is instead an example of the principle of superposition.

I wrote up the two derivations in #24 and they are quite distinct operations!

 

[hutchphd]

My guess is that the parallel axis theorem contains all of the information necessary to define the moments of inertia for any arbitrary object, rendering this discussion moot.
If not, then your point is taken...

 

[robphy]

In practice my use of the theorem always involves construction of a new object from lesser simpler pieces. If you enjoy a hundred special rules then you are wired different from me.

The construction of any object from two pieces is covered by the parallel axis theorem even if done sequentially in time. And the derivation of the parallel axis theorem must contain the case where the axes of the two objects coincide. I prefer one theorem not three.

In this problem given by the OP (adding of shells with a common center), the parallel-axis theorem is a vacuous step since the translation from the center of mass of the object is the zero vector.

Even if it were not the zero vector, the parallel axis theorem tells you about the moment of inertia of an object of mass M_1 about a parallel axis. Once armed with that information,
you still have to use superposition
to add that moment of inertia of the first object
to the moment of inertia of a second object with mass M_2 (possibly also obtained with help of the parallel axis theorem) .

So,

<br /> \begin{align*}<br /> I&#039;_{\underbrace{M_1+M_2}_{superposition}} &amp;=<br /> \underbrace{<br /> \underbrace{\left(M_1\Delta d_1^2+I_{M_1}\right)}_{\mbox{ parallel axis thm} }{{\Large +}}<br /> \underbrace{\left(M_2\Delta d_2^2+I_{M_2}\right)}_{\mbox{ parallel axis thm} }<br /> }_{\mbox{superposition}}<br /> \end{align*}<br />In the OP, d_1=0 and d_2=0 since concentric shells are being used.
So, the parallel axis theorem alone doesn't help the OP.

Update:
as @Ibix suggests below and @etotheipi suggested above,
the parallel axis theorem relies on superposition.

 

[Ibix]

The derivation of the parallel axis theorem that I'm familiar with is that for a body of density $\rho(x,y,z)$ rotating (without loss of generality) about an axis parallel to the z direction and passing through $x=x_0, y=0$ has moment of inertia$$\begin{eqnarray*}
I&=&\int_V\rho(x,y,z)\left((x-x_0)^2+y^2\right)dV\\
&=&\int_V\rho(x,y,z)(x^2+y^2)dV\\
&&-2x_0\int_V\rho(x,y,z)xdV\\
&&+x_0^2\int_V\rho(x,y,z)dV
\end{eqnarray*}$$
The first term is the moment of inertia about the origin, the second term is zero if the center of mass lies at the origin, and the third term is just $x_0^2M$, where $M$ is the mass of the object.

It seems to me that this derivation relies on an assumption that you can add moments of inertia for point particles rotating about a common axis - i.e. that you can write the integral at all. If you can do that then both the parallel axis theorem and the answer to the OP's question follow.

Edit: beaten to it by both @etotheipi and @robphy, I see

 

[etotheipi]

It seems to me that this derivation relies on an assumption that you can add moments of inertia for point particles rotating about a common axis - i.e. that you can write the integral at all.

And I suppose this really just comes from how the MoI is defined in the first place. E.g. for a rigid body rotating in the $x$-$y$ plane about the $\hat{z}$ axis, each particle has $\vec{L}_i = \vec{r}_i \times m_i (\vec{\omega} \times \vec{r}_i) = m_i {r_i}^2 \omega \hat{z}$, so it follows $\vec{L} = (\sum m_i r_i^2) \omega \hat{z}$ from which we define $I_z = \sum m_i r_i^2$, i.e. as a summation over mass elements. And same goes for the integral as the limit of this sum for mass elements $dm$.

 

[robphy]

Indeed, without superposition, one can't add vectors [and tensors].
(The moment of inertia is really a tensor.)