Talk:Moment of inertia/Archive 1

Moment of mass

Whats the difference between moment of inertia and moment of mass if there is any? --Light current 03:49, 23 December 2005 (UTC)

Late reponse I guess, but moment of mass usually refers to the first moment of mass, which is merely the product of a mass element with its distance from a particular reference point. (Usually split up into x, y and z components in order to calculate centers of mass) The moment of inertia is the second moment of mass, making use of perpendicular distances from axes. PrintfXh4 17:04, 6 June 2007 (UTC)

Rotational Inertia

"Rotational Inertia" is the term that has been adopted by the College Board for the Advanced Placement Physics exams, beginning at least in 2002. I use that term in my teaching, because "Moment" of inertia is confusing to my students, since most high school students that I've encountered can't separate 'moment' in this sense from 'moment' of time. jpsi 10:11 March 9 2007

"Rotational inertia" is the term used in a popular sophmore physics textbook by Halliday & Resnick (Physics, 2nd Ed.,Chapter 12) as "a measure of the resistance a body offers to a change in its rotational motion about a given axis." Yes it is the exact same thing as "moment of inertia". To Tokerboy and Cyp below, i would reply that any professor worth his salt would not laugh, but rejoice that a student had made a conceptual break-thru in understanding how "mass" or "translational inertia" relates to rotational motion. Kenny56 12:50, 29 September 2005 (UTC)

No, they are not the same, but can be easily confused. I will explain...

First, let's just think about linear motion. How hard it is to stop an object is related to it's mass and it's velocity. Mass times velocity is inertia. So, you could say that mass, velocity, and inertia are each "a measure of the resistance a body offers to a change in it's linear motion". However, this does NOT mean that mass, velocity, and inertia are all the same thing.

Now, let's look at circular motion. How hard it is to stop an object is related to it's rotational mass (moment of inertia) and it's rotational/angular velocity (spin rate). Moment of inertia times rotational velocity is rotational inertia. So, you could say that moment of inertia, rotational velocity, and rotational inertia are each "a measure of the resistance a body offers to a change in it's rotational motion". However, this does NOT mean that moment of inertia, rotational velocity, and rotational inertia are all the same thing. StuRat 20:52, 1 October 2005 (UTC)

Are you confusing inertia and momentum? Mass times velocity is the definition of momentum, not inertia. Mass is the amount of matter that an object contains; Inertia is a property of matter related to its resistance to change of state, either at rest or in motion, unless acted upon by an external force. Momentum is a measure of the change in velocity that results when an object with mass is acted upon by an external force, and it is also related to the kinetic energy of an object in motion.

Whether in rotation or translation, the mass of the object is the same. But it is the inertia property of the mass that is different in each case, and it is calculated by different formulas to account for either rotation or translation. The "rotational mass" is a confusing term and I am reverting back to Halliday and Resnick's usage quoted above. Kenny56 05:07, 3 October 2005 (UTC)

I think your "mass quantifies the inertia of an object" is more confusing than the phrase "rotational mass". StuRat 23:27, 5 October 2005 (UTC)

Okay, here's another way to consider this concept. Mass is a physical quantity; it measures the amount of matter that an object contains. Inertia is a physical property; it describes the persistant behaviour of matter in uniform motion or at rest.

The strength of this property is related to the quantity of matter, so an object of greater mass has more inertia than one of lesser mass. In this sense the units of measure of mass can be used to quantify inertia by providing numeric values and units for comparing the inertias of different objects.Kenny56 04:26, 6 October 2005 (UTC)

I still find the original intro much clearer:

Moment of inertia is a measure of the resistance of a physical object to increases or decreases in its rate of spin about its axis (angular acceleration). Moment of inertia is to rotational motion as mass is to linear motion. In this sense, the moment of inertia may be thought of as the rotational mass, because an increase in angular velocity causes it to behave as if it were more mass. In other words, rotation can be thought to increase the mass.

According to Keith R. Symon in his book Mechanics, Addison-Wesley, 1971, the analogy between mass and moment of inertia breaks down due to three differences:

1. Moment of inertia is a tensor, but mass is a scalar.

2. The inertia tensor is not constant with respect to axes fixed in space, but changes as the body rotates, whereas mass is constant.

3. There is no symmetrical set of three coordinates analogous to X-Y-Z with which to describe the orientation of a body in space. As a result the solution to the rotational equations of motion requires different methods than those used for linear motion.

I'm a big fan of using intuitive concepts in explanations whenever possible, however in the case of rotational motion I think it would be too misleading to use "mass" in place of "inertia" in light of Symon. Kenny56 03:51, 11 October 2005 (UTC)

1. As far as I can tell, mass and an INDIVIDUAL moment of inertia (for a given axis) are both scalars, which are considered to be the simplest type of tensor.

2. I assume this means the "moment of inertia tensor", not the "inertia tensor", which are different things. If this means there is a different moment of inertia about different axes of rotation, then I agree, that is a difference. It is explained in great detail later in the article, though, so any misconception here will be covered later. I'm not opposed to adding a disclaimer to this effect up top, however, if you think it would help.

3. I'm not sure what this is saying. There certainly seem to be similar methods to me, such as $F=ma\,$ and the rotational equivalent, $T=I{\alpha }\,$, or linear inertia, $mv\,$, and rotational equivalent, $I{\omega }\,$.

Also, the mere fact that he was listing differences means he also thought there were substantial similarites (you wouldn't bother to say things are different if it's obvious to everyone that they are). StuRat 16:11, 11 October 2005 (UTC)

---

What on earth does "Moment of inertia is the name for rotational inertia" mean? If there is no difference, then say they are synonyms. This makes it seem like the page should be moved to rotational intertia. Tokerboy

Suppose it's because, for some inexplicable lack of reason, rotational inertia isn't (officially) called rotational inertia. Perhaps, so that if a new person asks a bunch of professors about rotational inertia, they can look confused, and say there is no such thing, and later on laugh together behind the new persons back, about how that stupid little new person didn't even know that it's called moment of inertia, not rotational inertia. From then on, all the professors can mock and play practical jokes on the new person, with the new person having no idea why. -(Cyp)

I don't believe they are the same. Think of the case of linear inertia, which equals mass times velocity. Rotational inertia would then equal rotational mass (called the moment of inertia) times the rotational velocity. StuRat 01:49, 23 September 2005 (UTC)

---

I think the definition of I should only have one integral sign, i.e $I=\int r^{2}\,dm$ instead of $I=\int \int \int r^{2}\,dm$, because there is only one integral variable (dm). Or am I wrong? --Stw 22:03, 25 Jan 2004 (UTC)

I'm quite sure I'm right on this, so I changed it now --Stw 21:49, 1 Feb 2004 (UTC)

User:Stw you are wrong. dm is the mass density over the volume V. The triple integral is the customary notation for a volume integral.

I disagree with "User" above: I think "add up r^2 for all differential mass elements" is clearer, and more general than "add up r^2 for all differential elements of mass in all three directions". The triple-integral sign doesn't make sense until you I have the dxdydz (or whatever) that they go with, and you won't get those for line segments, or pieces of a plane; and I think it assumes too much about the coordinate system being used. . Sukisuki (talk) 15:59, 20 June 2009 (UTC)

Regarding the formula which follows the phrase "These quantities can be generalized to an object with continuous density":

$\mathbf {I} =\int _{V}\left[\left(\mathbf {r} \cdot \mathbf {r} \right)\mathbf {\delta } -\mathbf {r} \otimes \mathbf {r} \right]\ dm$

The delta is NOT explained, and in fact should be replaced with the 3x3 Identity matrix.

Ix, Iyz, Ixx ?

The notation "I_x= I_yz = moment of inertia about the axis parallel to the X-axis or in the plane parallel to the YZ plane" confused me at first - later on in the article, the notation I_yz is used for the non-diagonal elements of the tensor.

Wouldn't it be better to have I_x stand on its own in the initial paragraph, to be consistent? I am in doubt whether it would be useful to include I_x = I_yy + Izz ?

1st: done; 2nd: seems wrong, or again some other notation.--Patrick 00:47, 4 October 2005 (UTC)

I am familiar with $I_{x}\;$, not the $I_{xx}\;$ notation, for the moment of inertia about an axis parallel to the X-axis, thru the centroid of the object. Are you using $I_{x}\;$ to mean the moment of inertia directly about the X-axis ? If so, this can be simplified by moving the origin to the centroid of the object in question, in which case they are the same thing. StuRat 10:10, 5 October 2005 (UTC)

I used the same notation as later on. If they are the same we can add that.--Patrick 22:07, 5 October 2005 (UTC)

I added the simplified versions. StuRat 23:24, 5 October 2005 (UTC)

calculations

It might be important to know that the only way of calculating the moment of inertia of an arbatrary polygon is triangulate, calculate moment of each triangle about center of mass of the triangle, use off-axis equation to get it about center of mass of polygon, and add them up. This is complicated to implement however.

doubts

why moment of inertia is related to rotation.
- Cyborg 18:20 APRIL 01 2006 (UTC)

That's things only thing it's related to. It's not related to much else, except for perhaps mass if you compare linear motion and circular motion. --M1ss1ontomars2k4 ^{(T | C | @)} 22:30, 29 July 2006 (UTC)

Newton's second law

In the torque section the equation F = ma is given, but there is no mention or link provided to Newton's second law.

Product of Inertia

Product of inertia is a redirect to this page, yet this page does not mention it, at least I don't think it does. Wouldn't it help to have a section about this for those that are following this redirect? I know I would appreciate it, as I know little on the subject and was looking to learn more. Thanks. --Zsinj^Talk 17:24, 13 April 2006 (UTC)

Reorganization

I have reorganized and consolidated the article essentially under two headers: the scalar and the tensor. Within each the exposition is similar. Very little content was removed or added (except see below); some was changed in a copyedit way and some moved to reflect the reorganization. I also tweaked some math markup a bit.

I think it reads nicer now, and I hope everyone agrees. One section where I anticipate some may wish to revert somewhat is the notational comments about the scalar integration definition. I removed all notations about "infinitesimals", figuring if someone need to elaborate on that, they would consult integration references. But I do understand how the actual definition used, the one in terms of m, mass, may seem to be somewhat ambiguous. So have a go at it and feel free to discuss here. Baccyak4H 16:42, 26 October 2006 (UTC)

Could there be descriptions without calculus, here and in many physics articles?

I'm not sure we should remove calculus when it's pretty integral to the description (*resisting bad pun*). However, I would support alternate descriptions that don't require calculus knowledge. The issue is that many people who understand this subject don't know how to explain it without calc. - EndingPop 10:49, 1 November 2006 (UTC)

There are such descriptions at the top of the article. I am not sure there is a description that allows one to compute $\mathrm {I}$ that can do any better than calling it a weighted sum, which is merely a special case of an integral anyway. Baccyak4H 15:52, 1 November 2006 (UTC)

Moment of inertia# Angular momentum and torque

Out of the blue comes ${\boldsymbol {\omega }}\cdot \mathbf {r} _{i}=0$ for all i. This is only true in special cases. Is this on purpose or by mistake? --P.wormer 10:10, 31 January 2007 (UTC)

It's true for lamina, but not for general cases. Picture a triangle spinning on its own plane; the axis of rotation would be tangent to the axis of rotation, satisfying that condition. Modified to a tetrahedron with the same motion, this would be absolutely false. Erissian (talk) 02:28, 3 April 2008 (UTC)

Angular momentum

Please stop changing the angular momentum definition. Angular momentum is defined with respect to an origin, NOT an axis of rotation. Egendomligt 03:03, 5 March 2007 (UTC)

I second this--P.wormer 15:36, 5 March 2007 (UTC)

I agree that angular momentum is strictly a vector quantity and thus includes the axis in its definition, but beginners often think of it as a scalar quantity about an axis of rotation. This is equivalent for basic problems, and is a much simpler concept for a beginner. Could we not allow both (with a caveat)? Dbfirs 22:16, 14 March 2007 (UTC)

Because it's a tensor, you can assume that the moment of inertia is independent of the coordinate system. In other words, both are correct. It's my opinion, however, that explaining it in terms of an axis of rotation is more intuitive and useful to most people and that there's no harm in defining it wrt origin. Erissian (talk) 02:35, 3 April 2008 (UTC)

Request

Could someone please change the third example value of k because it is wrong?

According to this page the value is correct. Are you positive it is incorrect? - EndingPop 11:11, 26 April 2007 (UTC)

The page you link to is correct. But L (in that page) != R (in this page). This page states that I = (1/12)kMR^2 where R is the radius or distance from the axis of rotation. Which is wrong because the formula is I = (1/12)kML^2 where L is the length of the rod (as on that page). L=2R. 86.133.66.201 19:17, 29 April 2007 (UTC)

I think the section on principal moments could include a little more theroy on how the principal moments are derived and their how to find e1,e2,e3. The links provided are not much help. 138.64.2.77 (talk) 19:50, 11 November 2009 (UTC)

Heavy edits

I found a number of problems reading this article, and so decided to fix them. Below is a list of (major) changes and the justifications for each:

• Changed centroid references to center of mass in many locations. The centroid is only the center of mass when the density of the object in question is constant.
• Added "rotational" to various mentions of kinetic energy, where appropriate, since kinetic energy generally has both a translational and a rotational term.
• Removed a few mentions of mathematical symbols (e.g. for the axis of rotation and inertia tensor) when those symbols are not used in an adjacent derivation and disrupt the formatting.
• To better reflect the division of this article into scalar and tensor parts, I created a new subheading for the former.
• Removed a step from the scalar definition for a continuous body, since it said nonsensically that position and density are a function of volume.
• Moved the example of the spinning ice skater to the place where angular momentum is related to moment of inertia.
• Two of the values for 'k' were wrong, in that they used R as the distance to the center of mass when R should have denoted the length of the rod. I removed them and added k for a solid cylinder or disk to keep the table simple.
• The scalar parallel axis theorem is only a specialized form of Steiner's Thm.
• The angular momentum and torque section was fallacious. The expression L = Iw is (in scalar, not tensor form) only valid when L is parallel to w, which is only the case when w is directed along a principal axis. Instead of following the mystifying assumption that the object is like a flywheel, I scrapped most of this section and incorporated it into the previous one, as scalar expressions involving I, making sure to mention the assumptions under which the expressions are valid. There's no point in giving a derivation here, since principal axes of inertia are already beyond the scalar treatment. I also tightened the derivation of rotational kinetic energy.
• The moment of inertia tensor can be calculated about any point in space, not just the center of mass.
• Added the general definition of the components of the tensor, and mentioned that the off-diagonal terms are called the "products of inertia".
• The expression for the inertia tensor for a continuous object should have used the 3x3 identity matrix, not the unit matrix.
  • The "unit matrix" page at the time erroneously said that it was a matrix of ones.
• The tensor is diagonalizable because it is positive definite, not because it is real and symmetric.
  • I was wrong about this: being symmetric is indeed a sufficient criterion for diagonalizability.
• Added a note about the uniqueness of principal axes (in the nondegenerate case) to the "Principal Moments of Inertia" section.
• The final section wrongly said "dot product," which refers to a product of vectors, not a product of tensors and vectors. I know that in GR this product of tensors is called the "inner product," but that also might not be the right term here.

Anarchic Fox 21:52, 20 June 2007 (UTC)

GAOnHold

Speaking as a graduate physics student

• Physics and scientific corectness looks to be fine [unless I'm really stupid or something]
• The example should probably not be in the lead. The lead should be bulked up.
• Sources need to be inlined
• Should probably make it more accessible to the public. I guess digging up Resnick and Halliday and looking at one of the essays would be useful in having a generalised "chat" section.
• A few example calculations, eg of a calculation for a disk or a use of a Parallel Axis Theorem would be good.
• Pulling a diagram from one of the daughter articles would be good. If there is a picture of a diver or a figure skater pulling their arms and legs in to spin at greater rates, that would be even better.

Overall, this is very close. The inline cites is the main thing and only making a more talky section and more diagrams for the general public would make it even better. Blnguyen (bananabucket) 08:14, 23 July 2007 (UTC)

Alright, I've added more to the lead and overview, added a couple of pictures, and moved that example. I don't think it worthwhile to have an example for calculating moment of inertia, since any concise one would just be plugging in numbers, and an example for the parallel axis thm would probably take up too much space. (A good example would be moving from center to end of a rod, but then I'd have to explain where I got the original equation, using rod length instead of distance of edge from axis. Maybe it could be fit in without disruption... I'm not sure.) However, I don't know how to respond to your suggestion to put inline citations in. The current set of references are general citations, as mentioned at Wikipedia:Citing sources#Inline citations; they cover big swaths of the page. I don't know of any controversial statements in the main body that should have specific citations; if you could point some out, that would be good. Thank you for the review, anyway; please tell me how it looks now. Anarchic Fox 09:49, 23 July 2007 (UTC)

Well, there is nothing there which is controversial like a quote or a personal statement, but it's usually required to have at least one ref at the end of a paragraph to cover the info. just put <ref>Goldstein, pp. 101–102.</ref> and so forth as required. Blnguyen (bananabucket) 02:17, 24 July 2007 (UTC)

Hehheh, this'll teach me to edit when I should be packing. All the relevant physics books are now in boxes. Anyway, unless someone else puts them in, I'll make the changes and bring this article back up for GA status sometime in mid-August, after I've settled into the new apartment. Anarchic Fox 04:21, 26 July 2007 (UTC)

Well, when you are ready, simply ping my talk page. In the meantime I am simply going to procedurally remove the tag. Thanks, Blnguyen (bananabucket) 03:48, 1 August 2007 (UTC)

I've removed the GAC. Just ping me when you get back. Blnguyen (bananabucket) 01:50, 17 August 2007 (UTC)

Continuous mass distribution

The following equation

$\mathbf {I} =\int _{V}\left[\left(\mathbf {r} (m)\cdot \mathbf {r} (m)\right)\mathbf {I_{3}} -\mathbf {r} (m)\otimes \mathbf {r} (m)\right]\,dm,$

seems strange to me. I would expect

$\mathbf {I} =\int _{V}\left[\rho (\mathbf {r} )\left(r^{2}\mathbf {I_{3}} -\mathbf {r} \otimes \mathbf {r} \right)\right]\,dv,$

where ρ(r) is the mass distribution. If we would call that m, then I would expect m(r), not r(m). Further the notation of the identity matrix versus inertia tensor is somewhat unfortunate. Any source? --P.wormer 16:34, 28 July 2007 (UTC)

I do not have a source per se, but I have always assumed the dm definition was used to parallel the scalar integral in the earlier section, which was described in terms of infinitesimal masses. Either formulation can work; one can be thought of using a limiting grid of equal volumes (yours), one of equal masses (current). I would only question your ρ(r)—do you mean ρ(v), simply because v is the variable of integration?

I prefer the volume integral, since the mass integral is invariably converted into a volume one. The reason it was previously a mass integral is to form an analogy with the previous summation over point masses. My mechanics books are packed away, unfortunately, so I can't check them for a while.

The notation for the identity matrix is unfortunate, but I don't know any alternative conventions.

Also, the 'dv' doesn't give the variable of integration; it's shorthand for the 'dxdydz' (or its alternate forms) of a volume integral. (I do usually see it with a capital 'V', though). Anarchic Fox 05:29, 29 July 2007 (UTC)

dv is indeed short-hand for dxdydz and r=(x,y,z). You could use ${\boldsymbol {E}}_{3}$ for the unit matrix. And the earlier integral over m is strange too. See the classical mechanics books by Goldstein (p. 145), Whittaker (p.117), Arnold (p. 140), and Rutherford (p. 125) for the definition as I have given it. The least you must do is give a definition of the function r(m), which maps the scalar m onto a vector r. Personally, I've no idea what this function could be. Is m the total mass of the body, or is it a function of something? Finally, remember that by definition ρ(r)dv is an infinitesimal mass at point r. --P.wormer 09:33, 29 July 2007 (UTC)

r(m) is indeed nonsensical, unless you go into strange set-theoretical arguments. (The page had r(v) earlier, which had the same problem.) I only recall seeing a mass integral in elementary textbooks (Young & Freedman has it for sure), where it (as mentioned before) provides an analogy to the summed moment, and may also serve as a mnemonic for volume integrals. (If $m={\frac {4}{3}}\pi \rho r^{3}$, then taking differentials gives $dm=4\pi \rho r^{2}dr$, for instance). I have no objection to you replacing those equations. Anarchic Fox 09:17, 30 July 2007 (UTC)

Fine and dandy

The following form gives a matrix, not a scalar (for a programmer: a matrix is here a 3 x 3 array of reals, a scalar is a single real).

$I=\mathbf {{I}^{T}} \mathbf {\hat {n}} \cdot \mathbf {\hat {n}} .$

In this definition:

$I_{ij}=I_{ji}({\hat {n}}_{1}^{2}+{\hat {n}}_{2}^{2}+{\hat {n}}_{3}^{2})=I_{ji},\quad i,j=1,2,3,$

since ${\hat {\mathbf {n} }}$ is a unit vector. So , this relation states only that the inertia tensor is symmetric. If you program this, all you do is wasting computer cycles.

In the correct definition (double dot is double sum):

$I=\mathbf {I} :\mathbf {\hat {n}} \otimes \mathbf {\hat {n}}$

gives

$I=\sum _{i=1}^{3}\sum _{j=1}^{3}I_{ij}(\mathbf {\hat {n}} \otimes \mathbf {\hat {n}} )_{ij}\equiv \sum _{i=1}^{3}\sum _{j=1}^{3}I_{ij}{\hat {n}}_{i}{\hat {n}}_{j}=\sum _{i=1}^{3}{\hat {n}}_{i}\left(\sum _{j=1}^{3}I_{ij}{\hat {n}}_{j}\right).$

The rightmost expression shows that

$I=\mathbf {\hat {n}} ^{T}\;\mathbf {I} \;\mathbf {\hat {n}} .$

Personally, I would not put much trust in mathematically oriented programs written by Dndn1011. --P.wormer 08:00, 7 August 2007 (UTC)

Persnally, I would leave insults out of the equation (pun intended). Look, I have just implemented a physics simulation of a bouncing cube, and it appears accurate to the eye, and also in conservation of energy and momentum mathematically verified. So please do not insult me. Now to your points... it could be that I have misunderstood some notation, so I would welcome clarification. I should add that I verfified my formula by actually implementing it. If there is an error, it is an error in my understanding of the notation only.

$I=\mathbf {{I}^{T}} \mathbf {\hat {n}} \cdot \mathbf {\hat {n}} .$

This is:

Take the matrix I, transpose it, giving a matrix. The transpose of a matrix is not time consuming, and is part of any good math library. The matrix is multiplied by the axis vector n. This gives a vector as a result. The dot product of this resultant vector and n is then calculated. The result of this is a scalar. This is mathematically correct as far as I can see. It works, as far as I can see. It is the same as the original equation. If anyone wants to optimise this, they can do so by a variety of means, however everyone should know that premature optimisation is the root of all evil (Knuth).

Your statement that my formula gives a matrix, not a scalar, appears preposterous to me. Dndn1011 16:03, 7 August 2007 (UTC)

In mathematics certain notational conventions apply (as in natural language). When it is necessary to explain in words a simple equation like the present one, it means that you deviate from the common convention. In natural language you can do that too, you and your friends can call a "dog" a "cat", and whenever an outsider will not understand you, you can explain it. But this is not what Wikipedia is for.

Back to the equation. First convention is to have (mental) brackets like this:

$I=\mathbf {{I}^{T}} \left(\mathbf {\hat {n}} \cdot \mathbf {\hat {n}} \right).$

Second convention:

$\left(\mathbf {\hat {n}} \cdot \mathbf {\hat {n}} \right)=1.$

Third convention:

$\mathbf {{I}^{T}} \times 1=\mathbf {{I}^{T}}$

Now we are at:

$I=\mathbf {{I}^{T}}$

Fourth convention: Non-bold quantity is a scalar (single number), bold quantity is a matrix (in this case a 3 x 3 array).

Fifth convention: left and right hand side of equal sign must be equal (hence the name equal sign). A scalar cannot be equal to a matrix.

Writing this I started to understand your explanation of your equation. First you write (and this is where you call a dog a cat):

$I=\left(\mathbf {{I}^{T}} \mathbf {\hat {n}} \right)\cdot \mathbf {\hat {n}} .$

Again by convention this results in a column vector $\mathbf {\hat {n}} '$.

Then you say if I see the following

$I=\mathbf {\hat {n}} '\cdot \mathbf {\hat {n}} ,$

I couldn't care less whether I have two columns, two rows, or a row and a column: an inner product is an inner product is an inner product. In certain contexts this is true, but when matrices are involved, one is usally more careful in distinguishing row and columns. Thinking about it, I see now that your program may have been right, but the mathematical write-up is so much against the conventional notation, that it is confusing and bordering on wrong. I say wrong, because I showed you two interpretations of the same equation with different outcome.

Before we get into a long discussion: you will not be able to convince me that what I see as standard notational conventions are not that. So save your breadth, if you don't believe me I can live with that. --P.wormer 17:32, 7 August 2007 (UTC)

I stopped making contributions to wikipedia for a long time because of antagonistic corncobs. This is just a statement. Anything you draw from this statement is your of your own making. I did actually *state* that I might have an error with *notation* so a lot of what you just said seems superfluous. Additionally my purpose here is to present a useful arrangement of the formula that is more practically implemented by programmers, not an argument with someone who shoots first and ask questions later. The problem we have here is merely a question of operator precedence in mathematical notation. The best I could find on this on google is http://lfw.org/math/n-math.html. Clearly, dot products are listed as having a lower order of precedence there. But another reference is my trusty copy of mathcad, which seems to hav no problem understanding my forumla. Thus I submit that you are simply wrong. But to clarify, I shall add brackets. Incidentally, I did not show two interpretations. I showed one. The correct one. The other you created. Perhaps you should stick to biology, or at least not immediately assume incompetance just because you do not understand something that someone says. And yes, you do sound like someone who will never even consider for a moment the possibility that they are wrong. I can live with that, just please keep out of my face, thanks. Dndn1011 00:02, 8 August 2007 (UTC)

• To anybody who knows that for two arbitrary vectors of the same dimension a and b the following two formulations of the inner product are the same

$\mathbf {a} \cdot \mathbf {b} =\mathbf {a} ^{T}\mathbf {b}$

and knows that for two matrices A and B

$\left(\mathbf {A} \mathbf {B} \right)^{T}=\mathbf {B} ^{T}\mathbf {A} ^{T}\quad {\hbox{and}}\quad (\mathbf {A} ^{T})^{T}=\mathbf {A} ,$

I can show that the equation of Dndn1011 differs trivially from the original one. Indeed, using the two equations in succession and realizing that $\left(\mathbf {{I}^{T}} \mathbf {\hat {n}} \right)$ is a column vector and hence $\left(\mathbf {{I}^{T}} \mathbf {\hat {n}} \right)^{T}$ is a row vector, we get,

$I=\left(\mathbf {{I}^{T}} \mathbf {\hat {n}} \right)\cdot \mathbf {\hat {n}} =\left(\mathbf {{I}^{T}} \mathbf {\hat {n}} \right)^{T}\mathbf {\hat {n}} =\left(\mathbf {\hat {n}} ^{T}\mathbf {I} \right)\mathbf {\hat {n}} =\mathbf {\hat {n}} ^{T}\mathbf {I} \mathbf {\hat {n}} .$

In the notation preferred by Dndn1011 we can rewrite the first (original) equation as follows

$I=\mathbf {\hat {n}} ^{T}\;\mathbf {I} \;\mathbf {\hat {n}} =\mathbf {\hat {n}} \cdot \left(\mathbf {I} \mathbf {\hat {n}} \right).$

To me this formulation is the most natural for programming purposes. Subsequent transposition of this expression offers no advantages, because the left hand side is a scalar. With regard to programming either formulation: see also the present version of the article. Conclusion: in my humble opinion as a biologist (?) the equation of Dndn1011 can be deleted from the article (but I leave the decision to Dndn1011 himself, because I don't want to be an antagonistic corncob).--P.wormer 08:52, 8 August 2007 (UTC)

PS. The discussion was clarifying to me, because I made two mistakes: (i) At first I honestly believed that Dndn1011 tried his hand at an outer product and slipped in the notation. (ii) Then I did not immediately see that putting brackets into the formula made it correct (but trivial).--P.wormer 10:02, 8 August 2007 (UTC)

I think that we are sorted now... however I would point out that however trivial it might seem, to programmers that may not work with matrix algerba all the time, this might not seem trivial at all. Indeed I had to work out for myself the forumla I originally added (which I maintain was mathematically correct). However I do thnk you for pointing out that the transposition of an inertia tensor matrix is the same as the untransposed matrix. I have ammended the article in a manner that I hope if more concise and clear. The positive side to all of this is a very easy to implement equation.

Incidentally, the following:

$I=\mathbf {\hat {n}} ^{T}\;\mathbf {I} \;\mathbf {\hat {n}} =\mathbf {\hat {n}} \cdot \left(\mathbf {I} \mathbf {\hat {n}} \right).$

Is only correct if $\mathbf {I} \;$ is symmetrical. This should be stated clearly when writing out the equation to avoid much head scratching....Dndn1011 15:38, 8 August 2007 (UTC)

Make first paragraph simpler

I think the first paragraph should be made less technical for the general reader. To understand this para as it is, I need to know what inertia and angular momentum are. What about moving some of the excellent elementary explanations in the 2nd para to the 1st? At least, I think the simple statement that moment of inertia describes the resistance of an object to changes in it's rotation belongs in the 1st para. Also, maybe, the wonderful canonical example of the ice skater. --Chetvorno 23:37, 13 August 2007 (UTC)

I also think the angular form of Newton's 2nd law $\tau =I{\frac {d^{2}\theta }{dt^{2}}}\,$ should be introduced sooner, probably as the first equation in the article, instead of several sections down in Equations involving moment of inertia. The elementary mechanics applications (wheels, ice skaters, etc.) that bring most nontechnical readers to this page usually can be solved just by understanding that equation. --Chetvorno 19:23, 22 August 2007 (UTC)

Equal/ defined

Distinction between $=\,$, $\triangleq \,$, and ${\stackrel {\mathrm {def} }{=}}\,$ is not clear. —DIV (128.250.80.15 (talk) 01:33, 29 April 2008 (UTC))

Tensor Definition

The following equation is given as the definition of the Moment of Inertia Tensor

$I_{ij}\ {\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}(r_{k}^{2}\delta _{ij}-r_{ki}r_{kj})\,\!$

I found this indecipherable until comparing it to the products of inertia definition found below it. Particularly the subscripts to the final two r terms seem ambiguous at best and semantically void at worst. The following seems like a more standard representation.

$I_{ij}\ {\stackrel {\mathrm {def} }{=}}\ \sum _{k=1}^{N}m_{k}(r_{k}^{2}\delta _{ij}-r_{k_{i}}r_{k_{j}})\,\!$

Perhaps this might be combined with some additional explanatory text in the 'where' section. Something like:

r_ki is the i component of the distance of mass k from the point about which the tensor is calculated, where i is as described above

Waylonflinn (talk) 18:54, 22 July 2008 (UTC)

Analogy with covariance matrix

I appended a section giving an analogy with covariance matrix. It looks a bit funny on its own as it stands, but I believe it should be included somewhere. A reader who understands the other concepts in the analogy will immediately know what the inertia tensor is from that one sentence. --Tennenrishin (talk) 17:23, 25 July 2008 (UTC)

Good idea. I think the connection will gradually become clearer over time. Thermochap (talk) 18:49, 27 August 2008 (UTC)

I had been thinking about this too. I realized the analogy isn't as direct as I had thought and modified the page accordingly with I=1 tr(σ) - σ. These two matrices have the same eigenvectors but are distinct. Consider a football-shaped covariance. The long axis has the most variance so that eigenvalue of the covariance matrix is largest and the repeated eigenvalue is smaller, corresponding to the "waist" of the football. Conversely, the moment of inertia of the corresponding point cloud is smallest about the long axis, so that eigenvalue of the moment tensor is smallest and the repeated eigenvalue is larger than that one.

I'm a little confused about terminology, though. From Moment (mathematics), it says "The second central moment is the variance", yet the diagonal terms of the covariance matrix are clearly not of the same form as the diagonal terms of the moment-of-inertia matrix. —Ben FrantzDale (talk) 20:24, 27 August 2008 (UTC)

Here's a fragment of a thought: This relationship appears to be related by the fact that physicists want to be able to multiply a rotation vector by the matrix... I think the eye(3,3)*trace(C)-C form may come from the following: (See cross product#Conversion_to_matrix_multiplication for an explanation of the crossMat function.)

crossMat = @(x)([0 -x(3) x(2); x(3) 0 -x(1); -x(2) x(1) 0]);
X1 = crossMat([1 0 0]);
X2 = crossMat([0 1 0]);
X3 = crossMat([0 0 1]);
C = randn(3,3); C = .5*(C+C'); % Example covariance matrix.
% Note that the following two expressions are equal:
X1*C*X1'+X2*C*X2'+X3*C*X3', eye(3,3)*trace(C)-C
X1*C*X1'+X2*C*X2'+X3*C*X3' - (eye(3,3)*trace(C)-C)

I'm curious if anyone can say more about this... —Ben FrantzDale (talk) 20:30, 10 September 2008 (UTC)

Yes, the analogy is not as direct as I had thought, maybe not even worth including? If I am not mistaken, your correction implies that the pairs of corresponding eigenvalues (from Σ and from I) add up to tr(Σ), which explains the negative relationship you point out. So, I propose putting the following in, or scrapping the section:

"The inertia tensor of a rigid body has the same eigenspaces as the covariance matrix of a trivariate random vector whose probability density is proportional to the pointwise density of the body. If the density is normalized so that the body has unit mass, then for each eigenspace, the sum of the corresponding eigenvalues of the inertia tensor and the covariance matrix is $E[R^{2}\rho ]$ where R is the point/random vector and $\rho$ is the probability/physical density at R."

It is not as direct as the originally intended analogy but probably not useless. --Tennenrishin (talk) 15:16, 27 September 2008 (UTC)

I think the relationship should be made explicitly; had the same confusion myself. It seems that moment in mathematics and moment in physics measure the exact same thing, they just quantify it a little differently. If we don't make that clear people like you and I will continue get confused. —Ben FrantzDale (talk) 11:58, 29 September 2008 (UTC)

bad usage of "r"

In subsection "Detailed Analysis" of "Scalar moment of inertia", the scalar quantity "r" is defined as the distance to the rotation axis. However, the spatial representation of density using spherical coordinates uses "r" again, and here "r" is first used as a distance to the origin of the reference frame (spherical coordinates) and then as the (correct) distance to the rotation axis (and z denotes, I presume, the distance along that axis to the origin of the ref frame). 18:08, 20 January 2009. —Preceding unsigned comment added by 77.54.101.212 (talk)

Derivation

edit: now that I think about it a little more my cross product derivation, below, isn't that good: it only works in 3d, since cross products just aren't defined anywhere else. So I think we should stick with the derivation we have right now. But either way, I'm not a fan of the definition section: to me the derivation is the definition. (the Kronecker delta seems really artificial compared to the way it just happens in the derivation..) Sukisuki (talk) 14:06, 20 June 2009 (UTC)

I think we could significantly simplify/clean up the definition and derivation of the tensor, while making it more intuitive.

(If this goes in the article eventually we can probably cut out several steps. I'm just trying to be clear.)

The moment of inertia is the average square radius (from the axis of interest to to each mass element) of all differential elements of mass. For the 3d case, with as our axis unit vector, and our coordinate system being zeroed somewhere on the axis we have this. (which I think this is much clearer than the current version which projects the vector onto the axis and then subtracts the projection from the original vector...)

$I=\int (n\times r)\cdot (n\times r)\,dm$

The cross product giving the vector from the axis to the point, the dot product of that perpendicular vector with itself giving it's length squared.

The dot product is equivalent to matrix multiplying by the transpose of the vector.

$I=\int (n\times r)(n\times r)^{T}\,dm$

if we turn the cross products into matrix multiplication by replacing the radius vector with the equivalent lie group we get this

${I}=\int [n_{1},n_{2},n_{3}]{\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}([n_{1},n_{2},n_{3}]{\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}})^{T}dm$.

apply this property of transposes

$\left(\mathbf {AB} \right)^{\mathrm {T} }=\mathbf {B} ^{\mathrm {T} }\mathbf {A} ^{\mathrm {T} }\,$

to get this

${I}=\int [n_{1},n_{2},n_{3}]{\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}{\begin{bmatrix}0&z&-y\\-z&0&x\\y&-x&0\end{bmatrix}}{\begin{bmatrix}n_{1}\\n_{2}\\n_{3}\end{bmatrix}}dm$

Matrix multiplication is associative so we can multiply out the middle:

${I}=\int [n_{1},n_{2},n_{3}]{\begin{bmatrix}y^{2}+z^{2}&-xy&-xz\\-yx&x^{2}+z^{2}&-yz\\-zx&-zy&x^{2}+y^{2}\end{bmatrix}}{\begin{bmatrix}n_{1}\\n_{2}\\n_{3}\end{bmatrix}}dm$.

I don't know any matrix calculus but (I believe) if this is done around the center of mass then the axis unit vectors can be pulled out of the integral,

${I}=[n_{1},n_{2},n_{3}]\int {\begin{bmatrix}y^{2}+z^{2}&-xy&-xz\\-yx&x^{2}+z^{2}&-yz\\-zx&-zy&x^{2}+y^{2}\end{bmatrix}}dm{\begin{bmatrix}n_{1}\\n_{2}\\n_{3}\end{bmatrix}}$.

which brings us back to this:

$I=\mathbf {{\hat {n}}^{T}} \mathbf {I} \,\mathbf {\hat {n}}$

I'd start with it this way, so that you never have to mention the Kronecker delta, which in my opinion only confuses the issue.

The other way this is nice is when you compare it with the covariance matrix. When taken around the center of mass, one is the average square of the distance along the axis (dot product), the other is the average square of distance away from the axis (cross product).

Sukisuki (talk) 01:18, 3 March 2009 (UTC)

Edited:Sukisuki (talk) 01:46, 16 March 2009 (UTC)

Matlab/Octave code. Matrix notation.

For a set of particles with positions given by a 3xN matrix r, and masses given by a 1xN matrix m, the following code computes the matrix of inertia (if I'm not mistaken):

c=(m([1 1 1],:).*r)*r';
trace(c)*eye(3)-c

I don't have clear how you could express that in matrix notation (esp. the element by element .* product), but for me it gives a nice idea of how it works (plus it's very practical). If all particles have equal mass m, then it's even clearer:

c=r*r';
m*(trace(c)*eye(3)-c)

Franjesus (talk) 20:19, 12 March 2009 (UTC)

Two mathematical issues

I had two minor issues with this page:

• Wherever it says "continuous", I'm pretty sure the text really means "integrable". Consider a ball of radius one and density two embedded in a cube with side two of a material of density one. The formulas are applicable to this object, but the density distribution is discontinuous. The only restriction on the density distribution is that one should be able to integrate the formula; such a density distribution is called "integrable".

• I find the notation $r_{k}^{2}$ for $r_{k}\cdot r_{k}$ highly disturbing. Is this really standard notation for physicists and mechanical engineers? If yes, I propose to add a parenthetical definition of this notation for those not familiar with it; if no, I propose to just use $r_{k}\cdot r_{k}$ instead.

DutchCanadian (talk) 21:36, 9 April 2009 (UTC)

I agree on the first item. I'll fix that. On the second item, now that you mention it I remember being thrown by that too. I think the notation is generally pretty standard in that I don't see a superscript numeral two over any vectors; what is weird is that it implicitly has $r$ mean $\|\mathbf {r} \|$ in a number of places when $\|\mathbf {r} \|$ would do just fine. I'll fix that. —Ben FrantzDale (talk) 12:48, 10 April 2009 (UTC)

Sukisuki (talk) 15:08, 20 June 2009 (UTC)

Good point, continuous vs. integrable.

But I think where people made the 'mistake' they were thinking continuous vs. discrete. It's probably not mathematically rigorous but I think that's how most people describe the jump from sum to integrate.

For the second, I can't speak for all mech engineers, but to me squaring is something you do to a scalar, so in

$I=\int r^{2}\,dm\,\!$

it means we're dealing with a scalar radius. Using cylindrical coordinates comes naturally to many of these problems, why convert explicitly to Cartesian coordinates before we need to.

Definition

The basic definition of M of I that I find easiest to understand, explain, and apply conceptually to real world situations where the ratio of rotational inertia to mass (e.g. sailing!) is important is:

$I=\sum {mr^{2}}\,\!$

The meaning of the integral form is less obvious. I suggest using this expression either before introducing the integral form, or to replace it. GilesW (talk) 09:10, 7 May 2009 (UTC)

Conceptual entity description

Note that the moment of rotation element has 3 units of description and can therefor be considered to be a volume of something. We can draw 2 orthogonal (distance = radius) vectors and see that these can represent the sides of a square area. then we can draw a third perpendicular vector that represents the mass value and if it is considered to be the height of a cube we will have created a concept of a volume of space that represents a moment. And the moment volume concept keeps in mind the relative importance of the 2 conceptual entities that make up the moment value, since the radius vector has to be squared and the mass unit only involves 1 dimension.WFPM (talk) 12:39, 30 April 2010 (UTC) Therefor if you want to maximize the moment of something, you make it as thin and as wide in diameter as possible.WFPM (talk) 12:46, 30 April 2010 (UTC) See Flywheel.

Now note that the moment of inertia is involved with two properties of motion of the material entity. The first is that related to its quantity of inertia that it has as the result of its translation velocity without regard to direction, and is primarily a kinetic energy storage property. Therefor it has kinetic energy of motion, which is proportional to M times (linear v)squared and you add to it or take it out to speed it up or slow it down. The second property has to do with the directional orientation of this momentum in space and what it takes to change it's direction.WFPM (talk) 15:04, 30 April 2010 (UTC)

Explanation of small edit I just made

I saw the following sentence: The moment of inertia tensor is a convenient way to summarize all moments of inertia of an object with one quantity. My reaction was that the inertia tensor is not one quantity. Rather, it is one symbol that refers to an array of six quantities. So I edited it. Kimaaron (talk) 22:19, 12 May 2010 (UTC)

I disagree with this edit; the distinction that the moment of inertia of a body is a single tensor-valued quantity is an extraordinary powerful notion that should not be overlooked. The distinction between a "tensor" and a a symmetric matrix of six quantities is that a tensor really is a single "thing" invariant of coordinates; in a particular coordinate system, you can represent a tensor as a matrix. Rather than revert your edit, I rewrote those paragraphs from scratch. It could do with more cleanup, still. —Ben FrantzDale (talk) 02:02, 14 May 2010 (UTC)

Excellent. I like your new phrasing. By the way, I do appreciate that a tensor is a single "thing" (much like a vector is a single thing). I also understand that the representation as a matrix of values is a way of describing the components of the tensor in a particular coordinate system (much like using the three components of a 3D vector), but the underlying tensor is a physical thing that doesn't depend on the coordinate system. I just didn't like referring to a tensor as a single quantity. One "quantity" (i.e. one number) is not enough to describe a tensor.

Kimaaron (talk) 00:43, 18 May 2010 (UTC)

Cool. I'm glad we are on the same page. I think it's an important distinction and that clarifying it here can help people understand tensors in addition to helping them understand what moment is. —Ben FrantzDale (talk) 12:34, 20 May 2010 (UTC)

Requesting definition of vector r

It seems to me that in the following equation, we also need to define what the vector r is. At first, I was thinking it was simply the position vector from the origin to the point (x,y,z). I even edited it to say that. But then realized I was not sure that was correct, so I undid it. Earlier in the article, it defines r as the perpendicular distance from an axis of rotation, but here, we don't have an axis of rotation. So what is r? Maybe someone who understands this notation better than I do can add the definition at the tail end of the sentence where it says, " ... where $\mathbf {r} \otimes \mathbf {r}$ is their outer product, E₃ is the 3 × 3 identity matrix, and V is a region of space completely containing the object." Here's a copy and paste of the section I'm talking about:

These quantities can be generalized to an object with distributed mass, described by a mass density function, in a similar fashion to the scalar moment of inertia. One then has

$\mathbf {I} =\iiint _{V}\rho (x,y,z)\left(\|\mathbf {r} \|^{2}\mathbf {E} _{3}-\mathbf {r} \otimes \mathbf {r} \right)\,dx\,dy\,dz,$

where $\mathbf {r} \otimes \mathbf {r}$ is their outer product, E₃ is the 3 × 3 identity matrix, and V is a region of space completely containing the object.

A few minutes later: Oh, I was thinking this was deriving the tensor. I guess we *do* have an implied rotation axis. It seems that r is a perpendicular vector from the relevant axis to the point (x,y,z). Frankly, I do not find this notation helpful because I'm not terribly familiar with it. I'm sure that mathematicians like it, but for us simple engineers (arguably the people who actually use mass moments of inertia the most), the good old simple Ixx = integral (y^2 + z^2) dm is a lot easier to follow, albeit not as compact. Do we care more about conveying information to the most people or conveying it most compactly? The former, in my opinion.

Kimaaron (talk) 01:17, 18 May 2010 (UTC)

You are right. It wasn't clear. I rephrased it. $\mathbf {r}$ is simply the vector from the center of mass to each point. There is no implied axis. The fact that you can extract the moment about an axis comes from the $\|\mathbf {r} \|^{2}$ term. That lets us express the moment in terms of the outer product. It leaves the $I_{xx}$ term (ignoring density) as $\|\mathbf {r} \|^{2}-x^{2}=x^{2}+y^{2}+z^{2}-x^{2}=y^{2}+z^{2}$.

Let me know if it is still unclear. —Ben FrantzDale (talk) 12:51, 23 August 2010 (UTC)

^ “Mass moment of inertia” by Mehrdad Negahban, University of Nebraska

This deals with second momemt of area, not mass moment. J Westmore 20.133.0.13 (talk) 10:43, 23 August 2010 (UTC)

Figure skating

I admit I didn't read the whole article, but couldn't we have a little section about what happens when a figure skater is doing a spin?

• If you are spinning, and then you pull your arms in, does this increase or decrease your moment of inertia?
• How does this change in your moment of inertia affect your rotational speed?

I was guessing that by keeping your angular momentum the same while decreasing your moment of inertia (as you pull in your arms), you would automatically spin faster. Is this right?

• And if so, does this mean that Michelle Trachtenberg's character in "Ice Princess" got it wrong? --Uncle Ed (talk) 02:05, 15 November 2011 (UTC)

Pulling arms in can reduce your moment of inertia about the vertical axis by half or more. Seems unreasonable for such small masses but the distance moved is squared. When spinning it is difficult to change your angular momentum. Friction gradually reduces it but we like to consider this spin as conservative, thus angular speed is completely dependent upon how much momentum you started with and what is your instantaneous moment of inertia. Notice that figure skaters spread out after the spin. This is to slow their rotation so they can continue with skating. DGER (talk) 14:11 19 November 2011

Thanks; let's mention this in the article. By the way, the lead character of Ice Princess said it backwards, i.e., that pulling in her arms increased her moment of inertia. (Or did I hear it wrong when I watched the movie last week?) --Uncle Ed (talk) 21:26, 19 November 2011 (UTC)

Last paragraph of Examples section

This section: Moment_of_inertia#Examples last paragraph seems to be written in a very poor and repetitive style, specially the last sentence:

"When considering mechanisms like gear trains, worm and wheel, where there are more than one rotating element, more than one axis of rotation, an equivalent moment of inertia for the system should be found. Practically when a geared system is enclosed, equivalent moment of inertia can be measured by measuring the angular acceleration for a known torque or theoretically it can be estimated when the masses and dimensions of the rotating elements and shafts are known. In this practical the equivalent moment of inertia of a worm and wheel system is measured using above mention methods."

However I don't want to edit because it seems to refer to a specific engineering approach of angular momentum, which I don't know nothing about. Anyone else? --Skarmenadius (talk) 20:07, 25 November 2011 (UTC)

I'd bet the rent that the last sentence (at least: possibly the whole thing) is copied from instructions for an engineering school lab class! 86.176.166.181 (talk) 19:19, 25 April 2012 (UTC)

Figure Skating , Dimensions

I wouldn't dare edit this article, from which I've learned a lot. But I wonder if a more homely example than figure skating would be better. Get someone to spin you round in your office chair. Legs out you slow down, pull 'em in you speed up.

Secondly, I remember learning about Dimensional Analysis, and have found it most helpful over the years. So I wonder if the sentence "The easiest way to differentiate these quantities is through their units (kg·m² as opposed to m⁴)." would be better as "The easiest way to differentiate these quantities is through their Dimensions (M3 L T-2 as opposed to M4).".John Wheater (talk) 09:39, 23 January 2012 (UTC)

Not everyone know dimensiinal analysis. Units of measure are commonly understood (or should be). Dger (talk) 17:35, 5 March 2012 (UTC)

F = ma

I think it is potentially confusing that F = ma appears prominently on the right at the top of the article, but the relevant equivalent (T = I alpha) doesn't appear until halfway down this long article. I would think that that and the energy stored, would be key facts to put early on.Ccrrccrr (talk) 02:51, 13 April 2012 (UTC)

Inertia matrix

I misunderstood the organization of this article and unintentionally inserted a new section on the inertia matrix under the section on scalar moments of inertia. While it is an inertia tensor, I moved it to its own section, because the formulation is different from what is used in the section on inertia tensor. This formulation of the inertia matrix is actually presented in one line at the end of the article: "plugging in the definition of $[\cdot ]_{\times }$ the $[r]_{\times }[r]_{\times }^{\top }$ term leads directly to the structure of the moment tensor." However, I believe this deserves more attention. Prof McCarthy (talk) 18:10, 9 October 2012 (UTC)

Redundancies

What started as the introduction of the inertia matrix has grown to almost double the size of this article. I am sure there are some redundancies, but because of the complexity of the topic, it does not seem necessary to remove all of them. Various readers may come with different experiences and this may help with their understanding of the topic. Prof McCarthy (talk) 06:25, 29 October 2012 (UTC)

Moment of inertia tensor

I streamlined the presentation of the moment of inertia tensor. This was one of the earliest sections of this article and contained a variety of topics that are now covered in more detail in other sections. Prof McCarthy (talk) 17:28, 30 October 2012 (UTC)

Covariance matrix

I have read the various articles in Wikipedia on covariance matrices of various forms. It may be that the covariance matrix has a form similar to the moment of inertia matrix, but it is not shown explicitly anywhere that I could find, so it does not seem useful to refer to this relationship in this article on the moment of inertia matrix. Prof McCarthy (talk) 23:49, 30 October 2012 (UTC)

Introduction 2nd paragraph

Quote: The moment of the inertia force on a particle around an axis multiplies the mass of the particle by the square of its distance to the axis, and forms a parameter called the moment of inertia.

How to understand this example? Suppose I am familiar with inertia force on a particle, suppose I can imagine it's moment around an axis. But why should I imagine that this moment 'multiplies' mass by the square of it's distance? Why should I choose non-inertial reference frame and think that some fictious moment there multiplies something to achieve parameter indipendent from this non-inertial frame? For a guy without any knowledge over this matter this suggestion seems too puzzled . Maybe if all necessary pre-formulations are shown and the moment equation derived it could be seen how some part of it could be separated. Maybe someone who knows what is it all about could accept this imaginary step without pre-formulations, but what is the profit actually? Someone who knows the thing to get another example of a thing he already knows, or someone who does not know to find clear and simple step by step introduction in the subject?

Twowheelsbg (talk) 04:07, 1 November 2012 (UTC)

The inertia force on a particle is mass times acceleration. The moment of the inertia force multiplies this quantity by the distance to the pivot and the calculation of acceleration multiplies the distance by the angular acceleration around the pivot, thus mass is multiplied by the distance squared. This central role of mass times distance squared appears in most every section of the article. Prof McCarthy (talk) 04:43, 1 November 2012 (UTC)

Thank you for your kind clarification. Anyhow my comment was more assigned to the relative complexity for first example explaining the subject. Inertia force ( non-inertial frame ) seems not compulsory, a point-mass attached for rotation on an ideal link together with the equation of the regrouped terms would do better in my opinion. Twowheelsbg (talk) abt 11:30, 1 November 2012 (UTC) — Preceding unsigned comment added by 83.228.49.174 (talk)

I believe what you describe is in the section "Moment of inertia of a simple pendulum" which immediately follows the Overview.Prof McCarthy (talk) 11:05, 1 November 2012 (UTC)

Recent revisions

The section "scalar moment of inertia of a rigid body" has been revised. There seemed to be an error for the diatomic molecule, the wording for the thin rod was simplified, and the calculation for the thin disk was added. This made it easier to obtain the moment of inertia for the solid ball than the previous derivation. A section describing how to measure moment of inertia was added. Prof McCarthy (talk) 08:00, 15 November 2012 (UTC)

Bad maths

Editor Dai bach exposes a recurring difficulty that arises in teaching and writing about the dynamics of planar rigid body movement. Torque is a vector that is central to the study of rigid body dynamics in three dimensions, which when restricted to the plane is reduced to a single non-zero component that is directed perpendicular to the plane. This is awkward for the educator or editor who must now decide whether to present torque as a scalar using a mathematical formulation different from what is used in three dimensional space, or to appeal to the presence of a third dimension that is ignored everywhere else in the study of planar dynamics. In my opinion, the two dimensional formulation of planar dynamics is a convenience that simplifies the mathematics but does not exist in isolation from three dimensional dynamics. For this reason, I recommend presenting torque as a vector in the formulation of planar dynamics, even though Dai bach considers it to be "bad maths." Prof McCarthy (talk) 17:50, 23 November 2012 (UTC)

Thin disc

The restriction to "thin" in the calculation of moment of inertia of a disc is not needed, since the axis involved is the rotational axis of the disc. It would matter if the axis was parallel to the disc face (see the thin rod calc), but in this case a thick disc would be the same as a stack of thin discs. Note that this is used in the calculation for the ball. -John98.223.153.238 (talk) 07:32, 13 December 2012 (UTC)