Talk:Moment of inertia

NACA Technical Note No. 1629.
It seems that on February 6, 2013 User:Coretheapple recommended deletion of the Wikipedia article NACA Technical Note No. 1629, and nine days later it is gone. This article refers to perhaps one of the best presentations of an experimental method to determine physical properties that are critical to the dynamic analysis of mechanical systems. I have not found another reference that provides as detailed and practical a method for determining moment of inertia that is as complete and useful as this reference. One reviewer's comment caught my eye: "Delete. WP:not for source texts. As to transwiki, is Wikisource really interested in reproducing a random piece of technobabble from the 1940s, of no apparent importance of any kind? If you think so, go ahead and do it, but don't expect others to do the work." I have worked for a couple years now to provide clear and correct presentations to articles on machines, rigid body dynamics and even moment of inertia. However, I am beginning to realize that while Wikipedia has a lot of these articles, they may actually be out of place. A request to have the article moment of inertia evaluated as a good article yielded the response that Wikipedia is not a textbook. Perhaps it is correct that Wikipedia should be reserved for every detail of a sports franchise or video game series, and not waste space on the basic principles of science and mathematics that are important enough to appear somewhere else. Prof McCarthy (talk) 07:07, 5 March 2013 (UTC)

Moment of inertia in planar movement of a rigid body
Hi Professor McCarthy, I am quite confused about the following issues in the paragraph:


 * 1) If angular velocity ω is a scalar, then why can it be used in a cross product. If it is a vector, why isn't it in bold font?  What is the font style used for specifying a scalar or a vector?
 * 2) What does the x inside the equation $ti = kxei$ stand for? It's not defined in the paragraph.

Thanks for clearing up these issues.--LaoChen (talk)06:48, 6 April 2013 (UTC)


 * An equivalent to this question has been asked previously, so I have added a note in the text. In elementary presentations of planar movement, the cross product is not used.  However, the vector formulation of angular velocity and angular acceleration, even for planar movement, allows the use of the geometric properties of the cross product to present the interaction of mass and geometry that defines moment of inertia.  I hope this is helpful. Prof McCarthy (talk) 19:11, 6 April 2013 (UTC)

Thanks very much for the detailed explanation. According to the note added, we are using vectors to represent angular velocity and angular acceleration just for convenience of calculation. Now, why do we display these vectors differently, using the regular font, while other vectors are displayed using bold font? Is there a difference in characteristics between these vectors and other vectors? When this article is submitted for GA, the reviewers may question the consistency of the notation. If the notation used is not consistent, the readability of the article may be affected even though I think the content is excellent. --LaoChen (talk)06:50, 7 April 2013 (UTC)


 * This is a question about the use of fonts? I have made ω and α boldfaced. Prof McCarthy (talk) 13:01, 7 April 2013 (UTC)

Thanks for the modifications. It's a lot more readable now. I have also found the meanings of $$k$$ and $$\vec{k}$$ not quite clear beginning from the section "Scalar moment of inertia of a compound pendulum". Is the later variable some kind of a radius of gyration? How do you define this variable? To the editors and engineering students, this may be recollection of what they may have seen somewhere before. To other readers, some may have no idea what's going on here. We certainly don't want to confuse the readers at this early stage. They may get stuck and lose interest.--LaoChen (talk) 06:30, 8 April 2013 (UTC)


 * One minor point here. I see three exact copies of the statement "The moment of inertia $$I_C$$ about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia", yet the sources don't seem to support this definition. The more common definition seems to be equivalent to Second moment of area. RockMagnetist (talk) 22:10, 9 April 2013 (UTC)


 * This is often a source of confusion, so it seemed worthwhile to address it directly in the article. Prof McCarthy (talk) 07:53, 10 April 2013 (UTC)

Dragster length
Isn't the reason top-fuel dragsters are so long is more to resist the pitch moment rather than yaw? Funny cars seem to go straight enough without the length of top-fuel dragsters. Just asking. — Preceding unsigned comment added by Three d dave (talk • contribs) 18:00, 17 May 2013 (UTC)


 * You are correct that pitch moment is an important concern, as well as the yaw moments in steering. Prof McCarthy (talk) 18:19, 17 May 2013 (UTC)


 * Can we get a source for either of these assertions? The long chassis does not appear to be used to maximize the moment of inertia about either axis, since the engine and driver are clustered near the rear wheel. Perhaps the long wheelbase simply increases directional stability kinematically, instead of dynamically. A reliable source would settle the issue. -AndrewDressel (talk) 13:31, 8 July 2013 (UTC)


 * A dragster chassis must support the transformation of engine rotation into linear speed as quickly as possible. The chassis flexes as the longitudinal torque of the engine is rotated into transverse torque at the wheels, and this transverse torque lifts the chassis against gravity.  While this occurs friction on the front wheels provides steering forces to maintain a straight line trajectory. Prof McCarthy (talk) 15:37, 9 July 2013 (UTC)


 * This may all be true, but none of it explains why the wheelbase is so long. -AndrewDressel (talk) 18:44, 9 July 2013 (UTC)


 * The regulations for dragsters allow wheelbases ranging from 150 in to 300 in and most all dragsters are designed to 300 in even though the longer wheelbase means increased weight and increased flexibility. The longer wheelbase provides increased inertial forces that resist both the pitch moments and the steering moments. Prof McCarthy (talk) 15:37, 9 July 2013 (UTC)


 * This sounds pretty good, but as you know, sounding good is not sufficient. If the goal where merely to maximize the moment of inertia, why isn't the driver further forward? Wouldn't that increase the moment of inertia without the penalty of additional mass? On the other hand, I believe it is well established that a longer wheelbase improves yaw stability for purely kinematic reasons. The same lateral perturbation causes a smaller yaw angle on a vehicle with a longer wheelbase. Also, the sources I can find attribute wheelies to tire thrust and cm height vs cm forward distance from rear axle. (For example this Car and Driver article.) Doesn't moment of inertia only become a factor in the rate of rotation after the front wheels are unloaded, and that's when the wheelie bars come into play to prevent rotation? -AndrewDressel (talk) 18:44, 9 July 2013 (UTC)


 * The driver sits over the transmission to shift gears. The engine is positioned as close as possible to the rear wheels to minimize the length of the drive train.  My understanding is that stability is a dynamic property not a kinematic property.  A lateral perturbation is a force and the associated angular acceleration around the center of mass is proportional to the moment of that force divided by the moment of inertia vehicle.  The larger the moment of inertia the less the angular acceleration.  Lift of the body is angular acceleration that arises from the moment of the tire thrust about the center of mass.  Again the angular acceleration of the wheelie is proportional to the moment of the tire thrust divided by the moment of inertia of the vehicle.  Moment of inertia arises directly from Newton's second law of motion and comes into play whenever there is angular acceleration.  Angular acceleration must present before a body begins to rotate.  It does not matter whether the the front wheels are loaded or unloaded, or if the wheelie bars come into play. Prof McCarthy (talk) 00:47, 10 July 2013 (UTC)


 * According to this 2011 Popular Mechanics article and this [[NHRA] article], top fuel dragsters don't even have transmissions, so there is no shifting, just a clutch operated by timers. Perhaps you are thinking of the long-obsolete slingshot-style dragster. As the analysis given in Directional stability shows, the moment of inertia cancels and so does not appear in the final directional stability criteria, just wheelbase and center of mass location. As for wheelies, they cannot not begin until the front wheels bear no vertical load, and they usually cannot increase once the wheelie bars contact the ground. It now appears that there is no source available for the assertions about dragsters in this article, and so those assertions should be removed. -AndrewDressel (talk) 02:51, 10 July 2013 (UTC)


 * The point of the discussion of the position of the driver it is that it depends on considerations other than the dynamic performance of the dragster. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * The point of this discussion is whether the assertions about dragsters in the article are verifiable, and it is beginning to appear, unfortunately, that they are probably no more so than the assertion that "the driver sits over the transmission to shift gears." -AndrewDressel (talk)


 * The assertion that the driver sits over the transmission to shift gears has nothing to do with this article, and was raised only to address the odd statement that the driver should be moved further forward, which is as I have stated a distraction. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * I think the assertion that the driver sits over the transmission to shift gears suggests exactly how the dragster comments got into this article in the first place: it sounds reasonable unless someone checks the facts. -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * The facts are clear, your continued reference to the driver is not relevant to this discussion. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * At this point I don't know which is more humorous, that one would make such an easily-disproved assertion in a discussion about unverifiable assertions, or that the one making such an assertion would continually refuse to stop discussing it. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * Similarly, the reference to the directional stability analysis is incorrect because it refers to the relationship between steering forces in an actual turn, known as understeer and over steer, which do not make sense for a dragster. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * Actually, the article specifically states that "the hubs are aligned with the vehicle body, with the steering held central," so the analysis is not about "forces in an actual turn," and should be perfectly applicable to dragsters. -AndrewDressel (talk) 13:56, 10 July 2013 (UTC)


 * The article specifically states that the focus is on the stability of the tire slip angle and assumes the vehicle is in a turn which causes the direction of movement at the tire to be different than the steering direction of the tire. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * I cannot find that specific statement in the article. Would you please provide a quotation, as I have provided above indicating that the analysis is not of a vehicle in a turn. A non-zero slip angle does not indicate a turn. Instead, it merely indicates that the vehicle has experienced some disturbance that momentarily misaligns the direction it is pointing from the direction it is moving. Of course the slip angle of the tires and the forces produced is included in the analysis. Leaving it out would be equivalent to leaving out angle of attack and lift from the stability analysis of an aircraft. -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * That article derives a differential equation for tire slip and evaluates conditions when this slip angle can grow without bound. This is not lift. The role of moment of inertia is to resist the influence of applied torque on angular acceleration. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * That article derives a differential equation in "$$\beta$$ (beta), the slip angle for the vehicle as a whole", which is simply the difference between the direction it is pointed and the direction it is going, arguably a useful variable to track in the analysis of directional stability. Development of the equation happens to include consideration of tire slip angle, which is not the same thing, to improve fidelity. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * Furthermore, in that analysis, moment of inertia does not cancel, it is simply assumed to be the same for the vehicles because the focus is on variations in the location of the center of gravity. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * The focus of that analysis is no more on center of mass location than any other parameter, but the stability of the vehicle ends up hinging on the stiffness term, which in turn depends on the center of mass location. In calculating the critical speed, the moment of inertia, which had been included from the beginning, simply cancels out, just as mass does from the equations of motion of a pendulum. -AndrewDressel (talk) 13:56, 10 July 2013 (UTC)


 * The moment of inertia is clearly present as a parameter in all of the equations. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * Yes, right up until it cancels from the expression for the critical speed: $$V^2=\frac{2k(a+b)^2}{M(a-b)}$$ -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * The conditions for growth of the slip angle is a special result. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * How so? Is there some general result presented elsewhere or skipped over? -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * The focus in this article is the affect of moment of inertia on angular acceleration that results from applied torque. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * Yes, and top fuel dragsters appear to be poor examples of that phenomenon, especially when the examples remain unreferenced. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * These are misrepresentations of the laws of physics and do not contradict the central role of moment of inertia in the stability of a dragster movement. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * I don't know what "misrepresentations of the laws of physics" you mean. Are you suggesting that I am misrepresenting the laws of physics? -AndrewDressel (talk) 13:56, 10 July 2013 (UTC)


 * I am struggling to understand what you are actually saying. You seem to be saying that Newton's second law does not apply to a dragster, that yaw and pitch angular accelerations are not proportional to yaw and pitch moments,  and that the constant of proportionality is not the moment of inertia of the vehicle. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * I'm saying that there is no more evidence that top fuel dragsters are long and narrow in order to increase their moment of inertia and thereby increase roll and pitch stability than there is evidence that buses are long and narrow for the same reason. -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * When a bus accelerates in a straight line its long and narrow design has a large moment of inertia means which means pitch and yaw torque generate smaller angular acceleration than if it was shorter. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * Without a detailed analysis validated by physical testing, however, there is no reason to believe that this effect dominates or is even non-negligible, compared to tire and suspension effects, and so would be an equally poor example for this article. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * Furthermore, it is incorrect to say that wheelies cannot begin until the front wheels bear no vertical load. A wheelie begins with an upward rotational motion of the vehicle in the vertical plane, which must result from an applied torque.  This upward rotation begins with an angular acceleration that is proportional to the applied torque divided by moment of inertia.  This is simply Newton's second law and is what the article describes.  Without this angular acceleration the front wheels cannot unload and begin the upward rotation that becomes a wheelie. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * Now it appears that you are saying load transfer is not a real phenomenon or that it has no real effect. The rotation you are so anxious to get to simply cannot occur until the load borne by the front wheels has decreased to zero, and stability of the vehicle may be adversely effect by any of that decrease because of the accompanying loss of friction. -AndrewDressel (talk) 13:56, 10 July 2013 (UTC)


 * Again I am having trouble understanding your point. Load transfer is of course a real phenomenon, but it begins with the application of a torque not a specific value of the front wheel load forces.  You seem to be saying that angular acceleration of the body does not occur until torque on the body is enough to unload the front wheels.  This is of course contrary to the purpose of the article, which shows that Newton's laws require this angular acceleration to be equal to the applied torque divided by the moment of inertia even before the front wheels are unloaded. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * How odd. Perhaps the difficulty is that you are assuming that the torque applied by the rear wheels reaches its steady-state value instantaneously. I have been assuming that it builds up at a finite rate, and only lifts the front wheels when it is sufficiently large to do so. Thus, I am saying exactly that the angular acceleration of the body remains zero until the torque applied by the rear wheels is sufficiently large to unweight the front wheels. -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * There is no assumption, because a torque applied to the vehicle causes angular acceleration. The larger the moment of inertia the less the angular acceleration.  This is true no matter what the wheel loading.Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * Perhaps you mean a "net torque", "unbalanced torque", or "resultant torque" because if the "applied torque" is not large enough to unweight the front wheels no angular acceleration can occur, in the same way that linear acceleration would not occur due to a 1000 lb horizontal force applied to a 2500 lb dragster resting on drag strip with which it shares a coefficient of friction of 1 or more. Now you might also be thinking of deformable bodies, but then the body stiffness will likely dominate any inertial effect, and the situation is far more complicated than is appropriate for an example for this artlce. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * When the wheelie bars contact the ground they apply a torque to the vehicle that decelerates the angular velocity again in proportion to the moment of inertia of the vehicle. A discussion of this law of physics is the purpose of the article.  The assertions in this article are sourced to laws of physics, the error is the misuse of a collection of sources that distract from the actual phenomena of the acceleration of a dragster. Prof McCarthy (talk) 05:54, 10 July 2013 (UTC)


 * Being "sourced to laws of physics" is simply not sufficient, especially when the article specifically states that "a dragster is long and narrow to increase the moment of inertia to resist pitch and steering moments to keep it moving in a straight line." There is no more evidence that this is the reason a dragster is long and narrow than there is for jet liners, railway cars, stretch limousines, oil tankers, or any other long and narrow vehicle. "All material in Wikipedia must be attributable to a reliable, published source." I was initially intrigued by the assertion, and so looked for such a source, but could not find one. You are still welcome to provide one, but until one is found, those assertions belong here, and not in the article. Until such a source is found, I believe we've said all there is to say on this topic. -AndrewDressel (talk) 13:56, 10 July 2013 (UTC)


 * The point of the article is to illustrate the details of a fundamental law of physics which simply states that angular acceleration equals the applied torque divided by the moment of inertia. It is clearly correct that the long axes of jet liners, railway cars, stretch limosines, oil tankers, and other long and narrow vehicles provide a large moment of inertia, and therefore demand larger applied torques to achieve an angular acceleration than vehicles with shorter lengths and therefore less moment of inertia. If this law of physics is not a sufficient explanation, then I cannot imagine how to source it more appropriately. Prof McCarthy (talk) 05:40, 11 July 2013 (UTC)


 * The point is that none of these other long-narrow vehicles are long and narrow for this reason, and so they would be equally bad examples for an article about moment of inertia. Even if the statements were altered to no longer suggested that increasing the moment of inertia is the reason that top fuel dragsters are long, it would be a bad example because it is not clear how inertial effects compare to the stabilizing effects of other obvious features of such vehicles: tires, wheelie bars, and wings.


 * The article already contains three good examples of devices or actions specifically designed to maximize or manipulate moment of inertia: the long pole carried by tight-rope walkers, flywheels, and figure skater arms. Another might be the tuck position of gymnasts and divers. The reason for the long wheelbase of top fuel dragsters is simply not so clear cut, and without a specific reference, should not be included. -AndrewDressel (talk) 16:41, 11 July 2013 (UTC)


 * This discussion shows that the loading of the front wheels of a dragster during acceleration is a practical example of moment of inertia. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * Far from it. I have not yet seen a shred of evidence to support that assertion. Moment of inertia might play a role in the load transfer of vehicles with complaint, long-travel suspensions, but even 6 inches of travel represents just over a degree of rotation on a 300-inch-wheelbase vehicle. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * I will provide more details in this article. Prof McCarthy (talk) 14:51, 12 July 2013 (UTC)


 * I trust that they will be better researched and referenced than the assertion made above about driver location. (Sorry, but a whopper like that is just hard to ignore, especially when it is never retracted.) -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * I am familiar enough with Wikipedia to know that I am probably wasting my time. I also realize that the article Physics of Wheelstands has provided a mistaken understanding that may not be possible to change. But I will do my best. Prof McCarthy (talk) 15:08, 13 July 2013 (UTC)


 * I am curious to know what you believe that mistake to be. The weight transfer equation given is compatible with Cossalter's treatment of motorcycle wheelies -AndrewDressel (talk) 19:33, 13 July 2013 (UTC)


 * An independent treatment of wheelies, in Gray, Costanzo, and Plesha's 2010 Dynamics textbook begins with the full Newton-Euler equations and so necessarily includes the vehicle moment of inertia. They then calculate the maximum possible acceleration of the combined bike and rider rigid body such that the front wheel does not lift off the ground, in other words, the acceleration that causes vertical load on the front wheel to decreased to zero. With this constraint the angular acceleration is also zero, and the only term which includes moment of inertia is eliminated. Thus, their analysis seems to say exactly "that wheelies cannot begin until the front wheels bear no vertical load," and whether a vehicle, when treated as a rigid body, experiences a wheelie or not is independent of its moment of inertia. -AndrewDressel (talk) 16:04, 16 July 2013 (UTC)


 * The statement that the angular acceleration during a wheelie is zero until the front wheels are unloaded is simply wrong, and I will attempt to explain in a revision to this article. You can judge it then, and delete it if you so chose. Prof McCarthy (talk) 01:01, 19 July 2013 (UTC)


 * Well, I look forward to that. It will be interesting to see how you show up the various published authors I have already cited including a prominent high-performance vehicle dynamisist, and every dynamics textbook I can find. Besides Gray, Costanzo, and Plesha described above, Hibbeler also uses a motorcycle wheelie as an example with the normal force on the front wheel going to zero while the angular acceleration remains zero. -AndrewDressel (talk) 01:49, 19 July 2013 (UTC)


 * It will be easy because torque divided by moment of inertia is angular acceleration, as it says in all of your dynamics texts. It is not complicated, if there is no angular acceleration, then there is no torque. Prof McCarthy (talk) 23:14, 21 July 2013 (UTC)


 * But your statement seems to suggest that your are ignoring the important detail that it is only "net torque" or "resultant torque" or "sum of external torques" about a point that is equal to the rate of change of angular momentum about that point. Simply applying an external torque to a body is not necessarily sufficient to cause angular acceleration unless it is the only external torque or it is large enough to overcome whatever constraint forces and moments that may exist. In the case of the dragster we have been discussing, gravity and the ground apply significant forces and moments that prevent angular acceleration until the moment caused by the horizontal force at the drive wheels is sufficient to reduce the vertical force at the front wheels to zero. That is why vehicle acceleration and the imminent onset of a wheelie is an interesting problem in dynamics textbooks. It demonstrates the important role that angular momentum can play even though the angular acceleration is zero. As Ruina and Pratap state in their section on 1D motion w/ 2D & 3D forces "angular momentum balance is important even when there is no rotation." -AndrewDressel (talk) 13:10, 22 July 2013 (UTC)


 * If the torque is zero then there is no angular acceleration, and if there is no angular acceleration the torque is zero. There is no missing detail.  In order to have angular acceleration there must be torque, if there is angular acceleration then there is torque.  And the size of the angular acceleration is the torque divided by the moment of inertia. Prof McCarthy (talk) 20:39, 31 July 2013 (UTC)


 * Holy smokes! By that reasoning if I press my finger against this computer, it accelerates, no matter which direction I push or what the friction coefficient might be between it an the table on which it rests. Seriously? You can't (or won't) distinguish between a single applied torque and the net result of all the applied torques? How can I not question your competence to edit this article? -AndrewDressel (talk) 21:56, 31 July 2013 (UTC)
 * This is a relatively straight forward principle of mechanics. If a torque applied to a body is balanced by a constraint such as friction, then the torque on the body is zero, and it does not move.  If the torque on the body is not zero, then it must move with an angular acceleration, and the size of the angular acceleration decreases as the moment of inertia increases. Prof McCarthy (talk) 07:11, 1 August 2013 (UTC)


 * It would be helpful if you would use any of the conventional adjectives to distinguish "a single applied torque" from "the sum of all applied torques". Without them, sentences such as "If a torque applied to a body is balanced by a constraint such as friction, then the torque on the body is zero" sound like gibberish. Using the same unmodified word to mean two different things in the same sentence may be confusing to your readers.


 * Okay. Now that I understand that you mean "a torque" to be different from "the torque", it appears that you believe that when the horizontal force at the rear wheels of a dragster applies "a torque" to the body, "the torque" generated by all the external forces acting on the dragster must be non-zero. That analysis, however, goes exactly against the published examples about accelerating vehicles in the dynamics textbooks by
 * Gray, Costanzo, and Plesha: example of "a motorcycle popping a wheelie" on pages 545-548 of the 1st edition
 * Hibbeler: "Example 17.6" of the conditions for a motorcycle rider "to do a wheely" on page 406 of the 11th edition
 * Beer and Johnston: "Sample Problem 16.1" of the ground reaction forces of a braking truck on page 998 of the 6th edition
 * Bedford and Fowler: "Example 18.1" of the ground reaction forces of an accelerating airplane on pages 338-339 of the 3rd edition
 * Ruina and Pratap: "Sample 12.7" of a "a suitcase skidding on frictional ground" on pages 626-627 of the 1st edition
 * All of these examples show how the vertical force on each axle is a function of the acceleration and the center of mass location. A torque is applied, the vertical support forces change, the sum of all applied torques remains zero, and so the angular acceleration remains zero. This is exactly counter to your comments above from July 11:
 * You seem to be saying that angular acceleration of the body does not occur until torque on the body is enough to unload the front wheels. This is of course contrary to the purpose of the article, which shows that Newton's laws require this angular acceleration to be equal to the applied torque divided by the moment of inertia even before the front wheels are unloaded.


 * Perhaps you mean to invoke only the time during a drag race when the front wheels and the wheelie bars are not in contact with the ground, in which the rear wheels apply the only torque about the center of mass, and the simple relationship between net torque, angular acceleration, and moment of inertia holds, but even then it is not clear that top fuel dragsters are specifically designed to be "long and narrow to increase the moment of inertia to resist pitch and steering moments to keep it moving in a straight line." Causality has not been shown, the assertion is original research at best, and it remains a poor example for this article.


 * If you were to point out at least one published example to support your point or one published source that confirms your assertion, I would be happy to take a look. Short of that, however, this conversation appears to have run its course, and we both probably have better things to work on. -AndrewDressel (talk) 14:03, 1 August 2013 (UTC)


 * Again it is simple, the front wheels cannot become unloaded without the body of the dragster rotating. This rotation is generated by an angular acceleration, and this angular acceleration shows that there is an unbalanced torque on the body. Prof McCarthy (talk) 15:29, 1 August 2013 (UTC)
 * Well, you are welcome to believe that, and even welcome to keep asserting that on this talk page, but until you can explain the discrepancy between your assertion and the assertions of the published and widely-used textbooks listed above (one of which has even used recently by UCI to teach dynamics), it doesn't amount to a hill of beans. Wikipedia requires reliable sources, I have provided several for this discussion that directly contradict your point, and you have provided none. -AndrewDressel (talk) 16:13, 1 August 2013 (UTC)


 * In case you are unable to access any of the textbooks listed above, I have inserted below a free body diagram and simple 2D dynamic analysis of a representative top fuel dragster, based on the treatments in the textbooks listed above and Euler's second law. It is in two columns to show that summing the moments about a stationary point momentarily coincident with the rear tire contact patch produces the same results as summing the moments about the center of mass, and I believe it clearly shows how the vertical load on the front wheels is a simple function of mass, geometry, and horizontal acceleration. The front wheels remain loaded and the angular acceleration remains zero until the horizontal acceleration reaches the value so that ah = gb. Also, as long as ah < gb, there is no dependence on the moment of inertia. Once ah > gb, the front wheels become unloaded and the angular acceleration becomes non-zero. I will be fascinated to learn what you believe the "mistaken understanding" to be in this. -AndrewDressel (talk) 02:38, 2 August 2013 (UTC)




 * On second (or tenth) thought, I'm tempted to just say "never mind". Upon watching "Popular Mechanics: How... : How a Top Fuel Dragster Works" and looking at "How to Become an Indispensable Clutch Man" it becomes clear that when things are working correctly, even though ah is probably greater than gb almost instantaneously, the cars appear to wheelie very little, if at all, and the wheelie bars appear to play a complicated role in that. For example, the video shows that the fuel is stored in a tank forward of the driver. This probably provides the advantage of letting the center of mass drift rearward as fuel is consumed during a run so that ah stays close to gb as a inevitably decreases. Our simplistic analysis is not r for any of this interaction. These details do reinforce my opinion, however, that the design of these cars is far from simply "long and narrow to increase the moment of inertia to resist pitch and steering moments to keep it moving in a straight line." Based on the evidence available, they could just as well be long and narrow merely to fit all the parts into a package with the smallest frontal area. -AndrewDressel (talk) 17:01, 2 August 2013 (UTC)


 * Thank you for this analysis. It is similar to the one that I propose to provide.  I would modify it to include the front suspension, but it clearly states that angular acceleration depends on the torque divided by the moment of inertia.  This discussion of a wheelie has been to show the importance of moment of inertia on moderating the affect of torque on angular acceleration.  This is also true in the horizontal plane where it moderates the affect of steering torque. Prof McCarthy (talk) 13:43, 13 August 2013 (UTC)


 * If a rigid body experiences a "net torque" or "resultant torque" or "sum of (ALL) external torques" then it must change its angular momentum, in other words it has to experience angular acceleration. Dger (talk) 15:34, 22 July 2013 (UTC)


 * I believe "change its angular momentum" is only equivalent to "experience angular acceleration" if the point about which the angular momentum is calculated coincides with the center of mass. Okay, so well use the center of mass. -AndrewDressel (talk) 21:18, 22 July 2013 (UTC)


 * Gravity is a central force that can cause no angular impulse or change in angular momentum of the body. The ground reaction forces at a racing track will also be in balance when the dragster is at rest. The only way the GRFs can cause a rotation will be if an earthquake occurs or some external force gives the dragster a push or a lift. Dger (talk) 15:34, 22 July 2013 (UTC)


 * Not sure what point you are making here, but okay. -AndrewDressel (talk) 21:18, 22 July 2013 (UTC)


 * Your text above mentions gravity resisting a moment. Its moment on the whole dragster has to be zero. Dger (talk) 00:58, 23 July 2013 (UTC)


 * My exact words to Prof McCarthy are "gravity and the ground apply significant forces and moments," and I assumed that he would know which generated which. -AndrewDressel (talk) 02:03, 23 July 2013 (UTC)


 * When the engine kicks in, there will be a frictional force at the drive wheel that will cause a rotation of the dragster that may cause the front wheels to rise or unload since the centre of mass of the dragster is above ground level. Dger (talk) 15:34, 22 July 2013 (UTC)


 * Yes, of course, the driving force "may cause the front wheels to rise or unload". That point is not in dispute. The issue is what role moment of inertia plays in that possible unloading and whether the whole scenario makes for an effective or appropriate example in this article. -AndrewDressel (talk) 21:18, 22 July 2013 (UTC)


 * This is basic mechanics at work. Draw a free-body diagram and calculate the turning effect of the friction at the back wheels times the distance between the ground and the height of the centre of mass of the dragster.  Since gravity will not affect the angular momentum and the front wheels are bascally passive there can only be one result the dragster experience a rotation. Dger (talk) 15:34, 22 July 2013 (UTC)


 * But now it appears that you wish to assert that the dragster must experience rotation, which is simply incorrect unless you are considering frame flex and or suspension and tires compliance. If we are not modeling the dragster as a rigid body, as the directional stability article does, then the relative stabilizing effect of its large moment of inertia compared to all the other vehicle dynamics issues is far from clear, and it is a poor example for this article. -AndrewDressel (talk) 21:18, 22 July 2013 (UTC)


 * In any case, the free body diagram and 2D dynamic analysis is shown above, and it shows that there will be rotation only if ah > gb. -AndrewDressel (talk) 02:38, 2 August 2013 (UTC)


 * Increasing the length of the dragster increases the moment of inertia so that the amount of rotation is reduced. Of course, the dragster is not a rigid body so that what happens is that the body will deform and reduce the angular impulse of the drive wheels and can nullify the rotation. Just try turning a long bag of water. One end will turn but the other may not if the bag deforms enough in the middle. This is probably why we see no rotation of the front of the dragster. But having a long axis does help to reduce any turning effect. That's what the Euler's Law shows (simplified as M = Iα). Cheers. Dger (talk) 15:34, 22 July 2013 (UTC)


 * If the rotation is nullified, then the only term in Euler's second law that included the moment of inertia is also nullified, and the dragster is indeed a poor example to use in this article, just as a long bag of water would be a poor example of the stabilizing effect of a large moment of inertia. -AndrewDressel (talk) 21:18, 22 July 2013 (UTC)


 * Actually a more important role for the length of the dragster is so that some of the dragster's mass can be farther away from the drive wheels. The result being that again the moment of inertia of the dragster about an axis through the drive wheels will greatly increased; after all the front wheels are relatively light. Dger (talk) 00:58, 23 July 2013 (UTC)


 * I believe we already covered a similar supposition, equally unsourced, in the discussion above, with hilarious results. -AndrewDressel (talk) 02:03, 23 July 2013 (UTC)

The assertions about dragsters cut from the article are pasted below for safekeeping. -AndrewDressel (talk) 19:28, 12 July 2013 (UTC)


 * ConnieKalittaVSDanPastorini.jpg In the same way the long axis of a dragster resists turning forces which helps keep it moving in a straight line.

The Inertia Tensor
Looking at the dyadic construction of the inertia tensor, it seems to me it is type (2-0), qualified as a bivector generally at 'Tensor' article. Here is written 'binor' - I didn't find other reference in wikipedia for this term.

As for the bivectors matrix transformation it follows : I'=T*I*TT , same is here, but this article http://www.kwon3d.com/theory/moi/triten.html has derivation where I can't understand how we jump from inverse to transpose for last term. I could accept it only for ONB transformation. Otherwise it confuses me more, as I see I'=T*I*T-1 for liner transformation, or tensor of type (1-1). Would appreciate someone to help me about this catch. Twowheelsbg (talk) 09:31, 3 June 2013 (UTC)
 * The transformation T is a rotation matrix so its inverse is its transpose T-1=TT. I hope this helps. Prof McCarthy (talk) 14:24, 3 June 2013 (UTC)

Some of the maths is unclear
I'm not completely new to this topic, but it must be very hard going for those who are. This isn't a comprehensive list (it only really covers section 2), but here are some points that I might try to improve on if what I'm saying makes sense:


 * Some quantities don't seem to be introduced at their first instance or at all. They don't need to be explained, just a relevant wikilink will do. Notably, it is not apparent that $$I$$ is used for moment of inertia, the article's subject, until section 2.
 * I'm assuming that $$W$$ (from section 2.1) is weight. Is that not the same as the force, $$F$$, from earlier – or is that the component of weight in the direction of the acceleration? If so, wouldn't it be clearer if it was $$W$$ subscripted with something?
 * It isn't clear that natural frequency $$\left(\omega_n\right)$$ is an angular frequency measured in radians per second $$\left(\text{rad}\cdot \text{s}^{-1}\right)$$ rather than cycles per second $$\left(\text{Hz}\right)$$. Note that natural frequency redirects to resonance which doesn't even mention ‘angular frequency’.

Am I correct in saying that if one wanted to calculate the moment of inertia $$\left(I_P\right)$$ of a compound pendulum, one would need to measure its mass ($$m$$), and measure its natural ordinary frequency $$\left(f_n\right)$$ with respect to our gravity $$\left(g \approx 9.81\text{m}\text{s}^{-2}\right)$$, and calculate $$I_P$$ using this equation:
 * $$I_P = \frac{mg^2}{{\left(2\pi f_n\right)}^4}$$

Could this equation be included in section 2.1?

Worked example: For a clock pendulum which ticks 2 seconds in a cycle (typical Grandfather clock), $$f_n$$ should be tuned to almost exactly $$0.5\text{Hz}$$ and so $$\omega_n$$ should be about $$\pi\ \text{rad}\cdot \text{s}^{-1}$$. Given a pendulum mass of about $$0.68\text{kg}$$ $$\left(1.5\text{lb}\right)$$, $$I_P$$ works out to be about $$0.67\text{kg}\cdot\text{m}^2$$.

— James Haigh (talk) 2013-06-10 T 20:45:30Z (edited)


 * The natural frequency ωn of the compound pendulum is measured in radians/second. Cycles per second f is related to ωn by the formula
 * $$\omega_n = 2\pi f.$$
 * This means the formula for the moment of inertia of the compound pendulum can also be written in the form
 * $$ 2\pi f = \sqrt{\frac{mgr}{I_P}},$$
 * or
 * $$ I_P = \frac{mgr}{(2\pi f)^2}.$$
 * The distance r from the center of mass of the compound pendulum to its pivot point is an important parameter in this calculation.
 * Prof McCarthy (talk) 04:43, 11 June 2013 (UTC)


 * Thanks Dger, section 2 seems a lot clearer now.
 * As for my derived formula, it is not correct for a compound pendulum (which is what I meant to say; see edit), though it does hold for a simple pendulum. My error came from substituting $$r = \frac{g}{\omega_n^2}$$ which only holds for simple. Nevertheless, seeing as a Grandfather clock typically has a pendulum that's not far off being simple, the example isn't a bad one for the equation I derived. So for a simple pendulum, I think this is correct:
 * $$I_P = mr^2 = \frac{mgr}{{\left(2\pi f\right)}^2} = \frac{mg^2}{{\left(2\pi f\right)}^4}$$
 * but for a compound pendulum, only the middle one holds true:
 * $$I_P = \frac{mgr}{{\left(2\pi f\right)}^2}$$
 * For a simple pendulum, one may calculate $$I_P$$ without knowing both $$r$$ and $$f$$. (With the Grandfather clock, it is far easier to obtain an accurate value for $$f$$ than it is for $$r$$ due to accessibility of the pivot in the mechanism. Only needing one value may be useful, but I think in many applications, approximate simple pendula are relatively rare. Indeed, the application that I'm trying to apply this to is definitely compound.) For a compound pendulum, one must obtain both $$r$$ and $$f$$, unless they happen to know the exact distribution of mass of the object (shape, density of parts, etc.).
 * So in summary:
 * $$I_P = mr^2$$ ← Simple only; don't need $$f$$.


 * $$I_P = \frac{mgr}{{\left(2\pi f\right)}^2}$$ ← Holds for compound.


 * $$I_P = \frac{mg^2}{{\left(2\pi f\right)}^4}$$ ← Simple only; don't need $$r$$.
 * Can you confirm that this is correct?
 * Dger, after your commit I noticed that there is a Grandfather clock example commented out in section 2. It could be good to clean it up and use it again for the part about simple pendula. I haven't really looked at it yet though.
 * Dger, My reasoning behind using $$T$$ for period of oscillation was for consistency with other Wp articles, notably the frequency article which redirects from period (physics). Nevertheless, a bit of cool-off wouldn't be a bad idea since only yesterday the article had $$T$$ for kinetic energy, and $$\mathbf{T}$$ for torque.
 * — James Haigh (talk) 2013-06-11 T 17:36:11Z


 * A simple pendulum has a physical parameter r, its length, that can be measured precisely. It does not make sense to measure this length by first measuring the local gravitational acceleration, and then the natural frequency.  In fact, the approach is generally the opposite, since r can be measured precisely the natural frequency is measured to determine the local acceleration of gravity.  Given r, then the moment of inertia is mr2. Prof McCarthy (talk) 13:30, 12 June 2013 (UTC) (split and moved by James Haigh)


 * Okay, so application may be the wrong way round for high precision applications, but I'm guessing that seeing as you haven't corrected me about the equations and whether they hold for simple/compound, I must have got that bit right. — James Haigh (talk) 2013-06-12 T 16:16:05Z

For the 3rd paragraph of section 2, I'm not clear whether the quantities are vectors or magnitudes. The text does not say they're magnitudes, but the symbols are not bold. If they were bold and treated as vectors, I guess the maths would be something like this:
 * $$\boldsymbol\tau = \mathbf{r}\times\mathbf{F}$$
 * $$\mathbf{F} = m\mathbf{a}$$
 * $$\mathbf{a} = \boldsymbol\alpha\times\mathbf{r}$$
 * $$\therefore\boldsymbol\tau = \mathbf{r}\times\left(m\left(\boldsymbol\alpha\times\mathbf{r}\right)\right) = m\left(\mathbf{r}\times\left(\boldsymbol\alpha\times\mathbf{r}\right)\right)$$

which must somehow equal $$mr^2\boldsymbol\alpha = I\boldsymbol\alpha$$, but I don't yet understand what happens to the cross products and $$\mathbf{r}$$ (the displacement vector from the pivot to the mass). Presumably, there must be some relation:
 * $$\vec{a}\times\left(\vec{b}\times\vec{a}\right) = a^2\vec{b}$$

— James Haigh (talk) 2013-06-12 T 12:12:44Z


 * The vector cross products that you have formulated are for spatial movement of a body. You will see how these cross products are reassembled into the inertia matrix further down in the article. Prof McCarthy (talk) 13:30, 12 June 2013 (UTC) (split by James Haigh)


 * Prof McCarthy, I hope you don't mind me splitting out your reply to the relevant ‘thread branches’ so that I can more easily reply to the 2 points separately.
 * I was treating $$I$$ as a scalar quantity given a specified axis, hence it being non-bold. I do not yet understand tensors, and it may be weeks or months before I find the time to get my head round the maths lower down in the article. The application that I'm working with (an analysis of my StringBike) has fixed axes from which to easily define scalar moments of inertia. DavidCary's commit has thankfully cleared up my immediate dilemma of whether to change $$\boldsymbol\tau = rF$$ to $$\tau = rF$$ or $$\boldsymbol\tau = \mathbf{r}\times\mathbf{F}$$.
 * — James Haigh (talk) 2013-06-12 T 16:06:11Z


 * I still haven't got onto the inertia matrix/tensor, but it turns out that the left part of that relation is a ‘vector triple product’, and that it does hold when $$\vec{a}\perp\vec{b}$$:
 * $$\vec{a}\times\left(\vec{b}\times\vec{a}\right) = \vec{b}\left(\vec{a}\cdot\vec{a}\right)-\vec{a}\left(\vec{a}\cdot\vec{b}\right) = a^2\vec{b}-\vec{a}\left(\vec{a}\cdot\vec{b}\right)$$
 * $$\vec{a}\cdot\vec{b} = 0,\text{ when }\vec{a}\perp\vec{b}$$
 * $$\therefore\vec{a}\times\left(\vec{b}\times\vec{a}\right) = a^2\vec{b},\text{ when }\vec{a}\perp\vec{b}$$
 * So for scalar moment of inertia $$\left(I\right)$$ of a point mass around a specified axis (simple pendulum), the torque can be formulated entirely from vectors without going into inertia matrices yet:
 * When $$\mathbf{r}\perp\boldsymbol\alpha,\ \boldsymbol\tau = m\left(\mathbf{r}\times\left(\boldsymbol\alpha\times\mathbf{r}\right)\right) = m\boldsymbol\alpha\left(\mathbf{r}\cdot\mathbf{r}\right) = mr^2\boldsymbol\alpha = I\boldsymbol\alpha$$
 * I like this, I find this much cleaner than using $$\vec{k}$$.
 * — James Haigh (talk) 2013-06-22 T 01:29:21Z


 * I assume you are aware that, in this case,
 * $$\boldsymbol\alpha = \alpha\vec{k}$$.
 * Prof McCarthy (talk) 18:48, 22 June 2013 (UTC)

Comment on GA nomination
Prof McCarthy:

My math and science is probably just barely good enough to help with a review on this article. But having looked at the /GA1 and /GA2 review pages a bit, I thought I might make the following comments. Perhaps it might give you some ideas for rebooting this a bit, even before I would consider diving into the math.

I hope you find these suggestions helpful. StevenJ81 (talk) 17:45, 31 July 2013 (UTC)
 * 1) Simplify the sections you can simplify. In particular, I'm thinking of the article lead, the introduction, and the first paragraphs on the simple pendulum. There is language that just doesn't have to be so complicated. If you want anyone but physicists (so to speak) to be willing to read even that far, these would help. For example:
 * 2) *In the translation of the sentence of Principia, use remain instead of persevere. You are not bound by requirements of sourcing and verifiability to use the exact wording of someone else's translation from the Latin. If anything, I would argue that the most common sense of persevere today has an overtone of meaning wherein a person patiently remains in place. I do not think people use persevere much for inanimate bodies these days.
 * 3) *Next paragraph: You don't need extended body; use body. I would change mass is constrained to rotate to mass rotates.
 * 4) *In a couple of places: You talk about the mass and shape of a body "combining", resulting in the MoI. That's true, but only in the sense of integrating the point-s over the whole body. To a lay person, that "combination" isn't obvious. I might say something more like "The MoI of a body is a property that depends on the mass and shape of the body."
 * 5) *Similarly: "The MoI of a rotating flywheel is used in a machine to resist variations in applied torque ..." The MoI can't be used for that; it's a mathematical proportionality scalar (or vector or tensor). The fact that a flywheel has rotational inertia is what is used for that purpose. To me, it is confusing to say that a mathematical constant does these things.
 * I hope you get the general idea about this. If I were a formal reviewer, I would want these things changed before passing the article on prose.
 * 1) I very much agree with . There is just too much here. Focus on the core topic, and let side topics go. More to the point, though ...
 * 2) Consider the critiques of WP:NOTTEXTBOOK seriously.
 * 3) *If this is a general encyclopedia, I think an article like this deserves, at minimum, a reasonable treatment of practical examples of the subject. You've provided a number of them: the wire walker, the skater, and so forth. I'd put them up front, so that the general reader can get to them promptly and see why the subject is important, and not just so much academic windage.
 * 4) *I'm not sure, per WP:NOTTEXTBOOK, whether this is the place for a heavy mathematical treatment at all. But other articles have such, and who am I to argue that? But what I would do is put them in a second-level header entitled something like "Mathematical treatment". Then anyone who wants that can just look at it, and anyone else knows that the general descriptive matter is finished.

Comments continued post close of /GA3

 * I appreciate very much the recommendations. I agree that there is too much here.  I have already completely rewritten the article, changing and removing quite a bit of the original material.  I see no reason to avoid deleting everything else.  I am not sure how to hide it under a heading of "mathematical treatment."  Perhaps I can just delete everything after the discussion of the pendulum. Prof McCarthy (talk) 20:52, 31 July 2013 (UTC)


 * Believe me, I understand the frustration you felt when you wrote about that old N.A.C.A. Technical Note last spring. Personally, I couldn't give a flying, er, leap about all those video games and flash-in-the-pan pop singers. Some people here do agree with your thinking on that.
 * You might be able to build some of the mathematical treatment into a short paper/pamphlet/whatever you call it on Wikibooks, and you can even link to that from here. My thinking was more along the lines of leaving the rest in place (organization of material aside), but demoting all the 2nd level headers to 3rd level headers, and putting them under a 2nd level header called "Detailed descriptions", or some such language. I think putting all that in Wikibooks might be better, though, and you are certainly free to put a template like Wikibooks or Wikibookshas to point people (like your students) to it.
 * FWIW: I never learned to ride a bicycle. It was not until I really understood the physical manifestation of the MoI, or at least of the law of conservation of angular momentum, that I understood that some of my fears about riding a bicycle were completely crazy. I think topics like these aimed at the general public are really, really valuable.
 * Good luck. StevenJ81 (talk) 21:15, 31 July 2013 (UTC)


 * It has been very interesting to experience the Good Article process. I understand better how those who manage Wikipedia believe it should be used, which contrasts remarkably with how I see my students actually use it.  Trying to address the concerns of various reviewers, I modified this article far from its original form.  In my opinion, its original form was a remarkably good article, and its current form is as well.  The article provides a nice summary of moment of inertia, so I will leave it alone.  Prof McCarthy (talk) 06:58, 1 August 2013 (UTC)

The math in this article is much more complicated than it needs to be (plus where should the minus sign be for the product of inertia).
From a physicists point of view (who also teaches a few engineering courses) the math in this article is way more complicated (and obscure from a physics point of view) then it needs to be. Consider the following:
 * Using a skew operator [] that converts a vector to a skew matrix that can easily be avoided by using things like the BAC-CAB rule.
 * Using tensor product symbols when simple matrix multiplication will do.
 * Using the Identity matrix (I) which has the same symbol as the moment of inertia I and when it adds nothing to the physical meaning nor the ability to calculate I (the moment of inertia I.)
 * Bringing up too many obscure (from a physicist point of view) equations such as talking about how it relates to calculating the distance between a line and a point. (What does that tell us about I?)
 * Obscuring the tensor form that physicist and engineers actually use I = m[y^2+z^2, -xy, -xz // -xy, x^2+Z^2, -yz // -xz, -yz, x^2+y^2]
 * Product of inertia redirects here but there is almost no discussion about it and what discussion there is is obscure and hidden.

I am willing to change these things, assuming I can find the time. But I don't want to do this if there are any major objections. It would involve wholesale changing of much of the meat of the article.

p.s. As a side note does anyone have a preference for whether we should use Ixy = xy or Ixy = -xy. As a physicist I prefer the latter, like Goldstein, but engineers I think use the former.


 * - TStein (talk) 00:34, 8 November 2013 (UTC)
 * It's worth a try. Maybe you can find something useful in my aborted GA review, which I had to give up on because it was taking too much time. As for Ixy, how about sticking with Goldstein and adding a note about the engineering definition? RockMagnetist (talk) 18:14, 8 November 2013 (UTC)
 * Note, by the way, that some of the Introduction was moved from the lead, and some of it should go back (particularly the alternative names). RockMagnetist (talk) 18:16, 8 November 2013 (UTC)


 * Thanks. Sounds reasonable. I will see what I can do about getting that stuff back into the lead. TStein (talk) 20:38, 8 November 2013 (UTC)


 * I do not consider the new lead to be an improvement, and I find a number of these major changes to be ill advised. Prof McCarthy (talk) 11:10, 9 November 2013 (UTC)
 * More than just saying it is worse would be appreciated. It usually takes me several iterations for me to get something right, and I make no promises that it won't get worse in the middle. In the interest of coming quicker to the best solution, let me summarize my motivation:
 * explain things qualitatively first then get progressively more detailed.
 * keep lead short.
 * moment of inertia affects rate of change in angular velocity not 'change in angular velocity'. You can get any change of angular velocity you want for any given moment of inertia and torque if you wait long enough.
 * - TStein (talk) 17:13, 11 November 2013 (UTC)


 * I am not sure what the question is, but if you want to insert "rate" in the lead, you do not need anyone's permission. Prof McCarthy (talk) 03:47, 12 November 2013 (UTC)
 * My point is that we are editing without understanding each others point of view and that if we keep on doing that the article will only get worse. In my point of view, the article needs major revisions and that your corrections to my revisions made things worse. You obviously feel the same about my revisions. That does not mean that either one of us are wrong in principle. What is does mean is that we need to communicate better about why we are making the revisions that we are. That was what I was trying to do in my admittedly clumsy way. I was trying to clarify why I was doing the things that I was doing.


 * For example, one point of contention seems to be a working qualitative definition of moment of inertia that is not ambiguous. To me, my version was significantly better than yours, but you changed it back for some reason that is not obvious to me. I could simply add the words 'rate of' to the definition but that makes it too involved and clumsy for my taste. I want something simpler and more qualitative that is still true. We have to be careful not to loose readers in the lead.


 * A second point of contention is when to introduce qualitatively the tensor nature of the moment of inertia. In my opinion, the fact that a torque about one axis can lead to angular acceleration about another axis is an important feature of the moment of inertia that needs to be emphasized early in a qualitative fashion and later explicitly in the moment of inertia tensor. You have seemed to go out of your way to eliminate that point. I am certain that you have your reasons, and I am open to the possibility that I am wrong about the importance of this point.


 * There are quite a few other points as well that I am certain that we disagree about.


 * I apologize if I am being rude. I am frustrated that we are talking past each other and I am trying to fix that by better communicating some of the reasons why I am doing the things that I am doing.


 * - TStein (talk) 17:55, 12 November 2013 (UTC)


 * There are no points of contention. It is fine with me if you want to insert rate of change or call moment of inertia a tensor.  You have made a number of edits that are incorrect, I am simply trying correct them.  Prof McCarthy (talk) 23:57, 12 November 2013 (UTC)
 * So what do you feel is incorrect, and how should it be changed? -- Chetvorno TALK 00:38, 13 November 2013 (UTC)
 * Those are the edits that I made. Our colleague TStein likes "the fact that a torque about one axis can lead to angular acceleration about another axis" so the lead now has the sentence "The inertia matrix includes off-diagonal terms called products of inertia that couple torque around one axis to acceleration about another axis."  He also likes the idea that the inertia matrix is a tensor, so the lead now has the sentence "The inertia matrix is often described as a symmetric rank two tensor, having six independent components."  He wanted the definition of inertia for a three dimensional body to be N=Iα, but that was incorrect, so I rewrote the section so it is now correct.  Prof McCarthy (talk) 01:19, 13 November 2013 (UTC)
 * Reading my old posts sometimes makes me cringe. It is hard to write technical articles well (worse it is damn hard to write technical articles well). I don't know what I was thinking if I wrote that N=Iα. I remember knowing the Euler equations at that time. The only problem is that the lead and the article still has a lot of problems, mostly from IMHO trying to do too much at once. The main problem seems to be that we have two separate use cases that are tangled more than they need to be (the fixed axis of rotation case and the full 3d case). At the risk of fulfilling Einstein's definition of insanity, I will try to post an idea for revising the lead (and structure of this article) below soon. TStein (talk) 20:11, 1 October 2014 (UTC)

Definition
The definition has: "... angular acceleration along a principal axis of the object ..." Shouldn't that be "around a principle axis" or "about a principle axis" rather than "along"? — Preceding unsigned comment added by 139.68.134.1 (talk) 14:04, 12 March 2014 (UTC)

Article is not comprehensible to nontechnical readers
This article illustrates perfectly the problem with technical articles on Wikipedia. Virtually every elementary textbook introduces moment of inertia with conservation of momentum through a simple example such as an ice skater pulling in her arms to spin faster 1, 2, 3,  4. A simple scalar equation $$I\omega = L\,$$ which a general reader can understand. But why explain things simply when you can explain them in a complicated way? This article introduces moment of inertia through - wait for it - the angular momentum of a pendulum??? Using vector cross products??? Hope the introductory reader is familiar with the BAC-CAB rule for vector triple products. I'm sure he will be.

The advanced sections of this article offer plenty of opportunity for the editor to show off his vector and tensor skills. But Wikipedia is a general-reference encyclopedia, and the nontechnical readers, who are going to be the majority of readers coming to this article, deserved an introductory section in plain English, with a simple example such as the ice skater, or a diver curling into a ball, that could give them an idea of what moment of inertia means. After all, the introduction is the only part of this article they are going to be able to understand. But no. It seems ego triumphed over the duty to write a comprehensible article. -- Chetvorno TALK 01:18, 30 September 2014 (UTC)
 * Thanks, Professor. Looks good to me.  Sorry I was sarcastic.  I've been sensitive about this issue ever since a reader called an article I edited "...one of those Wikipedia articles only the person who wrote it can understand." -- Chetvorno TALK 19:06, 30 September 2014 (UTC)
 * Your recommendations were very good. Thank you. Prof McCarthy (talk) 06:32, 1 October 2014 (UTC)

Another attempt at the lead.
Moment of inertia is a property of a rigid body that describes how resistant that object is to changes in its rotation. The moment of inertia of an object depends on the amount and distribution of its mass relative to its axis of rotation. Moment of inertia may also be called mass moment of inertia, rotational inertia, polar moment of inertia, or angular mass.

The moment of inertia of a single particle around a given axis is defined as I = mb2, where m is the mass of particle and b is the shortest distance between the particle and the axis of rotation. The moment of inertia of more complicated objects is determined by summing up the moments of inertia of the particles that make up the object. For rotation around a fixed axis, many important relations depend on I such as L = Iω and N = Iα, where L is the angular momentum, ω is the object's angular velocity, N is the torque applied to the object and α is the angular acceleration of the object.

In the more general case, called rigid body dynamics, one number is not sufficient to describe the moment of inertia, though. The proper description of the moment of inertia in this case involves 6 quantities that form a 3 by 3 symmetric tensor. In this case, L = Iω remains valid using matrix multiplication of the moment of inertia tensor with the angular momentum vector. But, the torque equation can become significantly more complicated leading to phenomena such as precession and shaking due to tire imbalance or unbalanced loads in clothes washers during the high speed spinning stage.

TStein (talk) 21:25, 1 October 2014 (UTC)


 * I edited the above until I am starting to get confident in it. My goal was to keep it as simple as possible but not one bit simpler. TStein (talk) 20:04, 8 October 2014 (UTC)


 * I like it better than the existing introduction (which is pretty opaque) but I think it could use some work. It's a hard concept to explain, isn't it?  Even professional physics texts have trouble.   Quite a few introductory texts use the analogy with mass to explain moment of inertia, do you think that would help?  Something like:
 * The moment of inertia of a body is the property which determines its resistance to being rotated, just as a body's mass determines its resistance to linear acceleration.
 * One thing that bothers me is your sentence stating all the equations without explanation: "many important relations depend on I such as L = Iω and N = Iα..." I think it might be too much simplification to dispense with all mathematics in the intro except I = mb2.  It seems to me the rotational Newton's law τ = Iα as one of the defining equations of M of I, should probably be explained in the intro, linking moment of intertia to the important but elementary concepts of torque and accaleration:
 * The amount of torque τ (rotational force) required to give a rigid body an angular acceleration of α about an axis is proportional to its moment of inertia I about the axis: τ = Iα. 
 * Anyway, considering the importance of the equations L = Iω and N = Iα, from the point of view of WP:SUMMARY my feeling is something more should probably be said about them.-- Chetvorno TALK 00:59, 9 October 2014 (UTC)


 * Hopefully, this was of some use. I like the lead of the article as it now stands. As you said it is very hard to define something in a way that is both technically correct and understandable to people who are not experts. Kudos to all who improved the lead. My only difficulty with the lead is the sentence talking about m_a + m_b = m_c then I_a + I_b = I_c. I am not quite certain what it means. TStein (talk) 17:45, 13 November 2014 (UTC)


 * The RBD part is deeply confusing and completely misleading. For one thing, you lost me at "symmetric tensor", and I suggest that you think about it, despite your own confidence.


 * More importantly, you're creating a confusion between two distinct notions of moment of inertia -- the "elementary" definition, a scalar quantity dependent on the choice of axis, and the RBD representation, which captures moments of inertia about all possible axes in a tensor. To make it worse, you're calling RBD a "more general case", which it certainly isn't -- it's a framework to help keep tabs on number crunching, and just that.  Particularly, it is not a useful conceptual device to promote understanding and inquiry, and most definitely not something one can make physical sense of without connection to the elementary definitions.


 * Also, you're detracting an uninitiated reader from the single most important point concerning angular motion -- that all angular motion properties depend on the choice of axis, and that the axis is always a matter of choice. It is a very counter-intuitive notion that one just can't stress well enough, and without which angular motion just doesn't make any sense at all.  This very article often loses sight of it, falling into talking about moment of inertia about points, for example.

Conservation of angular momentum and figure skaters
Per BRD I'm bringing this here since we're already on the second round of reverts. An editor objects to the use of the term "conservation of angular momentum" in the description of why a figure skater spins faster after reducing moment of inertia on the grounds that there is friction with the ice.

Yes there is friction with the ice (and there would still be air friction if the skater were suspended in the air). It's an external force that does work on the system and so changes the angular momentum. We are not discussing a perfect, frictionless system with no work where the angular momentum stays constant forever (i.e., we are not saying that the skater's angular momentum will never decrease). We're discussing why a figure skater can change rotational speed by changing his or her moment of inertia. It's still the conservation of the angular momentum that causes the change in rotational speed as the moment of inertia is changed.

Does anyone else want to take a crack at this? It's so blindingly obvious to me that I'm having a hard time. Meters (talk) 23:47, 1 October 2014 (UTC)


 * The figure skater in the article is NOT experiencing conservation of angular momentum while her skates are in contact with the ice. Pulling her arms inwards does increase her spin rate while her total angular momentum decreases because of friction with the ice. There is no need to apply an inappropriate principle to this situation. If she was spinning in the air, the conservation principle would be appropriate (ignoring air resistance). It is far more important to give a correct analysis of the role of changing moment of inertia to increase rotation than it is to apply an incorrect principle. The diving example in the text is a better conservation example. This article is about moment of inertia so it is better to use a more general example than use one that only applies in very special situations (such as airborne motions). -:Dger (talk) 23:59, 1 October 2014 (UTC)


 * Convenient how you are willing to ignore the air resistance to claim that your scenario would demonstrate conservation of angular momentum but are not willing to ignore ice friction for the scenario currently in the article. Meters (talk) 00:52, 2 October 2014 (UTC)


 * I agree. Dger, this is rediculous. You say a skating spin is NOT an example of conservation but an object spinning in the air IS?  That's an arbitrary distinction.   If your argument is that the small friction of the skates against the ice disqualifies the skater as an example of conservation of angular momentum, then why is air friction okay?  Every spinning object has some friction, every spinning object with no drive force will eventually stop, so by your argument there could never be a demonstration of conservation of A. M.  Conservation of angular momentum is what a skating spin is all about.  The skater stands with her feet together so the frictional torque will be minimum, and does everything she can to minimize friction.  That is what's so memorable about it, how long she can spin with just the angular momentum she starts with.  That is why virtually every physics textbook uses the spinning skater as an example of what you say it is not an example of. -- Chetvorno TALK 02:05, 2 October 2014 (UTC)


 * This article is about moment of inertia, not about conservation of angular momentum. The example of the figure skater is valuable as an example of how reducing moment of inertia increases angular speed. Simply drop the conservation principle. It doesn't apply when the skates are on the ice. Air resistance is a function of speed. Since she is spinning on the spot the air resistance is smaller than the measurement error. Actually, I have no idea how it could even be measured in this situation. Ice friction could be measured. It clearly exists in this situation why bother assuming it doesn't exist when it does? The edit I made is correct and should be reinstated. It is better for people to understand that conservation of angular momentum has little to do with the relationship between angular velocity and angular speed. Their product is angular momentum but nothing says their product has to stay constant. Realistically there are few situations that aren't subject to friction and therefore not true conservation situations. Why is it necessary to insist that this image represents conservation of angular momentum when the scratches on the ice show that it cannot be? Conservation of energy, momentum, or angular momentum aren't causes of motion. They are the consequences of forces behaving in certain ways. If ice friction occurs there can be no conservation. But angular speed can be changed by changing the body's moment of inertia. You don't even need to add angular impulse to the system. That is the important principle here. Dger (talk) 03:06, 2 October 2014 (UTC)
 * Again, we're not claiming that there is a total conservation of angular momentum. The only reason the change in moment of inertia affects the angular velocity is because of the conservation of angular momentum. Your edit, simply saying "A figure skater can reduce her moment of inertia by pulling in her arms, allowing her to spin faster." is not as clear as the original "A figure skater can reduce her moment of inertia by pulling in her arms, allowing her to spin faster due to conservation of angular momentum." It's a concrete example that will be understood by many people, and I can't think of a better example. If you can, let's hear it. Meters (talk) 04:07, 2 October 2014 (UTC)
 * Exactly. Dger I don't understand how you claim that conservation of angular momentum is not involved.  The skater demonstration shows the result of a change in moment of inertia $$I\,$$ on the skater's angular velocity $$\omega\,$$.  It shows a tradeoff, with $$\omega\,$$ increasing as $$I\,$$ decreases according to the defining equation $$I\omega = L\,$$ where $$L\,$$ is the angular momentum.  If $$L\,$$ is not sufficiently constant the demonstration won't work; if her angular momentum is dropping fast enough she can pull in her arms and she will still slow down.  But why should her angular momentum be constant to any degree at all?  Why should its value at any time have any relation to previous values?  We need to explain that to readers.  The reason is conservation of angular momentum.  Just because the angular momentum isn't perfectly constant doesn't mean the conservation law isn't necessary. -- Chetvorno TALK 04:35, 2 October 2014 (UTC)

Conservation of angular momentum is not a law like conservation of energy. The angular momentum of a system depends on the application of external forces. If the resultant of the external forces acting on the body passes through the centre of gravity and there is no external moment of force (such as occurs to an airborne body) then conservation occurs. It is not a law that must occur. The above equation, for example, does not apply because a term is missing that represents the external friction of the skates and the ice surface. You would need to say "assuming minimal no ice friction" in the caption. A better example would be a diver piking or tucking after takeoff so that angular velocity was increased to perform a certain number of spins.

For this skater to speed up it is necessary that her skates be aligned appropriately. If not she could be braking and slowing down. We cannot tell if this is happening from a still picture. We could say that angular momentum was transferred from her arms to her body by pulling them close to her midline. That would be a cause of the increased speed. What I am uncomfortable with is saying that conservation of angular momentum is the cause of the increased speed when in fact it is the reduction of her total body moment of inertia (and no large changes in the external moments of force) that is the direct cause. The angular momentum of a complex body is the sum of the angular momenta of all its parts. Distal parts tend to have more angular momentum than proximal parts because their moments of inertia are relatively larger due to the squared relationship of their distances from the axis of rotation, i.e., I = mr2. Dger (talk) 13:54, 2 October 2014 (UTC)

PS. I am thinking of the general reader by trying to clarify this principle. It may appear to a layperson that to spin faster all that is necessary is to pull the arms in and let conservation of angular momentum do the rest. In fact, one would also have to apply their skates to the ice in a way that does not negate the reduction in the total body moment of inertia. Perhaps the caption should also include this requirement. With a diver or airborne skater example this wouldn't be necessary because conservation is almost 100%. Dger (talk) 14:02, 2 October 2014 (UTC)


 * As I said, "The only reason the change in moment of inertia affects the angular velocity is because of the conservation of angular momentum." Your premise that an air-borne, spinning skater could be used as an example of conservation of momentum because but not this case because you "have no idea how it [air friction] could even be measured in this situation." is flawed. It so happens that I do know how to measure the air friction. Does that mean that the air-borne case Is not a valid example for me ["for me" mistakenly left off, and added after subsequent replies] but it is for you? No, because we are not discussing conservation of momentum in a system that is free of external forces in either case. You have not convinced me that your change is an improvement. I don't think you have convinced User:Chetvorno. Let's see if anyone else wants to discuss this. Meters (talk) 16:51, 2 October 2014 (UTC)


 * I have to agree with Meters and Chetvorno. Dger, I believe your mistakes include:
 * "Conservation of angular momentum is not a law like conservation of energy." These two laws are quite analogous, in that they both apply to closed systems without external input to the conserved quantity.
 * "It doesn't apply when the skates are on the ice. Air resistance is a function of speed. Since she is spinning on the spot the air resistance is smaller than the measurement error." I don't know why you suppose in this case that one source of friction is significant and the other is negligible.
 * In any case, the coin of this realm is a reliable source, and here are three engineering mechanics textbooks that use a figure skater (or the equivalent) as an example of the conservation of angular momentum:
 * 1. An Introduction to Dynamics by McGill and King (1995) on pages 350-351:
 * "A skater spinning about a point on the ice draws in her arms and her angular speed increases. Is angular momentum conserved? ...If we neglect the small friction couple at the skates and the small drag moments caused by air resistance, then the answer is yes because $$\sum M$$ is then zero.
 * 2. Engineering Mechanics: Dynamics by Gray, Costanzo, and Plesha (2010) on page 388:
 * "The skater's ability to control the spin rate relies on the fact that the quantity (the component of the angular momentum along the spin axis) remains constant." (emphasis added by original authors)
 * 3. Engineering Mechanics: Dynamics by Bedford and Fowler (2002) on page 416:
 * "In a well-known demonstration of conservation of angular momentum, a person stands on a rotating platform holding a mass 'm' in each hand. ... If we neglect friction in the rotating platform, the total angular momentum of the person, platform, and masses about the vertical axis of rotation is conserved."
 * Any or all of these can be cited to support the claim in contention. -AndrewDressel (talk) 17:14, 2 October 2014 (UTC)


 * Notice that in each of the above examples there is an assumption that conservation occurs by assuming friction was negligible or that angular momentum was constant. As a compromise I would be happy if we add "assuming conservation of angular momentum" to the caption. Then there is no problem. Regarding a closed system being needed for conservation of angular momentum, that is true but totally unrealistic for humans in contact with the earth.  Dger (talk) 17:40, 2 October 2014 (UTC)


 * A caveat would be fine, I suppose. I wonder what system you are imagining in contact with the earth in which energy is perfectly conserved. -AndrewDressel (talk) 18:02, 2 October 2014 (UTC)


 * I don't agree with the caveat. We don't need to assume absolute conservation of angular momentum for the decrease in moment of inertia to result in an increase in rotational speed. This is the equivalent of saying we cannot use conservation of momentum to discuss why a moving car keeps going when the engine is turned off unless we first assume no air resistance, rolling resistance, bearing losses, etc. Meters (talk) 18:22, 2 October 2014 (UTC)


 * This is what I am imagining (it is in one of the laboratories at my university): Reardon, Francis D.; Leppik, Kalle E.; Wegmann, René; Webb, Paul; Ducharme, Michel B.; & Kenny, Glen P. (2006). The Snellen human calorimeter revisited, re-engineered and upgraded: design and performance characteristics. Med Bio Eng Comput, 44:721–728. A picture of it is in the article calorimetry. I have used "frictionless turntables" to illustrate conservation of angular momentum. The experiment has to end early before it becomes obvious that the turntable is slowly down.


 * Regarding the car, you are loading the dice in your favour. The linear momentum of the car is magnitudes greater than the friction so conservation could be applied. The skater's angular momentum it not very large compared to the ice friction. Try sliding a paint can along a rough surface. Would you say conservation occurs then? Rolling it would be closer but still not a good fit, IMHO. Sliding it along a perfectly polished surface covered with spherical ball bearings would also work.  Illustrating conservation principles are often best left to mind games and by making assumptions. That's is how we can best solve this dilemma. Dger (talk) 18:47, 2 October 2014 (UTC)


 * I think you are missing the point of why the figure skater is a favorite example of the conservation of angular momentum. First, it is accessible because a large fraction of people, at least in developed nations, have seen figure skaters spin and change their angular rate, seemingly at will. The performance begs the question of how is it accomplished. Second, the rate of change in angular speed can be clearly seen to correlate to arm and leg position. The eventual slowing down is a much smaller, and expected phenomenon. The only explanation for the increase in angular rate is conservation of angular momentum, and the fact that the angular rate does not increase quite as much as it would if there were no friction is irrelevant to the example. -AndrewDressel (talk) 19:10, 2 October 2014 (UTC)
 * I agree, that puts it in a nutshell. Although I understand your point of view, I am afraid none of your arguments are supported by the sources, Dger.  Unless you can find enough WP:reliable sources that say that a spinning skater is not an example of conservation of angular momentum to refute the many sources that say it is, I think we are done here. -- Chetvorno TALK 19:30, 2 October 2014 (UTC)
 * Yes, that was well put. Unless someone else shows up I don't think there's much point in continuing this discussion. Meters (talk) 20:04, 2 October 2014 (UTC)

I would like to take back my suggestion that "assuming" conservation be added to the caption. In fact, conservation is not necessary for a skater to speed up during a spin. It is always problematic to analyze motion based on a still picture. You need to know what precedes and follows the instant in time. If the skater was airborne there would be no problem since conservation of angular momentum could be assumed. But with both skates on the ice there will be friction. Most spins are done with only the point of one skate on the ice consequently there is little friction. Regardless, the more appropriate equation is $$M (t_2-t_1) = I_2\omega_2 - I_1\omega_1$$, which is a simplified version of Euler's Second Law, where $$M$$ is the vertical ground (ice in this case) reaction moment of force and $$t$$ is time. Since there is friction, both sides of the equation will be negative indicating a reduction of the total body's angular momentum. All that is necessary to make the angular velocity increase ($$\omega_2$$) is to make $$I_2$$ smaller than $$I_1$$. Consequently, there is no necessity to assume conservation of angular momentum. The skater can be losing angular momentum while simultaneously increasing her spin rate. Since skaters start a spin with the arms out and one leg away from the body, the total body moment of inertia starts large and therefore can be significantly reduced. This is a more accurate explanation about what is actually happening in the image.

Note also that many of the references that have been used in the caption and above explicit state that conservation was assumed or friction was negilible or the example shows the skater leaping into the air. Which was precisely my point.

I commend the other contributors for exploring this issue in a responsible and respectful manner. My suggestion still stands that "due to conservation of angular momentum" is not necessary for a figure skater (on the ice) from increasing rate of spin.

Dger (talk) 01:16, 6 October 2014 (UTC)

Integral variables
In the section named "Calculating moment of inertia about an axis", shouldn't the integral and text say
 * $$ I_P = \int_V \rho(\mathbf{r})\,\mathbf{R}^2 \, dV.$$

R is the distance to a point r in the body from the specified axis (nearest point of the axis), and $ρ$(r) is the mass density at each point r
 * instead of what it says? There it seems as if the r in $$ \rho(\mathbf{r})$$ and the r in $$ \mathbf{r}^2$$.--84.125.11.33 (talk) 12:53, 28 December 2014 (UTC)

Physics
What is mean by centre of mass? Muruveena (talk) 17:34, 15 July 2017 (UTC)


 * @Muruveena: I've linked the first mention of the term to the article center of mass to explain it. —C.Fred (talk) 17:38, 15 July 2017 (UTC)

Proposed image for this article
I propose that this image be posted after Moment_of_inertia. Note in the caption to Commons. --Guy vandegrift (talk) 15:30, 29 December 2017 (UTC)


 * Why we need this figure:  OpenStax has finally come out with four volumes of quality textbooks for Introductory Physics.  But the real cost of higher education is not the cost of textbooks.  Nor is it the cost of educating people, since the web (e.g. Wikipeda and the Kahn Academy) do such a good job of presenting material.  The real cost is the cost of certifying compentency.  We need an open source quizbank and a quality formula sheet for students to access as they take these exam questions.  I might add that we also need a private wiki where instructors can secretly write new questions that are not yet available to the students.--Guy vandegrift (talk) 16:02, 29 December 2017 (UTC)
 * I don't object to the image, but the above comments have nothing to do with this article. Please read WP:NOTAFORUM. Meters (talk) 20:17, 29 December 2017 (UTC)
 * Your point is well taken -- I suppose I was a bit on a soapbox. But to fit into a WP article the image needs to be both noteworthy, and also fit in with the rest of the article.  I can make a good case for the significance of the image, but to be honest, have serious doubts as to whether it fits in, since it repeats information given above.  Those doubts have grown in the past few hours and I think I will place it instead at Moment of inertia--Guy vandegrift (talk) 03:37, 30 December 2017 (UTC)
 * I disagree, anyway. If the foundation hadn't wasted so many resources on formats (like quizzes, textbooks and dictionaries) for which open editing is poorly suited, it wouldn't need to beg for money so often. User:Anarchic Fox
 * I support the addition of the image. Thanks, Guy. --ChetvornoTALK 23:00, 2 June 2018 (UTC)

Tensor or scalar?
The introduction isn't quite right: it is the moment of inertia tensor that is a tensor, while the moment of inertia itself is a scalar defined with respect to a specific axis of rotation. This is explained at http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html. Can someone fix this? --Brian Josephson (talk) 10:50, 18 August 2018 (UTC)

distance vs perpendicular distance in definition
Editor is attempting to remove "perpendicular" from the long-standing definition in the lead: "for a point mass the moment of inertia is just the mass times the square of the perpendicular distance to the rotation axis". Editor claims it is redundant.

"Perpendicular" is only redundant if one assumes that distance must be the shortest possible (i.e. perpendicular) distance. That may not be clear to general readers, and I think we should stick with the long-standing lead wording. Note that we also use "perpendicular" in the Definition section of the article. Meters (talk) 05:35, 19 February 2019 (UTC)

Distinction between calculations in 2 vs 3 dimensions
This article seems to me to be a really good one for Wikipedia, that is, currently comparable to a second draft of an undergraduate textbook chapter, with the great virtue of being written to be accessible to someone studying at that level.

I only want to point out that the distinction between 2-dimensional and 3-dimensional calculations is not always well maintained. Case in point: leading up to the section labeled "Definition," the text is developing a 3-dimensional framework involving matrices. But the Definition section immediately drops back to 2 dimensions without announcing that change.

Later in the article there is again considerable attention given to 3 dimensions.

I suggest that "Definition" should begin by clarifying that in that section the aim is to develop 2-dimensional equations and descriptions, and that thereafter the distinction between 2-d and 3-d analyses be maintained in the remainder of the article.

I refrain from making the change that I am suggesting, because lately my changes to technical articles are always reverted by some editor, perhaps (or perhaps not) in every case for good reason, but in any case wasting both my time and that of the editor. So where (even slightly-substantive) technical changes are concerned, I currently confine myself to suggestions rather than edits.

Dratman (talk) 16:50, 10 January 2022 (UTC)

Parallell axis theorem
For scalars Ir >= Ic making the minus sign strange. It seems as if the text uses some vector formulation which may be very elegant and useful and the minus sign may be correct in this formulation. However the vector formulation may be unnecessary and impractical for many wiki readers so the scalar formulation ought to be given too and the existing note on the minus sign extended to comment on the diefference between the scalar and vector formulations.150.227.15.253 (talk) 14:34, 15 June 2022 (UTC)

Compare the Moment of inertia of inertia of a various giant wheel types
Compare the moment of inertia of a various giant wheel types 182.73.209.66 (talk) 11:46, 27 January 2023 (UTC)
 * Why would we analyze include that specific case? Meters (talk) 12:01, 27 January 2023 (UTC)