Talk:List of moments of inertia

From Wikipedia, the free encyclopedia

WikiProject Physics This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
??? This article has not yet received a rating on the assessment scale. [FAQ]
??? This article has not yet received an importance rating within physics.

Help with this template Please rate this article, and then leave comments to explain the ratings and/or to identify its strengths and weaknesses.

Contents

[edit] Polygon

"Thin, solid, polygon shaped plate with vertices \vec{P}_{1}, \vec{P}_{2}, \vec{P}_{3}, ..., \vec{P}_{N} and mass m."

Does this work for ALL polygons? Convex/Non-convex etc. I think this should be made clear. --Smremde 10:16, 21 June 2007 (UTC)

If you consider a thin, flat sheet of material then it should work, since finding the moment of inertia is easily done if you know the surface area of the material. The formula given should have considered the formula for the surface area of an n-sided polygon. Anyway, if n → infinity the moment of inertia should tend towards that of the this flat circular disc. AquaDTRS (talk) 16:39, 22 March 2008 (UTC)


[edit]  ?

I know it's generally obvious from the picture, but isn't it technically necessary to specify the axis of rotation as well as the body under consideration? Hammerite 00:51, 23 October 2005 (UTC)

[edit] duplicate item in area moments of inertia

There are two rows called "an axis collinear with the base", and somewhat worryingly they have different formulae. Could somebody check and delete/clarify one or the other? Bathterror 09:56, 8 March 2006 (UTC)

The two rows referred to a rectangular area and a triangular area. Should be a little clearer now. Hemmingsen 14:16, 22 April 2006 (UTC)

[edit] Thick cylinder with open ends mistake in formula?

The formula I of z axis has the multiplier 1/2 but shouldnt it be 1/8 ?

No, the first formula is OK. \scriptstyle{{1\over 2}m(R_{ext}^2+R_{int}^2)} LPFR 12:39, 4 September 2006 (UTC)

This page does NOT display correctly on several on several of my machines (Mac OSX, Linux, and Windows). WHY are these colons in front of the math directive needed and why is it responsible to put in formating that causes parse errors to the casual observer of these pages?

It displays fine on mine (Windows XP. IE and Netscape). LPFR 12:39, 4 September 2006 (UTC)
If the colons your are talking about are " \,</math> ", this is a TeX/LaTeX directive that adds a small space in mathmode. It cannot create a parse error in any decent browser. I think that you have a problem with your machines or your browsers. I think that this page is OK and well written. But if you think that there are things that can be ameliorated, just do it. LPFR 12:58, 4 September 2006 (UTC)

[edit] Splitting the article

Would anyone have a problem if I split this into two seperate articles? I can't see any point in having these two distinct properties in the one article. I think it adds to the common confusion between these two quantities, and the article is a bit long also. Brendanfox 02:02, 25 April 2006 (UTC)

That sounds like a good idea to me. Hemmingsen 18:23, 25 April 2006 (UTC)
Well, this should be long enough time for people to voice their objections. Going ahead with the split... (to List of area moments of inertia) Hemmingsen 15:03, 30 December 2006 (UTC)

[edit] Formula for a thick cylinder

The correct formula is \scriptstyle{{1\over 2}m(R_{ext}^2+R_{int}^2)}.

Someone using IP 216.148.248.31 (CERFN California Education and Research Federation Network) wrongly "corrected" the equation changing the + sign for a minus sign. I suggest to this user to verify this formula in a physics book. If you did find this formula with the minus sign, verify in others books and correct the wrong formula.

As you do not seem to be able to calculate the inertia moment by yourself, I can demonstrate why the sign must be a + sign. If \scriptstyle{R_{int}} increases, but maintaining the total mass m unchanged, this means that you are "sending" some of the mass from a small radius to a larger one. As the moment of inertia is proportional to the squared distance to axis of rotation, the moment of inertia should increase. That is, if \scriptstyle{R_{int}} increases the moment of inertia increases. Then, the sign must be a plus sign. LPFR 07:59, 7 September 2006 (UTC)

The mistake is understandable - usually one would expect the mass to decrease if one made the hole larger, and if the hole was the same diameter as the outer then the mass and MoI would be zero! If the mass stays constant then the density must increase, but the positive sign is then correct. If the density stays constant (which is the practical case of most interest) then the formula becomes (I think, if my late-night integration is correct) (pi/2)*density*length*(R2^4 - R1^4). If you substitute mass/volume for density in this you get back to the correct formula given above. 217.147.104.220 10:51, 20 December 2006 (UTC)

The mass of the cylinder could be kept constant without increasing the density of the material, by increasing the length of the cylinder.Gregorydavid 11:23, 20 December 2006 (UTC)

The formula Iz = (pi/2)*density*length*(R2^4 - R1^4) couldn't possibly be correct, since R1 = R2 gives Iz = 0, if I'm not somehow mistaken. /Andreas —Preceding unsigned comment added by 129.16.53.239 (talk • contribs) 13:45, 5 April 2008 (UTC)

That formula differs from the other formula also discussed above by assuming a fixed density instead of a fixed total mass. With a fixed, finite density and R1 = R2 the mass will be zero because the volume will be zero, and a body with no mass will have no moment of inertia. The formula looks right to me. Hemmingsen 16:17, 5 April 2008 (UTC)
Ok, I see. However, setting that expression equal to \frac{1}{2} m\left({r_1}^2 + {r_2}^2\right) seems incorrect and misleading. I think some clarification is in order as to what assumptions correspond to each expression for I_z. /Andreas —Preceding unsigned comment added by 193.11.234.85 (talk) 19:00, 5 April 2008 (UTC)
That does seem like a good point. The two sets of assumptions are equivalent in all cases except for the one where R1 = R2 and calculating the density as mass/volume would be a division by zero, but that special case does seem to be a case people are interested in. I will attempt to rewrite that part of the list slightly; let me know if you think it is insufficient or have a better idea. Hemmingsen 19:40, 5 April 2008 (UTC)
That seems to me like a good solution! /Andreas

I agree. The equation shows itself incorrect when you realize that as the interior radius approaches the exterior radius, this formula would give you a moment of inertia of 0 while the correct value, m*r^2 comes about when you use the formula with the plus sign.
Dexter411 18:03, 20 December 2006 (UTC)

[edit] Formula for Sheet rotated about one end or through center

I came here looking for this formula, it is in any intro physics text, and is rather important. Someone should add this. I don't have time now. - Anonymous