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Mass moment of inertia calculations 1

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dharhay

Mechanical
Aug 21, 2007
29
I have a fundamental question about calculating the mass moment of inertia. Using the US units and then the Metric units. But then can one use the common units conversion programs, like UCONEER for an appropriate conversion back and forth?

Take an example of a solid shaft, 840kg weight, 280mm dia.
Using the formulas that are common, (hyperphysics.phy-astr.gsu.edu), I = 0.5 * m * r^2
I obtain 0.839 kgm^2 for the metric and use a gc of 9.81.
Next the US, the result is 6.05 lbft^2 using a gc of 32.2.

The convertors have a mass moment section, but the conversions are not consistent with the above. Should one expect to be able to convert? Or are my units at any step not consistent? I said this is fundamental, but I am not comfortable with my two methods. Thanks.

Dave Harhay
Bronx, Taylor-Wilson
North Canton, OH
 
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Go to
and type into the box:
convert 1 kg*m^2 to lbm*ft^2

The result will be:
1 kg*m^2 = 23.73036 lbm*ft^2
which is consistent with results reported by IRstuff and Katmar

There is nothing inherently wrong with using lbm as a unit of mass (although I would choose SI given a choice, it's not the only way). Errors can happen in any system of units if you don't apply it properly. And when solving the originally posted problem of this thread, there is no difference if you treat kg as weight or mass in the original post - it describes the same shaft as first mentioned by IRstuff.

I have always been taught to use lbm and lbf rather than "lb" to distinguish the two units from each other. (It is usually clear from context, but it is also logical that we should distinguish the two units which describe fundamentally different attributes) Since kgf seems widespread in certain circles, perhaps we should use kgm when referring to mass?!? (Just kidding... there is no doubt what a kg is in SI)

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(2B)+(2B)' ?
 
And by the way, go to the grocery store deli section and place your order in units of slugs... see what kind of response you get ;-)

lbm will be around for a long time. Deal with it.

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(2B)+(2B)' ?
 
Actually I was not intending to provoke a unit debate or philisophical discussion of the difference between mass and weight. Just asking for patience and tolerance of other approaches.

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(2B)+(2B)' ?
 
electricpete said:
And by the way, go to the grocery store deli section and place your order in units of slugs... see what kind of response you get ...

I never ask my butcher to work out the mass moment of inertia of stuff. Butchers are armed. Probably, it is best to not make fun of them.

When I took dynamics in college, we learned two set of equations, one for English units, and one for SI. Mass was measured in pounds and kilograms. When I got to do real work, I learned that there is more than one metric system. We have kilograms force and centimeters.

I want to use one set of equations. I treat pounds as force (weight) and kilograms as mass, and everything works out. You don't even have to use slugs. You can just bury w/g in your equations.

Critter.gif
JHG
 
I never ask my butcher to work out the mass moment of inertia of stuff. Butchers are armed. Probably, it is best to not make fun of them.
LOL. Good one.

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(2B)+(2B)' ?
 
D,

That's the beauty of using Mathcad or SMath. There is only ONE equation. You can put in any unit that's defined, kg, lb, m, ft, or even furlong. So, something like 0.5 * 854 kg * (1 furlong)^2 = 1.728x10^7 kg-m^2. It's absolutely sweet for problems of this nature; no conversions, hidden or otherwise.

TTFN

FAQ731-376
 
So, something like 0.5 * 854 kg * (1 furlong)^2 = 1.728x10^7 kg-m^2. It's absolutely sweet for problems of this nature; no conversions, hidden or otherwise.

I don't understand your point there. That seems to have a hidden conversion from furlong^2 to m^2 built in. Shouldn't the result be 1.728x10^5 anyway?

My preference is to work entirely in consistent units, but if you need to convert units, do a sanity check on the sanity check.

And if checking someone else's work and you see a constant quoted to more than 3 or 4 significant figures check that carefully, because it has probably been converted from a different set of units, and it is probably out by a factor of about 10 (or even on one occaision 64).

Doug Jenkins
Interactive Design Services
 
Shouldn't the result be 1.728x10^5 anyway?

Oops, didn't take my own advice.

I got my archaic units mixed up; I was thinking chains, not furlongs. 1.728x10^7 is correct.


But isn't the Smoot the preferred unit for these situations?

Doug Jenkins
Interactive Design Services
 
5.967*10^6 Smoot^2-kg more precisely ;-) (It's also a built-in unit in Mathcad)

Well, obviously, everyone prefers to use consistent units, but we're not always in control of our customers, are we? I often get specifications that use knots for airspeed, feet for altitude, and kilometers for range.

My point, of course, is that you can try to do the conversion each time yourself, and hope that you've done it correctly, each time, or, you can have tool that does the conversion the instant you enter the unit into the equation. And while Mathcad is hardly bulletproof, it's still way more consistent than the typical human, and will reliably convert furlongs into Mathcad's internal unit system, which is SI, and store the unit quantity normalized to 201.168 m, which is 1/8 mi, which is 2.126x10^-14 lightyears. Any error in the numbers will not be from the conversion, per se, but from my transcription to this posting.

If Excel, or even the Windows calculator, had the same units capability, dimensional analysis would be a snap, and transcription errors would almost be nonexistent.

TTFN

FAQ731-376
 
In an ideal world, we all use the same standardised set of units (SI = The International System of Units?)


and: my teacher used to hammer on the fact that "Weight is a force" and thus is expressed in Newton.

And I love that Uconeer program, which has proven to be very helpful more than often.
 
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