Units for moment of inertia of a circle

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Discussion Overview

The discussion revolves around the appropriate units for the moment of inertia of a circle, specifically addressing the use of mm4 versus R4 in calculations. Participants explore the implications of using different notations and the significance of the variable R in the context of their calculations.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that moment of inertia should be expressed in mm4, while others suggest that R4 is a more appropriate representation for a circle.
  • There is a discussion about whether R is a constant or a unit, with some arguing that it is a scale factor and should remain in the calculations.
  • One participant mentions that their answer of 1.35 mm4 is correct, but questions whether they should include R in their final answer.
  • Another participant emphasizes that R is not a unit but a variable that should be plugged into the formulas, suggesting that the answer should reflect this.
  • There are conflicting views on whether to include units in the final answer, with some advocating for the removal of mm4 in favor of a more general expression involving R.
  • Participants express uncertainty about how to handle the units when R is defined as 1, leading to further discussion on the implications of this choice.
  • One participant expresses frustration over the lack of clarity regarding the units and the need for consistency in their calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the appropriate units for moment of inertia, with multiple competing views remaining on whether to use mm4, R4, or a combination of both in their calculations.

Contextual Notes

There are limitations regarding the definitions and assumptions about R, as well as the context in which the units are applied. The discussion reflects varying interpretations of how to represent the moment of inertia based on the definitions provided in the problem.

Femme_physics
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We've always been using mm^4, but I see in my answer book (answer written below) that one solution says R^4. Is it the same thing, just that R^4 is used for a circle? The measurement in my exercise are in mm.

I did get the answer, I just wasn't aware I was supposed to write it in terms of R^4.
 

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I think R is just a constant used to give the radiuses, and is not a unit. Moment of inertia/second moment of area should have dimension [length]4.
 
So was I correct to put in my answer 1.35 mm^4? (that's what come out if you do the calculations for polar moment of inertia on this object)
 
Well, if the units of R are not given, I would leave things in terms of R.
 
The units of R are mm^4 :) So I was correct, yes?
 
Did you leave the R^4 in? (1.35mm^4 is not 1.35*R^4 mm^4)

If so, yes. Otherwise, (without having done the problem) I'd say you were off by a factor of R^4 based on what I presume is the answer key up there.
 
I see. I'm not quite sure I have the intuitive understanding of "why" I need to add that R, I thought that R is just a part of the formulas for polar moment of inertia, and then you just leave the result "as in" and add the units.
 
This is a different R, and is a constant used to give the lengths you need to put into the formula.

So, the radius for the big circle is r=R, so you'd plug in R for r in the formula, and similar for the other parts.

For that matter, let's do something simple, finding the area of the big circle. A = πr2, of course, and since the big circle is radius R, you'd get A = πR2. Same idea for area moment of inertia.

(As an aside, I really hate the ISO standards they used in that drawing :smile:)
 
For that matter, let's do something simple, finding the area of the big circle. A = πr2, of course, and since the big circle is radius R, you'd get A = πR2. Same idea for area moment of inertia.

For area I'd just do

r=1
pi x radius^2 = 3.1415926... mm^2

No need to use R in the answer.

This is a different R, and is a constant used to give the lengths you need to put into the formula.

So I add it to every moment of inertia of circular objets calculations? Is it just for polar moment of inertia?
 
  • #10
Femme_physics said:
For area I'd just do

r=1
pi x radius^2 = 3.1415926... mm^2

No need to use R in the answer.

But the radius is not 1, it is R. (R is a number here, it is not the r in the formula. You plug R into r.)

Think algebraically. If I gave you f(x)=x^2 and told you to find f(B), what would you do? It's the same with r and R.

Femme_physics said:
So I add it to every moment of inertia of circular objets calculations? Is it just for polar moment of inertia?

The various radiuses given in terms of R would still have Rs.
 
  • #11
Looking at your diagram the radii are not given in mm so your answer cannot be in mm4.

The outer radius of the disk is R

The pitch circle of the holes is 0.5R

The radius of the holes os 0.25R

Whatever R is.

If you like R is a scale factor.

This is not the usual engineering drawing convention where R stands for radius and the units are understood. If that was the case then R would not appear in the answer.
 
  • #12
But the radius is not 1, it is R. (R is a number here, it is not the r in the formula. You plug R into r.)
Oh, you're right. I forgot that I defined R as 1. I am allowed to do that, I believe.
 
  • #13
Looking at your diagram the radii are not given in mm so your answer cannot be in mm4.

The outer radius of the disk is R

The pitch circle of the holes is 0.5R

The radius of the holes os 0.25R

Whatever R is.

If you like R is a scale factor.

In my answer here I defined R as 1, so is that legit that I just leave it as mm^4?

Attached my answer
 

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  • #14
Your final answer is correct, but I'd leave the R in in your calculations. R is arbitrary, so if you want a general solution you can't redefine R.
 
  • #15
jhae2.718 said:
Your final answer is correct, but I'd leave the R in in your calculations. R is arbitrary, so if you want a general solution you can't redefine R.

I see what you mean, good idea. Thanks to you both.
 
  • #16
I forgot that I defined R as 1

Much haste, less speed.

:wink:
 
  • #17
Well put monsieur :)
 
  • #18
*bumps*

I Like Serena said:
My attention was also drawn to 6.7 which isn't quite right. The numbers are correct, but the units are not.
I recommend you remove the mm4 everywhere, because they really should not be there.

But, what other units am I supposed to write there? I am calculating moment of inertia after all, and these are the units.
 
  • #19
Femme_physics said:
*bumps*

But, what other units am I supposed to write there? I am calculating moment of inertia after all, and these are the units.

In your calculation you have set R to 1.
I assume that is something your teacher did as well, did he?
However, this is a bit tricky.
This is because R is an unknown quantity that has not been given.

R might for instance be R=1 mm, which is basically what you have done.
Then you would get the result you calculated in mm4.
That is, you would have:
I0 = 1.35 mm4.

Note that R is not included in the result!

However, suppose R=2 mm.
Then all the calculations will be basically the same, except that in the final answer there will be an extra factor 24.
That is, you would have:
I0 = 1.35 x 24 mm4

I could also write this as:
I0 = 1.35 x (2 mm)4

Or: :)
I0 = 1.35 x R4

Note that mm4 is not included in the result!
 
  • #20
To be fair my first intuition was actually NOT to write R, but jhae said

Your final answer is correct, but I'd leave the R in in your calculations. R is arbitrary, so if you want a general solution you can't redefine R.

So that's a bit conflicting! Unless he meant that I drop mm^4 and just leave R in? It's just really hard/painful for me to see a result without units!
 
  • #21
Femme_physics said:
To be fair my first intuition was actually NOT to write R, but jhae said

jhae said:
Your final answer is correct, but I'd leave the R in in your calculations. R is arbitrary, so if you want a general solution you can't redefine R.

So that's a bit conflicting! Unless he meant that I drop mm^4 and just leave R in? It's just really hard/painful for me to see a result without units!

Yes, he meant that you drop mm^4 and just leave R in.
The unit is "embedded" in R.

Consider that now you have
I0=1.35 x R4 mm4

If you wanted to calculate the result for say R=2 mm, you'd get:
I0=1.35 x (2 mm)4 mm4

Which is:
I0=1.35 x 24 mm8

Somehow that doesn't look quite right...? :rolleyes:
 
  • #22
You must leave your answer as 1.35R4.

Too late...
 
  • #23
Oh, I see what you both mean. Thanks. Still, force of habit to use mm^4. I'll just be removing the R. As long as my answer is true to my definition of R.
 
  • #24
You really shouldn't get to attached to one particular unit (mm4)...

What if you had to deal with cm4? Or m4? Or even furlongs4?

It's best to think of area moment of inertia having dimension [length]4.
 
  • #25
You're right, but all our calculations have been with mm^4 so until I run into some other type of measurements of moment of inertia, I'll stick with mm^4. I think it makes more sense to my classmates to see mm^4 (I post all these answers in a solution blog), I'd hate them to start asking too many questions-- they already piled enough I reckon.
 
  • #26
Femme_physics said:
(I post all these answers in a solution blog)

:eek:
 
  • #28
Oh, you're only a mechanical engineer, so who cares...:biggrin:

That's Hebrew, right?
 
  • #29
Right! That's Hebrew.

The degree I'm going for is mechatronics, actually.

And "only a mech. engineer"? Do I sense an air of superiority about thee, sir? Yes...smells like a physicist, that one.
 
  • #30
Femme_physics said:
And "only a mech. engineer"? Do I sense an air of superiority about thee, sir? Yes...smells like a physicist, that one.

Nope.

(I'll leave it at that since I kind of derailed the thread. Have to remember this is one of the serious forums...:blushing:)
 

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