Units for moment of inertia of a circle

[Femme_physics]

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.

 

[jhae2.718]

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.

 

[Femme_physics]

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)

 

[jhae2.718]

Well, if the units of R are not given, I would leave things in terms of R.

 

[Femme_physics]

The units of R are mm^4 :) So I was correct, yes?

 

[jhae2.718]

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.

 

[Femme_physics]

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.

 

[jhae2.718]

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 )

 

[Femme_physics]

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?

 

[jhae2.718]

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.

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.

 

[Studiot]

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.

 

[Femme_physics]

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.

 

[Femme_physics]

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

 

[jhae2.718]

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.

 

[Femme_physics]

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.

 

[Studiot]

I forgot that I defined R as 1

Much haste, less speed.

 

[Femme_physics]

Well put monsieur :)

 

[Femme_physics]

*bumps*

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.

 

[I like Serena]

*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!

 

[Femme_physics]

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!

 

[I like Serena]

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!

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...? :uhh:

 

[jhae2.718]

You must leave your answer as 1.35R⁴.

Too late...

 

[Femme_physics]

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.

 

[jhae2.718]

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.

 

[Femme_physics]

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.

 

[jhae2.718]

(I post all these answers in a solution blog)

 

[Femme_physics]

It's for my classmates. And kinda for my soul :)

http://ortmechatronics.blogspot.com/

 

[jhae2.718]

Oh, you're only a mechanical engineer, so who cares...

That's Hebrew, right?

 

[Femme_physics]

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.

 

[jhae2.718]

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...)