# I Why is "dr" not used for moment of inertia calculations?

1. Dec 1, 2016

### Ehden

When calculating the moment of inertia for a rotating object, why is "dr" not incorporated into the integral? However, dm is part of the integral, isn't the distance from the axis of rotation changing as well? Hence shouldn't the integral involve both dm and dr?

2. Dec 1, 2016

### Staff: Mentor

Can you show us an example of an moment of inertia calculation that you're wondering about? The simpler and more symmetrical it is, the better.

3. Dec 1, 2016

### ZapperZ

Staff Emeritus
It IS taken into consideration.

dm = ρ dV

So for, say, a spherical mass distribution, you get

dm = ρ r2 sinθ dr dθ dφ

Zz.

4. Dec 1, 2016

### Staff: Mentor

However, watch out for the different kinds of "r"s. The "r" that matters in a moment of inertia calculation (dI = dm⋅r2) is the perpendicular distance from the axis of rotation, whereas the "r" in spherical coordinates is the distance from the origin (usually the center of a spherical distribution).

5. Dec 1, 2016

### Ehden

So generally, I = Σmiri2 = ∫r2dm
Why is the r2 left as a constant? Isn't it also changing depending on where the dm is located?

I asked this question to a physics major, and he said I should just take it as it is because I only know up to single variable calculus. He said, the integral is a multivariable integral involving dr as well. Not sure if he's correct though.

Last edited: Dec 1, 2016
6. Dec 1, 2016

### ZapperZ

Staff Emeritus
Did you even read my post?!

Oy vey.

Zz.

7. Dec 1, 2016

### Ehden

Oh, I don't know how I skipped your post.
I don't if I'm correct though, but, isn't that multivariable calculus? I'm still taking single variable at the moment. So I guess, dr is taken into consideration, but, I should just take the general formula as it is given?

Last edited: Dec 1, 2016
8. Dec 1, 2016

### Student100

Because the prereq for the course probably didn't include multivariate calculus. Just asking why it's left constant doesn't mean a whole lot without context, but generally in these courses where they try to keep things simple they will use some sort of symmetry or constraints to avoid math students haven't had yet.

You'll see this a lot in E&M.

9. Dec 1, 2016

### Ehden

The course just requires calc 2, or a coreq in calc 2. Currently I'm taking calc 2, so I guess I should just take this general formula as is?

However, I do plan to take mechanics, which requires multivariable, I guess that's where the proof for r2 will be explained?

That reminds me, my professor showed the class a differential equation for harmonic motion, and the answer to it. Since it's a certain class of differential equations and the class hasn't done differential equations, he said we should just take the answer for what it is. I guess it's the same for this case?

Last edited: Dec 1, 2016
10. Dec 1, 2016

### Staff: Mentor

I second Nugatory's request. It would prevent a lot of confusing guessing if you could give us a specific example of the kind of calculation you're thinking of.

11. Dec 1, 2016

### Ehden

A simple one would be a uniform rod. The method would involve ∫r2dm. However, I was confused as to why r2 is left as a constant. Since r is changing compared to where dm is, so shouldn't r also be represented as dr?

But, Student100 told me to just take the equation for what it is, since it involves higher level math (multi-variable), which I haven't taken yet.

12. Dec 1, 2016

### Student100

It's not quite the same, since moment of inertia for things like thin rigid body rods with fixed axis of rotations that are highly symmetrical ($dm = \rho dl$ ... etc.) are fully accessible.

The differential equation for simple harmonic motion should also be accessible, but is easier with auxiliary equations which you probably haven't see yet either.

Edit: Tried to clean it up, think I'm confusing myself here.

Errr, uniform thin rod is perfectly accessible. So in that case, you're looking a specific example of a fixed axis of rotation with a rigid body.

The moment of inertia when we constrain ourselves to rigid bodies around fixed axis's of rotation depends on the mass distribution of in the body with respect to the axis of rotation. $I = \int r^2 dm$ where $dm = \rho dV$ and r is the perpendicular distance to the axis of rotation.

For a uniformly thin rigid rod with mass M and length L you have get $$I = \int r^2dm$$ where if we choose our coordinate system so that the rod lays in the x axis with the fixed axis of rotation around the origin at the middle of the rod it then becomes $$I = \int_{-L/2}^{L/2} x^2 dm$$ With $dm = \frac{M}{L} dx$ so $$I = \frac{M}{L} \int_{-L/2}^{L/2} x^2 dx$$

Basically, you're integrating over r. r (as the location of the axis of rotation) is also a constant (not changing, stays at origin) in this case. You can have other cases were r changes with time, etc, etc. Maybe it would make more sense to call the distance from the axis of rotation another variable.

Last edited: Dec 1, 2016
13. Dec 1, 2016

### Ehden

Okay, I think I found the solution to my dilemma. Since, the distance from each point mass is r distance away, hence, r is changing, but, the mass at each point is the same for a uniform rod. So we represent the small changes in mass by dm, so we can replace it with a density formula to represent the change in r?

14. Dec 1, 2016

### Student100

r isn't changing, the distance the mass elements are from r is.

15. Dec 2, 2016

### Ehden

Oh, I thought r represented the distance between the mass element and the axis of rotation. Thus, r should be different for every mass element.

16. Dec 2, 2016

### ZapperZ

Staff Emeritus
17. Dec 2, 2016

### Ehden

I actually read that page and it just added to my dilemma. From the diagram the distance for each xi is different, just as how the masses are different. So why is x represented as a constant in the integral?

The same goes for r2. Like Student100 mentioned, r is a constant and not the distance from the axis of rotation for each element of mass. I thought of r as x is show in the diagram. Except that origin point is the axis of rotation.

I'm just confusing myself now, I've had a lot of different responses.

18. Dec 2, 2016

### ZapperZ

Staff Emeritus
I still don't understand why you keep thinking that the coordinate variable is a "constant". I will say it once again, this time with emphasis, so that you pay attention:

LOOK AT THE EXPRESSION FOR dm!!

For the rod, dm can be written in terms of the linear mass density and..... get this.... dx!

The position of each mass element is contained in dm, because it has a "length" factor! So you are integrating the mass element over x!

This is no different than when I showed you what happened with a spherical mass distribution earlier where "dr" is actually part of dm!

You need to sit down and figure out why this is not sinking in for you even after you've been told of this several times on here.

Zz.

19. Dec 2, 2016

### Ehden

I've been receiving different responses that have been confusing me. I should've mentioned, my notion of using calculus may have been wrong. I forgot that dx represents a minute change, and the distance x is basically broken down to small elements of dx using the linear mass formula, as dm correlates with the width of dx.

20. Dec 2, 2016

### pixel

Check out this link: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html and scroll down to Moment of Inertia Examples. For a rigid body, we consider the body to be made up of small elements, dm, at distance r from the rotation axis. Each element contributes r2dm to the moment of inertia, which his found by integrating over all the elements: ∫r2dm. If all elements are not at the same distance from the axis, then as ZapperZ showed by example, the expression for dm will contain dr.

21. Dec 3, 2016

### pixel

Meant to add: in general you don't have a d- for every variable in an integral. You determine the elemental expression, such as r2dm, and then you integrate over whatever d- you have in that expression.

22. Dec 3, 2016

### Ehden

Oh man! I understand now! It just clicked for me! I was looking at the expression the wrong way!

Thank you all for the help! This feeling of struggling to comprehend something then finally understanding it, it's just impalpable!

Last edited: Dec 3, 2016