# Moment of Inertia

1. Oct 25, 2005

### chui

I couldnt find the Moment of inertia of the next tube:

hope you can help..thnks

2. Oct 25, 2005

### Tide

What, exactly, did you try to do?

3. Oct 26, 2005

### chui

I am not a physics student yet, so I dont really know.
By seeing diffrent exempls of moment of inertia I thoght it may be
0.25M(r1^2-r2^2) + 0.25ML^2
But this is only by intuation so I cant actually use it.
please, help me if you can.

4. Oct 26, 2005

### Tide

You will not find the moment of inertia by just guessing. You need a place to start. Do you have a (mathematical) definition for moment of inertia? Have you had any calculus classes? What was the actual assignment?

5. Oct 27, 2005

### Mk

6. Oct 27, 2005

### andrevdh

I do'nt think it is an assignment - he wants to build/construct something or work something out without having the necessary Physics background. so either we need to just give hime the answer or direct him to a tutorial.
Chui, start by looking at http://wps.aw.com/aw_young_physics_11. Have a look at Part 1:Mechanics - 7.6 Rotational Inertia.

Last edited: Oct 27, 2005
7. Oct 27, 2005

### chui

andrevdh

andrevdh rights! (thankyouverymuch)
All I need is the formula, with no other choice, I can develpoe it by myself but it will take time.
I want it for other calculation so I only need a formula.

If anyone know,(or have a book which know) please write it.
thanks.

8. Oct 27, 2005

### Tide

This is the homework section and we're not supposed to "give" the answer without the person showing some work. Otherwise, post in a different section! :)

9. Oct 27, 2005

### chui

alright, I used Wikipedia, to learn some things with Integration,
but it wasn't enought, I didn't understand too many things yet.
however, I looked here:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
and learn the next things:
1) an axis located at the center of a solid cylinder , its MomentumOfInertia should be this:
(1/4)MR^2 + (1/12)ML^2 (number 3 on the list)
2) a Rod with its axis on center its I should be:
(1/12)ML^2 (one befort the last one on the list)
3) a Rod with its axis on the end, like I need, its I should be:
(1/3)ML^2 (the last one on the list)
I put it all together with number 2 in the list to get this:
(1/4)M(R1^2-R2^2) + (1/3)ML^2
I still don't sure if I should put a minus or a plus between R1^2
to R2^2

10. Oct 27, 2005

### Tide

It will be $r_1^2 + r_2^2$. That part of the integration gives $r_1^4-r_2^4$ which factors into $(r_1^2-r_2^2)(r_1^2+r_2^2$ and the first factor (minus sign) ultimately gets absorbed into M. I don't have time right now but I need to check on that factor of 1/4.

11. Oct 27, 2005

### bluejay18

i'm sorry I don't mean to change the subject, but how would I show that the Rotational Inertia would be 2/5MR^2 of a uniform sphere about any axis. I tried to first calculate the R.I. of a disk but I am not getting anywhere. . . please help if you can? I just became a memeber today and don't know how to post a new thread and this is the only thread relavent to what I need help with. . .

12. Oct 28, 2005

### Tide

bluejay,

You need to integrate the square of the distance from the rotation axis. It looks like this:

$$I = 2\pi \rho \int_0^R r^2 dr \int_0^\pi d\theta r^2 \sin^3 \theta$$

where the $2\pi$ if from integrating about the azimuth (based on symmetry). $\rho = \frac {M}{4\pi R^3/3}$ is the mass density. This works out to the desired result.

13. Oct 28, 2005

### andrevdh

Last edited by a moderator: Apr 21, 2017