How Does Mathworld Calculate the Moment of Inertia for Spheres?

Click For Summary

Homework Help Overview

The discussion revolves around the calculation of the moment of inertia for spheres, specifically referencing methods presented on Mathworld. Participants are exploring various integration techniques and the implications of different approaches to deriving the moment of inertia tensor for a sphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the elemental sheet method for calculating moment of inertia and express curiosity about alternative methods shown on Mathworld. There are inquiries about deriving integrals and the feasibility of reducing elements to points for integration.

Discussion Status

Some participants have provided guidance on the integration methods, noting that using three integrals may not be necessary in an exam context. Others are questioning the validity of certain approaches and exploring different ways to conceptualize the problem.

Contextual Notes

There are mentions of exam constraints and preferences for simplification in calculations. Participants are also reflecting on the expectations of their instructors regarding the methods used in problem-solving.

ritwik06
Messages
577
Reaction score
0
I know how to take out the moment of inertia of a sphere about an axis passing through a diameter. My method is the same old one of choosing an elemental sheet and integrating. Can someone please explain to me the things I saw on Mathworld please. The method looks fascinating to me and I wish to learn it and use it for other 3 bodies.
The volume of the sphere, V=4/3 pi* r^3 , can be found in Cartesian, cylindrical, and spherical coordinates, respectively, using the integrals
http://mathworld.wolfram.com/images/equations/Sphere/Inline68.gif
The interior of the sphere of radius and mass has moment of inertia tensor
http://mathworld.wolfram.com/images/equations/Sphere/NumberedEquation7.gif
The original link is:
http://mathworld.wolfram.com/Sphere.html
regards,
Ritwik
 
Physics news on Phys.org
ritwik06 said:
I know how to take out the moment of inertia of a sphere about an axis passing through a diameter. My method is the same old one of choosing an elemental sheet and integrating …

Hi ritwik! :smile:

Equations (15) and (16) are using your elemental sheet method.

Equation (17) is using the same method, but with an elemental sphere instead of a flat sheet.

Equation (18) is just a way of writing the moment of inertia for all possible diameters instead of just one … but the method of getting them is the same. :smile:

(moment of inertia of a rigid body is a 3x3 tensor, so if you write it in coordinates, it looks like a 3x3 matrix)
 
Could you please derive what those guys had written. I now possesses an overwhelming desire to see that. Three integrals at a time seem so fascinating.

------------------------------------------

Apart from this here are one of my wildest thoughts:

What if we take the small element of sphere to be literally reduced to a point! The volume of this would be dy*dx*dz

therefore;
dM=\frac{3M}{4*pi*R^{3}}*dy*dx*dz

Coordinates of the sphere= R cos a cos b \hat{i} + R cos a sin b \hat{j} + R sin a \hat{k}

where a is the angle made by the position vector with its projection in xy plane. b is the angle made by this projection with the x axis.

Is it possible to integrate thus! If yes, please do tell me how.

tiny-tim said:
Hi ritwik! :smile:

Equations (15) and (16) are using your elemental sheet method.

Equation (17) is using the same method, but with an elemental sphere instead of a flat sheet.

Equation (18) is just a way of writing the moment of inertia for all possible diameters instead of just one … but the method of getting them is the same. :smile:

(moment of inertia of a rigid body is a 3x3 tensor, so if you write it in coordinates, it looks like a 3x3 matrix)
 
ritwik06 said:
Could you please derive what those guys had written. I now possesses an overwhelming desire to see that. Three integrals at a time seem so fascinating.

Hi Ritwik! :smile:

If you use three integrals in the exam when you could simplify it to two or one, you will lose marks.

The whole point of dividing a volume into slices (what you call "elemental sheets") is to make the integration easier … you choose a slice that you already know the area of, and that saves you two integrations, and leaves you with only one.
What if we take the small element of sphere to be literally reduced to a point! The volume of this would be dy*dx*dz

Sorry, but that's hopeless … you have to slice the whole volume in a genuine way. :frown:
 
tiny-tim said:
Hi Ritwik! :smile:

If you use three integrals in the exam when you could simplify it to two or one, you will lose marks.

The whole point of dividing a volume into slices (what you call "elemental sheets") is to make the integration easier … you choose a slice that you already know the area of, and that saves you two integrations, and leaves you with only one.


Sorry, but that's hopeless … you have to slice the whole volume in a genuine way. :frown:

Oh, that's not the way here. We are given ample time in an exam- specially the Sciences and Maths. And my teacher shall appreciate it if I do the question in numerous ways.

By the way it doesn't matter. Could you please explain it to me? Please! I shall be very grateful.

Oh! That was just a thought! I already had the inkling that it might be wrong!
 
ok … in (15), the sphere is divided into vertical slices of thickness dx, from x = -R to R,

then each vertical slice is divided into vertical strips of thickness dy, from y = -√(R2 - x2) to √(R2 - x2),

and each vertical strip is divided into horizontal blocks of thickness dz, from z = -√(R2 - x2 - y2) to √(R2 - x2 - y2).

Then you add the volumes of all the blocks, which is the same as the integral in (15).

(16) is very simi;ar and I expect you can work this out for yourself.

(17) divides the sphere into spherical shells of thickness dr, from r = 0 to R,

then divides each shell into vertical lunes of angular thickness dtheta, from theta = 0 to π,

then divides each lune into blocks of thicknes r sinphi, from phi = 0 to 2π
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 40 ·
2
Replies
40
Views
6K
Replies
16
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
2
Views
7K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K