Calculating Moment of Inertia for a Thin-Walled Spherical Object

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a thin-walled spherical object, specifically focusing on the surface area and mass distribution of the sphere.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the moment of inertia using spherical coordinates and integrates over the surface area. Participants question the correct surface element used in the calculations and discuss the implications of measuring angles in spherical coordinates.

Discussion Status

Participants are actively engaging in clarifying the correct expressions for the surface element in spherical coordinates. There is a focus on understanding the geometry involved in the calculations, with multiple interpretations of angle definitions being explored.

Contextual Notes

There is an ongoing discussion regarding the definitions of the angles θ and φ in spherical coordinates, which may affect the surface area calculations. Some participants note the potential confusion arising from different conventions in defining these angles.

Karol
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Homework Statement


How to calculat the moment of inertia of a thin wall, the surface area of a ball of mass m and radii a.

Homework Equations


I=mr2

The Attempt at a Solution


Spherical coordinates.

[tex]dm=\frac{m}{4\pi a^2}\cdot a^2\cdot d\theta d\phi[/tex]

See drawing.

[tex]I=\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} dm\cdot a^2[/tex]
[tex]I=\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} \frac{m}{4\pi a^2}\cdot a^2\cdot a^2 d\theta d\phi[/tex]
[tex]I=\frac{a^2m}{4\pi}\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} d\theta d\phi[/tex]
[tex]I=\frac{2\pi m a^2}{4}[/tex]

The answer should be:

[tex]I=\frac{2ma^2}{3}[/tex]
 

Attachments

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I understand about the axis, but Why is the surface element sinφdθdφ, not dθdφ?
 
Because the surface element from θ to θ+dθ and from φ to φ+dφ is a tiny rectangle whose sides are rdθ and rsinφdφ …

making an area of r2sinφdθdφ :smile:

(btw, mathematicians usually define θ and φ the other way round, so you'll see usually see sinθdθdφ instead :wink:)
 
I understand the rdθ, but why the sinφ in rsinφdφ?
 
Karol said:
I understand the rdθ, but why the sinφ in rsinφdφ?

oops! I got the sinφ in the wrong place … it should be rdφ and rsinφdθ. :redface:

(Because, at latitude φ, the circle of latitude has radius rsinφ, so a slight change from θ to θ+dθ only takes you a distance radius time angle = rsinφdθ …

so that tiny rectangle has sides rdφ and rsinφdθ)
 
I understand, but then it should be rcosφdθ!
 
We usually measure φ from the pole, ie 0 ≤ φ ≤ π, and then it's rsinφdθ … see http://en.wikipedia.org/wiki/Spherical_coordinates" :wink:

Only if you measure φ from the equator, ie -π/2 ≤ φ ≤ π/2 (not recommended), is it rcosφdθ.
 
Last edited by a moderator:
Thanks a lot, you are a charming man (or woman...)
 
  • #10
no, fish! :biggrin:
 

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