Calculating Moment of Inertia for a Thin-Walled Spherical Object

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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]
 
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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.
 
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Thanks a lot, you are a charming man (or woman...)