Moment of inertia of an ellipse

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Homework Help Overview

The discussion revolves around calculating the moment of inertia (I) of an ellipse with given mass (m) and radii (a and b). The original poster presents an integral approach to derive the moment of inertia using the ellipse's area and its defining equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various integral expressions and substitutions to compute the moment of inertia. There are discussions on the complexity of the resulting expressions, particularly those involving arcsin. Some participants suggest using polar coordinates or trigonometric substitutions to simplify the calculations.

Discussion Status

The discussion is active, with participants providing insights on integral evaluations and suggesting alternative methods. There is no explicit consensus on the best approach, but several productive lines of reasoning are being explored.

Contextual Notes

Participants reference integral tables and express concerns about the complexity of certain expressions. There is a mention of potential complications arising from trigonometric substitutions, indicating a need for clarity on the methods being discussed.

Karol
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Homework Statement


To calculate I, the moment of inertia of an ellipse of mass m.
The radius are a and b, according to the drawing.

Homework Equations


I=mr^2
Ellipse:
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow y=b\sqrt{1-\frac{x^2}{a^2}}
Area of an ellipse: \piab

The Attempt at a Solution


I=4\frac{m}{\pi a b} \int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} x^2+y^2 dy
I=4\frac{m}{\pi a b} \left(\int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} x^2dy+\int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} y^2 dy \right)
I=4\frac{m}{\pi a b} \left(\int_{x=0}^{a} x^2 \left[y\right]_{0}^{b\sqrt{1-\frac{x^2}{a^2}}}dx+\int_{x=0}^{a}\frac{1}{3}\left[y^3\right]_{0}^{b\sqrt{1-\frac{x^2}{a^2}}}dx \right)
I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3}\int_{x=0}^{a}\left(1-\frac{x^2}{a^2}\right)^{3/2}dx\right)
I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3a^3}\int_{x=0}^{a} (a^2-x^2)^{3/2}dx \right)
And it gives complicated expressions which include \arcsin.
The answer should be:
I=m(a^2+b^2)/4
 

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Karol said:
I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3a^3}\int_{x=0}^{a} (a^2-x^2)^{3/2}dx \right)
And it gives complicated expressions which include \arcsin.

You result so far looks good. Those integrals are not too difficult to carry out. I don't see how you are getting arcsin expressions.
 
From integrals tables:
\int x^2 \sqrt{a^2-x^2}dx=-\frac{x}{4}\sqrt{(a^2-x^2)^3}+\frac{a^2}{8}\left( x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}\right)
 
May I suggest you work in polar coordinates?
 
Karol said:
From integrals tables:
\int x^2 \sqrt{a^2-x^2}dx=-\frac{x}{4}\sqrt{(a^2-x^2)^3}+\frac{a^2}{8}\left( x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}\right)

OK. You can do it with just trig substitutions.

But you can use your expression from the tables. It will simplify very nicely when you substitute the limits.
 
What trig substitution? if i make y=a*sinx it's complicated also
 
Try ##\; x = a \sin \theta##
 

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