Moment of inertia of an ellipse

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SUMMARY

The moment of inertia (I) of an ellipse with mass (m) and semi-major and semi-minor axes (a and b) can be calculated using the formula I = m(a² + b²)/4. The derivation involves double integrals over the area of the ellipse, specifically integrating x² and y² in Cartesian coordinates. The discussion highlights the complexity of the integrals, which can be simplified using trigonometric substitutions, such as x = a sin(θ), to avoid complicated expressions involving arcsin.

PREREQUISITES
  • Understanding of moment of inertia calculations
  • Familiarity with double integrals in calculus
  • Knowledge of polar coordinates and trigonometric substitutions
  • Basic concepts of ellipse geometry and area calculation
NEXT STEPS
  • Study the derivation of moment of inertia for various shapes, focusing on ellipses
  • Learn about trigonometric substitutions in integral calculus
  • Explore the use of polar coordinates in solving area-related integrals
  • Review integral tables for common integrals involving square roots and trigonometric functions
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Students studying physics or engineering, mathematicians interested in calculus applications, and anyone needing to calculate the moment of inertia for elliptical shapes.

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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