Moment of inertia of an ellipse

  • Thread starter Karol
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  • #1
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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


[tex]I=mr^2[/tex]
Ellipse:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow y=b\sqrt{1-\frac{x^2}{a^2}}[/tex]
Area of an ellipse: [itex]\pi[/itex]ab

The Attempt at a Solution


[tex]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[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
And it gives complicated expressions which include [itex]\arcsin[/itex].
The answer should be:
[tex]I=m(a^2+b^2)/4[/tex]
 

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Answers and Replies

  • #2
TSny
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[tex]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)[/tex]
And it gives complicated expressions which include [itex]\arcsin[/itex].

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.
 
  • #3
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From integrals tables:
[tex]\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)[/tex]
 
  • #4
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May I suggest you work in polar coordinates?
 
  • #5
TSny
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From integrals tables:
[tex]\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)[/tex]

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.
 
  • #6
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What trig substitution? if i make y=a*sinx it's complicated also
 
  • #7
TSny
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Try ##\; x = a \sin \theta##
 

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