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Homework Statement
Find the moment of inertia of the right circular cone or radius r and height h with respect to its axis, and in terms of its mass.
*As of this point, I am supposed to use solids of revolution, and so I need to rotate a line about an axis, and find the moment of inertia with respect to that axis. I only know basic integration at this point.
Homework Equations
The textbook given equations for moment of inertia of the solid of revolution about the x-axis and y-axis are:
[tex]I_{y}=2\pi\rho\int_a^b yx^3\ dx [/tex]
[tex]I_{x}=2\pi\rho\int_c^d xy^3\ dy [/tex]
The Attempt at a Solution
I set up a line with the equation [tex]y=(\frac{h}{r} )x[/tex] to be rotated about the y-axis. Applying the formula i get:
[tex]I_{y}=2\pi\rho\int_a^b\ yx^3 dx [/tex]
[tex]I_{y}=2\pi\rho\int_0^r\ ((\frac{h}{r} )x)x^3 dx [/tex]
[tex]I_{y}=(\frac{2\pi\rho\ h}{r})\int_0^r\ x^4 dx [/tex] which simplifies to:
[tex]I_{y}=(\frac{2\pi\rho\ h}{r})\cdot\frac{r^5}{5} = \frac{2\pi\rho hr^4}{5}[/tex]
Since the volume of this cone is [tex]\frac{\pi\rho hr^2}{3}[/tex]
[tex]I_{y}= \frac{2\pi\rho hr^4}{5}= \frac{6r^2}{5}\cdot \frac{\pi\rho hr^2}{3}[/tex]
[tex]I_{y}= \frac{6mr^2}{5}[/tex] Which is wrong because it is too large, and because the correct answer is [tex]I_{y}= \frac{3mr^2}{10}[/tex].
I've done this a few times and feel stupid because I still can't find my mistake. I suspect it is an obvious one, or has to do with simply "plugging in" info into the formula.
I am learning some calculus on my own over the summer, so I don't have anyone to ask other than the members of this forum. As this is my first post on this forum, I hope some of the members will make any suggestions as to my posting/question. Thanks.
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