- #1
alex3
- 44
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I'm aware of many, many solutions to this on the web (and on the forum) that I can follow, but I'm trying a different way (there are many, after all) and I can't figure out why it's not working, and I'd love to know where my logic is flawed.
I have a solid sphere, and I'm splitting it up into infinitely many discs. Taking a table top, the surface is my x-y plane, and up and down (wrt gravity) is my z-axis. I'm using [tex]\theta[/tex] as the angle from the x-y plane to the z-axis, [tex]-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}[/tex].
[tex]a[/tex] is the radius of the sphere, [tex]r[/tex] is the radius of a disc. Then, the thickness of my disc, an angle [tex]\theta[/tex] above the x-y plane, is [tex]a\operatorname{d}\theta[/tex]. The radius of the disc is [tex]a \cos{\theta}[/tex]. Then, the volume of the disc is [tex]a^2\pi\cos^2{\theta}\operatorname{d}\theta[/tex] (the cross-sectional area times the thickness [tex]\pi r^2 \operatorname{d}\theta[/tex]).
The equation for the MoI of a continuous solid is
[tex]I = \int r^2 \operatorname{d}m[/tex]
And here, our [tex]\operatorname{d}m[/tex] is [tex]\rho \operatorname{d}V[/tex], where [tex]\rho[/tex] is the density of the sphere, and [tex]\operatorname{d}V[/tex] is as defined above. As [tex]r = a\cos{\theta}[/tex], we get
[tex]I = a^5 \rho \pi \int_\frac{-\pi}{2}^\frac{\pi}{2} \cos^4{\theta} \operatorname{d}\theta[/tex]
The integral evaluates to
[tex]\frac{3\pi}{8}[/tex]
giving
[tex]I = \frac{3\pi^2}{8} a^5 \rho[/tex]
And we know
[tex]\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3}[/tex]
We get the result
[tex]I = \frac{9\pi}{32} m a^2[/tex]
Which is wrong. Obviously there's a flaw, where is it?
I have a solid sphere, and I'm splitting it up into infinitely many discs. Taking a table top, the surface is my x-y plane, and up and down (wrt gravity) is my z-axis. I'm using [tex]\theta[/tex] as the angle from the x-y plane to the z-axis, [tex]-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}[/tex].
[tex]a[/tex] is the radius of the sphere, [tex]r[/tex] is the radius of a disc. Then, the thickness of my disc, an angle [tex]\theta[/tex] above the x-y plane, is [tex]a\operatorname{d}\theta[/tex]. The radius of the disc is [tex]a \cos{\theta}[/tex]. Then, the volume of the disc is [tex]a^2\pi\cos^2{\theta}\operatorname{d}\theta[/tex] (the cross-sectional area times the thickness [tex]\pi r^2 \operatorname{d}\theta[/tex]).
The equation for the MoI of a continuous solid is
[tex]I = \int r^2 \operatorname{d}m[/tex]
And here, our [tex]\operatorname{d}m[/tex] is [tex]\rho \operatorname{d}V[/tex], where [tex]\rho[/tex] is the density of the sphere, and [tex]\operatorname{d}V[/tex] is as defined above. As [tex]r = a\cos{\theta}[/tex], we get
[tex]I = a^5 \rho \pi \int_\frac{-\pi}{2}^\frac{\pi}{2} \cos^4{\theta} \operatorname{d}\theta[/tex]
The integral evaluates to
[tex]\frac{3\pi}{8}[/tex]
giving
[tex]I = \frac{3\pi^2}{8} a^5 \rho[/tex]
And we know
[tex]\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3}[/tex]
We get the result
[tex]I = \frac{9\pi}{32} m a^2[/tex]
Which is wrong. Obviously there's a flaw, where is it?