- #1
alex3
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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 \theta as the angle from the x-y plane to the z-axis, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.
a is the radius of the sphere, r is the radius of a disc. Then, the thickness of my disc, an angle \theta above the x-y plane, is a\operatorname{d}\theta. The radius of the disc is a \cos{\theta}. Then, the volume of the disc is a^2\pi\cos^2{\theta}\operatorname{d}\theta (the cross-sectional area times the thickness \pi r^2 \operatorname{d}\theta).
The equation for the MoI of a continuous solid is
I = \int r^2 \operatorname{d}m
And here, our \operatorname{d}m is \rho \operatorname{d}V, where \rho is the density of the sphere, and \operatorname{d}V is as defined above. As r = a\cos{\theta}, we get
I = a^5 \rho \pi \int_\frac{-\pi}{2}^\frac{\pi}{2} \cos^4{\theta} \operatorname{d}\theta
The integral evaluates to
\frac{3\pi}{8}
giving
I = \frac{3\pi^2}{8} a^5 \rho
And we know
\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3}
We get the result
I = \frac{9\pi}{32} m a^2
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 \theta as the angle from the x-y plane to the z-axis, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.
a is the radius of the sphere, r is the radius of a disc. Then, the thickness of my disc, an angle \theta above the x-y plane, is a\operatorname{d}\theta. The radius of the disc is a \cos{\theta}. Then, the volume of the disc is a^2\pi\cos^2{\theta}\operatorname{d}\theta (the cross-sectional area times the thickness \pi r^2 \operatorname{d}\theta).
The equation for the MoI of a continuous solid is
I = \int r^2 \operatorname{d}m
And here, our \operatorname{d}m is \rho \operatorname{d}V, where \rho is the density of the sphere, and \operatorname{d}V is as defined above. As r = a\cos{\theta}, we get
I = a^5 \rho \pi \int_\frac{-\pi}{2}^\frac{\pi}{2} \cos^4{\theta} \operatorname{d}\theta
The integral evaluates to
\frac{3\pi}{8}
giving
I = \frac{3\pi^2}{8} a^5 \rho
And we know
\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3}
We get the result
I = \frac{9\pi}{32} m a^2
Which is wrong. Obviously there's a flaw, where is it?
