MoI of a Solid Sphere Derivation

[alex3]

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?

 

[tiny-tim]

hi alex3!

you went wrong after …

The equation for the MoI of a continuous solid is …

… you're assuming a constant r, but the r for your disc varies across the disc, doesn't it?

(using vertical cylinders instead of horizontal discs would work )

 

[alex3]

Aha! The distance from the axis changes along the disc, of course, thank you. Is there a simple (ish) fix that can be applied to my method to compensate for my mistake, of is it best for me to learn another tried-and-tested derivation?

 

[Quinzio]

I have many doubts about what I read in you post.

You can sum the pile of disks, ok.
But you have to sum the MoI of the single disk, not it's volume.

 

[tiny-tim]

Aha! The distance from the axis changes along the disc, of course, thank you. Is there a simple (ish) fix that can be applied to my method to compensate for my mistake, of is it best for me to learn another tried-and-tested derivation?

you can use (or prove) the formula for the moment of inertia of a disc, and then integrate that

(but vertical cylindrical shells is easiest)

 

[alex3]

Brilliant. You've all been very helpful indeed, improved my basic understanding of MoIs no end, thank you!