# I Deriving the moment of inertia of solid sphere

1. Apr 28, 2017

### Arisylia

So i was going through derivations of moments of inertia of objects. For objects like the disk and rod, i was able to assume a relationship between mass and volume and integrate From there like
$$\frac{d_m}{m} = \frac{dl}{l} \\ d_m = \frac{dl*m}{l} \\ \int_{0}^{L}r^2\frac{dl*m}{l} \\ \frac{ml^2}{3}$$
thats for a rod on its end point.
i tried doing something similar with a sphere
$$\frac{d_m}{m} = \frac{4\pi r^2 dr}{\frac{4}{3}\pi R^3} \\ d_m = \frac{4 r^2 dr*m}{\frac{4}{3} R^3} \\ \int_{0}^{R}r^2\frac{4 r^2 dr*m}{\frac{4}{3} R^3} \\ \frac{3mR^2}{5}$$
but its supposed to be 2/5mr^2
i dont know if its because i cant apply this method or because i screwed something up. I looked at the derivation using the slices but I'm still curious about this.

Thanks for the help.
(its my first post here, sorry if im missing some part of etiquette or anything ! not sure if this is intermediate or basic?)

2. Apr 29, 2017

### andrewkirk

The moment of inertial is not $\int_0^R r^2dm$, as implied by the above. The incremental mass $dm$ is a thin spherical shell, which is not all at the same distance (radius) from the axis of rotation. Hence the integrand does not represent the momentum of inertia of that shell.

You need to set up an integration in which the solid sphere is split up into a series of incremental masses for which you know the moment of inertia. If you know the moment of inertia of a spherical shell, you can use the above approach and replace the integrand by the MoI of a spherical shell of radius $r$ (which is in the list here). Otherwise you need to split the sphere up a different way, eg as a stack of discs.

Last edited: Apr 29, 2017
3. Apr 29, 2017