Deriving the Moment of Inertia of a Hollow Sphere with Uniform Density

[doerame]

I've been working on deriving the moment of inertia of a hollow sphere (ie basketball) with uniform density for a while now with no success... Can anyone show me this derivation?

The moment of inertia is I = 2/3 MR^2 . If anyone can help me get to this step, it would be greatly appreciated.

 

[FredGarvin]

Take a look here:

http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html

 

[dextercioby]

Okay,that's a link to a good calculation for the moment of inertia of a full sphere.
You need for a hollow sphere.
$$I=\int r^{2} dm$$ (1)
,where 'r' is the distance between the mass element 'dm' and the axis of rotation.
$$dm=\rho_{surf.sphere}dS$$ (2)
,where
$$\rho_{surf.sphere}=\frac{M_{sphere}}{S_{sphere}}=\frac{M}{4\pi R^{2}}$$ (3)

The distance 'r' to the axis of rotation chosen as Oz is
$$r=R\sin \theta$$(4)
The sphere surface element is
$$dS=R^{2}d\Omega=R^{2}\sin\theta d\theta d\phi$$(5)

Then I becomes
$$I=\int_{0}^{2\pi} d\phi\int_{0}^{\pi} d\theta (R^{2}\sin^{2}\theta)(\frac{M}{4\pi R^{2}}) R^{2}\sin\theta$$ (6)

Make simplifications and integrate after $\phi$,simplify again and u'll get
$$I=\frac{MR^{2}}{2}\int_{0}^{\pi} d\theta \sin^{3}\theta =\frac{MR^{2}}{2}(-)\int_{1}^{-1} (1-u^{2}) du =\frac{MR^{2}}{2}(u-\frac{u^{3}}{3})|_{-1}^{+1}=\frac{MR^{2}}{2}\frac{4}{3}=\frac{2MR^{2}}{3}$$(7)
,where i made use of
$$\sin^{3}\theta=\sin\theta(1-\cos^{2}\theta)$$(8)
and the substitution
$$\cos\theta\rightarrow u$$ (9)
,under which the limits of integration transform in the prescribed way.

Daniel.

$$I=$$