# Moment of Inertia of Hollow Cylinder Derivation

1. Aug 19, 2014

For a uniform, hollow cylinder, why is this derivation wrong?

M = mass of whole solid cylinder
m = mass of missing cylindrical piece
R = radius of whole cylinder
r = radius of missing cylindrical piece

moment of inertia = moment of inertia of whole cylinder - moment of inertia of missing cylindrical piece

I = MR2/2 - mr2/2

m/M = pi*r2*h/pi*R2*h = r2/R2

m = M*r2/R2

I = MR2/2 - M*r4/2R2

I = MR4/2R2 - M*r4/2R2

I = M/2R2*(R4 - r4)

2. Aug 19, 2014

### Delta²

It seems correct, it is only that the result contains in its expression the mass M of the whole cylinder and not the mass of the corresponding hollow cylinder. If you substitute $M=\frac{M_{h}R^2}{R^2-r^2}$ you ll get the usual expression for the inertia of a hollow cylinder $I=\frac{1}{2}M_h(R^2+r^2)$.

Last edited: Aug 19, 2014
3. Aug 19, 2014

### olivermsun

It's fine as written. If you want to express $I$ using the mass of the outer shell only, call it $M^\prime,$ then you have to use $M^\prime = M - m = M (1 - r^2/R^2)$ to get the usual form for $I$.

4. Aug 21, 2014