My name is Alexander FufaeV and here I will explain the following topic:

(Hollow) Cylinder - Moment of Inertia

Hollow cylinder that rotates around its axis of symmetry.

In the following, the moment of inertia $I$ of a hollow cylinder of homogeneous mass $m$ is derived. The cylinder has an inner radius $r_{\text i}$ (${\text i}$ für internal), an outer radius $r_{\text e}$ (${\text e}$ für external) a a height $h$. In the end, we want to get the moment of inertia $I$, which depends only on these given quantities.

It is also assumed that the axis around which the cylinder rotates passes through the center of the cylinder, that is it rotates around its symmetry axis.

The moment of inertia $I$ can be determined in general by integrating $r_{\perp}^2 \, \rho(\boldsymbol{r})$ over the volume $V$ of the body:

$$ \begin{align} I ~=~ \int_V r_{\perp}^2 \, \rho(\boldsymbol{r}) \, \text{d}v \end{align} $$

Here $r_{\perp} $ is the perpendicular distance of a volume element $\text{d}v$ of the body from the selected axis of rotation (see illustration 1). And $ \rho(\boldsymbol{r})$ is the mass density of the body, which in general depends on the position vector $\boldsymbol{r}$.

In our case, the cylinder has a homogeneous mass distribution, so the mass density is independent of position: $ \rho = \text{const}$. We may place the mass density in front of the integral:

$$ \begin{align} I ~=~ \rho \int_V r_{\perp}^2 \, \text{d}v \end{align} $$

For the integration we can express the infinitesimal volume element $\text{d}v$ of the cylinder with $\text{d}r_{\perp}$ and integrate over $r_{\perp}$. Divide the cylinder into concentric, infinitely thin hollow cylinders, with thickness $\text{d}r_{\perp}$ and height $h$. You can think of this integration as starting at the inner radius and summing up the infinitely thin hollow cylinders over $r_{\perp}$ until we arrive at the outer radius. So then $\text{d}v$ is the volume of an infinitely thin hollow cylinder.

The infinitely thin hollow cylinder has the lateral surface $2\pi \, r_{\perp} \, h$. Multiplied by its infinitesimal thickness $ \text{d}r_{\perp} $, we can write the volume $\text{d}v$ of the infinitesimal thin cylinder as follows:

$$ \begin{align} \text{d}v ~=~ 2 \pi \, r_{\perp} \, h \, \text{d}r_{\perp} \end{align} $$

Substitute 3 into the moment of inertia integral 2:

$$ \begin{align} I ~=~ \rho \int^{r_{\text e}}_{r_{\text i}} r_{\perp}^2 \, ( 2 \pi \, r_{\perp} \, h \, \text{d}r_{\perp} ) \end{align} $$

All constants may be placed in front of the integral:

$$ \begin{align} I ~=~ 2\pi \, \rho \, h \int^{r_{\text e}}_{r_{\text i}} r_{\perp}^3 \text{d}r_{\perp} \end{align} $$

Thus we have transformed the integral 2 over the volume $V$ into an integral 5 over the radius $r_{\perp}$. The integration of 5 results in:

$$ \begin{align} I ~=~ 2\pi \, \rho \, h \begin{bmatrix} \frac{r_{\perp}^4}{4} \end{bmatrix}^{r_{\text e}}_{r_{\text i}} \end{align} $$

Insert the upper and lower integration limits:

$$ \begin{align} I ~=~ 2\pi \, \rho \, h \left( \frac{r_{\text e}^4 - r_{\text i}^4}{4} \right) \end{align} $$

Factor out $1/4$ and eliminate factor 2:

$$ \begin{align} I ~=~ \frac{1}{2}\pi \, \rho \, h \left( r_{\text e}^4 - r_{\text i}^4 \right) \end{align} $$

We still have to somehow bring the given mass $m$ into play. The mass density $\rho$ is not known. First we factorize $r_{\text e}^4 - r_{\text i}^4 $ (using binomial formula):

$$ \begin{align} I ~=~ \frac{1}{2}\pi \, \rho \, h \left( r_{\text e}^2 - r_{\text i}^2 \right) \, \left( r_{\text e}^2 + r_{\text i}^2 \right) \end{align} $$

The total mass $m$ of the cylinder is related to the constant mass density as follows (mass density = mass per volume):

$$ \begin{align} m ~=~ \rho \, V \end{align} $$

The cylinder volume $V$ in Eq. 10 is the volume $ \pi \, r_{\text e}^2 \, h $ of the outer solid cylinder minus the volume $ \pi \, r_{\text i}^2 \, h $ of the inner solid cylinder. Thus 10 becomes:

$$ \begin{align} m ~=~ \pi \, \rho \, h \left( r_{\text e}^2 - r_{\text i}^2 \right) \end{align} $$

With this, we can now substitute the cylinder mass 11 into the equation 9 for the moment of inertia. First, rearrange Eq. 11 for $\left( r_{\text e}^2 - r_{\text i}^2 \right)$ and substitute the result into Eq. 9:

$$ \begin{align} I ~=~ \frac{1}{2} \class{brown}{m} \, \left( {\class{purple}{r_{\text e}}}^2 + {\class{purple}{r_{\text i}}}^2 \right) \end{align} $$

This is the moment of inertia we are looking for $I$ expressed with the given quantities.

From the formula for the moment of inertia of a hollow cylinder, we can also easily determine the moment of inertia of a filled cylinder (solid cylinder). In the case of a solid cylinder, the inner radius is $ r_{\text i} = 0 $.

Solid cylinder that rotates around its axis of symmetry.

Since we then have only one radius in the formula, we can write $ r_{\text e} $ for short instead of $ r $ to beautify the formula. The $r$ is the radius of the solid cylinder. Then we get:

$$ \begin{align} \class{brown}{I} ~=~ \frac{1}{2} \, m \, {\class{purple}{r}}^2 \end{align} $$