# Proof 29: Cylinder’s moment-of-inertia

Holed cylinder (with a hole in the middle) spinning about its axis

With a single point having the moment-of-inertia $I_\text{point}=r^2m$ ($m$ and $r$ are mass and distance from the axis-of-rotation),1 infinitely many “particles” making up a rotating object gives us the infinite sum, written as an integral:

$$I=\int r^2\,\mathrm dm$$

Let’s imagine the cylinder cut into many, many thin cylinder sheets. Each of them carries a small piece $\mathrm dm$ of the total mass $m$ and covers a small volume $\mathrm dV$ of the total volume $V$. The mass is uniform (evenly spread), so $\mathrm dm$ covers the same amount of the total mass $m$ as $\mathrm d V$ covers of the total volume $V$. Meaning, they are both the same fraction of the total:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dV}{V}\quad\Leftrightarrow \quad \mathrm dm= m\frac{\mathrm dV}{V}$$

Plug this in:

$$I=\int r^2\,\mathrm dm= \int r^2 \underbrace{\frac{m}{V} \mathrm dV}_{\mathrm dm}$$

The volume $\mathrm d V$ of a thin ring is its thin area $\mathrm dA$ times the total length $L$. The thickness of the cylinder sheet is $\mathrm dr$; with an inner radius of $r$, its outer radius is thus $r+\mathrm dr$. Therefore the small area $\mathrm dA$ is simply the circle area inside the cylinder sheet, $\pi r^2$, subtracted from the circle area to the outside of the cylinder sheet, $\pi (r+\mathrm dr)^2$:

\begin{align} \mathrm dV&=\mathrm dAL= ( \pi (r+\mathrm dr)^2 \,- \pi r^2 ) L = ( \pi (r^2+\mathrm dr^2+2r \mathrm dr ) \,- \pi r^2 ) L\\ &=( \cancel{ \pi r^2}+ \pi \mathrm dr^2+2 \pi r \mathrm dr \,- \cancel{\pi r^2} ) L= ( \cancel{\pi \mathrm dr^2}+2 \pi r \mathrm dr ) L \approx 2 \pi L r \mathrm dr \end{align}

Note that an insignificantly small, negligible term like $\mathrm dr$ becomes insignificantly insignificant when squared. It therefor basically disappears (it is much, much, much smaller than any other terms involved), so we set it to zero above, $\mathrm dr\rightarrow 0$.

Let’s plug this in and integrate from $r_1$ to $r_2$ in order to get the influence from all cylinder sheets added together to make up the entire cylinder wall:

$$I= \int _{r_1}^{r_2} r^2 \frac{m}{V} \underbrace{ 2 \pi L r \mathrm dr }_{\mathrm dV}= 2\pi \frac{m}{V} L \int _{r_1}^{r_2} r^3 \mathrm dr= 2\pi \frac{m}{V} L \left[\frac14 r^4\right] _{r_1}^{r_2} = 2\pi \frac{m}{V} L \left(\frac14 r_2^4-\frac 14 r_1^4\right)$$

The total volume $V$ is naturally just the total area $A$ times the length $L$, and that area $A$ is again simply the circle area to the outer radius subtracted the circle area to the inner radius:

$$V=AL=(\pi r_2^1-\pi r_1^2)L$$

Plugged in:

\begin{align} I&= 2\pi \frac{m}{ \underbrace{(\pi r_2^2-\pi r_1^2)\cancel L}_V } \cancel L \left(\frac14 r_2^4-\frac 14 r_1^4\right)\\ &= \frac 14 \cdot 2\cancel \pi \frac{m}{ \cancel \pi r_2^2-\cancel \pi r_1^2} \left(r_2^4- r_1^4\right)\\ &= \frac 12\frac{m}{ r_2^2-r_1^2} \left( r_2^4- r_1^4\right)\\ &= \frac 12\frac{m}{ \underbrace{(r_2+r_1)(r_2-r_1)}_{ r_2^2-r_1^2 }} \left( \underbrace{ (r_2^2+r_1^2)(r_2^2-r_1^2)}_{ r_2^4- r_1^4 } \right) \quad\leftarrow \text{A sum times a difference, proof 15}\\ &= \frac 12\frac{m}{ \cancel{(r_2+r_1)(r_2-r_1)}} \left( (r_2^2+r_1^2) \underbrace{ \cancel{(r_2+r_1)(r_2-r_1)}}_{r_2^2-r_1^2}\right)\quad\leftarrow \text{A sum times a difference}\\ &= \frac 12 m \left( r_2^2+r_1^2\right)\qquad \blacksquare\\ \end{align}

Solid cylinder spinning about its axis

A solid cylinder is a holed cylinder where the “hole” in the middle has reduced to a radius of zero, $r_1=0$. We will then simply call the outer radius $r$ instead of $r_2$:

$$I= \frac 12 m \left( \underbrace{r_2^2}_{r^2}+\underbrace{r_1^2}_{0^2}\right)= \frac 12 m r^2 \qquad \blacksquare$$

Hollow (thin-walled) cylinder spinning about its axis

A hollow cylinder is a holed cylinder where the hole is almost as large as the cylinder itself. Meaning, the inner radius $r_1$ is basically the same as the outer diameter $r_2$. Then $r_1=r_2$ are equal, and we can simply call them both $r$:

$$I= \frac 12 m \left( \underbrace{r_2^2}_{r^2}+\underbrace{r_1^2}_{r^2}\right)= \frac 12 m (r^2+r^2)= \cancel{\frac 12} m \cancel 2r^2=mr^2 \qquad \blacksquare$$

## Proof 28: Plate’s moment-of-inerta

Plate rotating perpendicularly about its centre-of-mass With a single point having the moment-of-inertia $I_\text{point}=r^2m$ ($m$ and $r$ are mass and distance from the axis-of-rotation), infinitely…