# Proof 27: Rod’s moment-of-inertia

**A rod 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),^{1} infinitely many “particles” making up a rotating object gives us the infinite sum, written as an integral:

$$I=\int r^2\,\mathrm dm= \int \underbrace{r_x^2+r_y^2+r_z^2}_{r^2}\,\mathrm dm= \int r_x^2\,\mathrm dm + \int r_y^2\,\mathrm dm + \int r_z^2\,\mathrm dm $$

Rewritten with Pythagoras’ Theorem to its components along with the rod, $r_z$, and perpendicular to the rod, $r_x$ and $r_y$.

Let’s saw the rod into many very thin slices. There is perfect symmetry for each of these circular slices around the centre. This means that each point on one side has an equal point on the opposite side. Such pairs of opposite points will have a negative and a positive $r_x$ value. Same for $r_y$. When summed up, the total $r_x$ and $r_y$ will thus cancel out:

$$I=\int r^2\,\mathrm dm=\cancel{ \int r_x^2\,\mathrm dm} + \cancel{\int r_y^2\,\mathrm dm} + \int r_z^2\,\mathrm dm= \int r_z^2\,\mathrm dm $$

Each slice is a small piece $\mathrm dm$ of the total mass $m$ with a small width $\mathrm dr_z$ out of the total length $L$. The mass is uniform (evenly spread), so $\mathrm dm$ covers the same amount of the total mass $m$ as $\mathrm d r_z$ covers of the total length $L$. Meaning, they are both the same fraction of the total:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dr}{L}\quad\Leftrightarrow \quad \mathrm dm= m\frac{\mathrm dr_z}{L} = \frac{m}{L} \mathrm dr_z $$

Plug this in:

$$I=\int r_z^2\,\mathrm dm= \int r_z^2\, \frac{m}{L} \;\mathrm dr_z $$

We now integrate over the full length $L$, meaning $\frac 12L$ on either side from the axis-of-rotation. So, from $-\frac 12L$ to $\frac 12L$:

$$\begin{align}

I&= \int_{-\frac 12L}^{\frac 12L} r_z^2 \frac{m}{L} \;\mathrm dr_z = \left[\frac 13 r_z^3 \frac{m}{L}\right]_{-\frac 12L}^{\frac 12L} = \frac 13 \left(\frac 12L\right)^3 \frac{m}{L}- \frac 13 \left(-\frac 12L\right)^3 \frac{m}{L} \\

&=\frac 13 \frac 18L^3 \frac{m}{L}+ \frac 13 \frac 18L^3 \frac{m}{L}= \frac 1{24}L^2 m+ \frac 1{24}L^2 m = 2\frac 1{24}L^2 m= \frac 1{12}mL^2 \qquad\blacksquare

\end{align}$$

**A rod rotating perpendicularly about an axis at its end**

This axis is parallel to the axis through the centre-of-mass. With the parallel-axis theorem:

$$\begin{align}

I_\text{end}&= I_\text{com} +md^2= \frac 1{12}mL^2 +m\left(\frac12 L\right)^2\\

&= \frac 1{12}mL^2 +m\frac14 L^2 = \frac 1{12}mL^2 +m\frac3{12} L^2\\

&= \frac 4{12}mL^2= \frac 13mL^2 \qquad\blacksquare

\end{align}$$

## Responses