# Proof 26: Parallel-axis theorem

When rotating about the centre-of-mass, the moment-of-inertia $I_\text{com}$ is often easier to find. When the axis-of-rotation is somewhere else, we’ll below derive a formula that still uses $I_\text{com}$. Note three things:

- Each “particle” that makes up a rotating object has a position $\vec s=(s_x,s_y)$.
- In the the formula $I=\sum mr^2$,
^{1}$r$ is the distance not from the coordinate-system origin but from*the new*axis. - And this new axis is located at a position $\vec d=(d_x,d_y)$

The horizontal distance $r_x$ between axis and “particle” must thus be the total distance from origin to “particle” subtracted the distance from origin to axis: $r_x=s_x-d_x$. Same for vertical distance: $r_y=s_y-d_y$ (and same for the z-direction if we are in 3D). Let’s start the derivation:

$$\begin{align}

I&=\sum mr^2\\

&= \sum m\left(r_x^2+r_y^2\right)\qquad \leftarrow \text{Pythagoras Theorem, } r_x^2+r_y^2=r^2\\

&= \sum m\left(\underbrace{(s_x-d_x)^2}_{r_x}+\underbrace{(s_y-d_y)^2}_{r_y}\right)\\

&= \sum m\left(\underbrace{s_x^2+d_x^2-2s_xd_x}_{ (s_x-d_x)^2 }+\underbrace{s_y^2+d_y^2-2s_yd_y}_{ (s_y-d_y)^2 }\right) \qquad \leftarrow \text{Square of binomial}\\

&= \sum m s_x^2+ \sum m d_x^2- \sum m 2s_xd_x+ \sum m s_y^2+ \sum m d_y^2- \sum m 2s_yd_y\\

&= \sum (m s_x^2+ m s_y^2 )+ \sum (m d_x^2+ m d_y^2 )- 2\sum m s_xd_x-2\sum m s_yd_y\\

&= \sum m \underbrace{ (s_x^2+ s_y^2)}_{s^2}+ \sum m \underbrace{(d_x^2+ d_y^2 )}_{d^2}- 2\sum m s_xd_x-2\sum m s_yd_y \\

&= \underbrace{\sum m s^2}_{I_\text{com}}+ d^2\underbrace{\sum m}_{m_\text{total}} \,- 2\sum m s_xd_x-2\sum m s_yd_y \qquad \leftarrow d^2 \text{ is constant}\\

&= I_\text{com}+ m_\text{total}d^2 \,- 2 d_x \sum m s_x-2 d_y \sum m s_y \qquad \leftarrow d_x \text{ and } d_y \text{ are constant} \\

&= I_\text{com}+ m_\text{total}d^2 \,- 2 d_x \sum (m s_x) \cdot \frac{\sum m}{ \sum m } -2 d_y \sum (m s_y) \cdot \frac{\sum m}{ \sum m } \qquad \leftarrow \text{Multiplying with 1} \\

&= I_\text{com}+ m_\text{total}d^2 \,- 2 d_x \underbrace{\frac{\sum m s_x}{ \sum m }}_{x_\text{com}} \cdot \sum m -2 d_y \underbrace{\frac{\sum m s_y}{ \sum m }}_{y_\text{com}} \cdot \sum m \\

&= I_\text{com}+ m_\text{total}d^2 \,- 2 d_x x_\text{com} \cdot \sum m -2 d_y y_\text{com} \cdot \sum m \\

\end{align} $$

If we place our coordinate system with its origin at the centre-of-mass, then the coordinates to the centre-of-mass will of course be zero $( x_\text{com} , y_\text{com} )=(0,0)$:

$$\begin{align}

I&= I_\text{com}+ m_\text{total}d^2 \,- 2 d_x \cancel{x_\text{com}} \cdot \sum m -2 d_y \cancel{y_\text{com}} \cdot \sum m \\

&= I_\text{com}+ m_\text{total}d^2

\end{align} $$

If the symbol for the total mass is simply $m$, then the formula gets its short form:

$$I=I_\text{com}+ md^2\qquad\blacksquare$$

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