# Proof 22: Geometric bonds

Angle $\theta$ measured in radians is ‘the number of radius-lengths along the periphery’.^{1} Multiply that number with the radius $r$ to get the total length $r\theta$ along the periphery that this angle corresponds to.

If the periphery is geometrically tied/bound to a linear path (such as a wheel driving without slipping over a road), this periphery length is the same as the linear length moved $s$. And so:

$$s=r\theta\qquad\blacksquare$$

This is also true for tiny angles $\mathrm d\theta$ and displacements $\mathrm ds$: $\mathrm ds=r \,\mathrm d\theta $. Divide through with a tiny time change and we have the derivative to time:

$$\mathrm ds=r \,\mathrm d\theta\quad\Leftrightarrow\quad

\underbrace{\frac{\mathrm ds}{\mathrm d t}}_v=r\,\underbrace{\frac{\mathrm d\theta } {\mathrm d t}}_\omega \quad\Leftrightarrow \quad v=r\omega \qquad\blacksquare $$

Similarly, one more derivative:

$$\mathrm dv=r \,\mathrm d\omega\quad\Leftrightarrow\quad

\underbrace{\frac{\mathrm dv}{\mathrm d t}}_a=r\,\underbrace{\frac{\mathrm d\omega} {\mathrm d t}}_\alpha\quad\Leftrightarrow \quad a=r\alpha \qquad\blacksquare $$

Note the assumptions/requirements made in these proofs: The periphery *must* follow the linear surface without slipping, and the linear surface is not moving or accelerating.