# Proof 20: Circular motion

We will prove that inwards (radial/centripetal) acceleration is squared speed over radius:

$$ a_\perp=\frac{v^2}r $$

When driving around a roundabout (a circle) with constant speed, you have an inwards acceleration $a_\perp$ ($\perp$, because it is sideways/perpendicular to the driving path),^{1} which causes a brief inwards speed component $\mathrm dv_\perp$ at every moment. It is very, very brief – in fact infinitely small, because it only happens infinitely shortly before the turning has happened – so we chose to symbolise it with $\mathrm d$.

While this brief change in sideways speed happens, we are still moving. The distance moved in this brief moment is also basically infinitely small, so we’ll write it as $\mathrm d s_\parallel$ ($\parallel$, because this distance is along with/parallel to the path).

Do these two changes follow each other? Yes, because both before and after this brief change, the *velocity* has the same angle to the *position*. Both turn equally much. If one turns so that its “turning” component is, say, a tenth of the other component (if $\mathrm d v_\perp=\frac1{10} v_\parallel$), then the other must also change by a tenth of its other component (if $\mathrm ds_\parallel=\frac1{10} s_\perp$) in order to turn the same angle. The proportions between $\frac{\mathrm ds_\parallel}{s_\perp}$ and $\frac{\mathrm dv_\perp}{v_\parallel}$ are apparently the same:

$$ \frac{\mathrm ds_\parallel}{s_\perp} = \frac{\mathrm dv_\perp}{v_\parallel} \quad\Leftrightarrow\quad \mathrm dv_\perp=\mathrm d s_\parallel \frac{v_\parallel}{s_\perp}$$

The inwards acceleration causing the turning can now be reshaped, starting from its definition:

$$a_\perp=\frac{\mathrm dv_\perp}{ \mathrm dt}= \frac{\mathrm ds_\parallel \frac{v_\parallel}{s_\perp} }{ \mathrm dt} = \underbrace{\frac{\mathrm ds_\parallel}{ \mathrm d t}}_{v_\parallel}\frac{v_\parallel}{s_\perp}=v_\parallel\frac{v_\parallel}{s_\perp}= \frac{v_\parallel^2}{s_\perp} =\frac{v^2}r$$

Since $s_\perp$ is distance from the centre to any position at the circle, we here replaced it with the radius $r$. Also, $v_\parallel$ is the actual speed we drive with along our path ($\mathrm dv_\perp$ is so tiny, that the other component $v_\parallel$ basically is the entire speed), so we just replaced it with $v$. $\quad_\blacksquare$

References:

- ‘
**Sears and Zemansky’s Univesity Physics with Modern Physics**’ (book),*Hugh D. Young & Roger A. Freedman*, Pearson Education, 13th ed., 2012 - ‘
**Online Etymology Dictionary**’ (dictionary),*Douglas Harper*, www.etymonline.com

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