# 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:

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

## 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),…

## 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…

## Proof 23: Cross-product

$$\vec c=\vec a \times \vec b$$ A cross-product is defined with two features: It’s magnitude $c$ is the perpendicular part of one vector multiplied with…

## Proof 22: Geometric bonds

Angle $\theta$ measured in radians is ‘the number of radius-lengths along the periphery’. Multiply that number with the radius $r$ to get the total length…

## Proof 14: Motion equations

In motion along a path, $a$ is acceleration, $v_0$ and $v$ are initial and final (current) speed, and $t_0$ and $t$ are initial and final…