\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Arc-length

Given the path of particle $\boldsymbol{r}(t)$, where t $\in [t_i,t_f]$, the arc-length of the path is defined as, \(\begin{align} s(t) = \int_{t_i}^{t_f} \|\boldsymbol{v} \| (\xi) d\xi \end{align}\)

The $s$ above is the path covered by the particle between $t = t_i$ and $t$.

Speed

The rate of change of the arc-length is defined as speed $v$. Using the Leibnitz rule of differentiation, it is given as

\[\begin{align} \frac{ds(t)}{dt} &= v(t) = \frac{d}{dt}\left(\int_{t_i}^{t_f} \|\boldsymbol{v} \| (\xi) d\xi \right) \\ &= \|\boldsymbol{v} \|(t) \end{align}\]

**The speed is equal to the magnitude of the velocity vector.

Tangent vector $\boldsymbol{e}_t$

The tangent vector is defined as,

\[\begin{align} \boldsymbol{e}_t = \frac{d\boldsymbol{r}}{ds} \end{align}\]

Also, we have

\[\begin{align} \boldsymbol{e}_t &= \frac{d\boldsymbol{r}}{dt}\frac{dt}{ds} = \frac{d\boldsymbol{r}}{dt}\frac{1}{ds/dt} \\ &= \frac{\boldsymbol{v}}{v} \end{align}\]

Therefore, $\boldsymbol{e}_t$ is of unit magnitude. The $\boldsymbol{e}_t$ vector points in the direction of the velocity. We know that velocity vector is tangential to path.

Using (5), we can write

\[\begin{align} \boldsymbol{v} = v\boldsymbol{e}_t \end{align}\]

Radius of curvature

The radius of curvature of the path is defined as,

\[\begin{align} \frac{1}{\rho} = \left\|\frac{d\boldsymbol{e}_t}{ds}\right\| \end{align}\]

Normal vector $\boldsymbol{e}_n$

We define the normal vector $\boldsymbol{e}_n$ as the unit vector in the direction of $d\boldsymbol{e}_t/ds$.

\[\begin{align} \boldsymbol{e}_n = \frac{d\boldsymbol{e}_t}{ds}\frac{1}{\left\|\frac{d\boldsymbol{e}_t}{ds}\right\|} \end{align}\]

From the above equation, we can write

\[\begin{align} \boldsymbol{e}_n = \rho\frac{d\boldsymbol{e}_t}{ds} \end{align}\]

To show $\boldsymbol{e}_t\cdot\boldsymbol{e}_n = 0$

\(\begin{align} \boldsymbol{e}_t \cdot \boldsymbol{e}_t &=1 ,\\ \frac{d}{ds}(\boldsymbol{e}_t \cdot \boldsymbol{e}_t) &= \frac{d}{ds}[1],\\ \frac{d\boldsymbol{e}_t}{ds} \cdot \boldsymbol{e}_t + \boldsymbol{e}_t \cdot \frac{d\boldsymbol{e}_t}{ds} &= 0,\\ 2\frac{d\boldsymbol{e}_t}{ds} \cdot \boldsymbol{e}_t &= 0, \\ \frac{\boldsymbol{e}_n}{\rho} \cdot \boldsymbol{e}_t &= 0 ,\\ \implies \boldsymbol{e}_n \cdot \boldsymbol{e}_t &= 0 . \end{align}\)

Example

Consider the path

\[\begin{align} \boldsymbol{r}(t) = R(\cos[\omega t]\boldsymbol{e}_1 + \sin[\omega t]\boldsymbol{e}_2) \end{align}\]

Calculating all the required:

\[\begin{align} \boldsymbol{v}(t) &= R\omega(-\sin[\omega t]\boldsymbol{e}_1 + \cos[\omega t]\boldsymbol{e}_2) \\ v &= R\omega \\ s &= R\omega t \\ \boldsymbol{e}_t(t) &= -\sin[\omega t]\boldsymbol{e}_1 + \cos[\omega t]\boldsymbol{e}_2 \end{align}\]

Using the relation between $s$ and time $t$ as in (20) , we have that,

\[\begin{align} \boldsymbol{e}_t(s) &= -\sin[s/R]\boldsymbol{e}_1 + \cos[s/R]\boldsymbol{e}_2, \\ \frac{d\boldsymbol{e}_t}{ds}(s) &= -\frac{1}{R}(\cos[s/R]\boldsymbol{e}_1 + \sin[s/R]\boldsymbol{e}_2) ,\\ \frac{d\boldsymbol{e}_t}{ds}(s) &= -\frac{1}{R}(\cos[s/R]\boldsymbol{e}_1 + \sin[s/R]\boldsymbol{e}_2) ,\\ \left\|\frac{d\boldsymbol{e}_t}{ds}\right\| &= \frac{1}{R}. \end{align}\]

We therefore get that $\rho = R$, that is the radius of curvature is equal to the radius of the circle on which the particle moves.

Velocity in terms of $\boldsymbol{e}_t$ and $\boldsymbol{e}_n$

\[\begin{align} \boldsymbol{v} = v\boldsymbol{e}_t \end{align}\]

Differentiating the above equation, we get

\[\begin{align} \dot{\boldsymbol{v}} &= \dot{v}\boldsymbol{e}_t + v \dot{\boldsymbol{e}_t} \\ &= \dot{v}\boldsymbol{e}_t + v \frac{d\boldsymbol{e}_t}{ds}v \\ &= \dot{v}\boldsymbol{e}_t + \frac{v^2}{\rho} \boldsymbol{e}_n \end{align}\]