# What is constant in uniform circular motion When we talk about kinematics of a point particle, mainly the velocity and acceleration vectors, it would be reasonable to use differential geometry to understand what is going on. The motion of a point particle is described fully if it's given its position vector $\:\mathbf{r}(t)\:$ as function of time $\:t\:$. Then the velocity vector $\:\mathbf{v}(t)\:$ and the acceleration vector $\:\mathbf{a}(t)\:$ are the 1rst and 2nd derivative of the position vector $\:\mathbf{r}(t)\:$ with respect to $\:t\:$. Nothing more, nothing less.

\begin{equation} \mathbf{v}\left(t\right)=\dfrac{\mathrm{d}\mathbf{r}\left(t\right)}{\mathrm{d}t} \tag{01} \end{equation}

\begin{equation} \mathbf{a}\left(t\right)=\dfrac{\mathrm{d}\mathbf{v}\left(t\right)}{\mathrm{d}t}=\dfrac{\mathrm{d}^{2}\mathbf{r}\left(t\right)}{\mathrm{d}t^{2}} \tag{02} \end{equation}

An example is the plane motion of a point particle as in Figure. The particle describes first from left a semi-circle with uniform circular motion, then in the middle a semi-ellipses and finally to the right a semi-circle with nonuniform circular motion. Its position vector is

\begin{equation} \mathbf{r}\left(t\right) = \begin{cases} \left[-7-3 \cos\left( \dfrac{t}{2} \right), \;3 \sin \left( \dfrac{t}{2} \right) \right] & t \in [ 0,2\pi) \\ \left[4\cos\left( \dfrac{t}{2} \right), \;3 \sin \left( \dfrac{t}{2} \right) \right] & t \in [ 2\pi,4\pi)\\ \left[7-3 \cos \left(\dfrac{t^2-16 \pi^2}{16\pi}\right), \;3 \sin \left(\dfrac{t^2-16 \pi^2}{16\pi}\right) \right] & t \in [ 4\pi,\sqrt{32}\pi] \end{cases} \tag{03a} \end{equation}

The velocity vector is

\begin{equation} \mathbf{v}\left(t\right) = \begin{cases} \dfrac{3}{2}\left[ \sin\left( \dfrac{t}{2} \right), \; \cos \left( \dfrac{t}{2} \right) \right] & t \in [ 0,2\pi) \\ \dfrac{1}{2}\left[-4 \sin\left( \dfrac{t}{2} \right), \; 3\cos \left( \dfrac{t}{2} \right) \right] & t \in [ 2\pi,4\pi) \\ \dfrac{3t}{8\pi}\left[ \sin \left(\dfrac{t^2-16 \pi^2}{16\pi}\right), \; \cos \left(\dfrac{t^2-16 \pi^2}{16\pi}\right) \right] & t \in [ 4\pi,\sqrt{32}\pi] \end{cases} \tag{03b} \end{equation}

while the acceleration vector is

\begin{equation} \mathbf{a}\left(t\right) = \begin{cases} \dfrac{3}{4}\left[ \cos\left( \dfrac{t}{2} \right), \;- \sin \left( \dfrac{t}{2} \right) \right] & t \in [ 0,2\pi) \\ \dfrac{1}{4}\left[ -4\cos\left( \dfrac{t}{2} \right), \;-3 \sin \left( \dfrac{t}{2} \right) \right] & t \in [ 2\pi,4\pi) \\ \ \dfrac{3}{8\pi}\left[ \sin F\left(t\right) +\dfrac{t^2}{8\pi} \cos F\left(t\right) ,\; \cos F\left(t\right) -\dfrac{t^2}{8\pi} \sin F\left(t\right) \right] & t \in [ 4\pi,\sqrt{32}\pi] \end{cases} \tag{03c} \end{equation} where in the 3rd case of above equation (03c) and in order to shorten its length \begin{equation} F\left(t\right) \equiv \dfrac{t^2-16 \pi^2}{16\pi} \tag{04} \end{equation}

The acceleration vector is analyzed in two components :

1. The orbital component $\: \mathbf{a}_{\Vert}\:$ tangent to the curve, that is parallel to the velocity vector $\:\mathbf{v}\:$ and responsible only for the changes of the magnitude this vector.
2. The centripetal component $\: \mathbf{a}_{\perp}\:$ normal to the curve, that is normal to the velocity vector $\:\mathbf{v}\:$ and responsible only for the changes of the direction of this vector.

In this example and given the velocity and acceleration vector diagrams for each case :

1. On the left semi-circle the particle executes uniform circular motion. The velocity vector has constant magnitude (speed), the acceleration vector has constant magnitude and is always normal to the velocity. There is no orbital acceleration component.

2. On the middle semi-ellipses the velocity vector is changing continuously both in direction and magnitude because of the centripetal and orbital components of the acceleration respectively. The acceleration vector is changing also continuously both in direction and magnitude.

3. On the right semi-circle the particle executes nonuniform circular motion. Both vectors are changing continuously both in direction and magnitude.

Note that in the example the velocity vector is a continuous function of time, while the acceleration vector is discontinuous at points $\:\mathrm{E}\:$ and $\:\mathrm{I}\:$.

EDIT

The well known equation for the magnitude of the centripetal acceleration \begin{equation} \Vert\mathbf{a}_{\perp}\Vert = \dfrac{\Vert\mathbf{v}\Vert^2}{\rho}= \dfrac{v^2}{\rho} \tag{05} \end{equation} is valid not only for circular motion but for every curvilinear regular motion where $\:\rho\:$ the radius of curvature . In Figure below see details around point F of Figure above. $\endgroup$