For a particle in uniform circular motion, the acceleration $\overline{a}$  at any point  $P(R,\theta )$ on the circular path of radius R (when $\theta$ is measured from the + ve x-axis and v is uniform speed) is :  

$$\begin{gathered} \overrightarrow{r}=R\cos \theta \hat{j} \\ \therefore \frac{d\overrightarrow{r}}{dt}=-R\sin \theta \frac{d\theta }{dt}\hat{i}+R\cos \theta \frac{d\theta }{dt}\hat{j} \\ \Rightarrow \overrightarrow{V}=-R\omega \sin \theta \hat{i}+R\omega \cos \theta \hat{j} \end{gathered}$$

(as  $\theta =\frac{d\theta }{dt}$  and  $\frac{d\overrightarrow{r}}{dt}=\overrightarrow{V}$ )

$$\begin{gathered} \Rightarrow \frac{d\overrightarrow{V}}{dt}=\overrightarrow{a}=R\omega \cos \theta \frac{d\theta }{dt}\hat{i}+R\omega (-\sin \theta )\frac{d\theta }{dt}\hat{j} \\ \Rightarrow \frac{ds}{dt}=\frac{d\theta r}{dt}=r\frac{d\theta }{dt} \\ =r\omega \end{gathered}$$

$\therefore \omega$  or  $\frac{d\theta }{dt}$  is constant 
 $\therefore \overrightarrow{V}=-R{\omega }^2\cos \theta \hat{i}-R{\omega }^2\sin \theta \hat{j}$

Substituting   $\omega =\frac{V}{R}$

$\overrightarrow{a}=\frac{-v^2}{R}\cos \theta \hat{i}-\frac{v^2}{R}\sin \theta \hat{j}$