What is uniform circular motion? By using a proper figure,derive the equation for centripetal acceleration $a_c = \frac{v^2}{r}$ for uniform circular motion. Show that its direction is towards the center.

(N/A) Uniform circular motion is the motion of an object traveling at a constant speed along a circular path.
Consider an object moving with a constant speed $v$ in a circle of radius $r$. Since the direction of velocity changes continuously,the object undergoes acceleration.
Let $\vec{r}$ and $\vec{r}'$ be the position vectors,and $\vec{v}$ and $\vec{v}'$ be the velocity vectors of the object at points $P$ and $P'$ respectively.
The velocity vector at any point is tangent to the path at that point.
From the triangle law of vector addition,the change in velocity $\Delta \vec{v} = \vec{v}' - \vec{v}$ is shown in figure $(a_2)$.
Since the path is circular,$\vec{v}$ is perpendicular to $\vec{r}$,and $\vec{v}'$ is perpendicular to $\vec{r}'$. Thus,$\Delta \vec{v}$ is perpendicular to $\Delta \vec{r}$.
Since the average acceleration $\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$,the direction of $\vec{a}$ is the same as $\Delta \vec{v}$.
As $\Delta t \rightarrow 0$,the triangle formed by $\Delta \vec{v}$ becomes similar to the triangle formed by $\Delta \vec{r}$.
From the similarity of triangles,$\frac{|\Delta \vec{v}|}{v} = \frac{|\Delta \vec{r}|}{r}$.
Dividing by $\Delta t$,we get $\frac{|\Delta \vec{v}|}{\Delta t} = \frac{v}{r} \frac{|\Delta \vec{r}|}{\Delta t}$.
As $\Delta t \rightarrow 0$,$|\Delta \vec{r}| \approx v \Delta t$,so $a_c = \lim_{\Delta t \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v}{r} (v) = \frac{v^2}{r}$.
Thus,the acceleration is directed towards the center.