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Derivation of the centripetal acceleration

To derive the centripetal acceleration of an object in uniform circular motion, we consider an object moving on a circle of radius $r$. Let $v$ is the speed of the object, which is a constant.

velocity-vectors-at-two-points

Now, let us take two points on the circle, P and Q separated by an angle $\theta$ as in the figure above. Let $\Delta t$ is the time the object takes to move from P to Q. The velocity of the object is tangent to the circle at any moment of time. So, at the points, P and Q, the velocity vectors are as shown. Let us take, $\vec v_1$ is the velocity of the object at the point P and $\vec v_2$ is the velocity at the point Q.

Next, move the vector $\vec v_2$ and connect its tail with the tail of the vector $\vec v_1$. From the geometry of the first figure, you get the angle between the two vectors as $\theta$. Now, draw a vector (the red arrow) from the tip of the vector, $\vec v_1$ to the tip of the vector, $\vec v_2$.

velocity-vectors-change

According to the tip-to-tail method, the vector representing the red arrow is $\vec v_2-\vec v_1$, which is the change in velocity $\Delta \vec v$.

Since the speed is constant, $|\vec v_1|=v$ and $|\vec v_2|=v$. So, considering only the magnitudes of the vectors, we get the following triangle,

velocity-magnitudes-triangle

From the first figure, we have another triangle,

distances-triangle

The above two triangles are similar triangles. So, the ratio of the corresponding sides should be the same. Therefore, we can write

$\dfrac{\Delta v}{PQ}=\dfrac{v}{r}$

Solving for $v$,

$\Delta v=\dfrac{v\,PQ}{r}$

Dividing by $\Delta t$, we get

$\dfrac{\Delta v}{\Delta t}=\dfrac{v}{r}\dfrac{PQ}{\Delta t}$

The left hand side is the average acceleration, $\bar a$, and PQ is the displacement of the object in a time, $\Delta t$. So, $\dfrac{PQ}{\Delta t}$ is the average velocity, $\bar v$ of the object. Substituting these, we get

$\bar a =\dfrac{v}{r}\,\bar v$

This is the average acceleration of the object over the time interval, $\Delta t$. We are interested in the instantaneous acceleration. To get the instantaneous acceleration, we just need to take a small time interval because when $\Delta t$ is small, the average values become instantaneous values.

i.e., when $\Delta t \rightarrow\:0$, $\bar a =a$ and $\bar v =v$. And we get,

$a=\dfrac{v}{r}\,v$

 $=\dfrac{v^2}{r}$

This is the centripetal acceleration.