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Radial Acceleration

The velocity of a moving object can be changed by changing its speed or direction. If we consider an object which is moved with non-uniform velocity; for example, a moving train, its velocity is increasing continuously and after sometime it starts to move with constant velocity. These changes in velocity are explained on the basis of acceleration.

In the uniform circular motion, the object is moved with constant velocity with zero acceleration but its direction is changed. The direction of the velocity is in the tangential direction. The second law of newton gives the relation between the exerted force and the direction of the acceleration is in the direction of force. In uniform circular motion, the acceleration direction is towards the centre which is called centripetal or radial acceleration.

The law of newton’s states that radial acceleration depends on the radius of circular path and speed and it is essential for uniform circular motion. It is directed towards radius while tangential acceleration is in the direction of the tangent of the circle. Let’s discuss properties of centripetal acceleration and its mathematical representation.

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Radial Component of Acceleration

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As mentioned above exists when an object moves on a curve. Since the concept is same, for easier understanding let us consider a circular motion.
As the first exercise, let us recall the fundamentals of a circular motion. In case of circular motion, the velocity involved is called angular velocity. An Angular velocity is measured in terms of the angle covered by the object per unit time. Normally, an angular velocity is denoted by the Greek letter $\omega$ .

Hence, average angular velocity is defined as,
$\omega$ =$\frac{\theta}{t}$,where $\theta$ is the angle rotated in the time t

Although the overall motion is circular, at any instant the object has a linear velocity v in a direction that is tangent to the circle of radius, say r, at that point. Let $\iota$ be the actual distance moved by the object along the circumference. As per the geometry of circles,
$\theta$ = $\frac{\iota}{r}$
Therefore, Angular Velocity,
$\omega$ = $\frac{\theta}{t}$ = $\frac{\iota}{(rt)}$ = $[(\frac{\iota}{t})(\frac{1}{r})]$,or,$\omega$ = $\frac{v}{r}$, where V is the linear velocity of the object at any point.

Now, an angular acceleration is defined as the rate of change of angular velocity with respect to time. It is denoted by another Greek letter $\alpha$. Now considering an infinitesimal study,

$\omega$ =$\frac{d\theta}{dt}$
and hence,
$\alpha$ = $\frac{d\omega}{dt}$, or
$\alpha$ = $\frac{d^{2}\theta}{dt^{2}}$.

But an object is subjected to two types of acceleration in a circular motion. Look at the following diagram.

Radial Acceleration

Let's say an object is tied at A, one end of a string OA and rotated keeping the other end O as center. When the string is rotated fast, the string gets completely stressed out and the length of the string becomes the radius of rotation. It means a force is exerted on the object from the center to the end and there by an acceleration ao in a radial direction is faced by the object. To counter this force, a tension force develops in the string acting in the opposite direction. This tension force is called the centripetal force and the acceleration generated on the object is called the radial acceleration and denoted as ar.

Formula

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For finding the equation, let us study the following diagram.

Figure : (i) Figure (ii)
Radial Acceleration Equation
In figure (i), A and B are the positions of an object and the positions are infinitesimally close. Figure (ii) shows the translated vector diagram of the centripetal velocity vectors at A and B. As per the similar triangles property,

$\frac{AB}{OA}$=$\frac{\iota}{r}$

Since A and B are very close we can approximate AB, to the length of the arc AB and hence AB = v*dt
in fig (ii), since A and B are very close, Hence v+dv $\approx$ dv.
Therefore,
$\frac{AB}{OA}$ = $\frac{dv}{v}$

becomes as,
$\frac{(v \times dt)}{(r)}$ =$\frac{dv}{v}$

or, 

$(\frac{dv}{dt})$ = $\frac{(v^2)}{(r)}$

Since $(\frac{dv}{dt})$ is the radial acceleration, we arrive at the formula as,

ar = $\frac{v^{2}}{r}$
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