A particle is moving in a circle of radius R with a constant speed v. Its average acceleration over the time when it moves over half the circle is :

Concept:

The acceleration of a particle moving in a circle with constant speed is always directed toward the center of the circle. This is called centripetal acceleration, and its magnitude is given by \(a_c=\frac{v^2}{R}\) , where 
v is the speed of the particle, and R is the radius of the circle.

When considering the average acceleration over the time it takes for the particle to move over half the circle, we need to consider the change in velocity. Since velocity is a vector, we'll need to think about the direction as well.

If the particle has a constant speed, then the only change in the velocity vector is in its direction. Over half the circle, the direction of the velocity vector changes by 180°. 

The change in velocity, Δv, will have the same magnitude as the original velocity, but in the opposite direction. So the average acceleration, a̅, over the time interval T/2, where T is the period of the circular motion, is: 

\(\bar a=\frac{\Delta v}{T/2}=\frac{2v}{T}\)

We can relate T to v and R by using the fact that the distance covered in one period is the circumference of the circle, 2πR:

But the particle moves over half the circle, so the circumference of the circle, πR:

\(T=\frac{\pi R}{v}\)

Substituting this back in, we have:

\(\bar a=\frac{2v}{\pi R/v}=\frac{2v^2}{\pi R}\)

So the average acceleration over the time when it moves over half the circle is \(\bar a=\frac{2v^2}{\pi R}\)