Question

A particle moving with a constant speed $ v $ in a circle of radius $ R $ , what is the magnitude of average acceleration after half revolution?
(A) $ \dfrac{{{v^2}}}{{2R}} $
(B) $ \dfrac{{2{v^2}}}{{\pi R}} $
(C) $ \dfrac{{{v^2}}}{R} $
(D) $ \dfrac{{{v^2}}}{{\pi R}} $

Answer 1

Hint : The distance covered after a half revolution is half of the circumference of a circle. After a half cycle, the direction of the velocity has been reversed i.e. now it points in the opposite direction as the initial velocity.

Formula used: In this solution we will be using the following formula;
 $ a = \dfrac{{v - u}}{t} $ where $ a $ is the average acceleration, $ v $ is the final velocity, $ u $ is the initial velocity and $ t $ is the time taken to accelerate from $ u $ to $ v $ .
 $ v = \dfrac{d}{t} $ where $ v $ is the average speed, $ d $ is the distance covered and $ t $ is time.

Complete step by step answer
The particle is said to move at a constant speed only change in direction affects the acceleration.
To calculate the average acceleration, we recall the formula given by
 $ a = \dfrac{{v - u}}{t} $ where $ a $ is the average acceleration, $ v $ is the final velocity, $ u $ is the initial velocity and $ t $ is the time taken.
Now, after a half cycle, the direction of the velocity changes direction to opposite the direction before the half cycle, hence, we can say that initial velocity $ u = - v $ and final velocity $ v = v $
Hence, by substitution, we have that
 $ a = \dfrac{{v - ( - v)}}{t} = \dfrac{{v + v}}{t} $
 $ a = \dfrac{{2v}}{t} $
Now, we also know that
 $ v = \dfrac{d}{t} $
Hence $ t = \dfrac{d}{v} $
Substituting into $ a = \dfrac{{2v}}{t} $
We have
 $ a = 2v \div \dfrac{d}{v} = 2v \times \dfrac{v}{d} $
 $ \Rightarrow a = \dfrac{{2{v^2}}}{d} $
Since it is a half revolution of a circle, the total distance covered is given by
 $ d = \pi R $
Hence,
 $ a = \dfrac{{2{v^2}}}{{\pi R}} $
Hence, the correct answer is B.

Note
Also, one might have been expecting that the centripetal acceleration (which is $ \dfrac{{{v^2}}}{R} $ ) should be equal to the average acceleration, which isn’t the case. This is because, in actuality, the centripetal acceleration is the instantaneous acceleration at any point on the circle. The average acceleration is related to the centripetal acceleration by
 $ {a_{ave}} = \dfrac{1}{t}\int_0^t {\dfrac{{{v^2}}}{r}\hat rdt} $ where $ \hat r $ is the unit vector in the radial direction.