A particle is moving in a circle of radius R with a constant speed v.

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 :

[amp_mcq option1=”$\frac{v^2}{R}$” option2=”$\frac{\pi v^2}{2R}$” option3=”$\frac{2v^2}{\pi R}$” option4=”0″ correct=”option3″]

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UPSC NDA-1 – 2023

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The correct option is C.

Average acceleration is defined as the change in velocity divided by the time taken. For a particle moving in a circle with constant speed, the velocity vector changes direction continuously. Over half a circle, the initial and final velocity vectors are in opposite directions.

Let the particle start at $\theta=0$ (position vector $R\hat{i}$) and move counter-clockwise. The initial velocity is $\vec{v}_i = v\hat{j}$. After moving half a circle ($\pi$ radians), it is at $\theta=\pi$ (position vector $-R\hat{i}$). The final velocity is $\vec{v}_f = -v\hat{j}$. The change in velocity is $\Delta\vec{v} = \vec{v}_f – \vec{v}_i = -v\hat{j} – v\hat{j} = -2v\hat{j}$. The magnitude of the change in velocity is $|\Delta\vec{v}| = 2v$. The distance covered is half the circumference, $\pi R$. The time taken is $t = (\pi R) / v$. The average acceleration is $\vec{a}_{avg} = \Delta\vec{v}/t = (-2v\hat{j}) / (\pi R / v) = -\frac{2v^2}{\pi R}\hat{j}$. The magnitude of the average acceleration is $|\vec{a}_{avg}| = \frac{2v^2}{\pi R}$.

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