For a particle in uniform circular motion, the acceleration $\vec{a}$ at any point $P (R, θ)$ on the circular path of radius R is (when $θ$ is measured from the positive $x$ -axis and v is uniform speed) :

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$- \frac{v^{2}}{R} \sin θ \hat{i} + \frac{v^{2}}{R} \cos θ \hat{j}$

$- \frac{v^{2}}{R} \cos θ \hat{i} - \frac{v^{2}}{R} \sin θ \hat{j}$

$- \frac{v^{2}}{R} \hat{i} + \frac{v^{2}}{R} \hat{j}$

$- \frac{v^{2}}{R} \cos θ \hat{i} + \frac{v^{2}}{R} \sin θ \hat{j}$

answer is C.

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Detailed Solution

$a = | \vec{a} | = \frac{V^{2}}{R}$

$\begin{aligned} \vec{a} = - a \cos θ \hat{i} - a \sin θ \hat{j} \\ = - \frac{V^{2}}{R} \cos θ \hat{i} - \frac{V^{2}}{R} \sin θ \hat{j} \end{aligned}$

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