Question 187 of 280

For a particle in uniform circular motion the acceleration $\vec{a}$ at a point $P(R, \theta)$ on the circle of radius $\mathrm{R}$ is (here $\theta$ is measured from the $x$-axis)

A$-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j}$B$-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j}$C$-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}$D$\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}$

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Answer: (C) $-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}$

For a particle in uniform circular motion,
$$\begin{aligned} & \overrightarrow{\mathrm{a}}=\frac{\mathrm{v}^2}{\mathrm{R}} \text { towards centre of circle } \\ & \therefore \quad \overrightarrow{\mathrm{a}}=\frac{\mathrm{v}^2}{\mathrm{R}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}}) \\ & \text { or } \quad \vec{a}=-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j} \\ & \end{aligned}$$