Question 175 of 183

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

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

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

Acceleration of a particle in uniform circular motion is directed towards the centre of the circle and its magnitude is $\dfrac{{v}^{2}}{R}$. The radius vector is$\vec{r}=R\cos \theta \hat{i}+R\sin \theta \hat{j}$
Therefore, $\vec{a}=-\dfrac{{v}^{2}}{R}\hat{r}=\dfrac{{v}^{2}}{R}\left(-\cos \theta \hat{i}-\sin \theta \hat{j}\right)$