Question

For a particle in uniform circular motion the acceleration $\vec{a}$ at a point $P\left(R,\,\theta\right)$ on the circle of radius $R$ 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. $\frac{v^{2}}{R} \hat{i} + \frac{v^{2}}{R} \hat{j}$

Answer 1

Answer: (C)

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

Answer 2

For a particle in uniform circular motion, $\vec{a} = \frac{v^{2}}{R}$ towards centre of circle $\therefore\quad\vec{a} = -\frac{v^{2}}{R} \left(cos\,\theta \,\hat{i} - sin\,\theta \,\hat{j}\right)$ or$\quad\vec{a} = -\frac{v^{2}}{R} cos\,\theta \,\hat{i} - \frac{v^{2}}{R} sin\,\theta \,\hat{j}$