Question

For a particle in uniform circular motion, the acceleration $\vec{a}$ at a point $P(R, \theta)$ 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. $-\frac{v^2}{R} \sin \theta \hat{i}+\frac{\dot{v}^2}{R} \cos \theta \hat{j}$
• 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)

Clearly
$\vec{a}=a_c \cos \theta(-\hat{i})+a_c \sin \theta(-\hat{j}) $
$=\frac{-v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}$