Question

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

• A. $-\frac{v^2}{R}\sin \theta \hat{i}+\frac{v^2}{R}\cos \theta \hat{j}$
• B. $-\frac{v^2}{R}\cos \theta \hat{i}+\frac{v^2}{R}\sin \theta \hat{j}$
• 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)

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

$\vec{a}=-a\cos \theta \hat{i}-a\sin \theta \hat{j}$
$=-\frac{V^2}{R}\cos \theta \hat{i}-\frac{V^2}{R}\sin \theta \hat{j}$