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)

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

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