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)

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