Q. For a particle in uniform circular motion the acceleration a at a point $P(R,\theta)$ on the circle of the radius R is (here $\theta$ is measured from the $x -$axis)
A) $- \frac{v^{2}}{R}cos\ \theta\ \widehat{\dot{\mathbf{i}}} + \frac{v^{2}}{R}\sin{\theta\ \widehat{\dot{\mathbf{j}}}}$
B) $- \frac{v^{2}}{R}sin\ \theta\ \widehat{\dot{\mathbf{i}}} + \frac{v^{2}}{R}\cos{\theta\ \widehat{\dot{\mathbf{j}}}}$
C) $- \frac{v^{2}}{R}cos\ \theta\ \widehat{\dot{\mathbf{i}}} - \frac{v^{2}}{R}\sin{\theta\ \widehat{\dot{\mathbf{j}}}}$
D) $- \frac{v^{2}}{R}\ \ \widehat{\dot{\mathbf{i}}} + \frac{v^{2}}{R}\ \widehat{\dot{\mathbf{j}}}$
Explanation
(c) For a particle in uniform circular motion
[image:4_image82.emf]
$\mathbf{a} = \frac{v^{2}}{R} = towards\ center\ of\ circle\ $
$\mathbf{a} = \frac{v^{2}}{R}\left( - \cos\theta\widehat{\dot{\mathbf{i}}} - \sin\theta\ \widehat{\dot{\mathbf{j}}} \right)$
$or\ \mathbf{a} = \frac{v^{2}}{R}\cos\theta\widehat{\dot{\mathbf{i}}} - \frac{v^{2}}{R}\sin{\theta\ \widehat{\dot{\mathbf{j}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.