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

(C) $-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}$

(A) $-\frac{\mathrm{v}^{2}}{\mathrm{R}} \sin \theta \hat{\mathrm{i}}+\frac{\mathrm{v}^{2}}{\mathrm{R}} \cos \theta \hat{\mathrm{j}}$

(B) $-\frac{\mathrm{v}^{2}}{\mathrm{R}} \cos \theta \hat{\mathrm{i}}+\frac{\mathrm{v}^{2}}{\mathrm{R}} \sin \theta \hat{\mathrm{j}}$

(D) $-\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Kinematics

For a particle in uniform circular motion, the acceleration

a

at any point

P (R, θ)

on the circular path of radius

R

is (when

θ

is measured from the positive x -axis and v is uniform speed) :

(A) $- \frac{v ^{2}}{R} sin θ \hat{i} + \frac{v ^{2}}{R} cos θ \hat{j}$
(B) $- \frac{v ^{2}}{R} cos θ \hat{i} + \frac{v ^{2}}{R} sin θ \hat{j}$
(C) $- \frac{v ^{2}}{R} cos θ \hat{i} - \frac{v ^{2}}{R} sin θ \hat{j}$ ✓ Correct
(D) $- \frac{v ^{2}}{R} \hat{i} + \frac{v ^{2}}{R} \hat{j}$

Answer: (C)

In uniform circular motion, the centripetal acceleration has magnitude

\frac{v ^{2}}{R}

and points toward the center. At point

P (R, θ)

, the position vector makes angle

θ

with the x-axis, so the acceleration (pointing toward the origin) is

a = - \frac{v ^{2}}{R} cos θ \hat{i} - \frac{v ^{2}}{R} sin θ \hat{j}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option