For a particle in uniform circular motion the acceleration $\vec{a}$ at a point $P(R, \theta)$ on the circle of radius $R$ is (here $\theta$ is measured from the x-axis)

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

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

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

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

JEE Main · 2010 · 11 Apr 2010 · Offline · Kinematics

For a particle in uniform circular motion the acceleration

a

at a point

P (R, θ)

on the circle of radius

R

is (here

θ

is measured from the x-axis)

(A) $- \frac{V ^{2}}{R} cos θ \hat{i} + \frac{V ^{2}}{R} sin θ \hat{j}$
(B) $- \frac{v ^{2}}{R} sin θ \hat{i} + \frac{v ^{2}}{R} cos θ \hat{j}$
(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 only acceleration is centripetal: its size is

v^{2} / R

and it always points from the particle back to the centre of the circle. At the point

P (R, θ)

the outward radial direction is

cos θ \hat{i} + sin θ \hat{j}

, so the inward direction towards the centre is

- cos θ \hat{i} - sin θ \hat{j}

. Multiplying the magnitude by this inward unit vector gives

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

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option