Q: For a particle in a uniform circular motion, the acceleration at a point P (R, θ) on the circle of radius R is (here, θ is measured from the X-axis)

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

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

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

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

Ans: (c)

Sol: $ \displaystyle a = \frac{v^2}{R} $ (towards the centre of circle )

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

$ \displaystyle -\frac{v^2}{R}cos\theta\hat{i} – \frac{v^2}{R}sin\theta \hat{j} $