Question : For a particle in uniform circular motion the acceleration a at a point π(π , π) on the circle of the radius R is (here π is measured from the π₯ βaxis)
(A) \dfrac{-v^{2}}{R}\cos \theta \widehat{i}+\dfrac{v^{2}}{R}\sin \theta \widehat{j}
(B) \dfrac{-v^{2}}{R}\sin \theta \widehat{i}+\dfrac{v^{2}}{R}\cos \theta \widehat{j}
(C) \dfrac{-v^{2}}{R}\cos \theta \widehat{i}-\dfrac{v^{2}}{R}\sin \theta \widehat{j}
(D) -\dfrac{v^{2}}{R}\widehat{i}+\dfrac{v^{2}}{R}\widehat{j}
Answer : option (C)
Solution :
