question_answer

For a particle in uniform circular motion, the acceleration\[\overrightarrow{a}\]at a point\[P(R,\text{ }\theta )\]on the circle of radius R is (Here\[\theta \]is measured from the x-axis)                                            [AIEEE 2010]

A) \[\frac{{{v}^{2}}}{R}\hat{i}+\frac{{{v}^{2}}}{R}\hat{j}\]                

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

C) \[-\frac{{{v}^{2}}}{R}sin\theta \hat{i}+\frac{{{v}^{2}}}{R}\cos \theta \hat{j}\]           

D) \[-\frac{{{v}^{2}}}{R}\cos \theta \hat{i}-\frac{{{v}^{2}}}{R}sin\theta \hat{j}\]

Correct Answer: D

Solution :

[d]

\[\overrightarrow{a}={{a}_{c}}\cos \theta (-\hat{i})+{{a}_{c}}\sin \theta (-\hat{j})\]

\[\overrightarrow{a}=-\frac{{{V}^{2}}}{R}\cos \theta \hat{i}-\frac{{{V}^{2}}}{R}\sin \theta \hat{j}\]