JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 7)
For a particle in uniform circular motion, the acceleration $$\overrightarrow a $$ at any point P(R, $$\theta$$) on the circular path of radius R is (when $$\theta$$ is measured from the positive x-axis and v is uniform speed) :
$$ - {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j$$
$$ - {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j$$
$$ - {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$$
$$ - {{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$$
Explanation
_25th_June_Evening_Shift_en_7_1.png)
As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $$\omega$$2R or $${{{v^2}} \over R}$$ directed towards centre so from diagram,
$$\overrightarrow a = {{{v^2}} \over R}\cos \theta ( - \widehat i) + {{{v^2}} \over R}\sin ( - \widehat j)$$
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