Verified Solution • 2025–26 Session moderate

QJEEM22Ph0ffb: JEE Main 2022 Physics - Class 11 JEEM22Ph0ffb Solution: For a particle in uniform circular motion, the acceleration at…

For a particle in uniform circular motion, the acceleration a→\overrightarrow a at any point P(R, θ\theta ) on the circular path of radius R is (when θ\theta is measured from the positive x-axis and v is uniform speed) :

Multiple Choice Options

  • A −v2Rsin⁡θi^+v2Rcos⁡θj^- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j
  • B −v2Rcos⁡θi^+v2Rsin⁡θj^- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j
  • C −v2Rcos⁡θi^−v2Rsin⁡θj^- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j
  • D −v2Ri^+v2Rj^- {{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th June Evening Shift Physics - Circular Motion Question 38 English Explanation

As the particle in uniform circular motion experiences only centripetal acceleration of magnitude ω\omega 2R or v2R{{{v^2}} \over R} directed towards centre so from diagram,

a→=v2Rcos⁡θ(−i^)+v2Rsin⁡(−j^)\overrightarrow a = {{{v^2}} \over R}\cos \theta ( - \widehat i) + {{{v^2}} \over R}\sin ( - \widehat j)

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