For a particle in uniform circular motion, the acceleration a at a point P ( R, ) on the…
Physics · JEE Advanced · NTA Exams — Rotational Motion
For a particle in uniform circular motion, the acceleration \(\bar{a}\) at a point P (\(R, \theta\)) on the circle of radius R is
- \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)
- \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)
- \(-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}\)
- \(-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}\)
Answer
(D) - v^ 2 R i - v^ 2 R j
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