JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • 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 :

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

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