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

Correct Answer is : −v2Rcosθ∧i−v2Rsinθ∧j

-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \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}

\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}

AIEEE 2010 Solved Paper

Section: Physics

Question : 32 of 89

Marks: +1, -0

For a particle in uniform circular motion the acceleration

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at a point P(R,θ) on the circle of radius R is (here θ is measured from the x-axis)

[AIEEE 2010]

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