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Q.

For a particle in uniform circular motion the acceleration $\vec{a}$ at a point $P (R, θ)$ on the circle of radius $R$ is (here ' $v$ ' is the speed of the particle and $θ$ is measured from the $x$ axis) .

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a

− v 2 R cos θ i^+ v 2 R sin θ j^

b

− v 2 R sin θ i^+ v 2 R cos θ j^

c

− v 2 R cos θ i^− v 2 R sin θ j^

d

v 2 R i^+ v 2 R j^

answer is C.

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Detailed Solution

Concept- We'll begin by sketching the figure that represents any particle moving in a uniform circular motion. The particle's acceleration at the given point will be displayed and resolved into Cartesian components.
For a particle in uniform circular motion,

\vec{a} = \frac{v^{2}}{R}

towards the centre of the circle.

\vec{a} = \frac{v^{2}}{R} (- \cos θ \hat{i} - \sin θ \hat{j})

\vec{a} = - \frac{v^{2}}{R} \cos θ \hat{i} - \frac{v^{2}}{R} \sin θ \hat{j}

Hence, option 3 is the correct answer.

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