# For a particle in uniform circular motion, the acceleration $\dpi{100} \bar{a}$ at a point $\dpi{100} P(R,\Theta )$ on the circle of radius R is(Here $\dpi{100} \Theta$ is mesured by x-rays) Option 1) $\frac{\upsilon^{2} }{R}\hat{i}+\frac{\upsilon^{2} }{R}\hat{j}$ Option 2) $-\frac{\upsilon^{2} }{R}\cos \Theta \hat{i}+\frac{\upsilon^{2} }{R}\sin \Theta \hat{j}$ Option 3) $-\frac{\upsilon^{2} }{R}\sin \Theta \hat{i}+\frac{\upsilon^{2} }{R}\cos \Theta \hat{j}$ Option 4) $-\frac{\upsilon^{2} }{R}\cos \Theta \hat{i}-\frac{\upsilon^{2} }{R}\sin \Theta \hat{j}$

S Sabhrant Ambastha

As we learnt in

Centripetal acceleration -

When a body is moving in a uniform circular motion, a force is responsible to change direction of its velocity.This force acts towards the centre of circle and is called centripetal force.Acceleration produced by this force is centripetal acceleration.

$= \frac{-v^2}{r}\cos \Theta \hat{i} - \frac{-v^2}{r}\sin\Theta \hat{j}$

$a= \frac{v^{2}}{r}$

- wherein

Figure Shows Centripetal acceleration

For a  particle in uniform circular motion

Acceleration $a=\frac{v^2}{R}$

$a^{\rightarrow}= - a\cos\Theta\hat{i}-a\sin \Theta \hat{j}$

$= \frac{-v^2}{R}\cos \Theta \hat{i} - \frac{-v^2}{R}\sin\Theta \hat{j}$

Option 1)

$\frac{\upsilon^{2} }{R}\hat{i}+\frac{\upsilon^{2} }{R}\hat{j}$

Incorrect

Option 2)

$-\frac{\upsilon^{2} }{R}\cos \Theta \hat{i}+\frac{\upsilon^{2} }{R}\sin \Theta \hat{j}$

Incorrect

Option 3)

$-\frac{\upsilon^{2} }{R}\sin \Theta \hat{i}+\frac{\upsilon^{2} }{R}\cos \Theta \hat{j}$

Incorrect

Option 4)

$-\frac{\upsilon^{2} }{R}\cos \Theta \hat{i}-\frac{\upsilon^{2} }{R}\sin \Theta \hat{j}$

Correct

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