For a particle in uniform circular motion, the acceleration \bar{a} at a point P(R,\Theta ) on the circle of radius R is

(Here \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}

 

Answers (1)
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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