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Circular Motion question

2022 · 25 Jun · Shift 2 · Q52
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  5. /2022 · 25 Jun · Shift 2 · Q52

Circular Motion question

2022 · 25 Jun · Shift 2 · Q52

JEE MainPhysicsCircular MotionMCQ+4 / −1
For a particle in uniform circular motion, the acceleration a→\overrightarrow aa at any point P(R, θ\thetaθ) on the circular path of radius R is (when θ\thetaθ is measured from the positive x-axis and v is uniform speed) :
  1. A
    −v2Rsin⁡θi^+v2Rcos⁡θj^- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j−Rv2​sinθi+Rv2​cosθj​
  2. B
    −v2Rcos⁡θi^+v2Rsin⁡θj^- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j−Rv2​cosθi+Rv2​sinθj​
  3. C
    −v2Rcos⁡θi^−v2Rsin⁡θj^- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j−Rv2​cosθi−Rv2​sinθj​
  4. D
    −v2Ri^+v2Rj^- {{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j−Rv2​i+Rv2​j​
View written solutionFree

Correct answer: C

  1. Position vector at point P(R,θ)P(R,\theta)P(R,θ)

For a particle on a circle of radius RRR centered at the origin, the position vector is

r⃗=Rcos⁡θ i^+Rsin⁡θ j^\vec r = R\cos\theta\,\hat i + R\sin\theta\,\hat jr=Rcosθi^+Rsinθj^​

So the radial unit vector is

r^=cos⁡θ i^+sin⁡θ j^\hat r = \cos\theta\,\hat i + \sin\theta\,\hat jr^=cosθi^+sinθj^​
  1. Acceleration in uniform circular motion

In uniform circular motion, the speed vvv is constant, and the acceleration is always directed towards the center of the circle.

Hence,

a⃗=−v2R r^\vec a = -\frac{v^2}{R}\,\hat ra=−Rv2​r^

Substituting r^\hat rr^:

a⃗=−v2R(cos⁡θ i^+sin⁡θ j^)\vec a = -\frac{v^2}{R}(\cos\theta\,\hat i + \sin\theta\,\hat j)a=−Rv2​(cosθi^+sinθj^​)
  1. Resolve into components
a⃗=−v2Rcos⁡θ i^−v2Rsin⁡θ j^\vec a = -\frac{v^2}{R}\cos\theta\,\hat i - \frac{v^2}{R}\sin\theta\,\hat ja=−Rv2​cosθi^−Rv2​sinθj^​
  1. Match with the options
  • A: −v2Rsin⁡θ i^+v2Rcos⁡θ j^-\dfrac{v^2}{R}\sin\theta\,\hat i + \dfrac{v^2}{R}\cos\theta\,\hat j−Rv2​sinθi^+Rv2​cosθj^​ ❌
  • B: −v2Rcos⁡θ i^+v2Rsin⁡θ j^-\dfrac{v^2}{R}\cos\theta\,\hat i + \dfrac{v^2}{R}\sin\theta\,\hat j−Rv2​cosθi^+Rv2​sinθj^​ ❌
  • C: −v2Rcos⁡θ i^−v2Rsin⁡θ j^-\dfrac{v^2}{R}\cos\theta\,\hat i - \dfrac{v^2}{R}\sin\theta\,\hat j−Rv2​cosθi^−Rv2​sinθj^​ ✅
  • D: −v2Ri^+v2Rj^-\dfrac{v^2}{R}\hat i + \dfrac{v^2}{R}\hat j−Rv2​i^+Rv2​j^​ ❌

Therefore, the correct option is C.

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