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Question: For a particle in uniform circular motion the acceleration \(\overrightarrow{a}\) at a point \(P\lef...

For a particle in uniform circular motion the acceleration a→\overrightarrow{a} at a point P(R,θ)P\left( R,\theta \right) on the circle of radius R is (here ‘v’ is the speed of the particle and θ\theta is measured from the x-axis)
A.−v2Rcos⁡θi^+v2Rsin⁡θj^-\dfrac{{{v}^{2}}}{R}\cos \theta \widehat{i}+\dfrac{{{v}^{2}}}{R}\sin \theta \widehat{j}
B.−v2Rsin⁡θi^+v2Rcos⁡θj^-\dfrac{{{v}^{2}}}{R}\sin \theta \widehat{i}+\dfrac{{{v}^{2}}}{R}\cos \theta \widehat{j}
C.−v2Rcos⁡θi^−v2Rsin⁡θj^-\dfrac{{{v}^{2}}}{R}\cos \theta \widehat{i}-\dfrac{{{v}^{2}}}{R}\sin \theta \widehat{j}
D.v2Ri^+v2Rj^\dfrac{{{v}^{2}}}{R}\widehat{i}+\dfrac{{{v}^{2}}}{R}\widehat{j}

Explanation

Solution

You could recall the expression for the acceleration of the body that is undergoing circular motion. Now, you could represent the situation in a diagram by clearly marking the vertical and horizontal components of acceleration. Now you could substitute accordingly to get the right answer.

Formula used:
Centripetal acceleration,
ac=mv2R{{a}_{c}}=\dfrac{m{{v}^{2}}}{R}

Complete answer:
In the question, we are given the acceleration at a point P(R,θ)P\left( R,\theta \right) on the trajectory of a uniform circular motion asa→\overrightarrow{a}. The radius of the circular path is found to be RR, the speed of the particle is given as v and θ\theta is the angle made with the x-axis. The given situation can be depicted in a diagram as,

We know that acceleration of the body that is undergoing circular motion will be centripetal acceleration. And centripetal acceleration is given by,
a=mv2Ra=\dfrac{m{{v}^{2}}}{R} ……………………………………… (1)
Where, m is the mass of the body, v is its velocity and R is the circular path in which the body is moving.
From given diagram, we see that the given situation can be represented by,
a→=−acos⁡θi^−asin⁡θj^\overrightarrow{a}=-a\cos \theta \widehat{i}-a\sin \theta \widehat{j}
Now, we could substitute the value of a from (1) to get,
a→=−mv2Rcos⁡θi^−mv2Rsin⁡θj^\overrightarrow{a}=-\dfrac{m{{v}^{2}}}{R}\cos \theta \widehat{i}-\dfrac{m{{v}^{2}}}{R}\sin \theta \widehat{j}
Hence, we found that the uniform acceleration of the particle in the given question will be given by,
a→=−mv2Rcos⁡θi^−mv2Rsin⁡θj^\overrightarrow{a}=-\dfrac{m{{v}^{2}}}{R}\cos \theta \widehat{i}-\dfrac{m{{v}^{2}}}{R}\sin \theta \widehat{j}

Hence, we found that option C is the right answer.

Note:
We know that the centripetal acceleration of a body undergoing circular motion is always directed towards the centre of the circular path in which it is moving. Hence we could be sure that its horizontal and vertical components will be in negative x axis and negative y axis respectively. Thereby, we have represented the acceleration accordingly.