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  • For a particle in uniform circular motion, the acceleration <img alt="\mathrm{\underset{a}{\rightarrow}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cunderset%7Ba%7D%7B%5Cri

For a particle in uniform circular motion, the acceleration \mathrm{\underset{a}{\rightarrow}} at ay point \mathrm{P\left ( R,\theta \right )} on the circular path of radius \mathrm{R} is ( when \mathrm{\theta} is measured from the positive x-axis and \mathrm{v} is uniform speed ):

Option: 1

\mathrm{-\frac{v^{2}}{R}\sin\theta\hat{i}+\frac{v^{2}}{R}\cos\theta\hat{j}}


Option: 2

\mathrm{-\frac{v^{2}}{R}\cos\theta\hat{i}+\frac{v^{2}}{R}\sin\theta\hat{j}}


Option: 3

\mathrm{-\frac{v^{2}}{R}\cos\theta\hat{i}-\frac{v^{2}}{R}\sin\theta\hat{j}}


Option: 4

\mathrm{-\frac{v^{2}}{R}\hat{i}+\frac{v^{2}}{R}\hat{j}}


Answers (1)

best_answer

\mathrm{\bar{R}=R \cos \theta \hat{i}+R \sin \theta \hat{\jmath}}\\

\mathrm{\bar{a}=d^{2} \bar{R} / d t^{2}=-\left(\frac{d^{2} \theta}{d t^{2}} \right) R[\cos \theta \hat{i}+\sin \theta \hat{\jmath}]}

\mathrm{\bar{a}=-\omega^{2} R[\cos \theta \hat{i}+\sin \theta \hat{j}]}\\

\mathrm{\bar{a}=-\frac{V^{2}}{R}[\cos \theta \hat{i}+\sin \theta \hat{j}]}

Hence the correct answer is option 3.

Posted by

sudhir.kumar

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