For a particle in uniform circular motion the acceleration a at a point 𝑃(𝑅, πœƒ) on the circle of the radius R is (here πœƒ is measured from the π‘₯ βˆ’axis)

Question : For a particle in uniform circular motion the acceleration a at a point 𝑃(𝑅, πœƒ) on the circle of the radius R is (here πœƒ is measured from the π‘₯ βˆ’axis)

(A) \dfrac{-v^{2}}{R}\cos \theta \widehat{i}+\dfrac{v^{2}}{R}\sin \theta \widehat{j}

(B) \dfrac{-v^{2}}{R}\sin \theta \widehat{i}+\dfrac{v^{2}}{R}\cos \theta \widehat{j}

(C) \dfrac{-v^{2}}{R}\cos \theta \widehat{i}-\dfrac{v^{2}}{R}\sin \theta \widehat{j}

(D) -\dfrac{v^{2}}{R}\widehat{i}+\dfrac{v^{2}}{R}\widehat{j}

Answer : option (C)

Solution :

For a particle in uniform circular motion the acceleration a at a point 𝑃(𝑅, πœƒ) on the circle of the radius R is (here πœƒ is measured from the π‘₯ βˆ’axis)
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