Question 141: Work, Energy and Power & Circular Motion

Question

For a particle in uniform circular motion the acceleration $$\overrightarrow a $$ at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)

Accepted Answer

For a particle in uniform circular motion,
$${a_c} = {{{v^2}} \over R}$$ towards the center of the circle
From figure, $$\overrightarrow a  = {a_c}\cos 	heta \left( { - \widehat i} ight) + {a_c}\sin 	heta \left( { - \widehat j} ight)$$
$$ = {{ - {v^2}} \over R}\cos 	heta \widehat i - {{{v^2}} \over R}\sin 	heta \widehat j$$

Answer

$$ - {{{v^2}} \over R}\cos 	heta \widehat i + {{{v^2}} \over R}\sin 	heta \widehat j$$

Answer

$$ - {{{v^2}} \over R}\sin 	heta \widehat i + {{{v^2}} \over R}\cos 	heta \widehat j$$

Answer

$$ - {{{v^2}} \over R}\cos 	heta \widehat i - {{{v^2}} \over R}\sin 	heta \widehat j$$

Answer

$${{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$$

Work, Energy and Power & Circular Motion

For a particle in uniform circular motion the acceleration $\overrightarrow a $ at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)

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