Question

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (3.20 m, 3.50 m) with a velocity of –2.50 i^ m/s and an acceleration of +11.6 j^ m/s2. What are the (a) x and (b) y coordinates of the center of the circular path?

Answer 1

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

$a_c = \frac{v^2}{r}$

$11.6 = \frac{2.5^2}{R}$

$R = 0.54 m$

so the coordinates of the center will be

$(x, y) = (3.20 , (3.5 + 0.54))$

$(x, y) = (3.20, 4.04)$

so the coordinate is (3.20 m, 4.04 m)