Question

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (2.50 m, 2.50 m) with a velocity of -5.00i hat m/s and an acceleration of -10.0j m/s2. What are the coordinates of the center of the circular path?

Answer 1

Answer:

(2.5,0)

Explanation:

The particle can be described by the following equations:

$x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\\frac{dx}{dt}=-\omega Rcos(-\omega t)\\\frac{dy}{dt}=\omega Rsin(-\omega t)\\\frac{d^2x}{dt^2}=-\omega^2Rsin(-\omega t)\\\frac{d^2y}{dt^2}=-\omega^2Rcos(-\omega t)$

For R = 2.5, ω = 2 and t = 0:

$x=2.5\\y=2.5\\\\\frac{dx}{dt}=-5\\ \frac{d^2y}{dt^2}=-10$

The center of the circle would be at point (2.5,0)