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

(2.5,0)

Step-by-step explanation:

The particle can be described by the following equations:

`x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\(dx)/(dt)=-\omega Rcos(-\omega t)\\(dy)/(dt)=\omega Rsin(-\omega t)\\(d^2x)/(dt^2)=-\omega^2Rsin(-\omega t)\\(d^2y)/(dt^2)=-\omega^2Rcos(-\omega t)`

For R = 2.5, ω = 2 and t = 0:

`x=2.5\\y=2.5\\\\(dx)/(dt)=-5\\ (d^2y)/(dt^2)=-10`

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