- #1
ogehsim
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Homework Statement
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.70 s, it is at point (4.30 m, 5.50 m) with velocity (3.10 m/s)j and acceleration in the positive x direction. At time t2 = 13.0 s, it has velocity (–3.10 m/s)i and acceleration in the positive y direction. What are the (a)x and (b)y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.
Homework Equations
I honestly do not know... Anything to do with circular motion I guess.
The Attempt at a Solution
270degrees=4.712 radians
w=angular velocity
w=theta/t
w=4.712/(13-3.7)
w=.5066rad/s
ac=centripetal acceleration
ac=v^2/r
r=v^2/ac
r=3.10^2/ac
r=9.61/ac
ac=r*alpha
ac=r*(w/t)
ac=r*(.5066/(13-3.7))
r=9.61/(.5066r/9.3)
r=89.373/.5066r
r^2=176.41
r=13.28
so 4.3+13.28=17.58 and y would remain 5.5 y (add radius to x, y would remain same as it just is to the direct right of point 1)
17.58 is WRONG
5.5 is RIGHT
I don't get how to solve for r when you don't know ac.