Uniform circular motion of a particle problem

AI Thread Summary
A particle moves in uniform circular motion with specific velocity and acceleration vectors at two different times. At t1, the particle is at (5.10 m, 6.80 m) with a velocity of 3.90 m/s in the positive y-direction, indicating it is crossing the horizontal axis. The time interval between t1 and t2 is 6.60 seconds, allowing for the calculation of the distance traveled, which corresponds to a quarter of the circular path. The discussion concludes that the particle moves clockwise, confirming that it indeed travels three-quarters around the circle. The center's coordinates can be derived from the particle's position and the calculated radius.
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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 = 3.30 s, it is at point (5.10 m, 6.80 m) with velocity (3.90 m/s)## \hat j ## and acceleration in the positive x direction. At time t2 = 9.90 s, it has velocity (–3.90 m/s)## /hat 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


a=v2/r, v=2πr/T, [/B]

The Attempt at a Solution


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If all its speed is in j-hat while at (5.1, 6.8), then it is crossing the horizontal axis through the circle at that moment. Hence, the y-component of the center is the y-component of (5.1,6.8)

It’s tangential speed is 3.9 m/s. It takes 9.9-3.3 seconds to travel ¼ around its orbit. (I’m assuming it didn’t go 5/4, or 9/4 around, etc.). So now you have a speed and a time, so you can solve for distance (circumference). Then it’s easy to get radius. If you know the radius and you know its position when all its velocity in is the j hat direction, then your radius is the offset from you x-component of that position.
 
tony873004 said:
If all its speed is in j-hat while at (5.1, 6.8), then it is crossing the horizontal axis through the circle at that moment. Hence, the y-component of the center is the y-component of (5.1,6.8)

It’s tangential speed is 3.9 m/s. It takes 9.9-3.3 seconds to travel ¼ around its orbit. (I’m assuming it didn’t go 5/4, or 9/4 around, etc.). So now you have a speed and a time, so you can solve for distance (circumference). Then it’s easy to get radius. If you know the radius and you know its position when all its velocity in is the j hat direction, then your radius is the offset from you x-component of that position.
would'nt it travel 3/4 around the circle 3pi/2 if its velocity is positive jhat at the left side of the circle its moving clockwise?
 
Is it given that its moving clockwise?
 
tony873004 said:
Is it given that its moving clockwise?
no but both vectors are pointing in a direction that would indicate clockwise motion
 
You're right, it moves 3/4 around.
 
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