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Finding the Center of a circle

  1. Oct 13, 2007 #1
    The question is:

    A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (3.00 m, 3.00 m) with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2. What are the coordinates of the center of the circular path?

    I know centripical acceleration is a = (v^2)/r

    I know:
    ax = 0 m/s^2
    ay = 10 m/s^2

    and . . .

    vx = -5 m/s
    vy = 0 m/s

    where do i go from here?
     
  2. jcsd
  3. Oct 13, 2007 #2

    Doc Al

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    Staff: Mentor

    What will that allow you to calculate?

    I assume that the directions were given?

    Make use of that centripetal acceleration formula.
     
  4. Oct 13, 2007 #3
    ax = 0 m/s^2
    ay = 10 m/s^2

    and . . .

    vx = -5 m/s
    vy = 0 m/s

    I am just assuming that from reading from the question . . .is that a safe assumption?

    This is the question that I am asked:
    A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (3.00 m, 3.00 m) with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2. What are the coordinates of the center of the circular path?

    and what I did was find the accel. in the x and y direction and the velocity in the x and y velocity
     
  5. Oct 13, 2007 #4

    Doc Al

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    Staff: Mentor

    The question says "with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2", but what direction? You assume the velocity is in the -x direction, but I see nothing in the problem statement that tells you that.

    Perhaps this statement "A particle moves horizontally in uniform circular motion, over a horizontal xy plane" was supposed to read "A particle moves along the x-axis...". (If it's moving in a horizontal xy plane, then both x-axis and y-axis are horizontal.)

    OK, let's assume your directions are correct. Now make use of the centripetal acceleration formula. You have v and a; find r.
     
  6. Oct 13, 2007 #5
    ahhhhh . . .there is another piece i left out . . "-5.00 i(hat) m/s and an acceleration of +10.0 j(hat) m/s2 "
    . . .that is where I got my information from but even in that case would I still just do a direct plug in? so. . .

    10 = 25/r and r having a value of 2.5m?
     
  7. Oct 13, 2007 #6

    Doc Al

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    Staff: Mentor

    That makes all the difference! :smile:
    Right. Now use that to locate the center of the circle. (You know which way the acceleration points.)
     
  8. Oct 13, 2007 #7
    wow i was just making it way too hard. . .thank you so much Doc!
     
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