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Uniform Circular Motion of a particle

  1. Oct 8, 2005 #1
    Im new at physics and have been looking at this question all day, any help would be greatly appreciated,

    here it is

    A particle moves along a cirular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.00s, it is at point (5.oom, 6.00m) with velocity (3.00 m/s)j and acceleration in the positive x direction. At time t2 = 10.0s, it has velocity (-3.00 m/s)i and acceleration in the positive y direction. What are the x and y coordinates of the centre of the circular path.

    I have no idea where to start??? Please help!!
     
  2. jcsd
  3. Oct 8, 2005 #2

    mezarashi

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    The clues help you find two things

    1. The center of the circular motion
    2. The radius of motion

    Draw out a diagram of a particle under circular motion. At each quarter circle, draw out the vectors for velocity and acceleration. Compare it to your questions problem.

    The times give you a clue to finding the radius and subsequently the center of the motion. Within (10 - 4)seconds, the object moved from one part of the circle to another. You know d = vt. You also know that v^2/r defines acceleration in circular motion. A bit tricky, but I hope now you can start thinking about it.
     
  4. Oct 8, 2005 #3
    thanks for the post, but i'm still lost????
     
  5. Oct 8, 2005 #4

    mezarashi

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    Have you done what I said. Pictures will help you alot:

    Draw out a diagram of a particle under circular motion. At each quarter circle, draw out the vectors for velocity and acceleration. Compare it to your questions problem.
     
  6. Oct 8, 2005 #5
    from the diagram i see that at the first point its in the 2nd quarter and at the second point its in the 4th quarter( moving in a clockwise direction). Would the radius be the square root of (5^2 + 6^2). I still can't understand how this would relate to an xy centre coordinate(I thought the centre coordinate should be the origin of your plane).
     
  7. Oct 8, 2005 #6

    mezarashi

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    Aha, you're starting to understand a bit now. Your direction of motion is also correct. Not so fast about the radius just yet. You were drawing your own diagram based on the center of the motion being at 0,0. But in this case, the center isn't there, and you must find it. Can you find the corresponding point on the question's x-y coordinate for the first point? From there you can possibly estimate where the 2nd point would be, but you can't tell until you know the radius.

    Now to finding this radius. You know that points 1 and 2 are separated by 3/4 of a circle, do you agree? Refer back to the diagram you drew. The circumference of a circle can be described as [tex]2\pi r[/tex]. Now we know it took (10-4) seconds to make its way across three quarters of it, where d = vt. Getting closer?
     
  8. Oct 8, 2005 #7

    HallsofIvy

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    You know that the circle passes through (5, 6) and that the vector 3.00j is tangent to the circle there. Since 3j is vertical, you know that (5,6) is at one end of a horizontal diameter. Further, you know that its speed, at any time, is 3 m/s. You also know that, 6 seconds later later, "it has velocity (-3.00 m/s)i and acceleration in the positive y direction. " so at that time, when it has moved 3(6)= 18 m, it is at the top of the vertical diameter. That appears to mean that it has moved 3/4 of the entire circle. What is the circumference of the circle? What is the diameter of the circle? Knowing the radius and that one end of a horizontal diameter is (5,6), it should be easy to find the coordinates of the center.
    (Notice the word "appears". It is also possible that in those 6 seconds, it has gone completely around the circle and then one quarter of the circle.)
     
  9. Oct 8, 2005 #8
    so 3/4 of the trip was 18m , then the total circumference would be 24m , the diameter 7.6 and radius 3.8 m. Then the centre of the circle would be x = 8.8 y = 6
     
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