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Velocity and Acceleration of a particle around a circle

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moves with constant speed around a circle. When it is at the top of the circle, its coordinates are x=0 and y=2 and its velocity is 4(m/s) i. When it reaches the left hand of the circle, where its coordinates are now x=2 and y=0, what is its velocity and acceleration respectively?


    2. Relevant equations

    I know that a position vector consists of r= R (cos theta i + sin theta j), also velocity is the first derivative of the position vector and acceleration is the second derivative of the position vector.

    How do I take a derivative of position vectors i j k, or do i just leave them as they are?


    3. The attempt at a solution

    Ive tried taking derivatives to arrive at an answer with the proper position vectors but can't seem to get it right.
     
  2. jcsd
  3. Dec 5, 2015 #2

    PeroK

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    One of the simple things about cartesian coordinates is that the unit vectors i, j and k are independent of position and time. Hence, they do not affect derivatives with respect to time or x, y and z.

    The first step in this question is to find ##\textbf{r}(t)##: the position vector of the particle as it moves in time. Can you do that?
     
  4. Dec 5, 2015 #3

    Shouldn't the left hand of the circle be x = -2, y = 0? Do you know how to tell from the velocity at x = 0, y = 2 whether the particle is traveling clockwise or counter-clockwise?
     
  5. Dec 13, 2015 #4
    I believe I had done it correctly, I got r(t) = 0i + 2j
     
  6. Dec 13, 2015 #5
    Yes, that is correct and an error on my part. well, if the velocity at r(t)= 0 i +2j is 4(m/s)i, then I would assume it is going in the clockwise direction because of the i vector. if it had been -4(m/s)i, then it would be in the counter clockwise direction.
     
  7. Dec 13, 2015 #6
    Right. So if it is going clockwise, what is its velocity at -2,0 (the left hand side of the circle)? What is its acceleration at this point?
     
  8. Dec 13, 2015 #7
    well, I found an equation of a= (v^2)/2 i, which gave an acceleration of a = 8 m/s i. I don't understand how this equation was derived. Although, now that I have actually thought about this, the acceleration vector around a curve or a circle always points toward the center of the circle and the velocity vector is always tangent to the point on the curve, so I understand how the acceleration vector is i and the velocity vector is j for x=-2,y=0. The only problem I have is the equation I came across.

    I know that velocity is the derivative of the r vector with respect to time and acceleration is the derivative of the velocity with respect to time, but where does a = (v^2)/2 come from?
     
  9. Dec 13, 2015 #8
    What a great question!!!! As you said, the acceleration is the derivative of the velocity with respect to time, and the velocity has both magnitude and direction. So the equation for the acceleration a = v^2/r in this circular motion with constant velocity is related to how the direction of the velocity vector is changing with time. There are different ways of deriving this equation, so it depends on which method is used in your textbook.

    Chet
     
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