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Finding the radius based on time and velocit

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    I honestly do not know... Anything to do with circular motion I guess.


    3. The attempt at a solution
    physics.jpg

    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.
     
  2. jcsd
  3. Sep 7, 2010 #2
    With regards to your approach:
    w*r = v
    Here you are confusing centripetal acceleration with angular acceleration.
    Although there is a force (and thus an acceleration) maintaining the centripetal motion there is no acceleration of the speed of this motion as the acceleration is always tangent (in this case).
     
  4. Sep 7, 2010 #3
    Ohhh. So I could only use w*r=v if v at point 1 was 3.10 and if it was then different at point 2?

    Is there any way to solve this numerically, or is it purely a logic-out question like what I did for y? It doesn't give me the location of the 2nd point, so I don't know x2, and therefore can't logic my way to the x-coordinate of the origin.
     
  5. Sep 7, 2010 #4
    Additional Info on Problem (enough wrong answers allows you a hint): At each time, consider where the center of the circle is, and where the particle is along its path around the circle. From the known speed and times, the distance the particle travels can be found. This allows you to calculate the radius of the circle.

    13.0s-3.7s=9.3s
    9.3s*3.10m/s=28.83m

    ... can I assume that even though 3.7*4=14.8, not 13, that 3.7s is approximately the time to travel 1/4 of the circle's circumfrence, and 13.0s is the time it takes to travel all of it?
    In which case: 13.0s*3.10m/s=40.4m
    c=pi*2r
    40.4m=pi*2r
    r=6.43m

    x=4.3m+6.43m=10.73m

    10.73 is WRONG, and I only have 1 try left.
    I guess this means my assumption is wrong, but I don't understand how if my drawing is correct... So the particle must start b/w point 1 and point 2 and not at point 2... so how do I find the time to travel the entire circle... do I need to find that itme?
     
  6. Sep 7, 2010 #5
    13.0s-3.7s=9.3s
    9.3s*3.10m/s=28.83m

    c=pi*2*r
    .75c=.75*pi*2*r
    28.83m=.75*pi*2*r
    r=6.1179m

    x=4.3m+6.1179m=10.4179m

    (To see if this is within the 2% margin of error for my other answer. If it is, this is also wrong.
    10.4179*.02=.208
    10.4179+.208=10.626
    This does NOT encompass my previous guess of 10.73, making it a valid new answer.)

    Can anyone verify my math on this?
     
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