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Acceleration vector

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data

    There is a man sitting at the lowest point of a ferris wheel that is just beginning to move it has velocity 3m/s and is accelerating at a constant rate of .5m/s^2. find the magnitude and direction of the acceleration vector.

    2. Relevant equations



    3. The attempt at a solution

    I know that acceleration has two components, perpendicular and tangential, one controlling direction and the other magnitude.

    So if the acceleration is a constant 5m/s^2 then its components would be perpendicular .5sin(0) and parallel .5cos(0). this gives acceleration = arctan(1)= 45 degrees.

    But this would mean that the vector is pointing away from the center of ferris wheel, which should not be the case if this is uniform circular motion. But its not hes his speed is increasing. But if its increasing what's preventing him from being flung off the wheel?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 1, 2011 #2

    Dick

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    I think the acceleration is tangential. .5sin(0)=0. How are you getting 45 degrees out of this again?
     
  4. Jul 1, 2011 #3
    angle=arctan(y/x)=arctan(.5/.5)
     
  5. Jul 1, 2011 #4

    Dick

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    .5*sin(0)=0, .5*cos(0)=.5. I don't think the quotient is .5/.5.
     
  6. Jul 1, 2011 #5
    Wow, I can't believe I miss some of these mistakes. But would still mean that there is no perpendicular component correct?
     
  7. Jul 1, 2011 #6

    Dick

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    All of the acceleration due to the motion of the ferris wheel is towards the center of the wheel. Isn't it? Isn't it all perpendicular? I think what you are missing here (and that I have been ignoring since I got fixated on looking at more surface problems) is that the motion is circular, not linear. You can't even answer any questions about the magnitude of the acceleration unless you know the radius of the ferris wheel.
     
    Last edited: Jul 2, 2011
  8. Jul 2, 2011 #7
    So perpendicular accl would be equal to 3^2/14, if the diameter=14?
     
  9. Jul 2, 2011 #8

    Dick

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    Not if the diameter is 14. It would be if the RADIUS were 14. Specifying units would be helpful here. And you still have a tangential acceleration to worry about if you want the total acceleration.
     
  10. Jul 2, 2011 #9
    Alright I think im starting to find out where im getting confused. perp accl would be equal 9/7 and tangential would be equal to the derivative of the magnitude of velocity. in this case v=3+.5t, the derivative would be 3 and that would be the tangential acceleration.
     
  11. Jul 2, 2011 #10

    Dick

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    The derivative of v=3+.5t isn't 3. Come on, you can do better than that. And the acceleration due to the velocity, 9/7 m/s^2 is perpendicular to the path and the other component is tangential. Add them as vectors.
     
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