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Magnitude of acceleration in circular motion

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A ferris wheel with radius 14.0m is turning about a horizontal axis through its center. The linear speed of a passenger on the rim is constand and equal to 7.69m/s. What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion?

    2. Relevant equations



    3. The attempt at a solution
    Because the speed is constant I thought that the magnitude of the acceleration would be zero and only the direction of the acceleration would have a value, but zero isn't the right answer so there's a concept to answering this question that I'm completely missing and I have no idea how to go about finding it. Does the fact that the person is at the lowest point make a difference? All of this is very new to me and I'm just very confused.
     
  2. jcsd
  3. Nov 6, 2012 #2
    If you tie a rock to a string and swing it in a circle - like the ferris wheel - doesn't it apply tension to the string? What causes the tension?

    What are the relevant equations you have at your disposal?
     
  4. Nov 6, 2012 #3
    I'm given atan=(dlvl)/dt
     
  5. Nov 6, 2012 #4
    I have no idea what you have written. That said, what about formulas for centripetal acceleration such as V^2/R or R*omega^2 where omega is angular speed in radians/second?
     
  6. Nov 6, 2012 #5

    haruspex

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    Acceleration is a change in velocity. (Linear acceleration is a change in linear velocity, angular acceleration is a change in angular velocity.)
    Velocity is a vector. The speed is the magnitude of the velocity.
    If the speed stays constant but the direction changes then the velocity changes, hence the acceleration is not zero. Acceleration is also a vector. When the acceleration is at right angles to the velocity the speed stays constant. If an object moves in a circle at constant speed its acceleration is towards the centre of the circle.
     
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