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Centripetal Force And Acceleration

  1. Nov 18, 2005 #1
    1. Which of the following statements best applies to an object moving with uniform circular motion?(Answer: B)
    A. Acceleration is zero.
    B. Acceleration is directed outward.
    C. Acceleration is tangent to the path.
    D. Magnitude of acceleration is constant.

    I thought the answer was D but how can it be B? Centripetal acceleration is always towards the center of the circular path. How is that outward?

    2. A 50kg man is riding a ferris wheel at a constant speed. At the top of the wheel, the seat exerts an upward force of 420N on him. What is the centripetal force on the man at the top of the wheel? (Answer: 70N)

    Fnet = FN+Fg+Fc
    0 = 420+50(-9.8)+Fc
    70N [up] = Fc

    Again, this direction has me confused. Centripetal force is always towards the center of the circular path so why am I getting an up answer?
  2. jcsd
  3. Nov 18, 2005 #2
    For number two, if you consider mg negative, the upward force is positive. Since mg is directed at the center of the wheel and 50(9.8)>420, thus 420 - 50(9.8)>0. Since it is negative it is directed to the center of the wheel; this is the centripetal force. Your mistake was to consider F_c to be positive, but it is negative since it's in the same direction as mg.
  4. Nov 18, 2005 #3


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    Are you sure that you're transcribing these directly. There's an important difference between centripital acceleration and centrifugal force.
    For part 1:
    With the question as you have written it, of the choices listed, the only correct one is D.
    For part 2:
    Are you sure the question is asking for "centripetal" force? The correct answer, to the question as you have it written would probably be "the force of gravity".
    However, if the questions were asking about centrifugal force rather than centripetal acceleration, then the answers that are given would make more sense. (Although the magnittude of the centrifugal force is constant for uniform circular motion.)
  5. Nov 18, 2005 #4
    Yep, I copied these questions exactly as they were stated in my workbook.

    Werg22, I don't understand your explaination. I'm saying that positive is up and negative is down. 0 = 420-490+Fc. Fc has to be positive 70N which means its up. I don't think there's anything wrong with the algebra; it just doesn't make sense.
  6. Nov 18, 2005 #5
    You let F_c and the upward force be of the same sign, which is wrong since F_c is in the same direction as mg and thus mg and F_c need to be of the same sign.


    You start with

    F_c = Fnet

    F_c = 430 + mg

    Since F_c is in the same direction as mg, they are of the same sign

    If you consider mg negative, thus Fc is negative.

    0= 430 + mg -F_c, Since you multiplied F_c by -1, its sign changes. Your mistake was to consider F_c positive, but it is not since the sign changed. So -F_c = 70, and F_c = -70.
    Last edited: Nov 18, 2005
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