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Centripetal Force/Changing direction on ice - Turn length

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I am running to school at 6m/s when I hit a patch of icy sidealk. The coefficient of friction between the ice and my shoes is 0.16. I wish to turn my 55kg frame to the left. What is the smallest turn I can make?

    v = 6 m/s
    μ = 0.16
    m = 55 kg

    2. Relevant equations
    Fnet = mv^2/r
    Ff = μFn
    Fnet = ma

    3. The attempt at a solution
    Assume that up and left are positive.

    Fnet = ma
    Fg + Fn = ma
    -539 + Fn = 0
    Fn = 539

    Ff = μFn
    Ff = 0.16 (539)
    Ff = 86.24

    Fnet = mv^2/r
    Ff = mv^2/r
    86.24 = 55(6)^2/r
    86.24 = 1980/r
    r = 23 m

    I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving.
  2. jcsd
  3. Mar 1, 2012 #2


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    Homework Helper

    You got the radius of the circle (correct) the boy can turn without slipping outwards. He wants to turn to the left, change direction by 90 degrees. Instead of just turning round his body, he makes the turn along an arc. I think the problem asks the length of the arc.


    Attached Files:

  4. Mar 1, 2012 #3
    So you mean like a quarter of the circumference?

    C = 2∏r
    C = 46∏
    C = 145
    C/4 = 36.25m

    Therefore the smallest turn the boy can make is 36.25m?
  5. Mar 1, 2012 #4


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    Homework Helper

    That is, (pi/2)*R. . Yes, I think, that was the question.

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