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Minimum Coefficient of Friction calculation using Centripetal force

  1. Nov 3, 2011 #1
    A() 61 kg child stands at the rim of a merrygo-round of radius 1.05 m, rotating with an
    angular speed of 1.87 rad/s
    The acceleration of gravity is 9.8 m/s^2

    What is the child’s centripetal acceleration?
    Answer in units of m/s

    What minimum force between her feet and
    the floor of the carousel is required to keep
    her in the circular path?
    Answer in units of N


    What minimum coefficient of static friction is
    required?

    3. The attempt at a solution

    1)
    I found the centripetal acceleration since a = v^2/r and v = 1.9635 and r = 1.05.
    The acceleration is 3.671745

    2) I found the force which is F=ma
    F = 61(3.671745) = 223.976445 N

    3) I'm confused as to how I should solve for the minimum of coefficient of friction.

    Help is appreciated!!
     

    Attached Files:

  2. jcsd
  3. Nov 3, 2011 #2

    PeterO

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    Homework Helper

    Assumin it is the word "minimum"that is confusing you, remember that the usual formula F = μN refers to the maximum friction possible, so if the coefficient of friction was 2.3, only a fraction of the potential friction would be necessary.
    If the friction force required matches the maximum friction available, then we have the minimum coefficient.

    If it is the idea of friction that is confusing you, you better re-read the section of your text on friction
     
  4. Nov 3, 2011 #3
    But, I have to solve for the coefficient of static friction. The equation for static friction is f ≤ μN , the one you showed is for kinetic friction.
     
  5. Nov 3, 2011 #4

    PeterO

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    Homework Helper

    OK, replace the = in my response with ≤ and re-read.

    EDIT: I don't distinguish between static and kinetic - since I am always aware which one I am using.
     
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