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A wizard sits 3 m from the center of a rotating platform, coeefficient of static fric

  1. Jun 18, 2006 #1
    The wizard has a mass of 80 kg and sits 3 m from the center of a rotating platform. Due to the rotation his speed is increased from rest by [itex]\dot{v}\,=\,0.4\,\frac{m}{s^2}[/itex]. If the coefficient of static friction between his clothes and the platform is [itex]\mu_s\,=\,0.3[/itex], determine the time requiredto cause him to slip.


    Here is what I have, it is off by 0.4 seconds though:







    The book has 7.39s and the units don't match!
    Last edited: Jun 19, 2006
  2. jcsd
  3. Jun 19, 2006 #2
    You are neglecting the kinematics of this problem. Your work is perfect to the point where you have found the force of friction. Which can sustain a maximum acceleration of u*g. (2.94 m/s^2)

    Now think of the in plane components of acceleration. The problem states that he is initially accelerating at 0.4 m/s^2. Because he has no initial velocity this must mean that this acceleration is purely tangential. Notice that this component of acceleration is always tengential to the path the wizard. But remember, as the velocity increases to be non-zero we obtain a centripetal component of acceleration that will always act radially inwards with magnitue v^2/r.

    Now if we apply basic kinematics v = vdot*t and by application of the pythagorean theorem the acceleration as result of both the centripetal and tangential accelerations is then.

    a = sqrt([(vdot*t)^2/r]^2 + vdot^2)

    notice the first term corresponds to the centripetal acceleration
    the second term corresponds to the tangential acceleration

    since F = ma , it must then follow that

    u*(g/m) = sqrt([(vdot*t)^2/r]^2 + vdot^2)

    now we simply solve for t.

    You can check your dimensions and will find here that by solving for t we will have it such that dim[t] = T

    Also, this will yield an answer of 7.39 s when using your numbers along with g = 9.8 m/s^2.

    Good Luck!
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