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*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.2.

Respuesta :

Answer:

  w = 0.626 rad / s, v = 3.13 m/s

Explanation:

For this exercise let's use Newton's second law

        F = m a

Where the force is a friction force and the acceleration is centripetal,

      a = v² / r = w² r

The formula for friction force

     fr = μ N

  In a free body diagram

       N- W = 0

      W = N

The frictiμon outside goes from zero to the maximum value, let's calculate the speed for the maximum value of the friction force, replace

       μ m g = m w² r

       w = √ μ g / r

Let's calculate

     w = √(0.2 9.8 / 5)

     w = 0.626 rad / s

angular and linear velocity are related

      v = w r

      v = 0.626 5

      v = 3.13 m/s