Question

A small button placed on a horizontal rotating platform with diameter 0.326 mm will revolve with the platform when it is brought up to a rotational speed of 42.0 rev/minrev/min , provided the button is a distance no more than 0.149 mm from the axis.1) What is the coefficient of static friction between the button and the platform? 2) How far from the axis can the button be placed, without slipping, if the platform rotates at 57.0 rev/min?

Answer 1

Answer:

1) The coefficient of static friction between the button and the platform is   0. 294

2) The radius is 0.897

Explanation:

knowing that Centipetal force is provided by friction, so

mω²r =  μmg

then, the coefficient of friction μ is

μ = ω²r / g

   = (2π 42/60)² x 0.149 / 9.8

μ = 0. 294

mω²r = μmg

⇒ r = μg / ω²

    = (0.326 x 9.8) / (2π 57/60)²

  r = 3.1948 / 35.6293

    = 0.897