Question

"A particle of dust lands 41.0 mm from the center of a compact disc (CD) that is 120 mm in diameter. The CD speeds up from rest, and the dust particle is ejected when the CD is rotating at 84.0 revolutions per minute. What is the coefficient of static friction between the particle and the surface of the CD?"

Answer 1

Answer:

The coefficient of static friction is  $\mu = 0.474$

Explanation:

From the question we are told that

   The position of the particle is  $x = 41.00 \ mm = 0.041 \ m$

     The diameter of the CD is   $d =120 \ mm = 0.12 \ m$

      The radius of the CD is evaluated as  $r = \frac{d}{2} = \frac{0.12}{2} = 0.06$

     The angular velocity of the CD  when particle was ejected $w = 84.0 rpm = 84 .0 * \frac{2 * \pi}{60} = 8.7976 \ rad/s$

At the instant just before the particle is ejected from the CD

    The frictional force of the particle  =  centrifugal force on the particle

So  

         $\mu * m * g = mw^2 r$

=>       $\mu * g = w^2 r$

=>       $\mu = \frac{8.7976^2 * 0.06}{ 9.8}$

=>      $\mu = 0.474$