Question 1182825
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A small particle is placed at the edge of a phonograph turntable of diameter 30 cm. If the turntable rotates 
at 45 revolutions per minute, what must be the coefficient of friction so the particle will {{{highlight(cross(bot))}}} <U>not</U> slide out?
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<pre>
It is easy.


Simply EQUATE the centripetal force and the friction force.


The centripetal force acting to the particle is

    {{{F[c]}}} = {{{(m*v^2)/R}}}.        (1)


The friction force is  {{{F[fr]}}} = k*m*g,  where m is the mass of the particle, k is the friction coefficient and g is the gravity acceleration.


So, the equation takes the form

    {{{(m*v^2)/R}}} = k*m*g.     (2)


Cancel the mass "m" in both sides

    {{{v^2/R}}} = k*g.           (3) 


Use  v = {{{2*pi*R*(45/60)}}} = {{{2*3.14*R*(3/4)}}} = {{{2*3.14*0.15*(3/4)}}} = 0.7065  m/s  (the linear speed).


So, your equation (3) takes the form


    {{{0.7065^2/0.15}}} = 10k,   or   10k = 3.328


which gives for the friction coefficient k


    k = {{{3.328/10}}} = 0.3328   (dimensionless value).


<U>ANSWER</U>.  The friction coefficient should be at least 0.3328.
</pre>

Solved and carefully explained.