SOLUTION: 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

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Question 1182825: 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 bot slide out?
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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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%28cross%28bot%29%29 not slide out?
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It is easy.


Simply EQUATE the centripetal force and the friction force.


The centripetal force acting to the particle is

    F%5Bc%5D = %28m%2Av%5E2%29%2FR.        (1)


The friction force is  F%5Bfr%5D = 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

    %28m%2Av%5E2%29%2FR = k*m*g.     (2)


Cancel the mass "m" in both sides

    v%5E2%2FR = k*g.           (3) 


Use  v = 2%2Api%2AR%2A%2845%2F60%29 = 2%2A3.14%2AR%2A%283%2F4%29 = 2%2A3.14%2A0.15%2A%283%2F4%29 = 0.7065  m/s  (the linear speed).


So, your equation (3) takes the form


    0.7065%5E2%2F0.15 = 10k,   or   10k = 3.328


which gives for the friction coefficient k


    k = 3.328%2F10 = 0.3328   (dimensionless value).


ANSWER.  The friction coefficient should be at least 0.3328.

Solved and carefully explained.