contestada

A coin is placed 12.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 37.0 rpm (revolutions per minute) is reached, at which point the coin slides off.

Respuesta :

Explanation:

For the given situation,

          Frictional force = centripetal force

        [tex]\mu mg = m \omega^{2}r[/tex]

Cancel out the common terms, we get

       [tex]\mu = \frac{\omega^{2}r}{g}[/tex]

                   = [tex]\frac{(3.87)^{2} \times 0.12 m}{9.8}[/tex]  (as 1 rpm = 0.10472 rad/sec)

                   = [tex]\frac{1.797}{9.8}[/tex]

                   = 0.183

thus, we can conclude that coefficient of static friction in the given situation is 0.183.

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