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A coin is placed 19 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains stationary with respect to the turntable until a rate of 32 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Respuesta :

Answer:

0.22

Explanation:

Given that

Radius of turntable, R = 19 cm = 0.19 m

Speed of rotation the turntable, f = 32 rpm = 32/60 = 0.533 rev/sec

from free body diagram and forces resolution, we have

μ(s) * m * g = m * v²/R

μ(s) * g = v²/R

μ(s) = v² / gR, recall, v = wr, so

μ(s) = w²R² / gR

μ(s) = w²R / g, recall again, w = 2πf

μ(s) = (2πf)²R / g, on substituting,

μ(s) = (2 * 3.142 * 0.533)² * 0.19 / 9.8

μ(s) = 3.35² * 0.19 / 9.8

μ(s) = 2.13 / 9.8

μ(s) = 0.22

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