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A coin is placed 13.0 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 38.0 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?
A coin is placed 13.0 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 38.0 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?
Thecoefficientofstatic frictionbetween thecoinand theturntableisμs = 0.2098Calculation of the coefficient of static friction:SinceR = 13.0 cm = 0.13 mω = 38 r.p.m = (38 rev/min)(2π/1 rev)(1 min/60 s) = 3.98 rad/sNow here we appliedNewton's Second LawSo,∑ Fc = m*ac⇒ Ff = m*acHere Ff is theforce of friction:NowFf = μs*N = μs*(m*g) ⇒ Ff = μs*m*gAndac = ω²*RSo,μs*m*g = m*(ω²*R) ⇒ μs = ω²*R / gμs = (3.98 rad/s)²*(0.13 m) / (9.81 m/s²)μs = 0.2098learn more about thespeedhere:brainly.com/question/3902437...
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