A coin placed 30.0 cm from the center of a rotating, horizontal turntable slips when its speed is 50.0 cm/s. What is the coefficient of static friction between coin and turntable?
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Given: Distance from center, r = 0.3 m Speed, v = 0.5 m/s Mass, m = given Centrifugal force = m * v^2 / r Show more…
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A coin placed 30.0 $\mathrm{cm}$ from the center of a rotating hori- zontal turntable slips when its speed is 50.0 $\mathrm{cm} / \mathrm{s}$ . (a) What force causes the centripetal acceleration when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between the coin and turntable?
A coin placed 30.2 cm from the center of a rotating, horizontal turntable slips when its speed is 52.0 cm/s. What is the coefficient of static friction between coin and turntable? (Hint: When the coin just begins to slip, the force of static friction equals the centripetal force.)
Adi S.
(II) 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?
CIRCULAR MOTION; GRAVITATION
Kinematics of Uniform Circular Motion
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