Question
Consider a small object at 3.72cm from the axis of a phonograph turntable that operates at 22.8 rev/min . a) Assuming the object does not slip, what is its acceleration in cm/s^2? b) What is the minimum coefficient of friction between the object and the turntable ? c) Suppose that the turntable achieved its angular speed by starting from rest and undergoing a constant angular acceleration for 0.345s. Calculate the minimum coefficient of static friction required for the object not to slip during the startup period.
Explanation / Answer
Data: Distance, r = 3.72 cm Angular speed, = 22.8 rev/min = 22.8 * ( 2 / 60 ) rad/s = 2.39 rad/s Time, t = 0.345 s Solution: (a) Acceleration, a = ^2 r = (2.39)^2 * 3.72 = 21.2 cm/s^2 Ans: Acceleration, a = 21.2 cm/s^2 (b) Frictional force = Centripetal force m g = m a g = a = a / g = 21.2 / 980 = 0.02 Ans: Coefficient of friction = 0.02 (c) = o + t 2.39 = 0 + * 0.345 = 6.93 rad/s^2 Tangential acceleration, at = r = 3.72 * 6.93 = 25.8 cm/s^2 Centripetal acceleration, ac = 21.2 cm/s^2 [ from part(a) ] Net acceleration, a = [ (ac)^2 + (at)^2 ] = [ (21.2)^2 + (25.8)^2 ] = 33.37 cm/s^2 s m g = m a s g = a s = a / g = 33.37 / 980 = 0.034 Ans: Coefficient of static friction = 0.034 (b) Frictional force = Centripetal force m g = m a g = a = a / g = 21.2 / 980 = 0.02 Ans: Coefficient of friction = 0.02 (c) = o + t 2.39 = 0 + * 0.345 = 6.93 rad/s^2 Tangential acceleration, at = r = 3.72 * 6.93 = 25.8 cm/s^2 Centripetal acceleration, ac = 21.2 cm/s^2 [ from part(a) ] Net acceleration, a = [ (ac)^2 + (at)^2 ] = [ (21.2)^2 + (25.8)^2 ] = 33.37 cm/s^2 s m g = m a s g = a s = a / g = 33.37 / 980 = 0.034 Ans: Coefficient of static friction = 0.034