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A record turntable is rotating at 33 1/3 rev/min. A watermelon seed is on the turntable 6.1 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. .7432 Correct: Your answer is correct. m/s2 (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? .0758 Correct: Your answer is correct. (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.20 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
A record turntable is rotating at 33 1/3 rev/min. A watermelon seed is on the turntable 6.1 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. .7432 Correct: Your answer is correct. m/s2 (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? .0758 Correct: Your answer is correct. (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.20 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
Answer:0.075760.13235Explanation:= Angular speed =r = Distance from the center = 6.1 cmg = Acceleration due to gravity = 9.81 m/s²Acceleration of the seed would beThe acceleration of the seed isFrictional force is given byThe coefficient of friction is 0.07576Transverse acceleration is given byThe resultant acceleration is given byThe coefficient of friction is 0.13235...
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