A 27.7 g metal cylinder is placed on a turntable, with its center 84.7 cm from t

Question

A 27.7 g metal cylinder is placed on a turntable, with its center 84.7 cm from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is ?s= 0.645. A thin, massless string of length 84.7 cm connects the center of the turntable to the cylinder, and initially, the string has zero tension in it. Starting from rest, the turntable very slowly attains higher and higher angular velocities, but the turntable and the cylinder can be considered to have uniform circular motion at any instant. Calculate the tension in the string when the angular velocity of the turntable is 60 rpm (rotations per minute).

Explanation / Answer

Given Mass of the cylinder is , m = 0.0277 kg Distance from the mass to the table's center is , r = 0.847 m The coefficient of static friction , ? = 0.645 Angular velocity , ? = 60rpm = 6.28 rad/s Net force on the cylinder is T = mr?2 -?mg T = 0.0277 kg (0.847 m *(6.28 rad/s)^2 - 0.645 *9.8 m/s^2) T = 0.75N Thus, the tension in the string is 0.75N Given Mass of the cylinder is , m = 0.0277 kg Distance from the mass to the table's center is , r = 0.847 m The coefficient of static friction , ? = 0.645 Angular velocity , ? = 60rpm = 6.28 rad/s Net force on the cylinder is T = mr?2 -?mg T = 0.0277 kg (0.847 m *(6.28 rad/s)^2 - 0.645 *9.8 m/s^2) T = 0.75N Thus, the tension in the string is 0.75N T = 0.75N Thus, the tension in the string is 0.75N

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