A 18.0-g metal cylinder is placed on a turntable, with its center 84 cm from the

Question

A 18.0-g metal cylinder is placed on a turntable, with its center 84 cm from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is ?s = 0.75. A thin, massless string of length 84 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 40 rpm (rotations per minute).

Explanation / Answer

frequency = 40/60 Hz ? = 2pif = 2pi/40/60 = 3pi rad/s the centrifugal force is Fc = m?^2*R Fc = 0.018*(3pi)^2*0.84 N=1.343 the friction force is Ff = mgµ Ff = 0.018*9.81*0.75=0.1323 The tension is T = Fc - Ff T = 0.018*(3pi)^2*0.84- 0.018*9.81*0.75 T = 1.21 N

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