a 20g metal cylinder is placed on a turntable, with its center 80cm from the tur

Question

a 20g metal cylinder is placed on a turntable, with its center 80cm from the turntable's center. the coefficient of static friction between the cylinder and the turntable's surface is 0.8. a thin massless string of length 80cm 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.

Explanation / Answer

Given w = 60rpm = 60*2pi/60 = 2pi rad/s
On the bodies frame, centrifugal force = mw^2r = 0.02*(2pi)^2*0.8 = 0.631 N (away from centre)
Friction = mu*mg = 0.8*0.02*9.8 = 0.1568 N (towards centre)
Tension, T + f = 0.631
T = 0.631 - 0.1568 = 0.4742 N

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