4. A small cylinder rests on a circular turntable that isrotatingclockwise at a

Question

4. A small cylinder rests on a circular turntable that isrotatingclockwise at a constant speed. The cylinder is at adistance of r =12 cm from the center of the turntable. Thecoefficient of staticfriction between the bottom of the cylinderand the surface of theturntable is 0.45. What is the period of thecylinder if it istravevling at the maximum speed vmax that thecylinder can havewithout slipping off the turntable?
A. 1.42 s B. 1.03 s C. 0.69 s D. 0.10 s E. 0.014 s

Explanation / Answer

The rotational force (centripetal) is mv2/r = F The frictional force which opposes the centripetal = *m*g The two forces are equal. mv2/r = *m*g ==> v2/r = *g = 0.45 g = 9.81 m/s2 r = 12 cm = 0.12 m v2 / 0.12 = 0.45 * 9.81 v2 = 12 * 0.45 * 9.81 v = 0.7278 m/s Therefore T = 2 r / v                   = 2 * 3.14 * 0.12 / 0.7278                   = 1.03 s

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