Suppose that the metal cylinder in the last problem has a mass of0.10 kg and tha

Question

Suppose that the metal cylinder in the last problem has a mass of0.10 kg and that the coefficient of static friction between thesurface and the cylinder is 0.12. If the cylinder is 0.2 m from thecenter of the turntable, what is the maximum speed that thecylinder can move along its circular path without slipping off theturntable? A)0.5 m/s B)1 m/s C)1.5 m/s D)2.0 m/s E)2.5 m/s The centripetal force needed to keep the cylindermoving on a circular path is F = ma = mv^2/r. This force mustbe due to static friction between the bottom of cylinder and theturntable. Static friction prevents the cylinder from slidingoff the rotating table. The maximum force static friction canprovide is 0.12*N = 0.12*mg. = 0.1176 N. Setting this equalto mv^2/r = 0.5v^2 and solving for v yields the maximum speed thecylinder can have without slipping off the turntable.

Explanation / Answer

It's easier to write m v2 / R = mg        centripetal force =frictional force v = ( g R) = (.12 * 9.8 * .2) = .48 m/s˜ .5 m/s

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