The coefficient of friction between a certain brass block and a large revolving turntable is µ = 0.23. How far from the axis of rotation can the block be placed before it slides off the turntable if it is rotating at 33 1/3 rev/min?(adsbygoogle = window.adsbygoogle || []).push({});

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the only force on the block is the friction in the same direction as the centripetal acceleration, right?

so...

f = m * a(centripetal)

0.23 * N = m * v^2 / r = mRω^2, where ω = 2*pi*F(frequency), F = 1/(33 1/3) = 0.03 and N = mg.

so, simplifying,

0.23*m*g = 100/3*m*R*ω^2,

0.23*g = 100/3*R*w^2

0.0069*g = R(w^2)

R = 0.06762 / w^2, where w^2 = 0.03553

R = 1.9032 meters, which, apparently, is wrong. but why?

my work seems right, and the answer sounds right...

what have i done wrong?

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# Homework Help: Circular Motion and Friction of brass block

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