Solution to 2001 Problem 2

The force of friction provides the entire centripetal acceleration. So, we can find that maximum distance be equating the maximum frictional force with the centripetal force required to maintain circular motion:
\begin{align*}m \omega^2 r = m g \mu_s \Rightarrow r = \frac{g \mu_s}{\omega^2} = \frac{g \cdot 0.3}{\left(33.3 \cdot 2 \pi/(...
Therefore, answer (D) is correct.

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