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3 comments:
In this problem, the friction force F = mu*N = mu * W * cos(theta). The force opposing F is W * sin(theta). When mu > tan(theta), F will always be larger, meaning the block has acceleration up the bar. This contradicts the initial condition where the acceleration is toward O. Omega_min then does not exist.
An imaginary omega would allow an acceleration up the bar rather than down.
Good responses. There are no real values for omega that allow for the block to move inward. Therefore, if the block slips, it must slip outward.
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