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[Rotational dynamics] cube sliding on a dish
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[QUOTE="FranzDiCoccio, post: 5453603, member: 141434"] Sure, of course Yes, that's what I thought, and why I assumed the simplest situation, i.e. friction with the ground. This should be at least constant, since it depends on the weight of the cube. It gives no information about the details of the friction. I thought it might be with the wall just because the detailed solution concludes "at equilibrium only the centripetal force and the friction force act on the cube, so that [itex]F_a=m w_f r^2[/itex]". The book's solution only depends on the final angular velocity, which is determined by the conservation of the angular momentum. I mean, that velocity is the same irrespective of how long the cube runs before stopping. Suppose the friction coefficient is doubled. I expect that the cube runs half of the circumference before stopping. But my soluton depends on the length of the path of the cube. So either it is correct, or I'm missing something, and that length is useless. [/QUOTE]
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[Rotational dynamics] cube sliding on a dish
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