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Fall 2010 - Speliotopoulos - Final (solution)

# Fall 2010 - Speliotopoulos - Final (solution) - AtVo...

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Final Exam Solution: Problem 4 Physics 7A, UC Berkeley, Fall 2010, Prof. A. Speliotopoulos Grader: Aaron Alpert Force Considerations. The normal force is N = mg , and therefore, the maximum frictional force is f max = μ s N = μ s mg . The rolling without slipping condition 1 is ¯ v = r ω , which implies ¯ a = r α . (Simply take the time deriva- tive to accomplish this.) That implies 1 m F = r I ¯ τ 1 m F = r 2 5 mr 2 ¯ τ = 5 2 mr ¯ τ (1) The two relevant forces are friction and the spring force. When the friction is maximized, so is the spring force. At the maximal condition, f max F s,max m = 5 2 mr ( rf max ) F s,max = 7 2 f max (2) As previously stated, the maximum frictional force is μ s mg , which leads to F s,max = 7 2 μ s mg .
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Fall 2010 - Speliotopoulos - Final (solution) - AtVo...

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