Analytical Mech Homework Solutions 112

Analytical Mech Homework Solutions 112 - 10 40 l n lm m = 2...

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10 2 40 1 m m l nl m M l g α π =≅    Letting lm ; l ; we obtain 1.3 m = 1.0 = m 3 1.15 10 ≅⋅ 2 cm I ma ( l mass = 8.11 (a) al in rim) 2 22 2 rim I ma ma ma =+= 2 rim I a T mga g ππ == 2 zxy c m 2 I II m a =+ = & I = ( ) cm I = (b) 2 2 cm ma I = & hence 2 3 rim ma I ma ma & 3 2 rim I a T mga g & 8.12 cm mx mg T = ±± T G mg G a cm I aT ω = ± cm x a = ± 2 2 5 cm I ma = 2 12 2 55 cm cm cm cm Ix mx mg mg ma mg mx aaa =− ± 5 7 cm x g = x 8.13 When two men hold the plank, each supports
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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