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homework10sol.20100421.4bcf9b7c0c8001.03937942

homework10sol.20100421.4bcf9b7c0c8001.03937942 - TAM 412...

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TAM 412, Homework 10, Selected answers Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 1. Here, q 1 q 2 q 3 = ( C 1 e it + C 2 e - it ) 1 0 1 + ( C 3 e it + C 4 e - it ) 1 1 0 + parenleftBig C 5 e i 2 t + C 6 e - i 2 t parenrightBig 0 1 0 from which it follows that - C 3 C 4 q 2 3 + ( C 2 C 3 + C 1 C 4 ) q 3 ( q 1 - q 3 ) - C 1 C 2 ( q 1 - q 3 ) 2 = ( C 2 C 3 - C 1 C 4 ) 2 The special case when a = c = 1 and b = 0 correspond to the four sign combinations given by ( q 3 , q 1 - q 3 ) = ( ± cos t, ± sin t ) 3. Let the two end points of the rod have cartesian coordinates ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) relative to a coordinate system with origin at P and with gravity along the negative Z -axis. It follows that L = m 6 ( ˙ x 2 1 + ˙ x 1 ˙ x 2 + ˙ x 2 2 + ˙ y 2 1 + ˙ y 1 ˙ y 2 + ˙ y 2 2 + ˙ z 2 1 + ˙ z 1 ˙ z 2 + ˙ z 2 2 ) - mg 2 ( z 1 + z 2 ) The generalized forces corresponding to the constraints are then given by F x 1 = 2 λ 1 ( x 1 - x 2 ) + 2 λ 2 x 1 F y 1 = 2 λ 1 ( y 1 - y 2 ) + 2
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