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Unformatted text preview: ME 375 Homework 3 : Solution February 6, 2009 Problem 1 Part A bracketleftbigg M 1 J eff bracketrightbiggbracketleftbigg ¨ x 1 ¨ θ bracketrightbigg + bracketleftbigg B 1 B eff bracketrightbiggbracketleftbigg ˙ x 1 ˙ θ bracketrightbigg + bracketleftbigg 2 K 1 K 1 R 1 K 1 R 1 K eff bracketrightbiggbracketleftbigg x 1 θ bracketrightbigg = bracketleftbigg τ 1 bracketrightbigg Part B From equation: M 1 ¨ x 1 + B 1 ˙ x 1 + 2 K 1 x 1 K 1 R 1 θ 1 = 0 we can determine that : θ 1 = M 1 K 1 R 1 ¨ x 1 + B 1 K 1 R 1 ˙ x 1 + 2 K 1 K 1 R 1 x 1 The first and second order derivatives for θ 1 are calculated as follow: ˙ θ 1 = M 1 K 1 R 1 x (3) 1 + B 1 K 1 R 1 ¨ x 1 + 2 K 1 K 1 R 1 ˙ x 1 ¨ θ 1 = M 1 K 1 R 1 x (4) 1 + B 1 K 1 R 1 x (3) 1 + 2 K 1 K 1 R 1 ¨ x 1 Substituting in equation : J eff ¨ θ 11 + B eff ˙ θ 1 + K eff θ 1 K 1 R 1 x 1 = τ 1 1 We get the I/Oequation for the system with τ 1 as the input and x 1 as the output: J eff M 1 K 1 R 1 x (4) 1 + bracketleftbigg J eff B 1 K 1 R 1 + B eff M 1 K 1 R 1 bracketrightbigg x (3) 1 + bracketleftbigg J eff 2 K 1 K 1 R 1 + B eff B 1 K 1 R 1 + K eff M 1 K 1 R 1 bracketrightbigg...
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This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue.
 Spring '10
 Meckle

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