HW3soln

HW3soln - ME 375 Homework 3 Solution February 6 2009...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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/O-equation 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...
View Full Document

This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue.

Page1 / 7

HW3soln - ME 375 Homework 3 Solution February 6 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online