me304_hw2_sol - ME304/Spring 2005/Sections...

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Unformatted text preview: ME304/Spring 2005/Sections 02+03+04/HW2Solutions Page 1 / 6 SOLUTIONS TO HOMEWORK 2 for Sections 02, 03 & 04 Prepared by Gökhan BAYAR (C-204), Atilla BAYRAM (A-148), Erdinç İ Y İ AY (C-209) Problem 2 The free body diagrams for each element can be drawn as shown below. θ 1 I 1 T 1 T k Motor T k T k Elastic Shaft I 2 R θ 2 T k T b Drum P T b T b Torsional Damper mg v(t) P Load MIDDLE EAST TECHNICAL UNIVERSITY MECHANICAL ENGINEERING DEPARTMENT ME 304 CONTROL SYSTEMS SPRING 2005 ME304/Spring 2005/Sections 02+03+04/HW2Solutions Page 2 / 6 The equation of motion for the motor is 1 1 θ & & I T m = Σ ( 1 ) 1 1 1 θ & & I T T k = − ( 2 ) Note that ) ( 2 1 θ θ − = t k k T ( 3 ) Using (3) in (2), one obtains 1 1 2 1 1 ) ( θ θ θ & & I k T t = − − ( 4 ) Taking the Laplace transform of both sides of equation (4) yields ) ( ) ( ) ( ) ( 1 2 1 2 1 1 s s I s k s k s T t t θ θ θ = + − ( 5 ) Rearranging this equation ) ( ) ( ) ( ) ( 1 2 1 2 1 s T s k s k s I t t = − + θ θ ( 6 ) The equation of motion for the drum can be obtained as 2 2 θ & & I T d = Σ ( 7 ) 2 2 ) ( θ & & I R P T T b k = − − ( 8 ) Note that 2 θ & t b b T = ( 9 ) Using (3) and (7) in (6), one obtains 2 2 2 2 1 ) ( ) ( θ θ θ θ & & & I R P b k t t = − − − ( 1 ) Taking the Laplace transform of both sides of equation (8) yields...
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me304_hw2_sol - ME304/Spring 2005/Sections...

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