SolSec 4_7 - Problems and Solutions for Section 4.7(4.76...

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Problems and Solutions for Section 4.7 (4.76 through 4.79) 4.76 Use Lagrange's equation to derive the equations of motion of the lathe of Fig. 4.21 for the undamped case. Solution: Let the generalized coordinates be ! 1 , 2 and 3 . The kinetic energy is T = 1 2 J 1 ! 1 2 + 1 2 J 2 ! 2 2 + 1 2 J 3 ! 3 2 The potential energy is U = 1 2 k 1 2 " 2 ( ) 2 + 1 2 k 2 3 " 2 ( ) 2 There is a nonconservative moment M(t) on inertia 3. The Lagrangian is L = T ! U = 1 2 J 1 ! " 1 2 + 1 2 J 2 ! 2 2 + 1 2 J 3 ! 3 2 ! 1 2 k 1 2 ! 1 ( ) 2 ! 1 2 k 2 3 ! 2 ( ) 2 Calculate the derivatives from Eq. (4.136): ! L ! ! 1 = J 1 ! 1 d dt ! L ! ! 1 # \$ % ( = J 1 !! 1 ! L ! ! 2 = J 2 ! 2 d dt ! L ! ! 2 # \$ % ( = J 2 !! 2 ! L ! ! 3 = J 3 ! 3 d dt ! L ! ! 3 # \$ % ( = J 3 !! 3 ! L ! 1 = ) k 1 1 + k 1 2 ! L ! 2 = ) k 1 1 ) k 1 + k 2 ( ) 2 + k 2 3 ! L ! 3 = ) k 2 2 ) k 2 3 Using Eq. (4.136) yields J 1 !! 1 + k 1 1 " k 2 2 = 0 J 2 !! 2 " k 1 1 + k 1 + k 2 ( ) 2 " k 2 3 = 0 J 3 !! 3 " k 2 2 + k 2 3 = M t ( ) In matrix form this yields

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1 0 0 0 J 2 0 0 0 J 3 ! " # # # \$ % !! + k &
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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SolSec 4_7 - Problems and Solutions for Section 4.7(4.76...

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