hw6solution - Homework 6 - Solution 1. Draw the FBDs for...

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Homework 6 Solution 1. Draw the FBDs for the two blocks: m 1 m 2 x 1 x 2 r 11 kx 22 21 (2 ) cx x ±± F C θ Note that the displacement at the top of the disk is twice the displacement x 1 at the disk’s center. Apply Newton’s 2nd law: M C J C ̈ k 1 x 1 r c x ̇ 2 2 x ̇ 1  2 r J C ̈ F x ma x c x ̇ 2 2 x ̇ 1 k 2 x 2 m 2 x ̈ 2 The moment of inertia of the disk with respect to the point of contact is J C J m 1 r 2 1 2 m 1 r 2 m 1 r 2 3 2 m 1 r 2 From kinematics we have x 1 r The moment equation can then be rewritten as 3 2 m 1 r 2 x ̈ 1 r c x ̇ 2 2 x ̇ 1  2 r k 1 x 1 r 0 After some algebraic simplification, the equations of motion become 3/2 m 1 x ̈ 1 4 cx ̇ 1 2 cx ̇ 2 k 1 x 1 0 m 2 x ̈ 2 2 cx ̇ 1 cx ̇ 2 k 2 x 2 0 Recast the equations of motion in the matrix form: 3 m 1 /2 0 0 m 2 x ̈ 1 x ̈ 2 4 c 2 c 2 cc x ̇ 1 x ̇ 2 k 1 0 0 k 2 x 1 x 2 0 0 2. The FBDs of the rod and the block are drawn:
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O m x 2 () kxL θ M (2 ) cL ± 1 (3 ) kL
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This note was uploaded on 04/30/2008 for the course ME 3504 taught by Professor Tschang during the Spring '08 term at Virginia Tech.

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hw6solution - Homework 6 - Solution 1. Draw the FBDs for...

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