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hwk05_soln

# hwk05_soln - Solution for Homework 5.1 vG = vA rG/A = x i 2...

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Solution for Homework 5.1 ~v G = ~v A + × ~ r G/A = ˙ x ˆ i + L 2 ˙ ω ˆ k × sinθ ˆ i - cosθ ˆ j = ˙ x + L 2 ˙ θcosθ ˆ i + L 2 ˙ θsinθ ˆ j v 2 G = ˙ x + L 2 ˙ θcosθ 2 + L 2 ˙ θsinθ 2 = ˙ x 2 + ( Lcosθ ) ˙ x ˙ θ + L 2 4 ˙ θ 2 T = 1 2 (2 m ) v 2 A + 1 2 ( m ) v 2 G + 1 2 I G ˙ θ 2 = 1 2 (2 m ) ˙ x 2 + 1 2 ( m ) ˙ x 2 + ( Lcosθ ) ˙ x ˙ θ + L 2 4 ˙ θ 2 + 1 2 1 12 mL 2 ˙ θ 2 = 1 2 (3 m ) ˙ q 2 1 + 1 2 ( mcosθ ) ˙ q 1 ˙ q 2 + 1 2 1 3 m ˙ q 2 2 where q 1 = x and q 2 = . Therefore, [ M ] = [ m ] ~ θ =0 = 3 m m 2 cosθ m 2 cosθ m 3 ~ θ =0 = m 6 18 3 3 2 U = 1 2 kx 2 - mg L 2 cosθ = 1 2 kq 2 1 - mg L 2 cos q 2 L

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Therefore, [ K ] = 2 U ∂q i ∂q j ~ θ =0 = k 0 0 mg 2 L cosθ ~ θ =0 = k 0 0 mg 2 L = mg 2 L 2 α 0 0 1 where α = kL/mg . Eigenvalue problem : ~ q ( t ) = ~ Xe iωt - 18 μ + 2 α - 3 μ - 3 μ - 2 μ + 1 X 1 X 2 = 0 0 where μ = ω 2 ( L/ 3 g ). Characteristic equation and natural frequencies : 0 = - 18 μ + 2 α - 3 μ - 3 μ - 2 μ + 1 = ( - 18 μ + 2 α )( - 2 μ + 1) - 9 μ 2 = 27 μ 2 - (18 + 4 α ) μ + 2 α Solving: μ 1 , 2 = 18 + 4 α ± p (18 + 4 α ) 2 - 216 α 54 with ω j = p (3 g/L ) μ j .
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hwk05_soln - Solution for Homework 5.1 vG = vA rG/A = x i 2...

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