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ASE 365 - HW 2 Solutions

ASE 365 - HW 2 Solutions - Solutions Homework Set 2 1...

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Unformatted text preview: Solutions: Homework Set 2 1. Problem 2.4. Assuming small rotation θ (clockwise because of mg ) and measuring m ’s displacement x positive downward, spring forces are f S 1 = k 1 ( x- aθ ), f S 2 = k 2 bθ . From FBD of (massless) rigid bar, ∑ M O = k 1 ( x- aθ ) a- k 2 b 2 θ = 0. From FBD of m , ∑ F x = m ¨ x = mg- k 1 ( x- aθ ). Eliminate θ to get EOM: m ¨ x + k eq x = mg, k eq = k 1 k 2 b 2 k 1 a 2 + k 2 b 2 Equilibrium position (particular solution): x eq ′ m = mg k eq . If θ eq ′ m is of interest, obtain from ∑ M O equation: θ eq ′ m = k 1 a k 1 a 2 + k 2 b 2 x eq ′ m . Natural frequency: ω n = radicalbig k eq /m . 2. Problem 2.8. From Problem 1.8, EOM: 3( R- r ) ¨ θ +2 g sin θ = 0. For small θ , sin θ ≈ θ . Natural frequency: ω n = radicalbig 8 g/ 9 R . 3. Problem 2.15. EOM: m ¨ x + k eq x = 0 where k eq = parenleftbigg 1 k 1 + k 2 + 1 k 3 parenrightbigg- 1 . Therefore ω n = radicalbig k eq /m . Response to ICs: x ( t ) = x cos ω n t + v ω n...
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ASE 365 - HW 2 Solutions - Solutions Homework Set 2 1...

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