ADS Homework 5 - Y M Y M Y M sin 6 1 2 4 12 1 4 1 12 8 12 10 2 1 2-= t a p kY Y C Y M sin 6 1 2 4 33 4-= Equation of Motion Circular frequency of

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Homework 5: Write the equation of motion for the given system. Solution: The total work done by the forces and moments must be equal to zero. There will be no lateral displacement in the system because of the simple support. 0 3 sin 2 1 12 1 2 2 12 8 12 10 2 2 2 2 2 2 0 2 2 2 = - - - - - - - - - - - Y t a p a Y a Y k a Y a Y Ma Y Y M Y kY a Y a Y Ma a Y a Y Ma Y Y M Y Y M Y Y C δ ω θ It is given that; 2 ka k = t a p kY Y C Y M Y M Y M
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Unformatted text preview: Y M Y M Y M sin 6 1 2 4 12 1 4 1 12 8 12 10 2 1 2-= + + + + + + + t a p kY Y C Y M sin 6 1 2 4 33 . 4-= + + : Equation of Motion : Circular frequency of the external force 1...
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This note was uploaded on 11/23/2010 for the course DEP 503E 10852 taught by Professor Prof.zekaicelep during the Spring '10 term at Istanbul Technical University.

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