hwk04_soln

# hwk04_soln - ME 563 Fall 2011 Homework Prob 4.1 SOLUTION...

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ME 563 – Fall 2011 SOLUTION Homework Prob. 4.1 The system shown below is made up of a block A of mass m , a dashpot with damping coefficient c and a spring of stiffness k . Let x be the generalized coordinate used to describe the motion of block A. During impact testing of this system, block A is struck by a hammer that pivots about point O. The head of the hammer, P, is to be modeled as a particle with mass M , and the mass of the handle of the hammer is to be considered negligible. Immediately prior to impact, the handle is rotating at a rate of ω , as indicated in the figure below. The coefficient of restitution of impact of P with A is e . The duration of the impact of P with A is assumed to be very short. Prior to impact, A is at rest. a) Determine the speed of A immediately after impact. Consider using both conservation of moment of A+P during impact and the coefficient of restitution. Assume that the hammer contacts A only once. Ignore the influence of the spring and dashpot on the motion of the block during impact. b) Derive the EOM for the mass-spring-dashpot system governing motion for all time after impact in terms of the coordinate x . c) Determine the response x(t) for all time after impact. What is the maximum displacement of A during this response? Use the following parameters: M = 2 kg , m = 4 kg , k = 1600 N/m , c = 32 kg/sec , e = 0.8 , L = 0.2 meters and ω = 25 rad / sec . During impact: F x = 0 momentum conserved in x direction Mv P 1 = Mv P 2 + mv A 2 ; v P 1 = L ω Also, k c x L ω m M P A O P A O

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e = v A 2 v P 2 v P 1 v A 1 v P 2 = v A 2 ev P 1 = v A 2 eL ω Combining the above two equations gives: ML ω = M v A 2 eL ω ( ) + mv A 2 v A 2 = 1 + e ( ) M M + m L ω = 1 + 0.8 ( ) 2 2 + 4 0.2 ( ) 25 ( ) = 3 m / sec After impact (system = mass, spring and dashpot) T = 1 2 m x 2 R = 1 2 c x 2 U = 1 2 kx 2 Therefore, the EOM is given by: m  x + c x + kx = 0 dividing through by m ( )  x + c m x + k m x =  x + 2 ζω n x + ω n 2 x = 0 ω n = k m = 1600 4 = 20 rad / sec ζ = c 2 m ω n = 32 2 4 ( ) 20 ( ) = 0.2 underdamped ( ) Underdamped, free response: x t ( ) = e ζω n t c cos ω d t + s sin ω d t ( ) where ω d = ω n 1 ζ 2 = 20 1 0.2 2 = 8 6 rad / sec Enforcing ICs: 0 = e 0 c cos 0 + s sin 0 ( ) = c x 0 ( ) = ζω n e ζω n t c cos ω d t + s sin ω d t ( ) + ω d e ζω n t c sin ω d t + s cos ω d t ( ) t = 0 = ω d s s = v A 2 ω d = 3 8 6 Therefore, x t ( ) = se ζω n t sin ω d t = 3 8 6 e
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