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Unformatted text preview: 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 massspringdashpot 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 ∑ = ⇒ 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 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 = ⇒ dividing through by m ( ) x + c m x + k m x = x + 2 ζω n x + ω n 2 x = ⇒ 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...
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This document was uploaded on 12/23/2011.
 Fall '09

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