Consider an inclined bar made of two sections of length L₁ connected to a spring of constant ki and length L2 connected

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Consider an inclined bar made of two sections of length L₁ connected to a spring of constant ki and length L2 connected

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Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 1
Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 1 (152.79 KiB) Viewed 62 times
Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 2
Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 2 (18.47 KiB) Viewed 62 times
Consider an inclined bar made of two sections of length L₁ connected to a spring of constant ki and length L2 connected to a damper of constant c₁ such that the co-linear sections is initially at a rotation of fo to the horizontal that is connected to a pivot and then allowed to rotate by a small clockwise angle as shown below. C1 L2 L₁ Y L3 k3 L5 L4 Attached perpendicular to the intersection of sections L₁ and L₂ is another section of length L3 to which is attached at its end point a rectangular planar mass of mass m with a breadth of b and a width of w. When the mechanism rotates by a clockwise angle a wire rope connected to the center-of-mass of the planar rectangle pulls a spring of spring constant ką. Considering the above system, determine the following information: d²0 1.1 (3) 1.2 The equation of motion of the rotating rod in the form meg d2 + Cegat + keq in (15) terms of the available information using kinematics and Newtons equations; The natural angular frequency wn and damping factor of the rotating rod if 0 + 25w₁0 + w²/20 = = 0 by manipulating the equation of motion if L₁ = 0.400 m, L2 = 0.550 m, L3 = 0.720 m, L4 = 0.630 m, L5 0.420 m, ki = 800 N/m, k3 = 2450 N/m, C1 = 450 Ns/m, b = 0.065 m, w = 0.037 m, and o = 22° respectively. = 00 m W UV

Iy = 2m (b²+w²), I|| = Icom + md² mä+ci+ka= 0 +25wni+w²₁2x = 0,w₂ = 04 V << 1 ⇒ C 2mwn x = C[exp(-[wnt)] [sin (wat +)], wa = wn √1-5² v = [Cwa cos(wat +)-(Cwn sin(wat + y)] e-Sunt
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