- Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 1 (178.31 KiB) Viewed 68 times
- Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 2 (56.23 KiB) Viewed 68 times
- Consider An Inclined Bar Made Of Two Sections Of Length L Connected To A Spring Of Constant Ki And Length L2 Connected 3 (132.82 KiB) Viewed 68 times
1 I₂ = 1/2m (b² + w²), -m I = I com + md² C mx + cx+kx = 0 + x + 25w₁x + w/7 x = 0, wn = = V " 2mwn << 1 ⇒ x = C[exp (-Cwnt)] [sin (wat + y)], wa = wn√1-5² V= [Cwa cos(wat +)-(Cwn sin(wat +)] e=Sunt k
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 8o to the horizontal that is connected to a pivot and then allowed to rotate by a small clockwise angle as shown below. C1 L2 kı L₁ W 80 L3 LA Attached perpendicular to the intersection of sections L₁ and L₂ is another section of length Ls 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 k3. Considering the above system, determine the following information: 1.1 The equation of motion of the rotating rod in the form megdi + Ceqi + keqe in (15) terms of the available information using kinematics and Newtons equations; The natural angular frequency , and damping factor of the rotating rod if (3) 1.2 0 A +25w₁₂0 + ²0 = 0 by manipulating the equation of motion if L₁ = 0.400 m, L₂ = 0.550 m, Ls = 0.720 m, L₁= 0.630 m, L5 = 0.420 m, ki 800 N/m, ks = 2450 N/m, c₁ = 450 Ns/m, b = 0.065 m, w = 0.037 m, and 0o = 22° respectively. m W k3 www L5