- Iy 12m B W Mx Cx Kx 0 1 I Icom Md K C 25w X W 2 X 0 Wn 5 N M 2mwn X C Exp Wnt 1 (20.44 KiB) Viewed 55 times
- Iy 12m B W Mx Cx Kx 0 1 I Icom Md K C 25w X W 2 X 0 Wn 5 N M 2mwn X C Exp Wnt 2 (20.44 KiB) Viewed 55 times
- Iy 12m B W Mx Cx Kx 0 1 I Icom Md K C 25w X W 2 X 0 Wn 5 N M 2mwn X C Exp Wnt 3 (155.86 KiB) Viewed 55 times
Iy = 12m(b² + w²) mx + cx + kx = 0 $<1 I|| = Icom +md² k C +25w₁x + w/²/2 x = 0, wn = ₂5 'n m 2mwn ⇒ x = C [exp(-(wnt)] [sin (wat + v)], wa=wn √1-5² [Cwa cos(wat +) - (Cwn sin(wat +)] e-Sunt V = 7 =
Consider an inclined bar made of two sections of length L₁ connected to a spring of constant k₁ and length L2 connected to a damper of constant c₁ such that the co-linear sections is initially at a rotation of o to the horizontal that is connected to a pivot and then allowed to rotate by a small clockwise angle as shown below. C1 L2 k₁ L₁ WW 00 m L3 k3 L5 LA 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 k3. Considering the above system, determine the following information: d²0 1.1 The equation of motion of the rotating rod in the form meq d2 + Ceq dt+keq in (15) terms of the available information using kinematics and Newtons equations; 1.2 The natural angular frequency wn and damping factor of the rotating rod if (3) 0 + 25w₁0+w²₂0 = 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, k₁ = 800 N/m, = = k3 = 2450 N/m, c₁ = 450 Ns/m, b = 0.065 m, w = 0.037 m, and 00 = 22° respectively. x b W