Question

Let ? be the angular displacement of the rigid bar fromthe system's equilibrium position, as the generalized coordinate. Take m-5484 gr, L4 m, k-6450 N/m. (a) Derive the differential equation for the system. (b) For what value of e is the system critically damped? For Mo 0, suppose the bar is rotated 3 from equilibrium and released. Determine c-o?25cc ,(ii) C-ce, (iii) c-1.25c? ) if rigid bar, mass m (d) For Mo 0, how long will it take for the response to be permanently within 1° of the equilibrium position if () c 0.25ce, (ii) c-Cos (ii) c-1.25c (e) For MorO and do-80 rad/s, (i) take c=0.01cc and (ii) take c 0.3c determine corresponding magnification 2 factors of the rigid bar and discuss the results. (t) For o-80 rad/s and c-0.3ce determine the maximum value of Mo such that the steady-state amplitude of the angular oscillation does not exceed 6°
Mosinwt

Answer 1

Lets draw free body diagram of rod x_2 = x_3 = x x_1 = 3x Taking moments about point '0' I- -M_0 + kx_1 3l/4 + cx_2 l/4 + 3kx_3 R/4 = 0 I_rad = I_C.G + mh^2 = ml^2/12 + m(l/4)^2 = 7 ml^2/48 = x/l/4 = 4x/l, = 4x/l 7ml^2/48 4x/l -M_0 + 9kxl/4 + c xl/4 + 3knl/4 = 0 putting the values m = 5484 gram = 5.484 kg L = 4m, k = 6450 N/m 3.2x-M_0 + C x + 77400 x = 0 C for critically damped system considering M_0 = 0 x = -C/3.2 plusminus squareroot C^2 -4 times 3.2 rimes 77400/3.2 For critically damped solution C^2-4 times 3.2 times 77400 = 0 c^2 = 990720 C = 995.35 Ns/m