Question

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e.g.23 An electric motor of 10 kg is suffering from vibration due to unbalance of 0.01 kg. m2. The stiffness coefficient of the isolator is k = 105 N/m. It is required to choose a value of 3 from the following table such that minimum Ft to the floor is required and the maximum amplitude of vibration for the motor mass is 0.001 m. S = 0.1, 0.3, 0.5, 0.7,0.8 and 0.85.

Answer 1

con Tos 낙 100 rad/s 10 when. we take. 2 = 0.1 wna w. J-22 loo - 100.60 radis - z² 1-(0.172 4 Now, c = 2zon = 20 = 2(0.1) 100 2 F = J K² + c²w2 5 (105)2 + (20)2 x (100.60) 100020.2 N . Now, when 2 = 0.3 wn 100 = 105.26 sad/s Ji-(0.3)2 - 0.09 C = - 60 2 Zon = 2(0.3) X100 fr = J K2 + c2 w2 J Clas? +60)2x105.262 f = 100199.23 N

1 oo 19 Now, 4 when • z = 0.5 = 116.27 rodis JI-(0.5) 2 = 2 22n = 2(0.5) 100 = loo fi = 5K² + c²w2 J GOS) + (coo)? x 16.27) ? Ft 100 673.66 N when, فا 2 = 0.7 = 140.84 rad's Ji-(0.72 Now, e 2 (0.7) (100) = 140 F = ² +0²02 fr - Seose+ (140)2 #140.34² FT : 101925.52N ہے یا با 11 11 % 166,66 rad's Ji-(0.872 C = 2(0,8) x 100 = 160 unos 2 + (160) 2 x (166.6672 = 1034 94.498 N FT = Ly when = 0.85 = 189.86 rad's JI-(0.852 1 c = 200.85] x 10 fp = Sciol2 +(170)2 x 189.802 Fr 105079.81 N Ito

3 so for the minimum transmitted force (F) we take damping factor ? = 0.1 ans.