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Answer
Draw the rotational mechanical system shown TU) θ1(t) 50 kg-m N-30 N2-100 100 kg-m2 100 N-m/rad 100 N-m-s/rad The torque T(t) is multiplied by 2 as it is reflected by the gear N. with a gear ratio of 2 MT- 00)-333T(0) Calculate the value of equivalent moment of inertia J 50x50x Draw the modify diagram in frequency domain %(100)2 30 555.56 .33 T 2(s) Je 03(s) 555.56 kg-m2 : 100 kg-m2 100 N-m/rad 100 N-m-sec/rad Write the equations of motion. Calculate the equation of motion is written using the formula Sum of Sum of mpedanceS connected to the motion at θ impedances 2between and θ Sum of applied torques at θ 1)
Sum of impedances between 0, and e Sum of impedances connected to the motion at θ Sum of applied torques at 6, Write the equation of motion for 555.56 kg-m 555.56s 6, (s) +1000, (s)-1000, (s) 3.33T (s) (555.56s? +100) (s)-1000, (s)-3.337(s) Write the equation of motion for 100kg-m2. 100s'0 (s)+100se, (s)+1006, (s)-1000, (s)0 -1000, (s)+(100s+100s +100)e (s) 0 Consider the following 4 s02 (s)- 6, se, (s) 03 Rewrite the equations (3) and (4) 555.560, + 1006,-1000, = 3.33T 555.560,-1000, 1000, + 3.33 e,0.180, 0.180 0.006T 1000, +1000 +1000 +10000 1008 100a-1000,-1000,-0
Define the state variables Write the state equations using equations of motion and in terms of state variables Rewrite the equations (5) and (6) Ý, =-0.18x, +0.18x3 + 0.006T 0 t.| |-0.18 0 0.18 0 ||x.| |0.006 0 0 0x 1 0 0 0 0 1 00 -0.18 0 0.18 0 0 0 01 1 0-1 The output equation(y) is 0.006 t, y 3.33 0 0 0ll2 y=[3.33 0 0 01x
