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//Chapter 2:Dynamics of Electric Drives
//Example 1
clc;
//Variable Initialization
Jo=0.2 // inertia of the motor in kg-m2
a1=0.1 // reduction gear
J1=10 // inertia of the load in kg-m2
Tl1=10 // load torque
v=1.5 // speed of the translational load
M1=1000 // mass of the translational load
N=1420 // speed of the motor
n1=.9 // efficiency of the reduction gear
n1_=0.85 // efficiency of the translational load and the motor
F1=M1*9.81 // force of the translational load
//Solution
Wm=N*%pi/30 //angular speed
J=Jo+a1**2*J1+ M1*(v/Wm)**2 // total equivalent moment of inertia
Tl= a1*Tl1/n1+F1/n1_*(v/Wm) // total equivalent torque
//Result
mprintf("\nEquivalent moment of inertia is : %.1f kg-m2",J)
mprintf("\nEquivalent load torque : %.2f N-m",Tl)
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