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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)