In [4]:

```
import math
#variable declaration
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
#Calculation
Wm=N*math.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
print"\nEquivalent moment of inertia is :",round(J,1),"kg-m2"
print"\nEquivalent load torque :",round(Tl,2),"N-m"
```

In [7]:

```
import scipy
from scipy import integrate
import math
# variable declaration
J=10 #moment of inertia of the drive in kg-m2
print("Passive load torque during steady state is :Tl=0.05*N in N-m")
print("And load torque : T=100-0.1*N in N-m ")
print("load torque when the direction is reversed T=-100-0.1*N in N-m")
#Calculation
print("T-Tl=0")
print("100-0.1*N-0.05*N=0")
N=100/0.15 #Required speed of the motor in rpm during steady state
N2=-100/0.15 #During reversal speed is in opposite direction
print("\nJdWm/dt=-100-0.1*N-0.05*N during reversing")
print("dN/dt=30/(J*pi)*(-100-0.15*N)")
print("dN/dt=(-95.49-0.143*N)")
N1=N
N2=N2*0.95 #for speed reversal
x2 = lambda N: 1/(-95.49-0.143*N)
t=integrate.quad(x2, round(N1), round(N2))
#result
print"\nHence Time of reversal is :",round(t[0],2),"s"
```

In [12]:

```
import math
from __future__ import division
#variable declaration
Tlh=1000 # load torque in N-m
Tmax=700 # maximum motor torque
Tll=200 # light load for the motor to regain its steady state
Tmin=Tll # minimum torque
t_h=10 # period for which a load torque of 1000 N-m is apllied in sec
Jm=10 # moment of inertia of the motor in Kg-m2
No=500 # no load speed in rpm
Tr=500 # torque at a given no load speed in N-m
#Calculation
Wmo=No*2*math.pi/60 # angular no load speed in rad/sec
s=0.05 # slip at a torque of 500 N-m
Wmr=(1-s)*Wmo # angular speed at a torque of 500 N-m in rad/sec
y=math.log((Tlh-Tmin)/(Tlh-Tmax))
x=Tr/(Wmo-Wmr)
J=x*t_h/y
Jf=J-Jm
#Result
#answer in the book is wrong
print"\n\nMoment of inertia of the flywheel : ", round(Jf,1),"Kg-m2"
```