# Example 1, Page 28¶

In :
#Variable declaration
#a) INELASTIC
#for sphere 1 ,mass=m1 and initial velocity=u1
#for sphere 2 ,mass=m2 and initial velocity=u2
m1=100.#lb
u1=10.#ft/s
m2=50.#lb
u2=5.#ft/s

#Calculations
v=(m1*u1+m2*u2)/(m1+m2)
#change in kinetic energy
#initial kinetic energy = ke1
ke1=(m1*(u1**2)+m2*(u2**2))/(2*32.2)
#Kinetic Energy after inelastic colision = ke2
ke2=((m1+m2)*8.333**2)/(2*32.2)
#Change in Kinetic Energy =l
l=ke1-ke2
#b) Elastic
# for a very short time bodies will have a common velocity given by v=8.333 ft/s
# for a very short time bodies will have a common velocity given by v=8.333 ft/s
#immidiately after impact ends the velocities for both the bodies are given by v1 and v2
v1=2*v-u1
v2=2*v-u2
#c) Coeeficient of Restitution=0.6
e=0.6
ve1=(1+e)*v-e*u1
ve2=(1+e)*v-e*u2
ke3=(m1*(ve1**2)+m2*(ve2**2))/(2*32.2)
loss=ke1-ke3

#Results
print "kinetic energy before collisio0n is %.1f ft lb"%ke1
print "\na) INELASTIC"
print "velocity after collision is %.3f ft/s"%v
print "the Kinetic Energy after collision is %.1f ft lb"%ke2
print "the change in Kinetic Energy after collision is %.1f ft lb"%l
print "\nb) ELASTIC"
print "velocity of 1 after collision is %.3f ft/s"%v1
print "velocity of 2 after collision is %.3f ft/s"%v2
print "there is no loss of kinetic energy in case of elastic collision"
print "\nc) e=0.6"
print "velocity of 1 after collision is %.3f ft/s"%ve1
print "velocity of 2 after collision is %.3f ft/s"%ve2
print "the Kinetic Energy after collision is %.1f ft lb"%ke3
print "the change in Kinetic Energy after collision is %.2f ft lb"%loss

kinetic energy before collisio0n is 174.7 ft lb

a) INELASTIC
velocity after collision is 8.333 ft/s
the Kinetic Energy after collision is 161.7 ft lb
the change in Kinetic Energy after collision is 13.0 ft lb

b) ELASTIC
velocity of 1 after collision is 6.667 ft/s
velocity of 2 after collision is 11.667 ft/s
there is no loss of kinetic energy in case of elastic collision

c) e=0.6
velocity of 1 after collision is 7.333 ft/s
velocity of 2 after collision is 10.333 ft/s
the Kinetic Energy after collision is 166.4 ft lb
the change in Kinetic Energy after collision is 8.28 ft lb


## Example 2, Page 29¶

In :
#Variable declaration
m1=15.#tons
u1=12.#m/h
m2=5.#tons
u2=8.#m/h
k=2.#ton/in
e1=0.5#coefficient of restitution

#Calculations&Results
#conservation of linear momentum
v=(m1*u1+m2*u2)/(m1+m2)
print "velocity at the instant of  collision is %.2f mph"%v
e=(m1*m2*(88./60)**2*(u1-u2)**2)/(2*32.2*(u1+u2))
print "The difference between the kinetic energy before and during the impact is %.2f ft tons"%e
#energy stored in spring equals energy dissipated
#s=(1/2)*k*x**2
#s=e
#since there are 4 buffer springs ,4x**2=24 inches (2 ft=24 inches)
x=((e*12)/4)**.5
print "Maximum deflection of the spring is %.2f in"%x
# maximum force acting between pair of buffer = stiffness of spring*deflection
f=k*x
print "Maximum force acting between each buffer is %.2f tons"%f
#assuming perfectly elastic collision
v1=2*11-12
v2=2*11-8
print "Speed of loaded truck after impact %.2f mph"%v1
print "speed of unloaded truck after impact %.2f mph"%v2
#if coefficient of restitution =o.5
ve1=(1+.5)*11-.5*12
ve2=(1+.5)*11-.5*8
print "Speed of loaded truck after impact when e=0.5 %.2f mph"%ve1
print "Speed of unloaded truck after impact when e=0.5 %.2f mph"%ve2
#net loss of kinetic energy=(1-e**2)*energy stored in spring
l=(1-(e1**2))*2#ft tons
print "Net loss of kinetic energy is %.2f ft tons"%l

velocity at the instant of  collision is 11.00 mph
The difference between the kinetic energy before and during the impact is 2.00 ft tons
Maximum deflection of the spring is 2.45 in
Maximum force acting between each buffer is 4.90 tons
Speed of loaded truck after impact 10.00 mph
speed of unloaded truck after impact 14.00 mph
Speed of loaded truck after impact when e=0.5 10.50 mph
Speed of unloaded truck after impact when e=0.5 12.50 mph
Net loss of kinetic energy is 1.50 ft tons


## Example 3, Page 31¶

In :
import math

#Variable declaration
m1=500.#lb ft^2
m2=1500.#lb ft^2
k=150#lb ft^2
w1=150#rpm

#Calculations&Results
N=(w1*m1)/(m1+m2)
print "Angular velocity at the instant when speeds of the flywheels are equalised is given by %.f r.p.m"%N
#kinetic energy at this instance
ke1=(1./2)*((m1+m2)/32.2)*((math.pi*N)/30)**2
print "The kinetic energy of the system at this instance is %.2f ft lb"%ke1

#initial kinetic energy
ke0=(1./2)*((m1)/32.2)*((math.pi*w1)/30)**2
print "The initial kinetic energy of the system is %.2f ft lb"%ke0

#strain energy = s
s=ke0-ke1
print "strain energy stored in the spring is %.2f ft lb which is approximately 1435 ft lb"%s

x=((1435*2)/150)**.5
print "Maximum angular displacement is %.2f in radians which is equal to 250 degrees"%x
#na1 and na are initial and final speeds of the flywheel 1 and same nb1 and nb for flywheel 2
na=2*N-w1#w1=na1
nb=2*N-0#nb1=0
print "Speed of flywheel a and b when spring regains its unstrained position are %.2f rpm and %.2f rpm respectively"%(na,nb)

Angular velocity at the instant when speeds of the flywheels are equalised is given by 38 r.p.m
The kinetic energy of the system at this instance is 478.92 ft lb
The initial kinetic energy of the system is 1915.68 ft lb
strain energy stored in the spring is 1436.76 ft lb which is approximately 1435 ft lb
Maximum angular displacement is 4.36 in radians which is equal to 250 degrees
Speed of flywheel a and b when spring regains its unstrained position are -75.00 rpm and 75.00 rpm respectively


## Example 4, Page 36¶

In :
import math

#Variable declaration
m1=150 #lb
l=3#ft
#number of oscillation per second is given by n

#Calculations&Results
n=(50/92.5)
print "number of oscillation per second = %.3f"%n
#length of simple pendulum is given by L=g/(2*math.pi*n)**2
L=32.2/(2*math.pi*n)**2
print "length of simple pendulum = %.2f ft"%L
# distance of cg from point of suspension is given by a
a=25./12
moi=m1*k**2
print "The moment of inertia of rod is %.f lb ft**2"%moi

number of oscillation per second = 0.541
length of simple pendulum = 2.79 ft
The moment of inertia of rod is 221 lb ft**2


## Example 5, Page 37¶

In :
import math

#Variable declaration
n1=50/84.4
n2=50/80.3

#Calculations
L1=(32.2*12)*(84.4/(100*math.pi))**2
L2=(32.2*12)*(80.3/(100*math.pi))**2
#a1(L1-a1)=k**2=a2(L2-a2) and a1+a2=30 inches
#substituting and solving for a we get
a1=141/6.8
a2=30-a1
k=(a1*(L1-a1))**.5
moi=90*(149./144)#moi=m*k**2

#Results
print "length of equivalent simple pendulum when axis coincides with small end and big end respectively-"
print "L1=%.1f in"%L1
print "L2=%.1f in"%L2
print "distances of cg from small end and big end centers respectively are-"
print "a1=%.1f in"%a1
print "a2=%.1f in"%a2
print "Moment of inertia of rod =%.2f lb ft^2"%moi

length of equivalent simple pendulum when axis coincides with small end and big end respectively-
L1=27.9 in
L2=25.2 in
distances of cg from small end and big end centers respectively are-
a1=20.7 in
a2=9.3 in
Moment of inertia of rod =93.13 lb ft^2


## Example 6, Page 39¶

In :
import math

#Variable declaration
m1=150
l=8.5
g=32.2
a=83.2
n=25

#Calculations&Results
#k=(a/2*%pi*n)*(g/l)**0.5
k=(14*a*((g)**0.5))/(2*math.pi*n*(l**0.5))
k1=14.5/12
print "radius of gyration is %.2f inches which is equal to %.2f ft"%(k,k1)
moi=m1*(k1**2)
print "moment of inertia=%.f lb ft^2"%moi

radius of gyration is 14.43 inches which is equal to 1.21 ft
moment of inertia=219 lb ft^2


## Example 7, Page 41¶

In :
#Variable declaration
m=2.5#lb
a=6#in
k=3.8#in
l=9#in
c=3#in
w=22500

#Calculations&Results
#k^2=ab
#case a) to find equivalent dynamic system
b=(k**2)/a
ma=(2.5*6)/8.42#m*a/a+b
mb=m-ma
print "Mass ma =%.2f lb will be situated at 6 inches from cg \nand mb =%.2f lb will be situated at %.2f inches " \
"\nfrom cg in the equivalent dynamical system"%(ma,mb,b)

#if two masses are situated at the bearing centres
ma1=(2.5*6)/9
mb1=m-ma1
k1=(a*c)**.5
#t=m*((k1^2)-(k^2))*w
t=((2.5*(18-3.8**2))*22500)/(32.2*12*12)
print "\ncorrection couple which must be applied in order that the two mass system is dynamically equivalent to"\
"the rod is given by %.1f lb ft"%t

Mass ma =1.78 lb will be situated at 6 inches from cg
and mb =0.72 lb will be situated at 2.41 inches
from cg in the equivalent dynamical system

correction couple which must be applied in order that the two mass system is dynamically equivalent tothe rod is given by 43.2 lb ft


## Example 8, Page 44¶

In :
#Variable declaration
m=20.#lb
g=32.2
a=200#ft/s^2
k=7.#in

#Calculations
f=(m/g)*a#effective force appllied to the link
#this force acts parallel to the acceleration fg
t=(m/g)*(k/12)**2*w#couple required in order to provide the angular acceleration
#the line of action of F is therefore at a distance from G given by
x=t/f

#Results
print "Effective force applied to the link is %.3f lb and the line of action of F is therefore at a distance"\
"from G given by %.3f ft"%(f,x)
print "F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved along" \
"the appropriate lines of action to give the magnitudes of Fa and Fb"
print "From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb"

Effective force applied to the link is 124.224 lb and the line of action of F is therefore at a distancefrom G given by 0.204 ft
F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved alongthe appropriate lines of action to give the magnitudes of Fa and Fb
From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb


## Example 9, Page 46¶

In :
#Variable declaration
m=10#ton
m2=1000#lb
a=3#ft/s^2

#Calculations
#the addition to actual mass in order to allow for the rotational inertia of the wheels and axles
m1=2*(1000./2240)*(15./21)**2#m1=m2*k**2/r**2 and 1 ton=2240 lbs
M=m+m1
F=3*(10.46/32.2)#F=M.a
f=F*2240#lb
Fa=(2*1000./2240)*(3/32.2)*(15./21)**2#total tangential force required in order to provide the angular acceleration of the wheels and axles
#Limiting friction force =uW
#u*10>0.042
u=0.042/10

#Results
print "The total tangential force required in order to provide the angular acceleration of the wheels and axles is %.3f ton"%Fa
print "If there is to be pure rolling ,u>%.4f"%u

The total tangential force required in order to provide the angular acceleration of the wheels and axles is 0.042 ton
If there is to be pure rolling ,u>0.0042


## Example 10, Page 48¶

In :
#Variable declaration
Ia=200#lb ft2
Ib=15#lb ft2
G=5#wb==5*wa
m=150.#lb
r=8#in

#Calculations
#the equivalent mass of the geared system referred to the circumference of the drum is given by
#Me=(1./r)**2*(Ia+(G**2*Ib))
Me=(12./r)**2*(Ia+(G**2*Ib))
M=m+Me
a=(m/M)*32.2#acceleration
#if efficiency of gearing is 90% then Me=(1/r**2)*(Ia+(G**2*Ib)/n)
n=.9
Me1=(12./r)**2*(Ia+(G**2*Ib)/n)
M1=Me1+m
a1=(m/M1)*32.2

#Results
print "acceleration = %.2f ft/s2"%a
print "acceleration when gear efficiency is 0.9= %.2f ft/s2"%a1

acceleration = 3.35 ft/s2
acceleration when gear efficiency is 0.9= 3.14 ft/s2


## Example 11, Page 49¶

In :
#Variable declaration
#let
#S=displacement of car from rest with uniform acceleration a, the engine torque T assumed to remain ocnstant
#v=final speed ofcar
#G=gear ratio
#n=efficiency of transmission
#M=mass of the car
#Ia and Ib=moments of inertia of road whels and engine
#formulas => F=29.5nG ; Me= 1648+\$.54nG^2 ; a=32.2 F/Me
#given
G1=22.5
G2=12.5
G3=7.3
G4=5.4
n=.82#for 1st ,2nd and 3rd gear
n4=.9#for 4th gear

#Calculations
F1=29.5*n*G1
F2=29.5*n*G2
F3=29.5*n*G3
F4=29.5*n4*G4
#on reduction and putting values we get
Me1=1648+4.54*n*G1**2
Me2=1648+4.54*n*G2**2
Me3=1648+4.54*n*G3**2
Me4=1648+4.54*n4*G4**2
a1=32.2*F1/Me1
a2=32.2*F2/Me2
a3=32.2*F3/Me3
a4=32.2*F4/Me4

#Results
print "Maximum acceleration of car on top gear is %.2f ft/s^2"%a4
print "Maximum acceleration of car on third gear is %.2f ft/s^2"%a3
print "Maximum acceleration of car on second gear is %.2f ft/s^2"%a2
print "Maximum acceleration of car on first gear is %.2f ft/s^2"%a1

Maximum acceleration of car on top gear is 2.61 ft/s^2
Maximum acceleration of car on third gear is 3.08 ft/s^2
Maximum acceleration of car on second gear is 4.37 ft/s^2
Maximum acceleration of car on first gear is 4.96 ft/s^2


## Example 12, Page 57¶

In :
import math

#Variable declaration
I=40#lb ft2
n=500#rpm

#Calculations
w=math.pi*n/30#angular velocity
wp=2*math.pi/5#angular velocity of precession
I1=I/32.2
T=I1*w*wp#gyroscopic couple

#Result
print "The couple supplied to the shaft= %.2f lb ft"%T

The couple supplied to the shaft= 81.74 lb ft


## Example 13, Page 57¶

In :
import math

#Variable declaration
I=250#lb ft2
n=1600#rpm
v=150#mph
r=500#ft

#Calculations&Results
w=math.pi*160/3#angular velocity of rotation
wp=(150.*88)/(60*500)#angular velocity of precession
#T=I*w*wp
T=(250/32.2)*math.pi*(160./3)*wp

The magnitude of gyroscopic couple is given by 572 lb ft