# Chapter2-Angular Motion¶

## EX1-pg13¶

In [1]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##equation of motion, Mass of moment of inertia, percentage
##reduction in speed
##initialisation of variables
g=5.##ft
w=300.##rev/min
a=0.86##red/s^2
h=2240.##ft/s
q=4.##ft
g1=32.1##ft/s
k=3105000.##ft lbf
##CALCULATIONS
T=(w*(2.*math.pi/60.))/(a)##sec
M=(q*h*(g**2))/(g1)##slug ft^3
K=((1/2.)*M)*((w*2.*math.pi**2)/(60.))##ft lbf
P=((((w*2.*math.pi)/60.)-W)/((w*2.*math.pi)/60.))*100.##percent
##RESULTS
print'%s %.2f %s'%('The equation of motion= ',T,' sec')
print'%s %.2f %s'%('Mass of moment of inertia of = ',K,' ft lbf')
print'%s %.2f %s'%('the percentage reduction in speed= ',P,' percent')

The equation of motion=  36.53  sec
Mass of moment of inertia of =  344360.03  ft lbf
the percentage reduction in speed=  5.04  percent


## Ex2-pg15¶

In [2]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
m=2.58065##slug ft^3
w=2.144##in
##CALCULATIONS
R=math.sqrt(m/w)##ft
##RESULTS
print'%s %.2f %s'%('The radius of gyration= ',R,' ft')

The radius of gyration=  1.10  ft


## Ex3-pg15¶

In [3]:
##Solutions to Problems In applied mechanics
##A N Gobby
##distance travelled along incline before coming to rest
import math
##initialisation of variables
w1=10.##tonf
r=36.##in
w=1./4.##tonf
g=14.##in
t=30.##mile/h
s=100.##in
m=20.##lbf/tonf
h=2240.##lbf
q=44.##in
g1=32.2##ft
##CALCULATIONS
K=(w1*h*(q**2))/(2.*g1)##ft lbf
R=(2.*1./2.*(1./4.*h/g1)*(g/12.)**2)*L**2##ft lbf
T=K+R##ft lbf
M=m*w1##lbf
G=w1*h*(1./s)##lbf
S=K/(M+G)##ft
##RESULTS
print'%s %.2f %s'%('the distance travelled along incline before coming to rest= ',S,' ft')
#answer is different due to round off error

the distance travelled along incline before coming to rest=  1588.19  ft


## Ex4-pg16¶

In [4]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##percentage fluctuation in speed
##initialisation of variables
g=32.2##ft
p=275.##rev/min
w=1/2.*p##ft
d=15.##hp
h=33000.##ft
r=0.8##ft
h1=2240.##ft
##CALCULATIONS
W=(d*h)/w##ft lbf
E=r*W##ft lbf
I=(1.*h1*(3.)**2)/(g)##slug ft^2
Q=(E*100.)/(I*(m)**2*2.)##percent
##RESULTS
print'%s %.2f %s'%('the percentage fluctuation in speed= ',Q,' percent')

the percentage fluctuation in speed=  0.28  percent


## Ex5-pg17¶

In [5]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##weight of flywheel and the work done by frictional torque
##initialisation of variables
w=140.##rev
r=8.##in
g=12.##in
t=30.##mile/h
q=(1/4.)##tonf
I=0.99##slug ft^3
p=32.2##ft^2
##CALCULATIONS
W=(I*p)/(r/g)**2##lbf
T=(I*(2*math.pi)**2)/(2.*(2.*math.pi)*w)##lbf ft
##RESULTS
print'%s %.2f %s'%('The weight of flywheel=',W,'lbf')
print'%s %.2f %s'%('the work done by frictional torque in stopping flywheel= ',T,' lbf ft')

The weight of flywheel= 71.73 lbf
the work done by frictional torque in stopping flywheel=  0.02  lbf ft


## Ex6-pg18¶

In [6]:
##Solutions to Problems In applied mechanics
##A N Gobby
##mass moment of inertia, kinetic enrgy and shear blades
import math
##initialisation of variables
w=2.##tonf
t=250.##rev/min
g=32.2##ft
h=2240.##ft
f=0.8##percent
t1=60.##ft
s=1*(2./3.)##min
r=480.##ft
w1=20.##ft
##CALCULATIONS
M=(w*h*(w**2))/g##slug ft^2
I=M*A##lbf ft
K=1/2.*(M)*(2.*math.pi/t1)**2*r*w1##ft lbf
F=f*K/(3./12.)##lbf
##RESULTS
print'%s %.2f %s'%('the mass moment of inertia = ',I,' lbf ft')
print'%s %.2f %s'%('the kinetic energy= ',K,' ft lbf')
print'%s %.2f %s'%('the average force on the shear blades= ',F,' lbf')
#answer is different due to round off error

the mass moment of inertia =  161.89  lbf ft
the kinetic energy=  29294.13  ft lbf
the average force on the shear blades=  93741.22  lbf


## Ex7-pg19¶

In [7]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##frictional torque retarding and tangential braking acting
##initialisation of variables
h=2240.##ft
w=0.06##ft
w1=4.##ft
q=12.##ft
g=5.##ft
g1=32.2##ft
d=100.##rev/min
f=120.##sec
##CALCULATIONS
T=w*(w1*h)*(w1/q)##lbf ft
I=((w1*h*(g)**2)/g1)*d*(2.*math.pi/60.)##slug ft^2/s or lbf ft s
M=I/T##sec
P=430.8##lbf ft
R=(P/2.5)##lbf
##RESULTS
print'%s %.2f %s'%('the frictional torque retarding= ',M,' sec')
print'%s %.2f %s'%('the tangential braking acting= ',R,' lbf')

the frictional torque retarding=  406.52  sec
the tangential braking acting=  172.32  lbf


## Ex8-pg20¶

In [8]:
##Solutions to Problems In applied mechanics
##A N Gobby
##tangential force
import math
##initialisation of variables
I=179.2##lbf ft
h=2240.##ft
w=4.##ft
w1=5.##ft
r=120.##ft
g=32.2##ft
p=100.##ft
t=60.##ft
##CALCULATIONS
M=(w*h*(w1)**2)/g##slug ft^3
D=p*(2.*math.pi)/(t*T)##sec
T1=M*N##lbf ft
B=T1-I##lbf ft
F=B/2.*1/2.##lbf
##RESULTS
print'%s %.2f %s'%('the deceleration = ',D,' sec')
print'%s %.2f %s'%('the tangential force on brake rim= ',F,' lbf')
#answer is different due to round off error

the deceleration =  406.52  sec
the tangential force on brake rim=  106.97  lbf


## Ex9-pg20¶

In [9]:
##Solutions to Problems In applied mechanics
##A N Gobby
##friction of bearings is to to neglected
import math
##initialisation of variables
h=2240.##ft
g=32.2##ft
g1=15.##in
w=1200.##lbf
q=12.##ft
r=1.5##ft
t=3.28##tonf ft
t1=1.7##tonf ft
x=550.##ft
s=6.##ft
##CALCULATIONS
T=((w*(g1/q)**2)/(h*g))*(3./r)##tonf ft
T1=t-t1+T##tonf ft
W=(T1*h*s/(r))/(x)##ft lbf
##RESULTS
print'%s %.2f %s'%('the friction of bearings is to to neglected =',W,'')

the friction of bearings is to to neglected = 26.59


## Ex10-pg22¶

In [10]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##torque to acceleration drum and truck

##initialisation of variables
v=20.##ft/s
s=150.##ft
h=2240.##ft
g=32.2##ft
d=3.##ft
p=364.9##lbf
q=4##ft
##CALCULATIONS
A=v**2/(2.*s)##ft/s^2
T=(h*(d)**2/g)*(A/q)+p*q##lbf ft
##RESULTS
print'%s %.2f %s'%('the torque to acceleration drum and truck= ',T,' lbf ft')

the torque to acceleration drum and truck=  1668.30  lbf ft


## Ex11-pg22¶

In [11]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##gravitational force
##initialisation of variables
v=35.##hp
p=25.##percent
v1=30.##mile/h
q=28.##in
d=30.##in
w=3200.##lbf
t=33000.##lbf
s=88.##in
W=w*(1./v1)##lbf
m=0.364##mile/h
##CALCULATIONS
N=(v1*s/60.)/(14./12.)*(60./(2*math.pi))##rev/min
Ne=N*6.##rev/min
E=(v*t)/(2.*math.pi*Ne)##lbf ft
T=(v*0.75*t)/(2.*math.pi*N)##lbf ft
P=T/(14./12.)##lbf
V=math.sqrt((P-W)/m)##mile/h
##RESULTS
print'%s %.2f %s'%('the gravitational force= ',V,'mile/h')

the gravitational force=  24.67 mile/h