Chapter3-Motion in a circle

Ex1-pg27

In [1]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##axis of rotation thus balancing the flywheel
##initialisation of variables
w=2000.##lbf
q=0.01##in
f=600.##rev/min
r=18.##in
g=32.2##ft^2
d=12.##in
s=1.5##ft
##CALCULATIONS
F=(w/g)*(f*2*math.pi/60.)**2*(q/d)##lbf
W=w*(q/d)/s##lbf
##RESULTS
print'%s %.2f %s'%('the axis of rotation thus balancing the flywheel= ',W,' lbf')

the axis of rotation thus balancing the flywheel=  1.11  lbf

Ex2-pg29

In [2]:
##Solutions to Problems In applied mechanics
##A N Gobby
##speed and clutch will begin to transmit power and horsepower
import math
##initialisation of variables
w=4.##lbf
r=60.##lbf/in
d=13.##in
g=32.2##in
p=500.##rev/min
h=0.25##in
b=5.##in
q=1.##in
f=62.2##lbf
V=31.1##rad/s
k=6.5##in
s=33000.##ft
##CALCULATIONS
W=f/2.##rad/s
F=(w*w/g)*(p*(2.*math.pi/r))**2*1/2.##lbf
N=F-w*r##lbf
T=N*h*k/12.##lbf ft
H=2.*math.pi*p*T/s##lbf ft
##RESULTS
print'%s %.2f %s'%('The speed and clutch will begin to transmit power = ',W,' rad/s')
print'%s %.2f %s'%('the horsepower transmitted =',H,'')

The speed and clutch will begin to transmit power =  31.10  rad/s
the horsepower transmitted = 5.69

Ex3-pg29

In [3]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w1=10.##lbf
w2=5.##lbf\
g=32.2##ft
h=8.##ft
d=3.##ft
v=10.##lbf
q=15.##ft
V=13.9##ft/s
##CALCULATIONS
M=(v*V+w2)/(v+w2)##ft/s
K=(v*(V)**2/(2.*g))-(q*(M)**2/(2.*g))##lbf
H=(q*(M)**2/(2.*g))/q##ft
F=(v*(V)**2/(g*h))##lbf
T=F+v##lbf
##RESULTS
print'%s %.2f %s'%('The moment of bodies before impact= ',M,' ft/s')
print'%s %.2f %s'%('The loss of kinetic energy in impact = ',K,' ft/lbf')
print'%s %.2f %s'%('Gain in potential energy after impact = ',H,' ft')
print'%s %.2f %s'%('tension in string centrifugal force weight= ',T,' lbf')

The moment of bodies before impact=  9.60  ft/s
The loss of kinetic energy in impact =  8.54  ft/lbf
Gain in potential energy after impact =  1.43  ft
tension in string centrifugal force weight=  17.50  lbf

Ex4-pg30

In [4]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w1=8.##lbf
s=3.##ft
m=35.##lbf
g=32.2##ft/s
##CALCULATIONS
U=math.sqrt(g*s)##ft/s
T=w1+w1##lbf
P=m-w1##lbf
Umax=math.sqrt(P*g*s/w1)##ft/s
##RESULTS
print'%s %.2f %s'%('the centrifugal force= ',Umax,' ft/s')

the centrifugal force=  18.06  ft/s

Ex5-pg31

In [5]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=3.##lbf
v=5.##ft
a=60.##degree
g=32.2##ft
u=28.4##ft/s
t=25.4##ft/s
q=12.##ft
p=1.5##ft
##CALCULATIONS
U=math.sqrt(g*v)##ft/s
T=w*(t)**2/(2.*g)+w*math.cos(a)##lbf
W=q+p##lbf
##RESULTS
print'%s %.2f %s'%('the tension in the string at position C= ',W,' lbf')

the tension in the string at position C=  13.50  lbf

Ex6-pg32

In [6]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=30.##mile/h
r=500.##ft
h=2240.##ft
q=44.##ft
t=(88./60.)##ft
g=32.2##ft
##CALCULATIONS
Tan=(w*t)**2/(g*r)
W=h*math.cos(Tan)+(h*(q)**2*math.sin(Tan))/(g*r)##lbf
##RESULTS
print'%s %.2f %s'%('the car and resolve forces normal and parallel to the slope=',Tan,'')
print'%s %.2f %s'%('the total normal reaction = ',W,' lbf')

the car and resolve forces normal and parallel to the slope= 0.12 
the total normal reaction =  2256.14  lbf

Ex7-pg33

In [7]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
h=5.##ft
h1=3.##ft
r=200.##ft
f=0.5##ft
v=60.##ft/s
w=62.0##ft/s
q=1.5##ft/s
g=32.2##ft
##CALCULATIONS
V=math.sqrt(q)/(w/(g*r))/2.##ft/s
F=math.sqrt(f*g*r)##ft/s
T=(v)**2/(g*r)##degree
##RESULTS
print'%s %.2f %s'%('The value of the speed= ',V,' ft/s')
print'%s %.2f %s'%('The block is on the point of overturning =',F,' ft/s')
print'%s %.2f %s'%('the centrifugal force must just be equal to the frictional force= ',T,' degree')

The value of the speed=  63.61  ft/s
The block is on the point of overturning = 56.75  ft/s
the centrifugal force must just be equal to the frictional force=  0.56  degree

Ex8-pg34

In [8]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=20.##cwt
q=3.##ft
d=30.##ft/ss
w1=4.##ft
w2=6.##in
h=2240.##ft/s
g=32.2##ft
s=15.##ft
f=4.5##ft
c=2.25##ft
##CALCULATIONS
T=(h*(d)**2/(g*s*q))##lbf
G=T*q##lbf ft
W=h*f/2.##lbf ft
R=186.5##lbf
D=h-R##lbf
r=(q*h*d**2/(c*h)/g)##ft
##RESULTS
print'%s %.2f %s'%('the equal moment of the centrifugal force= ',r,' ft')

the equal moment of the centrifugal force=  37.27  ft

Ex9-pg35

In [9]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
l=3.##ft
w=8.##lbf
p=40.##rev/min
q=6.##ft
h=3.5##ft
g=32.2##ft
f=6.##in
t=15.33##lbf
##CALCULATIONS
F=q/t##in/lbf
R=w*q/t##in
D=(h*w)/t*10.##in
##RESULTS
print'%s %.2f %s'%('the distance horizantal circle= ',D,' in')

the distance horizantal circle=  18.26  in