Chapter 2: Units and Dimensions

Exa 2.1a

In [1]:
#According to newton's second law for angular motion, torque equals 
#the product of the mass moment of inertia and angular acceleration 
#By means of dimensonal symbolism determine the units of I in SI units
#initialisation of variables
F= 1					#N
L= 1					#m
T= 1					#s
I= 1					#N m s^2
N= 1					#Kg m s^-2
#CALCULATIONS
I= F*L*T*T 				#Kg m^2
#RESULTS
print '%s %.2f' % ('I in SI system (Kg m^2) = ',I)
raw_input('press enter key to exit')

I in SI system (Kg m^2) =  1.00
press enter key to exit
Out[1]:
''

Exa 2.1b

In [ ]:
#According to newton's second law for angular motion, torque equals 
#the product of the mass moment of inertia and angular acceleration 
#By means of dimensonal symbolism determine the units of I in British units
#initialisation of variables
F= 1					#lbf
L= 1					#ft
T= 1					#s
I= 1					#lbf ft s^2
lbf= 1					#slug ft s^-2
#CALCULATIONS
I= F*L*T*T 				#slug ft^2
#RESULTS
print '%s %.2f' % (' I in British Gravitational System (slug ft^2) = ',I)
raw_input('press enter key to exit')

 I in British Gravitational System (slug ft^2) =  1.00

Exa 2.2

In [1]:
#Determine the gravitational constant gc for the british absoulte system.
#initialisation of variables
F= 1 					#Pouunda
m= 1 					#lbm
g= 1 					#fts^-2
#CALCULATIONS
gc= m*g/F 				#Gravitation in British Units
#RESULTS
print '%s %.2f' %('gc (lbm ft/poundal^2) = ',gc)
raw_input('press enter key to exit')

gc (lbm ft/poundal^2) =  1.00
press enter key to exit
Out[1]:
''

Exa 2.3

In [2]:
#If a mercury barometer shows a height of 76 cmHg, express the atmospheric
#pressure in force units of the british absoulte system.
#initialisation of variables
h= 76. 					#cmhg
g= 32.2 				#ft/s^2
h= 76.0 				#cmHg
Dhg= 847. 				#lbm/ft^3
#CALCULATIONS
Pa= Dhg*g*h*0.33		#P in lbm/ft S^2
Pa1= Pa/1. 				#P in poundal/ft^2
#RESULTS
print '%s %.2f' % ('Pa (poundal/ft^2) = ',Pa1)
raw_input('press enter key to exit')

Pa (poundal/ft^2) =  684016.87
press enter key to exit
Out[2]:
''