# Chapter 3: Velocity and Acceleration¶

## Example 3, Page 90¶

In [4]:
import math

#Variable declaration
OC=6.#in
CP=24#in
N=240#rpm
X=45#degrees
XP=19#in
XC=6#in
YP=32#in
YC=9#in
#Scalling off lenghts from fig , we have
CI=2.77#in
PI=2.33#in
XI=2.33#in
YI=3.48#in

#Calculations
Vc=((math.pi*N)/30)*(OC/12)#changing OP into feets
print "\nw=%.2f ft/s"%Vc
#w=Vc/CI=Vp/PI=Vx/XI=Vy/YI
w=Vc/CI
Vp=w*PI
Vx=w*XI
Vy=w*YI

#Results
print "velocity of points P, X and Y are %.2f ft/s, %.2f ft/s and %.1f ft/s respectively"%(Vp,Vx,Vy)

w=12.57 ft/s
velocity of points P, X and Y are 10.57 ft/s, 10.57 ft/s and 15.8 ft/s respectively


## Example 4, Page 93¶

In [5]:
import math

#Variable declaration
OC=9.#inches
CP=36.#inches
XC=12#inches
X=40#degrees
CM=6.98#from the scaled figure
N1=240#rpm
N2=240#rpm (instantaneous) with angular aceleration (ao) 100 rad/s^2

#Calculations&Results
w=(math.pi*N1/30)
a=w**2*(OC/12)
print "Centripetal acceleration = %.f ft/s^2"%a
f1=Wr**2*(CP/12)#centripetal component of acceleration of p realtive to C
#Solution a)
#given from fig 58(a)
tp=296
cp=306
ox=422
f2=tp #Tangential component of acceleration of p realtive to C
f3=cp#acceleration of p realtive to C
fx=ox#acce;eration of x
ar=f2/(CP/12)#angular acceleration of rod
print "\nCase a) \nap= %.f ft/s^2,\nax= %.f ft/s^2 and\nar= %.1f rad/s^2"%(f3,fx,ar)

#Solution b)
#given from fig 58(b)
oc1=474
oc=480
pt=238.
pc=246
xo=452
f4=pt#Tangential component of acceleration of p realtive to C
f5=pc#acceleration of p realtive to C
Ar=f4/(CP/12)#angular acceleration of rod
f6=ao*(OC/12)#tangential component of acceleration realtive to C
Fx=xo#acce;eration of x
print "\nCase b) \nap= %.f ft/s^2,\nax= %.f ft/s^2 and\nar= %.1f rad/s^2"%(f4,Fx,Ar)

Centripetal acceleration = 474 ft/s^2

Case a)
ap= 306 ft/s^2,
ax= 422 ft/s^2 and

Case b)
ap= 238 ft/s^2,
ax= 452 ft/s^2 and


## Example 5, Page 98¶

In [6]:
import math

#Variable declaration
AB=2.5#inches
BC=7.#inches
CD=4.5#inches
DA=8#inches
N=100#rpm
X=60#degrees

#Calculations
w=(math.pi*N)/30
#From triangle ABM we have
AM=0.14#feet
BM=0.12#feet
Vb=w*AB/12#ft/s
Vc=w*AM#ft/s
Vcb=w*BM#ft/s
fb=w**2*(AB/12)#ft/s^2
bt=Vcb**2/(BC/12)#ft/s^2
os=Vc**2/(CD/12)#ft/s^2
#By measurement from acceleration diagram
sc=19.1#ft/s^2
tq=14.4#ft/s^2
Acd=sc/(CD/12)
Abc=tq/(BC/12)

#Results
print "Vb=%.2f ft/s \nVc=%.2f ft/s\nVcb=%.2f ft/s\nfb=%.2f ft/s^2\nbt=%.2f ft/s^2\nos=%.2f ft/s^2"%(Vb,Vc,Vcb,fb,bt,os)
print "Angular acceleration of CD(counter-clockwise)= %.1f rad/s^2"%Acd
print "Angular acceleration of BC(counter-clockwise)= %.1f rad/s^2"%Abc

Vb=2.18 ft/s
Vc=1.47 ft/s
Vcb=1.26 ft/s
fb=22.85 ft/s^2
bt=2.71 ft/s^2
os=5.73 ft/s^2
Angular acceleration of CD(counter-clockwise)= 50.9 rad/s^2
Angular acceleration of BC(counter-clockwise)= 24.7 rad/s^2


## Example 6, Page 106¶

In [7]:
#Variable declaration
OP=2#ft
f=4#ft/s^2
Vpq=3 #ft/s

#Calculations
r=OP
os=w**2*r#component 1
sq=a*r#component 2
qt=f#component 3
tp=2*w*Vpq#component 4
Apo=((os-qt)**2+(sq+tp)**2)**(1./2)

#Result
print "Acceleration of P realative to fixed point O is %.1f ft/s^2"%Apo

Acceleration of P realative to fixed point O is 22.4 ft/s^2


## Example 7, Page 110¶

In [8]:
import math

#Variable declaration
OC=8#inches
CP=4#inches
N=60#inches
ON=15#inches
RN=6.#inches
X=120#degrees
OP=10.6
OQ=OP

#Calculations
#from fig 65(a)
Vq=1.56#ft/s
Vrn=0.74#ft/s
#from fig 65(b)
ftq=3.74#ft/s^2
ftrn=2.03#ft/s^2
w1=(math.pi*N)/30
w=Vq/(OQ/12)
wrn=Vrn/(RN/12)
a=ftq/(OP/12)#Angular acceleration of ON
a1=ftrn/(RN/12)#angular acceleration of RN

#Results

W=1.77 rad/s
Angular acceleration of ON= 4.23 rad/s^2


## Example 8, Page 112¶

In [9]:
import math

#Variable declaration
OC=3#inches
CP=9.#inches
N=1200 #rpm (clockwise)
X=55 #degrees
#from the figure 66
OP=10.35#inches
PM=10.74#inches
OM=2.95#inches
PC=12.84#inches
PR=PC
RV=2.49#inches
UV=1.29#inches
OU=5.90#inches
PV=13.05#inches
OV=6.06#inches
OQ=OP

#Calculations
w=(math.pi*N)/30#the angular velocity of the cylinder line OP
Vq=w*(OP/12)#the velocity of Q
Vp=w*(PM/12)#The velocity of P
w1=Vp/(CP/12)#The angular velocity of CP
Vpq=w*(OM/12)#the velocity of sliding of the piston along the cylinder
fq=w**2*(OQ/12)#the centripetal acceleration of Q
Acp=w1**2*(PC/12)#The centripetal component of acceleration of P
Atp=w**2*(RV/12)#The tangential component of acceleration of P
acp=Atp/(CP/12)# The angular acceleration of the connecting rod CP
f=w**2*(UV/12)#component c
d=2*w*Vpq#component d
Ap=w**2*PV#the resultant acceleration of P
Apq=w**2*OV#the acceleration of P realative to Q

#Results
print "The velocity and acceleration of the piston along the cylinder are %.1f ft/s and %.f ft/s^2 respectively"\
"\nThe angular velocity and angular acceleration of the connecting rod cp are %.1f rad/s and %.f rad/s^2 respectively"\
"\nAnd the coriolis component of the acceleration of P is %.f ft/s^2"%(Vpq,f,w1,acp,d)

The velocity and acceleration of the piston along the cylinder are 30.9 ft/s and 1698 ft/s^2 respectively
The angular velocity and angular acceleration of the connecting rod cp are 150.0 rad/s and 4369 rad/s^2 respectively
And the coriolis component of the acceleration of P is 7764 ft/s^2