# CHAPTER 5:PLANE KINEMATICS OF RIGID BODIES¶

## SAMPLE PROBLEM 5/1,PAGE NUMBER:331¶

In [1]:
import math
import sympy
# Given data
n_1=1800;# rev/min
t_0=0;# s
# alpha=4t;
n_2=900;# rev/min

# Calculation
# (a)
# omega=-(60*math.pi)+2t**2
t=math.sqrt((omega_2-omega_1)/2);# s
# (b)
# The flywheel changes direction when its angular velocity is momentarily zero. Thus,
t_b=math.sqrt((0-omega_1)/2);# s
# (c)
t_0=0;# s
t_1=t_b;# s
N_1=abs(-theta_1/(2*math.pi));# rev(clockwise)
t_1=t_b;# s
t_2=14;# s
N_2=theta_2/(2*math.pi);# rev
N=N_1+N_2;# rev
print"\n(a)The time required for the flywheel to reduce its clockwise angular speed,t=%1.2f s \n(b)The time required for the flywheel to reverse its direction of rotation,t=%1.2f s \n(c)The total number of revolutions,N=%3.0f rev"%(t,t_b,N);

(a)The time required for the flywheel to reduce its clockwise angular speed,t=6.86 s
(b)The time required for the flywheel to reverse its direction of rotation,t=9.71 s
(c)The total number of revolutions,N= 81 rev


## SAMPLE PROBLEM 5/2,PAGE NUMBER:332¶

In [2]:
import math
# Given data
v=3;# ft/sec
s=4;# ft
d_C=48;# inch
d_B=36;# inch
d_A=12;# inch
r_A=d_A/2;# inch
r_C=d_C/2;# inch
r_B=d_B/2;# inch

# Calculation
# (a)
a=v**2/(2*s);# ft/sec**2
a_t=a;# ft/sec**2
a_n=v**2/(r_C/12);# ft/sec**2
a_C=math.sqrt(a_n**2+a_t**2);# ft/sec**2
# (b)
print"\n(a)The acceleration of point C on the cable in contact with the drum,a_C=%1.2f ft/sec**2 \n(b)The angular velocity and angular accelerationof the pinion A,omega_A=%1.1f rad/sec CW and alpha_A=%1.3f rad/sec**2 CW"%(a_C,omega_A,alpha_A);

(a)The acceleration of point C on the cable in contact with the drum,a_C=4.12 ft/sec**2
(b)The angular velocity and angular accelerationof the pinion A,omega_A=3.0 rad/sec CW and alpha_A=0.000 rad/sec**2 CW


## SAMPLE PROBLEM 5/3,PAGE NUMBER:332¶

In [3]:
import math
from numpy.linalg import norm
# Given data
x=0.4;# m
y=0.3;# m

# Calculation
# Using the right-hand rule gives
r=[x,y];# m
v=[-omega*r[1],omega*r[0]];# (i,j) (k*i=j)(k*j=-i) m/s
a_n=[-omega*v[1],omega*v[0]];# m/s**2
a_t=[-alpha*r[1],alpha*r[0]];# m/s**2
a=a_n+a_t;# m/s**2
print"\nThe vector expression for the velocity,v=%0.1fi+(%0.1f)j m/s \nThe vector expression for the acceleration of point A,a=%2.1fi+%0.1fj m/s**2"%(v[0],v[1],a[0],a[1]);
v=norm(v);# m/s
a=norm(a);# m/s**2

The vector expression for the velocity,v=0.6i+(-0.8)j m/s
The vector expression for the acceleration of point A,a=-1.6i+-1.2j m/s**2


## SAMPLE PROBLEM 5/5,PAGE NUMBER:340¶

In [4]:
import math
# Given data
r_1=4;# inch
r_2=4;# inch
# Case(a)
# Pulley 1:
alpha_1a=omega_dot;
# Pulley 2:
# Case(b)
# Pulley 1:
# Pulley 2:
ABbar=12;# inch
AObar=4;# inch

# Calculation
# Case (a)
v_D=r_2*omega_2a;# in/sec
a_D=r_2*alpha_2a;# in/sec
alpha=a_D/ABbar;# in/sec**2
print"\n(a)omega=%0.3f rad/sec (CCW)\n   alpha=%1.0f rad/sec**2 (CW) \n   v_O=%1.3f in/sec \n   a_O=%1.0f in/sec**2"%(omega,alpha,v_O,a_O);
# Case (b)
v_C=r_1*omega_1b;# in/sec
v_D=r_2*omega_2b;# in/sec
a_C=r_1*alpha_1b;# in/sec**2
a_D=r_2*alpha_2b;# in/sec**2
v_O=v_C+(AObar*omega);# in/sec
a_O=a_C+(AObar*alpha);# in/sec
print"\n(b)omega=%0.3f rad/sec (CCW)\n   alpha=%1.0f rad/sec**2 (CW) \n   v_O=%1.3f in/sec \n   a_O=%1.0f in/sec**2"%(omega,alpha,v_O,a_O);

(a)omega=0.667 rad/sec (CCW)
v_O=2.667 in/sec
a_O=-4 in/sec**2

v_O=5.333 in/sec
a_O=8 in/sec**2


## SAMPLE PROBLEM 5/6,PAGE NUMBER:341¶

In [5]:
import math
# Given data
v_A=0.3;# m/s
b=0.2;# m
theta=30;# degree

# Calculation
v_B=-v_A*math.tan(theta*math.pi/180);# m/s
a_B=-((v_A**2)/b)*(math.acos(theta*math.pi/180))**3;# m/s**2
print"\nThe velocity of the center of the roller B in the horizontal guide,v_B=%1.4f m/s \nThe acceleration of the center of the roller B in the horizontal guide,a_B=%0.3f m/s**2 \nThe angular velocity of edge CB,omega=%1.3f rad/s \nThe angular acceleration of edge CB,alpha=%1.3f rad/sec**2"%(v_B,a_B,omega,alpha);

The velocity of the center of the roller B in the horizontal guide,v_B=-0.1732 m/s
The acceleration of the center of the roller B in the horizontal guide,a_B=-0.477 m/s**2
The angular velocity of edge CB,omega=1.530 rad/s
The angular acceleration of edge CB,alpha=1.351 rad/sec**2


## SAMPLE PROBLEM 5/7,PAGE NUMBER:351¶

In [6]:
import math
from numpy.linalg import det,norm
# Given data
r=0.300;# m
v_O=3;# m/s
theta=30;# degree
r_0=0.200;# m
ACbar=0.436;# m
OCbar=0.300;# m

# Calculation
# Solution I (Scalar-Geometric)
v_AO=r_0*omega;# m/s
v_A=math.sqrt(v_O**2+v_AO**2+(2*v_O*v_AO*math.cos(theta*math.pi/180)));# m/s
v_AC=(ACbar/OCbar)*v_O;# m/s
v_A=v_AC;# m/s
print"\nThe velocity of point A on the wheel,v_A=%1.2f m/s"%v_A;
# Solution II (Vector)
r_0=[(r_0*-math.cos(theta*math.pi/180)),(r_0*math.sin(theta*math.pi/180)),0];# m
v_O=[v_O,0,0];# m/s
v_AO1=det([(omega[1],omega[2]),(r_0[1],r_0[2])]);# m/s
v_AO2=-det([(omega[0],omega[2]),(r_0[0],r_0[2])]);# m/s
v_AO3=det([(omega[0],omega[1]),(r_0[0],r_0[1])]);# m/s
v_AO=[v_AO1,v_AO2,v_AO3];# m/s
v_A=v_O+v_AO;# m/s
print"\nThe velocity of point A on the wheel,v_A=%1.0fi+%1.3fj m/s"%(v_A[0],v_A[1]);
v_A=norm(v_A);# m/s

The velocity of point A on the wheel,v_A=4.36 m/s

The velocity of point A on the wheel,v_A=3i+0.000j m/s


## SAMPLE PROBLEM 5/8,PAGE NUMBER:352¶

In [7]:
import math
# Given data
OCbar=0.250;# m
OAbar=0.100;# m
OBbar=0.050;# m
ABbar=0.075;# m

# Calculation
# Solution I (Vector)
r_A=[0,0.100,0];# (i,j,k) m
r_B=[-0.75,0,0];# (i,j,k) m
r_AB=[-0.175,0.50,0];# (i,j,k) m
# omega_OA*r=(omega_CB*r_B)+(omega_AB*r_AB);
# omega_OA=omega_OA*k
# omega_CB=2k
# omega_AB=omega_ABk
# Matching coefficients of the respective i- and j-terms gives
print"\n(I)The angular velocity of OA,omega_OA=%0.3f rad/s \n   The angular velocity of AB,omega_AB=%0.3f rad/s"%(omega_OA,omega_AB);
# Solution II (Scalar-Geometric)
r_A=0.100;# m
r_B=0.075;# m
v_B=r_B*omega;# m/s
tantheta=(OAbar-OBbar)/(OCbar-r_B);
# v_AB=v_B/ cos(theta);
# ABbar= (OCbar-r_AB)/ cos(theta);
v_A=v_B*tantheta;# m/s
print"\n(II)The angular velocity of OA,omega_OA=%0.3f rad/s \n    The angular velocity of AB,omega_AB=%0.3f rad/s"%(omega_OA,omega_AB);

(I)The angular velocity of OA,omega_OA=-0.429 rad/s
The angular velocity of AB,omega_AB=-0.857 rad/s

(II)The angular velocity of OA,omega_OA=0.429 rad/s
The angular velocity of AB,omega_AB=0.857 rad/s


## SAMPLE PROBLEM 5/9,PAGE NUMBER:353¶

In [8]:
import math
# Given data
n=1500;# rev/min
theta=60;# degree
r=5;# inch
d_AG=10;# The distance from A to G in inch
d_GB=4;# The distance from G to B in inch
d_AB=14;# The distance from A to B in inch

# Calculation
v_B=(r*12**-1)*((2*math.pi*n)/60);# ft/sec
# From the law of sines,
beta=math.asin(r/(d_AB/math.sin(theta*math.pi/180)));# degree
theta_3=30;# degree
theta_1=90-beta;# degree
theta_2=180-theta_3-theta_1;# degree
v_A=(v_B*math.sin(theta_2*math.pi/180))/math.sin(theta_1*math.pi/180);# ft/sec
v_AB=(v_B*math.sin(theta_3*math.pi/180))/math.sin(theta_1*math.pi/180);# ft/sec
ABbar=d_AB/12;# ft
GBbar=d_GB/12;# ft
v_GB=(GBbar/ABbar)*v_AB;# ft/sec
# From velocity diagram
v_G=64.1;# ft/sec
print"\nThe velocity of the piston A,v_A=%2.1f ft/sec \nThe velocity of point G on the connecting rod,v_G=%2.1f ft/sec \nThe angular velocity of the connecting rod,omega_AB=%2.1f rad/sec"%(v_A,v_G,omega_AB);

The velocity of the piston A,v_A=56.9 ft/sec
The velocity of point G on the connecting rod,v_G=64.1 ft/sec
The angular velocity of the connecting rod,omega_AB=32.7 rad/sec


## SAMPLE PROBLEM 5/10,PAGE NUMBER:354¶

In [9]:
import math
# Given data
v_B=0.8;# The velocity in m/s
theta=30;# degree
d_co=18;# The distance in inch

# Calculation
v_A=v_B*math.cos(theta*math.pi/180);# ft/sec
OAbar=(d_co/12)/(math.cos(theta*math.pi/180));# ft
print"\nThe angular velocity of the slotted arm,omega=%0.3f rad/sec CCW"%omega;

The angular velocity of the slotted arm,omega=0.600 rad/sec CCW


## SAMPLE PROBLEM 5/11,PAGE NUMBER:364¶

In [10]:
import math
# Given data
r=0.300;# m
r_0=0.200;# m
v_o=3;# m/s
OCbar=r;# m
theta=120;# degree

# Calculation
ACbar=math.sqrt(r**2+r_0**2-(2*r*r_0*math.cos(theta*math.pi/180)));# m
v_A=ACbar*omega;# m/s
print"\nThe velocity of point A for the position indicated,v_A=%1.2f m/s"%v_A;

The velocity of point A for the position indicated,v_A=4.36 m/s


## SAMPLE PROBLEM 5/12,PAGE NUMBER:364¶

In [11]:
import math
# Given data
theta=45;# degree
OBbar=(6*math.sqrt(2))/12;# ft
BCbar=(14*math.sqrt(2))/12;# ft
ACbar=14/12;# ft
CDbar=15.23/12;# ft

# Calculation
v_A=ACbar*omega_BC;# ft/sec
v_D=CDbar*omega_BC;# ft/sec
print"\nThe velocity of A,v_A=%1.2f ft/sec \nThe velocity of D,v_D=%1.2f ft/sec \nThe angular velocity of link AB,omega_AB=%1.2f rad/sec CCW"%(v_A,v_D,omega_BC);

The velocity of A,v_A=4.29 ft/sec
The velocity of D,v_D=5.44 ft/sec
The angular velocity of link AB,omega_AB=4.29 rad/sec CCW


## SAMPLE PROBLEM 5/14,PAGE NUMBER:375¶

In [12]:
import math
from scipy.optimize import fsolve
# Given data
r_A=100;# mm
r_B=75;# mm
OCbar=250;# mm

# Calculation
# The acceleration equation is a_A=a_B+(a_A/B)_n+(a_A/B)_t;
# a_A=(alpha_OA*r_A)+(omega_OA*(omega_OA*r_A))
# a_A=(-100*alpha_OA)i-((100)*(3/7)**2)j mm/s**2
# a_B=(alpha_CB*r_B)+(omega_CB*(omega_CB*r_B)) mm/s**2
# a_B=300i mm/s**2
# (a_A/B)n=omega_AB*(omega_AB*r_AB)
# (a_A/B)n=(6/7)**2*(175i-50j) mm/s**2
# (a_A/B)t= alpha_AB*r_A/B
# (a_A/B)t=(-50*alpha_AB)i-(175*alpha_AB)j mm/s**2
# Equate separately the coefficients of the i-terms and the coefficients of the j-terms to give
def eqn(p):
alpha_OA,alpha_AB=p
X1=(-100*alpha_OA)-(429-(50*alpha_AB));
X2=(-18.37)-(-36.7-(175*alpha_AB));
return(X1,X2)
alpha_OA,alpha_AB=fsolve(eqn,(1,1));
print"\nThe angular acceleration of link AB,alpha_AB=%0.4f rad/s**2 \nThe angular acceleration of link OA,alpha_OA=%1.2f rad/s**2"%(alpha_AB,alpha_OA);

The angular acceleration of link AB,alpha_AB=-0.1047 rad/s**2
The angular acceleration of link OA,alpha_OA=-4.34 rad/s**2


## SAMPLE PROBLEM 5/15,PAGE NUMBER:376¶

In [13]:
import math
from scipy.optimize import fsolve
# Given data
N=1500;# rev/min
theta_1=60*math.pi/180;# degree
r=5*12**-1;# ft
ABbar=14*12**-1;# ft

# Calculation
a_B=r*omega**2;# ft/sec**2
a_AB_n=ABbar*omega_AB**2;
# If we adopt an algebraic solution using the geometry of the acceleration polygon, we first compute the angle between AB and the horizontal. With the law of sines, this angle becomes 18.02 degree.
theta_2=18.02*math.pi/180;# degreB
def equations(p):
a_AB,a_A=p;
return((((a_B*math.cos(theta_1)+(a_AB_n*math.cos(theta_2)-(a_AB*math.sin(theta_2)))-a_A),(((a_B*math.sin(theta_1))-(a_AB_n*math.sin(theta_2))-(a_AB*math.cos(theta_2)))-0))));
a_AB,a_A=fsolve(equations,(1,1))
r=ABbar;# ft
print"\nThe acceleration of the piston A,a_A=%4.0f ft/sec**2 \nThe angular acceleration of the connecting rod AB,alpha_AB=%4.0f rad/sec**2"%(a_A,alpha_AB);

The acceleration of the piston A,a_A=3312 ft/sec**2
The angular acceleration of the connecting rod AB,alpha_AB=7742 rad/sec**2


## SAMPLE PROBLEM 5/16,PAGE NUMBER:391¶

In [14]:
import math
from numpy.linalg import norm
# Given data
r=6;# in
rdot=5;# in/sec
rdotdot=81;# in/sec**2

# Calculation
# Velocity
v_rel=rdot;# (k) in/sec
v_A=[v_rel,(omega*r)];# in/sec
print"\nv_A=%1.0fi+%2.0fj in/sec"%(v_A[0],v_A[1]);
v_A=norm(v_A);# in/sec
print"\nv_A=%2.1f in/sec"%v_A;
# Acceleration
O=-(omega*(omega*r));# in/sec**2
O_2=2*(omega)*(v_rel);# in/sec**2
a_rel=rdotdot;# in/sec**2
a_A=[(a_rel+O),(O_2+O_1)];# in/sec**2
print"\na_A=%2.0fi+(%2.0f)j in/sec**2"%(a_A[0],a_A[1]);
a_A=norm(a_A);# in/sec**2
print"\na_A=%2.0f in/sec"%a_A;

v_A=5i+24j in/sec

v_A=24.5 in/sec

a_A=-15i+(-20)j in/sec**2

a_A=25 in/sec


## SAMPLE PROBLEM 5/17,PAGE NUMBER:392¶

In [15]:
import math
# Given data
theta=45;# degree
OCbar=450;# mm
CAbar=225;# mm

# Calculation
# v_A=omega_CA*r_CA;
# v_A=(225/math.sqrt(2))omega_CA*(i-j)
OPbar=math.sqrt((OCbar-CAbar)**2+(CAbar)**2);# mm
r=OPbar;# mm
O=omega*r;# mm/s
# Substitution into the relative-velocity equation gives
# (225/math.sqrt(2))omega_CA*(i-j)=(450*math.sqrt(2)j+xdoti)
# Equating separately the coefficients of the i and j terms yields
omega_CA=O/(225/math.sqrt(2));# mm/s
xdot=(225/math.sqrt(2))*omega_CA;# mm/s
v_rel=xdot;# mm/s
v_A=CAbar*abs(omega_CA);# mm/s
v_P=OPbar*omega;# mm/s
v_AP=abs(v_rel);# mm/s
print"\nThe actual angular velocity of CA,omega_CA=%1.0f rad/s \nThe velocity of A relative to the rotating slot in OD,xdot=v_rel=%3.2f mm/s \nThe velocity of pin A,v_A=%3.0f mm/s"%(omega_CA,xdot,v_A);

The actual angular velocity of CA,omega_CA=4 rad/s
The velocity of A relative to the rotating slot in OD,xdot=v_rel=636.40 mm/s
The velocity of pin A,v_A=900 mm/s


## SAMPLE PROBLEM 5/18,PAGE NUMBER:393¶

In [16]:
import math
# Given data
theta=45;# degree
OCbar=450;# mm
CAbar=225;# mm

# Calculation
r=CAbar*math.sqrt(2);# mm
v_rel=(-OCbar*math.sqrt(2));# mm/s
O_1=0;# mm/s**2
O_2=omega*(omega*r);# mm/s**2
O_2=2*omega*v_rel;# mm/s**2
# a_rel=xdotdot;
print"\nThe angular acceleration of AC,omega_CA=%2.0f rad/s \nThe acceleration of A relative to the rotating slot in OD,xdotdot=%4.0f mm/s"%(omegadot_CA,xdotdot);

The angular acceleration of AC,omega_CA=32 rad/s
The acceleration of A relative to the rotating slot in OD,xdotdot=8910 mm/s


## SAMPLE PROBLEM 5/19,PAGE NUMBER:394¶

In [17]:
import math
# Given data
v_B=150;# (i) m/s
v_A=100;# (i) m/s
rho=400;# m
r=-100;# m

# Calculation

The instantaneous velocity,v_rel=-87.5i m/s