# CHAPTER 4:KINETICS OF SYSTEMS OF PARTICLES¶

## SAMPLE PROBLEM 4/4,PAGE NUMBER:279¶

In [1]:
import math
# Variable declaration
m=20;# kg
u_z=300;# m/s
g=9.81;# m/s**2
m_a=5;# kg
m_b=9;# kg
m_c=6;# kg
theta=45;# degree
s=4000;# m
x=3;# m
y=4;# m
r=5;# m
h_a=500;# m

# Calculation
t=(u_z*y)/(r*g);# The time required for the shell to reach P in s
h=u_z**2/(2*g);# The verticl rise in m
v_a=math.sqrt(2*g*h_a);# m/s
v_b=s/t;# m/s
v_c=[((m*u_z*(x/r))-(m_b*v_b*math.cos(theta*math.pi/180)))/6,(m_b*v_b*math.sin(theta*math.pi/180))/6,(m_a*v_a)/6];# m/s
v_c=math.sqrt((v_c[0])**2+(v_c[1])**2+(v_c[2])**2);# m/s
print"\nThe velocity which fragment C has immediately after the explosion,v_C=%3.0f m/s"%v_c;

The velocity which fragment C has immediately after the explosion,v_C=259 m/s


## SAMPLE PROBLEM 4/5,PAGE NUMBER:280¶

In [2]:
import math
# Variable declaration
g=32.2;# The acceleration due to gravity in ft/sec**2
n_12=80;# rev/min
n_34=100;# rev/min
W_a=32.2;# lb
W_b=3.22;# lb
n=4;# Number of balls
vbar=4;# m/s
r_12=18*12**-1;# ft
r_34=12*12**-1;# ft

# Calculation
# (a)Kinetic energy
v_rel12=r_12*((2*math.pi*n_12)/60);# ft/sec
v_rel34=r_34*((2*math.pi*n_34)/60);# ft/sec
ke=(((W_a/g)+(n*(W_b/g)))/2)*(vbar)**2;# ft-lb
ke_r=(2*((W_b/g)*v_rel12**2)/2)+(2*((W_b/g)*v_rel34**2)/2);# The rotational part of the kinetic energy in f
T=ke+ke_r;# The total kinetic energy in ft-lb
# (b)Linear momentum
G=((W_a/g)+(n*(W_b/g)))*vbar;# ft-lb-sec

(a)The kinetic energy,T=38 ft-lb