Chapter 22: Kinetics Of Rigid Body:Force And Acceleration¶

import math

# Initilization of variables

N=1500 # r.p.m
r=0.5 # m # radius of the disc
m=300 # N # weight of the disc
t=120 # seconds # time in which the disc comes to rest
omega=0 
pi=3.14 
g=9.81 # m/s^2

# Calculations

omega_0=(2*pi*N)*0.01666 # rad/s #1/60=0.01666

# angular deceleration is given as,
alpha=-(omega_0/t) # radian/second^2
theta=(omega_0**2)/(2*(-alpha)) # radian

# Let n be the no of revolutions taken by the disc before it comes to rest, then
n=theta/(2*pi)

# Now,
I_G=((0.5)*m*r**2)/g

# The frictional torque is given as,
M=I_G*alpha # N-m

# Results

print"(a) The no of revolutions executed by the disc before coming to rest is ",round(n,2)
print"(b) The frictional torque is ",round(M),"N-m"

import math

# Initilization of variables

s=1 # m
mu=0.192 # coefficient of static friction
g=9.81 # m/s^2

# Calculations

# The maximum angle of the inclined plane is given as,
theta=arctan(3*mu)*(180/pi) # degree
a=(2/3)*g*sin(theta*(pi/180)) # m/s^2 # by solving eq'n 4
v=(2*a*s)**0.5 # m/s

# Let the acceleration at the centre be A which is given as,
A=g*sin(theta*(pi/180)) # m/s^2 # from eq'n 1

# Results

print"(a) The acceleration at the centre is ",round(A,3),"m/s^2"
print"(b) The maximum angle of the inclined plane is ",round(theta),"degree"

import math

# Initilization of variables

W_a=25 # N 
W_b=25 # N 
W=200 # N # weight of the pulley
i_g=0.2 # m # radius of gyration
g=9.81 # m/s^2

# Calculations

# Solving eqn's 1 & 2 for acceleration of weight A (assume a)
a=(0.15*W_a*g)/(((W*i_g**2)/(0.45))+(0.45*W_a)+((0.6*W_b)/(3))) # m/s^2

# Results

print"The acceleration of weight A is ",round(a,2),"m/s^2"

import math

# Initilization of variables

r_1=0.075 # m
r_2=0.15 # m
P=50 # N
W=100 # N
i_g=0.05 # m
theta=30 # degree
g=9.81 # m/s^2

# Calculations

# The eq'n for acceleration of the pool is given by solving eqn's 1,2 &3 as,
a=(50*g*(r_2*cos(theta*(pi/180))-r_1))/(100*((i_g**2/r_2)+r_2)) # m/s^2

# Results

print"The acceleration of the pool is ",round(a,2),"m/s^2"

import math

# Initilization of variables

L=1 # m # length of rod AB
m=10 # kg # mass of the rod
g=9.81 
theta=30 # degree

# Calculations

# solving eq'n 4 for omega we get,
omega=(2*16.82*sin(theta*(pi/180)))**0.5 # rad/s

# Now solving eq'ns 1 &3 for alpha we get,
alpha=(1.714)*g*cos(theta*(pi/180)) # rad/s

# Components of reaction are given as,
R_t=((m*g*cos(theta*(pi/180)))-((m*alpha*L)/4)) # N
R_n=((m*omega**2*L)/4)+(m*g*sin(theta*(pi/180))) # N
R=(R_t**2+R_n**2)**0.5 # N 

# Results

print"(a) The angular velocity of the rod is ",round(omega,2),"rad/sec"
print"(b) The reaction at the hinge is ",round(R,1),"N"