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{
"metadata": {
"name": "chapter22.ipynb"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 22: Kinetics Of Rigid Body:Force And Acceleration"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22.22-1,Page No:562"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"N=1500 # r.p.m\n",
"r=0.5 # m # radius of the disc\n",
"m=300 # N # weight of the disc\n",
"t=120 # seconds # time in which the disc comes to rest\n",
"omega=0 \n",
"pi=3.14 \n",
"g=9.81 # m/s^2\n",
"\n",
"# Calculations\n",
"\n",
"omega_0=(2*pi*N)*0.01666 # rad/s #1/60=0.01666\n",
"\n",
"# angular deceleration is given as,\n",
"alpha=-(omega_0/t) # radian/second^2\n",
"theta=(omega_0**2)/(2*(-alpha)) # radian\n",
"\n",
"# Let n be the no of revolutions taken by the disc before it comes to rest, then\n",
"n=theta/(2*pi)\n",
"\n",
"# Now,\n",
"I_G=((0.5)*m*r**2)/g\n",
"\n",
"# The frictional torque is given as,\n",
"M=I_G*alpha # N-m\n",
"\n",
"# Results\n",
"\n",
"print\"(a) The no of revolutions executed by the disc before coming to rest is \",round(n,2)\n",
"print\"(b) The frictional torque is \",round(M),\"N-m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) The no of revolutions executed by the disc before coming to rest is 1499.4\n",
"(b) The frictional torque is -5.0 N-m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22.22-2,Page No:563"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"s=1 # m\n",
"mu=0.192 # coefficient of static friction\n",
"g=9.81 # m/s^2\n",
"\n",
"# Calculations\n",
"\n",
"# The maximum angle of the inclined plane is given as,\n",
"theta=arctan(3*mu)*(180/pi) # degree\n",
"a=(2/3)*g*sin(theta*(pi/180)) # m/s^2 # by solving eq'n 4\n",
"v=(2*a*s)**0.5 # m/s\n",
"\n",
"# Let the acceleration at the centre be A which is given as,\n",
"A=g*sin(theta*(pi/180)) # m/s^2 # from eq'n 1\n",
"\n",
"# Results\n",
"\n",
"print\"(a) The acceleration at the centre is \",round(A,3),\"m/s^2\"\n",
"print\"(b) The maximum angle of the inclined plane is \",round(theta),\"degree\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) The acceleration at the centre is 4.896 m/s^2\n",
"(b) The maximum angle of the inclined plane is 30.0 degree\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22.22-5,Page No:568"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"W_a=25 # N \n",
"W_b=25 # N \n",
"W=200 # N # weight of the pulley\n",
"i_g=0.2 # m # radius of gyration\n",
"g=9.81 # m/s^2\n",
"\n",
"# Calculations\n",
"\n",
"# Solving eqn's 1 & 2 for acceleration of weight A (assume a)\n",
"a=(0.15*W_a*g)/(((W*i_g**2)/(0.45))+(0.45*W_a)+((0.6*W_b)/(3))) # m/s^2\n",
"\n",
"# Results\n",
"\n",
"print\"The acceleration of weight A is \",round(a,2),\"m/s^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The acceleration of weight A is 1.08 m/s^2\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22.22-8,Page No:571"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"r_1=0.075 # m\n",
"r_2=0.15 # m\n",
"P=50 # N\n",
"W=100 # N\n",
"i_g=0.05 # m\n",
"theta=30 # degree\n",
"g=9.81 # m/s^2\n",
"\n",
"# Calculations\n",
"\n",
"# The eq'n for acceleration of the pool is given by solving eqn's 1,2 &3 as,\n",
"a=(50*g*(r_2*cos(theta*(pi/180))-r_1))/(100*((i_g**2/r_2)+r_2)) # m/s^2\n",
"\n",
"# Results\n",
"\n",
"print\"The acceleration of the pool is \",round(a,2),\"m/s^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The acceleration of the pool is 1.62 m/s^2\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 22.22-10,Page No:574"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Initilization of variables\n",
"\n",
"L=1 # m # length of rod AB\n",
"m=10 # kg # mass of the rod\n",
"g=9.81 \n",
"theta=30 # degree\n",
"\n",
"# Calculations\n",
"\n",
"# solving eq'n 4 for omega we get,\n",
"omega=(2*16.82*sin(theta*(pi/180)))**0.5 # rad/s\n",
"\n",
"# Now solving eq'ns 1 &3 for alpha we get,\n",
"alpha=(1.714)*g*cos(theta*(pi/180)) # rad/s\n",
"\n",
"# Components of reaction are given as,\n",
"R_t=((m*g*cos(theta*(pi/180)))-((m*alpha*L)/4)) # N\n",
"R_n=((m*omega**2*L)/4)+(m*g*sin(theta*(pi/180))) # N\n",
"R=(R_t**2+R_n**2)**0.5 # N \n",
"\n",
"# Results\n",
"\n",
"print\"(a) The angular velocity of the rod is \",round(omega,2),\"rad/sec\"\n",
"print\"(b) The reaction at the hinge is \",round(R,1),\"N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) The angular velocity of the rod is 4.1 rad/sec\n",
"(b) The reaction at the hinge is 103.2 N\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
