{ "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": {} } ] }