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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": {}

+  }

+ ]

+}
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