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diff --git a/Theory_Of_Machines/ch4.ipynb b/Theory_Of_Machines/ch4.ipynb new file mode 100755 index 00000000..845b97ee --- /dev/null +++ b/Theory_Of_Machines/ch4.ipynb @@ -0,0 +1,601 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cff1e23e14e854a300136760b464ae325e3afc2c7094a19b255704b9aeba45ea" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Simple Harmonic Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 120. \t\t\t#rpm\n", + "r = 1. #m\n", + "x = 0.75 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating Angular Velocity\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating Velocity of the Piston\n", + "v = omega*math.sqrt(r**2-x**2) \t\t\t#m/s\n", + "#Calculating Acceleration of the Piston\n", + "a = omega**2*x\n", + "\n", + "#Results:\n", + "print \" The Velocity of the Piston, v = %.2f m/s.\"%(v)\n", + "print \" The Acceleration of the Piston, a = %.2f m/s**2.\"%(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Velocity of the Piston, v = 8.31 m/s.\n", + " The Acceleration of the Piston, a = 118.44 m/s**2.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "x1 = .75\n", + "x2 = 2. \t\t\t#m\n", + "v1 = 11.\n", + "v2 = 3.\t\t\t #m/s\n", + "\n", + "#Solution:\n", + "#We have11 = omega*math.sqrt(r**2-.75**2) and 3 = omega*math.sqrt(r**2-2**2).\n", + "#These upon solving yield r**2-(121/omega**2)-0.5625 = 0 and r**2-(9/omega**2)-4 = 0.\n", + "#Take r**2 = x and (1/omega**2) = y and the equation become x-121y = 0.5625 and x-9y = 4.\n", + "#Variables Matrix\n", + "A = [[1, -121],[ 1, -9]]\n", + "#Constants Matrix\n", + "B = [.5625, 4]\n", + "V = linalg.solve(A,B)\n", + "#Calculating Amplitude of the Particle\n", + "r = math.sqrt(V[0]) \t \t\t#m\n", + "#Calculating Angular Velocity of the Particle\n", + "omega = math.sqrt(1./V[1]) \t\t\t#rad/s\n", + "#Calculating Periodic Time\n", + "tp = 2*math.pi/omega \t\t \t#seconds\n", + "#Calculating Maximum Acceleration\n", + "amax = omega**2*r \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The Angular Velocity omega = %.1f rad/s.\"%(omega)\n", + "print \" The Periodic Time tp = %.1f s.\"%(tp)\n", + "print \" The Maximum Acceleration amax = %.2f m/s**2.\"%(amax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Angular Velocity omega = 5.7 rad/s.\n", + " The Periodic Time tp = 1.1 s.\n", + " The Maximum Acceleration amax = 67.38 m/s**2.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 60. \t\t\t#kg\n", + "r = 0.0125 #m\n", + "x = 0.005 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Extension of the Spring\n", + "delta = (.25/1.5)*60*10**-3 \t\t\t#m\n", + "#Calculating the Frequency of the System\n", + "n = 1./(2*math.pi)*math.sqrt(9.81/delta) \t\t\t#Hz\n", + "#Calculating the Angular Velocity of the Mass\n", + "omega = math.sqrt(9.81/delta) \t\t\t#rad/s\n", + "#Calculating the Linear Velocity of the Mass\n", + "v = omega*math.sqrt(r**2-x**2)\n", + "\n", + "#Results:\n", + "print \" The Frequency of Natural Vibration n = %.2f Hz.\"%(n)\n", + "print \" The Velocity of the Mass v = %.2f m/s.\"%(v)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Frequency of Natural Vibration n = 4.98 Hz.\n", + " The Velocity of the Mass v = 0.36 m/s.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 1. #kg\n", + "m1 = 2.5 \t\t\t#kg\n", + "s = 1.8*10**3 \t\t#N/m\n", + "l = (300.+300)*10**-3\t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Mass Moment of Inertia of the System\n", + "IA = (m*l**2/3)+(m1*l**2) \t\t\t#kg-m**2\n", + "#Calculating the Ratio of Alpha to Theta\n", + "#delta = 0.3*theta\n", + "#Restoring Force = s*delta = 540*theta\n", + "#Restoring torque about A = 540*theta*0.3 = 162*theta N-m ...(i)\n", + "#Torque about A = IA*alpha = 1.02*alpha N-m ...(ii)\n", + "#Equating (i) and (ii) 1.02*alpha = 162*theta \n", + "alphabytheta = 162/1.02\n", + "#Calculating the Frequency of Oscillation\n", + "n = 1/(2*math.pi)*math.sqrt(alphabytheta)\n", + "\n", + "#Results:\n", + "print \" The Frequency of Oscillation n = %.2f Hz.\"%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Frequency of Oscillation n = 2.01 Hz.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 85. \t\t\t#kg\n", + "h = 0.1 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 100./145 \t\t\t#Hz\n", + "#Calculating the Equivalent Length of Simple Pendulum\n", + "L = (1/(2*math.pi)/.69*math.sqrt(9.81))**2\n", + "#Calculating the Radius of Gyration\n", + "kG = math.sqrt((L-h)*h)\n", + "#Calculating the Moment of Inertia of the Flywheel through the Centre of Gravity\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Moment of Inertia of the Flywheel Through its c.g. I = %.1f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Moment of Inertia of the Flywheel Through its c.g. I = 3.6 kg-m**2.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "m = 60. \t\t\t#kg\n", + "d1 = 75.\n", + "d2 = 102. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the Frequencies of Oscillation\n", + "n1 = 100./190\n", + "n2 = 100./165 \t\t\t#Hz\n", + "\n", + "#Calculating the Equivalent Lengths of Simple Pendulum\n", + "L1 = 9.81/(2*math.pi*n1)**2 \t\t\t#m\n", + "L2 = 9.81/(2*math.pi*n2)**2 \t\t\t#m\n", + "\n", + "#Calculating Dismath.tance of c.g. from the Small and Big End Centres (h1 and h2) and the Radius of Gyration\n", + "def f(x):\n", + " h1 = x[0]\n", + " h2 = x[1]\n", + " kG = x[2]\n", + " y = [0,0,0]\n", + " y[0] = L1*h1-h1**2-kG**2\n", + " y[1] = L2*h2-h2**2-kG**2\n", + " y[2] = h1+h2-1\n", + " return y\n", + "\n", + "z = fsolve(f,[1,1,1])\n", + "\n", + "h1 = z[0]\n", + "h2 = z[1]\n", + "kG = z[2]\n", + "\n", + "#Calculating the Mass Moment of Inertia of the Rod\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Moment of Inertia of the Rod I = %d kg-m**2.\"%(I)\n", + "print \" The C.G is at a Distance of h1 = %.3f m from the Small End Centre.\"%(h1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Moment of Inertia of the Rod I = 6 kg-m**2.\n", + " The C.G is at a Distance of h1 = 0.759 m from the Small End Centre.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 1.2 \t\t\t#m\n", + "theta = 3*math.pi/180 \t\t\t#rad\n", + "\n", + "#Solution:\n", + "#Calculating the Distance Between the Knife Edge and C.G. of the Rod\n", + "h = 1.2/2-.05 \t\t\t#m\n", + "#Calculating the Radius of Gyration of the Rod About C.G.\n", + "kG = l/math.sqrt(12) \t\t\t#m\n", + "#Calculating the Time of Swing of the Rod\n", + "tp = 2*math.pi*math.sqrt((kG**2+h**2)/(9.81*h)) \t\t\t#seconds\n", + "#Calculating the Minimum Time of Swing\n", + "tpmin = 2*math.pi*math.sqrt((2*kG)/9.81) \t\t\t#seconds\n", + "#Calculating Angular Velocity\n", + "omega = 2*math.pi/tp \t\t\t#rad/s\n", + "#Calculating Maximum Angular Velocity\n", + "omegamax = omega*theta \t\t\t#rad/s\n", + "#Calculating Maximum Angular Acceleration\n", + "alphamax = omega**2*theta \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The Time of Swing of the Rod tp = %.2f seconds.\"%(tp)\n", + "print \" The Minimum Time of Swing tpmin = %.2f seconds.\"%(tpmin)\n", + "print \" The Maximum Angular Velocity omegamax = %.4f rad/s.\"%(omegamax)\n", + "print \" The Maximum Angular Acceleration (alphamax) = %.3f rad/s**2.\"%( alphamax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Time of Swing of the Rod tp = 1.76 seconds.\n", + " The Minimum Time of Swing tpmin = 1.67 seconds.\n", + " The Maximum Angular Velocity omegamax = 0.1871 rad/s.\n", + " The Maximum Angular Acceleration (alphamax) = 0.669 rad/s**2.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "m = 30. \t\t\t#kg\n", + "OG = 1.05 #m\n", + "h = OG #m\n", + "AG = 0.15 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 20/43.5 \t\t\t#Hz\n", + "#Calculating the Equivalent Length of Simple Pendulum\n", + "L = 9.81/(2*math.pi*n)**2 \t\t\t#m\n", + "#Calculating the Distance of Centre of Percussion (C) from the Centre of Gravity (G)\n", + "CG = L-OG \t\t\t#m\n", + "#Calculating the Distance of Centre of Percussion (C) from the Knife Edge A\n", + "AC = AG-CG \t\t\t#m\n", + "#Calculating the Radius of Gyration of the Pendulum About O\n", + "kO = math.sqrt(L*h) \t\t\t#m\n", + "h1 = h*(1-math.cos(60*math.pi/180)) \t\t\t#m\n", + "#Calculating the Angular Velocity of the Pendulum\n", + "omega = math.sqrt(2*m*9.81*h1/(m*kO**2)) \t\t\t#rad/s\n", + "OA = OG+AG\n", + "#Calculating the Velocity of Striking\n", + "v = omega*(OA) \t\t\t#Velocity of Striking\n", + "#Calculating the Angular Velocity of the Pendulum Immediately After Impact\n", + "I = m*kO**2\n", + "LKE = 55. \t\t\t#Loss of Kinetic Energy N-m\n", + "omega1 = math.sqrt(omega**2-LKE*2/I)\n", + "#Calculating the Impulses at Knife Edge A and at Pivot O (P and Q)\n", + "CLM = m*h*(omega-omega1) \t\t\t#Change of Linear Momentum\n", + "CAM = m*(kO**2-h**2)*(omega-omega1) \t\t\t#Change of Angular Momentum\n", + "#P+Q = Change of Linear Momentum and 0.15P-1.05Q = Change of Angular Momentum.\n", + "#i.e. P+Q = CLM and 0.15P-1.05Q = CAM\n", + "#Variables Matrix\n", + "A = [[1, 1],[0.15,-1.05]]\n", + "B = [CLM, CAM]\n", + "V = linalg.solve(A,B)\n", + "P = V[0]\n", + "Q = V[1]\n", + "#Calculating the Change in Axis Reaction When the Pendulum is Vertical\n", + "CAR = m*(omega**2-omega1**2)*h \t\t\t#Change in Axis Reaction, N\n", + "\n", + "#Results:\n", + "print \" The Distance of Centre of Percussion, AC = %.3f m.\"%(AC)\n", + "print \" The Velocity of Striking = %.2f m/s.\"%(v)\n", + "print \" The Impulse at the Knife Edge P = %.1f N-s.\"%(P)\n", + "print \" The Impulse at the Pivot Q = %.2f N-s.\"%(Q)\n", + "print \" The Change in Axis Reaction When the Pendulum is Vertical = %d N.\"%(CAR)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Distance of Centre of Percussion, AC = 0.024 m.\n", + " The Velocity of Striking = 3.47 m/s.\n", + " The Impulse at the Knife Edge P = 17.6 N-s.\n", + " The Impulse at the Pivot Q = 0.37 N-s.\n", + " The Change in Axis Reaction When the Pendulum is Vertical = 93 N.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "m = 1.5 \t\t\t#kg\n", + "l = 1.25 #m\n", + "x = 120.*10**-3 #m\n", + "y = x \t\t\t #m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 20./40 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration of the Connecting Rod\n", + "kG = 1/(2*math.pi*n)*math.sqrt(9.81*x*y/l) \t\t\t#m\n", + "#Calculating the Moment of Inertia of the Connecting Rod\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Radius of Gyration kG = %d mm.\"%(kG*1000)\n", + "print \" The Mass Moment of Inertia I = %.3f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Radius of Gyration kG = 107 mm.\n", + " The Mass Moment of Inertia I = 0.017 kg-m**2.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10 Page No : 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 2.5\n", + "r = 250.*10**-3 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 50./170 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration of the Wheel\n", + "kG = r/(2*math.pi*n)*math.sqrt(9.81/l) \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" The Radius of Gyration kG = %d mm.\"%(kG*10**3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Radius of Gyration kG = 267 mm.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 5.5 #kg\n", + "m2 = 1.5 \t\t\t#kg\n", + "l = 1.25 #m\n", + "r = 125.*10**-3 \t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 10./30 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration About an Axis Through the c.g.\n", + "kG = r/(2*math.pi*n)*math.sqrt(9.81/l) \t\t\t#m\n", + "#Calculating the Mass Moment of Inertia About an Axis Through its c.g.\n", + "m = m1+m2 \t\t\t#Total Mass kg\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \"The Mass Moment of Inertia About an Axis Through its c.g. I = %.3f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Mass Moment of Inertia About an Axis Through its c.g. I = 0.196 kg-m**2.\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
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