Added(A)/Deleted(D) following books

 A  The_Theory_of_Machines_by_T._Bevan/ch10.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch11.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch12.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch13.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch14.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch15.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch2.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch3.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch4.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch5.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch6.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch7.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch8.ipynb
A  The_Theory_of_Machines_by_T._Bevan/ch9.ipynb
A  The_Theory_of_Machines_by_T._Bevan/screenshots/amp_forced_vibr.png
A  The_Theory_of_Machines_by_T._Bevan/screenshots/couple_sup_shaft_2.png
A  The_Theory_of_Machines_by_T._Bevan/screenshots/vel,disp,acc.png
A  sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta_2.ipynb

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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Motion Inertia"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#a) INELASTIC\n",
+      "#for sphere 1 ,mass=m1 and initial velocity=u1 \n",
+      "#for sphere 2 ,mass=m2 and initial velocity=u2\n",
+      "m1=100.#lb\n",
+      "u1=10.#ft/s\n",
+      "m2=50.#lb\n",
+      "u2=5.#ft/s\n",
+      "\n",
+      "#Calculations\n",
+      "v=(m1*u1+m2*u2)/(m1+m2)\n",
+      "#change in kinetic energy\n",
+      "#initial kinetic energy = ke1\n",
+      "ke1=(m1*(u1**2)+m2*(u2**2))/(2*32.2)\n",
+      "#Kinetic Energy after inelastic colision = ke2\n",
+      "ke2=((m1+m2)*8.333**2)/(2*32.2)\n",
+      "#Change in Kinetic Energy =l\n",
+      "l=ke1-ke2\n",
+      "#b) Elastic\n",
+      "# for a very short time bodies will have a common velocity given by v=8.333 ft/s\n",
+      "# for a very short time bodies will have a common velocity given by v=8.333 ft/s\n",
+      "#immidiately after impact ends the velocities for both the bodies are given by v1 and v2\n",
+      "v1=2*v-u1\n",
+      "v2=2*v-u2\n",
+      "#c) Coeeficient of Restitution=0.6\n",
+      "e=0.6\n",
+      "ve1=(1+e)*v-e*u1\n",
+      "ve2=(1+e)*v-e*u2\n",
+      "ke3=(m1*(ve1**2)+m2*(ve2**2))/(2*32.2)\n",
+      "loss=ke1-ke3\n",
+      "\n",
+      "#Results\n",
+      "print \"kinetic energy before collisio0n is %.1f ft lb\"%ke1\n",
+      "print \"\\na) INELASTIC\"\n",
+      "print \"velocity after collision is %.3f ft/s\"%v\n",
+      "print \"the Kinetic Energy after collision is %.1f ft lb\"%ke2\n",
+      "print \"the change in Kinetic Energy after collision is %.1f ft lb\"%l\n",
+      "print \"\\nb) ELASTIC\"\n",
+      "print \"velocity of 1 after collision is %.3f ft/s\"%v1\n",
+      "print \"velocity of 2 after collision is %.3f ft/s\"%v2\n",
+      "print \"there is no loss of kinetic energy in case of elastic collision\"\n",
+      "print \"\\nc) e=0.6\"\n",
+      "print \"velocity of 1 after collision is %.3f ft/s\"%ve1\n",
+      "print \"velocity of 2 after collision is %.3f ft/s\"%ve2\n",
+      "print \"the Kinetic Energy after collision is %.1f ft lb\"%ke3\n",
+      "print \"the change in Kinetic Energy after collision is %.2f ft lb\"%loss"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "kinetic energy before collisio0n is 174.7 ft lb\n",
+        "\n",
+        "a) INELASTIC\n",
+        "velocity after collision is 8.333 ft/s\n",
+        "the Kinetic Energy after collision is 161.7 ft lb\n",
+        "the change in Kinetic Energy after collision is 13.0 ft lb\n",
+        "\n",
+        "b) ELASTIC\n",
+        "velocity of 1 after collision is 6.667 ft/s\n",
+        "velocity of 2 after collision is 11.667 ft/s\n",
+        "there is no loss of kinetic energy in case of elastic collision\n",
+        "\n",
+        "c) e=0.6\n",
+        "velocity of 1 after collision is 7.333 ft/s\n",
+        "velocity of 2 after collision is 10.333 ft/s\n",
+        "the Kinetic Energy after collision is 166.4 ft lb\n",
+        "the change in Kinetic Energy after collision is 8.28 ft lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m1=15.#tons\n",
+      "u1=12.#m/h\n",
+      "m2=5.#tons\n",
+      "u2=8.#m/h\n",
+      "k=2.#ton/in\n",
+      "e1=0.5#coefficient of restitution\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#conservation of linear momentum\n",
+      "v=(m1*u1+m2*u2)/(m1+m2)\n",
+      "print \"velocity at the instant of  collision is %.2f mph\"%v\n",
+      "e=(m1*m2*(88./60)**2*(u1-u2)**2)/(2*32.2*(u1+u2))\n",
+      "print \"The difference between the kinetic energy before and during the impact is %.2f ft tons\"%e\n",
+      "#energy stored in spring equals energy dissipated\n",
+      "#s=(1/2)*k*x**2\n",
+      "#s=e\n",
+      "#since there are 4 buffer springs ,4x**2=24 inches (2 ft=24 inches)\n",
+      "x=((e*12)/4)**.5\n",
+      "print \"Maximum deflection of the spring is %.2f in\"%x\n",
+      "# maximum force acting between pair of buffer = stiffness of spring*deflection\n",
+      "f=k*x\n",
+      "print \"Maximum force acting between each buffer is %.2f tons\"%f\n",
+      "#assuming perfectly elastic collision\n",
+      "#for loaded truck \n",
+      "v1=2*11-12\n",
+      "#for unloaded truck \n",
+      "v2=2*11-8\n",
+      "print \"Speed of loaded truck after impact %.2f mph\"%v1\n",
+      "print \"speed of unloaded truck after impact %.2f mph\"%v2\n",
+      "#if coefficient of restitution =o.5\n",
+      "#for loaded truck \n",
+      "ve1=(1+.5)*11-.5*12\n",
+      "#for unloaded truck \n",
+      "ve2=(1+.5)*11-.5*8\n",
+      "print \"Speed of loaded truck after impact when e=0.5 %.2f mph\"%ve1\n",
+      "print \"Speed of unloaded truck after impact when e=0.5 %.2f mph\"%ve2\n",
+      "#net loss of kinetic energy=(1-e**2)*energy stored in spring\n",
+      "l=(1-(e1**2))*2#ft tons\n",
+      "print \"Net loss of kinetic energy is %.2f ft tons\"%l"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity at the instant of  collision is 11.00 mph\n",
+        "The difference between the kinetic energy before and during the impact is 2.00 ft tons\n",
+        "Maximum deflection of the spring is 2.45 in\n",
+        "Maximum force acting between each buffer is 4.90 tons\n",
+        "Speed of loaded truck after impact 10.00 mph\n",
+        "speed of unloaded truck after impact 14.00 mph\n",
+        "Speed of loaded truck after impact when e=0.5 10.50 mph\n",
+        "Speed of unloaded truck after impact when e=0.5 12.50 mph\n",
+        "Net loss of kinetic energy is 1.50 ft tons\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3, Page 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=500.#lb ft^2\n",
+      "m2=1500.#lb ft^2\n",
+      "k=150#lb ft^2\n",
+      "w1=150#rpm\n",
+      "\n",
+      "#Calculations&Results\n",
+      "N=(w1*m1)/(m1+m2)\n",
+      "print \"Angular velocity at the instant when speeds of the flywheels are equalised is given by %.f r.p.m\"%N\n",
+      "#kinetic energy at this instance \n",
+      "ke1=(1./2)*((m1+m2)/32.2)*((math.pi*N)/30)**2\n",
+      "print \"The kinetic energy of the system at this instance is %.2f ft lb\"%ke1\n",
+      "\n",
+      "#initial kinetic energy\n",
+      "ke0=(1./2)*((m1)/32.2)*((math.pi*w1)/30)**2\n",
+      "print \"The initial kinetic energy of the system is %.2f ft lb\"%ke0\n",
+      "\n",
+      "#strain energy = s\n",
+      "s=ke0-ke1\n",
+      "print \"strain energy stored in the spring is %.2f ft lb which is approximately 1435 ft lb\"%s\n",
+      "\n",
+      "x=((1435*2)/150)**.5\n",
+      "print \"Maximum angular displacement is %.2f in radians which is equal to 250 degrees\"%x\n",
+      "#na1 and na are initial and final speeds of the flywheel 1 and same nb1 and nb for flywheel 2 \n",
+      "na=2*N-w1#w1=na1\n",
+      "nb=2*N-0#nb1=0\n",
+      "print \"Speed of flywheel a and b when spring regains its unstrained position are %.2f rpm and %.2f rpm respectively\"%(na,nb)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angular velocity at the instant when speeds of the flywheels are equalised is given by 38 r.p.m\n",
+        "The kinetic energy of the system at this instance is 478.92 ft lb\n",
+        "The initial kinetic energy of the system is 1915.68 ft lb\n",
+        "strain energy stored in the spring is 1436.76 ft lb which is approximately 1435 ft lb\n",
+        "Maximum angular displacement is 4.36 in radians which is equal to 250 degrees\n",
+        "Speed of flywheel a and b when spring regains its unstrained position are -75.00 rpm and 75.00 rpm respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4, Page 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=150 #lb\n",
+      "l=3#ft\n",
+      "#number of oscillation per second is given by n\n",
+      "\n",
+      "#Calculations&Results\n",
+      "n=(50/92.5)\n",
+      "print \"number of oscillation per second = %.3f\"%n\n",
+      "#length of simple pendulum is given by L=g/(2*math.pi*n)**2\n",
+      "L=32.2/(2*math.pi*n)**2\n",
+      "print \"length of simple pendulum = %.2f ft\"%L\n",
+      "# distance of cg from point of suspension is given by a\n",
+      "a=25./12\n",
+      "k=(a*(L-a))**.5#radius of gyration\n",
+      "moi=m1*k**2\n",
+      "print \"The moment of inertia of rod is %.f lb ft**2\"%moi"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of oscillation per second = 0.541\n",
+        "length of simple pendulum = 2.79 ft\n",
+        "The moment of inertia of rod is 221 lb ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5, Page 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1=50/84.4\n",
+      "n2=50/80.3\n",
+      "\n",
+      "#Calculations\n",
+      "L1=(32.2*12)*(84.4/(100*math.pi))**2\n",
+      "L2=(32.2*12)*(80.3/(100*math.pi))**2\n",
+      "#a1(L1-a1)=k**2=a2(L2-a2) and a1+a2=30 inches\n",
+      "#substituting and solving for a we get \n",
+      "a1=141/6.8\n",
+      "a2=30-a1\n",
+      "k=(a1*(L1-a1))**.5\n",
+      "moi=90*(149./144)#moi=m*k**2\n",
+      "\n",
+      "#Results\n",
+      "print \"length of equivalent simple pendulum when axis coincides with small end and big end respectively-\"\n",
+      "print \"L1=%.1f in\"%L1\n",
+      "print \"L2=%.1f in\"%L2\n",
+      "print \"distances of cg from small end and big end centers respectively are-\"\n",
+      "print \"a1=%.1f in\"%a1\n",
+      "print \"a2=%.1f in\"%a2\n",
+      "print \"Moment of inertia of rod =%.2f lb ft^2\"%moi"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "length of equivalent simple pendulum when axis coincides with small end and big end respectively-\n",
+        "L1=27.9 in\n",
+        "L2=25.2 in\n",
+        "distances of cg from small end and big end centers respectively are-\n",
+        "a1=20.7 in\n",
+        "a2=9.3 in\n",
+        "Moment of inertia of rod =93.13 lb ft^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6, Page 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=150\n",
+      "l=8.5\n",
+      "g=32.2\n",
+      "a=83.2\n",
+      "n=25\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#k=(a/2*%pi*n)*(g/l)**0.5\n",
+      "k=(14*a*((g)**0.5))/(2*math.pi*n*(l**0.5))\n",
+      "k1=14.5/12\n",
+      "print \"radius of gyration is %.2f inches which is equal to %.2f ft\"%(k,k1)\n",
+      "moi=m1*(k1**2)\n",
+      "print \"moment of inertia=%.f lb ft^2\"%moi"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radius of gyration is 14.43 inches which is equal to 1.21 ft\n",
+        "moment of inertia=219 lb ft^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7, Page 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=2.5#lb\n",
+      "a=6#in\n",
+      "k=3.8#in\n",
+      "l=9#in\n",
+      "c=3#in\n",
+      "w=22500\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#k^2=ab\n",
+      "#case a) to find equivalent dynamic system\n",
+      "b=(k**2)/a\n",
+      "ma=(2.5*6)/8.42#m*a/a+b\n",
+      "mb=m-ma\n",
+      "print \"Mass ma =%.2f lb will be situated at 6 inches from cg \\nand mb =%.2f lb will be situated at %.2f inches \" \\\n",
+      "      \"\\nfrom cg in the equivalent dynamical system\"%(ma,mb,b)\n",
+      "\n",
+      "#if two masses are situated at the bearing centres \n",
+      "ma1=(2.5*6)/9\n",
+      "mb1=m-ma1\n",
+      "k1=(a*c)**.5\n",
+      "#t=m*((k1^2)-(k^2))*w\n",
+      "t=((2.5*(18-3.8**2))*22500)/(32.2*12*12)\n",
+      "print \"\\ncorrection couple which must be applied in order that the two mass system is dynamically equivalent to\"\\\n",
+      "      \"the rod is given by %.1f lb ft\"%t\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass ma =1.78 lb will be situated at 6 inches from cg \n",
+        "and mb =0.72 lb will be situated at 2.41 inches \n",
+        "from cg in the equivalent dynamical system\n",
+        "\n",
+        "correction couple which must be applied in order that the two mass system is dynamically equivalent tothe rod is given by 43.2 lb ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8, Page 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=20.#lb\n",
+      "g=32.2\n",
+      "a=200#ft/s^2\n",
+      "w=120#rad/s^2\n",
+      "k=7.#in\n",
+      "\n",
+      "#Calculations\n",
+      "f=(m/g)*a#effective force appllied to the link\n",
+      "#this force acts parallel to the acceleration fg\n",
+      "t=(m/g)*(k/12)**2*w#couple required in order to provide the angular acceleration\n",
+      "#the line of action of F is therefore at a distance from G given by\n",
+      "x=t/f\n",
+      "\n",
+      "#Results\n",
+      "print \"Effective force applied to the link is %.3f lb and the line of action of F is therefore at a distance\"\\\n",
+      "      \"from G given by %.3f ft\"%(f,x)\n",
+      "print \"F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved along\" \\\n",
+      "      \"the appropriate lines of action to give the magnitudes of Fa and Fb\"\n",
+      "print \"From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Effective force applied to the link is 124.224 lb and the line of action of F is therefore at a distancefrom G given by 0.204 ft\n",
+        "F is the resultant of Fa and Fb, using x as shown in figure.25 , the force F may then be resolved alongthe appropriate lines of action to give the magnitudes of Fa and Fb\n",
+        "From the scaled diagram shown in figure we get,Fa=65 lb and Fb=91 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9, Page 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m=10#ton\n",
+      "m2=1000#lb\n",
+      "a=3#ft/s^2\n",
+      "\n",
+      "#Calculations\n",
+      "#the addition to actual mass in order to allow for the rotational inertia of the wheels and axles\n",
+      "m1=2*(1000./2240)*(15./21)**2#m1=m2*k**2/r**2 and 1 ton=2240 lbs\n",
+      "M=m+m1\n",
+      "F=3*(10.46/32.2)#F=M.a\n",
+      "f=F*2240#lb\n",
+      "Fa=(2*1000./2240)*(3/32.2)*(15./21)**2#total tangential force required in order to provide the angular acceleration of the wheels and axles\n",
+      "#Limiting friction force =uW \n",
+      "#u*10>0.042\n",
+      "u=0.042/10\n",
+      "\n",
+      "#Results\n",
+      "print \"The total tangential force required in order to provide the angular acceleration of the wheels and axles is %.3f ton\"%Fa\n",
+      "print \"If there is to be pure rolling ,u>%.4f\"%u\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total tangential force required in order to provide the angular acceleration of the wheels and axles is 0.042 ton\n",
+        "If there is to be pure rolling ,u>0.0042\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10, Page 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ia=200#lb ft2\n",
+      "Ib=15#lb ft2\n",
+      "G=5#wb==5*wa\n",
+      "m=150.#lb\n",
+      "r=8#in\n",
+      "\n",
+      "#Calculations\n",
+      "#the equivalent mass of the geared system referred to the circumference of the drum is given by\n",
+      "#Me=(1./r)**2*(Ia+(G**2*Ib))\n",
+      "Me=(12./r)**2*(Ia+(G**2*Ib))\n",
+      "M=m+Me\n",
+      "a=(m/M)*32.2#acceleration\n",
+      "#if efficiency of gearing is 90% then Me=(1/r**2)*(Ia+(G**2*Ib)/n)\n",
+      "n=.9\n",
+      "Me1=(12./r)**2*(Ia+(G**2*Ib)/n)\n",
+      "M1=Me1+m\n",
+      "a1=(m/M1)*32.2\n",
+      "\n",
+      "#Results\n",
+      "print \"acceleration = %.2f ft/s2\"%a\n",
+      "print \"acceleration when gear efficiency is 0.9= %.2f ft/s2\"%a1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "acceleration = 3.35 ft/s2\n",
+        "acceleration when gear efficiency is 0.9= 3.14 ft/s2\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11, Page 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#let\n",
+      "#S=displacement of car from rest with uniform acceleration a, the engine torque T assumed to remain ocnstant\n",
+      "#v=final speed ofcar\n",
+      "#G=gear ratio\n",
+      "#r=effective radius\n",
+      "#n=efficiency of transmission\n",
+      "#M=mass of the car\n",
+      "#Ia and Ib=moments of inertia of road whels and engine \n",
+      "#formulas => F=29.5nG ; Me= 1648+$.54nG^2 ; a=32.2 F/Me\n",
+      "#given\n",
+      "G1=22.5\n",
+      "G2=12.5\n",
+      "G3=7.3\n",
+      "G4=5.4\n",
+      "n=.82#for 1st ,2nd and 3rd gear\n",
+      "n4=.9#for 4th gear\n",
+      "\n",
+      "#Calculations\n",
+      "F1=29.5*n*G1\n",
+      "F2=29.5*n*G2\n",
+      "F3=29.5*n*G3\n",
+      "F4=29.5*n4*G4\n",
+      "#on reduction and putting values we get\n",
+      "Me1=1648+4.54*n*G1**2\n",
+      "Me2=1648+4.54*n*G2**2\n",
+      "Me3=1648+4.54*n*G3**2\n",
+      "Me4=1648+4.54*n4*G4**2\n",
+      "a1=32.2*F1/Me1\n",
+      "a2=32.2*F2/Me2\n",
+      "a3=32.2*F3/Me3\n",
+      "a4=32.2*F4/Me4\n",
+      "\n",
+      "#Results\n",
+      "print \"Maximum acceleration of car on top gear is %.2f ft/s^2\"%a4\n",
+      "print \"Maximum acceleration of car on third gear is %.2f ft/s^2\"%a3\n",
+      "print \"Maximum acceleration of car on second gear is %.2f ft/s^2\"%a2\n",
+      "print \"Maximum acceleration of car on first gear is %.2f ft/s^2\"%a1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum acceleration of car on top gear is 2.61 ft/s^2\n",
+        "Maximum acceleration of car on third gear is 3.08 ft/s^2\n",
+        "Maximum acceleration of car on second gear is 4.37 ft/s^2\n",
+        "Maximum acceleration of car on first gear is 4.96 ft/s^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12, Page 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "I=40#lb ft2\n",
+      "n=500#rpm\n",
+      "\n",
+      "#Calculations\n",
+      "w=math.pi*n/30#angular velocity\n",
+      "wp=2*math.pi/5#angular velocity of precession\n",
+      "I1=I/32.2\n",
+      "T=I1*w*wp#gyroscopic couple\n",
+      "\n",
+      "#Result\n",
+      "print \"The couple supplied to the shaft= %.2f lb ft\"%T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The couple supplied to the shaft= 81.74 lb ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13, Page 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "I=250#lb ft2\n",
+      "n=1600#rpm\n",
+      "v=150#mph\n",
+      "r=500#ft\n",
+      "\n",
+      "#Calculations&Results\n",
+      "w=math.pi*160/3#angular velocity of rotation\n",
+      "wp=(150.*88)/(60*500)#angular velocity of precession\n",
+      "#a) with three bladed screw\n",
+      "#T=I*w*wp\n",
+      "T=(250/32.2)*math.pi*(160./3)*wp\n",
+      "#b)with two bladed air screw\n",
+      "#T1=2*I*w*wp*sin(o)\n",
+      "print \"The magnitude of gyroscopic couple is given by %.0f lb ft\"%T\n",
+      "#Tix=T(1-cos(2o)) lb ft\n",
+      "#T1y=Tsin(2o)) lb ft\n",
+      "print \"The component gyroscopic couple in the vertical plane =%.0f(1-cos(2x)) lb ft\"%T\n",
+      "print \"The component gyroscopic couple in the horizontal plane =%.0f(sin(2x)) lb ft\"%T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnitude of gyroscopic couple is given by 572 lb ft\n",
+        "The component gyroscopic couple in the vertical plane =572(1-cos(2x)) lb ft\n",
+        "The component gyroscopic couple in the horizontal plane =572(sin(2x)) lb ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file