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diff --git a/Theory_Of_Machines/ch3.ipynb b/Theory_Of_Machines/ch3.ipynb new file mode 100755 index 00000000..281d238d --- /dev/null +++ b/Theory_Of_Machines/ch3.ipynb @@ -0,0 +1,1380 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:102997571ab69dde264cca49f55aff87dce1c51132ce271fb2095c0f1179d655" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Kinetics of Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "k = 1. \t\t\t#m\n", + "m = 2500. \t\t\t#kg\n", + "T = 1500. \t\t\t#N-m\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the flywheel\n", + "alpha = T/I \t\t\t#rad/s**2\n", + "#The angular speed at start\n", + "omega1 = 0\n", + "t = 10. \t\t\t#seconds\n", + "#Calculating the angular speed of the flywheel after t = 10 seconds from start\n", + "omega2 = omega1+alpha*t \t\t\t#rad/s\n", + "#Calculating the kinetic energy of the flywheel after 10 seconds from start\n", + "E = 1./2*I*omega2**2/1000 \t\t\t#kJ\n", + "\n", + "#Results:\n", + "print \" The angular acceleration of the flywheel, alpha = %.1f rad/s**2.\"%(alpha)\n", + "print \" The kinetic energy of the flywheel, E = %d kJ.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular acceleration of the flywheel, alpha = 0.6 rad/s**2.\n", + " The kinetic energy of the flywheel, E = 45 kJ.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "mC = 500. #kg\n", + "mD = 250. \t\t\t#kg\n", + "s = 100. #m\n", + "r = 0.5 #m\n", + "k = 0.35 \t\t\t#m\n", + "m = 3. \t\t\t#kg/m\n", + "\n", + "#Solution:\n", + "#Velocities of the cage\n", + "u1 = 0.\n", + "v1 = 10.\n", + "v2 = 10.\n", + "u3 = 10.\n", + "v3 = 0. \t\t\t#m/s\n", + "#Accelerations of the cage\n", + "a1 = 1.5\n", + "a3 = -6. \t\t\t#m/s**2\n", + "s = 100. \t\t\t#m\n", + "\n", + "#Calculating the time taken by the cage to reach the top\n", + "t1 = (v1-u1)/a1 \t\t\t#seconds\n", + "#Calculating the dismath.tance moved by the cage during time t1\n", + "s1 = (v1+u1)/2*t1 \t\t\t#m\n", + "#Calculating the time taken for the cage from initial velocity u3 = 10 m/s to final velocity of v3 = 0\n", + "t3 = (v3-u3)/a3 \t\t\t#seconds\n", + "#Calculating the dismath.tance moved by the cage during time t3\n", + "s3 = (v3+u3)/2*t3 \t\t\t#m\n", + "#Calculating the dismath.tance travelled during consmath.tant velocity of v2 = 10 m/s\n", + "s2 = s-s1-s3 \t\t\t#m\n", + "#Calculating the time taken for the cage during consmath.tant velocity\n", + "t2 = s2/v2 \t\t\t#seconds\n", + "#Calculating the time taken for the cage to reach the top\n", + "t = t1+t2+t3 \t\t\t#seconds\n", + "#Calculating the total mass of the rope for 100 metres\n", + "mR = m*s \t\t\t#kg\n", + "#Calculating the force to raise the cage and rope at uniform speed\n", + "F1 = (mC+mR)*9.81 \t\t\t#N\n", + "#Calculating the torque to raise the cage and rope at uniform speed\n", + "T1 = F1*r \t\t\t#N-m\n", + "#Calculating the force to accelerate the cage and rope\n", + "F2 = (mC+mR)*a1 \t\t\t#N\n", + "#Calculating the torque to accelerate the cage and rope\n", + "T2 = F2*r \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the drum\n", + "I = mD*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alpha = a1/r \t\t\t#rad/s**2\n", + "#Calculating the torque to accelerate the drum\n", + "T3 = I*alpha \t\t\t#N-m\n", + "#Calculating the total torque which must be applied to the drum at starting\n", + "T = T1+T2+T3 \t\t\t#N-m\n", + "#Calculating the mass of 33.35 m rope\n", + "m1 = m*33.35 \t\t\t#kg\n", + "#Calculating the reduction of torque\n", + "T4 = (m1*9.81+m1*a1)*r \t\t\t#N-m\n", + "#Calculating the angular velocity of drum\n", + "omega = v2/(2*math.pi*r) \t\t\t#rad/s\n", + "#Calculating the power\n", + "P = T4*omega/1000 \t\t\t#Power kW\n", + "\n", + "#Results:\n", + "print \" The time taken for the cage to reach the top t = %.2f s.\"%(t)\n", + "print \" The total torque which must be applied to the drum during starting T = %.1f N-m.\"%(T)\n", + "print \" The power required is %.3f kW.\"%(P)\n", + "#Answers differ due to rounding-off values in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The time taken for the cage to reach the top t = 14.17 s.\n", + " The total torque which must be applied to the drum during starting T = 4615.9 N-m.\n", + " The power required is 1.801 kW.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 4.*1000 \t\t\t#W\n", + "I = 140. \t\t\t#kg-m**2\n", + "N1 = 240. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular acceleration at the commencement of operation\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Calculating the energy supplied by the motor (E1) and the energy consumed in closing a revet in 1 second\n", + "E1 = 4000.\n", + "E2 = 10000. \t\t\t#N-m\n", + "#Calculating the loss of kinetic energy of the flywheel during the operation\n", + "E = E2-E1 \t\t\t#N-m\n", + "#Calculating the kinetic energy of the flywheel at the commencement of operation\n", + "KEc = 1./2*I*omega1**2 \t\t\t#Kinetic energy at the commencement N-m\n", + "#Calculating the kinetic energy of the flywheel at the end of operation\n", + "KEe = KEc-E \t\t\t#Kinetic energy at the end N-m\n", + "#Calculating the angular speed of the flywheel immediately after closing a revet\n", + "omega2 = math.sqrt(KEe*2/I) \t\t\t#rad/s\n", + "#Calculating the reduction of speed\n", + "ReductionofSpeed = (omega1-omega2)*60/(2*math.pi) \t\t\t#rpm\n", + "#Calculating the maximum rate at which the revets can be closed per minute\n", + "Rate = P*60/E2 \t\t\t#Maximum rate at which the revets can be closed per minute\n", + "\n", + "#Results:\n", + "print \" The reduction of speed is %.1f rpm.\"%(ReductionofSpeed)\n", + "print \" The maximum rate at which rivets can be closed per minute is %d.\"%(Rate)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The reduction of speed is 16.9 rpm.\n", + " The maximum rate at which rivets can be closed per minute is 24.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 14.*1000 #kg\n", + "m1 = 1.25*1000 #kg\n", + "m2 = 110. \t\t\t#kg\n", + "d = 1. #m\n", + "r = d/2 #m\n", + "k1 = 450./1000 #m\n", + "k2 = 125./1000 \t\t#m\n", + "F = 1.2*1000 \t\t#N\n", + "eta = 0.85\n", + "v = 1.8 \t\t\t#m/s\n", + "a = 0.1 \t\t\t#m/s**2\n", + "\n", + "#Solution:\n", + "#Calculating the forces oppomath.sing the motion\n", + "P1 = m*9.81*1/20+m*a+F \t\t\t#N\n", + "#Calculating the torque on the drum shaft to accelerate the load\n", + "T1 = P1*r \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the drum\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alpha1 = a/r \t\t\t#rad/s\n", + "#Calculating the torque on the drum to accelerate the drum shaft\n", + "T2 = I1*alpha1 \t\t\t#N-m\n", + "#Calculating the torque on the armature to accelerate drum and load\n", + "T3 = (T1+T2)/(40*eta) \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the armature\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the armature\n", + "alpha2 = a/r*40 \t\t\t#rad/s**2\n", + "#Calculating the torque on the armature to accelerate armature shaft\n", + "T4 = I2*alpha2 \t\t\t#N-m\n", + "#Calculating the torque on the motor shaft\n", + "T = T3+T4 \t\t\t#N-m\n", + "#Calculating the angular speed of the motor\n", + "omega = v/r*40 \t\t\t#rad/s\n", + "#Calculating the power developed by the motor\n", + "P = T*omega/1000 \t\t\t#Power developed by the motor kW\n", + "\n", + "#Results:\n", + "print \" The torque on the motor shaft T = %.2f N-m.\"%(T)\n", + "print \" The power developed by the motor is %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The torque on the motor shaft T = 154.46 N-m.\n", + " The power developed by the motor is 22.24 kW.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 12.0*1000\n", + "m1 = 2.0*1000\n", + "m2 = 2.5*1000 \t\t\t#kg\n", + "k1 = 0.4\n", + "d1 = 1.2\n", + "r1 = d1/2.0\n", + "k2 = 0.6\n", + "d2 = 1.5\n", + "r2 = d2/2.0\n", + "s = 6.0 \t\t\t#m\n", + "v = 9.0*1000/3600 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the front roller\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the rear axle together with its wheels\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular speed of the front roller\n", + "omega1 = round(v/r1,2) \t\t\t#rad/s\n", + "#Calculating the angular speed of rear wheels\n", + "omega2 = round(v/r2,1) \t\t\t#rad/s\n", + "#Calculating the kinetic energy of rotation of the front roller\n", + "E1 = 1.0/2*I1*4.16**2 \t\t\t#N-m\n", + "#Calculating the kinetic energy of rotation of the rear axle with its wheels\n", + "E2 = 1.0/2*I2*omega2**2 \t\t\t#N-m\n", + "#Calculating the total kinetic energy of rotation of the wheels\n", + "E = round(E1+E2,-1) \t\t\t#N-m\n", + "#Calculating the kinetic energy of translation of the road roller\n", + "E3 = 1.0/2*m*v**2 \t\t\t#N-m\n", + "#Calculating the total kinetic energy of the road roller\n", + "E4 = E3+E \t\t\t#N-m\n", + "#Calculating the braking force to bring the roller to rest\n", + "F = E4/s \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" The total kinetic energy of rotation of the wheels E = %.f N-m.\"%(E)\n", + "print \" The total kinetic energy of the road roller E4 = %d N-m.\"%(E4)\n", + "print \" The braking force required to bring the roller to rest F = %.1f N.\"%(F)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The total kinetic energy of rotation of the wheels E = 7670 N-m.\n", + " The total kinetic energy of the road roller E4 = 45170 N-m.\n", + " The braking force required to bring the roller to rest F = 7528.3 N.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page no : 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import Symbol,solve\n", + "import math\n", + "\n", + "# variables\n", + "s = 4. # N/m\n", + "m = 4000. # N/m ; \n", + "x1 = 0.075 # m \n", + "x2 = 0.03 # m;\n", + "m = 5. # kg ; \n", + "R = 70. # N\n", + "g = 9.81\n", + "\n", + "# calculations\n", + "Q = m*(x1 - x)\n", + "P = Q + m*g - R\n", + "x = 0.045\n", + "t = Symbol(\"t\")\n", + "ans = -solve(0.07*(1 - math.cos(math.sqrt(800)) * t) - 0.045)[0]\n", + "a = math.acos(ans)\n", + "t = a/math.sqrt(800)\n", + "\n", + "# result\n", + "print \"t = %.4f s\"%t\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t = 0.0426 s\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page No : 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "r = 500./1000\n", + "k = 450./1000 \t\t\t#m\n", + "m1 = 500.\n", + "m2 = 1250. \t\t\t#kg\n", + "u = 0.75 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of drum\n", + "I2 = m2*k**2 \t\t\t#kg-m**2\n", + "#Calculating the speed of truck\n", + "#Impulse\n", + "#F = m1*v or\n", + "#F-m1*v = 0 .....(i)\n", + "#Moment of impulse\n", + "#F*r = I2*(omega2-omega2) or\n", + "#F*r = I2*(u-v)/r or\n", + "#F*r+I2*v/r = I2*u/r .....(ii)\n", + "#Solving (i) and (ii)\n", + "A = [[1, -m1],[ r, I2/r]]\n", + "B = [0, I2*u/r]\n", + "V = linalg.solve(A,B)\n", + "v = V[1]\n", + "#Calculating the energy lost to the system\n", + "E = 1./2*I2*(u**2-v**2)/r**2-1./2*m1*v**2 \t\t\t#Energy lost to the system, N-m\n", + "\n", + "#Results:\n", + "print \" The speed of the truck when the motion becomes steady, v = %.3f m/s.\"%(v)\n", + "print \" The energy lost to the system is %d N-m.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The speed of the truck when the motion becomes steady, v = 0.502 m/s.\n", + " The energy lost to the system is 94 N-m.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "s = 0.7*10**6 \t\t\t#N/m\n", + "m1 = 10.*10**3\n", + "m2 = 15.*10**3 \t\t\t#kg\n", + "v1 = 1.8\n", + "v2 = 0.6 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the common velocity when moving together during impact\n", + "v = (m1*v1+m2*v2)/(m1+m2)\n", + "#Calculating the kinetic energy lost to the system\n", + "E = (1./2*m1*v1**2+1./2*m2*v2**2)-1./2*(m1+m2)*v**2\n", + "#Calculating the compression of each buffer spring\n", + "x = math.sqrt(E/(2*s))\n", + "#Calculating the velocity of each truck on separation\n", + "#Final KE after separation = KE at common velocity+Half of energy stored in springs.\n", + "#And initial and final momentum must be equal.\n", + "#Simplifying the two equations\n", + "# we get\n", + "\n", + "#1/2*m1*v3**2+1/2*m2*v4**2 = 1/2*(m1+m2)*v**2+1/2*E .....(i)\n", + "#m1*v3+m2*v4 = (m1+m2)*v\n", + "def f(x):\n", + " v3 = x[0]\n", + " v4 = x[1]\n", + " y = [0,0]\n", + " y[0] = 1./2*m1*v3**2+1./2*m2*v4**2-1./2*(m1+m2)*v**2-1./2*E\n", + " y[1] = m1*v3+m2*v4-(m1+m2)*v\n", + " return y\n", + " \n", + "z = fsolve(f,[1,1])\n", + "v3 = z[1]\n", + "v4 = z[0]\n", + "\n", + "#Results:\n", + "print \" The common velocity when moving together during impact, v = %.2f m/s.\"%(v)\n", + "print \" The kinetic energy lost to the system is %.2f kN-m.\"%(E/1000.)\n", + "print \" The compression of each buffer spring, x = %.f mm.\"%(x*1000.)\n", + "print \" The velocity of separation for 10 tonnes truck, v3 = %.1f m/s.\"%(v3)\n", + "print \" The velocity of separation for 15 tonnes truck, v4 = %.1f m/s.\"%(v4)\n", + "\n", + "# note : rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The common velocity when moving together during impact, v = 1.08 m/s.\n", + " The kinetic energy lost to the system is 4.32 kN-m.\n", + " The compression of each buffer spring, x = 56 mm.\n", + " The velocity of separation for 10 tonnes truck, v3 = 0.7 m/s.\n", + " The velocity of separation for 15 tonnes truck, v4 = 1.6 m/s.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 300. #kg\n", + "m2 = 500. \t\t\t#kg\n", + "s = 1. #m\n", + "x = 150./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the velocity with which mass m1 hits the pile\n", + "u = 0\n", + "v1 = math.sqrt(2*9.81*s+u**2) \t\t\t#m/s\n", + "#Calculating the common velocity after impact\n", + "v2 = 0\n", + "v = (m1*v1+m2*v2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the kinetic energy before impact\n", + "KEb = m1*9.81*s \t\t\t#Kinetic energy before impact N-m\n", + "#Calculating the kinetic energy after impact\n", + "KEa = 1./2*(m1+m2)*v**2 \t\t\t#Kinetic energy after impact N-m\n", + "#Calculating the energy lost in the blow\n", + "E = KEb-KEa \t\t\t#Energy lost in the blow N-m\n", + "#Calculating the average resistance against the pile\n", + "R = KEa/x+m1*9.81+m2*9.81\n", + "\n", + "#Results:\n", + "print \" The energy lost in the blow is %d N-m.\"%(E)\n", + "print \" The average resistance against the pile R = %.3f kN.\"%(R/1000)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The energy lost in the blow is 1839 N-m.\n", + " The average resistance against the pile R = 15.206 kN.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No : 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "m1 = 0.7\n", + "m2 = 2.4 \t\t\t#kg\n", + "k1 = 270./1000\n", + "k2 = 185./1000\n", + "h1 = 0.25\n", + "DL = 0.2\n", + "CM = 0.275 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocity of hammer just before impact\n", + "h = h1*(1-math.cos(20*math.pi/180))\n", + "omega = math.sqrt(m1*9.81*h*2/(m1*k1**2)) \t\t\t#rad/s\n", + "#Calculating the relative linear velocity\n", + "RLV = 0.8*omega*CM\n", + "#Calculating the values of angular velocities\n", + "#The two equations we get in terms of omegaA and omegaB are\n", + "#DL*omegaA-CM*omegaB = RLV .....(i)\n", + "#m1*k1**2*(omega-omegaB) = .275/.2*m2*k2**2*omegaA or\n", + "#2.21*omegaA+omegaB = 2.01 .....(ii)\n", + "A = [[DL, -CM],[ 2.21, 1]]\n", + "B = [RLV,2.01]\n", + "V = linalg.solve(A,B)\n", + "\n", + "#Results:\n", + "print \" The angular velocity of the anvil A, omegaA = %.2f rad/s.\"%(V[0])\n", + "print \" The angular velocity of the hammer B, omegaB = %.2f rad/s, i.e. %.2f rad/s, in the reverse direction.\"%(V[1],\n", + "V[1]*-1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular velocity of the anvil A, omegaA = 1.23 rad/s.\n", + " The angular velocity of the hammer B, omegaB = -0.71 rad/s, i.e. 0.71 rad/s, in the reverse direction.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "m = 30. \t\t\t#kg\n", + "AG = 1\n", + "GB = 150./1000\n", + "k1 = 1.1\n", + "k2 = 350./1000 \t\t\t#m\n", + "theta = 60.*math.pi/180 \t\t\t#rad\n", + "t = 0.005 \t\t\t#s\n", + "a = AG\n", + "b = GB\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the pendulum about the point of suspension A\n", + "IA = m*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia ofthe pendulum about centre of gravity G\n", + "IG = m*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of the pendulum\n", + "h1 = a-a*math.cos(theta)\n", + "omega = math.sqrt(2*m*9.81*h1/IA) \t\t\t#rad/s\n", + "#Calculating the striking velocity of the pendulum\n", + "v = omega*(a+b) \t\t\t#m/s\n", + "#Calculating the angular velocity of the pendulum just after the breakage of the specimen\n", + "omega1 = math.sqrt(omega**2-2*54/IA)\n", + "#Calculating the linear velocity of G just before the breakage of specimen\n", + "vG = omega*AG \t\t\t#m/s\n", + "#Calculating the linear velocity of G just after the breakage of specimen\n", + "vGdash = omega1*AG \t\t\t#m/s\n", + "#Calculating the impulses at pivot A and knife edge B\n", + "#F1+F2 = m*(vG-vGdash) .....(i)\n", + "#b*F2-a*F1 = IG*(omega-omega1) .....(ii)\n", + "A = [[1, 1],[-a, b]]\n", + "B = [[m*(vG-vGdash)], [IG*(omega-omega1)]]\n", + "V = linalg.solve(A,B)\n", + "F1 = V[0]\n", + "F2 = V[1]\n", + "\n", + "#Calculating the angle of swing of the pendulum after impact\n", + "theta1 = math.cos(a-1./2*IA*omega1**2/(m*9.81))/a \t\t\t#radians\n", + "#Calculating the average force exerted at the pivot\n", + "Fp = F1/t \t\t\t#N\n", + "#Calculating the average force exerted at the knife edge\n", + "Fk = F2/t \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" The striking velocity of the pendulum, v = %.2f m/s.\"%(v)\n", + "print \" Impulse at the pivot A, F1 = %.1f N.\"%(F1)\n", + "print \" Impulse at the knife edge B, F2 = %.1f N.\"%(F2)\n", + "print \" Angle of swing of the pendulum after impact, theta = %.2f degree.\"%(theta1*180/math.pi)\n", + "print \" Average force exerted at the pivot is %d N.\"%(round(Fp,-1))\n", + "print \" Average force exerted at the knife edge is %d N.\"%(Fk)\n", + "\n", + "# note : python linalg solve gives slightly different answer but accurate. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The striking velocity of the pendulum, v = 3.27 m/s.\n", + " Impulse at the pivot A, F1 = 0.4 N.\n", + " Impulse at the knife edge B, F2 = 17.0 N.\n", + " Angle of swing of the pendulum after impact, theta = 44.43 degree.\n", + " Average force exerted at the pivot is 80 N.\n", + " Average force exerted at the knife edge is 3407 N.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "T = 150. \t\t\t#N-m\n", + "m1 = 60. #kg\n", + "m2 = 20. \t\t\t#kg\n", + "k1 = 140./1000 #m\n", + "k2 = 80./1000 \t\t#m\n", + "N1 = 750. #rpm\n", + "N2 = 0. \t\t\t#rpm\n", + "\n", + "#Sloution:\n", + "#Calculating the angular speeds\n", + "omega1 = 2*math.pi*N1/60\n", + "omega2 = 0 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the rotor on motor\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the parts attached to machine\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the speed after engagement of the clutch and the time taken\n", + "#We know that impulsive torque = change in angular momentum\n", + "#T*t = I1*(omega1-omega) or I1*omega+T*t = I1*omega1 .....(i)\n", + "#T*t = I2*(omega-omega2) or I2*omega-T*t = I2*omega2 .....(ii)\n", + "A = [[I1, T],[ I2, -T]]\n", + "B = [I1*omega1,I2*omega2]\n", + "V = linalg.solve(A,B)\n", + "omega = V[0] \t\t\t#rad/s\n", + "t = V[1] \t\t\t#s\n", + "#Calculating the kinetic energy lost during the operation\n", + "E = I1*I2*(omega1-omega2)**2/(2*(I1+I2)) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The speed after engagement, omega = %.1f rad/s.\"%(omega)\n", + "print \" The time taken, t = %.2f s.\"%(t)\n", + "print \" The kinetic energy lost during the operation, E = %d N-m.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The speed after engagement, omega = 70.8 rad/s.\n", + " The time taken, t = 0.06 s.\n", + " The kinetic energy lost during the operation, E = 356 N-m.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "M = 75. \t\t\t#kg\n", + "r = 0.3 \t\t\t#m\n", + "G = 6.\n", + "IA = 100. #kg-m**2\n", + "IB = 5. \t\t\t#kg-m**2\n", + "eta = 90./100 \t\t\n", + "\n", + "#Solution:\n", + "#Calculating the equivalent mass of the geared system\n", + "me = 1/r**2*(IA+G**2*IB) \t\t\t#kg\n", + "#Calculating the total equivalent mass to be accelerated\n", + "Me = me+M \t\t\t#kg\n", + "#Calculating the acceleration when it is allowed to fall freely\n", + "F = M*9.81 \t\t\t#Accelerating force provided by the pull of gravity N\n", + "a = F/Me \t\t\t#m/s**2\n", + "#Calculating the equivalent mass of the geared system when the efficiency is 90%\n", + "me1 = 1/r**2*(IA+G**2*IB/eta) \t\t\t#kg\n", + "#Calculating the total equivalent mass to be accelerated\n", + "Me1 = me1+M \t\t\t#kg\n", + "#Calculating the acceleration when the efficiency is 90%\n", + "F1 = M*9.81 \t\t\t#Accelerating force by the pull of gravity N\n", + "a1 = F1/Me1 \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The acceleration of the mass M if it is allowed to fall freely, a = %.3f m/s**2.\"%(a)\n", + "print \" The acceleration of the mass M when the efficiency of the gearing system is 0.9, a = %.3f m/s**2.\"%(a1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The acceleration of the mass M if it is allowed to fall freely, a = 0.231 m/s**2.\n", + " The acceleration of the mass M when the efficiency of the gearing system is 0.9, a = 0.216 m/s**2.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14 pageno : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "T = 100. #N-m ; \n", + "IA = 2. # kg-m**2 ;\n", + "IB = 32. # kg-m**2\n", + "\n", + "# calculations\n", + "G = math.sqrt(IB/IA)\n", + "alphaB = G*100/(IA*G**2 + 32)\n", + "alphaA = G*alphaB\n", + "\n", + "# results\n", + "print \"G = %.f\"%G\n", + "print \"alpha B = %.2f rad/s**2\"%alphaB\n", + "print \"alpha A = %.f rad/s**2\"%alphaA" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G = 4\n", + "alpha B = 6.25 rad/s**2\n", + "alpha A = 25 rad/s**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 page no : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "m = 1500. #kg ; \n", + "d = 600. \n", + "mm = 0.6 #m \n", + "r = 0.3 #m ; \n", + "IA = 8. #kg-m 2 ;\n", + "IB = 1. # kg-m 2 ; \n", + "n = 0.85 \n", + "v = 24. #km/h ; \n", + "F = 300. #N ; \n", + "TB = 200. #N-m ; \n", + "sintheta = 0.25\n", + "\n", + "# calculatins\n", + "tW = n * TB\n", + "P = 170./r\n", + "G = round((14 + math.sqrt(14**2 + 4 * 168.4))/2)\n", + "Amax = (P*G - 3980)/ (1590+9.44*G**2)\n", + "v = 6.67 #m/s\n", + "speed = v/r\n", + "W = G*speed\n", + "power = TB*W/1000\n", + "\n", + "#Result\n", + "print \"G = %.f \"%G\n", + "print \"maximum acceleration : %.2f m/s\"%Amax\n", + "print \"speed of the road wheels : %.f rad/s\"%W\n", + "print \"power of the engine : %.1f kW\"%power" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G = 22 \n", + "maximum acceleration : 1.38 m/s\n", + "speed of the road wheels : 489 rad/s\n", + "power of the engine : 97.8 kW\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 page no : 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "TB = 1000. # N-m ; \n", + "v1 = 27.8 #m/s ; \n", + "v2 = 76.4 #m /s ; \n", + "d = 0.9 # m \n", + "r = 0.45 # m ; \n", + "G = 3.3 ; \n", + "v = 47.2 # m/s ; \n", + "P = 50 * 10**3 # W ;\n", + "M = 1000 #kg ; \n", + "m = 40. # kg ; \n", + "k = 0.25 #m ; \n", + "IB = 1. # kg-m**2\n", + "\n", + "# calculation\n", + "IA = 4 * m* k**2\n", + "F = round(P/v,-1)\n", + "Tw = G*1000\n", + "FT = Tw/r\n", + "C1 = round(-(2325./(2*121.5)*math.log((121.5+v1)/(121.5-v1))),1)\n", + "t = round((2325./(2*121.5)*math.log((121.5+v2)/(121.5-v2)))-4.5,1)\n", + "\n", + "# result\n", + "print \"the time taken for the speed to rise : %.1f s\"%t\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the time taken for the speed to rise : 9.6 s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17 pageno : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "G = 9\n", + "IA = 0.6 #kg-m2 \n", + "IB = 45. # kg-m 2 ; \n", + "TB = 100. # N-m;\n", + "n = 0.95; \n", + "N = 160. # r.p.m. ; \n", + "N1 = 0 \n", + "N2 = 60. # r.p.m. ; \n", + "TA = 30. # N-m\n", + "W1 = 0\n", + "# calculations\n", + "P = 2*math.pi*N*TB/(60*n)\n", + "\n", + "t = 60. # time\n", + "TA = 30.\n", + "TB1 = G*TA*n\n", + "B = TB1 - TB\n", + "alphaB = B/91.2\n", + "W2 = 2*math.pi*N2/60\n", + "t = (W2 - W1)/alphaB\n", + "G1 = (7.27 + math.sqrt(7.27**2+4*78.95))/2\n", + "\n", + "# result\n", + "print \"the power which the motor must develop : %.f W\"% P\n", + "print \"final angular speed : %.1f s\"%t\n", + "print \"maximum angular acceleration : %.3f \"%G1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the power which the motor must develop : 1764 W\n", + "final angular speed : 3.7 s\n", + "maximum angular acceleration : 13.235 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18 Page No : 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 1.5 # m\n", + "r = d/2 # m\n", + "d1 = 1. # m\n", + "kM = 90./1000\n", + "kI = 225./1000\n", + "kD = 600./1000\n", + "kP = 450./1000 \t\t\t#m\n", + "NM = 900. # rpm\n", + "N1 = 275. # rpm\n", + "ND = 50. \t\t\t #rpm\n", + "mM = 200. # kg\n", + "mI = 375. # kg\n", + "mD = 2250. # kg\n", + "mP = 200. # kg\n", + "m1 = 1150. # kg\n", + "m2 = 650. \t\t\t#kg\n", + "FI = 150. # N-m\n", + "FD = 1125. # N-m\n", + "FP = 150. \t\t\t#N-m\n", + "F1 = 500. # N\n", + "F2 = 350. \t\t\t#N\n", + "a = 0.9 \t\t\t#m/s**2\n", + "\n", + "#Solution:\n", + "#Calculating the speed of guide pulley\n", + "NP = ND*d/d1 \t\t\t#rpm\n", + "#Calculating the gear ratio for intermediate gear and motor\n", + "G1 = N1/NM\n", + "#Calculating the gear ratio for drum and motor\n", + "G2 = round(ND/NM,3)\n", + "#Calculating the gear ratio for the guide pulley and motor\n", + "G3 = NP/NM\n", + "#Calculating the mass moment of inertia of the motor\n", + "IM = mM*kM**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the intermediate gear\n", + "II = mI*kI**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the drum and shaft\n", + "ID = mD*kD**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the guide pulley\n", + "IP = mP*kP**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alphaD = a/r \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of the motor\n", + "alphaM = alphaD*NM/ND \t\t\t#rad/s**2\n", + "#Calculating the equivalent mass moment of inertia of the system\n", + "I = IM+G1**2*II+G2**2*ID+2*G3**2*IP \t\t\t#kg-m**2\n", + "#Calculating the torque at motor to accelerate the system\n", + "T1 = round(I*alphaM,1) \t\t\t#N-m\n", + "#Calculating the torque at motor to overcome friction at intermediate gear\n", + "#drum and two guide pulleys\n", + "T2 = round(G1*FI+G2*FD+2*G3*FP,1) \t\t\t#N-m\n", + "#Calculating the tension in the rimath.sing rope between the pulley and drum\n", + "Q1 = m1*9.81+m1*a+F1 \t\t\t#N\n", + "#Calculating the tension in the falling rope between the pulley and drum\n", + "Q2 = m2*9.81-m2*a-F2 \t\t\t#N\n", + "#Calculating the torque at drum\n", + "TD = round((Q1-Q2)*r) \t\t\t#N-m\n", + "#Calculating the torque at motor to raise and lower cages and ropes and to overcome frictional resistance\n", + "T3 = G2*TD \t\t\t#N-m\n", + "#Calculating the total motor torque required\n", + "T = T1+T2+T3 \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The total motor torque required, T = %.1f N-m.\"%(T)\n", + "\n", + "# rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The total motor torque required, T = 583.8 N-m.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19 Page No : 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 50. #kg\n", + "m2 = 25. \t\t\t#kg\n", + "u1 = 3. #m/s\n", + "u2 = 1.5 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#When the impact is inelastic\n", + "#Calculating the common velocity after impact\n", + "v = (m1*u1+m2*u2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy during impact\n", + "EL = m1*m2/(2*(m1+m2))*(u1-u2)**2 \t\t\t#N-m\n", + "#When the impact is elastic\n", + "#Calculating the velocity of the first sphere immediately after impact\n", + "v1 = 2*v-u1 \t\t\t#m/s\n", + "#Calculating the velocity of the second sphere immediately after impact\n", + "v2 = 2*v-u2 \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy\n", + "EL1 = 0\n", + "#When the coefficient of restitution e = 0.6\n", + "e = 0.6\n", + "#Calculating the velocity of the first sphere immediately after impact\n", + "v12 = (1+e)*v-e*u1 \t\t\t#m/s\n", + "#Calculating the velocity of the second sphere immediately after impact\n", + "v22 = (1+e)*v-e*u2 \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy\n", + "EL2 = m1*m2/(2*(m1+m2))*(u1-u2)**2*(1-e**2) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The common velocity after impact when the impact is inelastic, v = %.1f m/s.\"%(v)\n", + "print \" The loss of kinetic energy during impact, EL = %.2f N-m.\"%(EL)\n", + "print \" The velocity of the first sphere immediately after impact when the impact is elastic, v1 = %d m/s.\"%(v1)\n", + "print \" The velocity of the second sphere immediately after impact, v2 = %.1f m/s.\"%(v2)\n", + "print \" The loss of kinetic energy, EL = %d.\"%(EL1)\n", + "print \" The velocity of the first sphere immediately after impact When the coefficient of \\\n", + "restitution is 0.6, v1 = %.1f m/s.\"%(v12)\n", + "print \" The velocity of the second sphere immediately after impact, v2 = %.1f m/s.\"%(v22)\n", + "print \" The loss of kinetic energy during impactm EL = %d N-m.\"%(EL2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The common velocity after impact when the impact is inelastic, v = 2.5 m/s.\n", + " The loss of kinetic energy during impact, EL = 18.75 N-m.\n", + " The velocity of the first sphere immediately after impact when the impact is elastic, v1 = 2 m/s.\n", + " The velocity of the second sphere immediately after impact, v2 = 3.5 m/s.\n", + " The loss of kinetic energy, EL = 0.\n", + " The velocity of the first sphere immediately after impact When the coefficient of restitution is 0.6, v1 = 2.2 m/s.\n", + " The velocity of the second sphere immediately after impact, v2 = 3.1 m/s.\n", + " The loss of kinetic energy during impactm EL = 12 N-m.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20 Page No : 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 15.*1000 #kg\n", + "m2 = 5.*1000 \t\t\t#kg\n", + "u1 = 20.*1000/3600 #m/s\n", + "u2 = 12.*1000/3600 \t\t#m/s\n", + "s = 1000.*10**3 \t\t#N/m\n", + "e = 0.5\n", + "\n", + "#Solution:\n", + "#Calculating the common speed\n", + "v = (m1*u1+m2*u2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the difference in kinetic energies before impact and during impact\n", + "d = m1*m2/(2*(m1+m2))*(u1-u2)**2 \t\t\t#Difference in kinetic energies N-m\n", + "#Equating the difference between kinetic energies to the strain energy stored in the springs\n", + "x = math.sqrt(d*2/(4*s))*1000 \t\t\t#mm\n", + "#Calculating the speed of the loaded wagon immediately after impact ends\n", + "v11 = 2*v-u1 \t\t\t#m/s\n", + "#Calculating the speed of the empty wagon immediately after impact ends\n", + "v21 = 2*v-u2 \t\t\t#m/s\n", + "#Calculating the speeds of the wagons taking into account the coefficient of restitution e = 0.5\n", + "v12 = (1+e)*v-e*u1 \t\t\t#m/s\n", + "v22 = (1+e)*v-e*u2 \t\t\t#m/s\n", + "#Calculating the amount of energy dissipated during impact\n", + "EL = m1*m2/(2*(m1+m2))*(u1-u2)**2*(1-e**2) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The magnitude of common speed v = %d m/s.\"%(v)\n", + "print \" The maximum deflection of each buffer spring during impact x = %d mm.\"%(x)\n", + "print \" The speed of the loaded wagon immediately after the impact ends v1 = %.2f m/s.\"%(v11)\n", + "print \" The speed of the empty wagon immediately after the impact ends v2 = %.2f m/s.\"%(v21)\n", + "print \" When coefficient of restitution is taken into account v1 = %.3f m/s.\"%(v12)\n", + "print \" When coefficient of restitution is taken into account v2 = %.3f m/s.\"%(v22)\n", + "print \" The amount of energy dissipated during impact EL = %d N-m.\"%(EL)\n", + "\n", + "# rounding off error. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The magnitude of common speed v = 5 m/s.\n", + " The maximum deflection of each buffer spring during impact x = 68 mm.\n", + " The speed of the loaded wagon immediately after the impact ends v1 = 4.44 m/s.\n", + " The speed of the empty wagon immediately after the impact ends v2 = 6.67 m/s.\n", + " When coefficient of restitution is taken into account v1 = 4.722 m/s.\n", + " When coefficient of restitution is taken into account v2 = 5.833 m/s.\n", + " The amount of energy dissipated during impact EL = 6944 N-m.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21 Page No : 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "IA = 22.5 #kg-m**2\n", + "IB = 67.5 \t\t\t#kg-m**2\n", + "q = 225. \t\t\t#N-m/rad\n", + "NA = 150. #rpm\n", + "NB = 0. \t\t\t#rpm\n", + "\n", + "#Calculating the angular speed of the flywheel\n", + "omegaA = 2*math.pi*NA/60 \t\t\t#rad/s\n", + "#Calculating the angular speed of both the flywheels at the instant their speeds are equal\n", + "omega = IA*omegaA/(IA+IB) \t\t\t#rad/s\n", + "#Calculating the kinetic energy of the system at that instant\n", + "E2 = 1./2*(IA+IB)*omega**2 \t\t\t#N-m\n", + "#Calculating the kinetic energy of the flywheel A\n", + "E1 = 1./2*IA*omegaA**2 \t\t\t#N-m\n", + "#Calculating the strain energy stored in the spring\n", + "E = E1-E2 \t\t\t#Strain energy stored in the spring N-m\n", + "#Calculating the maximum twist of the spring\n", + "theta = math.sqrt(E*2/q) \t\t\t#radians\n", + "thetad = theta*180/math.pi \t\t\t#Maximum twist degrees\n", + "#Calculating the speed of each flywheel when the spring regains its initial unstrained condition\n", + "N = 60*omega/(2*math.pi)\n", + "NA1 = 2*N-NA \t\t\t#rpm\n", + "NB1 = 2*N-NB \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" The strain energy stored in the spring is %d N-m.\"%(E)\n", + "print \" The maximum twist of the spring theta = %.1f degrees.\"%(thetad)\n", + "print \" The speed of flywheel A when the spring regains its initial unstrained condition NA1 = %d rpm \\\n", + " \\ni.e. %d rpm in the opposite direction.\"%(NA1,-NA1)\n", + "print \" The speed of flywheel B when the spring regains its initial unstrained condition NB1 = %d rpm.\"%(NB1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The strain energy stored in the spring is 2081 N-m.\n", + " The maximum twist of the spring theta = 246.5 degrees.\n", + " The speed of flywheel A when the spring regains its initial unstrained condition NA1 = -75 rpm \n", + "i.e. 75 rpm in the opposite direction.\n", + " The speed of flywheel B when the spring regains its initial unstrained condition NB1 = 75 rpm.\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
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