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| author | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
| commit | db0855dbeb41ecb8a51dde8587d43e5d7e83620f (patch) | |
| tree | b95975d958cba9af36cb1680e3f77205354f6512 /Engineering_Mechanics_by_Tayal_A.K./chapter16_3.ipynb | |
| parent | 5a86a20b9de487553d4ef88719fb0fd76a5dd6a7 (diff) | |
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diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter16_3.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter16_3.ipynb new file mode 100755 index 00000000..7bd00420 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter16_3.ipynb @@ -0,0 +1,471 @@ +{
+ "metadata": {
+ "name": "chapter16.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16: Kinetics Of A Particle : Work And Energy"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-1,Page No:432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "k=1000 # N/m # stiffness of spring\n",
+ "x_1=0.1 # m # distance upto which the spring is stretched\n",
+ "x_2=0.2 # m \n",
+ "x_0=0 # initial position of spring\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Work required to stretch the spring by 10 cm from undeformed position is given as,\n",
+ "U_10=-(k/2)*(x_1**2-x_0**2) # N-m \n",
+ "# Work required to stretch from 10 cm to 20 cm is,\n",
+ "U=-(0.5)*k*(x_2**2-x_1**2) # N-m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The work of the spring force is \",round(U_10),\"N-m\"\n",
+ "print\"The work required to stretch the spring by 20 cm is \",round(U),\"N-m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The work of the spring force is -5.0 N-m\n",
+ "The work required to stretch the spring by 20 cm is -15.0 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-3,Page No:436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M_A=100 # kg # mass of block A\n",
+ "M_B=150 # kg # mass of block B\n",
+ "mu=0.2 # coefficient of friction between the blocks and the surface\n",
+ "x=1 # m # distance by which block A moves\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D\n",
+ "# Applying the principle of work and energy to the system of blocks A&B and on simplifying we get the value of v as,\n",
+ "v=(((-mu*M_A*g)+(M_B*g))/(125))**0.5 # m/s \n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of block A is \",round(v,2),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of block A is 3.19 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-4,Page No:440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M=500*10**3 # kg # mass of the train\n",
+ "u=0 # m/s # initial speed\n",
+ "v=90*1000*3600**-1 # m/s # final speed\n",
+ "t=50 # seconds\n",
+ "F_r=15*10**3 # N # Frictioal resistance to motion\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Acceleration is given as,\n",
+ "a=v*t**-1 # m/s^2\n",
+ "# The total force required to accelerate the train is,\n",
+ "F=M*a # N\n",
+ "# The maximum power required is at, t=50s & v=25 m/s\n",
+ "P=(F+F_r)*v*(10**-6) # MW\n",
+ "# At any time after 50 seconds, the force required only to overcome the frictional resistance of 15*10^3 N is,\n",
+ "P_req=F_r*v*(10**-3) # kW\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"(a) The maximum power required is \",round(P,3),\"MW\"\n",
+ "print\"(b) The power required to maintain a speed of 90 km/hr is \",round(P_req,2),\"kW\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The maximum power required is 6.625 MW\n",
+ "(b) The power required to maintain a speed of 90 km/hr is 375.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-5,Page No:440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "W=50 # N # Weight suspended on spring\n",
+ "k=10 # N/cm # stiffness of the spring\n",
+ "x_2=15 # cm # measured extensions\n",
+ "h=10 # cm # height for position 2\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the required F.B.D.\n",
+ "\n",
+ "# POSITION 1: The force exerted by the spring is,\n",
+ "F_1=W # N\n",
+ "\n",
+ "# Extension of spring from undeformed position is x_1,\n",
+ "x_1=F_1/k # cm\n",
+ "\n",
+ "# POSITION 2: When pulled by 10 cm to the floor. P.E of weight is,\n",
+ "P_E_g=-W*h # N-cm # (P_E_g= P_E_gravity)\n",
+ "\n",
+ "# P.E of the spring with respect to position 1\n",
+ "P_E_s=(0.5)*k*(x_2**2-x_1**2) # N-cm # (P_E_s= P_E_ spring)\n",
+ "\n",
+ "# Total P.E of the system with respect to position 1\n",
+ "P_E_t=P_E_g+P_E_s # N-cm # (P_E_t= P_E_total)\n",
+ "\n",
+ "# Total energy of the system,\n",
+ "E_2=P_E_t # N-cm\n",
+ "\n",
+ "# Total energy of the system in position 3 w.r.t position 1 is:\n",
+ "x=-(100)**0.5 # cm\n",
+ "x=+(100)**0.5 # cm\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The potential energy of the system is \",round(E_2),\"N-cm\"\n",
+ "print\"The maximum height above the floor that the weight W will attain after release is \",round(x),\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy of the system is 500.0 N-cm\n",
+ "The maximum height above the floor that the weight W will attain after release is 10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-6,Page No:442"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "m=5 # kg # mass of the ball\n",
+ "k=500 # N/m # stiffness of the spring\n",
+ "h=10 # cm # height of drop\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D.\n",
+ "# In eq'n 1 substitute the respective values and simplify it further. In this eq'n of 2nd degree a=1 b=-0.1962 & c=-0.01962. Thus the roots of the eq'n is given as,\n",
+ "a=1 \n",
+ "b=-0.1962\n",
+ "c=-0.01962\n",
+ "delta=((-b+(((b**2)-(4*a*c))**0.5))/(2*a))*(10**2) # cm # We consider the +ve value of delta\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The maximum deflection of the spring is \",round(delta,2),\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum deflection of the spring is 26.91 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-7,Page No:444"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "m=5 # kg # mass of the ball\n",
+ "k=500 # N/m # stiffness of the spring\n",
+ "h=0.1 # m # height of vertical fall\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D\n",
+ "# On equating the total energies at position 1 & 2 we get eq'n of delta as,\n",
+ "delta=((2*m*g*h)/(k))**0.5 # m \n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum compression of the spring is 0.14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-9,Page No:445"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "m=5 # kg # mass of the collar\n",
+ "k=500 # N/m # stiffness of the spring\n",
+ "AB=0.15 # m # Refer the F.B.D for AB\n",
+ "AC=0.2 # m # Refer the F.B.D for AC\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D\n",
+ "\n",
+ "# POSITION 1: \n",
+ "P_E_1=m*g*(AB)+0 \n",
+ "K_E_1=0\n",
+ "E_1=P_E_1+K_E_1 #\n",
+ "\n",
+ "# POSITION 2 : Length of the spring in position 2\n",
+ "CB=(AB**2+AC**2)**0.5 # m \n",
+ "# x is the extension in the spring\n",
+ "x=CB-AC # m\n",
+ "# On substuting and Equating equations of total energies for position1 & position 2 we get the value of v as,\n",
+ "v=(((E_1-((0.5)*k*x**2))*2)/m)**0.5 # m/s\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of the collar will be \",round(v,2),\"m/s\"\n",
+ "# The answer given in the text book (v=16.4 m/s) is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the collar will be 1.64 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-10,Page No:446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "m=5 # kg # mass of the block\n",
+ "theta=30 # degree # inclination of the plane\n",
+ "x=0.5 # m # distance travelled by the block\n",
+ "k=1500 # N/m # stiffness of the spring\n",
+ "mu=0.2 # coefficient of friction between the block and the surface\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the F.B.D of the block\n",
+ "# Applying the principle of work and energy between the positions 1 & 2 and on further simplification we get the generic eq'n for delta as, 750*delta^2-16.03*delta-8.015=0. From this eq'n e have values of a.b & c as,\n",
+ "a=750\n",
+ "b=-16.03\n",
+ "c=-8.015\n",
+ "# Thus the roots of the eq'n are given as,\n",
+ "delta=(-b+((b**2-(4*a*c))**0.5))/(2*a) # m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The maximum compression of the spring is \",round(delta,3),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum compression of the spring is 0.115 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16-11,Page No:448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "M=10 # kg # Here M=M_1=M_2\n",
+ "g=9.81 # m/s^2 # acc due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Consider the respective F.B.D\n",
+ "# Applying the principle of conservation of energy and by equating the total energies at position 1 & position 2 we get v as,\n",
+ "v=((M*g*4)/(25))**0.5 # m/s\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The velocity of mass M_2 is \",round(v,2),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of mass M_2 is 3.96 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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