From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- .../chapter15_6.ipynb | 775 +++++++++++++++++++++ 1 file changed, 775 insertions(+) create mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter15_6.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter15_6.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter15_6.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter15_6.ipynb new file mode 100755 index 00000000..db992973 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter15_6.ipynb @@ -0,0 +1,775 @@ +{ + "metadata": { + "name": "chapter15.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Curvilinear Motion Of A Particle" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-2,Page No:386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "r=200 # m # radius of the curved road\n", + "v_1=72*1000*3600**-1 # m/s # initial speed of the car\n", + "v_2=36*1000*3600**-1 # m/s # speed of the car after 10 seconds\n", + "t=10 # seconds\n", + "\n", + "# Calculations\n", + "\n", + "A_n=v_1**2/r # m/s^2 # normal component of acceleration\n", + "A_t=0 # since dv/dt=0 # tangential component of acceeration\n", + "delv=v_1-v_2\n", + "delt=t-0\n", + "a_t=delv/delt # m/s^2 # tangential component of deceleration after the brakes are applied\n", + "a_n=v_1**2/r # m/s^2 # normal component of deceleration after the brakes are applied\n", + "\n", + "# Results\n", + "\n", + "print\"The normal component of acceleration is \",round(A_n),\"m/s^2\"\n", + "print\"The tangential component of acceleration is \",round(A_t),\"m/s^2\"\n", + "print\"The normal component of deceleration is \",round(a_n),\"m/s^2\"\n", + "print\"The tangential component of deceleration is \",round(a_t),\"m/s^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The normal component of acceleration is 2.0 m/s^2\n", + "The tangential component of acceleration is 0.0 m/s^2\n", + "The normal component of deceleration is 2.0 m/s^2\n", + "The tangential component of deceleration is 1.0 m/s^2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-3,Page No:387" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Iintilization of variables\n", + "\n", + "r=250 # m # radius of the curved road\n", + "a_t=0.6 # m/s^2 # tangential acceleration\n", + "a=0.75 # m/s^2 # total acceleration attained by the car\n", + "\n", + "# Calculations\n", + "\n", + "a_n=(a**2-a_t**2)**0.5 # m/s^2\n", + "v=sqrt(a_n*r) # m/s\n", + "\n", + "# Using v=u+a*t\n", + "u=0\n", + "t=v/a_t # seconds\n", + "\n", + "# Now using v^2-u^2=2*a*s\n", + "s=v**2/(2*a_t) # m\n", + "\n", + "# Results\n", + "\n", + "print\"The distance traveled by the car is \",round(s,2),\"m\"\n", + "print\"The time for which the car travels is \",round(t,2),\"seconds\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance traveled by the car is 93.75 m\n", + "The time for which the car travels is 17.68 seconds\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-5,Page No:388" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "v=10 # m/s # speed of the car\n", + "r=200 # m # radius of the road\n", + "t=15 # seconds\n", + "\n", + "# Calculations\n", + "\n", + "omega=(v*r**-1) # radian/seconds # angular velocity of the car\n", + "\n", + "# Velocity in x & y direction is given by eq'n\n", + "v_x=omega*r*sin(omega*t) # m/s # value of v_x is -ve but we consider it to be +ve for calculations\n", + "v_y=omega*r*cos(omega*t) # m/s\n", + "\n", + "# Acceleration in x & y direction is given by\n", + "a_x=(omega**2)*r*cos(omega*t) # m/s^2 # value of a_x is -ve but we consider it to be +ve for calculations\n", + "a_y=(omega**2)*r*sin(omega*t) # m/s^2 # value of a_y is -ve but we consider it to be +ve for calculations\n", + "\n", + "# Results\n", + "\n", + "print\"The component of velocity in X direction (v_x) is \",round(v_x,2),\"m/s\"\n", + "print\"The component of velocity in Y direction (v_y) is \",round(v_y,2),\"m/s\"\n", + "print\"The component of acceleration in X direction (a_x) is \",round(a_x,3),\"m/s^2\"\n", + "print\"The component of acceleration in Y direction (a_y) is \",round(a_y,3),\"m/s^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The component of velocity in X direction (v_x) is 6.82 m/s\n", + "The component of velocity in Y direction (v_y) is 7.32 m/s\n", + "The component of acceleration in X direction (a_x) is 0.366 m/s^2\n", + "The component of acceleration in Y direction (a_y) is 0.341 m/s^2\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-6,Page No:392" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "t=1 # seconds\n", + "pi=3.14\n", + "\n", + "# Calculations\n", + "\n", + "# From the equations of r and theta given we find 1st & 2nd derative and substitute t=1sec Here we consider the 1st derative as r_1 & theta_1 and so on...\n", + "r=(1.25*t**2)-(0.9*t**3) # m\n", + "r_1=(1.25*(2*t))-(0.9*(3*t**2)) # m/s\n", + "r_2=2.5-(0.9*3*(2*t)) # m/s^2\n", + "theta=(pi/2)*(4*t-3*t**2) # radian\n", + "theta_1=(pi/2)*(4-(6*t)) # rad/second\n", + "theta_2=(pi/2)*(0-(6*t)) # rad/second^2\n", + "\n", + "# Velocity of collar P\n", + "v_r=r_1 # m/s\n", + "v_theta=r*theta_1 # m/s\n", + "v=(v_r**2+v_theta**2)**0.5 # m/s\n", + "alpha=arctan(v_theta/v_r) # degree\n", + "\n", + "# Acceleration of the collar P\n", + "a_r=r_2-(r*theta_1**2) # m/s^2\n", + "a_theta=(r*theta_2)+(2*r_1*theta_1) # m/s^2\n", + "a=(a_r**2+a_theta**2)**0.5 # m/s^2\n", + "beta=arctan(a_theta/a_r) # degree\n", + "\n", + "# Acceleration of collar P relative to the rod. Let it be a_relative\n", + "a_relative=r_2 # m/s^2 # towards O\n", + "\n", + "# Calculations\n", + "\n", + "print\"The velocity of the collar is \",round(v,3),\"m/s\"\n", + "print\"The accelaration of the collar is \",round(a,3),\"m/s^2\"\n", + "print\"The acceleration of the collar relative to the rod is \",round(a_relative,1),\"m/s^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of the collar is 1.117 m/s\n", + "The accelaration of the collar is 6.671 m/s^2\n", + "The acceleration of the collar relative to the rod is -2.9 m/s^2\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-7,Page No:394" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Consider the eq'ns of motion from the book\n", + "# The notations have been changed for the derivatives of r & theta\n", + "# (1) At t=0 s\n", + "theta_0=0\n", + "theta_1=2*pi # rad/s\n", + "theta_2=0\n", + "r_0=0\n", + "r_1=10 # cm/s\n", + "r_2=0\n", + "# At t=0.3 s\n", + "t=0.3 # sec\n", + "theta=2*pi*t # rad\n", + "theta1=2*pi # rad/s\n", + "theta2=0\n", + "r=10*t # cm\n", + "r1=10 # cm/s\n", + "r2=0\n", + "# (i) \n", + "#Velocity\n", + "v_r=r_1 # cm/s\n", + "v_theta=r_0*theta_1\n", + "v=sqrt(v_r**2+v_theta**2) # cm/s\n", + "# Acceleration\n", + "a_r=r_2-(r_0*theta_1**2) # cm/s^2\n", + "a_theta=(r_0*theta_2)+(2*r_1*theta_1) # cm/s^2\n", + "a=sqrt(a_r**2+a_theta**2) # cm/s^2\n", + "# (ii)\n", + "# Velocity\n", + "V_R=r1 # cm/s\n", + "V_theta=r*theta1 # cm/s\n", + "V=sqrt(V_R**2+V_theta**2) # cm/s\n", + "# Acceleration\n", + "A_r=r2-(r*theta1**2) # cm/s^2\n", + "A_theta=(r*theta2)+(2*r1*theta1) # cm/s^2\n", + "A=sqrt(A_r**2+A_theta**2) # cm/s^2\n", + "\n", + "# Results\n", + "\n", + "print\"The velocity and the acceleration of the partice at t=0 s is \",round(v),\"cm/s\",\"&\",round(a,1),\"cm/s^2\"\n", + "print\"The velocity and the acceleration of the partice at t=0.3 s is \",round(V,2),\"cm/s\",\"&\",round(A,2),\"cm/s^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity and the acceleration of the partice at t=0 s is 10.0 cm/s & 125.7 cm/s^2\n", + "The velocity and the acceleration of the partice at t=0.3 s is 21.34 cm/s & 172.68 cm/s^2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-9,Page No:404" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Calculations\n", + "# Tension in the wire before it is cut\n", + "T_ab=1/((2.747*0.643)+(0.766)) # From eqn's 1 & 2..Here T_ab is multiplied with W (i.e weight of small ball) \n", + "T_AB=cos(40*(pi/180)) # Tension in the wire after the wire is cut. Again T_AB is multiplied with W.\n", + "# Results\n", + "\n", + "print\"The tension in the wire before and after it is cut is respectively \",round(T_ab,3),\"W\",\"&\",round(T_AB,3),\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The tension in the wire before and after it is cut is respectively 0.395 W & 0.766 W\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-10,Page No:405" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "W_A=120 # N # Weight of block A\n", + "W_B=80 # N # Weight of block B\n", + "mu_a=0.4 # coefficient of friction under block A\n", + "n=40 # r.p.m # rotation of frame\n", + "r_a=1.2 # m # distance from block A to the axis of rotation\n", + "r_b=1.6 # m # distance from block A to the axis of rotation\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "pi=3.14 # constant\n", + "\n", + "# Calculations\n", + "\n", + "# Consider the F.B.D of block A.\n", + "N=W_A # N # Sum F_y=0\n", + "\n", + "# Now here, a_n=omega^2*r\n", + "omega=(2*pi*n)*60**-1 # rad/sec\n", + "a_n=(omega**2)*r_a # m/s^2\n", + "\n", + "# sum F_x=0 gives the eq'n of T as,\n", + "T=((W_A*g**-1)*a_n)-(mu_a*N) # N \n", + "\n", + "# Now consider the F.B.D of block B\n", + "A_n=(omega**2)*r_b # m/s^2\n", + "N_1=(W_B*g**-1)*A_n # N # sum F_x=0\n", + "mu=(T-W_B)*N_1**-1 \n", + "\n", + "# Results\n", + "\n", + "print\"The coefficient of friction of block B is \",round(mu,3) \n", + "#answer may vary due to decimal variance in python\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coefficient of friction of block B is 0.565\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-12,Page No:408" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "W=10000 # N # Weight of the locomotive\n", + "# Calculations\n", + "# Consider the various derivations given in the textbook\n", + "R_max=W*20**-1 # N # eq'n for max reaction\n", + "# The position of occurence of maximum thrust cannot be defined here. Refer textbook for the answer\n", + "# Results\n", + "\n", + "print\"The maximum lateral thrust is \",round(R_max),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum lateral thrust is 500.0 N\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-13,Page No:411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "W=10 #N # Weight of the ball\n", + "# Calculations\n", + "# consider the eq'n derived to find the reaction, given as\n", + "R=W*(1+((2*pi**2)*9**-1)) # N \n", + "# Results\n", + "\n", + "print\"The value of the reaction is \",round(R,1),\"N\"\n", + "#answer may vary due to decimal variance\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of the reaction is 31.9 N\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-15,Page No:412" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "P=50 # N # Weight of ball P\n", + "Q=50 # N # Weight of ball Q\n", + "R=100 # N # Weight of the governing device\n", + "l=0.3 # m # length of each side\n", + "theta=30 # degree\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "pi=3.14 # constant\n", + "\n", + "# Calculations\n", + "\n", + "# Consider the respective F.B.D\n", + "r=l*sin(theta*(pi/180)) # m # Radius of circe\n", + "# On solving eqn's 1,2 &3 we get the value of v as,\n", + "v=(((Q+R)*g*r)/(3**0.5*Q))**0.5 # m/s \n", + "# But the eq'n v=omega*r we get the value of N as,\n", + "N=(60*v)/(2*pi*r) # r.p.m \n", + "\n", + "# Results\n", + "\n", + "print\"The speed of rotation is \",round(N,1),\"r.p.m\"\n", + "# NOTE: In the text book (A.K. Tayal) this sum is numbered as 'EXAMPLE 15.14' which is incorrect.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of rotation is 101.7 r.p.m\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-16,Page No:414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "Q=20 # N # Weight of the governor device\n", + "W=10 # N # Weight of the fly balls\n", + "theta=30 # degree # angle between the vertical shaft and the axis AB\n", + "l=0.2 # m # length of the shaft\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "pi=3.14 # constant\n", + "\n", + "# Calculations\n", + "\n", + "# Consider the respective F.B.D\n", + "# Radius of the circle is given as,\n", + "r=Q*sin(theta*(pi/180))*(10**-2) # m \n", + "\n", + "# Solving eq'n 1 & 2 for v. The eq'n for v is given as,\n", + "v=(((W*l*0.5)+(0.05*Q))/((W*0.2*(3)**0.5)/(2*g*r)))**0.5 # m/s\n", + "\n", + "# But, v=r*omega=2*pi*N*r/60. From this eq'n we get N as,\n", + "N=(v*60)/(2*pi*r) # r.p.m.\n", + "\n", + "# Results\n", + "\n", + "print\"The speed of the fly-balls is \",round(N,1),\"r.p.m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the fly-balls is 101.7 r.p.m\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-18,Page No:421 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "r=50 # m # radius of the road\n", + "mu=0.15 # coefficient of friction between the wheels and the road\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# The eq'n fo max speed of the vehicle without skidding is \n", + "v=(mu*g*r)**0.5 # m/s\n", + "\n", + "# The angle theta made with the vertical while negotiating the corner is \n", + "theta=arctan(v**2/(g*r))*(180/pi) # degree\n", + "\n", + "# Results\n", + "\n", + "print\"The maximum speed with which the vehicle can travel is \",round(v,2),\"m/s\"\n", + "print\"The angle made with the vertical is \",round(theta,2),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum speed with which the vehicle can travel is 8.58 m/s\n", + "The angle made with the vertical is 8.54 degree\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-19,Page No:422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "v=100*1000*3600**-1 # m/s # or 100 km/hr\n", + "r=250 # m # radius of the road\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# The angle of banking is given by eq'n,\n", + "theta=arctan((v**2)/(g*r))*(180/pi) # degree \n", + "\n", + "# Results\n", + "\n", + "print\"The angle of banking of the track is \",round(theta,2),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle of banking of the track is 17.47 degree\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-20,Page No:422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "W=10000 # N # Weight of the car\n", + "r=100 # m # radius of the road\n", + "v=10 # m/s # speed of the car\n", + "h=1 # m # height of the C.G of the car above the ground\n", + "b=1.5 # m # distance between the wheels\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# The reactions at the wheels are given by te eq'ns:\n", + "R_A=(W/2)*(1-((v**2*h)/(g*r*b))) # N # Reaction at A\n", + "R_B=(W/2)*(1+((v**2*h)/(g*r*b))) # N # Reaction at B\n", + "\n", + "# The eq'n for max speed to avoid overturning on level ground is,\n", + "v_max=((g*r*(b/2))/(h))**0.5 # m/s\n", + "\n", + "# Results\n", + "\n", + "print\"The reaction at Wheel A (R_A) is \",round(R_A,1),\"N\"\n", + "print\"The reaction at Wheel B (R_B) is \",round(R_B,1),\"N\"\n", + "print\"The maximum speed at which the vehicle can travel without the fear of overturning is \",round(v_max,2),\"m/s\" \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at Wheel A (R_A) is 4660.2 N\n", + "The reaction at Wheel B (R_B) is 5339.8 N\n", + "The maximum speed at which the vehicle can travel without the fear of overturning is 27.12 m/s\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15-21,Page No:423" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "W=1 # N # Weight of the bob\n", + "theta=8 # degree # angle made by the bob with the vertical\n", + "r=100 # m # radius of the curve\n", + "g=9.81 # m/s^2 # acc due to gravity\n", + "\n", + "# Calculations\n", + "\n", + "# from eq'n 1 & 2 we get v as,\n", + "v=((g*r*tan(theta*(pi/180)))**0.5)*3600*1000**-1 # km/hr \n", + "T=W/cos(theta*(pi/180)) # N # from eq'n 2\n", + "\n", + "# Results\n", + "\n", + "print\"The speed of the cariage is \",round(v,2),\"km/hr\"\n", + "print\"The tension in the chord is \",round(T,2),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the cariage is 42.26 km/hr\n", + "The tension in the chord is 1.01 N\n" + ] + } + ], + "prompt_number": 47 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit