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diff --git a/Engineering_Mechanics_by_A._K._Tayal/Chapter18.ipynb b/Engineering_Mechanics_by_A._K._Tayal/Chapter18.ipynb new file mode 100644 index 00000000..bce7ef40 --- /dev/null +++ b/Engineering_Mechanics_by_A._K._Tayal/Chapter18.ipynb @@ -0,0 +1,532 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18 Impact Collision of Elastic Bodies" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.1 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of ball A after impact is 0.000000 m/s\n", + "The velocity of ball B after impact is 1.000000 m/s\n" + ] + } + ], + "source": [ + "import numpy\n", + "# Initilization of variables\n", + "m_a=1 # kg # mass of the ball A\n", + "v_a=2 # m/s # velocity of ball A\n", + "m_b=2 # kg # mass of ball B\n", + "v_b=0 # m/s # ball B at rest\n", + "e=1/2 # coefficient of restitution\n", + "# Calculations\n", + "# Solving eqn's 1 & 2 using matrix for v'_a & v'_b,\n", + "A=numpy.matrix('1 2;-1 1')\n", + "B=numpy.matrix('2;1')\n", + "C=numpy.linalg.inv(A)*B\n", + "# Results\n", + "print('The velocity of ball A after impact is %f m/s'%C[0])\n", + "print('The velocity of ball B after impact is %f m/s'%C[1])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.2 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of ball A after impact on ball B is 0.000000 m/s\n", + "The velocity of ball B after getting impacted by ball A is 8.000000 m/s\n", + "The final velocity of ball B is 0.000000 m/s\n", + "The velocity of ball C after getting impacted by ball B is 6.000000 m/s\n" + ] + } + ], + "source": [ + "import numpy\n", + "# Initilization of variables\n", + "m_a=2 # kg # mass of ball A\n", + "m_b=6 # kg # mass of ball B\n", + "m_c=12 # kg # mass of ball C\n", + "v_a=12 # m/s # velocity of ball A\n", + "v_b=4 # m/s # velocity of ball B\n", + "v_c=2 # m/s # velocity of ball C\n", + "e=1 # coefficient of restitution for perfectly elastic body\n", + "# Calculations\n", + "# (A)\n", + "# Solving eq'n 1 & 2 using matrix for v'_a & v'_b,\n", + "A=numpy.matrix('2 6;-1 1')\n", + "B=numpy.matrix('48;8')\n", + "C=numpy.linalg.inv(A)*B\n", + "# Calculations\n", + "# (B)\n", + "# Solving eq'ns 3 & 4 simultaneously using matrix for v'_b & v'_c\n", + "P=numpy.matrix('1 2;-1 1')\n", + "Q=numpy.matrix('12;6')\n", + "R=numpy.linalg.inv(P)*Q\n", + "# Results (A&B)\n", + "print('The velocity of ball A after impact on ball B is %f m/s'%C[0]) # here the ball of mass 2 kg is bought to rest\n", + "print('The velocity of ball B after getting impacted by ball A is %f m/s'%C[1])\n", + "print('The final velocity of ball B is %f m/s'%R[0]) # here the ball of mass 6 kg is bought to rest\n", + "print('The velocity of ball C after getting impacted by ball B is %f m/s'%R[1])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.3 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ball was dropped from a height of 13.500000 m\n", + "The coefficient of restitution between the glass and the floor is 0.816497 \n" + ] + } + ], + "source": [ + "import math\n", + "# Initilization of variables\n", + "h_1=9 # m # height of first bounce\n", + "h_2=6 # m # height of second bounce\n", + "# Calculations\n", + "# From eq'n (5) we have, Coefficient of restitution between the glass and the floor is,\n", + "e=math.sqrt(h_2/h_1)\n", + "# From eq'n 3 we get height of drop as,\n", + "h=h_1/e**2 # m\n", + "# Results\n", + "print('The ball was dropped from a height of %f m'%h)\n", + "print('The coefficient of restitution between the glass and the floor is %f '%e)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.4 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of ball A after impact is 8.353604 m/s\n", + "The velocity of ball B after impact is 15.179186 m/s\n", + "The direction of ball A after impact is 36.765696 degree\n", + "The direction of ball B after impact is 58.848472 degree\n" + ] + } + ], + "source": [ + "import math\n", + "# Initilization of variables\n", + "e=0.90 # coefficient o restitution\n", + "v_a=10 # m/s # velocity of ball A\n", + "v_b=15 # m/s # velocity of ball B\n", + "alpha_1=30 # degree # angle made by v_a with horizontal\n", + "alpha_2=60 # degree # angle made by v_b with horizontal\n", + "# Calculations\n", + "# The components of initial velocity of ball A:\n", + "v_a_x=v_a*math.cos(alpha_1*math.pi/180) # m/s\n", + "v_a_y=v_a*math.sin(alpha_1*math.pi/180) # m/s\n", + "# The components of initial velocity of ball B:\n", + "v_b_x=-v_b*math.cos(alpha_2*math.pi/180) # m/s\n", + "v_b_y=v_b*math.sin(alpha_2*math.pi/180) # m/s\n", + "# From eq'n 1 & 2 we get,\n", + "v_ay=v_a_y # m/s # Here, v_ay=(v'_a)_y\n", + "v_by=v_b_y # m/s # Here, v_by=(v'_b)_y\n", + "# On adding eq'n 3 & 4 we get,\n", + "v_bx=((v_a_x+v_b_x)+(-e*(v_b_x-v_a_x)))/2 # m/s # Here. v_bx=(v'_b)_x\n", + "# On substuting the value of v'_b_x in eq'n 3 we get,\n", + "v_ax=(v_a_x+v_b_x)-(v_bx) # m/s # here, v_ax=(v'_a)_x\n", + "# Now the eq'n for resultant velocities of balls A & B after impact are,\n", + "v_A=math.sqrt(v_ax**2+v_ay**2) # m/s\n", + "v_B=math.sqrt(v_bx**2+v_by**2) # m/s\n", + "# The direction of the ball after Impact is,\n", + "theta_1=math.degrees(math.atan(-(v_ay/v_ax))) # degree\n", + "theta_2=math.degrees(math.atan(v_by/v_bx)) # degree\n", + "# Results\n", + "print('The velocity of ball A after impact is %f m/s'%v_A)\n", + "print('The velocity of ball B after impact is %f m/s'%v_B)\n", + "print('The direction of ball A after impact is %f degree'%theta_1)\n", + "print('The direction of ball B after impact is %f degree'%theta_2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.5 Motion of ball" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the ball is 0.661438 v\n", + "The direction of the ball is 49.106605 degrees\n" + ] + } + ], + "source": [ + "# Initiization of variables\n", + "theta=30 # degrees # ange made by the ball against the wall\n", + "e=0.50\n", + "# Calculations\n", + "# The notations have been changed\n", + "# Resolving the velocity v as,\n", + "v_x=math.cos(theta*math.pi/180)\n", + "v_y=math.sin(theta*math.pi/180)\n", + "V_y=v_y\n", + "# from coefficient of restitution reation\n", + "V_x=-e*v_x\n", + "# Resultant velocity\n", + "V=math.sqrt(V_x**2+V_y**2)\n", + "theta=math.degrees(math.atan(V_y/(-V_x))) # taking +ve value for V_x\n", + "# NOTE: Here all the terms are multiplied with velocity i.e (v).\n", + "# Results\n", + "print('The velocity of the ball is %f v'%V)\n", + "print('The direction of the ball is %f degrees'%theta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.6 Principle of conservation of Energy" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The height from which the ball A should be released is 0.149705 m\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "e=0.8 # coefficient of restitution\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calcuations\n", + "# Squaring eqn's 1 &2 and Solving eqn's 1 & 2 using matrix for the value of h\n", + "A=numpy.matrix([[-1,(2*g)],[-1,-(1.28*g)]])\n", + "B=numpy.matrix([[0.945**2],[(-0.4*9.81)]])\n", + "C=numpy.linalg.inv(A)*B # m\n", + "# Results\n", + "print('The height from which the ball A should be released is %f m'%C[1])\n", + "# The answer given in the book i.e 0.104 is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.7 Principle of conservation of Energy" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle through which the sphere B will swing after the impact is 28.955024 degree\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "theta_a=60 # degree # angle made by sphere A with the verticle\n", + "e=1 # coefficient of restitution for elastic impact\n", + "# Calculations\n", + "# theta_b is given by the eq'n cosd*theta_b=0.875, hence theta_b is,\n", + "theta_b=math.degrees(math.acos(0.875)) # degree\n", + "# Results\n", + "print('The angle through which the sphere B will swing after the impact is %f degree'%theta_b)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.8 Principle of conservation of Energy" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is 18.188288 degree\n", + "(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is 22.088290 degree\n", + "(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is 14.674584 degree\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "m_a=0.01 # kg # mass of bullet A\n", + "v_a=100 # m/s # velocity of bullet A\n", + "m_b=1 # kg # mass of the bob\n", + "v_b=0 # m/s # velocity of the bob\n", + "l=1 # m # length of the pendulum\n", + "v_r=-20 # m/s # velocity at which the bullet rebounds the surface of the bob # here the notation for v'_a is shown by v_r\n", + "v_e=20 # m/s # velocity at which the bullet escapes through the surface of the bob # here the notation for v_a is shown by v_e\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# Momentum of the bullet & the bob before impact is,\n", + "M=(m_a*v_a)+(m_b*v_b) # kg.m/s......(eq'n 1)\n", + "# The common velocity v_c ( we use v_c insted of v' for notation of common velocity) is given by equating eq'n 1 & eq'n 2 as,\n", + "# (a) When the bullet gets embedded into the bob\n", + "v_c=M/(m_a+m_b) # m/s\n", + "# The height h to which the bob rises is given by eq'n 3 as,\n", + "h_1=(1/2)*(v_c**2/g) # m\n", + "# The angle (theta_1) by which the bob swings corresponding to the value of height h_1 is,\n", + "theta_1=math.degrees(math.acos((l-h_1)/l)) # degree\n", + "# (b) When the bullet rebounds from the surface of the bob\n", + "# The velocity of the bob after the rebound of the bullet from its surface is given by equating eq'n 1 & eq'n 4 as,\n", + "v_bob_rebound=M-(m_a*v_r) # m/s # here v_bob_rebound=v'_b\n", + "# The equation for the height which the bob attains after impact is,\n", + "h_2=(v_bob_rebound**2)/(2*g) # m\n", + "# The corresponding angle of swing \n", + "theta_2=math.degrees(math.acos((l-h_2)/l)) # degree\n", + "# (c) When the bullet pierces and escapes through the bob\n", + "# From eq'n 1 & 5 the velocity attained by the bob after impact is given as,\n", + "v_b_escape=M-(m_a*v_e) # m/s # here we use, v_b_escape insted of v'_b\n", + "# The equation for the height which the bob attains after impact is,\n", + "h_3=(v_b_escape**2)/(2*g) # m\n", + "# The corresponding angle of swing \n", + "theta_3=math.degrees(math.acos((l-h_3)/(l))) # degree\n", + "# Results\n", + "print('(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is %f degree'%theta_1)\n", + "print('(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is %f degree'%theta_2)\n", + "print('(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is %f degree'%theta_3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.9 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compression of the spring over and above caused by the static action of weight W_a is 7.502999 cm \n", + "\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "W_a=50 # N # falling weight\n", + "W_b=50 # N # weight on which W_a falls\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "m_a=W_a/g # kg # mass of W_a\n", + "m_b=W_b/g # kg # mass of W_b\n", + "k=2*10**3 # N/m # stiffness of spring\n", + "h=0.075 # m # height through which W_a falls\n", + "# The velocity of weight W_a just before the impact and after falling from a height of h is given from the eq'n, ( Principle of conservation of energy)\n", + "v_a=math.sqrt(2*g*h) # m/s\n", + "# Let the mutual velocity after the impact be v_m (i.e v_m=v'), (by principle of conservation of momentum)\n", + "v_m=(m_a*v_a)/(m_a+m_b) # m/s\n", + "# Initial compression of the spring due to weight W_b is given by,\n", + "delta_st=(W_b/k)*(10**2) # cm\n", + "# Let the total compression of the spring be delta_t, Then delta_t is found by finding the roots from the eq'n:\n", + "#delta_t**2-0.1*delta_t-0.000003=0. In this eq'n let,\n", + "a=1\n", + "b=-0.1\n", + "c=-0.000003\n", + "delta_t=((-b+(math.sqrt(b**2-(4*a*c))))/2*a)*(10**2) # cm # we consider the -ve value\n", + "delta=delta_t-delta_st # cm\n", + "# Results\n", + "print('The compression of the spring over and above caused by the static action of weight W_a is %f cm \\n'%delta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.10 Principle of conservation of Momentum" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The distance through which the block is displaced from its initial position is 0.908325 m\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "v_a=600 # m/s # velocity of the bullet before impact\n", + "v_b=0 # m/s # velocity of the block before impact\n", + "w_b=0.25 # N # weight of the bullet\n", + "w_wb=50 # N # weight of wodden block\n", + "mu=0.5 # coefficient of friction between the floor and the block\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "m_a=w_b/g # kg # mass of the bullet\n", + "m_b=w_wb/g # kg # mass of the block\n", + "# Let the common velocity be v_c which is given by eq'n (Principle of conservation of momentum)\n", + "v_c=(w_b*v_a)/(w_wb+w_b) # m/s\n", + "# Let the distance through which the block is displaced be s, Then s is given by eq'n\n", + "s=v_c**2/(2*g*mu) # m\n", + "# Results\n", + "print('The distance through which the block is displaced from its initial position is %f m'%s)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 18.11 Principle of conservation of Energy Momentum and work and energy" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance to penetration to the pile is 79.022132 kN\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "M=750 # kg # mass of hammer\n", + "m=200 # kg # mass of the pile\n", + "h=1.2 # m # height of fall of the hammer\n", + "delta=0.1 # m # distance upto which the pile is driven into the ground\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Caculations\n", + "# The resistance to penetration to the pile is given by eq'n,\n", + "R=(((M+m)*g)+((M**2*g*h)/((M+m)*delta)))*(10**-3) # kN \n", + "# Results\n", + "print('The resistance to penetration to the pile is %f kN'%R)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |