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diff --git a/Engineering_Mechanics_by_A._K._Tayal/Chapter20.ipynb b/Engineering_Mechanics_by_A._K._Tayal/Chapter20.ipynb new file mode 100644 index 00000000..1f3a63ab --- /dev/null +++ b/Engineering_Mechanics_by_A._K._Tayal/Chapter20.ipynb @@ -0,0 +1,448 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20 Motion of Projectile" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.1 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the projectile is 435.272891 m/s\n", + "The direction of the projectile is 5.841405 degree\n" + ] + } + ], + "source": [ + "import math\n", + "# Initilization of variables\n", + "v_o=500 # m/s # velocity of the projectile\n", + "alpha=30 # angle at which the projectile is fired\n", + "t=30 # seconds\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "v_x=v_o*(math.cos(alpha*math.pi/180)) # m/s # Initial velocity in the horizontal direction\n", + "v_y=v_o*(math.sin(alpha*math.pi/180)) # m/s # Initial velocity in the vertical direction\n", + "# MOTION IN HORIZONTA DIRECTION:\n", + "V_x=v_x # m/s # V_x=Horizontal velocity after 30 seconds\n", + "# MOTION IN VERTICAL DIRECTION: # using the eq'n v=u+a*t\n", + "V_y=v_y-(g*t) # m/s # -ve sign denotes downward motion\n", + "# Let the Resultant velocity be v_R. It is given as,\n", + "v_R=math.sqrt((V_x)**2+(-V_y)**2) # m/s\n", + "theta=math.degrees(math.atan((-V_y)/V_x)) # degree # direction of the projectile\n", + "# Results\n", + "print('The velocity of the projectile is %f m/s'%v_R) \n", + "# The answer of velocity is wrong in the text book.\n", + "print('The direction of the projectile is %f degree'%theta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.2 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The velocity of body B for horizontal range is 9.306049 m/s\n", + "(b) The velocity of body B for the maximum height is 12.247449 m/s\n", + "(c) The velocity of body B for equal time of flight is 12.247449 m/s\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "v_A=10 # m/s # velocity of body A\n", + "alpha_A=60 # degree # direction of body A\n", + "alpha_B=45 # degree # direction of body B\n", + "# Calculations\n", + "# (a) The velocity (v_B) for the same range is given by eq'n;\n", + "v_B=math.sqrt((v_A**2*math.sin(2*alpha_A*math.pi/180))/(math.sin(2*alpha_B*math.pi/180))) # m/s\n", + "# (b) Now velocity v_B for the same maximum height is given as,\n", + "v_b=math.sqrt((v_A**2)*((math.sin(alpha_A*math.pi/180))**2/(math.sin(alpha_B*math.pi/180))**2)) # m/s\n", + "# (c) Now the velocity (v) for the equal time of flight is;\n", + "v=(v_A*math.sin(alpha_A*math.pi/180))/(math.sin(alpha_B*math.pi/180)) # m/s\n", + "# Results\n", + "print('(a) The velocity of body B for horizontal range is %f m/s'%v_B)\n", + "print('(b) The velocity of body B for the maximum height is %f m/s'%v_b)\n", + "print('(c) The velocity of body B for equal time of flight is %f m/s'%v)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.3 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The least velocity with which the ball can be thrown is 9.779954 m/s\n", + "The angle of projection for the same is 60.255119 degree\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "y=3.6 # m # height of the wall\n", + "x_1=4.8 # m # position of the boy w.r.t the wall\n", + "x_2=3.6 # m # distance from the wall where the ball hits the ground\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# The range of the projectile is r, given as,\n", + "r=x_1+x_2 # m\n", + "# Let the angle of the projection be alpha, which is derived and given as,\n", + "alpha=math.degrees(math.atan((y)/(x_1-(x_1**2/r)))) # degree\n", + "# Now substuting the value of alpha in eq'n 3 we get the least velocity (v_o) as;\n", + "v_o=math.sqrt((g*r)/(math.sin(2*alpha*math.pi/180))) # m/s\n", + "# Results\n", + "print('The least velocity with which the ball can be thrown is %f m/s'%v_o)\n", + "print('The angle of projection for the same is %f degree'%alpha)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.5 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time difference between the two hits is 67.908579 seconds\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "v_o=400 # m/s # initial velocity of each gun\n", + "r=5000 # m # range of each of the guns\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "pi=180 # degree \n", + "# Calculations\n", + "# now from eq'n 1\n", + "theta_1=math.degrees(math.asin((r*g)/(v_o**2)))/(2) \n", + "# from eq'n 3\n", + "theta_2=(pi-(2*theta_1))/2 # degree \n", + "# For 1st & 2nd gun, s is\n", + "s=r # m\n", + "# For 1st gun \n", + "v_x=v_o*math.cos(theta_1*math.pi/180) # m/s\n", + "# Now the time of flight for 1st gun is t_1, which is given by relation,\n", + "t_1=s/(v_x) # seconds\n", + "# For 2nd gun\n", + "V_x=v_o*math.cos(theta_2*math.pi/180)\n", + "# Now the time of flight for 2nd gun is t_2\n", + "t_2=s/(V_x) # seconds\n", + "# Let the time difference between the two hits be delta.T. Then,\n", + "deltaT=t_2-t_1 # seconds\n", + "# Results\n", + "print('The time difference between the two hits is %f seconds'%deltaT)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.6 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pilot should release the bomb from a distance of 3028.912664 m\n", + "The angle at which the target would appear is 33.436901 degree\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "h=2000 # m/ height of the plane\n", + "v=540*(1000/3600) # m/s # velocity of the plane\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# Time t required to travel down a height 2000 m is given by eq'n,\n", + "u=0 # m/s # initial velocity\n", + "t=math.sqrt((2*h)/(g)) # seconds\n", + "# Now let s be the horizonta distance travelled by the bomb in time t seconds, then\n", + "s=v*t # m\n", + "# angle is given as theta,\n", + "theta=math.degrees(math.atan(h/s)) # degree\n", + "# Results\n", + "print('The pilot should release the bomb from a distance of %f m'%s)\n", + "print('The angle at which the target would appear is %f degree'%theta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.7 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The maximum height to which the bullet will rise above the soldier is 127.420999 m\n", + "(b) The velocity with which the bullet will hit the target is 104.790267 m/s\n", + "(c) The time required to hit the target is 11.111111 seconds\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "theta=30 # degree # angle at which the bullet is fired\n", + "s=-50 # position of target below hill\n", + "v=100 # m/s # velocity at which the bullet if fired\n", + "g=9.81 # m/s**2 \n", + "# Calculations\n", + "v_x=v*math.cos(theta*math.pi/180) # m/s # Initial velocity in horizontal direction\n", + "v_y=v*math.sin(theta*math.pi/180) # m/s # Initial velocity in vertical direction\n", + "# (a) Max height attained by the bullet\n", + "h=v_y**2/(2*g) # m\n", + "# (b)Let the vertical Velocity with which the bullet will hit the target be V_y. Then,\n", + "V_y=math.sqrt((2*-9.81*s)+(v_y)**2) # m/s # the value of V_y is +ve & -ve\n", + "# Let V be the velocity with wich it hits the target\n", + "V=math.sqrt((v_x)**2+(V_y)**2) # m/s\n", + "# (c) The time required to hit the target\n", + "a=g # m/s**2\n", + "t=(v_y-(-V_y))/a # seconds\n", + "# Results\n", + "print('(a) The maximum height to which the bullet will rise above the soldier is %f m'%h)\n", + "print('(b) The velocity with which the bullet will hit the target is %f m/s'%V)\n", + "print('(c) The time required to hit the target is %f seconds'%t)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.8 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The distance x where the hammer hits the round is 5.159708 m\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "W=30 # N # Weight of the hammer\n", + "theta=30 # degree # ref fig.20.12\n", + "mu=0.18 # coefficient of friction\n", + "s=10 # m # distance travelled by the hammer # fig 20.12\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# The acceleration of the hammer is given as,\n", + "a=g*((math.sin(theta*math.pi/180))-(mu*math.cos(theta*math.pi/180))) # m/s**2\n", + "# The velocity of the hammer at point B is,\n", + "v=math.sqrt(2*a*s) # m/s\n", + "# Let the initial velocity of the hammer in horizontal direction be v_x & v_y in vertical direction, Then,\n", + "v_x=v*math.cos(theta*math.pi/180) # m/s\n", + "v_y=v*math.sin(theta*math.pi/180) # m/s\n", + "# MOTION IN VERTICAL DIRECTION\n", + "# Now, let time required to travel vertical distance (i.e BB'=S=5 m) is given by finding the roots of the second degree eq'n as,\n", + "# From the eq'n 4.9*t**2+4.1*t-5=0,\n", + "a=4.9\n", + "b=4.1\n", + "c=-5\n", + "# The roots of the eq'n are,\n", + "t=((-b)+(math.sqrt(b**2-(4*a*c))))/(2*a)\n", + "# MOTION IN HORIZONTAL DIRECTION\n", + "# Let the horizotal distance travelled by the hammer in time t be s_x.Then,\n", + "s_x=v_x*math.cos(theta*math.pi/180)*t # m\n", + "x=1+s_x # m\n", + "# Results\n", + "print('The distance x where the hammer hits the round is %f m'%x)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.9 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The velocity of shell just before bursting is 500.255037 m/s\n", + "(b) The velocity of first part immediately after the shell burst is 1000.510074 m/s\n", + "(c) The velocity of second part immediately after the shell burst is 1000.510074 m/s\n", + "(b) The distance between the firing point & the point where the second part of the shell hit the ground is 3000.000000 m\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "s=1000 # m # distance OB (ref fig.20.13)\n", + "h=19.6 # m # height of shell from ground\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# MOTION OF ENTIRE SHELL FROM O to A.\n", + "v_y=math.sqrt(2*(g)*h) # m/s # initial velocity of shell in vertical direction\n", + "t=v_y/g # seconds # time taken by the entire shell to reach point A\n", + "v_x=s/t # m/s # velocity of shell in vertical direction\n", + "# VELOCITIES OF THE TWO PARTS OF THE SHELL AFTER BURSTING AT A:\n", + "# Let v_x2 be the horizontal velocity of 1st & the 2nd part after bursting which is given as,\n", + "v_x2=v_x*2 # m/s\n", + "# Now distance BC travelled by part 2 is\n", + "BC=v_x2*t # m\n", + "# Distance from firing point OC\n", + "OC=s+BC # m\n", + "# Results\n", + "print('(a) The velocity of shell just before bursting is %f m/s'%v_x)\n", + "print('(b) The velocity of first part immediately after the shell burst is %f m/s'%v_x2)\n", + "print('(c) The velocity of second part immediately after the shell burst is %f m/s'%v_x2)\n", + "print('(b) The distance between the firing point & the point where the second part of the shell hit the ground is %f m'%OC)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 20.10 Motion of Projectile" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The range covered (i.e BC) is 7650.920998 m\n" + ] + } + ], + "source": [ + "# Initilization of variables\n", + "v_o=200 # m/s # initial velocity\n", + "theta=60 # degree # angle of the incline\n", + "y=5 # rise of incline\n", + "x=12 # length of incline\n", + "g=9.81 # m/s**2 # acc due to gravity\n", + "# Calculations\n", + "# The angle of the inclined plane with respect to horizontal\n", + "beta=math.degrees(math.atan(y/x)) # degree\n", + "# The angle of projection with respect to horizontal\n", + "alpha=90-theta # degree\n", + "# Range is given by eq'n (ref. fig.20.14)\n", + "AB=(2*v_o**2*(math.sin((alpha-beta)*math.pi/180))*math.cos(alpha*math.pi/180))/(g*(math.cos(beta*math.pi/180))**2) # m\n", + "# Range AC when the short is fired down the plane\n", + "AC=(2*v_o**2*(math.sin((alpha+beta)*math.pi/180))*math.cos(alpha*math.pi/180))/(g*(math.cos(beta*math.pi/180))**2) # m\n", + "BC=AB+AC # m\n", + "# Results\n", + "print('The range covered (i.e BC) is %f m'%BC)" + ] + } + ], + "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 +} |