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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 22 Kinetics of Rigid Body Force and Acceleration"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 22.1 Relation between the translatory motion and rotary motion of a body in plane motion"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The no of revolutions executed by the disc before coming to rest is 1500\n",
+ "(b) The frictional torque is -5.003811 N-m\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "#Initialization of variables\n",
+ "N=1500 # r.p.m\n",
+ "r=0.5 # m , radius of the disc\n",
+ "m=300 # N , weight of the disc\n",
+ "t=120 #seconds , time in which the disc comes to rest\n",
+ "omega=0 \n",
+ "g=9.81 #m/s**2\n",
+ "#Calculations\n",
+ "omega_0=(2*math.pi*N)/60 #rad/s\n",
+ "#angular deceleration is given as,\n",
+ "alpha=-(omega_0/t) #radian/second**2\n",
+ "theta=(omega_0**2)/(2*(-alpha)) #radian\n",
+ "#Let n be the no of revolutions taken by the disc before it comes to rest, then\n",
+ "n=theta/(2*math.pi)\n",
+ "#Now,\n",
+ "I_G=((1/2)*m*r**2)/g\n",
+ "#The frictional torque is given as,\n",
+ "M=I_G*alpha #N-m\n",
+ "#Results\n",
+ "print('(a) The no of revolutions executed by the disc before coming to rest is %d'%n)\n",
+ "print('(b) The frictional torque is %f N-m'%M)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 22.2 Relation between the translatory motion and rotary motion of a body in plane motion"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The acceleration at the centre is 4.896389 m/s**2\n",
+ "(b) The maximum angle of the inclined plane is 29.941943 degree\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Initilization of variables\n",
+ "s=1 # m\n",
+ "mu=0.192 # coefficient of static friction\n",
+ "g=9.81 # m/s**2\n",
+ "# Calculations\n",
+ "# The maximum angle of the inclined plane is given as,\n",
+ "theta=math.degrees(math.atan(3*mu)) # degree\n",
+ "a=(2/3)*g*math.sin(theta*180/math.pi) # m/s**2 # by solving eq'n 4\n",
+ "v=math.sqrt(2*a*s) # m/s\n",
+ "# Let the acceleration at the centre be A which is given as,\n",
+ "A=g*math.sin(theta*math.pi/180) # m/s**2 # from eq'n 1\n",
+ "# Results\n",
+ "print('(a) The acceleration at the centre is %f m/s**2'%A)\n",
+ "print('(b) The maximum angle of the inclined plane is %f degree'%theta)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 22.5 Relation between the translatory motion and rotary motion of a body in plane motion"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The acceleration of weight A is 1.081102 m/s**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Initilization of variables\n",
+ "W_a=25 # N \n",
+ "W_b=25 # N \n",
+ "W=200 # N # weight of the pulley\n",
+ "i_g=0.2 # m # radius of gyration\n",
+ "g=9.81 # m/s^2\n",
+ "# Calculations\n",
+ "# Solving eqn's 1 & 2 for acceleration of weight A (assume a)\n",
+ "a=(0.15*W_a*g)/(((W*i_g**2)/(0.45))+(0.45*W_a)+((0.6*W_b)/(3))) # m/s^2\n",
+ "# Results\n",
+ "print('The acceleration of weight A is %f m/s**2'%a)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 22.8 Relation between the translatory motion and rotary motion of a body in plane motion"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The acceleration of the pool is 1.615819 m/s**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Initilization of variables\n",
+ "r_1=0.075 # m\n",
+ "r_2=0.15 # m\n",
+ "P=50 # N\n",
+ "W=100 # N\n",
+ "i_g=0.05 # m\n",
+ "theta=30 # degree\n",
+ "g=9.81 # m/s^2\n",
+ "# Calculations\n",
+ "# The eq'n for acceleration of the pool is given by solving eqn's 1,2 &3 as,\n",
+ "a=(50*g*(r_2*math.cos(theta*math.pi/180)-r_1))/(100*((i_g**2/r_2)+r_2)) # m/s**2\n",
+ "# Results\n",
+ "print('The acceleration of the pool is %f m/s**2'%a)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Example 22.10 Relation between the translatory motion and rotary motion of a body in plane motion"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The angular velocity of the rod is 4.101219 rad/sec\n",
+ "(b) The reaction at the hinge is 103.227964 N\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Initilization of variables\n",
+ "L=1 # m # length of rod AB\n",
+ "m=10 # kg # mass of the rod\n",
+ "g=9.81 \n",
+ "theta=30 # degree\n",
+ "# Calculations\n",
+ "# solving eq'n 4 for omega we get,\n",
+ "omega=math.sqrt(2*16.82*math.sin(theta*math.pi/180)) # rad/s\n",
+ "# Now solving eq'ns 1 &3 for alpha we get,\n",
+ "alpha=(12/7)*g*math.cos(theta*math.pi/180) # rad/s\n",
+ "# Components of reaction are given as,\n",
+ "R_t=((m*g*math.cos(theta*math.pi/180))-((m*alpha*L)/4)) # N\n",
+ "R_n=((m*omega**2*L)/(4))+(m*g*math.sin(theta*math.pi/180)) # N\n",
+ "R=math.sqrt(R_t**2+R_n**2) # N \n",
+ "# Results\n",
+ "print('(a) The angular velocity of the rod is %f rad/sec'%omega)\n",
+ "print('(b) The reaction at the hinge is %f N'%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",
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+}