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+{
+ "cells": [
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "%plot --format svg"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "setdefaults"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "pkg load odepkg"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Archimedian Spiral\n",
+ "A particle is moving in an Archimedian Spiral, so the position of the particle is\n",
+ "defined by $\\bar{r}=at\\cos(\\theta)\\hat{i}+at\\sin(\\theta)\\hat{j}$ meters and $\\theta=t/\\pi$\n",
+ "radians with t measured in seconds. Derive the velocity and acceleration as a function of\n",
+ "time. Use a plotting software to make three plots: plot velocity in the x- and\n",
+ "y-directions vs time, acceleration in the x- and y-direction vs time and the position of\n",
+ "the particle ($\\hat{i}-$ and $\\hat{j}-$ components) from 0 to $10\\pi$ seconds. Use a=3 m/s\n",
+ "for the plots. "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "t=linspace(0,10*pi);\n",
+ "a=1;\n",
+ "r=[a*t.*cos(t/pi);a*t.*sin(t/pi)];\n",
+ "v=[a*cos(t/pi)-a*t/pi.*sin(t/pi);a*sin(t/pi)+a*t/pi.*cos(t/pi)];\n",
+ "a=[-a*t/pi^2.*cos(t/pi)-2*a/pi*sin(t/pi); -a*t/pi^2.*sin(t/pi)+2*a/pi*cos(t/pi)];\n",
+ "\n",
+ "figure(1)\n",
+ "plot(r(1,:),r(2,:))\n",
+ "xlabel('x-position (m)')\n",
+ "ylabel('y-position (m)')"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "figure(2)\n",
+ "title('velocity')\n",
+ "plot(t,v(1,:),t,v(2,:))\n",
+ "xlabel('time (s)')\n",
+ "ylabel('velocity (m/s)')\n",
+ "legend('x-component','y-component')"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ ""
+ ],
+ "text/plain": [
+ ""
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "figure(3)\n",
+ "title('acceleration')\n",
+ "plot(t,a(1,:),t,a(2,:))\n",
+ "xlabel('time (s)')\n",
+ "ylabel('acceleration (m/s/s)')\n",
+ "legend('x-component','y-component')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Foucalt's pendulum\n",
+ "\n",
+ "Foucalt's pendulum is a design to demonstrate the Coriolis effect. It consists of a\n",
+ "mass on a long string connected to a ball joint. As the Earth rotates, the path of the\n",
+ "pendulum changes direction. The analytical solution for the rotation of Foucalt's pendulum\n",
+ "is based upon the coupled ordinary differential equations for the Coriolis Effect:\n",
+ "\n",
+ "$\\ddot{x}-2\\omega_{e}\\dot{y}\\sin\\lambda=\\frac{F_{x}}{m}$\n",
+ "\n",
+ "$\\ddot{y}+2\\omega_{e}(\\dot{x}\\sin\\lambda-\\dot{z}\\cos\\lambda)=\\frac{F_{y}}{m}$\n",
+ "\n",
+ "$\\ddot{z}+2\\omega_{e}\\dot{y}\\cos\\lambda=\\frac{F_{z}}{m}-g$\n",
+ "\n",
+ "Where x,y, and z describe the position of the pendulum, $\\lambda$ is the latitude of the\n",
+ "pendulum, $F_{x},~F_{y},$ and $F_{z}$ are the applied forces in the x-, y-, and\n",
+ "z-directions, $\\omega_{e}$ is the rotation of the Earth. Show that these equations can be\n",
+ "reduced for a swinging pendulum to:\n",
+ "\n",
+ "$\\ddot{x}=2\\omega_{e}\\sin\\lambda\\dot{y}-\\frac{g}{l}x$ (eq. 1)\n",
+ "\n",
+ "$\\ddot{y}=-2\\omega_{e}\\sin\\lambda\\dot{x}-\\frac{g}{l}y$ (eq. 2)\n",
+ "\n",
+ "$z=-(l^{2}-x^{2}-y^{2})^{1/2}$ (eq. a)\n",
+ "\n",
+ "The analytical solution returns $y=-x\\tan(\\omega_{e}\\sin\\lambda t)$. \n",
+ "\n",
+ "Rewrite (1) and (2) as 4 first order differential equations for\n",
+ "$[\\dot{x},~\\ddot{x},~\\dot{y},~\\ddot{y}]$ and create a function `myode.m` that returns `dr`\n",
+ "when `r` and `t` are given `dr=myode(r,t)`. Use $l=7$ m, $\\lambda$=41.8084$^{\\circ}$\n",
+ "\n",
+ "Use `[t,r]=ode45(myode,[0 30000],[1 0 0 0]);` to solve the two coupled ordinary\n",
+ "differential equations from 0 to 30000 seconds and initial conditions $x(0)$=1 m;\n",
+ "$\\dot{x}(0)=y(0)=\\dot{y}(0)=0$. \n",
+ "\n",
+ "Plot a comparison of the numerically computed $-y/x$ with the analytically derived,\n",
+ "$-y/x=\\tan(\\omega_{e}\\sin\\lambda t)$.\n",
+ "\n",
+ "If the analytical solution and numerical solution agree, plot the 3D path for the first\n",
+ "100 time steps, 100 time steps in the middle, and the last 100 time steps.\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "function ddr = myode(t,r,l,L)\n",
+ " g=9.81; % acceleration due to gravity m/s^2\n",
+ " R=6378.1e3; % radius of Earth in m\n",
+ " we=2*pi/23.934/3600; % rotation of Earth (each day is 23.934 hours long)\n",
+ " ddr=zeros(4,1); % initialize ddr\n",
+ "\n",
+ " ddr(1) = r(2); % x North(+) South (-)\n",
+ " ddr(2) = 2*we*r(4).*sin(L)-g/l*r(1); % dx/dt \n",
+ " ddr(3) = r(4); % y West (+) East (-)\n",
+ " ddr(4) = -2*we*(r(2).*sin(L))-g/l*r(3); % dy/dt\n",
+ "end"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "L=41.8084;\n",
+ "L=L*pi/180;\n",
+ "[t,r]=ode45(@(t,r) myode(t,r,7,L),[0 30000], [1 0 0 0 ]);\n",
+ "figure()\n",
+ "z=-sqrt(10^2-r(:,1).^2-r(:,3).^2);\n",
+ "we=2*pi/23.934/3600; % rotation of Earth (each day is 23.934 hours long)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "data": {
+ "image/svg+xml": [
+ "