How things work college course/Air drag drop experiment

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How much does air drag slow a falling object?
This is a high level investigation that can be appreciated at all levels. We will investigate this experimentally using no equipment except a ruler. Theoretical investigations will be at the advance Matlab/Calculus level.

Initial results
We dropped a variety of objects from the top second floor guardrail outside the Physics lab.
 * The estimated height was either 17 or 20 feet, made using only the one foot squares on floor, body height, and a belief that floors are typically 10 feet.
 * Just for fun, each class will estimate that height using only paper and the one-foot squares on the floor.


 * Clay, steel, and styrofoam balls were dropped. Also dropped was a ping-pong ball.
 * The first question is this: Is a clay ball an acceptable proxy for a steel ball?  (We don't want to drop steel balls because they damage the floor.  In our investigation a cushion was placed on the floor for the steel ball.
 * The answer was "definitely"  bot one inch and half inch diameter clay ball fell as fast as the 3/4 inch steel ball.
 * The required "head start" was about 1 meter for the ping pong ball.
 * Mass = 2.4 gm = .0024 kg
 * Diameter = 40 mm = 0.04 m

Copied from Wikipedia
Drag depends on the properties of the fluid and on the size, shape, and speed of the object. One way to express this is by means of the drag equation:
 * $$F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$$

where
 * FD is the drag force,
 * ρ is the density of the fluid,
 * v is the speed of the object relative to the fluid,
 * A is the cross section area, and
 * CD is the drag coefficient – a dimensionless number.

The drag coefficient depends on the shape of the object and on the Reynolds number:
 * $$R_e=\frac{vD}{\nu}$$

where D is some characteristic diameter or linear dimension and ν is the kinematic viscosity of the fluid (equal to the viscosity μ divided by the density). At low Reynolds number, the drag coefficient is asymptotically proportional to the inverse of the Reynolds number, which means that the drag is proportional to the speed. At high Reynolds number, the drag coefficient is more or less constant. The graph to the right shows how the drag coefficient varies with Reynolds number for the case of a sphere.

Matlab simulation code
This code simulates the situation with and without air drag. function [] = odeNumericalSolver(~) %ODENUMERICALSOLVER numerically solves a non-stiff ode. % It solves and plots the ode associated with an object falling under high % Reynolds number air drag.

% It can be reconfigured if this function is to be called (i.e., if it is % to be called to return Y and T) % There is no need for the next line if this function is to be called: clc;close all; clear all; % The following and all other such parameters must be made global because % the defining function (called odeMat) can take only to input arguments. global drag; drag=.1; timeRange =[0 1.5]; % zero to one seconds initCnds = [17*12*2.54/100,0]; % position, velocity lift=.5; options = odeset('RelTol',1e-4,'AbsTol',[1e-4 1e-4]); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% [T,Y] = ode45(@odeMat,timeRange,initCnds,options); [ynd, vnd]=nodrag(T, initCnds,lift); % does the no-drag case, lifted a bit figure % Is this needed? subplot(1,2,1) % (2 rows, 1 column, plot 1) plot(T,Y(:,1),'.','Markersize',10); hold on plot(T,ynd,'-','Markersize',10); hold off grid; title('position versus time');% subplot(1,2,2) % (2 rows, 1 column, plot 2) plot(T,Y(:,2),'.','Markersize',10); hold on plot(T,vnd,'-','Markersize',10); hold off grid; title('velocity versus time');% end %%%%%%%%%% function dq = odeMat(t,q) % q = [x v] global drag; dq = zeros(2,1);   % a column vector dq(1) = q(2); % dx/dt = v dq(2) = -9.8 + drag*q(2)^2; end

function [ynd vnd]=nodrag(T, initCnds,lift) %calculates y if there is no drag but the ball was lifted higher by lift. y0=initCnds(1)+lift; v0=initCnds(2); ynd=zeros(size(T)); ynd=y0+v0*T-.5*9.8*T.^2; vnd=v0-9.8*T; end