# How things work college course/Air drag drop experiment

THIS PAGE NEEDS TO BE MOVED TO A SUPPATE OF Physics and Astronomy Labs

## 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[1]

Drag coefficient Cd for a sphere as a function of Reynolds number Re, as obtained from laboratory experiments. The dark line is for a sphere with a smooth surface, while the lighter line is for the case of a rough surface.

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:

${\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A}$

where

FD is the drag force,
ρ is the density of the fluid,[2]
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:

${\displaystyle R_{e}={\frac {vD}{\nu }}}$

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