Problem 2.6

By Nasser Abbasi

Part (a)

SOURCE CODE

% nma_balle_HW_2_6 - Program to analyse Galilo 2 inches claim

%by Nasser Abbasi, HW 2.6 (a)

%PHYS 240, San Jose State Univ.

%

% Algorithm:

%

% Drop the heavy ball, from rest, from a height of Y

% meters, vertically downwards. record the time it takes

% to reach the ground using Cd=0.5.

%

% Then at this time, according to Galilo, the small ball was

% 2 inches above the ground.

%

% So, use this time and run the simulatiom again using the small ball

% and find what Cd is needed to cause the small ball to be 2 inches above

% the ground when the time, found in the first simulation, has elapsed.

%

% i.e. drop the small ball, and keep adjusting the Cd, such that

% after the above amount of time, the small ball would have travled

% all the way to the ground less 2 inches.

%

% determine if such a Cd is realistic.

%

clear; help nma_balle_HW_2_6; % Clear memory and print header

DELTA = 0.0508; % 2 inches expressed in meter units.

tau = input('Enter timestep, tau (sec): '); % (sec)

%*Set initial position and velocity of the baseball

initialY = input('Height to drop the balls from (Meter):');

initialPos = [0, initialY]; % Initial vector position

initialVelosity = [0 0]; % drop from rest

% Get the mass of the two balls, and find A for each.

%use SI units everywhere

mass_1 = input('first ball mass (lb):');

mass_2 = input('second ball mass (lb):');

% keep mass_1 as the larger one

if(mass_1 < mass_2)

temp = mass_1;

mass_1 = mass_2;

mass_2 = temp;

end

mass_1 = 0.4536 * mass_1 %convert from lb to KG

mass_2 = 0.4536 * mass_2 %convert from lb to KG

densityOfMetal = 7.8*1000; % in KG/M^3

volume_1 = mass_1/densityOfMetal; % in M^3

volume_2 = mass_2/densityOfMetal; % in M^3

%find r from Volume= 4/3 PI r^3

%

% V

% r^3= --------

% (4/3 PI)

radius_1= ( volume_1 / ((4/3)*pi) )^(1/3); % in M

radius_2= ( volume_2 / ((4/3)*pi) )^(1/3); % in M

% find drag area, which is the cross sectional area of the sphere

area_1 = pi*(radius_1^2) % in m^2

area_2 = pi*(radius_2^2) % in m^2

% Set physical parameters (mass, Cd, etc.)

Cd = 0.5; % Drag coefficient (dimensionless)

g = 9.81; % Gravitational acceleration (m/s^2) near surface of earth

rho = 1.2; % density of air

airConst = -0.5*Cd*rho*area_1; % Air resistance constant

istep = 0;

currentPos = initialPos;

currentVelosity = initialVelosity; % Set initial position and velocity

while( currentPos(2) > 0 )

istep = istep+1;

currentTime = (istep-1)*tau;

% Calculate the acceleration of the ball

accel = (1/mass_1)*airConst*norm(currentVelosity)*currentVelosity; % Air resistance

accel(2) = accel(2)-g; % Gravity

% Calculate the new position and velocity using Euler method

currentPos = currentPos + (tau* currentVelosity); % Euler step

currentVelosity = currentVelosity + tau*accel;

end

% this is the time it took the heavy ball to reach the ground.

recordedTime = currentTime;

fprintf('Time for heavy ball to reach the ground is %g (sec)\n', recordedTime);

%Now drop the small ball, force it to reach 2 inchess above the ground

%at the same time the large ball above reached the ground. change Cd to

%force this to happen.

isFoundCd = 0;

currentCd = 0.51;

while( ~isFoundCd )

% we keep reducing air resistance, untill the small ball

% reaches 2 inches above the graound at the same time as the

% the large ball

currentCd = currentCd-0.01;

airConst = -0.5*currentCd*rho*area_2; % Air resistance constant

istep = 0;

currentTime = 0;

currentVelosity = initialVelosity;

currentPos = initialPos;

accel = [0 0];

while( currentTime <= recordedTime )

istep = istep+1;

currentTime = (istep-1)*tau; % Current time

%* Calculate the acceleration of the ball

accel = (1/mass_2)*airConst*norm(currentVelosity)*currentVelosity; % Air resis

accel(2) = accel(2)-g; % Gravity

%* Calculate the new position and velocity using Euler method

currentPos = currentPos + (tau* currentVelosity); % Euler step

currentVelosity = currentVelosity + tau*accel;

end

if( currentPos(2) <= DELTA )

isFoundCd=1;

end

end

fprintf('Cd that will cause the small ball to reach the ground 2 inches behind the largers one is=%g\n',currentCd);

fprintf('Speed the small ball was at when it reached 2 inches before the ground =%g (M/S)\n',currentVelosity(2));

fprintf('diameter of small ball=%g Meter\n', radius_1*2);

RUN OUTPUT

» nma_balle_HW_2_6

nma_balle_HW_2_6 - Program to analyse Galilo 2 inches claim

Enter timestep, tau (sec): 0.001

Height to drop the balls from (Meter):50

first ball mass (lb):100

second ball mass (lb):1

mass_1 =

45.3600

mass_2 =

0.4536

area_1 =

0.0391

area_2 =

0.0018

Time for heavy ball to reach the ground is 3.2 (sec)

Cd that will cause the small ball to reach the ground 2 inches behind the largers one is=0.16

Speed the small ball was at when it reached 2 inches before the ground =-31.0135 (M/S)

diameter of small ball=0.223114 Meter

»

Analysis

Cd, the drag coefficient, for a smooth sphere, had to be reduced all the way to 0.16 to cause the small ball to reach the ground 2 inches behind the heavy ball.

Looking at the chart, for Cd=0.16, Reynold number is about 300,000

But

Reynold number = V*diameter/Viscosity.

The small ball had a speed of 31 M/S when it reached 2 inches above the ground, and diameter of the ball is 0.223114 Meter. So Viscosity of air had to be

Viscosity = V*diameter/300,000= 31*0.223114/300,000= 2.3055 x 10^-5 M^2 per second.

Which is about 65% higher than the expected value of 1.5x10^-5 M^2 per second. So, this means the small ball being 2 inches behind the large ball when the large ball reached the ground does not agree with the given number for viscosity of air.

Part ( C )

Cd is 0.16, which is derived in part (a). Please see part(a) for how Cd was found.