terminal velocity. wy not ) dies à If not, E2.5 A small sphere moving though a l
ID: 3677158 • Letter: T
Question
terminal velocity. wy not ) dies à If not, E2.5 A small sphere moving though a luid at a slow velocity will have a drag force acting on it which is described by Stokes' law. The sphere could be a dust particle or a raindrop moving in air or a ball bearing moving in oil. The drag force described by Stokes' law is (2.15) where D is the drag R is the radius of the sphere is the viscosity of the fluid Vis the velocity of sphere Let us consider a ball bearing dropped in oil (see Figure 2.13) with an initial velocity of zero. We want to determine the velocity and position as a function of time. The forces acting on the ball bearing are the gravitational force, W buoyancy force, B; and the drag force, D. The buoyancy force is equal the weight of the fluid displaced.Explanation / Answer
matlab program:
%Declare all variables
u=3.85;
R=0.01;
psteel=7910;
poil=899;
g=9.81;
%Evaluate formulas
B=poil*g*(4/3)*pi*R^3
W=psteel*g*(4/3)*pi*R^3
VT=((g*4*R^2)/(18*u))*(psteel-poil)
i=1;
%run for loop for vector V
for t=0:0.005:0.05
V(i)=VT*(1-exp(((-6*pi*R*u*g)/W)*t));
i=i+1;
end
t=0:0.005:0.05;
A = [t;V];
%store the t,V in table formate in a file
fileID = fopen('exp.txt','w');
fprintf(fileID,'%6s %12s ','t','V');
fprintf(fileID,'%6.4f %12.8f ',A);
A = [B;W;VT];
fprintf(fileID,'%6s %12s %15s ','B','W','VT');
fprintf(fileID,'%6.4f %12.8f %12.8f ',A);
fclose(fileID);
%view the content of a file
type exp.txt
output:
B =
0.0369
W =
0.3250
VT =
0.3970
t V
0.0000 0.00000000
0.0050 0.04117930
0.0100 0.07808709
0.0150 0.11116644
0.0200 0.14081448
0.0250 0.16738713
0.0300 0.19120341
0.0350 0.21254924
0.0400 0.23168086
0.0450 0.24882797
0.0500 0.26419642
B W VT
0.0369 0.32503797 0.39698649
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