Download group quiz 1 solution

Group problem:
When investigating the motion of objects dropped in fluids, you need to determine how the
position and velocity of the object depend on time. You know the ratio of the density of the fluid
to the density of the object, and the terminal velocity. You don’t know whether the resistance
force depends on velocity or velocity squared so you do both cases to compare. If you can get an
analytic solution for either case, show it, describe how you got it, and sketch plot of position and
velocity as a function of time for the first 3 seconds. Use the following numbers to make your
plots: g = 10 m/s2; object density = 3 times fluid density; terminal velocity = 5 m/s. If you
cannot get an analytic solution, get a numerical solution to make your plots. For any part of this
problem that you use Matlab, give the Matlab script including comments.
v
R
B
W
+y
W = weight = mg
B = buoyant force = weight of fluid displaced
R = fluid resistance = b1v or b2v2
vT = terminal velocity
B = mfluid g = rfluid Vobject g
Vobject = m/ robject
B = rfluid (m/ robject) g
B = (rfluid / robject) m g= r m g
F = ma
W − B − R = my
mg– rmg − b/ y = my for a resistance force proportional to v.
Terminal velocity when a = 0
mg– rmg − b/ v1 = 0
mg– rmg
= b/ v1
Equation of motion for a resistance force proportional to v.
mg– rmg
mg– rmg − y = my
v1
mg
mg 1 − r –
1 − r y = my
v1
g
g 1 − r –
1 − r y = y
v1
Call g 1 − r = K
K
y = y
v1
Need a function where the 2nd derivative has the same form as the first derivative and a constant.
K–
Guess: y = AeBt +Ct +D
y = ABe78 + C
y = AB ; e78
K
ABe78 + C = AB ; e78
v1
K
K
K– ABe78 − C = AB ; e78
v1
v1
K–
K
K
C = AB ; e78 + ABe78
v1
v1
Can only be true if
K−
K −
K
C = 0
v1
1 −
And
/
<=
C = 0
C = v1
K
ABe78
v1
>
0 = B + 0 = AB ; e78 + <=
>
B = − <=
?
y = Ae
>
8
<=
+ v1 t + D
Initial conditions @t=0: y(0) = 0, v(0) = 0
0=𝐴+0+𝐷
D = -A
y = −A
K ?<> 8
e = + v1
v1
0 = −A
K
+ v1
v1
A=
y=
v1;
>
>
? 8
e <=
K
Position
y=
<D=
+ v1 t −
v1;
K
v1; ?<> 8
e = − 1 + v1 t
K
Velocity
?
v = −v1 e
>
8
<=
?
v = v1 1 − e
+ v1
>
8
<=
Make a graph using g = 10 m/s2 and r = 1/3 which means K = 20/3, and vT = 5 m/s
Checked by using dsolve in Matlab
Matlab got
y = t*vT - vT^2/(g - g*r) + (vT^2*exp(-(t*(g - g*r))/vT))/(g - g*r)
Changing the notation
y = v1 t −
E /?F
v1;
v1;
?
8
+
e <=
g 1−r
g 1−r
y = v1 t −
v1; v1; ?>8
+ e
K
K
y=
<D=
>
e?>8 − 1 + v1 t agrees with my solution.
Matlab script
%fluid drop v resistance
%specify symbols
syms g r vT y(t) y1(t) y2(t);
%specify first derivative
Dy=diff(y,t);
%specify second derivative
D2y=diff(y,t,2);
%specify differential equation to be solved
%air resistance proportional to velocity
%force equation mD2y = mg (1- fluid density/object density)(1-Dy/terminal
%velocity)
equ=D2y==g*(1-r)*(1-Dy/vT);
%specify specific conditions (initial position = 0,
%initial velocity = 0)
cond=[y(0)==0,Dy(0)==0];
%solve the differential equation
dsolve(equ,cond)
%plot result, need numbers
%gravitational acceleration = 10m/s^2
g=10;
%buoyancy fluid density/object density = 3
r=1/3;
%terminal velocity = 5 m/s
vT=5;
equn=D2y==g*(1-r)*(1-Dy/vT);
y2=dsolve(equn,cond)
fplot(y2,[0,3],'-b')
xlabel('time in seconds')
ylabel('position in meters in green/blue, velocity in m/s in red/magenta')
hold on
y1=diff(y2,t);
fplot(y1,[0,3],'-m')
hold on
Now get the equation of motion for a v2 fluid resistance – getting the equation of motion is
almost the same.
mg– rmg − b/ y ; = my for a resistance force proportional to v2.
Terminal velocity when a = 0
mg– rmg − b/ v1 ; = 0
mg– rmg
= b/ v1 ;
Equation of motion for a resistance force proportional to v.
mg– rmg ;
mg– rmg − y = my
v1 2
mg
mg 1 − r – 2 1 − r y ; = my
v1
g
g 1 − r – 2 1 − r y ; = y
v1
Call g 1 − r = K
K–
K
y ; = y
2
v1
Not easy to guess the solution, use Matlab
From Matlab
y = (vT^2*log(1 - tanh((g*t*(r - 1))/vT)^2))/(2*g*(r - 1))
Changing to my notation
v1;
tanh −Kt
y=−
log 1 −
2K
v1;
and
v = -vT*tanh((g*t*(r - 1))/vT)
v = −v1 tanh
−Kt
v1
Make a graph using g = 10 m/s2 and r = 1/3 which means K = 20/3, and vT = 5 m/s
Matlab script
%fluid drop v^2 resistance
%specify symbols
syms g dr vT y(t) y1(t) y2(t);
%specify first derivative
Dy=diff(y,t);
%specify second derivative
D2y=diff(y,t,2);
%specify differential equation to be solved
%air resistance proportional to velocity squared
%force equation mD2y = mg (1- fluid density/object density)(1-Dy^2/terminal
%velocity^2)
equ=D2y==g*(1-dr)*(1-Dy^2/vT^2);
%specify specific conditions (initial position = 0,
%initial velocity = 0)
cond=[y(0)==0,Dy(0)==0];
%solve the differential equation
y1=dsolve(equ,cond)
y2=diff(y1,t)
%plot result
%need numbers for plots
%gravitational acceleration = 10m/s^2
g=10;
%buoyancy object density/fluid density = 3
dr=1/3;
%terminal velocity = 5 m/s
vT=5;
equn=D2y==g*(1-dr)*(1-Dy^2/vT^2);
y1=dsolve(equn,cond);
y2=diff(y1,t);
fplot(y1,[0,3],'-b')
xlabel('time in seconds')
ylabel('position in meters in green/blue, velocity in m/s in red/magenta')
hold on
fplot(y2,[0,3],'-m')
hold on