Math 2280-001
Mon Feb 2
2.3 Improved velocity models: velocity-dependent drag forces
For particle motion along a line, with
position x t (or y t ,
velocity x# t = v t , and
acceleration x## t = v# t = a t
We have Newton's 2nd law
m v# t = F
where F is the net force.
, We're very familiar with constant force F = m a , where a is a constant:
v# t = a
v t = a t C v0
1
x t = a t2 C v0 t C x0 .
2
Examples we've seen a lot of:
, a =Kg near the surface of the earth, if up is the positive direction, or a = g if down is the positive
direction.
, boats or cars or "particles" subject to constant acceleration or deceleration.
New today !!! Combine a constant force with a velocity-dependent drag force, at the same time. The text
calls this a "resistance" force:
m v# t = m a C FR
Empirically/mathematically the resistance forces FR depend on velocity, in such a way that their magnitude
is
FR z k v p , 1 % p % 2 .
, p = 1 (linear model, drag proportional to velocity):
m v# t = m a Kk v
This linear model makes sense for "slow" velocities, as a linearization of the frictional force function,
assuming that the force function is differentiable with respect to velocity...recall Taylor series for how the
velocity resistance force might depend on velocity:
1
FR v = FR 0 C FR # 0 v C
F ## 0 v2 C ...
2! R
FR 0 = 0 and for small enough v the higher order terms might be negligable compared to the linear term,
so
FR v z FR # 0 v zKk v .
We write Kk v with k O 0, since the frictional force opposes the direction of motion, so sign opposite of
the velocity's.
http://en.wikipedia.org/wiki/Drag_(physics)#Very_low_Reynolds_numbers:_Stokes.27_drag
Exercise 1a: Rewrite the linear drag model as
v# t = a K r v
k
where the r =
. Construct the phase diagram for v . Notice that v t has exactly one constant
m
(equilibrium) solution, and find it. Its value is called the terminal velocity. Explain why terminal velocity
is an appropriate term of art, based on your phase diagram.
1b) Solve the IVP
v# t = a K r v
v 0 = v0
and verify your phase diagram analysis. (This is, once again, our friend the first order constant coefficient
linear differential equation.)
1c) integrate the velocity function above to find a formula for the position function y t .
, p = 2 , for the power in the resistance force. This can be an appropriate model for velocities which are
not "near" zero....described in terms of "Reynolds number". Accounting for the fact that the resistance
opposes direction of motion we get
m v# t = m a Kk v2 if v O 0
m v# t = m a C k v2 if v ! 0.
http://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_velocity
k
we can rewrite the DE's as
m
v# t = a Kr v2 if v O 0
v# t = a C r v2 if v ! 0.
2a) Consider the case in which a =Kg , so we are considering vertical motion, with up being the positive
direction. Draw the phase diagrams. Note that each diagram contains a half line of v-values. Make
conclusions about velocity behavior in case v0 O 0 and v0 % 0. Is there a terminal velocity?
Exercise 2) Once again letting r =
2b) Set up the two separable differential equation IVPs for the cases above, so that you will be able to
complete finding the solution formulas (in your homework)....Of course, once you find the velocity
function you'll still need to integrate that, if you want to find the position function!
Application: We consider the bow and deadbolt example from the text, page 93-95. It's shot vertically into
m
the air (watch out below!), with an initial velocity of 49
. In the no-drag case, this could just be the
s
vertical component of a deadbolt shot at an angle. With drag, one would need to study a more complicated
system of DE's for the horizontal and vertical motions, if you didn't shoot the bolt straight up.
Exercise 3: First consider the case of no drag, so the governing equations are
m
v# t =Kg zK9.8 2
s
v t =Kg t C v0 =Kg t C 5 g
1
1
x t =K g t2 C v0 t C x0 =K g t2 C 5 g t .
2
2
Find when v = 0 and deduce how long the object rises, how long it falls, and its maximum height.
Maple check:
> restart :
Digits d 5 :
> g d 9.8;
v0 d 49.0;
v1 d t/Kg$t C v0;
1
y1 d t/K $g$t2 C v0$t;
2
Exercise 4: Now consider the linear drag model for the same deadbolt, with the same initial velocity of
m
m
5 g = 49
. We'll assume that our deadbolt has a measured terminal velocity of vt =K245
=K25 g,
s
s
g
so vt = 25 g =
0 r = .04 (convenient). So, from our earlier work:
r
g
g
v =K C v0 C
eKr t =K245 C 294 eK.04 t .
r
r
When does the object reach its maximum height, what is this height, and how long does the object fall?
Compare to the no-drag case with the same initial velocity, in Exercise 3.
Maple check, and then work:
> with DEtools :
> r d .04;
dsolve v# t =Kg K r$v t , v 0 = v0 , v t
1
t
25
K
> v2 d t/K245.0 C 294 e
;
;
t
> y2 d t/0 C
v2 s ds;
0
> v2 t ;
y2 t ;
> solve v2 t = 0, t ;
solve y2 t = 0, t ;
picture:
> with plots :
plot1 d plot y1 t , t = 0 ..10, color = green :
plot2 d plot y2 t , t = 0 ..9.4110, color = blue :
display plot1, plot2 , title = `comparison of linear drag vs no drag models` ;
comparison of linear drag vs no drag models
100
60
0
0
2
4
6
t
>
8
10
Escape Velocity (p. 99-100 text) This is an improvement on the constant acceleration model of vertical
motion, and takes into account the real law of graviational attraction.
Near the surface of earth (and neglecting friction), we tend to use
x## t =Kg
for the acceleration of an object moving vertically, with up being the positive direction for x . This is
actually an approximation to the universal inverse square law of gravitational attraction, which says the
attractive force between two objects of mass m, M has magnitude
GMm
F =
r2
where r is the distance between their centers of mass and G is a universal constant.
Write R for the radius of the earth; write r t for the distance of a projectile object of mass m from the
center of the earth, and consider the initial value problem in which this object starts at r0 z R and is given
an initial vertical velocity v0 . If we write r t = r0 C x t and assume x t is small compared to R, then
Newton's second law and the universal graviational law yields
GMm
GMm
m r ## t = m x## t =KF =K
zK 2 =Km g
2
r t
R
GM
i.e. g = 2 .
R
So our familiar constant gravity acceleration model is a "constantization" model of the truth (as opposed to
a linearization). If x t approaches a significant fraction of R so that r t z R no longer holds, then it
would be more accurate to use the universal law of attraction. Thus, for a projectile shot from the surface
of the earth we'd have the second order DE initial value problem
GM
r ## t =K 2
r
r 0 =R
r # 0 = v0 .
As r t increases, v t will decrease because of the negative acceleration, so the velocity v will be a
decreasing function r , i.e. we can consider
v=v r t .
By the chain rule,
dv
dv dr
dv
r ## t =
=
$
=
$v
dt
dr dt
dr
So we can convert the tKdifferential equation and IVP into the 1st -order separable r-differential equation
dv
GM
v
=K 2
dr
r
v R = v0
Exercise 5
5a) Use separation of variables to find the implicit solution to the IVP for v = v r .
Hint: The answer is an equation equivalent to
1
1
v2 = v20 C 2 GM
K
.
r
R
5b) The equation above is equivalent to a conservation of energy law involving kinetic and potential
energy. If you've had this discussion in a physics mechanics class, or of conservative vector fields in Calc
III, you should be able to derive the equation above in that way too. Give it a try!
5c) How large does v0 have to be so that the implicit equation for v
1
1
v2 = v20 C 2 GM
K
r
R
can be solved explicitly for a differentiable function v r defined for all r R R ? (This value of v0 is called
the escape velocity.)
2 GM
(Hint: vescape =
R
5d) If v0 ! vescape , how high above the surface of the earth does the object get? Start with the same
equation
1
1
v2 = v20 C 2 GM
K
.
r
R
2e) Could you add drag considerations to this discussion?