Math 2250-1
Mon Sept 10
2.3 Improved velocity models.
Recap of Friday:
We were studying particle motion along a line, with velocity v t satisfying the linear drag model of
acceleration:
m v# t = m a Kk v
v# t = a K r v
a
, The equilibrium solution for v t is
and the phase diagram is
r
a
////
))))
r
so no matter the initial velocity v0 , we know
a
d vt
r
which we call the terminal velocity. For example, in the case of vertical motion near the surface of the
Kg
earth and if we call "up" the positive direction, then a =Kg and the terminal velocity vt =
is negative.
r
lim v t =
t /N
,
Integrating the IVP
v# t = a K r v
v 0 = v0
yields
a
a Kr t
C v0 K
e
r
r
v t = vt C v0 K vt eKr t .
v t =
This formula agrees with the phase diagram analysis.
,
Integrating once more, with y 0 = y0 we found
y t = y0 C vt t C
vt K v0
r
eKr t K 1 .
Then I tried rushing through the example below in the last five minutes. Today, let's give it the time it
deserves, because there are some important things to consider.
Application: We consider the bow and deadbolt example from the text, page 102-104. It's shot vertically
m
into the air (watch out below!), with an initial velocity of 49
= 5 g s. (In the no-drag case, this could
s
just be the 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.)
In the case of no drag, the governing equations are
v# t =Kg zK9.8
m
s2
v t =Kg t C 5 g
1
1
t
x t =K g t2 C v0 t C x0 =K g t2 C 5 g t = g t K C 5 .
2
2
2
Looking at these equations we see that the velocity is zero at t = 5 s, the maximum height is x 5 =
25
g,
2
and the deadbolt lands after 10 s. Using g = 9.8 we can rewrite the formulas above as
v t =K9.8 t C 49
x t =K4.9 t2 C 49 t .
Now consider the linear drag model for the same deadbolt, with the same initial velocity of 5 g = 49
m
.
s
m
=K25 g s . Equate this to the formula for terminal velocity
s
Kg
vt =
r
to deduce r = 0.04. In this case, call the height y t . Our formulas are
v t = vt C v0 K vt eKr t =K245 C 294 eK.04 t
Assume the terminal velocity is vt =K245
y t = y0 C vt t C
vt K v0
r
294 K.04 t
e
K1 .
.04
eKr t K 1 =K245 tK
v t = vt C v0 K vt eKr t =K245 C 294 eK.04 t
y t = y0 C vt t C
vt K v0
294 K.04 t
e
K1 .
.04
eKr t K 1 =K245 tK
r
Exercise 1 - technology may be appropriate!
a) When does the object reach its maximum height. How does this compare to the no-drag case?
b) What is this height, and how does it compare to the height in the no-drag case?
c) How long does the object fall? Compare to the no-drag case.
d) Does the object spend more time going up, or more time coming down?
e) Graph the height functions for the drag and for the no-drag case.
> v d t/K245 C 294$eK.04$t :
294
> y d t/K245$t K
$ eK.04$t K 1 :
.04
e) picture:
> with plots :
x d t/K4.9$t2 C 49$t :
plot1 d plot x t , t = 0 ..10, color = green : #no drag height function
plot2 d plot y 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. 105-107 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 2
2a) 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
2b) 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!
2c) 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
2d) 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?