Transcript Linear Air Resistance

Physics 430: Lecture 3
Linear Air Resistance
Dale E. Gary
NJIT Physics Department
2.1 Air Resistance
When a projectile moves through the air (or other medium—such as gas or
liquid), it experiences a drag force, which depends on velocity and acts in
the direction opposite the motion (i.e. it always acts to slow the projectile).
ˆ, where the function
 Quite generally, we can write this force as f   f (v ) v
f(v) can in general be any function of velocity.
 At relatively slow speeds, it is often a good approximation to write
f (v)  bv  cv 2  f lin  f quad
where flin and fquad stand for the linear and quadratic terms, respectively:

f lin  bv
and
f quad  cv 2
 The physical reasons for these two different terms are as follows:
As an aside, we introduce the Taylor Series expansion (see inside front cover
 The linear term arises due to the viscous drag of the medium, and is proportional
of
the text). Any function f(x) can be expanded about the point x = a by
to the viscosity of the medium and the linear size D of the projectile.
 The quadratic term arises from the projectile’s
having
to accelerate
f(x)
= f(a) + f’(a)(x  a) + 1/2! f’’(a)(x  a)2 + 1/3!
f’’’(a)(x
 a)3 + …the mass of
air with which it is continually colliding, and is proportional to the density of the
medium and the cross-sectional area D2 of the projectile.
2
Therefore, expanding f(v) about v = 0 gives f(v) = f(0) + f’(0)v + f’’(0)v + …
Since the force f(0) = 0, the above expression for f(v) can beSeptember
seen as just
an
8, 2008
expansion of the drag force into its leading terms.
2.1 Air Resistance, cont’d

It is convenient to have parameters that do not depend on the projectile’s size
or area, but rather on intrinsic properties of the medium. We therefore write
b  bD
and
c  gD 2

For a spherical projectile in air at STP (Standard Temperature and Pressure),
for example, the approximate values of b and g are:
b  1.6 104 Ns/m 2
g  0.25 Ns2 /m 4


Although these values are only strictly valid for a sphere in air at STP,
nevertheless they give a good idea of the relative importance of the linear
and quadratic force terms even for non-spherical bodies moving through
gases other than air.
When it comes time to do problems with air resistance, we are going to want
to neglect one or the other of these terms. We can tell their relative
importance by looking at the ratio f
cv 2 gD
s 

quad


v  1.6 103 2  Dv
f lin
bv
b
m 

September 8, 2008
Example 2.1

A Baseball and Some Drops of Liquid


Assess the relative importance of the linear and quadratic drag forces on a
baseball of diameter D = 7 cm , traveling at a modest v = 5 m/s. Do the same for
a drop of rain (D = 1 mm and v = 0.6 m/s) and for a tiny droplet of oil used in the
Millikan oildrop experiment (D = 1.5 mm and v = 5x105 m/s).
Baseball
f quad
f lin

shows that flin is completely negligible for a baseball. Use
Rain
f quad
f lin

s 

 1.6 103 2 (0.07 m)(5 m/s )  600
m 

f  cv 2 vˆ
s 

 1.6 103 2 (0.001 m)(0.6 m/s )  1
m 

shows that both are needed. Must use full expression
Oil Drop
f  (bv  cv 2 ) vˆ
s 

 1.6 103 2 (1.5 10 6 m)(5 10 5 m/s )  10 7
f lin 
m 
shows that fquad is completely negligible for the oil drop. Use f  bv vˆ
f quad
September 8, 2008
Linear vs. Quadratic Drag
The moral of the previous example is clear. There are some objects for
which the linear drag force dominates—namely very small liquid drops in
air, or somewhat larger objects in a very viscous liquid (e.g. a ball bearing
in molasses).
 For most projectiles we will meet, however, including baseballs, cannon
balls, even humans in free-fall, the appropriate drag force to use is the
quadratic one.
 This is unfortunate, since the linear drag force is much easier to solve
mathematically, and we will start with linear case because it is easier, and
it allows us to introduce some useful mathematics.
 There is a branch of physics called Fluid Mechanics that makes use of a
dimensionless number called the Reynolds number, which is closely
related to the ratio fquad/flin. It is R = Dvr/h, where r is the gas or fluid
density and h is the viscosity (see Problem 2.3). When the Reynolds
number is large, the quadratic term is important, and when the Reynolds
number is small, the linear term is important.

September 8, 2008
2.2 Linear Air Resistance
Let’s put the linear and quadratic drag forces into Newton’s 2nd Law to see
what the character of the solutions are. As always, we write Newton’s 2nd
Law as the equation of motion:
v
mr   Forces
flin=bv
 For a projectile with linear drag, the projectile experiences
y
both gravity and the drag force, the latter directed in the
mg
opposite direction of its motion. Newton’s 2nd Law becomes
mr  mg  bv
 But r  v
 , so we can write this as a first-order differential equation for v:
mv  mg  bv


This vector equation represents (in two dimensions) two separate equations
for the x and y components mvx  bvx
mv y  mg  bv y
 Notice that the two equations do not depend on one another.
September 8, 2008
x
Contrast with Quadratic Air
Resistance
For a projectile with quadratic drag, the situation is not so simple:
v
mr  mg  cv 2 vˆ
v
 But since v
, v 2 vˆ  vv. We can also write v via
ˆ
fquad  cv 2 vˆ
v
the Pythagorean Theorem
v  vx2  v y2


x
y
mg
When separated into its two equations for x and y components
mvx  c vx2  v y2 vx
mv y  mg  c vx2  v y2 v y
Now these two equations do depend on one another—they are coupled
equations. That makes them considerably harder to solve, which is why we
are going to start with the simpler, linear drag case.
 Let’s go back to the linear case and solve the horizontal and vertical
equations separately. First the horizontal case.

September 8, 2008
Horizontal Motion with Linear Drag-1

Consider an object such as the cart in the figure, coasting horizontally in a
linearly resistive medium. Now gravity is not important, so we can deal
v
with the equation for the x component alone.
flin=bv
mvx  bvx

We solved this first-order differential equation in Lecture 1 (Problem 1.24).
To refresh your memory, we write
dv x
dv x
b
b
  vx 
  dt
dt
m
vx
m

Then integrate both sides to get
b
t c
m
where c is an arbitrary constant of integration. Taking the inverse log of
both sides, and writing b/m = k, we have v x  Ae  kt  v xo e  kt where the
arbitrary constant of integration has morphed into v = vxo at t = 0.
log v x  
September 8, 2008
Horizontal Motion with Linear Drag-2
 kt
All the solution v x  vxo e says is that the cart starting out with some
velocity vxo slows down exponentially, approaching zero velocity only after
infinite time has passed.
 Since the argument of the exponential must have no units, the units of the
constant k must be inverse time, so 1/k can be considered a time constant
  1 / k  m / b [for linear drag]
 The solution is an equation for velocity. To find the equation for the
position of the cart, we just integrate. The left side is
t
t dx
x (t )


v
d
t

d
t

0 xo 0 dt 
x(0) dx  x(t )  x(0)  x(t )
 Here we assume that the position at t = 0 is x(0) = 0. The right side is


t
0


vxo et / dt    vxo et /

t
0

 vxo 1  et /



The final solution for the position is x(t )  x 1  e t / where we have
introduced the parameter x  vxo , the value of x as t   .
September 8, 2008
Horizontal Motion with Linear Drag-3

The final solutions for v(t) and x(t) are:
v x (t )  v xo e  t /

x(t )  x 1  e t /

  m/b

[for linear drag]
x  vxo
Graphs of these functions are:
x
vx
v x0
x
Make your own cart in
Phun and try it with/without
air resistance.

t

t
This behavior should certainly not be surprising. But hopefully this helps you
get an appreciation for the power of mathematics for describing physical
behavior.
September 8, 2008
Vertical Motion with Linear Drag-1
Let’s now consider the equation we derived for vertical (y-direction) motion.
mv y  mg  bv y
v
flin=bv
 In this case, because of the opposite directions of the forces,
you can see that if vy is small, gravity will dominate and the
mg
projectile will accelerate downward, making the drag force grow
until eventually it equals the gravity force. At that point, the net force goes
to zero and the projectile falls with a constant terminal speed vter given
by:
mg
vy 
 vter
b
 Looking at the dependence of the terminal speed, you can see that a more
massive object has a larger terminal speed. Conversely, if air resistance is
great (value of b is large), the terminal speed is small.

September 8, 2008
Example 2.2—Terminal Speed of
Small Liquid Drops

Statement of the problem:


Find the terminal speed of a tiny oildrop in the Millikan oildrop experiment
(diameter D = 1.5 mm and density r = 840 kg/m3). Do the same for a small drop of
mist with diameter D = 0.2 mm.
Solution:

vter 
mg mg

b
bD, we need the mass of the drop.
To calculate the terminal speed
However, we are only given the density and size, from which we have to calculate
the mass.
3
m  rV  r 43  D / 2


The terminal speed is then
rD 2 g
vter 
6b
[for linear drag]
Putting in numbers for the oil drop, we get vter  6.1 10 5 m/s [for oil drop]
and for the drop of mist
vter  1.3 m/s [for drop of mist]
where we have used the previous value for beta
b  1.6 104 Ns/m 2
More massive drops fall faster
September 8, 2008
Vertical Motion with Linear Drag-2
Writing the original equation in terms of vter, we have mv y  bv y  vter 
which is again a first-order differential equation which is not so different
from the one for horizontal motion.
 We can most easily see this by making a change of variable and writing
u  v y  vter  u  v y

Then our equation becomes mu  bu, which is identical to our old equation
for vx: mvx  bvx, with the same solution: u  Ae kt .
 When we put back the original variable, and again use  = 1/k, this
becomes:
v y  vter  Aet /
 To determine the integration constant A, as usual we need initial conditions.
If the projectile starts with velocity vy = vyo at t = 0, then A  v yo  vter , so

v y  v yoet /  vter (1  et / )

As t  , v y (t )  vter as before (and as we expect).
September 8, 2008
Vertical Motion with Linear Drag-3

Let’s take a look at the solution for vyo = 0 (dropping the projectile from
rest). In this case, the equation is just
A word about the
v y  vter (1  et / )
“characteristic time” .
which is plotted below.
Note that by the time
vy
v yo
t = , the projectile
vy
has already reached
vter
vter
v y  vter (1  e1 )  0.63vter
v yo
By the time t = 3, the
v yo  vter
v yo 
 v0ter
projectile velocity is at
95% of vter.
t
t



Note that it is not enough to simply derive an equation. To really
understand the motion you need to sketch such plots, or look at limiting
behavior (e.g. position and velocity as t  ).
September 8, 2008
Example 2.3—Characteristic Time for
Two Liquid Drops

Statement of the problem:


Find the characteristic times, , for the oildrop and drop of mist in Example 2.2.
Solution:

The characteristic time was defined as  = m/b, while vter = mg/b, so we have the
useful relation:
vter  g


This can be interpreted as saying that vter is the velocity the drop would attain if it
were accelerated for a time, , at constant acceleration equal to g. In fact, the
acceleration is less than g, because of the variable drag force acting in the
opposing direction, so the drop does not quite attain speed vter after time .
For the Milliken oildrop, we found that vter  6.110 5 m/s , so
vter 6.1105


 6.2 106 s
g
9.8


After falling only 20 ms, the oildrop attains 95% of its terminal speed!
For the drop of mist, the characteristic time is 0.13 s. After about 0.4 s, the drop
should have attained 95% of its terminal speed.
September 8, 2008
Vertical Motion with Linear Drag-4
We have obtained the general equation for the velocity as
v y  v yoet /  vter (1  et / )
 To get the projectile’s position, we need to integrate this equation to get

t


y (t )   v yo e t  /  vter (1  e t  / ) dt 
0
 vtert  v yo  vter  (1  e t / )
where we have assumed the initial position y(0) = 0.
 We now have the equations for the projectile position for horizontal and
vertical motion, separately, as:

x(t )  vxo 1  e t /

y(t )  vtert  v yo  vter  (1  et / )
for x(0) = y(0) = 0, and y downward.
September 8, 2008
2.3 Trajectory and Range
in a Linear Medium



To get a trajectory including BOTH horizontal and vertical motion, we should
consider y position upward. The corresponding equation for y(t) is the same as
before, but we must reverse the sign of vter (convince yourself that is the case).
Thus, our two equations are: x(t )  vxo 1  e t / 
y(t )  v yo  vter  (1  et / )  vtert
We can combine these into a single equation by solving the first for t

x 


t   ln 1 
y
 vxo 
and substituting into the second:
v yo  vter

x 

y
x  vter ln 1 
R
vxo
 vxo 
This is rather too complicated to understand easily,
air drag
but here is a plot of the trajectory compared with one
without air resistance.
no air drag
vxo
September 8, 2008
Rvac
x
Horizontal Range-1

You have already seen the method for finding the range of a trajectory without
air resistance, in Physics I. To remind you of the solution, it is
2vxo v yo
Rvac 
[ no air resistance ]
g

What is it in the case of linear air resistance? Recall that the range R is the
value of x when y as given by the range equation is zero:
v yo  vter

R 
  0
y
R  vter ln 1 
vxo
 vxo 
This is a transcendental equation (because of the ln term) and cannot be
solved in terms of elementary functions. You need to use a computer to solve
it numerically (which you will do for the homework). Meanwhile, you can solve
it approximately by assuming the argument of the ln function is small.
To do this, you can use a Taylor expansion as we did earlier (and which you
will also do in the HW):
R
ln 1       12  2  13  3   where  
v xo


September 8, 2008
Horizontal Range-2
We can now substitute this approximation of the ln term into the range
equation to get:
2
3
 R
 v yo  vter 




1 R
1 R 
  
   0
 

 R  vter 
 vxo 2  vxo  3  vxo  
 vxo 
 We can simplify this by noting that the second term in the first bracket cancels
the first term in the second bracket, and after factoring out a common R,
 v yo 1 vter R 1 vter  R  2 


 R  0

2
3 
v
2

3

vxo
vxo   
 xo
g
 We can simplify further by recalling that vter/ = g, and dividing by
2:
2vxo
 2vxo v yo
2 R2 
R

R  0
3 vxo 
 g
 Right away we see that R = 0 is a solution, but not an interesting one, hence
we have
2vxo v yo 2 R 2
R

g
3 vxo

September 8, 2008
Horizontal Range-3

The text goes through a rather unilluminating argument that for small enough
air resistance the range R  Rvac, hence we can get away with substituting
2vxo v yo
2
Rvac 
Rvac
g
for R2, and finally get:
 4 v yo 
2 2vxo v yo

R  Rvac 
Rvac  Rvac 1 
3vxo
g
 3 vter 
This is only valid for low air resistance (v << vter).
 Example 2.4:



I flick a tiny metal pellet with diameter D = 0.2 mm and v = 1 m/s at 45o. Find its
horizontal range assuming the pellet is gold (density r = 16 g/cm3). What if it is
aluminum (density r = 2.7 g/cm3)?
Solution:

Without air resistance, both pellets would have the same range:
Rvac 
2vxo v yo
g
2v 2 sin( 45) cos( 45)

 v 2 / g  10.2 cm
g
September 8, 2008
Example 2.4, Solution, Cont’d

Recall from example 2.2 that the terminal speed in air is given by
rD 2 g
vter 
 21 m/s [gold]
6b

Thus, the range correction term is
4 v yo 4 0.707

 0.05
3 vter 3 21


and the range is about 5% less than in vacuum.
The density of aluminum is about 6 times smaller, so the terminal velocity is likewise
6 times smaller so the correction is more like 30%.
Thus, the gold pellet will sail about 9.6 cm while the aluminum pellet will only go
about 7 cm. Because the air resistance for aluminum is larger, we can expect that
the equation for the range is not so accurate.
September 8, 2008