```Physics 430: Lecture 4
Dale E. Gary
NJIT Physics Department
Linear Air Resistance Recap-1
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
 For linear air resistance, the equation of motion is mr  mg  bv or in terms
of velocity, it is a first-order differential equation mv  mg  bv, which has
component equations:
mvx  bvx
mv y  mg  bv y
v y
vx
b
b
 ;

 Equations of this form can be written:
vx
m
v y  vter
m
mg
where vter 
is the terminal velocity.
b
September 10, 2009
Linear Air Resistance Recap-2

Such equations are said to be in separable form (terms involving v on one
side, and no dependence on v on the other side). Solutions of this
dv y
particular form, e.g. dv x
b
b
  dt ;
  dt
vx
m
v y  vter
m
have exponential solutions:
v x  v xo e t / v y  v yoet /  vter (1  et / )
  m / b
which we then integrate to get x and y positions:
t / 
x(t )  x 1  e t / y(t )  vtert  v yo  vter  (1  e )
 We can then combine these equations by eliminating t, to get a single
equation for the trajectory:
v yo  vter

x 

y
x  vter ln 1 
vxo
v

xo 

 Finally, we solved this for the range R, i.e. the value of x for which y = 0,
valid for low air resistance:
 4 v yo 

R  Rvac 1 
3
v
ter 

 Linear air resistance applies only to tiny projectiles or viscous fluids.


September 10, 2009






For more normal size projectiles (baseball, cannon ball), it is the quadratic
drag force that applies.
We are now going to follow exactly the same procedure, but starting with the
The equation of motion (in terms of v) then becomes: mv  mg  cv 2 vˆ
with component equations: mvx  cv 2  c vx2  v y2 vx
mv y  mg  cv 2  mg  c vx2  v y2 v y
As we noted last time, these two equations are coupled, and are generally
not solvable analytically (in terms of equations), although they can be solved
numerically.
However, we can solve these equations for special cases of either solely
horizontal motion (vy = 0), or solely vertical motion (vx = 0), in which case the
equations become
mv  cv 2
[horizonta l motion]
mv  mg  cv 2 [vertical motion]
Let’s look at these one at a time.
September 10, 2009

As before, we write the equation in separable form (move the terms involving
v to one side). For the horizontal equation, it is trivial: dv   c dt
v2
m

This equation is called a non-linear differential equation because one of the
derivatives (the zeroth one, in this case) has a non-linear dependence. Such
equations are significantly harder to solve, in general. In this case, however,
the separable form allows us to integrate both sides directly v dv
c t
vo v2   m 0 dt 
 1 1
vo
ct
to get     
or v(t ) 
, where I have introduced the
1

t
/

m
 vo v 
m
characteristic time, , in terms of constants:  
.
cvo

To find the position, we again integrate the velocity equation to get
v0
dt   vo ln( 1  t /  )
0 1  t / 
x(t )  xo  
t
[if xo  0]
September 10, 2009


The final solutions for v(t) and x(t) are:
vo
  m / cvo
v(t ) 
1 t /
x(t )  vo ln( 1  t /  )
Graphs of these functions are:
x
vx
vo

t

t
They may look similar at first to the linear case, but now the velocity as t  
approaches zero much more slowly, like 1/t, so the position does not approach
some limiting value x like in the linear case, but rather continues to increase
forever. If this sounds impossible, you are right. What really happens is that
as the speed drops, quadratic drag gets swamped by linear drag.
September 10, 2009
We now consider motion solely in the vertical direction, governed by the
equation of motion: mv  mg  cv 2 [vertical motion]
 Before we write the vertical equation in separated form, however, we notice
as before that the gravity force mg is balanced by the drag force cvy2 at
terminal velocity
mg
vter 
c
after which v y  0 , i.e. the velocity becomes constant. In terms of vter, the
separated form for the vertical equation is:
dv
 g dt
2
1  v 2 / vter

In this separated form, we can integrate both sides directly (assuming vo = 0).
v
t
dv 

g
0 1  v2 / vter2
0 dt 
dx
 Looking at the inside front cover of the book we find  1  x 2  arctanh x
which is what we have if we write x = v/vter. What the heck is arctanh?

September 10, 2009
Hyperbolic Functions—Problem 2.33(a)

Statement of the Problem:

The hyperbolic functions cosh z and sinh z are defined as follows:
e z  e z
e z  e z
cosh z 
and sinh z 
2
2
for any z, real or complex. (a) Sketch the behavior of both functions over a suitable
range of real values of z.
cosh z
e z
2
sinh z
ez
2
1
1
2
ez
2
1
2
z
 12
z

z
e
2
September 10, 2009
Hyperbolic Functions—Problem 2.33(b)

Statement of the Problem, cont’d:


(b) Show that cosh z = cos(iz). What is the corresponding relation for sinh z?
Solution:

To do this part, you have to know the relations:
eix  e ix
cos x 
2

and
eix  e ix
sin x 
2i
Then the solution is very easy:
ei (iz )  e i (iz ) e  z  e z
cos iz 

 cosh z
2
2
ei (iz )  e i (iz ) e  z  e z
e z  e z
sin iz 

i
 i sinh z
2i
2i
2

So
sinh z 
sin iz
 i sin iz
i
September 10, 2009
Hyperbolic Functions—Problem 2.33(c)

Statement of the Problem, cont’d:


(c) What are the derivatives of cosh z and sinh z? What about their integrals?
Solution:

The derivatives are:
d cosh z 1 d z
 2 (e  e  z )  12 (e z  e  z )  sinh z
dz
dz
d sinh z 1 d z
 2 (e  e  z )  12 (e z  e  z )  cosh z
dz
dz

The integrals are equally straightforward:
 cosh z dz  sinh z
 sinh z dz  cosh z
September 10, 2009
Hyperbolic Functions—Problem 2.33(d)

Statement of the Problem, cont’d:


(d) Show that cosh2 z – sinh2 z = 1.
Solution:

2
 e z  ez
2
Since cosh z  
 2

  14 (e 2 z  2e z e  z  e  2 z )  14 (e 2 z  2  e  2 z )

 e z  ez
2
sinh z  
 2

  14 (e 2 z  2e z e  z  e  2 z )  14 (e 2 z  2  e  2 z )

and
the difference is
2
cosh 2 z  sinh 2 z  12  12  1
September 10, 2009
Hyperbolic Functions—Problem 2.33(e)

Statement of the Problem, cont’d:
dx
 (e) Show that
 1  x 2  arcsinh x . [Hint: One way to do this is to make the
substitution x = sinh z.]

Solution:

Making that substitution, we have dx = cosh z dz, so:

cosh z dz
1  sinh z
2

cosh z dz
cosh z  sinh z  sinh z
2
2
2
  dz  z
but
sinh z  x  z  arcsinh x
so we have shown that

dx
1 x
2
 arcsinh x
September 10, 2009

Likewise, you can do Problem 2.34, which gives the definition: tanh z 
and leads you through the steps needed to show

Now back to our equation:

The left side is
dv 
0 1  v2 / vter2  g
v
sinh z
cosh z
dx
 1  x 2  arctanh x .
t
 dt 
0
v / vter
dv
dx

v
0 1  v2 / vter2 ter 0 1  x 2
v
 vter arctanh x 0
v / vter
 v 

 vter arctanh 
 vter 
 gt 

v(t )  vter tanh 
while the right side is just gt, so solving for v, we get
v
 ter 
 To get the position, integrate (see Prob. 2.34) to get
2

 gt 
vter

y (t ) 
ln cosh 
g 
 vter 
September 10, 2009
Example 2.5

A Baseball Dropped from a High Tower


Find the terminal speed of a baseball (diameter D = 7 cm , mass m = 0.15 kg).
Make plots of its velocity and position for the first six seconds after it is dropped
from a tall tower.
Solution


Recall that the constant c can be written c = gD2., where g = 0.25 Ns2/m2. So
mg
(0.15 kg)(9.8 m/s 2 )
vter 

 35 m/s
2
2
2
c
(0.25 Nm /s )(0.07 m)
which is nearly 80 mph.
You can sketch the velocity and position, or you can calculate it in Matlab. Here
are the plots. As expected, the velocity increases more slowly than it would in a
vacuum under gravity (dashed line), and approaches vter = 35 m/s (dotted line).
As a consequence, the position is less than the parabolic dependence in vacuum.
September 10, 2009
Vertical Motion



As we said before, the general problem of combined horizontal and
vertical motion yields a set of coupled equations
mvx  c vx2  v y2 vx
mv y  mg  c vx2  v y2 v y
where now we take y positive upward.
The projectile does not follow the same x and y equations we just derived,
because the drag in the x direction slows the projectile and changes the
drag in the y direction, and vice versa. In fact, these equations cannot be
solved analytically at all! The best we can do is a numerical solution, but
that requires specifying initial conditions. That means we cannot find the
general solution—we have to solve them numerically on a case-by-case
basis.
Let’s take a look at one such numerical solution.
September 10, 2009
Example 2.6

Trajectory of a baseball


The baseball of the previous example is thrown with velocity 30 m/s (about 70
mph) at 50o above the horizontal from a high cliff. Find its trajectory for the first
8 s of flight and compare with the trajectory in a vacuum. If the same baseball
were thrown on the same trajectory on horizontal ground, how far would it travel
(i.e. what is its horizontal range)?
Solution

First, what are the initial conditions for the position and velocity? For the
position, we are free to choose our coordinate system, so we certainly would
choose xo = 0 and yo = 0 at t = 0. For the velocity, the statement of the problem
gives the initial conditions vxo = vocos q = 19.3 m/s, vyo = vosin q = 23.0 m/s. Using
these values, we need to write a program in Matlab that performs a numerical
solution to the equations
2
2
mvx  c vx  v y vx
mv y  mg  c vx2  v y2 v y
for the time range 0 &lt; t &lt; 8 s. We will use the routine ode45 (ode stands for
‘ordinary differential equation’).
September 10, 2009
Example 2.6, cont’d

Solution, cont’d


First we have to write a function that will be called by ode45. The heart of that
routine is quite simple, just write expressions for the two equations:
Vdot_x = -(c/m)*sqrt(v(1)^2+v(2)^2)*v(1);
= [block 2]
Vdot_y = -g-(c/m)*sqrt(v(1)^2+v(2)^2)*v(2);
Here, v is the velocity vector, so v(1) is the horizontal velocity vx and v(2) is the
vertical velocity vy. Before these equations will work, we have to define the
constants, g, c, and m. Recall that c = gD2.
m = 0.15;
% Mass of baseball, in kg
g = 9.8;
% Acceleration of gravity, in m/s
diam = 0.07;
% Diameter of baseball, in m
gamma = 0.25; % Coefficient of drag in air at STP, in Ns^2/m^2
c = gamma*diam^2;

= [block 1]
The last step is to name the function and indicate the inputs and outputs.
ODE45 specifies that the function must have two inputs—the limits of the
independent variable (time in this case), and the array of initial conditions (start
velocity in x and y in this case). function vdot = quad_drag(t,v)
…
[block 1]
[block 2]
…
vdot = [vdot_x; vdot_y];

After saving this function as quad_drag.m, we call ODE45 with
September 10, 2009
Example 2.6, cont’d

Solution, cont’d

The arrays T and V that are returned are the times and x and y velocities, but
what we need is the trajectory, i.e. the x and y positions. For those, we have to
integrate the velocities. There is probably an elegant way to do this in Matlab,
but I wrote a simple (and rather inaccurate) routine to do that, given the T and
V arrays:
function y = int_yp(t,yp)
n = length(t);
y = yp;
y(1,:) = [0 0];
for i=1:n-1
dt = t(i+1)-t(i);
dy = yp(i,:)*dt;
y(i+1,:) = y(i,:)+dy;
end
y = y(1:n-1,:);

Save this as int_yp.m, and then call it by
pos = int_yp(T,V);

which returns the position array [pos(1,:) is x, pos(2,:) is y].
All that remains is to plot the trajectory ( i.e. pos(1,:) vs. pos(2,:) ).
plot(19.3*T,23.0*T-4.9*T.^2); % Plot vacuum case
hold on
plot(pos(:,1),pos(:,2),'color','red'); % Overplot quadratic drag case
hold off
September 10, 2009
Example 2.6, cont’d

Solution, cont’d




Here is the resulting plot (somewhat improved by labels).
Note that the range is about
60 m, much shorter than the
equivalent trajectory in a
vacuum.
Note also that the baseball
does not reach quite as high
as in a vacuum, and reaches
its peak earlier.
You will be given homework
problems in which I will ask
you to try your hand at such
numerical solutions and plotting.
useful skills, or you can make
use of the Matlab helpers.
September 10, 2009
2.5 Motion of a Charge in a
Uniform Magnetic Field

You may recall from Physics 121 that the force on a charge moving in a
magnetic field is
F  qv  B
where q is the charge and B is the magnetic field strength. The equation of
motion then becomes
mv  qv  B
which is a first-order differential equation in v.
 In this type of problem, we are often free to choose our coordinate system
so that the magnetic field is along one axis, say the z-axis:
B  (0,0, B )
and the velocity can in general have any direction v  (vx , v y , vz ). Hence,
v  B  (v y B,  vx B, 0)
and the three components of the equation of motion are: mvx  qv y B
mv y  qvx B
mvz  0
September 10, 2009
Motion of a Charge in a Uniform
Magnetic Field-2
This last equation simply says that the component of velocity along B,
vz = const. Let’s now focus on the other two components, and ignore the
motion along B. We can then consider the velocity as a two-dimensional
vector (vx, vy) = transverse velocity.
 To simplify, we define the parameter w = qB/m:, so the equations of motion
become:
vx  w v y



v y  w vx
We will take the opportunity provided by these two coupled equations to
introduce a solution based on complex numbers.
As you should know, a complex number is a number like z = x + iy, where
imaginary
i is the square root of 1. Let us define:
part
h = vx + ivy
and then plot the value of h as a vector in the
vy
complex plane whose components are vx and vy.
real part
vx
September 10, 2009
Motion of a Charge in a Uniform
Magnetic Field-3

Next, we take the time derivative of h:
h  vx  iv y  wv y  iwvx  iw (vx  iv y )
or
h  iwh
So the equation in terms of this new relation has the same form we saw in
the previous lecture for linear air resistance, with the familiar solution
h  Aeiwt
 The only difference is that this time the argument of the exponential is
imaginary, but it turns out that this makes a huge difference.
 Before we can discuss the solution in detail, however, we need to introduce
some properties of complex exponentials, which we will do next time.

September 10, 2009
```