Physics
430: Lecture 4
Quadratic Air Resistance
Dale
E. Gary
NJIT Physics Department
September 10, 2009
- 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).
- Quite generally,
we can write this force as
, where the function f(v) can in general be any function
of velocity.
- At relatively
slow speeds, it is often a good approximation to write
where flin and fquad stand for the linear and quadratic
terms, respectively:
- For linear
air resistance, the equation of motion is
or in terms of velocity, it is a first-order differential equation
, which has component equations:
- Equations
of this form can be written:
where
is the terminal velocity.
Linear
Air Resistance Recap-1
September 10, 2009
- 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 particular form, e.g.
have
exponential solutions:
which
we then integrate to get x and y positions:
- We can then
combine these equations by eliminating t, to get a single equation for
the trajectory:
- Finally,
we solved this for the range R, i.e. the value of x for which y
= 0, valid for low
air resistance:
- Linear air
resistance applies only to tiny projectiles or viscous fluids.
Linear
Air Resistance Recap-2
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 quadratic
form of the drag force:
- The equation
of motion (in terms of v) then becomes:
with
component equations:
- 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
- Let’s
look at these one at a time.
2.4 Quadratic
Air Resistance
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:
- 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
to
get
or
, where I have introduced the
characteristic
time, t, in terms of constants:
.
- To find the
position, we again integrate the velocity equation to get
Horizontal
Motion with Quadratic Drag-1
September 10, 2009
- The final
solutions for v(t) and x(t) are:
- Graphs of
these functions are:
Horizontal
Motion with Quadratic Drag-2
They may look similar
at first to the linear case, but now the velocity as
approaches zero much
more slowly, like 1/t, so the position does not approach
some limiting value
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:
- 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
after
which , i.e. the velocity
becomes constant. In terms of vter, the separated form for the
vertical equation is:
- In this separated
form, we can integrate both sides directly (assuming vo
= 0).
- Looking at
the inside front cover of the book we find
which
is what we have if we write x
= v/vter.
What the heck is arctanh?
Vertical
Motion with Quadratic Drag
September 10, 2009
- Statement
of the Problem:
- The hyperbolic
functions cosh
z and sinh z are defined as follows:
for
any z, real or complex. (a) Sketch
the behavior of both functions over a suitable range of real values
of z.
Hyperbolic
Functions—Problem 2.33(a)
September 10, 2009
- 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:
- Then the
solution is very easy:
Hyperbolic
Functions—Problem 2.33(b)
September 10, 2009
- Statement
of the Problem, cont’d:
- (c) What
are the derivatives of cosh
z and sinh z? What about their integrals?
- Solution:
- The integrals
are equally straightforward:
Hyperbolic
Functions—Problem 2.33(c)
September 10, 2009
- Statement
of the Problem, cont’d:
- (d) Show
that cosh2
z – sinh2 z = 1.
- Solution:
Hyperbolic
Functions—Problem 2.33(d)
September 10, 2009
- Statement
of the Problem, cont’d:
- (e) Show
that
. [Hint: One way to do this is to make the
- Solution:
- Making that
substitution, we have dx
= cosh z dz,
so:
Hyperbolic
Functions—Problem 2.33(e)
September 10, 2009
- Likewise,
you can do Problem 2.34, which gives the definition:
and
leads you through the steps needed to show
.
- Now back
to our equation:
while
the right side is just gt, so solving for v, we get
- To get the
position, integrate (see Prob. 2.34) to get
Vertical
Motion with Quadratic Drag
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
- 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
Quadratic
Drag with Horizontal and Vertical Motion
- As we said
before, the general problem of combined horizontal and vertical motion
yields a set of coupled equations
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
for
the time range 0
< t < 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:
- 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.
- 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).
- After saving
this function as quad_drag.m, we call ODE45 with
Vdot_x = -(c/m)*sqrt(v(1)^2+v(2)^2)*v(1);
Vdot_y = -g-(c/m)*sqrt(v(1)^2+v(2)^2)*v(2);
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;
function vdot = quad_drag(t,v)
…
[block 1]
[block 2]
…
vdot = [vdot_x; vdot_y];
[T,V] = ode45('quad_drag',[0
8],[19.3; 23.0]);
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:
- Save this
as int_yp.m, and then call it by
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,:) ).
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,:);
pos = int_yp(T,V);
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
- Note also
that the baseball
as
in a vacuum, and reaches
- You will
be given homework
useful skills, or you can make
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
where q is the charge and B is the magnetic field strength.
The equation of motion then becomes
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:
and
the velocity can in general have any direction
. Hence,
and
the three components of the equation of motion are:
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:
- 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 i is the square root of -1. Let us define:
and
then plot the value of h as a vector in the
complex
plane whose components are vx and vy.
imaginary
part
real
part
vx
vy
h = vx+
ivy
h = vx
+ ivy
September 10, 2009
Motion
of a Charge in a Uniform Magnetic Field-3
- Next, we
take the time derivative of h:
or
- 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
- 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.
