Home > Physics 430: Lecture 4 Quadratic Air Resistance Dale E. Gary NJIT Physics Department September 10, 2009 When a projectile

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 *f*_{lin} and *f*_{quad} 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 (*v*_{y}= 0), or solely vertical motion (*v*_{x}= 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*cv*_{y2}at terminal velocity

after
which , i.e. the velocity
becomes constant. In terms of *v*_{ter}, the separated form for the
vertical equation is:

- In this separated
form, we can integrate both sides directly (assuming
*v*_{o }= 0).

- Looking at
the inside front cover of the book we find

which
is what we have if we write *x*
= *v*/*v*_{ter}.
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:

- So

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 derivatives are:

- The integrals
are equally straightforward:

Hyperbolic Functions—Problem 2.33(c)

**September 10, 2009**

- Statement of the Problem, cont’d:
- (d) Show
that cosh
^{2 }*z*– sinh^{2 }*z*= 1. - Solution:
- Since

and

the
difference is

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

substitution *x*
= sinh *z*.]

- Solution:
- Making that
substitution, we have
*dx*= cosh*z**dz*, so:

but

so
we have shown that

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:

- The left
side is

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*=*g**D*^{2}., where*g*= 0.25 Ns^{2}/m^{2}. So

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
*v*_{ter}= 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 50
^{o}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
*x*_{o}= 0 and*y*_{o}= 0 at*t*= 0. For the velocity, the statement of the problem gives the initial conditions*v*_{xo}=*v*_{o}cos*q*= 19.3 m/s,*v*_{yo}=*v*_{o}sin*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:

= [block 2]

- Here, v is
the velocity vector, so v(1) is the horizontal velocity
*v*_{x}and v(2) is the vertical velocity*v*_{y}. Before these equations will work, we have to define the constants,*g*,*c*, and*m*. Recall that*c*=*g**D*^{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).

- 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

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.

I will help you learn these very

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

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**,

*v*_{z}
= 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 (*v*_{x}, *
v*_{y}) =
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 *v*_{x} and *v*_{y}.

imaginary

part

real
part

*v*_{x}

*v*_{y}

*h** = v*_{x}*+
iv*_{y}

*h** = v*_{x}*
+ iv*_{y}

**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.

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