Physics 430: Lecture 3 Linear Air Resistance Dale E. Gary NJIT Physics Department.

Physics 430: Lecture 3 Linear Air Resistance

Dale E. Gary

NJIT Physics Department

September 8, 2008

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:

The physical reasons for these two different terms are as follows: The linear term arises due to the viscous drag of the medium, and is

proportional 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 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.

2.1 Air Resistance

vf ˆ)(vf

quadlin2)( ffcvbvvf

2quadlin and cvfbvf

As an aside, we introduce the Taylor Series expansion (see inside front coverof the text). Any function f(x) can be expanded about the point x = a by

f(x) = f(a) + f’(a)(x a) + 1/2! f’’(a)(x a)2 + 1/3! f’’’(a)(x a)3 + …

Therefore, expanding f(v) about v = 0 gives f(v) = f(0) + f’(0)v + f’’(0)v2 + …Since the force f(0) = 0, the above expression for f(v) can be seen as just an expansion of the drag force into its leading terms.

September 8, 2008

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

For a spherical projectile in air at STP (Standard Temperature and Pressure), for example, the approximate values of and are:

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

2.1 Air Resistance, cont’d

2 and DcDb

42

24

/mNs 25.0

Ns/m 106.1

DvvD

bv

cv

f

f

lin

quad

23

2

m

s106.1

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 m and v = 5x105 m/s).

Baseball

shows that flin is completely negligible for a baseball. Use Rain

shows that both are needed. Must use full expression Oil Drop

shows that fquad is completely negligible for the oil drop. Use

600)m/s5)(m07.0(m

s106.1

23

lin

quad

f

f

1)m/s6.0)(m001.0(m

s106.1

23

lin

quad

f

f

7562

3 10)m/s105)(m105.1(m

s106.1

lin

quad

f

f

vf ˆ2cv

vf ˆ)( 2cvbv

vf ˆbv

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 = Dv/, where is the gas or fluid density and 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

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:

For a projectile with linear drag, the projectile experiencesboth gravity and the drag force, the latter directed in theopposite direction of its motion. Newton’s 2nd Law becomes

But , so we can write this as a first-order differential equation for v:

This vector equation represents (in two dimensions) two separate equations for the x and y components

Notice that the two equations do not depend on one another.

2.2 Linear Air Resistance

Forcesmr

vgr bmm

v

mg

flin=bv

vgv bmm vr

yy

xx

bvmgvm

bvvm

x

y

September 8, 2008

For a projectile with quadratic drag, the situation is not so simple:

But since , . We can also write v via the Pythagorean Theorem

When separated into its two equations for x and y components

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.

Contrast with Quadratic Air Resistance

22yx vvv

vgr ˆ2cvmm

yyxy

xyxx

vvvcmgvm

vvvcvm

22

22

x

y

v

mg

vf ˆ2quad cvvv

vv vv

v ˆ ˆ 2

September 8, 2008

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 with the equation for the x component alone.

We solved this first-order differential equation in Lecture 1 (Problem 1.24). To refresh your memory, we write

Then integrate both sides to get

where c is an arbitrary constant of integration. Taking the inverse log of both sides, and writing b/m = k, we have where the arbitrary constant of integration has morphed into v = vxo at t = 0.

Horizontal Motion with Linear Drag-1

vflin=bv

xx bvvm

dtm

b

v

dvv

m

b

dt

dv

x

xx

x

ctm

bvx log

ktx

ktx evAev o

September 8, 2008

All the solution 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

The solution is an equation for velocity. To find the equation for the position of the cart, we just integrate. The left side is

Here we assume that the position at t = 0 is x(0) = 0. The right side is

The final solution for the position is where we have introduced the parameter , the value of x as .

ktxx evv o

drag]linear [for //1 bmk

)()0()(0

)(

)0(0 o txxtxdxtdtd

dxtdv

t tx

x

t

x

/o0

/o0

/o 1 t

x

ttx

t tx evevtdev

/1)( textx

oxvx t

Horizontal Motion with Linear Drag-2

September 8, 2008

The final solutions for v(t) and x(t) are:

Graphs of these functions are:

/o)( t

xx evtv drag]linear [for / bm

/1)( textx oxvx

Horizontal Motion with Linear Drag-3

0xvxv

t

t

xx

This behavior should certainly not be surprising. But hopefully this helps youget an appreciation for the power of mathematics for describing physical behavior.

Make your own cart in Phun and try it with/without

air resistance.

September 8, 2008

Let’s now consider the equation we derived for vertical (y-direction) motion.

In this case, because of the opposite directions of the forces,you can see that if vy is small, gravity will dominate and the

projectile will accelerate downward, making the drag force growuntil 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:

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.

Vertical Motion with Linear Drag-1

yy bvmgvm

tervb

mgvy

vflin=bv

mg

September 8, 2008

Statement of the problem: Find the terminal speed of a tiny oildrop in the Millikan oildrop

experiment (diameter D = 1.5 m and density = 840 kg/m3). Do the same for a small drop of mist with diameter D = 0.2 mm.

Solution: To calculate the terminal speed , we need the mass of

the drop. However, we are only given the density and size, from which we have to calculate the mass.

The terminal speed is then

Putting in numbers for the oil drop, we getand for the drop of mist where we have used the previous value for beta

Example 2.2—Terminal Speed of Small Liquid Drops

D

mg

b

mgv

ter

334 2/DVm

drag]linear [for 6

2

ter gD

v

drop] oil[for m/s 101.6 5ter

vmist] of drop[for m/s 3.1ter v

24 Ns/m 106.1 More massive drops fall faster

September 8, 2008

Writing the original equation in terms of vter, we have

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

Then our equation becomes , which is identical to our old equation for vx: , with the same solution: .

When we put back the original variable, and again use = 1/k, this becomes:

To determine the integration constant A, as usual we need initial conditions. If the projectile starts with velocity vy = vyo at t = 0, then , so

As , as before (and as we expect).

Vertical Motion with Linear Drag-2

tervvbvm yy

yy vuvvu ter

buum xx bvvm ktAeu

/ter

ty Aevv

tero vvA y

)1( /ter

/o

ttyy evevv

ter)( vtvt y

September 8, 2008

Vertical Motion with Linear Drag-3

t

terv

yv

oyv

t

terv

yvoyv

Let’s take a look at the solution for vyo = 0 (dropping the projectile from rest). In this case, the equation is just

which is plotted below.

)1( /ter

ty evv

tero vvy tero vvy 0o yv

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

A word about the“characteristic time” . Note that by the time t = , the projectile has already reached

By the time t = 3, the projectile velocity is at 95% of vter.

ter1

ter 63.0)1( vevvy

September 8, 2008

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:

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 , so

After falling only 20 s, 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.

Example 2.3—Characteristic Time for Two Liquid Drops

gv ter

m/s 101.6 5ter

v

s 102.68.9

101.6 65

ter

g

v

September 8, 2008

Vertical Motion with Linear Drag-4

We have obtained the general equation for the velocity as

To get the projectile’s position, we need to integrate this equation to get

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:

)1( /ter

/o

ttyy evevv

)1(

)1()(

/teroter

0

/ter

/o

ty

t tty

evvtv

tdevevty

/o 1)( t

x evtx

)1()( /teroter

ty evvtvty

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:

We can combine these into a single equation by solving the first for t

and substituting into the second:

This is rather too complicated to understand easily, but here is a plot of the trajectory compared with one without air resistance.

/o 1)( t

x evtx tvevvty t

y ter/

tero )1()(

o

1lnxv

xt

oter

o

tero 1lnxx

y

v

xvx

v

vvy

y

xR vxo Rvac

no air drag

air drag

September 8, 2008

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

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:

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):

]resistanceair no [ 2 oo

vac g

vvR yx

01lno

tero

tero

xx

y

v

RvR

v

vvy

3312

211ln

oxv

Rwhere

September 8, 2008

Horizontal Range-2 We can now substitute this approximation of the ln term into the

range equation to get:

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,

We can simplify further by recalling that vter/ = g, and dividing by :

Right away we see that R = 0 is a solution, but not an interesting one, hence we have

03

1

2

13

o

2

ooter

o

tero

xxxx

y

v

R

v

R

v

RvR

v

vv

03

1

2

12

3o

ter2

o

ter

o

o

R

R

v

vR

v

v

v

v

xxx

y

2

o2 xv

g

03

22

o

2oo

R

v

RR

g

vv

x

yx

2o o

o

2 2

3x y

x

v v RR

g v

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

for R2, and finally get:

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 = 16 g/cm3). What if it is aluminum (density = 2.7 g/cm3)?

Solution: Without air resistance, both pellets would have the same range:

ter

ovacvac

oo

ovac 3

41

2

3

2

v

vRR

g

vv

vRR yyx

x

vacoo2

vac

2R

g

vvR yx

cm 2.10/)45cos()45sin(22 2

2oo

vac gvg

v

g

vvR yx

September 8, 2008

Example 2.4, Solution, Cont’d Recall from example 2.2 that the terminal speed in air is given by

Thus, the range correction term is

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.

05.021

707.0

3

4

3

4

ter

o v

vy

[gold] m/s 216

2

ter

gDv