Newton’s Laws of Motion

Claude A PruneauPhysics and AstronomyWayne State University

PHY 5200 Mechanical Phenomena

Click to edit Master title style

Click to edit Master subtitle style

1

Projectile MotionPHY 5200Mechanical Phenomena

Claude A PruneauPhysics and Astronomy DepartmentWayne State UniversityDec 2005.

ContentContent

• Projectile Motion– Air Resistance– Linear Air Resistance– Trajectory and Range in a Linear Medium– Quadratic Air Resistance

• Charge Particle Motion– Motion of a Charged Particle in a Uniform Field– Complex Exponentials– Motion in a Magnetic Field

Description of Motion with F=ma

• F=ma, as a law of Nature applies to a very widerange of problems whose solution vary greatlydepending on the type of force involved.

• Forces can be categorized as being “fundamental” or“effective” forces.

• Forces can also be categorized according to thedegree of difficulty inherent in solving the 2nd orderdifferential equation F = m a.– Function of position only– Function of speed, or velocity– Separable and non-separable forces

• In this Chapter– Separable forces which depend on position and velocity.– Non separable forces.

Air Resistance

• Air Resistance is neglected in introductory treatmentof projectile motion.

• Air Resistance is however often non-negligible andmust be accounted for to properly describe thetrajectories of projectiles.– While the effect of air resistance may be very small in some

cases, it can be rather important and complicated e.g.motion of a golf ball.

• One also need a way/technique to determine whetherair resistance is important in any given situation.

Air Resistance - Basic Facts

• Air resistance is known under different names– Drag– Retardation Force– Resistive Force

• Basic Facts and Characteristics– Not a fundamental force…– Friction force resulting from different atomic phenomena– Depends on the velocity relative to the embedding fluid.– Direction of the force opposite to the velocity (typically).

• True for spherical objects, a good and sufficient approximationfor many other objects.

• Not a good approximation for motion of a wing (airplane) -additional force involved called “lift”.

– Here, we will only consider cases where the force is anti-parallel to the velocity - no sideways force.

Air Resistance - Drag Force

• Consider retardation force strictly anti-parallel to the velocity.

• Where

f(v) is the magnitude of the force.• Measurements reveal f(v) is

complicated - especially near the speedof sound…

• At low speed, one can write as a goodapproximation:

!w = m

!g

!f = ! f (v)v̂

v̂

!f = ! f (v)v̂

v̂ =

!v !v

f (v) = bv + cv2= flin + fquad

Air Resistance - Definitionsf (v) = bv + cv

2= flin + fquad

flin ! bv Viscous drag• Proportional to viscosity of the medium

and linear size of object.

Inertial• Must accelerate mass of air which is in

constant collision.• Proportional to density of the medium

and cross section of object.

fquad ! cv2

For a spherical projectile (e.g. canon ball, baseball, drop of rain):

b = !D

c = ! D2

Where D is the diameter of the sphereβand γ depend on the nature of the mediumAt STP in air:

! = 1.6 "10#4N is / m

2

$ = 0.25N is / m4

Air Resistance - Linear or Quadratic

• Often, either of the linear or quadratic terms can be neglected.

• To determine whether this happens in a specific problem, consider

• Example: Baseball and Liquid Drops

• A baseball has a diameter of D = 7 cm, and travel at speed of order v=5 m/s.

• A drop of rain has D = 1 mm and v=0.6 m/s

• Millikan Oil Drop Experiments, D=1.5 mm and v=5x10-5 m/s.

fquad

flin=cv

2

bv=! D"v = 1.6 #103 s

m2( )Dv

!1: linear case

" 1: quadratic case

$

%&

'&

fquad

flin! 600

fquad

flin! 1

fquad

flin! 10

"7

!f = !cv

2v̂

Neither term can be neglected.

!f = !b

!v

Air Resistance - Reynolds Number

• The linear term drag is proportional to the viscosity, η• The quadratic term is related to the density of the

fluid, ρ.• One finds

fquad

flin! R !

Dv"

#Reynolds Number

Case 1: Linear Air Resistance

• Consider the motion of projectile for whichone can neglect the quadratic drag term.

• From the 2nd law of Newton:

• Independent of position, thus:

• Furthermore, it is separable in coordinates (x,y,z).

• By contrast, for f(v)~v2, one gets coupled y vs x motion

m!""r =!F = m

!g ! b

!v

!w = m

!g

!f = !b

!v

v̂

m!"v = m

!g ! b

!v A 1st order differential equation

y

x

m!vx = !bvx

m!vy = mg ! bvyTwo separate differential equationsUncoupled.

!f = !cv

2v̂ = !c vx

2+ vy

2 !v

m!vx= !c v

x

2+ v

y

2vx

m!vy= mg ! c v

x

2+ v

y

2vy

Case 1: Linear Air Resistance - Horizontal Motion

• Consider an object moving horizontally in a resistive linear medium.• Assume vx = vx0, x = 0 at t = 0.• Assume the only relevant force is the drag force.

• Obviously, the object will slow down

• Define (for convenience):

• Thus, one must solve:

• Clearly:

• Which can be re-written:

!f = !b

!v

!vx= !

b

mvx

k =b

m

!vx=dv

x

dt= !kv

x

dvx

vx

= !kdtdv

x

vx

! = "k dt! lnvx= !kt + C

vx(t) = v

x0e! t /" with ! = 1 / k = m / b

Velocity exhibits exponential decay

Case 1: Linear Air Resistance - Horizontal Motion(cont’d)

• Position vs Time, integrate

• One gets

dx

d !td !t

0

t

" = x(t) # x(0)

x(t) = x(0) + vx0e! "t /#

d "t0

t

$

= 0 + !vx0#e! "t /#%& '(o

t

x(t) = x!1" e

" t /#( )x!$ v

x0#

x(t) = x!1" e

" t /#( )x!$ v

x0#

vx(t) = v

x0e! t /"

Vertical Motion with Linear Drag

• Consider motion of an object thrownvertically downward and subject to gravityand linear air resistance.

• Gravity accelerates the object down, thespeed increases until the point when theretardation force becomes equal inmagnitude to gravity. One then hasterminal speed.

m!vy = mg ! bvy

!w = m

!g

!f = !b

!v

v̂y

x

0 = mg ! bvy vter = vy (a = 0) =mg

b

Note dependence on mass and linear drag coefficient b.Implies terminal speed is different for different objects.

Equation of vertical motion for linear drag

• The equation of vertical motion is determined by

• Given the definition of the terminal speed,

• One can write instead

• Or in terms of differentials

• Separate variables

• Change variable:

m!vy = mg ! bvy

vter =mg

b

m!vy = !b vy ! vterm( )

mdvy = !b vy ! vterm( )dt

dvy

vy ! vterm= !

bdt

m

u = vy ! vterm

du = dvy

du

u= !

bdt

m= !kdt

k =b

mwhere

Equation of vertical motion for linear drag (cont’d)

• So we have …

• Integrate

• Or…

• Remember

• So, we get

• Now apply initial conditions: when t = 0, vy = vy0

• This implies

• The velocity as a function of time is thus given by

du

u= !

bdt

m= !kdt

du

u! = "k dt! lnu = !kt + C

u = vy! v

term

u = Ae!kt

vy ! vter = Ae! t /"

with ! = 1 / k = m / b

vy0 ! vter = Ae!0 /"

= A

vy= v

ter+ v

y0! v

ter( )e! t /"

vy= v

y0e! t /"

+ vter1! e

! t /"( )

Equation of vertical motion for linear drag (cont’d)

• We found

• At t=0, one has

• Whereas for

• As the simplest case, consider vy0=0,I.e. dropping an object from rest.

vy= v

y0e! t /"

+ vter1! e

! t /"( )

vy= v

y0

t!" vy= v

y0

vy= v

ter1! e

! t /"( )time

t/tau

percent of

vter

0 0.0

1 63.2

2 86.5

3 95.0

4 98.2

5 99.3

Equation of vertical motion for linear drag (cont’d)

• Vertical position vs time obtained by integration!• Given

• The integration yields

• Assuming an initial position y=y0, and initial velocity vy = vy0.One gets

• The position is thus given by

vy= v

ter+ v

y0! v

ter( )e! t /"

y = vtert ! " vy0 ! vter( )e! t /" + C

y0= !" vy0 ! vter( ) + C

C = y0+ ! vy0 " vter( )

y = y0+ v

tert + ! v

y0" v

ter( ) 1" e" t /!( )

!w = m

!g

!f = !b

!v

v̂y

x

Equation of vertical motion for linear drag (cont’d)

• Note that it may be convenient to reverse thedirection of the y-axis.

• Assuming the object is initially thrown upward, theposition may thus be written

!w = m

!g

!f = !b

!v

v̂y

xy = y

0! v

tert + " v

y0+ v

ter( ) 1! e! t /"( )

Equation of motion for linear drag (cont’d)

• Combine horizontal and vertical equations to get the trajectory of a projectile.

• To obtain an equation of the form y=y(x), solve the 1st equation for t, andsubstitute in the second equation.

x(t) = vx0! 1" e

" t /!( )y(t) = y

0! v

tert + " v

y0+ v

ter( ) 1! e! t /"( )

y(t) = y0 +vy0 + vter

vx0

x + vter! ln 1"

x

vx0!

#

$%&

'(

Example: Projectile Motionm 5 b 0.1

tau 50 vx0*tau 100

vx0 2 (vy0+vter)*tau 34500

vy0 200

vter 490 vter*tau 24500

x (m)

y (m

)No friction

Linear friction

Horizontal Range

• In the absence of friction (vacuum), one has

• The range in vacuum is therefore

• For a system with linear drag, one has

x(t) = vxot

y(t) = vyot ! 0.98

2t2

Rvac

=2v

xovyo

g

0 =vy0+ v

ter

vx0

R + vter! ln 1"

R

vx0!

#

$%&

'(

A transcendental equation - cannot be solved analytically

Horizontal Range (cont’d)

• If the the retardation force is very weak…

• So, consider a Taylor expansion of the logarithm in

• Let

• We get

• Neglect orders beyond

• We now get

• This leads to

R! v

xo!

0 =vy0+ v

ter

vx0

R + vter! ln 1"

R

vx0!

#

$%&

'(

! =R

vxo"

ln(1! ") = ! " + 1

2"2+ 1

3"3+ ...( )

!3

0 =vy0+ v

ter

vx0

R ! vter"

R

vx0"+1

2

R

vx0"

#

$%&

'(

2

+1

3

R

vx0"

#

$%&

'(

3)

*++

,

-..

R = 0

R =2v

xovyo

g!

2

3vxo"R2

R ! Rvac

"2

3vxo#Rvac

2= R

vac1"

4

3

vyo

vter

$

%&'

()

Quadratic Air Resistance

• For macroscopic projectiles, it is usually a better approximation toconsider the drag force is quadratic

• Newton’s Law is thus

• Although this is a first order equation, it is NOT separable in x,y,zcomponents of the velocity.

!f = !cv

2!v

m!"v = m

!g ! cv

2!v

Horizontal Motion with Quadratic Drag

• We have to solve

• Rearrange

• Integration

• Yields

• Solving for v

• Note: for t=τ,

mdv

dt= !cv

2

mdv

v2= !cdt

Separation of v and t variables permitsindependent integration on both sides ofthe equality…

md !v

!v 2

vo

v

" = #c d !t0

t

" where v = vo

at t = 0.

m !1

"v#$%

&'(v

0

v

= m1

v0

!1

v

)

*+,

-.! ct

v(t) =v0

1+ cv0t / m

=v0

1+ t / !with ! =

m

cvo

v(! ) =v0

1+ ! / != v

0/ 2

Horizontal Motion with Quadratic Drag (cont’d)

• Horizontal position vs time obtained by integration …

• Never stops increasing

• By contrast to the “linear” case.

• Which saturates…

• Why? ! ?

• The retardation force becomesquite weak as soon as v<1.

• In realistic treatment, one must include both the linear and quadratic terms.

x(t) = x + v( !t )d !t0

t

"

= v0# ln(1+ t / # )

v(t) =v0

1+ t / !

x(t) = v0! ln(1+ t / ! )

x(t) = vx0! 1" e

" t /!( )

• Measuring the vertical position, y, down.

• Terminal velocity achieved for

• For the baseball of our earlier example, this yields ~ 35 m/s or 80 miles/hour

• Rewrite in terms of the terminal velocity

• Solve by separation of variables

• Integration yields

• Solve for v

• Integrate to find

Vertical Motion with Quadratic Drag

mdv

dt= mg ! cv

2

vter=

mg

c

dv

dt= g 1!

v2

vter2

"

#$%

&'

dv

1!v2

vter2

= gdt

vter

garctanh

v

vter

!

"#$

%&= t

v = vtertanh

gt

vter

!

"#$

%&

y =vter( )

2

gln cosh

gt

vter

!

"#$

%&'

()

*

+,

Quadratic Draw with V/H motion

• Equation of motion

• With y vertically upward

m!""r = m

!g ! cv

2v̂

= m!g ! cv

!v

m!vx= !c v

x

2+ v

y

2vx

m!vy= !mg ! c v

x

2+ v

y

2vy

Motion of a Charge in Uniform Magnetic Field

• Another “simple” application of Newton’s 2nd law…• Motion of a charged particle, q, in a uniform magnetic field, B, pointing

in the z-direction.• The force is

• The equation of motion

• The 2nd reduces to a first order Eq.

• Components of velocity and field

!F = q

!v !!B

Z

x

y

!B

!v

m!"v = q

!v !!B

!v = v

x,v

y,v

z( )

!B = 0,0,B( )

!v = v

yB,!v

xB,0( )

Motion of a Charge in Uniform Magnetic Field (cont’d)

• Three components of the Eq of motion

• Define

• Rewrite

m!vx= qBv

y

m!vy= !qBv

x

m!vz= 0 v

z= constant

! =qB

m

vx,v

y( ) ! transverse velocity

Cyclotron frequency

!vx=!v

y

!vy= "!v

x

Coupled Equations

Solution in the complex plane …

Complex Plane

O x (real part)

y(imaginarypart)

! = vx + ivy

vy

vx

Representation of the velocity vector

i = !1

Why and How using complex numbers for this?

• Velocity

• Acceleration

• Remember Eqs of motion

• We can write

• Or

! = vx + ivy

!vx=!v

y

!vy= "!v

x

!! = !vx + i !vy

!! = !vx + i !vy ="vy # i"vx = #i" vx + ivy( )

!! = "i#!

Why and How using … (cont’d)

• Equation of motion

• Solution

• Verify by substitution

!! = "i#!

! = Ae" i# t

d!

dt= "i#Ae

" i# t= "i#!

Complex Exponentials

• Taylor Expansion of Exponential

• The series converges for any value of z (real or complex, large orsmall).

• It satisfies

• And is indeed a general solution for

• So we were justified in assuming η is a solution of the Eqs of motion.

ez= 1+ z +

z2

2!+z3

3!+!

d

dzAe

kz( ) = k Aekz( )

df (z)

dz= kf (z)

Complex Exponentials (cont’d)

The exponential of a purely imaginary number is

Separation of the real and imaginary parts - since i2=-1, i3=-I

We get Euler’s Formula

e!= 1+ i! +

i!( )2

2!+i!( )

3

3!+i!( )

4

4!+!

where θ is a real number

e!= 1"

! 2

2!+! 4

4!"!

#

$%

&

'( + i ! "

! 3

3!+!

#

$%

&

'(

cos! sin!

ei!= cos! + i sin!

Complex Exponentials (cont’d)

• Euler’s Formula implies eiθ lies on a unit circle.

O

x

y

ei!

cos!

sin!

1!

ei!= cos! + i sin!

cos2! + sin

2! = 1

Complex Exponentials (cont’d)

• A complex number expressed in the polar form

O

x

y

A = aei!

acos!

asin!

a !

A = aei!= acos! + iasin!

a2cos

2! + a

2sin

2! = a

2

where a and θ are real numbers

!i"t

! = Ae" i# t

! = Ae" i# t

= aei $ "# t( )

Amplitude

Phase

Angular Frequency

Solution for a charge in uniform B field

• vz constant implies

• The motion in the x-y plane best represented by introduction ofcomplex number.

• The derivative of ξ

• Integration of η

z(t) = zo+ v

zot

! = x + iy

Greek letter “xi”

!! = !x + i!y = vx + ivy = "

! = "dt# = Ae$ i% tdt#

! =iA

"e# i" t

+ constant

x + iy = Ce! i" t

+ X+iY( )

Solution for a charge in uniform B field (cont’d)

x + iy = Ce! i" t

+ X+iY( )

Redefine the z-axis so it passes through (X,Y)

x + iy = Ce! i" t

which for t = 0, impliesC = xo + iyo

! =qB

m

Motion frequencyO

x

y

xo + iyo

!t

x + iy

xo

2+ y

o

2

Solution for a charge in uniform B field (cont’d)

x(t) + iy(t) = Ce! i" t

z(t) = zo+ v

zot

! =qB

m

O

x

y

xo + iyo

!t

x + iy

xo

2+ y

o

2

Helix Motion