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PHY
5200 Mechanical
Phenomena
Projectile
Motion
PHY 5200
Mechanical Phenomena
Newton’s Laws of Motion
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MasterA title
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Pruneau
Physics and Astronomy Department
Wayne State University
Dec
2005.
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A Pruneau
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Physics and Astronomy
Wayne State University
1
Content
• 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 wide
range of problems whose solution vary greatly
depending on the type of force involved.
• Forces can be categorized as being “fundamental” or
“effective” forces.
• Forces can also be categorized according to the
degree of difficulty inherent in solving the 2nd order
differential 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 treatment
of projectile motion.
• Air Resistance is however often non-negligible and
must be accounted for to properly describe the
trajectories 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 whether
air 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 approximation
for 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 antiparallel to the velocity - no sideways force.
Air Resistance - Drag Force
•
Consider retardation force strictly antiparallel to the velocity.
!
f = ! f (v)v̂
•
Where
!
v
 v̂ =
!
v
 f(v) is the magnitude of the force.
•
•
Measurements reveal f(v) is
complicated - especially near the speed
of sound…
At low speed, one can write as a good
approximation:
f (v) = bv + cv 2 = flin + fquad
v̂
!
f = ! f (v)v̂
!
!
w = mg
Air Resistance - Definitions
f (v) = bv + cv 2 = flin + fquad
flin ! bv
Viscous drag
• Proportional to viscosity of the medium
and linear size of object.
fquad ! cv 2
Inertial
• Must accelerate mass of air which is in
constant collision.
• Proportional to density of the medium
and cross section of object.
For a spherical projectile (e.g. canon ball, baseball, drop of rain):
b = !D
c = ! D2
Where D is the diameter of the sphere
βand γ depend on the nature of the medium
At STP in air:
! = 1.6 " 10 #4 Nis / m 2
$ = 0.25Nis / m 4
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
fquad
flin
=
cv
!D
=
v = 1.6 # 10 3 ms2
bv
"
(
2
)
$ ! 1: linear case
&
Dv %
&" 1: quadratic case
'
• Example: Baseball and Liquid Drops
• A baseball has a diameter of D = 7 cm, and travel at speed of order v=5 m/s.
fquad
flin
! 600
!
f = !cv 2 v̂
• A drop of rain has D = 1 mm and v=0.6 m/s
fquad
flin
Neither term can be neglected.
!1
• Millikan Oil Drop Experiments, D=1.5 mm and v=5x10-5 m/s.
fquad
flin
! 10
"7
!
!
f = !bv
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 which
one can neglect the quadratic drag term.
From the 2nd law of Newton:
x
y
v̂
!
!
f = !bv
!
!
w = mg
!"" !
!
!
mr = F = mg ! bv
•
•
Independent of position, thus:
!"
!
!
mv = mg ! bv
Furthermore, it is separable in coordinates (x,y,z).
m!vx = !bvx
m!vy = mg ! bvy
•
A 1st order differential equation
Two separate differential equations
Uncoupled.
By contrast, for f(v)~v2, one gets coupled y vs x motion
!
!
f = !cv 2 v̂ = !c vx2 + vy2 v
m!vx = !c vx2 + vy2 vx
m!vy = mg ! c vx2 + vy2 vy
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): k =
•
Thus, one must solve:
•
Clearly:
•
Which can be re-written:
f = !bv
v!x = !
b
vx
m
b
m
dv
v!x = x = !kvx
dt
dvx
= !kdt
vx
vx (t) = vx 0 e!t /"
dvx
! vx = "k ! dt
with
! =1/ k = m /b
Velocity exhibits exponential decay
ln vx = !kt + C
Case 1: Linear Air Resistance - Horizontal Motion
(cont’d)
• Position vs Time, integrate
t
dx
"0 dt ! dt ! = x(t) # x(0)
• One gets
t
x(t) = x(0) + $ vx 0 e! t " /# dt "
0
= 0 + %& !vx 0# e
(
! t " /#
'( o
x(t) = x! 1 " e"t /#
x! $ vx 0#
t
(
x(t) = x! 1 " e"t /#
)
x! $ vx 0#
)
vx (t) = vx 0 e!t /"
Vertical Motion with Linear Drag
•
Consider motion of an object thrown
vertically downward and subject to gravity
and linear air resistance.
!
!
f = !bv
x
y
v̂
!
!
w = mg
m!vy = mg ! bvy
•
Gravity accelerates the object down, the
speed increases until the point when the
retardation force becomes equal in
magnitude to gravity. One then has
terminal speed.
0 = mg ! bvy
mg
vter = vy (a = 0) =
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
m!vy = mg ! bvy
•
Given the definition of the terminal speed,
•
One can write instead
mg
vter =
b
(
m!vy = !b vy ! vterm
•
Or in terms of differentials
(
)
)
mdvy = !b vy ! vterm dt
•
Separate variables
dvy
vy ! vterm
•
Change variable:
bdt
=!
m
u = vy ! vterm
du = dvy
du
bdt
=!
= !kdt
u
m
where
k=
b
m
Equation of vertical motion for linear drag (cont’d)
•
So we have …
du
bdt
=!
= !kdt
u
m
•
Integrate
du
! u = "k ! dt
•
Or…
u = Ae! kt
•
Remember
u = vy ! vterm
•
So, we get
vy ! vter = Ae!t /"
•
•
Now apply initial conditions: when t = 0, vy = vy0
This implies
v ! v = Ae!0 /" = A
•
The velocity as a function of time is thus given by
y0
ln u = !kt + C
! =1/ k = m /b
with
ter
(
)
vy = vter + vy0 ! vter e!t /"
(
vy = vy0 e!t /" + vter 1 ! e!t /"
)
Equation of vertical motion for linear drag (cont’d)
(
vy = vy0 e!t /" + vter 1 ! e!t /"
•
We found
•
At t=0, one has
•
Whereas for
•
As the simplest case, consider vy0=0,
I.e. dropping an object from rest.
)
vy = vy0
t!"
vy = vy0
(
vy = vter 1 ! e!t /"
)
time percent of
t/tau 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
(
)
vy = vter + vy0 ! vter e!t /"
(
)
y = vter t ! " vy0 ! vter e!t /" + C
•
Assuming an initial position y=y0, and initial velocity vy = vy0.
One gets
(
)
y0 = !" vy0 ! vter + C
(
C = y0 + ! vy0 " vter
)
x
•
The position is thus given by
(
y
)(
y = y0 + vter t + ! vy0 " vter 1 " e"t /!
)
!
!
f = !bv
v̂
!
!
w = mg
Equation of vertical motion for linear drag (cont’d)
•
•
!
!
f = !bv
Note that it may be convenient to reverse the
direction of the y-axis.
Assuming the object is initially thrown upward, the
position may thus be written
(
)(
y = y0 ! vter t + " vy0 + vter 1 ! e
!t / "
)
v̂
y
x
!
!
w = mg
Equation of motion for linear drag (cont’d)
•
Combine horizontal and vertical equations to get the trajectory of a projectile.
(
)
x(t) = vx 0! 1 " e"t /!
y(t) = y0 ! vter t + " vy0 + vter 1 ! e!t /"
•
(
)(
)
To obtain an equation of the form y=y(x), solve the 1st equation for t, and
substitute in the second equation.
y(t) = y0 +
vy0 + vter
vx 0
#
x &
x + vter! ln % 1 "
vx 0! ('
$
Example: Projectile Motion
5
50
2
200
490
b
vx0*tau
(vy0+vter)*tau
0.1
100
34500
vter*tau
24500
y (m)
m
tau
vx0
vy0
vter
Linear friction
No friction
x (m)
Horizontal Range
• In the absence of friction (vacuum), one has
x(t) = vxot
y(t) = vyot !
0.98 2
2
t
• The range in vacuum is therefore
Rvac =
2vxo vyo
g
• For a system with linear drag, one has
0=
vy0 + vter
vx 0
#
R &
R + vter! ln % 1 "
vx 0! ('
$
A transcendental equation - cannot be solved analytically
Horizontal Range (cont’d)
R ! vxo!
•
If the the retardation force is very weak…
•
So, consider a Taylor expansion of the logarithm in 0 =
•
Let
•
We get ln(1 ! " ) = ! " + 12 " 2 + 13 " 3 + ...
•
Neglect orders beyond
!=
R
vxo"
•
We now get
•
This leads to
(
0=
vx 0
)
!3
2
3
) R
1# R &
1# R & ,
R ! vter" +
+ %
(' + 3 %$ v " (' .
v
"
2
v
"
$
+* x 0
.x0
x0
vy0 + vter
vx 0
R=0
R=
vy0 + vter
2vxo vyo
g
R ! Rvac "
!
2
R2
3vxo"
$
2
4 vyo '
2
Rvac
= Rvac & 1 "
3vxo#
3 vter )(
%
#
R &
R + vter! ln % 1 "
vx 0! ('
$
Quadratic Air Resistance
•
For macroscopic projectiles, it is usually a better approximation to
consider the drag force is quadratic
!
2!
f = !cv v
•
Newton’s Law is thus
•
Although this is a first order equation, it is NOT separable in x,y,z
components of the velocity.
!"
!
2!
mv = mg ! cv v
Horizontal Motion with Quadratic Drag
•
•
•
We have to solve
m
dv
= !cv 2
dt
Separation of v and t variables permits
independent integration on both sides of
the equality…
dv
= !cdt
v2
Rearrange
m
Integration
dv!
m " 2 = #c " dt !
v!
vo
0
Yields
) 1 1,
# 1&
m % ! ( = m + ! . ! ct
* v0 v $ v" ' v0
v
t
where
v = vo
at t = 0.
v
•
•
Solving for v
v(t) =
•
Note: for t=τ,
v(! ) =
v0
v0
=
1 + cv0t / m 1 + t / !
v0
= v0 / 2
1+! /!
with
!=
m
cvo
Horizontal Motion with Quadratic Drag (cont’d)
•
Horizontal position vs time obtained by integration …
t
x(t) = x + " v(t ! )dt !
0
= v0# ln(1 + t / # )
v(t) =
•
Never stops increasing
•
By contrast to the “linear” case.
(
x(t) = vx 0! 1 " e"t /!
•
Which saturates…
•
Why? ! ?
v0
1+ t /!
)
x(t) = v0! ln(1 + t / ! )
•
The retardation force becomes
quite weak as soon as v<1.
•
In realistic treatment, one must include both the linear and quadratic terms.
Vertical Motion with Quadratic Drag
dv
= mg ! cv 2
dt
•
Measuring the vertical position, y, down. m
•
Terminal velocity achieved for vter =
•
For the baseball of our earlier example, this yields ~ 35 m/s or 80 miles/hour
•
2
Rewrite in terms of the terminal velocity dv = g " 1 ! v %
$#
dt
v 2 '&
mg
c
ter
dv
= gdt
v2
1! 2
vter
•
Solve by separation of variables
•
Integration yields
! v $
vter
arctanh #
=t
g
" vter &%
Solve for v
! gt $
v = vter tanh # &
" vter %
•
•
Integrate to find
2
vter ) '
! gt $ *
(
y=
ln cosh
g
)
(
#" v &% ,
ter +
Quadratic Draw with V/H motion
• Equation of motion
!
!
m""
r = mg ! cv 2 v̂
!
!
= mg ! cvv
• With y vertically upward
m!vx = !c vx2 + vy2 vx
m!vy = !mg ! c vx2 + vy2 vy
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.
Z
The force is
!
! !
F = qv ! B
!
B
The equation of motion
!"
! !
mv = qv ! B
!
v
The 2nd reduces to a first order Eq.
x
•
Components of velocity and field
(
!
v = v x , v y , vz
!
B = ( 0, 0, B )
)
(
!
v = vy B, !vx B, 0
)
y
Motion of a Charge in Uniform Magnetic Field (cont’d)
• Three components of the Eq of motion
m!vx = qBvy
m!vy = !qBvx
m!vz = 0
• Define
( v , v ) ! transverse velocity
x
!=
• Rewrite
vz = constant
y
qB
m
v!x = ! vy
v!y = "! vx
Cyclotron frequency
Coupled Equations
Solution in the complex plane …
Complex Plane
y
(imaginary
part)
Representation of the velocity vector
! = vx + ivy
vy
O
vx
i = !1
x (real part)
Why and How using complex numbers for this?
• Velocity
! = vx + ivy
• Acceleration
!! = v!x + iv!y
• Remember Eqs of motion
v!x = ! vy
v!y = "! vx
• We can write
• Or
(
!! = v!x + iv!y = " vy # i" vx = #i" vx + ivy
!! = "i#!
)
Why and How using … (cont’d)
• Equation of motion
!! = "i#!
• Solution
! = Ae"i# t
• Verify by substitution
d!
= "i# Ae"i# t = "i#!
dt
Complex Exponentials
•
Taylor Expansion of Exponential
2
3
z
z
ez = 1 + z + + + !
2! 3!
•
•
The series converges for any value of z (real or complex, large or
small).
It satisfies
(
) (
d
Aekz = k Aekz
dz
•
)
And is indeed a general solution for df (z) = kf (z)
dz
•
So we were justified in assuming η is a solution of the Eqs of motion.
Complex Exponentials (cont’d)
The exponential of a purely imaginary number is
2
3
4
i
!
i
!
i
!
(
)
(
)
(
)
e! = 1 + i! +
+
+
+!
2!
3!
4!
where θ is a real number
Separation of the real and imaginary parts - since i2=-1, i3=-I
# !2 !4
& #
&
!3
e = %1 "
+
" !( + i %! " + !(
2!
4!
3!
$
' $
'
!
cos!
sin !
We get Euler’s Formula
i!
e = cos! + i sin !
Complex Exponentials (cont’d)
• Euler’s Formula implies eiθ lies on a unit circle.
i!
e = cos! + i sin !
y
cos!
ei!
1
O
sin !
!
x
cos 2 ! + sin 2 ! = 1
Complex Exponentials (cont’d)
• A complex number expressed in the polar form
i!
A = ae = a cos! + ia sin !
where a and θ are real numbers
a cos!
y
A = ae
a
!
i!
a 2 cos 2 ! + a 2 sin 2 ! = a 2
a sin !
Amplitude
Phase
O
x
! = Ae
"i# t
!i" t
! = Ae
"i# t
= ae
Angular Frequency
i ($ " # t )
Solution for a charge in uniform B field
z(t) = zo + vzot
•
vz constant implies
•
The motion in the x-y plane best represented by introduction of
complex number.
! = x + iy
Greek letter “xi”
•
The derivative of ξ
•
Integration of η
!! = x! + i!y = vx + ivy = "
! = # "dt = # Ae$i% t dt
iA #i" t
! = e + 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
y
which for t = 0, implies
xo + iyo
C = xo + iyo
Motion frequency
!=
qB
m
xo2 + yo2
O
x
x + iy
!t
Solution for a charge in uniform B field (cont’d)
x(t) + iy(t) = Ce!i" t
z(t) = zo + vzot
!=
qB
m
y
xo + iyo
xo2 + yo2
O
!t
x
x + iy
Helix Motion