Transcript ppt - RHIG - Wayne State University

PHY
5200 Mechanical
Phenomena
Projectile
Motion
PHY 5200
Mechanical Phenomena
Newton’s Laws of Motion
Click to editClaude
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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 r

v̂  v r
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  cv2  flin  fquad
v̂
f   f (v)v̂
r
w  mg
Air Resistance - Definitions
f (v)  bv  cv2  flin  fquad
flin  bv
fquad  cv2
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.
For a spherical projectile (e.g. canon ball, baseball, drop of rain):
b  D
c   D2
Where D is the diameter of the sphere
and  depend on the nature of the medium
At STP in air:
  1.6  10 4 Ngs / m 2
  0.25Ngs / 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  cv2 v̂
• A drop of rain has D = 1 mm and v=0.6 m/s
fquad
flin
1
Neither term can be neglected.
• Millikan Oil Drop Experiments, D=1.5 mm and v=5x10-5 m/s.
fquad
flin
 10 7
r
f  bv
Air Resistance - Reynolds Number
• The linear term drag is proportional to the viscosity, 
• The quadratic term is related to the density of the
fluid, .
• One finds
fquad
flin
: R
Dv

Reynolds Number
Case 1: Linear Air Resistance
•
•
x
v̂
y
r
f  bv
Consider the motion of projectile for which
one can neglect the quadratic drag term.
From the 2nd law of Newton:
r
w  mg
r&& r
r
r
mr  F  mg  bv
•
•
Independent of position, thus:
r&
r
r
mv  mg  bv
A 1st order differential equation
Furthermore, it is separable in coordinates (x,y,z).
m&
vx  bvx
m&
vy  mg  bvy
•
Two separate differential equations
Uncoupled.
By contrast, for f(v)~v2, one gets coupled y vs x motion
r
f  cv v̂  c v  v v
2
2
x
2
y
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.
r
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
vx  
b
vx
m
b
m
dv
vx  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)
Position and Velocity vs Time
2.5
• One gets
2.5
t
x(t)  x(0)   vx 0 e t  / dt 
2
2
 0   vx 0 e

 t  /
x(t)  x 1  e
x  vx 0
t
 o
t / 

v(t) and x(t)
0

x(t)  x 1  et /
1.5

1.5
x  vx 0
1
1
0.5
vx (t)  vx 0 e
0.5
t /
0
0
0
1
2
3
t/tau
4
5
Vertical Motion with Linear Drag
•
r
f  bv
x
y
Consider motion of an object thrown
vertically downward and subject to gravity
and linear air resistance.
v̂
r
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

mv&y  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
y0
•
lnu  kt  C
with
  1/ k  m / b
ter
The velocity as a function of time is thus given by


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
•

vy  vy0
t
vy  vy0
As the simplest case, consider vy0=0,
I.e. dropping an object from rest.

vy  vter 1  et /

Vertical Velocity
v(m/s)
•
120
100
80
time percent of
t/tau vter
0
0.0
1
63.2
2
86.5
3
95.0
4
98.2
5
99.3
60
40
20
0
0
1
2
3
4
5
t/tau
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 /

r
f  bv
v̂
r
w  mg
Equation of vertical motion for linear drag (cont’d)
•
•
Note that it may be convenient to reverse the
direction of the y-axis.
r
f  bv
v̂
Assuming the object is initially thrown upward, the
position may thus be written


y  y0  vter t   vy0  vter 1  e
t /

y
x
r
w  mg
Equation of motion for linear drag (cont’d)
•
Combine horizontal and vertical equations to get the trajectory of a projectile.


x(t)  vx0 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
vx0

x 
x  vter ln  1 
v  

x0
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
3000
No friction
2000
1000
0
-10
-1000
-2000
-3000
-4000
-5000
10
30
50
70
90
110
130
150
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
vx0

R 
R  vter ln  1 
v  

x0
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

vy0  vter
vx0
R
vxo
•
We now get
•
This leads to

0

3
2
3
 R
1 R 
1 R  
R  vter 
 
  3  v   
v

2
v


 x0

x0
x0
vy0  vter
vx0
R0
R
2vxovyo
g
R  Rvac 

2
R2
3vxo

2
4 vyo 
2
Rvac
 Rvac  1 
3vxo
3 vter 


R 
R  vter ln  1 
vx0 

Quadratic Air Resistance
•
For macroscopic projectiles, it is usually a better approximation to
consider the drag force is quadratic
r
f  cv v
2
•
Newton’s Law is thus
•
Although this is a first order equation, it is NOT separable in x,y,z
components of the velocity.
r&
r
2r
mv  mg  cv v
Horizontal Motion with Quadratic Drag
•
•
We have to solve
m
dv
 cv 2
dt
Rearrange
Integration
dv
m  2  c  dt 
v
vo
0
Yields
 1 1
 1
m     m     ct
 v0 v 
 v v0
v
•
Separation of v and t variables permits
independent integration on both sides of
the equality…
dv
m 2  cdt
v
t
where
v  vo
at t = 0.
v
•
•
Solving for v
v(t) 
•
Note: for t=,
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 
12
10
0
 v0 ln(1  t /  )
•
8
6
Never stops increasing
v(t) 
4
2
•
By contrast to the “linear” case.

x(t)  vx0 1  e
•
t /

Which saturates…
v0
1 t /
0
0
5
10
15
20
25
30
35
40
45
50
70
60
50
40
•
30
Why? ! ?
20
•
•
x(t)  v0 ln(1  t /  )
10
The retardation force becomes
quite weak as soon as v<1.
0
0
5
10
15
20
25
30
35
40
45
In realistic treatment, one must include both the linear and quadratic terms.
50
Vertical Motion with Quadratic Drag
dv
 mg  cv 2
dt
•
Measuring the vertical position, y, down.
•
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
v2 
m
mg
c
ter
•
•
•
dv
 gdt
v2
1 2
vter
Solve by separation of variables
Integration yields
Solve for v
40
35
 v 
vter
arctanh    t
g
 vter 
30
25
20
 gt 
v  vter tanh  
 vter 
15
10
5
•
Integrate to find
2
vter  
 gt  

y
ln cosh
g


 v  
ter 
0
0
5
10
15
20
25
30
Quadratic Draw with V/H motion
• Equation of motion
r
r
m&
r& mg  cv 2 v̂
r
r
 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
r r
F  qv  B
•
•
B
The equation of motion
r& r r
mv  qv  B
y
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

Motion of a Charge in Uniform Magnetic Field (cont’d)
• Three components of the Eq of motion
mv&x  qBvy
mv&y  qBvx
mv&z  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    L
2! 3!
•
•
The series converges for any value of z (real or complex, large or
small).
It satisfies

 
d
Ae kz  k Ae kz
dz

df (z)
 kf (z)
dz
•
And is indeed a general solution for
•
So we were justified in assuming  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 


L
2!
3!
4!
where  is a real number
Separation of the real and imaginary parts - since i2=-1, i3=-I
 2 4
 

3
e  1 

 L   i    L 
2! 4!
3!

 


cos
sin
We get Euler’s Formula
e  cos  i sin 
i
Complex Exponentials (cont’d)
• Euler’s Formula implies ei lies on a unit circle.
ei  cos  i sin 
cos
y
ei
1
O
sin

x
cos 2   sin 2   1
Complex Exponentials (cont’d)
• A complex number expressed in the polar form
A  ae  a cos  iasin 
i
where a and  are real numbers
y
acos
i
A  ae
a 2 cos2   a 2 sin 2   a 2
asin
a
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 
Integration of 
  x& i&
y  vx  ivy  
   dt   Aei t dt

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
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