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Intermediate Mechanics
Physics 321
Richard Sonnenfeld
New Mexico Tech
:00
Lecture #1 of 25
Course goals

Physics Concepts / Mathematical Methods
Class background / interests / class photo
Course Motivation

“Why you will learn it”
Course outline (hand-out)
Course “mechanics” (hand-outs)
Basic Vector Relationships
Newton’s Laws
Worked problems
Inertia of brick and ketchup III-3,4
2 :02
Physics Concepts
Classical Mechanics









Study of how things move
Newton’s laws
Conservation laws
Solutions in different reference frames (including
rotating and accelerated reference frames)
Lagrangian formulation (and Hamiltonian form.)
Central force problems – orbital mechanics
Rigid body-motion
Oscillations
Chaos
3 :04
Mathematical Methods
Vector Calculus





Differential equations of vector quantities
Partial differential equations
More tricks w/ cross product and dot product
Stokes Theorem
“Div, grad, curl and all that”
Matrices


Coordinate change / rotations
Diagonalization / eigenvalues / principal axes
Lagrangian formulation


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Calculus of variations
“Functionals”
Lagrange multipliers for constraints
General Mathematical competence
4 :06
Class Background and Interests
Majors

Physics?
EE?
CS? Other?
Preparation




Assume
Assume
Assume
Assume
Math 231 (Vector Calc)
Phys 242 (Waves)
Math 335 (Diff. Eq) concurrent
Phys 333 (E&M) concurrent
Year at tech

2nd 3rd 4th 5th
Graduate school?
Greatest area of interest in mechanics
5 :08
Physics Motivation
Physics component

Classical mechanics is incredibly useful
 Applies to everything bigger than an atom and slower
than about 100,000 miles/sec


Lagrangian method allows “automatic” generation
of correct differential equations for complex
mechanical systems, in generalized coordinates,
with constraints
Machines and structures / Electron beams /
atmospheric phenomena / stellar-planetary
motions / vehicles / fluids in pipes
6 :10
Mathematics Motivation
Mathematics component




Hamiltonian formulation transfers DIRECTLY to
quantum mechanics
Matrix approaches also critical for quantum
Differential equations and vector calculus
completely relevant for advanced E&M and wave
propagation classes
Functionals, partial derivatives, vector calculus.
“Real math”. Good grad-school preparation.
7 :12
About instructor
15 years post-doctoral industry experience



Materials studies (tribology) for hard-drives
Automated mechanical and magnetic
measurements of hard-drives
Bringing a 20-million unit/year product to market
Likes engineering applications of physics

Will endeavor to provide interesting problems that
correspond to the real world
8 :16
Course “Mechanics”
WebCT / Syllabus and Homework
Office hours, Testing and Grading
9 :26
Vectors and Central forces
r1  r2
r1  r2
Vectors

r1
r2
r2

Many forces are of
form F ( r1  r2 )
Remove dependence
of result on choice of
origin
r1
Origin 1
Origin 2
10 :30
Vector relationships

Vectors
dr dx
dy
dz

xˆ 
yˆ  zˆ
 Allow ready
dt dt
dt
dt
representation of 3


 
(or more!)
x
r  r  r r
xˆ 
components at once.
x
 Equations written in
 
r  s  r s cos( )
vector notation are
more compact
3
  ri si
i 1
11 :33
Vector Relationships -- Problem #L1-1
“The dot-product trick”
Given vectors A and B which correspond to
symmetry axes of a crystal:
B

A  2 xˆ

B  3xˆ  3 yˆ  3zˆ
Calculate:

A, B , 
A
Where theta is angle between A and B
12 :38
Vector relationships II – Cross product
  
q  r  s  r s sin(  )
 xˆ
 

r  s  det  rx
sx

qi 
yˆ
ry
sy
zˆ 

rz 
s z 
3
r s 
j , k 1
 ijk  0
j k ijk
For any two indices equal
 ijk   1
I,j,k even permutation of 1,2,3
 ijk   1
I,j,k odd permutation of 1,2,3
Determinant

Is a convenient
formalism to
remember the signs
in the cross-product
Levi-Civita Density
(epsilon)

Is a fancy notation
worth noting for
future reference
(and means the
same thing)
13
Newton’s Laws
1. A Body at rest remains at rest, while a
body in motion at constant velocity
remains in motion
Unless acted on by an external force


dP
F 
dt
2. The rate of change of momentum is
directly proportional to the applied
force.


F12   F21
3. Two bodies exert equal and opposite


dP1
dP2

dt
dt
<--- Using 2 and 3 Together
forces on each other
14 :42
Newton’s Laws imply momentum conservation


dP1
dP2

dt
dt
d  
P1  P2   0
dt
 
P1  P2  C
 In absence of external force,
momentum change is equal
and opposite in two-body
system.
 Regroup terms
 Integrate.
Q.E.D.
Newton’s laws are valid in all
inertial (i.e. constant velocity)
reference frames
15 :45
Two types of mass?
Gravitational mass mG
mG
W= mGg
Inertial mass mI
g
F=mIa
mI
a=0
a>0
“Gravitational forces and acceleration are
fundamentally indistinguishable” – A.Einstein
16 :65
Momentum Conservation -- Problem #L1-2
“A car crash”
James and Joan were drinking straight tequila
while driving two cars of mass 1000
 kg and
2000 kg with velocity
vectors 30 x m/s

and 10 x  60 y m/s
Their vehicles collide “perfectly inelastically” (i.e.
they stick together)
Assume that the resultant
wreck slides with

velocity vector v final
Friction has not had time to work yet. Calculate
v final and v final
17 :55
Two types of mass -- Problem #L1-3 a-b
“Galileo in an alternate universe”
A cannonball (mG = 10 kg) and a
golf-ball (mG = 0.1 kg) are
simultaneously dropped from
a 98 m tall leaning tower in
Italy.

g  9.8m / s 2
Neglect air-resistance
How long does each ball take to
hit the ground if:
a) mI=mG
b) mI =mG
*
mG
18 :65
Lecture #1 Wind-up


dP
.F 
dt
First homework due in class Thursday
8/29
Office hours today 3-5
Get on WebCT
19
Lecture #2 of 25
Questions on Assignment #1?
Homework Expectations /Office
hours/Reserve

(REA Mechanics Problem Solver, ISBN 0-87891-519-2 --$25)
Cross-product problem
Impulse and momentum conservation
Rocket propulsion
Worked problems
Saturn V launch I-4
About Newton
(http://scienceworld.wolfram.com/biography/Newton.html)
Inclined plane
20 :02
Lecture #2 of 25
Questions on Assignment #1?
Homework Expectations /Office
hours/Reserve

(REA Mechanics Problem Solver, ISBN 0-87891-519-2 --$25)
Impulse and momentum conservation
Rocket propulsion
Worked problems
Cross-product problem
Saturn V launch I-4
21 :10
Homework Bliss
GOOD
NOT GOOD
mr  mass of cannon  2.1103 kg
m1  2.1103
mc  mass of chicken  2.4 kg
m2  2.4
J  cannon impulse  (1000 Nt  s ) xˆ
impulse  1000
vic , v fc  initial , final velocity of chicken
vi , v f
mcannon  2.110 kg ; mchicken  2.4kg
3
vi chicken , v f chicken
Brief description
# of
points
No attempt
Just started /
confused
Well along
Nearly perfect
Physics Bliss
0
4
7
9
10
22 :17
Definition of Impulse
J  F (t ) t
J 

t2
F (t ) dt
Impulse J is a useful concept in
the study of collisions.
(e.g. Balls and bats, automobiles,
comets and planets)
Impulse is the average force acting
W  F ( r ) r
over a time period multiplied by
r2
the time period.
W 
F ( r ) dr
r1
It may also be written as an integral
Note the difference between impulse J
and work W.
t1

23 :22
Impulse and momentum change
t2
J   F (t )dt
t1
J 
t2
t1
dp(t )
dt  p (t2 )  p (t1 )  p
dt
Two equivalent impulses with different Fmax
1
Force (N)
0.8
0.6
0.4
0.2
0
0
2
4
6
Time (s)
8
10
Impulse is useful
because it
directly
allows
expression of
momentum
change.
24 :27
Impulse I -- Problem #L2-2
“A car crash”
James and Joan were drinking straight tequila
while driving two cars of mass 1000
 kg and
2000 kg with velocity
vectors 30 x m/s

and 10 x  60 y m/s
We may look at their collision in terms of impulse.
Joan’s car applied an impulse to James’ car as
3
follows:
J  (13.33xˆ  40 yˆ ) 10 Nt  s
What is the final momentum of James’ car
only? What is his final velocity?
Calculate
pJames  final , vJames  final and vJames  final
25 :32
Rocket Science I


F  ma
 Not the whole story


dP
 More general – get used to it
F 
dt
 Case where m not constant
 d
F  mv
dt
 v)
 (mv  m
 vexhaust defined as " thrust"
m

marocket  thrust  Instantaneous acceleration, but
rocket keeps getting lighter!!
26 :40
Rocket Science II

mr v  ve m
In absence of other
forces
v
m f dm
dv  ve
Separable ODE.
0
mi
m
Integrate both
sides
 mf 

v f  ve ln 

Rocket
velocity
in
 mi 
terms of lost mass
 m f  and exhaust
 velocity
v f  2.3ve log 10 


 mi 
27 :45
Rocket Science III
 mf
v f  2.3ve log 10 
 mi



If a rocket is
60% fuel, then
it limits at ve
If a rocket is
90% fuel, then
limit at 2.3 ve
At 99% fuel,
limit is 4.6 ve
Achieving escape velocity with Single-Stage to Orbit rocket
70000
Exhaust velocity 3350 mph
Exhaust velocity 6700 mph
Exhaust velocity 13400
Escape velocity
65000
60000
55000
50000
Velocity (mph)
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
90
80
70
60
50
40
30
Percentage of rocket mass remaining
20
10
1
28 :50
Specific Impulse
I sp
Engine Type
Best Chemical
Best Nuclear
Xenon Ion
Linear Accel.
EM Catapult
ve

g
Specific
Impulse
500
2,000
3,800
1,000,000
N/A
 In practice of rocketry, the “specific
impulse” is often quoted
Exhaust Velocity
m/s
mph
4,900
10,363
19,600
41,452
37,240
78,759
9,800,000
20,726,119
45,000,000
95,170,955
Rocket
Vfinal
mph
23,835
95,340
181,146
47,670,074
95,170,955
Travel to
Alpha-C
(yrs)
106,477
26,619
14,010
53
27
29 :45
:55
Problem T3.7
The first couple of minutes after a space shuttle
launch can be described as follows: The
initial mass is 2x10E6 kg, the final mass (after
2 minutes) is about 1x10E6 kg, the average
exhaust speed is about 3000 m/s. If all this
were taking place in outer space, with
negligible gravity, what would be the shuttle’s
speed at the end of this stage? What is the
thrust during this same period and how does
it compare with the total initial weight of the
shuttle (on earth)?
30 :65
Impulse II -- Problem #L2-3
“Another car crash”
James and Joan are partly recovered from their previous
injuries, and haven’t learned from their experience.
They are drinking Jack Daniels and not wearing seatbelts.
James’ vehicle has velocity vector 30 xˆ m / s
Joan’s vehicle has 30 xˆ m / s
Both vehicles’ mass=M. Both people’s mass=70 kg.
Solve for case of inelastic and elastic collisions of vehicles.
Joan has an airbag in her vehicle. It takes her 100 millisec
to reduce her velocity relative to her vehicle to zero.
James stops 5 millisec after impacting the steering wheel.
What impulse does each person experience? What is
31
average force for each? How many “g’s” do they feel.:75
Cross product – Problem #L2-1
  
q  r  s  r s sin(  )
 xˆ
 

r  s  det  rx
sx

qi 
yˆ
ry
sy
zˆ 

rz 
s z 
Calculate
A) Use the
determinant
method where
r  (2,3, 4)
s  (4,3, 2)
3
r s 
j , k 1
 ijk  0
j k ijk
For any two indices equal
q  r s
B) Use the Levi-Civita
density where
 ijk   1
I,j,k even permutation of 1,2,3
 ijk   1
I,j,k odd permutation of 1,2,3
r  (4,3, 2)
s  (2,3, 4)
32 :Bonus
Lecture #2 Wind-up
. Impulse J  t F (t )dt  p
t
. mv
exhaust defined as " thrust "
.
 mf 
v f  ve ln 

 mi 
Got on WebCT? / Got Books?
Office hours Friday 5-6
Homework problems in Taylor, Handout,
Galileo problem – check the web.
Second homework due in class Thursday 9/5
2
1

(Includes introducing gravity into rocket equation)
33 :72
Lecture #3 of 25
Homework #1 expectations
Questions on Assignment #2
Center of Mass


Defined
Relation to momentum
Polar and Spherical Coordinates
Worked problems
DVD Demonstration on momentum cons. and
CM motion
34 :10
Center of Mass
Center of Mass and Center of gravity happen
to be equivalent
For a multi-particle discrete mass-distribution
N
RCM 
N
m r  m r



1
N
m



1
M total
1
For a continuous mass-distribution
RCM
rdm  rdm  r (r )dV  r (r )dA





M
M
dm

(
r
)
dV


total
total
35 :15
Linear Momentum and CM
N
R CM 
m r


1
M total
N
 M R CM   m r
 1
N
Ptotal   m r  Ptotal  M R CM
 1
Ptotal  Fexternal , Ptotal  M R CM  M R CM
0
Fexternal  M R CM
36 :20
Spherical Coordinates and Earth
Radius on surface of earth is fixed, so two
coordinates are sufficient to specify any point
on Earth
Coordinates are:

Latitude (North-south)
 “Co-latitude” is 90 degrees minus latitude

Longitude (East-West)
Spherical coordinates


“Phi” or “Fee” j – East-west same as longitude
“Theta”  – North-south, same as Colatitude
  is 0 at north pole, 180 at south pole, 90 at equator
For third dimension, add “r” (radius)
37 :25
Cylindrical and Spherical Coordinates
Table of
Area and
Volume Elements
System
dA
dV
Cartesian
dxdy
dxdydz
Cylindrical
(rdj )dz
(rdj )drdz
Spherical
Polar
(r sin  dj )( rd ) (r sin  dj )( rd ) dr
(rdj )dr
(rdj )drdz
38 :30
Impulse II -- Problem #L2-3
“Another car crash”
James and Joan are partly recovered from their previous
injuries, and haven’t learned from their experience.
They are drinking Jack Daniels and not wearing seatbelts.
James’ vehicle has velocity vector 30 xˆ m / s
Joan’s vehicle has 30 xˆ m / s
Both vehicles’ mass=M. Both people’s mass=70 kg.
Solve for case of inelastic and elastic collisions of vehicles.
Joan has an airbag in her vehicle. It takes her 100 millisec
to reduce her velocity relative to her vehicle to zero.
James stops 5 millisec after impacting the steering wheel.
What impulse does each person experience? What is
39
average force for each? How many “g’s” do they feel.:40
Worked Example L3-1 – Discrete masses
Given m1 to m10
2 units
y
O2
y
x
O1
x
m= m
m = 3m
Calculate
1 unit
N
RCM 
N
m r  m r



1
N
m


1

1
M total
RCM
Given origin O1
For homework
given O2
40 :50
Worked Example L3-2 – Continuous mass
2 km
Given quarter circle with
uniform mass-density 
and radius 2 km:

j
O1


r

Calculate M total
Write r in polar coords
Write out double integral,
xˆ and yˆ components in
terms of r and phi.
Solve integral
Calculate
R CM
rdm  r (r )dA



M total
M total
RCM
Given origin O1
dA  (rdj )dr
41 :40
Lecture #3 Wind-up
. Fexternal  M R CM
.
rdm
rdm
. RCM 



M
dm

total
Office hours Today 2:30-4, tomorrow 4-5:30.
Homework problems in Taylor, Handout,
Galileo problem – check the web.
Second homework due in class Thursday 9/5
42 :72
Lecture #4 of 25
Questions on Assignment #2
Angular Momentum


And torque
And Central force
Moment of Inertia
DVD Demonstration on momentum
cons. and CM motion
43 :10
Impulse II -- Problem #L2-3
“Another car crash”
James and Joan are partly recovered from their previous
injuries, and haven’t learned from their experience.
They are drinking Jack Daniels and not wearing seatbelts.
James’ vehicle has velocity vector 30 xˆ m / s
Joan’s vehicle has 30 xˆ m / s
Both vehicles’ mass=M. Both people’s mass=70 kg.
Solve for case of inelastic and elastic collisions of vehicles.
Joan has an airbag in her vehicle. It takes her 100 millisec
to reduce her velocity relative to her vehicle to zero.
James stops 5 millisec after impacting the steering wheel.
What impulse does each person experience? What is
44 :20
average force for each? How many “g’s” do they feel.
Worked Example L3-1 – Discrete masses
Given m1 to m10
2 units
y
O2
y
x
O1
x
m= m
m = 3m
Calculate
1 unit
N
RCM 
N
m r  m r



1
N
m


1

1
M total
RCM
Given origin O1
For homework
given O2
45 :30
Worked Example L3-2 – Continuous mass
2 km
Given quarter circle with
uniform mass-density 
and radius 2 km:

j
O1


r

Calculate M total
Write r in polar coords
Write out double integral,
xˆ and yˆ components in
terms of r and phi.
Solve integral
Calculate
R CM
rdm  r (r )dA



M total
M total
RCM
Given origin O1
dA  (rdj )dr
46 :40
Equivalence of Linear and Angular motion
equations
Fext
dPtotal

dt
Ptotal  M cm vcm
 ext
dLtotal

dt
Ltotal  I 
L  r  p;   r  F
vtangential    r
47 :45
Moment of Inertia vs. Center of Mass
For a multi-particle discrete mass-distribution
N
N
I   m r r
R CM 
 1
m r


1
M total
For a continuous mass-distribution
2
2
2
I   r dm   r  dV   r  dA
.
R CM
rdm  r  (r )dV  r (r )dA




M total
M total
M total
48 :45
Angular Momentum and Torque
dL d
Lrp
 (r  p )
dt dt
d
(r  p)  (r  p )  (r  p)
dt
p  mr  (r  p )  (r  mr )  0
d
(r  p )  (r  p )  (r  F )
dt
dL
 (r  F )  
dt
49 :50
Parallel Axis Theorem
d
Axis 2 (Parallel to
axis 1)
CM
Axis 1 (through CM)
I Parallel  I CM  M total d 2
50 :55
Angular Momentum and Central Forces
Fcentral  F ( r )rˆ
dL
 (r  F )  (r  F (r ) rˆ)  0 because r  rˆ  0
dt
L  constant : Angular momentum conserved by Central forces
51 :55
Moment of Inertia worked problem
2
L
1
Calculate the moment
of inertia and kinetic
energy of a wire of
uniform mass-density
lambda, mass M, and
length L.

A) If rotated about axis
at midpoint at angular
velocity

1
B) If rotated about axis
at endpoint at angular
velocity
2
52 :60
Lecture #4 Wind-up

2
. I  r dm
.
dLsystem
.
  external
dt
I Parallel  I CM  M total d 2
We are done w/ chapters 1 and 3, read all.
Assignment includes Taylor /Supplement /
Lecture probs
Next assignment is longer – Give it more
time.
53 :72
Challenge questions
Given a cylinder with rho of r, what
function of should rho be to maximize
the moment of inertia of the body?
54
Lecture #5 of 25
Questions on Assignment #2
Angular Momentum


And torque
And Central force
Moment of Inertia
DVD Demonstration on impulse and
collisions
55
Angular Momentum and Central Forces
Fcentral  F ( r )rˆ
dL
 (r  F )  (r  F (r ) rˆ)  0 because r  rˆ  0
dt
L  constant : Angular momentum conserved by Central forces
Table of
Position vectors
System
Cartesian
Cylindrical
r
xxˆ  yyˆ  zzˆ
rrˆ  jjˆ  zzˆ
rrˆ  jjˆ  ˆ
Spherical
56 :45
Angular Momentum and Central Forces
Fcentral  F ( r )rˆ
dL
 (r  F )  (r  F (r ) rˆ)  0 because r  rˆ  0
dt
L  constant : Angular momentum conserved by Central forces
Table of
Position vectors
System
Cartesian
Cylindrical
dr
dxxˆ  dyyˆ  dzzˆ
drrˆ  rdjjˆ  dzzˆ
Spherical
drrˆ  r sin  djjˆ  rdˆ
57 :45
Velocity Dependent Force

  
F  F (r , r , t )  Forces are generally dependent on
velocity and time as well as position
 

 2

Fr (r )  br  cr  Fluid drag force can be
approximated with a linear and a
quadratic term
b   k D
c  kA
drag coefficient
b = Linear
(Stokes Law, Viscous or “skin” drag)
c
= Quadratic drag coefficient
( Newton’s Law, Inertial or “form” drag)
58 :15
Lecture #5 of 25
Moment of inertia
Retarding forces







Stokes Law (viscous drag)
Newton’s Law (inertial drag)
Reynolds number
Plausibility of Stokes law
Projectile motions with viscous drag
Plausibility of Newton’s Law
Projectile motions with inertial drag
59 :10
Moment of inertia L5-1
R
Given a solid quarter
disk with uniform massdensity  and radius R:

j
O1


r

Calculate I total
Write r in polar coords
Write out double integral,
both r and phi
components
Solve integral
Calculate
2
I CM   r 2 dm   r  (r )dA
I O1
Given that CM is
located at (2R/3, p/4
Calculate ICM
60 :10
Velocity Dependent Force

  
F  F (r , r , t )  Forces are generally dependent on
velocity and time as well as position
 

 2

Fr (r )  br  cr  Fluid drag force can be
approximated with a linear and a
quadratic term
Ratio
f quad
f lin
is important
drag factor
b = Linear
(Stokes Law, Viscous or “skin” drag)
c
= Quadratic drag factor
( Newton’s Law, Inertial or “form” drag)
61 :15
The Reynolds Number
density 
viscosity 
R
R < 10 – Linear drag
1000< R < 300,000 – Quadratic
R > 300,000 – Turbulent
D
vD
R

inertial (quad ) drag
viscous (linear ) drag
v
62
:20
vD
R

The Reynolds Number II
D
vD
R

R < 10 – Linear drag
Fd
1
2
v

2
CD 
v
1000< R < 300,000 – Quadratic
R > 300,000 – Turbulent
Linear Regime
kD v
CD  1 2
v A
2
density 
viscosity 
1
D

2
 vD Re
Quadratic Regime
CD 
kA 12  v 2
1
2
v A
2
k
63
:25
Defining Viscosity

Fdrag
y
A
u xˆ 
y
Fdrag
u
A
y
x
Two planes of Area “A” separated by gap  y
Top plane moves at relative velocity u xˆ

u defines viscosity  (“eta”)
F A
y
2

N

s
/
m
MKS Units of
are Pascal-seconds
Only CGS units (poise) are actually used
2
1 poise=0.1 N  s / m
64
:30
Viscous Drag I

Fdrag
A
u xˆ 
Fdrag
du
  A
xˆ
dy
An object moved through a fluid is surrounded by a
“flow-field” (red).
Fluid at the surface of the object moves along with the
object. Fluid a large distance away does not move
at all.
We say there is a “velocity gradient” or “shear field”
near the object.
We are changing the momentum of the nearby fluid.
This dp/dt creates a force which we call the viscous
drag.
65
:35
Viscous Drag II

Fdrag
D
u xˆ 
Fdrag   k D u xˆ
“k” is a “form-factor” which depends on the
shape of the object and how that affects the
gradient field of the fluid.
“D” is a “characteristic length” of the object
The higher the velocity of the object, the
larger the velocity gradient around it.
Thus drag is proportional to velocity
66
:40
Viscous Drag III – Stokes Law

Fdrag
D
u xˆ


Fdrag  b r

Fdrag   3p D u xˆ
Form-factor k becomes 3p
“D” is diameter of sphere
Viscous drag on walls of
sphere is responsible for
retarding force.
George Stokes [1819-1903] 
(Navier-Stokes equations/ Stokes’ theorem)
67
:45
Falling raindrops L5-2
A small raindrop falls through a cloud. It has a 10 mm
radius. The density of water is 1 g/cc. The
viscosity of air is 180 mPoise.
a) Draw the free-body diagram.
b) Quantify the force on the drop for a velocity of 10
mm/sec.
c) What is the Reynolds number of this raindrop
d) What should be the terminal velocity of the
raindrop?
Work the same problem with a 100 mm drop.
68 :50
Stokes Dynamics
69 :10
Lecture #5 Wind-up
. R  vD

.
g
 t

.
v
vz  v 1  e


mg
vt 
3pD



Read sections 2.1-2.4
70 :72
Class #6 of 30
Homework #2 –


“HAP”
Rocket with gravity
Retarding forces



Projectile motions with viscous drag
Plausibility of Newton’s Law
Projectile motions with inertial drag
Homework #3 –

CM of a sphere
Demo – High Speed photography
71 :15
Falling raindrops I

Fdrag
, 

mg
Problems:
A small raindrop falls through a
cloud. At time t=0 its
velocity v=0.
Describe it’s velocity vs.
time.
Raindrop is 10 mm diameter,
density is 1 g/cc, viscosity
of air is 180 mPoise
z
x
72 :18
Falling raindrops II

F


1) Newton
mr  mgzˆ  br
Assume vertical motion

mg
2) On z-axis



mz  mg  bz
z
dvz
b
3) Rewrite in terms
 g  vz
of v
x
dt
m
b
b 4) Variable
Define u  g  vz  u   vz
substitution
m
m
b

t
b
5) Solve by
u   u  u  u (0)e m
inspection
m
drag
73 :23
Falling raindrops III

b
t
m
u  u (0)e
b
mg m
u  g  vz  vz 
 u
m
b
b
b

t
mg m
vz 
 u (0)e m
b b b
mg

t


mg
m  v 

vz 
1 e 
3p D

b 
3pD

t
mg 
1  e m 
vz 

3pD 

g
 t

v
vz  v 1  e 




1) Our solution
2) Substitute
original
variable
3) Apply boundary
conditions
4) Expand “b”
5) Define vterminal
74 :30

Inertial Drag I
An
v t

 2
Fdrag  cr

2
Fdrag  kAn v
Plate with area “An” moves a distance v  t through
fluid with density 
The mass of the fluid displaced is M  An v t
Mass “M” must acquire a velocity “v” to move out of
the way of the plate.
The moving plate is causing p  Mv  ( An v t )  v
Rearranging we get p
t
 An v
2
75 :35
Inertial Drag II – A sphere

2
Fdrag  kAnv
1
2
An  p D
4
1
k
4

p
2 2
Fdrag  D v vˆ
16
Previously
demonstrated
“An” means “A
normal to velocity”
 Form factor for sphere
 Plug ‘n’ play
76 :40
Falling raindrops redux

Fdrag


mg
z
Problems:
A small raindrop falls through a
cloud. At time t=0 its
velocity is zero.
Describe it’s velocity vs.
time.
Raindrop is 1000 mm diameter,
density is 1 g/cc.
x
77
Falling raindrops redux II

Fdrag
mr  mgzˆ  cr 2
Assume vertical motion
mz  mg  cz 2
dvz
c 2
 g  vz
dt
m

mg
z
x
1) Newton
2) On z-axis
3) Rewrite in terms
of v
dvz
vz 2
mg
Define v 

 g (1  2 ) 4) Rearrange terms
c
dt
v
dvz
 vz 2  g  dt
(1  2 )
v
5) Separate
variables
78 :45
Falling raindrops redux III
v (t )
t
dvz
dvz
 g  dt  gt
2
 vz 2  g  dt 0
0
vz
(1 
(1  2 )
2)
vt
v


dx
vz
u   dvz  v du
  (1  x 2 )  arctanh( x) 
v


v du
0 (1  u 2 )  v arctanh(v(t ) / v )  gt
v(t )  v tanh( gt / v )
v ( t ) / v
79 :50
Tanh and sinh and cosh, oh my
 ix
e e
cos x 
2
ix
 ix
e e
sin x 
2
eix  eix
tan x  ix ix
e e
sin, sinh and tanh
5
sin(x)
sinh(x)
tanh(x)
4.5
4
3.5
3
f(x)
ix
2.5
2
1.5
x
e e
cosh x  cos ix 
2
e x  e x
tanh x  tan ix  x  x
e e
x
1
0.5
0
0
50
100
150
200
x [degrees]
250
300
80 :55
350
L6-1 Liquid water drops on Io
A small water droplet is traveling sideways on a cloud
on Jupiter’s moon Io. At time t=0 its velocity is
purely horizontal v0  3 xˆ m / s 2
Io is tiny. Neglect gravity.
Draw forces on the droplet.
Write down differential equation for velocity assuming
Stokes-law force.
Solve to get raindrop’s velocity vs. time.
Calculate the “decay time” for droplet velocity
assuming R=100 um and  for Io’s atmosphere
=20 m-poise.
81 :60
Falling raindrops L6-2
A small raindrop falls through a cloud. It has a
1000 mm radius. The density of water is 1
g/cc. The viscosity of air is 180 mPoise.
The density of air is 1.3 g/liter at STP.
a) Draw the free-body diagram.
b) What should be the terminal velocity of the
raindrop, using quadratic drag?
c) What should be the terminal velocity of the
raindrop, using linear drag?
d) Which of the previous of two answers
should we use??
e) What is the Reynolds number of this
82
raindrop?
:70
Integration in Different Coordinates
Table of
Position vectors
System
r
Cartesian
xxˆ  yyˆ  zzˆ
Cylindrical
r cos(j ) xˆ  r sin(j ) yˆ  zzˆ
Spherical r cos(j )sin( ) xˆ  r sin(j )sin( ) yˆ  r cos( ) zˆ
83 :45
Lecture #6 Wind-up

ˆ Linear Drag
. Fdrag   3p D u x
g

t

v
vz  v 1  e 





p
2 2
Fdrag  D v vˆ
16
v(t )  v tanh( gt / v )
Quadratic Drag
Read sections 2.4-2.7
Matlab Basics
84 :72
Class #7 of 30
Integration of vector quantities

CM problems revisited
Energy and Conservative forces




Stokes Theorem and Gauss’s Theorem
Line Integrals
Curl
Work done by a force over a path
Angular momentum demo
85 :03
Test #1 of 4
Thurs. 9/26/02 – in class
Four problems


Bring an index card 3”x5”. Use both sides. Write
anything you want that will help.
All calculations to be written out and numbers
plugged in BEFORE solving with a calculator. Full
credit requires a units check.
Linear and Angular momentum / Impulse
Moment of Inertia / Center of Mass
Retarding forces (Stokes / Newton etc.)
Conservative forces / Line integrals / curls /
energy conservation
86 :08
Integration in Different Coordinates
System
Cartesian
Cylindrical
Spherical
Position vector r
xxˆ  yyˆ  zzˆ
r cos(j ) xˆ  r sin(j ) yˆ  zzˆ
r cos(j ) sin( ) xˆ  r sin(j ) sin( ) yˆ  r cos( ) zˆ
87 :12
Worked Example L7-1 – Continuous mass
Given hemisphere with
uniform mass-density 
and radius 5 m:




Calculate M total
Write r in polar coords
Write out triple integral,
x.ˆ , yˆ and zˆ components
in terms of r and phi
Solve integral
Calculate
R CM
rdm  r (r )dV



M total
M total
RCM
Given origin O1
dV  (rd )(r sin  dj )dr
88 :20
Worked Example L3-2 – Continuous mass
2 km
Given quarter disk with
uniform mass-density 
and radius 2 km:

j
O1


r

Calculate M total
Write r in polar coords
Write out double integral,
xˆ and yˆ components in
terms of r and phi.
Solve integral
Calculate
R CM
rdm  r (r )dA



M total
M total
RCM
Given origin O1
dA  (rdj )dr
89 :30
Conservative Forces
A force is conservative iff:
F  F (r )
F  F (r , r , t )

P2
P1
F dr  Const
( over all paths)

P2
P1
1. The force depends only on
r
Dependence on r , t not allowed .
2. For any two points P1, P2 the
work done by the force is
independent of the path taken
between P1 and P2.
F dr  Work ( P1  P 2)
r
U (r )  W (r0  r )    F (r ) dr 
r0
90 :35
Line integral and Closed loop integral
Conservative force
P1
 F dr  Const
 F dr  0
P2



a) P1 and P2 with two
possible integration paths.
b) and c) P1 and P2 are
brought closer together.
d) P1 and P2 brought
together to an arbitrarily
small distance .
Geometric argument that
conservative force implies
zero closed-loop path
integral.
91 :40
Integration by eyeball
a
b
Conservative force
 F dr  0
c
d
e
f
92 :50
y
L7-2 – Path integrals
Q(0,1)
Taylor 4.3
c
F (r )   yxˆ  xyˆ
b
O
Calculate, along (a)
Calculate, along (b)
Calculate, along (c)
Calculate
x
P(1,0)
a

OQP
F dr
(modified)
Q
 F dr
 F dr
 F dr
P
P
Q
P
Q
93 :65
Angular Momentum and Central Forces
Fcentral  F ( r )rˆ
dL
 (r  F )  (r  F (r ) rˆ)  0 because r  rˆ  0
dt
L  constant : Angular momentum conserved by Central forces
Taylor 3-25
Given m, 0 , r0 and central force
Calculate  given new shorter r
Fcentral  F ( r )rˆ
94 :70
Lecture #7 Wind-up
r
. (r )   F (r ) dr 
U

r0

P1
P2
For conservative
forces
F dr  C  F dr  0
Read Chapter 4
First test 9/26
95 :72
Lecture #8 of 24
Homework #3 –


Moment of Inertia of disk w/ hole
Buoyancy
Energy and Conservative forces


Force as gradient of potential
Curl
 Stokes Theorem and Gauss’s Theorem



Line Integrals
Energy conservation problems
Demo – Energy Conservation
96
:02
Disk with Hole
R, M
d
Axis 2 (Parallel to axis 1)
Axis 1 (through CM)
I Parallel  I CM  M total d 2
:10
97
Buoyancy
Archimedes 


Impure crown?
King Hiero commissions
”Mission oriented Research”

Taking bath
Noticing water
displaced by his body
as he got in the tub
Running starkers thru the
streets shouting “Eureka”
The buoyant force on
an object is equal to the
WEIGHT of the fluid
displaced by that object
Fb  Vg
98 :15
Force as the gradient of potential
Scalar
Vector
x
r
0
r0
F ( x)   f ( x) dx U (r )    F (r ) dr 
dF
f ( x) 
dx x x
F (r )  U r r



  xˆ  yˆ  zˆ
x y
z

1  ˆ
1 
  rˆ 

jˆ
r r 
r sin  j
U  Axy  B sin Cz
3
99 :20
Gravitational Potential
GMm
U (r )  
r
1
F  U  GMm
r
 1 GMm
Fr  GMm
 2
r r
r
F  Fj  0



  xˆ  yˆ  zˆ
x y
z

1  ˆ
1 
  rˆ 

jˆ
r r 
r sin  j
100 :25
Curl as limit of tiny line-integrals
F dr

(curl F ) nˆ  lim
ai 0
ai
101 :25
Stokes and Gauss’s theorem’s
Gauss – Integrating divergence over a volume is
equivalent to integrating function over a surface
enclosing that volume.
Stokes – Integrating curl over an area is equivalent
to integrating function around a path enclosing
that area.
102 :30
The curl-o-meter (by Ronco®)
a
b
Conservative force
 F dr  0   F  0
c
 xˆ
e



  f  det
 x

 f x
d
f
yˆ

y
fy
zˆ 

 
z

f z 
103 :35
Maxwell’s Equations
m0
 B  I
A
Curl is zero EXCEPT
Where there is a current I.

 E  ; B 0
0
B
 E  
t
E
 B  m0 J   0 m0
t
104 :35
y
L8-1 – Area integral of curl
Q(0,1)
Taylor 4.3
c
F (r )   yxˆ  xyˆ
b
O
a
x
P(1,0)
Calculate, along a,c
Calculate, along a,b
Calculate, inside a,c
Calculate, inside a,b
(modified)

OQP
F dr
  F dA
OQP
105 :50
L8-2 Energy problems
y
h
y

O
x

x
A block of mass “m” starts from rest and slides down
a ramp of height “h” and angle “theta”.
a) Calculate velocity “v” at bottom of ramp
b) Do the same for a rolling disk (mass “m”, radius r)
c) Do part “a” again, in the presence of friction “m”
106 :65
Retarding forces summary
Fdrag 
3p Duxˆ  buxˆ
g
 t

vz  v  1  e v


Fdrag 
p
Linear Drag
on a sphere (Stokes)

mg
 ; v 

b Falling from rest w/ gravity

 D v vˆ  cv vˆ Quadratic Drag on a
16
2 2
v
2
2

mg

c
v(t )  v tanh( gt / v )
v (t ) 
v0
cv0
1
t
m
Sphere (Newton)
Falling from rest w/ gravity
Decelerating from v without gravity
107 :72
Lecture #8 Wind-up

. F dr 
Path

Surface
( F ) dA  Stokes’ Theorem
 F  0
Read Chapter 4
9/24 Office hours 4-6
pm
First test 9/26
HW posted midafternoon
For conservative
forces
108 :72
Lecture #9 of 24
Test advice
Review problems




Moment of Inertia of disk w/ hole
Line Integrals
Energy conservation problems
Others of interest
Energy


Pendulum / Simple and solid
2nd derivative as a spring constant
109
:02
Assumed to know
Resolving a vector into two components
Drawing a free body diagram and using it to
setup differential equation of motion
Selecting an appropriate coordinate system
and reference frame
Solving a separable differential equation
Applying energy conservation, momentum
conservation and understanding basic laws of
the trajectory of flying objects
Integrating in polar / cylindrical / spherical
coords.
Doing line integrals
:05
110
Best Advice
Draw a picture!!!
Check that signs and units make sense
Check that magnitude of answer makes sense
Try limiting conditions
:08
111
Richard’s index card (front)
I Parallel  I CM  M total d
r  s  r s cos( ) 
2
3
rs
i 1
i

Path
F dr  
Surface
( F ) dA
i
 xˆ


  f  det 
 x

 f x
yˆ

y
fy
zˆ 

 
z

f z 



xˆ  yˆ  zˆ
x y
z
Impulse J 
F (t )dt  p
t1

1  ˆ
1 
Conservative force   F dr  0   F  0
ˆ
 r

jˆ
r
r r 
r sin  j
 mf 


U
(
r
)


F
(
r
)
dr
v f  ve ln 



t2

 mi 
r0
mvexhaust  " thrust "
N
RCM 
m r


1
M total
Fexternal  M RCM
RCM
F (r )  U rr
rdm


 dm L  r  p  dL  (r  F )  
Torques and forces on an
Extended object operate on its center
of mass as if it were a point object.
dt
Ltotal   ri  pi 
i
dLtotal
 ext
dt
112
:12
Richard’s Index card (back side)
Fdrag 
3p Duxˆ  buxˆ
g
 t

vz  v  1  e v


Fdrag 
p
16
Linear Drag
on a sphere (Stokes)

mg
 ; v 

b Falling from rest w/ gravity

 D v vˆ  cv vˆ Quadratic Drag on a
2 2
1 poise  0.1N  s / m2
v
2
2

mg

c
v(t )  v tanh( gt / v )
v (t ) 
v0
cv0
1
t
m
Sphere (Newton)
Falling from rest w/ gravity
Decelerating from v without gravity
113 :15
Analogy of 1-D system to roller coaster
E1
E2
E3
E4
x
T(x)=E-U(x) <- General 1-D system
T(x)=E-mgx <- Roller Coaster
114 :20
Potential Wells
mx  kx

x(t )  A sin 

mx  kx
x(t )  Ae

 

k 
t
m 
K<0

k
t   
m

 Be



K>0
k 
t
m 
Mass m
Spring constant k
115 :25
Taylor Series Expansion
Taylor series -- Generic
f ( x  x0 )  f ( x0 )  f ( x0 )( x  x0 ) 
1
2!
f ( x0 )( x  x0 ) 2  ...
Taylor series -- Potential
1
U ( x  x0 )  U ( x0 )  U ( x0 )( x  x0 )  U ( x0 )( x  x0 ) 2  ...
2!
Can be ignored or set to zero … “gauge invariance”
1
Is already zero for potential evaluated
about a critical point x-0
Thus U ( x  x0 )  U ( x0 )( x  x0 ) 2  ...
2
1
 keffective ( x  x0 )2
2
keffective  U ( x0 )
116 :35
L9-1
Potential
Wells
Equivalent curvatures
30
U ( x)  20  10 x  x2  x3  x4  x5
20
Y
10
Evaluate keffective
0
-10
What is equation of parabola
of equivalent curvature?
-20
-30
-2
-1
0
X
1
2
keffective  U ( x0 )
10
0
-10
-20
-30
-40
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
117 :35
L 9-2 Potential Wells
-21
x 10
Lennard-Jones potential
Van Der Waals attraction (R-6)
-12
Pauli repulsion (R )
1.5
1
Energy (joules)
  12   6 
U (r )  4       
 r   r  
  1.05 1021 joule / molecule
  0.316 nm
Lennard-Jones potential for water
0.5
Calculate the
Equilibrium
separation of
two water
molecules
0
-0.5
-1
-1.5
0
0.5
1
1.5
Distance (nm)
118 :45
y
L8-1 – Area integral of curl
Q(0,1)
Taylor 4.3
c
F (r )   yxˆ  xyˆ
b
O
a
x
P(1,0)
Calculate, along a,c
Calculate, along a,b
Calculate, inside a,c
Calculate, inside a,b
(modified)

OQP
F dr
  F dA
OQP
119 :55
L8-2 Energy problems
y
h
y

O
x

x
A block of mass “m” starts from rest and slides down
a ramp of height “h” and angle “theta”.
a) Calculate velocity “v” at bottom of ramp
b) Do the same for a rolling disk (mass “m”, radius r)
c) Do part “a” again, in the presence of friction “m”
120 :65
Lecture #9 Wind-up
d2 f
keffective  2
dx
critical po int.
Office hours today 4-6

Wed 4-5:30
Thursday


Hand-written index card
Official physics league 3’x5’ size
121 :72