Transcript Testing

Intermediate Mechanics
Physics 321
Richard Sonnenfeld
New Mexico Tech
:00
Two particles with central forces
L
L
m2
2m
m
m1
L
m
2m
2 :35
y
h
y

O
x

x
3
Class #12 of 30
Finish up problem 4 of exam
Status of course
Lagrange’s equations

Worked examples
 Atwood’s machine
4 :72
Physics Concepts
Classical Mechanics
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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
5 :04
Mathematical Methods
Vector Calculus
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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

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Coordinate change / rotations
Diagonalization / eigenvalues / principal axes
Lagrangian formulation
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Calculus of variations
“Functionals”
Lagrange multipliers for constraints
General Mathematical competence
6 :06
Joseph LaGrange
Giuseppe Lodovico Lagrangia
The reader will find no figures in this work. The methods which I set forth do not require
either constructions or geometrical or mechanical reasonings: but only algebraic
operations, subject to a regular and uniform rule of procedure.
Preface to Mécanique Analytique.
Rescued from the guillotine by Lavoisier – who was himself killed. Lagrange Said:
“It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”
If I had not inherited a fortune I should probably not have cast my lot with mathematics.
I do not know. [summarizing his life's work]
Joseph Lagrange [1736-1813] 
(Variational Calculus, Lagrangian
Mechanics, Theory of Diff. Eq’s.)
Greatness recognized by Euler and
D’Alembert
1788 – Wrote “Analytical Mechanics”.
You’re taking his course.
7
:08
:45
Lagrange’s Equation
d  L 
F  ma 
 
0
 q i dt   q i 
L
Works best for conservative systems
Eliminates the need to write down forces of
constraint
ASSUMES THAT q-I are consistent with
forces of constraint!!
Automates the generation of differential
equations (physics for mathematicians)
Is much more impressive to parents, employers,
and members of the oppositeLsex T  U 8
:12
Lagrange’s Kitchen
Mechanics “Cookbook” for
Lagrangian Formalism
1) Write down T and U in any convenient
coordinate system.
It is better to pick “natural coords”, but isn’t necessary.
2) Write down constraint equations
Reduce 3N or 5N degrees of freedom to smaller number.
3) Define the generalized coordinates q i
One for each degree of freedom
4)
5)
6)
7)
8)
Rewrite L  T  U in terms of q i , q i
L
L
d L
and
Calculate and
q
q
dt  q
L
Plug into  q
Solve ODE’s
Substitute back original variables
i
d  L 
 
0
dt   q i 
9
:17
Degrees of Freedom for Multiparticle
Systems
5-N for multiple rigid bodies
3-N for multiple particles
10 :20
Atwood’s Machine
Reverend George Atwood – Trinity College, Cambridge / 1784
Two masses are hung from a frictionless,
massless pulley and released.
A) Describe their acceleration and motion.
B) Imagine the pulley is a disk of radius R
and moment of inertia I. Solve again.
:25
11
:45
Atwood’s Machine
Lagrangian recipe
1
1) T 
m 1 y1 
2
1
m 2 y 2 ; U  m1 gy1  m 2 gy 2
2
2
2
2) 2  particles  10 degrees of freedom
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
y1
y2
m1
m2
y1  y 2  k  y 2  k  y1  y 2   y1
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q
4) L 
5)
L
q
L
d  L 
6)
 
0
q
dt   q 
L
q
1
2
( m1  m 2 ) q  g ( m1  m 2 ) q
2
L  T U
  g ( m1  m 2 )
 ( m1  m 2 ) q ;
d L
dt  q
 g ( m1  m 2 )  ( m1  m 2 ) q  0
 ( m1  m 2 ) q
12
:40
Atwood’s Machine
Lagrangian recipe
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q L  T  U
4) L 
1
2
( m1  m 2 ) q  g ( m1  m 2 ) q
2
d  L 
6)
 
0
y 2  q dt   q 
L
y1
q
m1
m2
q
g ( m1  m 2 )
 ( m1  m 2 )
1
2
a
 g ( m1  m 2 )  ( m1  m 2 ) q  0
*
a t  q0t  q0  y 
* 2
1
1
2
a t  y 0t  y
* 2
1
1
0
13
:45
Atwood’s Machine
Simulation
14
:45
:50
The simplest Lagrangian problem
M ass m ; Velocity v 0 xˆ
v0
m
U  m gz
T 
1
2
mx 
2
1
2
my 
2
1
mz
2
2
A ball is thrown at v0 from a tower of height s.
g
Calculate the ball’s subsequent motion
1)
2)
3)
4)
5)
6)
7)
8)
Write down T and U in any convenient coordinate system.
Write down constraint equations
Define the generalized coordinates q i
Rewrite L  T  U in terms of q i , q i
L
L
d L
and
and
Calculate  q
q
dt  q
Plug into  L
q
Solve ODE’s
Substitute back original variables
i

d  L 

0
dt   q i 
15 :65
Class #12 Windup
d  L 
F  ma 
 
0
 q i dt   q i 
L
L  T U
Office hours today 4-6

Wed 4-5:30
16 :72
Atwood’s Machine with massive pulley
Lagrangian recipe
3
I
1) T 
R
1
2
m1 x 
2
1
2
m2 y 
2
1
2
I  ; U  m1 gx  m 2 gy
2
2) 3  objects  15 degrees of freedom
y1
y2
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
x 3 , y 3 , z 3 ,  3  const .;  3  
m1
m2
y1
R
y1  y 2  k  y 2  k  y1  y 2   y1
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q
17
:70
Class #13 of 30
Lagrange’s equations

Worked examples
 Pendulum with sliding support

You solve it
 T7-17 Atwood’s machine with
massive pulley
 T7-4, 7-16 Masses on ramps
18 :02
x1
Pendulum with sliding support-I
1) T 
1
m1 x1 
2

What is the frequency of small
oscillations of the
pendulum?
What is the center of mass
motion?
2
2
y1 , z1 ,  1 ,  1  const .; y 2 ,  2 ,  2  const .
m2
x 2  x1  L sin( )  x 2  x1  L cos( )
x2
A pendulum with mass m2 and
length L is suspended from
a block of mass m1 resting
on a frictionless plane.
2
m 2 ( x 2  z 2 ); U  m 2 gz 2
2) 2  objects  10 degrees of freedom
L
z2
1
2
z 2  L (1  cos( ))  z 2  L sin( )
10 degrees  9 constraints  1 new variable
 2 generalized coords
3) q1  x1 , q1  x1 ; q 2   ; q 2  
4) x 2  x1  L  cos   2 x1 L cos 
2
2
2
2
2
x 2  q1  L q 2 cos q 2  2 q1 q 2 L cos q 2
2
2
2
2
z 2  L q 2 sin q 2
2
2
2
2
2
19 :10
Pendulum with sliding support-II
1
1) T 
2
1
m1 x1 
2
2
m 2 ( x 2  z 2 ); U  m 2 gz 2
2
2
3) q1  x1 , q1  x1 ; q 2   ; q 2  
4) x 2  x1  L  cos   2 x1 L cos 
2
2
2
2
2
x 2  q1  L q 2 cos q 2  2 q1 q 2 L cos q 2
2
2
2
2
2
z 2  L q 2 sin q 2
2
2
2
1
2
m1 x1 
1
L = ( m 1  m 2 ) q1 
1
4) L  T 
2
2
1
2
2
1
2
2
m 2 ( x 2  z 2 ); U  m 2 gz 2
2
2
m 2 ( q1  L q 2 cos q 2  2 q1 q 2 L cos q 2 )
2
2
2
2
 m 2 L q 2 sin q 2  m 2 gL (1  cos q 2 )
2
2
2
2
20 :15
Pendulum with sliding support-III
1
4)L = ( m1  m 2 ) q1 
2
2
5)
L
 q1
0
d L
dt  q1
L
q2
L
q2
d L
dt  q 2
L
 q1
1
2
m 2 ( L q 2  2 q1 q 2 L cos q 2 )  m 2 gL cos q 2
2
2
 ( m1  m 2 ) q1  m 2 q 2 L cos q 2
d L
 ( m1  m 2 ) q1  m 2 q 2 L cos q 2  m 2 q 2 L sin q 2
2
dt  q1
L
  m 2 q1 q 2 L sin q 2  m 2 gL sin q 2
q2
L
 m 2 L q 2  m 2 q1 L cos q 2
2
q2
d L
 m 2 L q 2  m 2 q1 L cos q 2  m 2 q1 q 2 L sin q 2
2
dt  q 2
 ( m 1  m 2 ) q1  m 2 q 2 L
  m 2 gL sin q 2
 m 2 L q 2  m 2 q1 L
2
 m 2 L q 2  m 2 q1 L
2
21 :20
Pendulum with sliding support-IV
1
L = ( m 1  m 2 ) q1 
2
2
1
2
m 2 ( L q 2  2 q1 q 2 L cos q 2 )  m 2 gL cos q 2
2
2
6) 0  ( m1  m 2 ) q1  m 2 q 2 L
m 2 L q 2  m 2 q1 L   m 2 gL sin q 2
2
0  ( m 1  m 2 ) q1  m 2 q 2 L
L q 2  q1 L   gLq 2
2
0  ( m1  m 2 ) x1  m 2 L
5)
d L
dt  q1
L
q2
L
q2
d L
dt  q 2
 ( m 1  m 2 ) q1  m 2 q 2 L
  m 2 gL sin q 2
 m 2 L q 2  m 2 q1 L
2
 m 2 L q 2  m 2 q1 L
2
L  x1   g 
22 :25
x1
Pendulum with sliding support-V
0  ( m 1  m 2 ) x1  m 2 L

L  x1   g 
L
z2
m2
x2
A pendulum with mass m2 and
length L is suspended from
a block of mass m1 resting
on a frictionless plane.
What is the frequency of small
oscillations of the
pendulum?

m2
( m1  m 2 )
L  x1


m2
L  1 
   g
 ( m1  m 2 ) 
 m1
L 
 m1  m 2
g  m1  m 2 
  
     fixed
L  m1 

g  m1  m 2
   g     
L  m1

m1  m 2
2
pendulum
m1
23 :40



x1
Pendulum with sliding support-VI
0  ( m 1  m 2 ) x1  m 2 L

L
z2
x2
 ( m
m2
1
 m 2 ) x1 dt dt    m 2 L   dt dt 
( m 1  m 2 ) x1   m 2 L   C 1 t  C 2
A pendulum with mass m2 and
length L is suspended from
a block of mass m1 resting
on a frictionless plane.
x1  
What is the center of mass
motion?
R cm
m2
m1  m 2
L  
m2
m1  m 2
x2


m2
m1  
x2   m 2 x2
m 1 x1  m 2 x 2
 m1  m 2 


m1  m 2
m1  m 2
R cm 
m
2
2
m1  m 2
x2
0
24 :45
Atwood’s Machine
Lagrangian recipe
1) T 
y1
y2
4) L 
1
2
1
2
m 1 y1 
2
1) T 
1
2
m 1 y1 
2
1
2
m2 y2 
2
1
2
2
2
q  y1  g 
m2
2
m 2 y 2 ; U  m1 gy1  m 2 gy 2
( m1  m 2 ) q  g ( m1  m 2 ) q
*
m1
1
 g ( m1  m 2 )
( m1  m 2 )
I 3 3 ; U  m1 gy1  m 2 gy 2
2
3
I
R
y1
y2
m1
m2
25
:60
T7-17 Atwood’s Machine with massive pulley
Lagrangian recipe
3
I
1) T 
2
R
y1
1
m 1 y1 
2
1
2
m2 y2 
2
1
2
I 3 3 ; U  m1 gy1  m 2 gy 2
2
2) 3  objects  15 degrees of freedom
y2
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
x 3 , y 3 , z 3 ,  3  const .;  3  ?
m1
m2
y1  y 2  k  y 2  k  y1  y 2   y1
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q ;  ??
1)
2)
3)
4)
5)
6)
7)
Write down T and U in any convenient coordinate system.
Write down constraint equations
Define the generalized coordinates q i
Rewrite L  T  U in terms of q i , q i
L
L
d L
Calculate  q and  q and dt  q
Plug into  L
q
Solve ODE’s
i

d  L 

0
dt   q i 
26
:70
T7-4, T7-16 Masses on ramps
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 acceleration “a” at top of ramp, time “t” to get to
bottom of ramp and velocity “v” at bottom of ramp. Use
the Lagrangian formalism.
b) Do the same for a rolling disk (mass “m”, radius “r”) 27
:65
Class #13 Windup
d  L 
F  ma 
 
0
 q i dt   q i 
L
New variables can be introduced so
long as add additional constraints
Generalized coordinates do not
need to be of same type (e.g.
angle / position).
Office hours today 3-5

Wed 4-5:30
28 :72
Test #2 of 4
Thurs. 10/17/02 – in class


Bring an index card 3”x5”. Use both sides. Write
anything you want that will help.
You may bring your last index card as well.
Lagrangian method (most of exam)
Line integrals / curls
Generalized forces / Lagrange multipliers /
constraint forces
Tuesday 10/15 – Real-time review / problem
session
29 :08
Class #14 of 30
Extensions of the Lagrangian Method

Generalized force and momenta
 Plausibility of the Lagrange equations

Lagrange Multipliers
 Calculating forces of constraint
 Worked example
 You solve it (Taylor 7-50).
30 :72
Atwood’s Machine
Lagrangian recipe
1
1) T 
m 1 y1 
2
1
m 2 y 2 ; U  m1 gy1  m 2 gy 2
2
2
2
2) 2  particles  10 degrees of freedom
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
y1
y2
m1
m2
y1  y 2  k  y 2  k  y1  y 2   y1
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q
4) L 
5)
L
q
L
d  L 
6)
 
0
q
dt   q 
L
q
1
2
( m1  m 2 ) q  g ( m1  m 2 ) q
2
L  T U
  g ( m1  m 2 ) generalized force
 ( m1  m 2 ) q generalized m om entum
 g ( m1  m 2 )  ( m1  m 2 ) q  0
31
:40
T7-16 Rolling mass on ramp
y
1) T 
1
mx 
2
2
1
my 
2
2
1
I  ; U  m gy   m gq sin 
2
2
2) 1  particle  5 degrees of freedom  new coordinate q
q
z ,   const .; x  q cos  ; y  q sin  ;   q / R
6 coords  5 constraints  1 generalized coordinate

x
3) x  y  q ;     q / R ; T 
2
2
2
1
mq 
2
4) L 
1
2
5)
L
q
mq 
2
1
2
I
q
2
R
2
2
1
I
2
q
2
R
2
; U   mgq sin 
 m gq sin 
 mg sin  generalized force ;
L
q
 (m 
I
R
2
) q generalized momentum
32 :65
Generalized Force and Momentum
Traditional
Force
Momentum
Generalized
F    U ; Fi  
U
 xi
 1
2 
p i  m vi 
 m vi 
 vi  2

Fi 
pi 
L
qi
L
qi
 1
2 
Li  I  i 
I

i 

 i  2

Newton’s
Law
Fi 
d
dt
Pi
L
qi

d L
d t qi
33 :72
Lagrange Multipliers
d  L 
f
 
   0
 q i dt   q i 
qi
Lagrangian method as practiced so far
eliminates the need to write down forces
of constraint BECAUSE it assumes that q i
are consistent with forces of constraint!!
Forces of constraint of form f ( q1 , q 2 , q3 ...)  const
may be found by method of Lagrange
multipliers
L
34
:12
Lagrange Multipliers
d  L 
f
 
  
 q i dt   q i 
qi
Lagrangian method as practiced so far
eliminates the need to write down forces
of constraint BECAUSE it assumes that q i
are consistent with forces of constraint!!
Forces of constraint of form f ( q1 , q 2 , q3 ...)  const
may be found by method of Lagrange
multipliers
L
35
:12
Lagrange’s Dining Room
Mechanics “Cookbook” for
Lagrange Multipliers
1) Write down T and U in any convenient
2)
3)
4)
5)
6)
7)
8)
coordinate system.
Write down constraints of form f ( q1 , q 2 , q 3 ...)  const
Define the generalized coordinates q i
Rewrite L  T  U in terms of q i , q i
f
L
L
d L
and
and
and
Calculate  q
q
q
dt  q
L
d  L 
f




Plug into  q dt   q 
q


Solve ODE’s and solve for lambda
f
The terms 
give the constraint forces
i
i
i
i
i
i
i
qi
36
:17
Atwood’s Machine
Lagrangian recipe
1
1) T 
m 1 y1 
2
1
m 2 y 2 ; U  m1 gy1  m 2 gy 2
2
2
2
2) 2  particles  10 degrees of freedom
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
y1
y2
m1
m2
y1  y 2  k  y 2  k  y1  y 2   y1
3) q  y1 , q  y1 ; y 2  k  q ; y 2   q
4) L 
5)
L
q
L
d  L 
6)
 
0
q
dt   q 
L
q
1
2
( m1  m 2 ) q  g ( m1  m 2 ) q
2
L  T U
  g ( m1  m 2 )
 ( m1  m 2 ) q ;
d L
dt  q
 g ( m1  m 2 )  ( m1  m 2 ) q  0
 ( m1  m 2 ) q
37
:40
Atwood’s Machine
Lagrange Multiplier Recipe
1) T 
1
m 1 y1 
2
1
m 2 y 2 ; U  m1 gy1  m 2 gy 2
2
2
2
2) 2  particles  10 degrees of freedom
x1 , z1 ,  1 ,  1  const .; x 2 , z 2 ,  2 ,  2  const .
y1
y2
m1
3) q1  y1 , q1  y1 ; q 2  y 2 , q 2  y 2 ;
4) L 
m2
d  L 
f
 




 q i dt   q i 
qi
1
2
5)
L
f ( y1 , y 2 )  y1  y 2  k
L
qi
  mi g ;
m 1 y1 
2
L
qi
1
2
m 2 y 2  m1 gy1  m 2 gy 2
 mi qi ;
2
d L
dt  q i
 mi qi ;  
f
qi
 
6)  m i g  m i q i    for i  1, 2
38
:40
Atwood’s Machine
Lagrange Multiplier Recipe
2) f ( y1 , y 2 )  y1  y 2  k  0
5)
y1
y2
m1
7) y1 
1
2
1
2
a1t  0  0; y 2 
a2t  k 
2
1
2
a1 t  a 2 
2
qi
1
2
2k
t
2
  mi g ;
L
qi
 mi qi ;
d L
dt  q i
 mi qi ;  
f
qi
 
6)  m1 ( g  y1 )    ;  m 2 ( g  y 2 )   
y1 
m2
2
L

m1
 g  a1 ; y 2 
a 2   a1 ;
a 2 t  0  0  k  y1
2
 a1 for all t  0

m2

m1
 g  a2
gg

m2
 1
 m1 m 2 
1 
2g   

    2g 

m
m
m

m
2 
2 
 1
 1
39
:40
Atwood’s Machine
Lagrange Multiplier Recipe
6) y1 

f
 y1


m1
 g  a1 ; y 2 
m2
 g  a2
 m1 m 2 
  2g 

m

m
2 
 1
f
y2
 m1 m 2
2g 
m1
m2
 m1  m 2
6) y1 
m1
f

 constraint force  T1 
 y1
 m1 m 2 
2g 
  m g ( for m 1  m 2 )
 m1  m 2 
y1

y2



 m 2  m1
g
 m1  m 2

g

40
:40
Taylor 7-50
A mass m1 rests on a frictionless horizontal table. Attached to it is a
string which runs horizontally to the edge of the table, where it
passes over a frictionless small pulley and down to where it
supports a mass m2. Use as coordinates x and y, the distances
of m1 and m2 from the pulley. These satisfy the constraint
equations f(x,y)=x+y=const. Write down the two modified
Lagrange equations and solve them for x’’, y’’ and the Lagrange
multiplier Lambda. Find the tension forces on the two masses.
41
:12
Class #14 Windup
d  L 
f
 
  
 q i dt   q i 
qi
f
L
FC  
q
qi
L
 generalized force
i
L
q
 generalized m om entum
Exam next Thursday
42 :72
Class #15 of 30
Exam Review

Taylor 7.50, 7.8, 7.20, 7.29,
ODE’s, 4.4, 4.7, 4.15, 4.19, 4.20
43 :72
Test #2 of 4
Thurs. 10/17/02 – in class


Bring an index card 3”x5”. Use both sides. Write
anything you want that will help.
You may bring your last index card as well.
Anything on last 3 homeworks
Lagrangian method (most of exam)
Line integrals / curls
Generalized forces / Lagrange multipliers /
constraint forces
Tuesday 10/15 – Real-time review / problem
session
44 :08
Taylor 7-50
A mass m1 rests on a frictionless horizontal table. Attached to it is a
string which runs horizontally to the edge of the table, where it
passes over a frictionless small pulley and down to where it
supports a mass m2. Use as coordinates x and y, the distances
of m1 and m2 from the pulley. These satisfy the constraint
equations f(x,y)=x+y=const. Write down the two modified
Lagrange equations and solve them for x’’, y’’ and the Lagrange
multiplier Lambda. Find the tension forces on the two masses.
x

m1
f
y

y
m2
f
x
f  x y
  g
m1 m 2
m1  m 2
45
:12
Atwood’s Machine
Lagrange Multiplier Recipe -- CORRECTED
6) y1 

f
 y1


m1
 g  a1 ; y 2 
m2
 g  a2
 m1 m 2 
  2g 

m

m
2 
 1
f
y2
 m1 m 2
2g 
m1
m2
 m1  m 2
6) y1 
m1
f

 constraint force  T1  
 y1
 m1 m 2 
  2g 
  m g ( for m 1  m 2 )
 m1  m 2 
y1

y2



 m 2  m1
g
 m1  m 2

g

46
:40
Class #14 Windup - CORRECTED
d  L 
f
 
  
 q i dt   q i 
qi
f
L
FC  
q
qi
L
 generalized force
i
L
q
 generalized m om entum
Exam next Thursday
47 :72
ODE Summary
Math
F or a , b  0
x  ax  x  x 0 e
at
Physics
v   3 D  v  v  v 0 e
x   ax  x  x 0 e
 3  D t
 at
m x   kx  x  x 5 sin(
x  bx  x  x1 e
bt
x   bx  x  x1 e
 x2e
i bt
 bt
 x2e
 x 3 sin( bt )  x 4 cos( bt )
 x 5 sin( bt   )
i b t
l   g      6 sin(
k
t)
m
g
t)
l
48 :55
Class #15 Windup
HW due
Thursday
Tues 3-5, Wed
4-5:30
Bring up to two
index cards
Midterm grades
will be posted
on web
Only one (hard)
HW problem
Happy 49ers!
49 :72
Class #17 of 30
Central Force Motion







Gravitational law
Properties of Inverse-square forces
Center of Mass motion
Angular momentum conservation
Lagrangian for Central forces
Reduced Mass and CM reference
frame
Central Forces DVD
BRIEF Exam Review
50 :72
Tides
Why are
there bulges
on BOTH
sides of the
earth?
51 :05
Gravity and Electrostatics
Gravity
Universal
Constant
1
G  6 .6 7 3(7 5)  1 0
Nm
kg
Force Law
U  G
4  0
 6 .9 8 7 5 5 1 ...  1 0
Nm
2
F  G
11
2
2
C o u lo m b
m1 m 2
r
Potential
Electrostatics
2
m1 m 2
r
rˆ
F 
U 
1
2
q1 q 2
4  0 r
2
1
q1 q 2
4  0
r
rˆ
52 :08
9
Potential of a spherical shell
Unique property
of inversesquare forces.
For r>rshell, treat
the entire shell
mass as if
concentrated at
a point at the
center of the
shell
F o r r  rsh ell ; U   G
m 1 m sh ell
F o r r  rsh ell ; U   G
m 1 m sh ell
r
rsh ell
53 :72
Two particles with central forces
m2
r1  r2
F ( r1  r2 )   G
U ( r1  r2 )   G
m1
L 
1
2
r1
r2
m 1 r1 r1 
R  RCM 
m1 m 2
( r1  r2 )
3
( r1  r2 )
m1 m 2
r1  r2
1
2
m 2 r2 r2  G
m 1 r1  m 2 r2
M
m1 m 2
r1  r2
; M  m1  m 2
r  r1  r2
Origin
r1  R 
m2
M
r ; r2  R 
m1
M
r
54 :30
Two particles with central forces
m2
L  L CM + L rel
r

m1
1
(M )R R 
2
L rel 
RC M
 
1
r r  G
m1 m 2
2
1
r
 (r  r  )  G
2
2
2
m1 m 2
2
m1 m 2
M
r
1
; L rel 
 (r  r  )  U (r )
2
2
  r
2
2
2
RC M : M RC M  0
G
m1 m 2
r
G
M
r
r :  r  G
m1 m 2
r
2
55 :35
Potential of a sphere
F o r r  rsh ell ; U   G
m 1 m sh ell
F o r r  rsh ell ; U   G
m 1 m sh ell




r
rsh ell
Using above formulae –
1) What is potential at radius r
outside a solid sphere of radius R and
mass M?
2) What is potential inside a solid
sphere of radius R (at radius r<R)?
3) What is force? Give answers in
terms of r, R, and M
56 :15
Potential and Force for solid spheres
F or r  R sphere ; F   G
F or r  R sphere ;
F  G
m1 M sphere
r
2
rˆ
m1 r M sphere
R
3
sphere
57 :18
rˆ
Vectors and Central forces
r1  r2
r1  r2 
Vectors

r2 
r1
r2

Many forces are of
form F ( r1  r2 )
Remove dependence
of result on choice of
origin
r1
Origin 1
Origin 2
58 :20
Two particles with central forces
m2
r1  r2
F ( r1  r2 )   G
U ( r1  r2 )   G
m1
L 
1
2
r1
r2
m 1 r1 r1 
R  RCM 
m1 m 2
( r1  r2 )
3
( r1  r2 )
m1 m 2
r1  r2
1
2
m 2 r2 r2  G
m 1 r1  m 2 r2
M
m1 m 2
r1  r2
; M  m1  m 2
r  r1  r2
Origin
r1  R 
m2
M
r ; r2  R 
m1
M
r
59 :30
Two particles with central forces
m2
L  L CM + L rel
r

m1
1
(M )R R 
2
L rel 
RC M
 
1
r r  G
m1 m 2
2
1
r
 (r  r  )  G
2
2
2
m1 m 2
2
m1 m 2
M
r
1
; L rel 
 (r  r  )  U (r )
2
2
  r
2
2
2
RC M : M RC M  0
G
m1 m 2
r
G
M
r
r :  r  G
m1 m 2
r
2
60 :35
 
m1 m 2
m1  m 2

m1
1
Earth and Moon
m1
m2
If
m1
 1; 
m2
G
m1
m 1 (1 
m E arth
6.0  10
24
m m oon
7.3  10
22
6.67  10
1.
 11
m2
)
2.
kg
kg
Nm
kg
2
3.
4.
2
5.
L unar m onth  28 days6.
T  2, 419, 200 sec
7.
  ?
Write down the relative Lagrangian
for the earth-moon system
Use it to get the relative radial
equation for the earth-moon system
What is the reduced mass to use (in
kg)?
How many percent different is it
than the lunar mass
What is theta-dot?
What is the radius of a circular
orbit?
How would this change if the earth
were fixed in space by the hand of
61 :30
God or a Borg tractor beam?
 
m1 m 2
m1  m 2

m1
1
Earth and Moon
m1
m2
If
m1
 1; 
m2
G
m1
m 1 (1 
m E arth
6.0  10
24
m m oon
7.3  10
22
6.67  10
1.
 11
m2
)
2.
kg
kg
Nm
kg
2
3.
4.
2
5.
L unar m onth  28 days6.
T  2, 419, 200 sec
7.
  ?
Write down the relative Lagrangian
for the earth-moon system
Use it to get the relative radial
equation for the earth-moon system
What is the reduced mass to use (in
kg)?
How many percent different is it
than the lunar mass
What is theta-dot?
What is the radius of a circular
orbit?
How would this change if the earth
were fixed in space by the hand of
62 :50
God or a Borg tractor beam?
Tides
Why are there
bulges on BOTH
sides of the earth?
63 :55
Class #17 Windup
L
L
gravity
rel
 


1
2
1
2
r r  G
M
r
<- Gravitational
Lagrangian
2
2
2
 ( r  r  )  U ( r ) <- General Central Force “1-D eqn”
2
m1 m 2
M
(M )R R 
1
F or r  R sphere ; F   G
F or r  R sphere ; F   G
m1 m sphere
r
2
rˆ
m1 m sphere r
R
3
sphere
Office - Tues 3-5, Wed 4-5:30
Midterm grades will be posted on web
immed. after lunch
64 :60
rˆ
Exam 2-3
In Figure (a), a spring with spring constant “k” has
been added to an Atwood machine that is glued to
the top of the pulley. [If the pulley turned a large
fraction of a revolution, the spring would no longer
be horizontal, which makes it pretty messy. Ignore
this complication.] Assume the wheel only turns a
few degrees from the position in which it is shown
so that we may assume that the string compresses or
stretches but remains horizontal.
a) Define an appropriate generalized coordinate.
Write down constraints. Write down T and U and
finally the Lagrangian for this case. The pulley is
massless and has a radius “R” (if you feel you need
to know). (10)
b) Write down Lagrange’s equation(s) to find the
differential equation for the acceleration of m1 (8)
c) Initially, at time t=0, we are told that the spring is
unstretched (it has length L), and that the masses are
at the same height (just as drawn). What is the
acceleration of mass m1 at t=0? (5)
d) The system has an equilibrium point at which the
spring length is not L. How far are the masses
displaced from their initial positions shown in the
figure at this equilibrium point? (You don’t need to
solve the differential equation to get this answer!)
(5)
65 :72
Exam 2-2, 2-4
t

[A hoop and a sphere both start from rest at height h,
and race down two identical ramps as shown below.
They have identical radius “R” and identical mass “m”
(and the mass is distributed uniformly as you should
expect in a physics problem!)]
a) Define the generalized coordinate you will use.
Define which direction is positive. Write the equation of
constraint (6).
b) Write down the Lagrangian for the hoop AND the
Lagrangian for the sphere in terms of the generalized
coordinate. You may use "I-hoop" and “I-sphere” as the
moments of inertia (8 pts)
c) Write down Lagrange's equations,
using I as the 2
2 still
2
I

M
R
and
I hoop  M R
sphere
moment of inertia. (6)
5
d) Use Lagrange's equation to get the acceleration of the
hoop and the sphere along the ramp using. . [You are
not required to solve for position as a function of time.
Just get acceleration.] 5)
e) What is the outcome of the race? (3)
f) Why? (One sentence) (2)
A bead of mass “m” is constrained to slide on a rod. The rod is rotating
about the z-axis such that its angle “phi” at any time is omega-t. The
rod is not horizontal or vertical, it is inclinded at an angle “theta”
(which doesn’t change).
 ,  and s
a) The correct generalized coordinates for this problem are where is
the angle the rod makes with respect to the z-axis (the axis of rotation),
as shown and s is the position “s” of the bead defined as its distance on
the rod from the origin. Write x, y and z in terms of . (hint …this is
very similar to using spherical coordinates). (5)  ,  and s
b) Write down the Lagrangian for the system. (Gravity matters) (5)
c) Write down Lagrange’s equation for “s” (only).(5)
  t
Extra Credit –
66
There is an equilibrium position for the bead at any . What is it? (5)
Class #18 of 30
Central Force Motion?

Gravity assist
Non-inertial
67 :72
Class #19 of 30
Homework problem 7-13 and
Hamilton’s principle
68 :72
Class #18 of 30
Celestial engineering



Central Forces DVD
The power of 
Equivalent 1-D problem and
Pseudopotential
 Kepler’s 3rd law

Orbits and Energy
 The earth-moon flywheel

Energy and Eccentricity
69 :02
The Story so far
Solid Sphere
For r>Rsphere,
entire mass as if r  R
sphere
concentrated at a
point at the center
r  R sphere
of the shell
L  L CM + L rel ;  
L rel 
1
2
r
F  G
M
2
m1 m 2
2
r :  r  G
2
m 1 r M sphere
R
m1 m 2
 (r  r  )  G
2
F  G
m1 M sphere
3
Spherical Shell
rˆ
F  G
m1 M shell
r
rˆ
2
rˆ
0
sphere
Lagrangian breaks into a
relative piece and a center
of mass piece.
r
m1 m 2
r
2
  r
2
70 :12
Reduced two-body problem
r2
r1
m moon
m Earth
rrelative


71
:15
The power of
T w o real m asses
O ne reduced m ass
 I 1  I 2 
T
T 
1
2
r  r1  r2
miv
ri
I 1 
2
2
G
1
2
I 2
m1 m 2
r
2

 I    r 
2
2
T 
1
I  
2
2
v
r
2
G
1
r 
2
2
2
m1 m 2
r
2
72
:20
The power of

- part 2
 I 1  I 2   m 1 r1   m 2 r2 
2
2
2
2
 m2 
 m1 
 m1 
r    m2 
r 
 M 
M 

m2
m1
  m1
 m2
2
2
M
M

2
m 2  m1

  m1 m 2
2
M

  r
2

2

r


T w o real m asses
O ne reduced m ass
 I 1  I 2 
T 
 m1 m 2

2
r  

 m 2  m1
1
2
miv
I 1 
2
2
G
ri
1
2
I 2
 I    r 
2
2
m1 m 2
r
2
T 
1
I  
2
2
v
r
2
G
1
r 
2
2
m1 m 2
r
2

2

r


2
r2
r1
m E a rth
m m oon
73 :25
2
Equivalent 1-D problem
L rel 
1
 (r  r  )  G
2
2
2
2
r :  r  G
r
m1 m 2
r
 r  G
Relative
Lagrangian
m1 m 2
2
m1 m 2
r
2
  r
2
Radial equation
 
2

r
3
Total Radial Force Ftotal
  Ftotal dr  U
  r
U
p seu d o
 G
m1 m 2
r
2
2

2 r
2
pseudo
74 :30
Pseudopotential and Energy
U
p seu d o
 G
m1 m 2
2

r
L rel  T  U 
 T U 
1
H
2
2 r
2 r
2
1
rel
E H
E 
1
rel

1
2
m1 m 2
2
r
 (r  r  )  G
2
2
2
m1 m 2
E 
r
2
r 
2
2
r U
2 r
2
G
m1 m 2
r
2
2
G
2
2
2
U pseudo
 (r  r  )  G
2
1
2
 r
pseudo
2
m1 m 2
r
75 :35
Pseudopotential and Energy
U
p seu d o
 G
m1 m 2
r
2

2 r
2
E   0  unbounded orbit
E 
1
2
E 
 r  U pseudo
2
1
2
 r
2
E   0  bounded orbit
76 :37
Potential Wells
K<0
m x   kx

x ( t )  A sin 


m x  kx
x (t )  A e




k 
t
m 

t  

m

k
 Be




K>0
k 
t
m 
Mass m
Spring constant k
77 :25
Taylor Series Expansion
Taylor series -- Generic
f ( x  x 0 )  f ( x 0 )  f ( x 0 )( x  x 0 ) 
1
2!
f ( x 0 )( x  x 0 )  ...
2
Taylor series -- Potential
U ( x  x 0 )  U ( x 0 )  U ( x 0 )( x  x 0 ) 
1
2!
U ( x 0 )( x  x 0 )  ...
2
Can be ignored or set to zero … “gauge invariance”
Is already zero for potential evaluated
about a critical point x-0
1
Thus U ( x  x 0 ) 

1
2
k effective ( x  x 0 )
2
2
U ( x 0 )( x  x 0 )  ...
2
k effective  U ( x 0 )
78 :35
Orbital Energy
m E arth
6.0  10
24
kg
m m oon
7.3  10
22
kg
  7.2  10
22
  2.6  10
6
r  3.9  10 m
8
kg
1.
2.
rad / s
3.
Last class we derived values for
omega, mu and r for the earthmoon system
Total energy consumption on Earth
is 1000 Terajoules/day
If we could power human activites
by stealing the angular kinetic
energy from the earth-moon
system, how much should omega
change to give us 100 years of
power?
79 :47
Converting r ( t ) to r (. )
 r  G
m1 m 2
r
r
1
u
;
d
dt

2
2

d d
dt d 
u  ( )   u ( ) 
r
3

u
2
d
 d
G m1 m 2 
2
2
r ( ) 
G m 1 m 2  (1   cos  )
80 :30
Properties of ellipses
r ( ) 
2
G m1 m 2  (1   co s  )
2
c
G m1 m 2 
rm in 
b

c
1 
1 
; rm ax 
c
1 
2
a
E 
 G m1 m 2  
2
2
2

2
1

81 :30
Energy and Eccentricity
r ( ) 
E 
2
G m1 m 2  (1   cos  )
 G m1 m 2  
2
2
2
Eccentricity Energy
 0
 1
E<0
0    1 E<0
   1
2
Orbit
Circle
 1
0  1
 0
Ellipse
 1
E=0
Parabola
 1
E>0
hyperbola
82 :30
Class #18 Windup
 r 
2
T 
1
U
r 
2
r
2
G
m1 m 2
r
 G
E 
1
2
2

r
2
2
v
p seu d o
m1 m 2
2 r
2
 r  U pseudo
2
2
HW Assignment #10 posted – Do
supplement first.
83 :60
Class #19 of 30
Celestial engineering - II





Reduced 2-body problem
Kepler 1st, 2nd and 3rd laws
Cometary collision prediction
A whiff of scattering theory
DVD – Gravitational slingshot
84 :02
Reduced two-body problem
r2
r1
m moon
m Earth
rrelative


85
:15
Equivalent 1-D problem
L rel 
1
 (r  r  )  G
2
2
2
2
r :  r  G
r
m1 m 2
r
 r  G
Relative
Lagrangian
m1 m 2
2
m1 m 2
r
2
  r
2
Radial equation
 
2

r
3
Total Radial Force Ftotal
  Ftotal dr  U
  r
U
p seu d o
 G
m1 m 2
r
2
2

2 r
2
pseudo
86 :30
Kepler’s 1st, 2nd and 3rd laws
4 r
2
 
2
3
GM
 
  r  const .
2
So  r  const .
2
dA
dt

1
bh 
2
1
2
 const .
'
(1610)
1st Law – Planets move in ellipses
with sun at one focus
Third law demonstrated previously
relates period to semi-minor radius
2nd law is direct consequence of
momentum conservation
“Equal areas are swept out in
equal times”
True for ALL central forces
( r )( r )
b1
h1
h2
b2
"
87 :37
Predicting collisions
U
p seu d o
m1 m 2
 G
r

2 r
E 
2
1
2
E 
2
2 r
2
2
U pseudo
G
m1 m 2
r
 r  U pseudo
2
1
2
 r
2
At turning point
r  0
 r p
Case
 0
All objects hit the earth,
regardless of initial
distance.
88 :37
Predicting collisions
U
p seu d o
 G
m1 m 2
r
2

2 r
2
Case  0
Need rmin  6400 km
 r  p  rim pact p 
rm in
E
rim p a ct
 v
89 :37
Predicting collisions
E 
1
r  G
2
M
2
E 
1
2
2
 v ;
2
 v   G
2


2
r

2
(  v  rim pact )
2  rm in
rm in
2
2 r
  v  rim pact

M
v  rm in  2 G M rm in  v
2

r
1
 r  p  rim pact p 
2
2
im pact
2
2
0

2

r
v
G
M
 2 G M  4 G M  4 v  r im p a ct
 1  im pact 
r


m in
rm in 
2
 GM

2
v
 
2v 


If G M is w eak rm in  rim pact ; If G M is strong rm in  0
2
2
4
2
2

 1


90 :37
Predicting collisions
For v-infinity comparable to v-escape, R-impact is
approx same as R-earth.
For v-infinity < v-escape, R-impact is many times Rearth
120.0
112.5
R-impact / R-Earth
100.0
80.0
60.0
40.0
37.3
20.0
11.3
3.9
0.0
0
1.5
0.5
1
1.1
1.5
V-infinity / Escape Velocity
2
2.5
3
91 :37
Properties of ellipses
r ( ) 
2
G m1 m 2  (1   co s  )
2
c
G m1 m 2 
rm in 
b

c
1 
1 
; rm ax 
c
1 
2
a
E 
 G m1 m 2  
2
2
2

2
1

92 :30
E, L and Eccentricity
r ( ) 
E 
2
G m1 m 2  (1   cos  )
 G m1 m 2  
2
 
E
 1
2
2
2

2
 1
2
 G m1 m 2  
2
1
 1
0  1
 0
The physics is in E and L. Epsilon is purely a
geometrical factor.
Epsilon equation applies to ALL conic sections
(hyperbolae, ellipses, parabolas).
93 :30
E, L and Eccentricity
 
E
2
2
 G m1 m 2 
2

 1
1
 1
0  1
 0
The physics is in E and L, and it transfers to
quantum mechanics.
94 :30
Planetary Scattering Angle
Epsilon
r ( ) 
1
1.001
1.02
1.05
1.1
1.3
2
5
20
50
100
2
G m1 m 2  (1   cos  )
r ( )

  1 8 0  arcco s  1
rim p a ct

-1/eps 2*arccos Omega
-1.00
-1.00
-0.98
-0.95
-0.91
-0.77
-0.50
-0.20
-0.05
-0.02
-0.01
360.0
354.9
337.3
324.5
310.8
280.6
240.0
203.1
185.7
182.3
181.2
0.0
5.1
22.7
35.5
49.2
79.4
120.0
156.9
174.3
177.7
178.8

r ( )

95 :37
Strong Force Scattering Angle
Quantum mechanics is needed,
but…
Particle physicists probe strong
interactions by scattering
particles just as we can probe
96 :37
gravity by scattering meteors
Problem L19-1
r ( ) 
2
G m1 m 2  (1   cos  )
Two rocks are scattering off the Earth’s
gravitational field
One has orbit with Epsilon=1.0015
Other has orbit with Epsilon=3.0
How much are their paths deflected by the Earth?
Sketch the two orbits.
97 :30
Problem L19-2
r ( ) 
 
E
2
G m1 m 2  (1   cos  )
2
2
 G m1 m 2 
2

1
A 1000 kg rock is aimed in
the vicinity of Earth
Out beyond the orbit of
pluto, it has a kinetic
energy of 50 billion joules
and an angular
momentum of 96 trillion
kg-m^2/s.
Calculate epsilon
Will the rock hit earth?
If not, what is the
minimum radius?
98 :30
Class #19 Windup
Epsilon is function of L and E and GMm
KEPLER
1st Law – Planets move in ellipses
2nd law dA/dt=const
3rd law Period goes as semi-minor
radius to 3/2 power
r ( ) 
 
E
2
G m1 m 2  (1   cos  )
2
2
 G m1 m 2 
2

1
Office hours today 3-5, tomorrow 45:30
99 :60
Class #20 of 30
Celestial engineering

Clarification on scattering angles
Rotating reference frames



Vector angular velocity
Newton’s laws on rotating frame
Two “fictitious forces”
100 :02
Class #19 Windup
Epsilon is function of L and E and GMm
KEPLER
1st Law – Planets move in ellipses
2nd law dA/dt=const
3rd law Period goes as semi-minor
radius to 3/2 power
r ( ) 
 
E
2
G m1 m 2  (1   cos  )
2
2
 G m1 m 2  
2
1
101 :60
Class #20 Windup
No class or office hours on 11/5 and
11/6
Office hours Thurs 11/7. HW 11 due
11/12 in class. HW 12 due 11/14 as
usual.
v  r
r
S
r
S0
r
m r  Fexternal  2 m r    m (   r )  
102 :60
Planetary Scattering Angle – New …
improved!!
r ( ) 
Epsilon
2
1
1.001
1.02
1.05
1.1
1.3
2
5
20
50
100
G m1 m 2  (1   co s  )
r ( m ax )    1   co s  m ax  0
r ( )

  3 6 0  2 arcco s  1

rim p a ct


r ( )
-1/eps 2*arccos Omega
-1.00
-1.00
-0.98
-0.95
-0.91
-0.77
-0.50
-0.20
-0.05
-0.02
-0.01
360.0
354.9
337.3
324.5
310.8
280.6
240.0
203.1
185.7
182.3
181.2
0.0
5.1
22.7
35.5
49.2
79.4
120.0
156.9
174.3
177.7
178.8
Sketch for epsilon=2
2 m ax
103 :37
Vector angular velocity
v

r
v  r
Theta=90-Latitude = “CoLatitude”
Earth obeys right-hand rule
Omega points North Earth spins
clockwise Sun rises in east
104 :60
Vector velocity – 1 minute problem
v

r
v  r
5
earth  7.3  10 rad / s
Latitudes –
Socorro – 34 N
Nome, Alaska – 64 N
Hilo, Hawaii – 19 N
Hobart, Tazmania – 43 S
What is circumferential
velocity of each of these
105
four cities?
:60
Vector velocity – 1 minute problem
v

r
v  r
5
earth  7.3  10 rad / s
Latitudes –
Socorro – (.38 km/s=880 mph)
Nome (.20 km/s=473 mph)
Hilo – (.44 km/s=1021 mph)
Hobart (.34 km/s=790 mph)
106 :60
Vectors expressed
in rotating frames
r
S

 r eˆ
i i
i
r

S0

dri

dt
i
r
r
S


i
dri
dt
eˆi
eˆi 
r
i
i
deˆi
dt
B ecause eˆi is any arbitrary F IX E D
vector in fram e S 0, it t ransform s
deˆi
Imagine same axes
in S as
   eˆi
dt
(x,y,z) expressed in
two frames S0
dri
r 
eˆi   ri   eˆi
S
stationary) and S (fixed
dt
i
i
to earth).
ˆ
ˆ
 r e
i
i
    ri e i    r 
i
r
i
S


i
dri
dt
eˆi    r  r
S0
r
107 :60
Derive Newton in rotating frame
W e know
r
S


dri
i
dt
eˆi    r
W e differentiate again w ith
sam e procedure .
r
r
r
r
S
S
S
S
dri
d 


eˆi    r 

dt  i dt

r
r
r
S0
S0
S0
r 
d
(  r )
dt
   r  (  r )    (  r )
 2  r    (  r )
108 :60
Newton in a rotating frame
m r  Fexternal  2 m r    m (   r )  
Newtonian Term
Coriolis Term
Centrifugal Term
109 :60
Alternate form of Left-Hand Rule
m r  Fexternal  2 m r    m (   r )  
r

r
110 :60
Class #20 Windup
No class or office hours on 11/5 and
11/6
Office hours Thurs 11/7. HW 11 due
11/12 in class. HW 12 due 11/14 as
usual.
v  r
r
S
r
S0
r
m r  Fexternal  2 m r    m (   r )  
111 :60
Class #22 of 30
Review of rotating reference
frames

Coriolis force
Non-axi-symmetric rotations




The inertia tensor
Torques for off-axis rotations
Diagonalization and eigenvalues
Principal axes
112 :02
Vector velocity – last time
v

r
v  r
5
earth  7.3  10 rad / s
Latitudes –
Socorro 34N – (.38 km/s=880 mph)
Nome 64N – (.20 km/s=473 mph)
Hilo 19N – (.44 km/s=1021 mph)
Hobart 43S – (.34 km/s=790 mph)
Velocities are all relative to
INERTIAL frame.
113 :60
Derive Newton in rotating frame
W e know
r
S


dri
i
dt
eˆi    r
W e differentiate again w ith
sam e procedure .
r
r
r
r
S
S
S
S
dri
d 


eˆi    r 

dt  i dt

r
r
r
S0
S0
S0
r 
d
(  r )
dt
   r  (  r )    (  r )
 2  r    (  r )
114 :60
Newton in a rotating frame
m r  Fexternal  2 m r    m (   r )  
Newtonian Term
Coriolis Term
Centrifugal Term
115 :60
Coriolis force for a rocket going up
r 
r

m r  Fexternal  2 m r    m (   r )  
116 :60
Problem L22-1
5
A lbuquerque latitude  35 N ; earth  7.3  10 rad / s
A 100 kg weather rocket is launched straight up
from Langmuir Lab. At t=10 sec, it’s velocity is
100 m/s. At t=100 sec, it’s velocity is 1000 m/s.
What is the magnitude and direction of the
Coriolis force at t=10 and 100 sec?
An airplane flies due North from Albuquerque.
Because of Coriolis force, the airplane will be
deflected from its true North path
A) Which direction will the deflection be in?
B) How many miles off course will the airplane
be after one hour? Assume v=300 m/s.
117 :30
You thought you knew all about angular
momentum…
P=mv … what could
p  m v  m (  r )
be more familiar?
Let’s just relax
Lr p
ˆ ( y rz   z ry )
xm
xˆ
yˆ
p  m x
y
ˆ ( z rx   x rz )
 z   ym
rx
ry
rz
L m
zˆ
Cross product
ˆ ( x ry   y rx )
 zm
…and again
xˆ
yˆ
zˆ
rx
ry
rz
 y rz   z ry
 z rx   x rz
 x ry   y rx
118 :02
Inertia Tensor
L
m
2
2
xˆ   x ( ry  rz )  ( y rx ry   z rx rz ) 
I xx
I xx
I xz  x
  yˆ   y ( rx  rz )  ( x ry rx   z ry rz ) 
L  I yx
I yy
I yz  y
 zˆ   z ( rx  ry )  ( x rz rx   y rz ry ) 
I zx
I zy
I zz  z
2
2
2
2
L  I
r r
 rx ry
I  m  ry rx
r r
 rz rx
 rz ry
2
y
2
z
2
x
2
z
 rx rz
 ry rz Inertia tensor -single particle
2
2
rx  ry
119 :02
Charles “Chuck” Hermite
(1822-1901)
French mathematician who did brilliant work in
many branches of mathematics, but was plagued by poor performance
in exams as a student. However, on his own, he mastered Lagrange's
memoir on the solution of numerical equations and Gauss's
Disquisitiones Arithmeticae. He was forced to leave Ecole
Polytechnique after one year when it was decided that his congenitally
deformed right leg would not allow him to take a commission in the
military, making him not worth their time.
In 1869, he became a professor at École Normale, and in 1870 at
Sorbonne. All during his career, was generous in his help of young
mathematicians. He showed that e was a transcendental
number
Hermite also discovered some of the properties of
Hermitian matrices .
120 :02
Inertia Tensor is Hermitian
r r
 rx ry
 rx rz
I  m  ry rx
r r
 ry rz
 rz rx
 rz ry
r r
2
y
2
z
2
x
I I
I  I
A
1
i
1 i
2
2
z
T
2
x
2
y
Hermitian matrices have
real eigenvalues
If the eigenvalues differ,
the eigenvectors are
orthogonal
Inertia tensor is symmetric
Inertia tensor is Hermitian
*
 A 
T
1 1 i
i
2
 A 
*
1
1 i
i
2
121 :02
Inertia Tensor
r r
 rx ry
I  m  ry rx
r r
 rz rx
 rz ry
2
y

2
z
2
x
m  ( r y  r z )
2
2

I 
2
z
 rx rz Derived all this for
a single particle
 ry rz
Below, for multiple
2
2
rx  ry
particles m 
  m  r x r y
  m  r x r z
...
etc .

  m  r y r x


...
...
 m ( r
2
x
 r y )
2

122 :02
Angular momentum of non-axi-symmetric
object
z
   0 zˆ
m 1, r1
Inclined bar-bell
Bar length 2L
y
x
  t
m 2 , r2
r1  ( L cos  t sin  , L sin  t sin  , L cos  )
r2  (  L cos  t sin  ,  L sin  t sin  ,  L cos  )
123 :02
r1  ( L cos  t sin  , L sin  t sin  , L cos  )
r2  (  L cos  t sin  ,  L sin  t sin  ,  L cos  )

m  ( r y  r z )
2
2

I 
Non-axisym. object II
  m  r x r y
  m  r x r z

  m  r y r x
...
...
...
 Let 

 45
for sim plicity !!
etc .


m  ( r x  r y )
2
2

I xx  2 m L (cos   sin  sin  )  m L (1  sin  )
2
2
2
2
2
2
I yy  2 m L (cos   sin  cos  )  m L (1  cos  )
2
2
2
I zz  2 m L (sin  )  m L
2
2
2
2
2
I xz  I zx   2 m L sin  cos  cos   m L cos 
2
2
2
124 :60
Non-axisym. object III
1  sin 
2
I  m L  sin  cos 
2
 sin  cos 
 cos 
1  cos 
 sin 
 sin 
1
2
 cos 
1  sin 
2
L  I   m L  sin  cos 
2
 cos 
  t
 sin  cos 
1  cos 
2
 sin 
 cos   0 
 
 sin  0
 
 
1
 0
  cos  t 
  cos  t    sin  t 
d 


 

L  I    sin  t  
 sin  t    cos  t



 

dt



 
1
1
0 125 :60
But that’s not all!!
z
   0 zˆ
x
  t
m 2 , r2
m 1, r1 Principal Axes
The ones about
which rotations do
y
not produce torques
These are the
eigenvectors of the
inertia tensor
How do we find
them?
r1  ( L cos  t sin  , L sin  t sin  , L cos  )
r2  (  L cos  t sin  ,  L sin  t sin  ,  L cos  )
126 :02
Diagonalization
1  sin 
2
I  m L  sin  cos 
2
2
1
0
1
0
2
0
1
0
1
I   
 cos 
1  cos 
 sin 
 sin 
1
2
 cos 
I  mL
 sin  cos 
Set phi=0
If 
I   0
is an eigenvector
127 :60
I  mL
2
1
0
1
0
2
0
1
0
1
Diagonalization
mL  
0
mL
0
2mL  
0
0
mL  
2
I  
mL
2
2
2
det I    0
det
1  0
mL  
0
mL
0
2mL  
0
0
mL  
2
mL
2
2
2
2
2
 2  3  2 m L
2
 0 " D eg en era te "
eig en va lu es
128 :60
Problem L22-2
z
m 1, r1
   0 zˆ
I  mL
y
2
1
0
1
0
2
0
1
0
1
mL  
0
mL
0
2mL  
0
0
mL  
2
x
  t
I  
m 2 , r2
mL
1  0
 2  3  2 m L
" D eg en era te "
eig en va lu es
2
2
2
2
2
Given all the above, what are
the principal axes (the
eigenvectors) for this
problem?
129 :30
Class #22 Windup
Tuesday bring a tennis racket or
Halliday & Resnick and a rubber band.
Exam 11/19 Central force / rotating
reference frames
Office hours Th/Fri 11/7-8. HW 11 due
11/12 in class. HW 12 due 11/14 as
usual.
130 :60