Transcript L - Morgan

Chapter 7: Rigid Body Dynamics
Chapter 7 Goals:
• To define the rigid body as a system of point masses
• Through use of forces and force moments, to
introduce the physics of static equilibrium
• To distinguish the force moment from the special case
of the torque
• To learn about the moment of inertia of a rigid body
about a given axis, and to consider various situations
• To apply N2 and its rotational equivalent to the
rotation of a rigid body about a fixed axis
• To discuss how conservation laws for energy and
angular momentum may be applied fruitfully
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What is a rigid body?
• extended in space, has mass M, and has a c.o.m.
• conceive of it as a collection of very large number
of infinitesimal point masses dm
• the distance between any pair of point masses is
fixedit is rigid (for N masses, there are therefore
N(N–1)/2 kinematic constraints on system
• Its motion may be ‘broken down’ into pure
translation of the c.o.m. combined with pure
rotation about the c.o.m.
• many of the summations we have been
considering, as sums over all the bodies from i=1 to
N, are going to become, at worst, volume integrals
• YIKES!!!
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AN couple of those integrals
 body
mass M :
 mi  M :  dm
i 1
 center of
body
1
mass location R C : M
N

i 1
1
ri mi  R C : M
 r dm
body
1
 of course the c.o.m. equation is really 3; e.g. X C : M
 x dm
body
• in practice, the multiple integrals can be handled
pretty easily once one identifies how to express dm
• the key tasks are to think about the mass density of
the stuff, and whether the object is one-dimensional
(like a thin stick, or an arc), or 2d (a plate or a shell)
or 3d (a brick; a ball; a cylinder)
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1d Example Ia
A piece of wire is bent into the arc of a circle of radius R
which subtends an angle q in radians. Find the mass and the
c.o.m. of the object. The linear mass density of the material
is l, in units of kg/m.
y
• Put origin at the center of the arc
• Let the wire subtend an angle 2q0, so
that is lies between – q0 and + q0
x
• The length of the wire is the arc
length: L = 2 R q0, where angles are in
radians
• Consider a very small arc length ds,
which contains mass dm, and subtends
an angle dq0 (see next slide)
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• Infinitesimal mass dm ‘lives’ at angle q
• It subtends an infinitesimal angle dq
• Its infinitesimal arc length is ds = R dq
• Since in 1d any mass is “density x length”
 dm = l R dq [key step in the solution!]
1d Example Ib
y
dq
dm
q

M  dm 
x
XC 

q0
q0
q
0
l
Rd
q

l
R
d
q

l
R
q


q
q 0
q 0
0
 lRq 0  (q 0 )   2lRq 0  density x length
• Obviously YC = 0 by ‘up-down’ symmetry’
• dm is located at (x,y) = (R cosq, R sinq)
q0
q0
q0
2
2
2
xdm  lR cosq dq  lR cosq dq  lR sinq q
0
q 0
q 0


 lR sin q 0  sin( q 0 )   2lR sin q 0 
2
2
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MR sin q
q
0
1d Example IIa
A stick of length L is fabricated
from a material with a variable
lR
linear mass density, that increases
from the value lL on the left end to
the value lR on the right end. Find lL
the mass and locate the c.o.m.
• Using slope-intercept ideas
l ( x )  lL 
l(x) vs. x
L
lR  lL
L
x
l
x : lL 
x
L
• dm = l dx for the infinitesimal bit of mass located at x
M   dm   l ( x) dx   l
L
L
L
0

l
2L
x
2
L
0
 lL L 

l
L
x  dx   l
L
L
0
l
2L
( L )  lL L 
2
L
dx 
0
l L
R
2


l
L
x
L
dx  lL x 0
0
l L
L
2
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
lL lR L
2
: laveL
1d Example IIb
XC 
1
M

lR
0
lL
 x dm   l ( x) x dx
1
M
 l x 
L
1
M
L
L
l
L

x 2 dx
0


l(x) vs. x
L x
This result is a bit
L
2
L
hard to interpret, but
l
l
L

l

l
2
3
3
L x
L

x


(
L
)
if we take lL = 0 as a
2M
3 LM
2
M
3
L
M
0
0
special case, we get
M = lRL/2 and so
l L2
l L2
l L2  l l  L2
L
R
L
XC = lRL2/3M =


  L  R 
2M
3M
3M
3 M
 6
2L/3
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Review of system ‘momentous’ quantities
L :
N
N
N
i 1
i 1
i 1
 Li   ri  pi   mi R C  r'i  VC  ui 
 M R C  VC  
N

mi r 'i u i
M  r  F so M net  r  Fnet 
i 1
dL
:
dt
N

i 1
dr

mi  dti

 v i  ri 




ri   Fi ext 


i 1 

N

dv
i
dt



N
 mi vi  vi  ri  ai 
i 1

f j i    M EXT 

j i

N

N
N
 ri  f ji
i 1 j i
• assume N3 and central
dL
: M EXT
forces: double sum zero!!
dt
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dL
dt
The idea of the ‘center of gravity’
• a system (may or may not be rigid) is acted on by
gravity because all of its masses have weight
• these forces act at different places, obviously, and
we regard them as ‘external’
N
• the total external force is clearly FEXT   mi g  gM
i 1
• what about moment of force? Use ri = RC + r’i
M EXT 
N
N
N
N
i 1
i 1
i 1
i 1
 ri  mi g    mi R C  r'i  g    mi R C  g    mir'i g 
 N

  mi R C  g 


 i 1 

N
 mir'i  g  MR C  g  0 [c.o.m. is at 0 as seen from it]
i 1
M EXT  R C  Mg
• for gravity force moment, gravity
acts at c.om c.o.g. IS the c.o.m.
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The science of statics
• system feels no net force: FEXT = 0  system
momentum is constant  c.o.m does not accelerate,
and we assume that it is not translating either
• system feel no net force moment : MEXT = 0 
system angular momentum is constant, and we no
rotational motion either
• This state of affairs is called static equilibrium
• upside: the right sides of the two N2 laws are zero
• downside: choice of origin is important, and one
must count equations carefully so that system is is not
‘overdetermined’
• example of an overdetermined system: 4-legged
table!!
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Example: the massive lever
250 N
100 N
50 kg
d
4m
S
W
FL
FR
Two forces act as shown.
The lever itself has a mass
of 50 kg, and has a length of
4 m. The whole system is
supported by the fulcrum, a
distance d from the left end.
Find support force and d.
• Take + to be UP and IN
L
• put all forces where they act (c.o.g.)
• Forces: – FL – W – FR + S = 0
•  S = FL + W + FR = 850 N
• Moments: put origin at (say) left end
• 0∙250 – d∙850 + 2∙500 + 4∙100 = 0
•  d = 1400/850 = 1.65 m
• a massless lever FL/FR = dR/dL
•‘The lever equation’
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Example: the
building sign
3
0
c
m
q
Fargo
40
cm
f
S
q
W
A heavy sign (20 kg) hangs from a
massless bar and a support wire. Find
the tension, and find both components
of the bar support force S
• Take + to be UP, to RIGHT, and IN
• put all forces where they act (c.o.g.)
• 3:4:5q  37°; sin q = .60; cos q = .80
• Forces UP: S cos f – W + T sin q = 0
• Forces RIGHT: S sin f – T cos q = 0
• put origin at left so S does not appear in
T equation – clever!!
• Moments IN: 0S + .20∙200 – .4T sin 143° = 0
40
40
T

 167 N

.4.6
.4 sin 143
S cosf  W  T sin q  200  100  100 N
S sin f  T cosq  133 N
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Not in book: the idea of stability in gravity
• as can tips, c.o.g. rises: stable
• once c.o.g. is directly above pivot: tipping point!
• from then on, object tips over and c.o.g. descends
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Torque as distinct from/related to force moment
• a system of point masses (not necessarily a rigid body)
with positions relative to O written ri feeling external
forces Fi,ext (again, we assume internal forces are central
and cancel in pairs)
• R is position vector relative to O of a second origin O’
• positions of the masses relative to O’ : r’i := ri – R
• express the force moments using ri = r’i – R
M EXT 
 ri  Fi,ext   r'i  Fi,ext   R  Fi,ext 
i
M 'EXT  R 
i
i
 Fi,ext  M'EXT  R  FEXT
i
• so, if there is no external force on the system,
the external force moment is origin-independent!!
• if this is the case, we call it the torque T
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Review of rigid body dynamics
• now let the second origin be at the c.o.m.  RC
dVC
dP
FEXT 
M
usual system N2; no origin dependence
dt
dt
dVC
dL
M EXT 
 M'EXT  R C  FEXT  M'EXT M R C 
dt
dt
• The system external force causes the c.o.m. to
accelerate – this is also manifest in the moment equaton
• the external force moments cause the body to change its
angular moment um with respect to (‘about’) the c.o.m.
• if the system external force is zero, the c.o.m. does not
accelerate, and all that happens is rotational acceleration
about the c.o m. We therefore have, in this case,
dVC
dP
FEXT 
M
dt
dt
dL
 M EXT  M'EXT : T
dt
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Kinetic energy in terms of moment of inertia
• a rigid body is fixed to an axis, not necessarily through
its c.o.m. and is currently turning at angular speed w
• treat it as a collection of point masses: with the origin on
the axis (normal to the screen), vi = w si where si is the
distance from the axis to mass i
mi vi2
miw 2 si2 I Aw 2
2
K   Ki  
i
i
2

i
2

2
where I A :  mi si
i
• IA is called the moment of inertia or rotational inertia
• it is the rotational analog to the mass – it depends on the
masses; also quadratically on the distances from the axis
• It is crucial to bear in mind that the axis must be
described clearly – and it may or may not even penetrate
the body
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How to calculate the moment of inertia
IA 

mi si2  I A 
i

s 2 dm which gives rise to some fun integrals!
body
• the moment of inertia captures the resistance of the body
to being angularly accelerated about that axis
• it is the rotational analog to the mass – it depends on the
masses; also quadratically on the distances from the axis
Example of calculation of the moment of inertia
Find the moment of inertia for a long thin bar of mass M and length
L about an axis through its c.o.m. and perpendicular to the bar
L/2

I A  s 2 dm 
x
O
IA 
M
L

x 2 ldx where l  M/L
L / 2
L/2

L / 2
x dx 
2
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M
L
x

 3
3
L/2
ML2



12
 - L/2
Example of calculation of the moment of inertia
Find the moment of inertia for a hoop of mass M and radius R
about an axis through its c.o.m. and along the hoop’s axis

IA 

s 2 dm 
body
R 2 dm  R 2
body

dm  MR 2
body
This one is trivial since all of the matter is at the same distance
from the axis: hoops are special!!
Example of calculation of the moment of inertia
Find the moment of inertia for a solid cylinder of mass M and
radius R about an axis through its c.o.m. and along the axis
L/2

I A  s 2 dm 

x 2 ldx where l  M/L
L / 2
L/2
L/2
ML2
x 
IA 
x dx   

3
 - Las/2Pearson12
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Addison-Wesley.
L / 2
M
L

2
M
L
3
Kinetic energy of c.o.m. and referred to c.o.m.:
the parallel-axis theorem
• a second axis through c.o.m. point, parallel to the first
• let l be the displacement vector from first to c.o.m. axis
(perpendicular to both axes, as always in geometry)
• re-express origin-to-axis vector si = l + ri where ri is
the perpendicular vector from c.o.m. axis to mass i
1
1
2
2
K  2 miw si  2 miw 2 l  ρi   l  ρi 



i
i
1
2

miw 2l 2 
i
 2 w 2l 2
1

miw 2 l  ρi  2
1
i

i
mi  w 2 l 
i

1
mi ρi  2
i

miw 2 r i2
i
 2 Mw 2l 2  w 2 l  (0)  2 w 2
1

miw 2 r i2
1

mi r i2 : 2 Mw 2l 2  2 I Cw 2
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i
1
1
Parallel-axis and perpendicular-axis theorems
• interpret this by comparing to the previous expression
for K in terms of IA and w
K : I Aw 2  12 Mw 2l 2  12 I Cw 2
 I A  Ml 2  I C [and so it must be that I A  I C ]
• there is also a perpendicular-axis theorem
• for a thin plate-like object lying in xy plane, let Izz be the
moment of inertia for an axis through a point along z
I zz :  m s  mi x  y
2
i i
2
i
2
i

i
• for an axis
along x ior y, the distances
are trivial: just y2 or x2 respectively
I xx :

i
2
mi yi ; I yy
:

2
mi xi
I zz I xx  I yy
i
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z
y
x
Angular momentum in terms of moment of inertia
• angular velocity vector w points along the axis by RHR
• However, L is parallel to that vector ONLY if there is no
external force moment on the body
• consider the ‘dumbbell’ comprising two masses m and a
O
massless stick of length d,
2 
2
• axis through the c.o.m.

md
md
I C  2


4
2
perpendicular to the stick;

origin at c.o.m.
• L = L1 + L2 and both masses contribute an identical
angular momentum; total magnitude is md2w/2 and it
points up, exactly along w -- and it is a constant
• Therefore, there is no need for a force moment on system
• we have that L = IC w in this case
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What happens if we alter the axis?
O
• put axis at x, not d/2; still ┴
• angular velocity vector w
points along the axis by RHR
I A  mx 2  mx  d 2 and L  I A ω (constant)
• note now that because the c.o.m. is accelerating, there
must be a net horizontal force on the system… the thing is
wobbly… but there is still no need for a net force moment
• construct L = r x p for left mass
L
• get same result for right mass
O
• clearly L CHANGES as dumbell
r
whirls around and L is NOT
p
parallel to w: L ≠ IC w
• we need a torque in this case!! FEXT is however zero
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Example IA
O
An object is made from a thin stick that is 75 cm in length and of
mass 2.4 kg. At one end there is a solid sphere of radius 5.0 cm
and mass 4.0 kg. The object turns about its center at 8 rad/sec.
Find the c.o.m. and find the contribution to I and K for each part.
Where is center of mass of system? Put origin at left end:
1
4(0)  2.4(37.5) 90 kg - cm
XC 
mi xi 

 14.1 cm
M
4  2.4
6.4 kg

What is I for left mass? Use parallel axis theorem and IC for ball:
I AL  I CL  m  mL R  mL  
2
2
5
2
L
2
2
5
4.05  4.375
2
2
 .0040  .563  .567 kg - m 2
2
2
2



I

I

m
L

2
.
4
.
75

.
113
kg
m
What is I for stick? AS CS 12 S S 12
1
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1
Example IB
O
I A  I AL  I AS  .57  .11  .68 kg - m 2
Question: can we regard
this K as the K of c.o.m.
+ K referred to c.o.m.?
1
2
M
V
C
2

1
2 2
M
w

2
KA 
1
2
I
w
2 A
 K A ?  ? 2 MVC2  2 I C w 2
1
1
6.4(8) 2 .234 2

 11.21 J
2

L2
2 


 .141   mS  12  .234 






 324.0010  .0199   2.4.0469  .0548 
 32.084  ..244   10.50 J
1
2
I
w
2 C
.68(8) 2

 21.76 J
2
 2 RL2
64 
 mL 
 5
2 
2

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11.21 J
+ 10,50 J
= 21.71 J
YES!!
Example II with a shocking result?!!
A heavy door (1.25 m x 2.0 m) is hung by hinges
at one end. Its mass is 40 kg. A force of 20 N is
applied on the outer edge in a direction normal to
the door. Find the angular acceleration of the door
and the force of the hinges on the door during the
acceleration. Hint: put an origin at the hinges
O
FH
b
F
What is net moment on the door as seen from the origin?
M EXT  r  F  0  M EXT is out of screen and M EXT  bF  25 N - m
 since M EXT  Ia  a  M EXT 1
2
M
b
3
 25 1
2




40
1
.
25
3
 1.20 rad/s2
the c.o.m translates due to net force: a = ab/2 = .75 m/s2
F  FH  Ma  FH  F  Ma  20  (40)(.75)  10 N
THE HINGES ALSO PUSH TO THE RIGHT ON THE DOOR!
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Example III
A mass of 5 kg hangs from a string that is wrapped several
times around a hoop (radius 60 cm) with a horizontal
frictionless fixed axle. The mass has a = 4 m/s2
5
Find the angular
atan 4 m/s 2
2
acceleration of the atan  aR  a  R  .60 m  6.7 rad/s
hoop, and its mass.
Make FBDs for both objects. Put
T
origin at axle. Choose UP & IN  +
S
O
Force on m : T  mg  ma
T
W
W
 T  m( g  a)  5(10  4)  30 N
T
30
Moment on M : RT  I Ca  MR a  M 

 7.5 kg
Ra .6(6.7)
2
Force on M :  T  S  Mg  0  S  Mg  T  105 N
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Kinematics of rolling motion I
w(t)
• Any object of circular cross-section
• c.o.m. is at the center for smooth rolling
• no sliding (so if a ≠ 0, need static friction)
• let v = center of mass speed
• in one rotation, c.o.m. and point of contact (p.o.c.) move
a distance of one circumference 2pR
• time is rotation period: T = distance/speed = 2pR/v
•  v = 2pR/T = wR: same formula as for linear speed as
related to angular speed for a turning object, where R is
distance from axis!!
• p.o.c appears to move… but that bit of roller can’t slide
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
w(t)
Kinematics of rolling motion II
• c.o.m. moves at speed v = wR, as established
• p.o.c. cannot move unless slippage: v = 0
• thus p.o.c. acts as an instantaneous axis of rotation,
that slides along at speed v
• top of roller is distance 2R from the instantaneous
axis, so it moves forward at speed 2v (!!)
• rotation (left)
• translation
(mid)
• both t+r (right)
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
w(t)
Kinematics of rolling motion III
• c.o.m. moves forward at v in steady motion
• a dot on the edge moves on a path cycloid
• for an instant at the very bottom, it stops
• at the top, it is moving forward at 2v
• top of roller is distance 2R from the instantaneous
axis, so it moves forward at speed 2v (!!)
• what does a flanged wheel do??
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Energetics of rolling motion I
• a solid ball [IC = (2/5)MR2]
rolls down a ramp of height h
• as it speeds up, it is both
translating and rotating
• find c.o.m. speed at bottom
• two approaches A and B give
same result: different origins
• A: origin at p.o.c.  pure rotation
• by parallel-axis thm, IA = (2/5)MR2 +MR2 = (7/5)MR2
K
1
2
I
w
2 A

2
17
 v 
2
 MR  
25
 R 
 10 Mv 2
7
• B: origin at c.o.m.  translation + rotation
K
1
2
M
v
2

1
2
I
w
2 C

1
2
M
v
2

2
12
 v 
2
 MR  
25
 R 
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
 10 Mv 2
7
Energetics of rolling motion II
• The ball rolls down in gravitational field
• Two forces act, N and W  Only W does work
• W has a potential energy: U = Mgy = Mgh
• Since E is conserved, in the process K = – U
• assume ball released from rest, so
K final  K  10 Mv 2 and U  Mgy  Mgh
7
 10 Mv 2   Mgh  v  1.4 gh (not slide result 2 gh )
7
• thus, in a race, a fricitionless slider beats a roller
• hoops, having the largest I, fare the worst
• static friction is necessary to provide the net
external moment that allows angular acceleration
{show Active Figure 1-_26}
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example III revisited
A mass of 5 kg hangs from a string that is wrapped several
times around a hoop (radius 60 cm) of mass 7.5 kg with a
horizontal frictionless fixed axle. Find the speed when the
mass has fallen 2.0 m by conserving energy.
1
5
1
Ei  E f  K i  U i  K f  U f  0  2 mv 2  2 I Cw 2  mgh
but v  wR and I C  MR 
2
 m  M v 2  2mgh  v 
1
2
mv
2

 v  2
1
2
MR  
 R 
2 

2 m gh
 mM 

2 (5)(10)(2 )
 mgh
 5 7 .5 
If it were a different object--say, a solid cylinder--the
answer would be different. In that case M would appear
as M /2 – do you see why?
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
 4 m/s
• conservation of energy is assumed
• falling mass is m, spool radius is r, time of
fall of m is t, distance of fall of m is h
• final velocity of falling mass is v
• final angular velocity of object is w = v/r
The lab
equation
example
2h
v  gt and h  gt  eliminate g to get v 
t
1
2
2
2h
v  gt and h  gt  eliminate g to get v 
t
2
2 2


2
gt
mr
gt
1
1
v
2
2

 mgh  2 mv  2 I
 I  mr
1 
 2h

r
2h


1
2
2

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A taste of angular momentum conservation
Example
We model an ice skater as a
cylinder (head/torso/legs) of
mass 50 kg and radius 25
cm) along with two
cylindrical arms, each of
mass 4 kg, radius 5 cm and
length 80 cm)
a) Find moment of inertia with arms DOWN
and with arms OUT
b) Argue that angular momentum should be
conserved when a skater spins, and then
assuming an initial rotation rate of 2
rotations/sec with arms OUT, find the final
rotation rate with arms DOWN.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
I C,axial  2 MR 2
1
example
II
I A,normal@end  4 MR 2  3 ML2
1
1
  .05 m 2
2
 I C, DOWN  50 kg .25 m   24 kg  2  .30 m  


1
2
2
 1.56 kg - m 2  .73 kg - m 2  2.29 kg - m 2
• first term is torso axial c.o.m. moment of inertia
• factor of 2 x mass is for two arms of mass 4 kg
• first term in big parenthesis is arm’s axial c.o.m. IC
• second term in big parenthesis is parallel-axis
part
2
2
 I C,OUT 
1
2
50 kg .25 m 2  24 kg  .054m 


 .80 m 
3


.25 m 2 

 1.56 kg - m 2  2.21 kg - m 2  3.77 kg - m 2
• L is conserved because the external force moment is zero!!
 Linit  Lfin  I initwinit  I finwfin  wfin 

winit  2 rev
2p
s
rad
rev
  4p
rad
 wfin
s

I
init w
init
I
fin
3.77 
rad
rad 
 4p
  6.6p
2.29 
s
rev 
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.