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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. CornwellLecture Notes:
Brian P. SelfCalifornia Polytechnic State University
CHAPTER
2013 The McGraw-Hill Companies, Inc. All rights reserved.
12Kinetics of Particles:
Newtons Second Law
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Contents
12 - 2
Introduction
Newtons Second Law of
Motion
Linear Momentum of a Particle
Systems of Units
Equations of MotionDynamic Equilibrium
Sample Problem 12.1
Sample Problem 12.3
Sample Problem 12.4Sample Problem 12.5
Sample Problem 12.6
Angular Momentum of a Particle
Equations of Motion in Radial &
Transverse Components
Conservation of Angular Momentum
Newtons Law of Gravitation
Sample Problem 12.7Sample Problem 12.8
Trajectory of a Particle Under a
Central Force
Application to Space Mechanics
Sample Problem 12.9
Keplers Laws of Planetary Motion
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Kinetics of Particles
2 - 3
We must analyze all of the forcesacting on the wheelchair in order
to design a good ramp
High swing velocities can
result in large forces on a
swing chain or rope, causing
it to break.
T
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Introduction
12 - 4
Newtons Second Law of Motion
m F a
If the resultant force acting on a particle is not
zero, the particle will have an acceleration
proportional to the magnitude of resultant
and in the direction of the resultant.
Must be expressed with respect to aNewtonian (or inertial)
frame of reference, i.e., one that is not accelerating or rotating.
This form of the equation is for a constant mass system
TE
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Linear Momentum of a Particle
12 - 5
Replacing the acceleration by the derivative of the velocity
yields
particletheofmomentumlinear
L
dt
Ldvm
dt
d
dt
vdmF
Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the linear momentum
of the particle remains constant in both magnitude and direction.
TE
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Systems of Units
12 - 6
Of the units for the four primary dimensions (force,
mass, length, and time), three may be chosen arbitrarily.
The fourth must be compatible with Newtons 2nd Law.
International System of Units (SI Units): base units are
the units of length (m), mass (kg), and time (second).
The unit of force is derived,
22
s
mkg1
s
m1kg1N1
U.S. Customary Units: base units are the units of force(lb), length (m), and time (second). The unit of mass is
derived,
ft
slb1
sft1
lb1slug1
sft32.2
lb1lbm1
2
22
TE
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Equations of Motion
12 - 7
Newtons second lawamF
Can use scalar component equations, e.g., for
rectangular components,
zmFymFxmF
maFmaFmaF
kajaiamkFjFiF
zyx
zzyyxx
zyxzyx
TE
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Dynamic Equilibrium
12 - 8
Alternate expression of Newtons second law,
ectorinertial vamamF
0
With the inclusion of the inertial vector, the system
of forces acting on the particle is equivalent to
zero. The particle is in dynamic equilibrium.
Methods developed for particles in static
equilibrium may be applied, e.g., coplanar forces
may be represented with a closed vector polygon.
Inertia vectors are often called inertial forces as
they measure the resistance that particles offer tochanges in motion, i.e., changes in speed or
direction.
Inertial forces may be conceptually useful but are
not like the contact and gravitational forces found
in statics.
fTE
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Free Body Diagrams and Kinetic Diagrams
12 - 9
The free body diagram is the same as you have done in statics; we
will add the kinetic diagram in our dynamic analysis.
2. Draw your axis system (e.g., Cartesian, polar, path)
3. Add in applied forces (e.g., weight, 225 lb pulling force)4. Replace supports with forces (e.g., normal force)
1. Isolate the body of interest (free body)
5. Draw appropriate dimensions (usually angles for particles)
x y225 N
FfN mg
25o
V t M h i f E i D iTE
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Free Body Diagrams and Kinetic Diagrams
12 - 10
Put the inertial terms for the body of interest on the kinetic diagram.
2. Draw in the mass times acceleration of the particle; if unknown,
do this in the positive direction according to your chosen axes
1. Isolate the body of interest (free body)
x y225 N
FfN mg
25o
may
max
m F a
V t M h i f E i D iTE
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Free Body Diagrams and Kinetic Diagrams
2 - 11
Draw the FBD and KD for block A (note that the
massless, frictionless pulleys are attached to block A
and should be included in the system).
V t M h i f E i D iTE
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Free Body Diagrams and Kinetic Diagrams
2 - 12
1. Isolate body
2. Axes3. Applied forces
4. Replace supports with forces
5. Dimensions (already drawn)
x
y
mg
Ff-1N1
T
T
T
T
T
Ff-B
NB
may = 0
max
6. Kinetic diagram
=
V t M h i f E i D iTe
E
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Free Body Diagrams and Kinetic Diagrams
2 - 13
Draw the FBD and KD for the collar B. Assumethere is friction acting between the rod and collar,
motion is in the vertical plane, and qis increasing
V t M h i f E i D iTe
E
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Free Body Diagrams and Kinetic Diagrams
2 - 14
1. Isolate body
2. Axes3. Applied forces
4. Replace supports with forces
5. Dimensions
6. Kinetic diagram
mg
Ff
N
mar
maqeq er
=q
q
V t M h i f E i D iTe
E
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Sample Problem 12.1
12 - 15
A 200-lb block rests on a horizontal
plane. Find the magnitude of the forceP required to give the block an
acceleration of 10 ft/s2 to the right. The
coefficient of kinetic friction between
the block and plane is mk 0.25.
SOLUTION:
Resolve the equation of motion for the
block into two rectangular component
equations.
Unknowns consist of the applied force
P and the normal reaction N from theplane. The two equations may be
solved for these unknowns.
V t M h i f E i D iTe
Ed
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Sample Problem 12.1
12 - 16
N
NF
g
Wm
k
25.0
ftslb21.6
sft2.32
lb200
2
2
m
x
y
O
SOLUTION:
Resolve the equation of motion for the blockinto two rectangular component equations.
:maFx
lb1.62
sft10ftslb21.625.030cos 22
NP
:0 yF
0lb20030sin PN
Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
lb1.62lb20030sin25.030cos
lb20030sin
PP
PN
lb151P
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Vector Mechanics for Engineers: Dynamicsenth
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Sample Problem 12.3
12 - 17
The two blocks shown start from rest.
The horizontal plane and the pulley arefrictionless, and the pulley is assumed
to be of negligible mass. Determine
the acceleration of each block and the
tension in the cord.
SOLUTION:
Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
Write the equations of motion for the
blocks and pulley. Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.
Vector Mechanics for Engineers D namicsTe
Ed
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Vector Mechanics for Engineers: Dynamicsenth
dition
Sample Problem 12.3
12 - 18
Write equations of motion for blocks and pulley.
:AAx amF AaT kg1001
:BBy amF
B
B
BBB
aT
aT
amTgm
kg300-N2940
kg300sm81.9kg300
2
22
2
:0 CCy amF
0212
TT
SOLUTION:
Write the kinematic relationships for the dependentmotions and accelerations of the blocks.
ABAB aaxy 21
21
x
y
O
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Vector Mechanics for Engineers: Dynamicsenth
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Sample Problem 12.3
12 - 19
N16802
N840kg100
sm20.4
sm40.8
12
1
221
2
TT
aT
aa
a
A
AB
A
Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
ABAB aaxy 21
21
AaT kg1001
AB
a
aT
21
2
kg300-N2940
kg300-N2940
0kg1002kg150N2940
02 12
AA aa
TT
x
y
O
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Vector Mechanics for Engineers: Dynamicsenth
dition
Sample Problem 12.4
12 - 20
The 12-lb blockB starts from rest and
slides on the 30-lb wedgeA, which is
supported by a horizontal surface.Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
SOLUTION:
The block is constrained to slide downthe wedge. Therefore, their motions are
dependent. Express the acceleration of
block as the acceleration of wedge plus
the acceleration of the block relative to
the wedge.
Write the equations of motion for the
wedge and block.
Solve for the accelerations.
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Vector Mechanics for Engineers: Dynamicsenth
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Sample Problem 12.4
12 - 21
SOLUTION:
The block is constrained to slide down the
wedge. Therefore, their motions are dependent.
ABAB aaa
Write equations of motion for wedge and block.
x
y
:AAx
amF
AA
AA
agWN
amN
1
1
5.0
30sin
:30cos ABABxBx aamamF
30sin30cos
30cos30singaa
aagWW
AAB
ABABB
:30sin AByBy amamF
30sin30cos1 ABB
agWWN
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Vector Mechanics for Engineers: Dynamicsnthdition
Sample Problem 12.4
12 - 22
AAagWN
15.0
Solve for the accelerations.
30sinlb12lb302
30coslb12sft2.32
30sin2
30cos
30sin30cos2
30sin30cos
2
1
A
BA
BA
ABBAA
ABB
a
WW
gWa
agWWagW
agWWN
2sft07.5Aa
30sinsft2.3230cossft07.5
30sin30cos
22AB
AAB
a
gaa
2sft5.20ABa
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Group Problem Solving
2 - 23
The two blocks shown are originally at
rest. Neglecting the masses of the pulleys
and the effect of friction in the pulleys and
between block A and the horizontal
surface, determine (a) the acceleration of
each block, (b) the tension in the cable.
SOLUTION:
Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
Write the equations of motion for theblocks and pulley.
Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.
Draw the FBD and KD for each block
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Ed
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Vector Mechanics for Engineers: Dynamicsnthdition
Group Problem Solving
2 - 24
xA
yB
const nts3 aA Bx y L
3 0A Bv v
3 0A Ba a
3A Ba a
SOLUTION:
Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
This is the same problem worked last
chapter- write the constraint equation
Differentiate this twice to get the
acceleration relationship.
(1)
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Vector Mechanics for Engineers: Dynamicsnthition
Group Problem Solving
2 - 25
: x A AF m a
A BT m a
3 A BT m a
y B BF m a
3(3 )B A B B Bm g m a m a
229.81 m/s 0.83136 m/s
30 kg1 91 9
25 kg
BA
B
ga
m
m
22.49 2.49 m/s
A a
23 30 kg 0.83136 m/sT
74.8 NT
Draw the FBD and KD for each block
mAg
T
NA
maAx
mBg
2T T
maBy
Write the equation of motion for each block
==
From Eq (1)(2) (3)3B B BW T m a
(2)(3)
B A
Solve the three equations, 3 unknowns
+y
+x
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Vector Mechanics for Engineers: Dynamicsnthition
Concept Quiz
2 - 26
The three systems are released from rest. Rank the
accelerations, from highest to lowest.a) (1) > (2) > (3)
b) (1) = (2) > (3)
c) (2) > (1) > (3)
d) (1) = (2) = (3)
e) (1) = (2) < (3)
(1) (2) (3)
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Vector Mechanics for Engineers: Dynamicsnthition
Kinetics: Normal and Tangential Coordinates
2 - 27
Aircraft and roller coasters can both experience largenormal forces during turns.
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Vector Mechanics for Engineers: Dynamicsnthition
Equations of Motion
12 - 28
For tangential and normal components,
t t
t
F ma
dvF mdt
Newtons second law amF
2
n n
n
F ma
vF m
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Vector Mechanics for Engineers: Dynamicsnthtion
Sample Problem 12.5
12 - 29
The bob of a 2-m pendulum describes
an arc of a circle in a vertical plane. If
the tension in the cord is 2.5 times the
weight of the bob for the positionshown, find the velocity and accel-
eration of the bob in that position.
SOLUTION:
Resolve the equation of motion for thebob into tangential and normal
components.
Solve the component equations for the
normal and tangential accelerations.
Solve for the velocity in terms of the
normal acceleration.
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Vector Mechanics for Engineers: Dynamicsnthtion
Sample Problem 12.5
12 - 30
SOLUTION:
Resolve the equation of motion for the bob into
tangential and normal components.
Solve the component equations for the normal and
tangential accelerations.
:tt maF
30sin
30sin
ga
mamg
t
t
2sm9.4ta
:nn maF
30cos5.2
30cos5.2
ga
mamgmg
n
n
2
sm03.16na
Solve for velocity in terms of normal acceleration.
22
sm03.16m2 nn avv
a
sm66.5
v
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Vector Mechanics for Engineers: Dynamicsnthtion
Sample Problem 12.6
12 - 31
Determine the rated speed of a
highway curve of radius= 400 ft
banked through an angle q= 18o. The
rated speed of a banked highway curveis the speed at which a car should
travel if no lateral friction force is to
be exerted at its wheels.
SOLUTION:
The car travels in a horizontal circularpath with a normal component of
acceleration directed toward the center
of the path.The forces acting on the car
are its weight and a normal reaction
from the road surface.
Resolve the equation of motion for
the car into vertical and normal
components.
Solve for the vehicle speed.
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Vector Mechanics for Engineers: Dynamicsthtion
Sample Problem 12.6
12 - 32
SOLUTION:
The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the centerof the path.The forces acting on the
car are its weight and a normal
reaction from the road surface.
Resolve the equation of motion for
the car into vertical and normal
components.
:0 yF
q
q
cos
0cos
WR
WR
:nn maF
q
q
q
2
sincos
sin
v
g
WW
agWR n
Solve for the vehicle speed.
18tanft400sft2.32
tan
2
2 qgv
hmi1.44sft7.64 v
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Vector Mechanics for Engineers: Dynamicsthtion
Group Problem Solving
2 - 33
The 3-kg collarB rests on the
frictionless armAA. The collar is
held in place by the rope attached to
drumD and rotates about O in a
horizontal plane. The linear velocityof the collar B is increasing according
to v= 0.2 t2 where v is in m/s and t is
in sec. Find the tension in the rope
and the force of the bar on the collar
after 5 seconds ifr= 0.4 m.
v SOLUTION:
Write the equations of motion for the
collar.
Combine the equations of motion with
kinematic relationships and solve.
Draw the FBD and KD for the collar.
Determine kinematics of the collar.
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Vector Mechanics for Engineers: Dynamicsthtion
Group Problem Solving
2 - 34
SOLUTION: Given: v= 0.2 t2, r= 0.4 m
Find: T and N at t= 5 sec
Draw the FBD and KD of the collar
manT N
et
en
mat
Write the equations of motion
=
n nF ma t tF ma 2v
N m
dv
T m
dt
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Group Problem Solving
2 - 35
manT N
et
en
mat
=2 2
25 62.5 (m/s )0.4
n
va
20.4 0.4(5) 2 m/stdv
a tdt
Kinematics : find vt, an, at2 20.2 0.2(5 ) =5 m/stv t
Substitute into equations of motion
3.0(62.5)N 3.0(2)T
187.5 NN 6.0 NT
q
n nF ma t tF ma
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Group Problem Solving
2 - 36
manT N
et
en
mat
=How would the problemchange if motion was in the
vertical plane?
mg
q
You would add an mgterm
and would also need to
calculate q
When is the tangential force greater than the normal force?
Only at the very beginning, when starting to accelerate.
In most applications, an >> at
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Vector Mechanics for Engineers: Dynamicsthion
Concept Question
2 - 37
D
B CA
A car is driving from A to D on the curved path shown.
The driver is doing the following at each point:
Agoing at a constant speed Bstepping on the acceleratorCstepping on the brake Dstepping on the accelerator
Draw the approximate direction of the cars acceleration
at each point.
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Vector Mechanics for Engineers: Dynamicsthion
Kinetics: Radial and Transverse Coordinates
2 - 38
Hydraulic actuators and
extending robotic arms are
often analyzed using radial
and transverse coordinates.
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Vector Mechanics for Engineers: Dynamicshion
Eqs of Motion in Radial & Transverse Components
12 - 39
qqq
rrmmaF
rrmmaF rr
2
2
Consider particle at rand q, in polar coordinates,
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Vector Mechanics for Engineers: Dynamicshon
Sample Problem 12.7
12 - 40
A blockB of mass m can slide freely ona frictionless arm OA which rotates in a
horizontal plane at a constant rate .0q
a) the component vrof the velocity ofB
along OA, and
b) the magnitude of the horizontal force
exerted onB by the arm OA.
Knowing thatB is released at a distance
r0
from O, express as a function ofr
SOLUTION:
Write the radial and transverseequations of motion for the block.
Integrate the radial equation to find an
expression for the radial velocity.
Substitute known information into the
transverse equation to find an
expression for the force on the block.
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Vector Mechanics for Engineers: DynamicshonSample Problem 12.7
12 - 41
SOLUTION:
Write the radial and transverse
equations of motion for the block.
:
:
qq amF
amFrr
q
rrmF
rrm
2
0 2
Integrate the radial equation to find an
expression for the radial velocity.
r
r
v
rr
rr
rr
rrr
drrdvv
drrdrrdvv
dr
dvv
dt
dr
dr
dv
dt
dvvr
r
0
20
0
202
q
dr
dvv
dt
dr
dr
dv
dt
dvvr rr
rrr
20
220
2rrvr q
Substitute known information into the
transverse equation to find an expression
for the force on the block.
21202202 rrmF q
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Vector Mechanics for Engineers: DynamicshonGroup Problem Solving
2 - 42
Write the equations of motion for the
collar.
Combine the equations of motion withkinematic relationships and solve.
Draw the FBD and KD for the collar.
Determine kinematics of the collar.
SOLUTION:
The 3-kg collarB slides on the frictionless armAA. The arm is attached to
drumD and rotates about O in a horizontal plane at the rate whereand tare expressed in rad/s and seconds, respectively. As the arm-drum
assembly rotates, a mechanism within the drum releases the cord so that the
collar moves outward from O with a constant speed of 0.5 m/s. Knowing that
at t= 0, r= 0, determine the time at which the tension in the cord is equal to
the magnitude of the horizontal force exerted onB by armAA.
0.75tq q
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Vector Mechanics for Engineers: DynamicshonGroup Problem Solving
2 - 43
mar
T N
eqer
maq
=
r rF ma BF m aq q
Draw the FBD and KD of the collar
Write the equations of motion
SOLUTION: Given:
Find: time when T = N
2( )T m r r q ( 2 )N m r rq q
0.75tq
5 m/sr
(0) 0r
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Vector Mechanics for Engineers: DynamicshonGroup Problem Solving
2 - 44
3 2: (3 kg)( 0.28125 ) m/sr rF ma T t
2: (3 kg)(1.125 ) m/sBF m a N tq q
0 00.5r tdr dt
(0.5 ) mr t
0r
2
(0.75 ) rad/s
0.75 rad/s
tq
q
2 2 3 20 [(0.5 ) m][(0.75 ) rad/s] (0.28125 ) m/sra r r t t t q
2
2
2 [(0.5 ) m][0.75 rad/s ] 2(0.5 m/s)[(0.75 ) rad/s]
(1.125 ) m/s
a r r t t
t
q q q
Kinematics : find expressions for r and q
Substitute values into ar , aq
Substitute into equation of motion3(0.84375 ) (3.375 )t t
2 4.000t
2.00 st
SetT = N
5 m/sr
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Vector Mechanics for Engineers: DynamicshonConcept Quiz
2 - 45
Top View
e1
e2
w
v
The girl starts walking towards the outside of the spinning
platform, as shown in the figure. She is walking at a constant
rate with respect to the platform, and the platform rotates at aconstant rate. In which direction(s) will the forces act on her?
a) +e1 b) - e1 c) +e2 d) - e2
e) The forces are zero in the e1
and e2
directions
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Vector Mechanics for Engineers: DynamicshonAngular Momentum of a Particle
2 - 46
Satellite orbits are analyzed using conservation
of angular momentum.
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Vector Mechanics for Engineers: DynamicshonEqs of Motion in Radial & Transverse Components
12 - 47
qqq
rrmmaF
rrmmaF rr
2
2
Consider particle at rand q, in polar coordinates,
q
q
q
q
rrmF
rrrm
mrdt
dFr
mrHO
2
22
2
2
This result may also be derived from conservationof angular momentum,
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ecto ec a cs o g ee s y a cshonAngular Momentum of a Particle
12 - 48
moment of momentum or the angular
momentum of the particle about O.
VmrHO
Derivative of angular momentum with respect to time,
O
O
M
Fr
amrVmVVmrVmrH
It follows from Newtons second law that the sum of
the moments about O of the forces acting on the
particle is equal to the rate of change of the angular
momentum of the particle about O.
zyx
O
mvmvmv
zyx
kji
H
is perpendicular to plane containingOH
Vmr
and
q
q
2
sin
mr
vrm
rmVHO
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g yonConservation of Angular Momentum
12 - 49
When only force acting on particle is directed
toward or away from a fixed point O, the particle
is said to be moving under a central force.
Since the line of action of the central force passes
through O, and0 OO HM
constant OHVmr
Position vector and motion of particle are in a
plane perpendicular to .OH
Magnitude of angular momentum,
000 sin
constantsin
Vmr
VrmHO
massunit
momentumangular
constant
2
2
hrm
H
mrH
O
O
q
q
or
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g yonConservation of Angular Momentum
12 - 50
Radius vectorOPsweeps infinitesimal area
qdrdA 221
Define qq
2212
21 r
dt
dr
dt
dAareal velocity
Recall, for a body moving under a central force,
constant2 qrh
When a particle moves under a central force, its
areal velocity is constant.
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g yonNewtons Law of Gravitation
12 - 51
Gravitational force exerted by the sun on a planet or by
the earth on a satellite is an important example ofgravitational force.
Newtons law of universal gravitation - two particles of
massMand m attract each other with equal and opposite
force directed along the line connecting the particles,
4
49
2
312
2
slb
ft104.34
skg
m1073.66
ngravitatioofconstant
G
r
MmGF
For particle of mass mon the earths surface,
222 s
ft2.32
s
m81.9 gmg
R
MGmW
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g ynSample Problem 12.8
12 - 52
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 18820 mi/h from
an altitude of 240 mi. Determine the
velocity of the satellite as it reaches it
maximum altitude of 2340 mi. The
radius of the earth is 3960 mi.
SOLUTION:
Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
atA andB and solve for the velocity atB.
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g ynSample Problem 12.8
12 - 53
SOLUTION:
Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
atA andB and solve for the velocity atB.
mi23403960mi2403960
hmi18820
constantsin
B
AAB
BBAA
O
r
rvv
vmrvmrHvrm
hmi12550Bv
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g ynTrajectory of a Particle Under a Central Force
12 - 54
For particle moving under central force directed towards force center,
022 qqqq FrrmFFrrm r
Second expression is equivalent to from which,,constant2
hr q
rd
d
r
hr
r
h 1and
2
2
2
2
2
q
q
After substituting into the radial equation of motion and simplifying,
ru
umh
Fu
d
ud 1where
222
2
q
IfFis a known function ofroru, then particle trajectory may be
found by integrating foru = f(q), with constants of integration
determined from initial conditions.
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g ynApplication to Space Mechanics
12 - 55
constant
1where
22
2
2
2222
2
h
GMu
d
ud
GMmur
GMmF
ru
umh
Fu
d
ud
q
q
Consider earth satellites subjected to only gravitational pull
of the earth,
Solution is equation of conic section,
tyeccentricicos11 2
2
GM
hC
h
GM
ru q
Origin, located at earths center, is a focus of the conic section.
Trajectory may be ellipse, parabola, or hyperbola depending
on value of eccentricity.
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g ynApplication to Space Mechanics
12 - 56
tyeccentricicos112
2 GMhC
hGM
rq
Trajectory of earth satellite is defined by
hyperbola, > 1 orC > GM/h2. The radius vector
becomes infinite for
21111 cos1cos0cos1
hCGM
parabola, = 1 orC = GM/h2. The radius vector
becomes infinite for
1800cos122
ellipse, < 1 orC < GM/h2. The radius vector is finite
forqand is constant, i.e., a circle, for< 0.
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g ynApplication to Space Mechanics
12 - 57
Integration constant Cis determined by conditions
at beginning of free flight, q=0, r = r0,
20002
0
2
20
11
0cos11
vr
GM
rh
GM
rC
GM
Ch
h
GM
r
0
0
200
2
2
or1
r
GMvv
vrGMhGMC
esc
Satellite escapes earth orbit for
Trajectory is elliptic forv0 < vesc and becomes
circular for= 0 orC= 0,
0r
GMvcirc
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g ynApplication to Space Mechanics
12 - 58
Recall that for a particle moving under a central
force, the areal velocity is constant, i.e.,
constant212
21 hr
dt
dAq
Periodic time or time required for a satellite to
complete an orbit is equal to area within the orbit
divided by areal velocity,
h
ab
h
ab
2
2
where
10
1021
rrb
rra
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g ynSample Problem 12.9
12 - 59
Determine:
a) the maximum altitude reached by
the satellite, and
b) the periodic time of the satellite.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 36,900 km/h at an
altitude of 500 km.
SOLUTION:
Trajectory of the satellite is described by
qcos1
2C
h
GM
r
Evaluate C using the initial conditions
at q = 0.
Determine the maximum altitude by
finding rat q= 180o.
With the altitudes at the perigee and
apogee known, the periodic time canbe evaluated.
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g ynSample Problem 12.9
12 - 60
SOLUTION:
Trajectory of the satellite is described by
qcos12
Ch
GM
r
Evaluate C using the initial conditions
at q = 0.
2312
2622
29
3600
3
0
6
0
sm10398
m1037.6sm81.9
sm104.70
sm1025.10m106.87
sm1025.10
s/h3600
m/km1000
h
km36900
m106.87
km5006370
gRGM
vrh
v
r
1-9
22
2312
6
20
m103.65
sm4.70
sm10398
m1087.6
1
1
h
GM
rC
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n
Sample Problem 12.9
12 - 61
Determine the maximum altitude by finding r1
at q= 180o.
km66700m107.66
m
1103.65
sm4.70
sm103981
61
9
22
2312
21
r
Ch
GM
r
km60300km6370-66700altitudemax
With the altitudes at the perigee and apogee known,
the periodic time can be evaluated.
sm1070.4
m1021.4m1036.82
h
2
m1021.4m107.6687.6
m1036.8m107.6687.6
29
66
66
10
66
21
1021
ab
rrb
rra
min31h19s103.70 3
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enth
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dition
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Keplers Laws of Planetary Motion
Results obtained for trajectories of satellites around earth may also be
applied to trajectories of planets around the sun.
Properties of planetary orbits around the sun were determined
astronomical observations by Johann Kepler (1571-1630) before
Newton had developed his fundamental theory.
1) Each planet describes an ellipse, with the sun located at one of itsfoci.
2) The radius vector drawn from the sun to a planet sweeps equal
areas in equal times.
3) The squares of the periodic times of the planets are proportional tothe cubes of the semimajor axes of their orbits.