Dynamics FE Review
For rectilinear motion - motion in a
straight line - where the position is
defined by s:
Mechanics
Response of mass (body)
to mechanical disturbance
Statics
Dynamics
Analysis of body
at rest
Kinematics
Video
2011
v
Analysis of body
in motion
Kinetics
Geometry of motion—
no concern for forces
that caused motion
1
Find: v(t) and a(t)
Solution:
dv
dt
Where v is the instantaneous velocity, a is
the instantaneous acceleration, and t is
time.
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3
Given: Position of a car is described by
s  3t 3  t 2 m.
For a particle whose position is
defined by the vector r:
a
a ds  v dv
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• Fundamental equations of motion
dr
dt
dv
dt
Example Problem:
Kinematics of Particles
v
a
Note: Instead of s, the position could
be defined by x, y, etc.
Relation between
force, mass, and
motion
Figures and problems taken from the textbook Dynamics, 5th
edition, Meriam and Kraige, Wiley.
ds
dt
2
2011
v
ds
:
dt
v  9t 2  2t m/s
a
dv
:
dt
a  18t  2
m/s 2
4
1
Example Problem:
So, we start with
Given: Acceleration of a car is given by:
a(t ) 
dv
dt
Note we wrote the
acceleration as a(t) to
emphasize the fact that
a is a function of t.
Then: dv  a( t ) dt
a( t )  3t 2  5t  1 m/
m/s2
v
t
v0
0
 dv   a(t ) dt
At t = 0, v0 = 4 m/s
Find: Velocity (v) when t=3 s.
t
v  v0   a( t ) dt
0
This gives us v as a
function of time or v(t)
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2011
t
v( t )  v0   a(t ) dt
0
7
t
First, look for a fundamental equation that
contains both a, t and v.
v( t )  v0   a(t ) dt
0
So for a(t) = 3t2+5t+1
5t 1, v0 = 4 m/s, and t = 3 s
v
ds
dt
a
dv
dt
t
v  4   (3t 2  5t  1) dt
a ds  v dv
0
3
 3t 3 5t 2 

 t
v  4
2
 3
0
v  56.5 m/s
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2011
8
2
For uniformly accelerated rectilinear
(UARM) motion (a=constant) the following
equations apply:
– If the acceleration is constant, we can
apply the UARM equations in the x and y
directions.
For the x direction
v  v0  a (t  t0 )
a (t  t0 )
s  s0  v0 (t  t0 ) 
2
vy   vy   ayt
vx   vx 0  axt
2
at
x  x0   vx 0 t  x
2
v  v  2a  s  s0 
2
For the y direction
0
2
y  y0   v y  t 
ayt 2
0
2
vx 2   vx 0  2ax  x  x0  v y 2   v y 0  2a y  y  y0 
2
0
2
2
THESE ONLY APPLY IF THE
ACCELERATION IS CONSTANT!!!
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9
2011
Note: This is for t0 = 0. If t0 = 0
then replace t with t-t0
– Projectile Motion using Rectangular Coordinates
• Curvilinear Motion Using Rectangular
Coordinates (x-y)
– Useful when the position (r) is given in
rectangular coordinates
ax = 0
ay = -g
Fig 2/7 Meriam and Kraige
v
11
dr
 xi  y j
dt
Fig 2/8 Meriam and Kraige
For the x direction
vx   vx 0
dv
a
 
xi  
yj
dt
x  x0   vx 0 t
For the
y direction
v y   v y   gt
0
y  y0   v y  t 
0
gt 2
2
v y 2   v y   2 g  y  y0 
2
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10
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0
12
3
• Curvilinear Motion Using NormalTangential (Path) Coordinates (n-t)
Example:
Given: Projectile fired off a cliff as shown
– Useful when the path is given, especially
the curvature of the path
y
et
180 m/s
o
ymax
30º
x
en
150 m
x at impact
Fig
g 2/9 Meriam and Kraige
g
Find: x at impact and
v  vet
ymax
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13
a
v2

en  vet
an 
v2

at  v
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15
– Special Case: Circular motion using n-t
coordinates
 = constant = r
Angular position
given
i
b
by 
v  r
v2
 r 2  v
r
at  v  r
an 
note:    and   
Fig 2/12 Meriam and Kraige
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4
Example: Car on the circular part of track
For Curvilinear Motion Using Polar Coordinates:
Given: FAS, r = 200 m, v = 50 m/s, at = 2 m/s2
v  rer  reθ
et


er
e
en

For circular
motion:
r  0

r 0
r
note:   
Find: a
Fig P2/144 Meriam and Kraige
2011

a  
r  r 2 er  r  2r eθ
17
and   
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Kinetics of Particles
• Kinetics: Relations between forces and
motion.
• Newton’s Second Law: The acceleration of
a particle is proportional to the resulting
force acting on it and is in the direction of
this force.
force
(assumes m is constant)
F  ma

• FBD: You must be able to draw good free
body diagrams!
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5
• Rectangular Coordinates (Cartesian)
• Polar Coordinates (Radial/Transverse)
For particle P:
y
F2
For particle A:
 F  ma
 F  ma i  ma j
F3
x
F3
F2
y
 F  ma
 F  ma e
r r
 ma e
P
F1
Scalar components:
x
F
F
x
y
 ma y
r
Fig 2-13 Meriam and Kraige
21
• Normal/Tangential Coordinates (Path)
F2
F1
n n
r
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23
– Units
For particle C:
 F  ma
 F  ma e
 F  ma  m  r  r 
 F  ma  m  r  2r 
2
 ma x
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F3
Scalar components:
F1
Force
A
Acceleration
l
ti
Mass
g
 mat et
Scalar components:
SI
US
N
m/s
/ 2
kg
9.81 m/s2
lb
2
ft/
ft/sec
slug (32.2 lbm)
32.2 ft/sec2
Fig 2-9 Meriam and Kraige
F
n
 ma n  m
 F  ma
t
2011
t
v2
a = 1 m/s2

F=1N
m=1 kg
 mv
22
2011
a = 1 ft/sec2
F
F = 1 lb
m=1 slug
24
6
Example Problem: Textbook 3/12
Example Problem: Textbook 3/1
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2011
25
3/1
Given:
Find:
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27
3/12
FAS
v0 = 7 m/s at x0 = 0
k = 0.4, m = 50 kg
Given:
FAS, W = 100 lb
ft/sec
c2 up incline
nc n
a = 5 ft/s
k = 0.25
Find:
P
t and x when v = 0
26
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7
Example Problem: Textbook 3/54
Example Problem:Textbook 3/50
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3/50
Given:
FAS, m = 2 kg
vB = 3.5 m/s
 = 2.4
24m
Find:
NB and vA such that
NA = 0
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31
3/54
30
Given:
FAS, W = 0.2 lb
 = 30
 = 3 rad/sec ccw
r = -4 ft/sec
Find:
N
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8
Example Problem: Textbook 3/56
Work/Energy
• We have been using the direct application of
Newton’s Second Law to solve kinetics
problems.
 F  ma
Forces
Acceleration
Motion
This method of solution can be very difficult sometimes!
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3/56
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35
• Work/Energy methods:
– These methods will make it MUCH EASIER
to solve some kinetics problems!
– Definition of work:
Given:
FAS,, W = 3000 lb
r = 100 ft, v = 35 mi/hr
Find:
aN, FN
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U
• Component of force acting in the direction of
motion times the displacement.
• Units: SI N•m = J
US
34
2011
ft•lb
36
9
– Conservative force: Work done by a
conservative force is independent of path.
Consider a particle moving
along the path from A to A
r2
• The work only depends on the starting and
ending positions!
• When a p
particle moves under the influence
f
of
fa
conservative force:
SCALAR!
U   F  dr
r1
If we let |dr| = ds
U 1 2  V1  V2
s2
U   Ft ds
Sign convention:
Positive if active force (Ft)
is in the direction of motion
and negative if it is in the
opposite direction
2011
Then:
37
V g  mgh
from the datum.
+ if above datum
- if below datum
Ve 
39
T1  V g 1  Ve1  U 1 2  T2  V g 2  Ve 2
h is measured
Spring:
T1  V g 1  Ve1  U 1 2  T2  V g 2  Ve 2
This is the fundamental equation for
applying the work/energy method.
• Energy available due to position
datum
U12
2011
– Potential Energy: (V)
Gravity:
T1  V1  T2  V2
•Work done by nonconservative forces:
s1
Fig 3-2 Meriam and Kraige
and
The FE reference handbook gives the equation
in this form:
1 2
kx
2
T1  U 1  W1 2  T2  U 2
Fig 33-6
6 Meriam and Kraige
- Kinetic Energy: (T)
2011
T 
Whenever you have conservative forces doing work -- gravity and
springs -- consider using the work/energy method.
1
mv 2
2
38
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10
Example Problem: Textbook Sample 3/17
Example Problem: Textbook 3/104
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2011
41
Sample 3/17
3/104
Given: FAS, vA = 5 m/s,
hA = 0,
hB = 0.8 m
Find: vB
Given: FAS, mg = 6 lb, k = 2 lb/in,
unstretched length = 24 in,
vA = 0
Find:
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42
vB
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11
Example Problem: Textbook 3/144
Impulse/Momentum
• In some situations, the FORCES are
described as acting over an interval of
TIME. Impulse/Momentum methods work
well in these cases.
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3/144
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47
•Linear Impulse - Linear Momentum
Define:
t2
  Fdt  Linear Impulse
VECTOR!
t1
m v  G  Linear Momentum
Then:
Given: FAS, m = 4 kg, vA = 0
unstretched length = 24 in
Find:
vB, x at max deformation
t2
  Fdt  G
2
VECTOR!
 G 1  G
t1
Linear Impulse = Change of Linear Momentum
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48
12
Rearranging:
3/179
t2
G 1    Fdt  G 2
t1
Scalar Components:
t2
G x1    Fx dt  G x2
Given:
FAS, For projectile:
M=75 g, v1=600 m/s
For block: M=50 kg
v1 = 0
Find:
E during impact
t1
t2
G y1    Fy dt  G y2
t1
Conservation of
Linear Momentum:
If
If F  0

then G 1  G 2
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2011
Example Problem: Textbook 3/179
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Example Problem: Textbook 3/188
50
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13
3/188
y
Given: FAS, For tanker: M=10.43x106 slugs, v1=0,
Cable tension = 50,000 lb
Find: Time required to bring speed of tanker to 1 knot
O
x
Fig 3-11 Meriam and Kraige
For plane motion in the
x-y plane:
H o  r  mv
 mvr sin  k
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53
• Angular Impulse - Angular Momentum
2011
Define
Angular
Impulse:
– Angular Momentum: H0
• The moment of the linear momentum
about a point
It can be shown:
55
t2
M
o
dt  Angular Impulse
o
dt  H O2  H O1   H O
t1
t2
M
t1
Rearranging:
O
t2
H O1    M o dt  H O2
t1
Conservation of
Angular Momentum:
If
M
O
 0 then H O 1  H O 2
Fig 3-11 Meriam and Kraige
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56
14
Kinetics of Particles Impact
Example Problem: Textbook 3/227
• Impact: Collision between two bodies.
bodies
Direct Central
Impact: Centers
of mass located on
Line
L
ne of Impact
(LOI). Velocities
in direction of
LOI.
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3/277
LOI
LOI
Fig 3-14 Meriam and Kraige
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59
If we have no external impulsive forces, TOTAL
linear momentum of the system is conserved.
G Before  G After
Along the LOI:
m 1 v1  m 2 v 2  m 1 v1'  m 2 v 2'
Coefficient of Restitution (e)
Given: FAS, N1 = 0
N2 = 150 rpm
T = 20 N
Find:
2011
e
t
58
2011
Relative velocity after
Relative velocity before

v2'  v1'
v1  v2
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15
Plane Kinematics of Rigid
Bodies - Plane Motion
Example Problem: Textbook 3/247
• Rigid Body: System of particles for
which the distances between the
particles remain unchanged.
• Plane Motion: All parts of the body move
in parallel planes.
planes
– Plane of Motion: Plane that contains the
center of mass.
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3/247
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63
• Types of Plane Motion:
Translation: All points
on the rigid body have
the same velocity and
acceleration.
l
Kinematic concepts
from Chapter 2 apply
Gi
Given:
FAS e = 0
FAS,
0.6
6
Find:
2011
v1’ and v2’
62
2011
Fig 5-1 Meriam and Kraige
64
16
– Rotation Concepts:
Example Problem: Textbook 5/2
d
 
dt
d

   
dt
 d    d

Fig 5-2 Meriam and Kraige
Angular velocity
Angular acceleration
For  = constant:
   O   (t  t 0 )
   O2  2    O 
1
2
   O   (t  t 0 )   (t  t 0 ) 2
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5/2
– Fixed Axis Rotation:
For any point on the rigid body,
v  r
a n  r 2  v
2
a t  r
r
 v
In vector form,
Given: FAS, =10 rad/s
v  r  ω  r
Fig 5-3 Meriam and Kraige
2011
Find:
F
a n  ω   ω  r    2 r
vA an
and aA
at  α  r
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17
Relative Motion -Translating Axes (Velocity)
Plane Kinematics of Rigid Bodies Relative Motion Method
•G
Generall Plane
Pl
Motion
M ti can be
b considered
id
d
as Translation + Rotation
(Translating
Frame)
Absolute
velocity of A
wrt fixed
frame (X,Y)
y
vA  vB  vA B
A/B
Y
x
(Fixed Frame)
X
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Relative Motion:
rA  rB  rA
rA  rB  rA

rA  
rB  
rA
If A and B are on the same
rigid body, vA/B =   rA/B
y
((Translating
ranslat ng
Frame)
A/B
B
B
Absolute
acceleration of
A wrt fixed
f
frame
(X,Y)
x
B
or v A  v B  v A
or a A  a B  a A
B
aA  aB  aA B
Y
(Fixed Frame)
B
X
2011
Relative
R
l
velocity
l
of
f A wrt a
translating frame (x,y)
attached to B.
71
Relative Motion -Translating Axes (Acceleration)
XY Fixed
xy Translating with B
Fig.
g 2-17 Meriam and Kraige
g
Fig 5/6 Meriam and Kraige
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69
Absolute
velocity of origin
of the
translating
frame at B wrt
fixed frame
(X,Y)
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2011
Fig 5/9 Meriam and Kraige
Absolute
acceleration of
origin of the
t
translating
l ti
frame at B wrt
fixed frame
(X,Y)
Relative acceleration of A
wrt a translating frame
(x,y) attached to B. Due to
rotation about B
72
18
Relative Motion Translating Axes.
To summarize:
5/120
vA  vB  vA B
or
v A  v B  ω  rA B
y
(Translating
Frame)
or
A/B
x
v A  v B  ω  rrel
aA  aB  aA B
or
Y
(Fixed Frame)
X
a A  a B  ω   ω  rA B   α  rA B
Given:
FAS,  = 2 rad/s,
 = 0, a0=3 m/s2
Find:
aA when  = 0º, 90º,
and 180º
or
a A  a B  ω   ω  rrel   α  rrel
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Example Problem: Textbook 5/120
75
Example Problem: Textbook 5/141
100
60
180
Y
80
80
X
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19
5/141
• Relative Motion - Rotating Axes
– This method works best when sliding occurs
relative to two rigid bodies.
– Consider the following:
100
180
60
80Y
80
A
Given:
FAS,
P
OA = 10 rad/s
CCW
Find:
AB
Fig 5/11 Meriam and Kraige
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5/141
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79
Relative Motion -Rotating Axes (Velocity)
100
180
60
80Y
v A  v B  v P B  v A/ P
80
Or:
v A  v B  ω  rrel  v rel
Given:
FAS,
OA = 10 rad/s
CCW
Find:
AB
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2011
80
20
5/174
Relative Motion -Rotating Axes (Acceleration)
a A  aB  aP B  a A/ P
Given:
Or:
a A  a B  α  rrel  ω   ω  rrel   2ω  v rel  a rel
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81
Find:
FAS,
OA = 10 rad/s
CW
 = 30
BC , arel
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83
Example Problem: Textbook 5/174
Given:
Find:
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2011
FAS,
OA = 10 rad/s
CW
 = 30
BC , arel
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21
IC Method Steps:
1. Identify directions of
velocity vectors of two points.
2. At these two points draw lines
perpendicular to the velocity
vectors.
3. These lines intersect at the
IC point C.
Given:
Find:
Fig 5/7 Meriam and Kraige
FAS,
OA = 10 rad/s
CW
 = 30
BC , arel
2011
4. If we know the magnitude of
vA or vb you can solve for .
85
v A  rA
v B  rB 
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87
Example Problem: Textbook 5/97
Plane Kinematics of Rigid Bodies
- Instantaneous Center of Zero
Velocity Method
For plane motion, at any instant, the motion may
be considered as pure rotation about a point
called the instantaneous center of zero velocity
(IC)
Works well if we know the directions of the
velocity vectors of two points on the rigid body
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22
5/97
For general Plane Motion:
Video
 F  ma
 M  Iα :
 M  I
G
G
Given:
Find:
Or:
FAS,
FAS
OB = 0.8 rad/sec
cw
vA and vC
2011
 F  ma
M  I
Where C refers
to the center of
mass, shown as G
in the figures
89
C
C
C
α
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91
Plane Kinetics of Rigid Bodies Newton’s Second Law
If we take
moments about an
arbitrary point, P,
then the moment
equation becomes:
• All of the concepts of particle kinetics apply
to kinetics of rigid bodies.
M
M
• However, we must account for the rotational
effects of the rigid body.
P
 Iα  ρ  m a
P
 I   mad
Or:
• We will use the overbar to indicate a quantity
reverenced to the center of mass, G.
M
P
 I C α  ρ PC  m a C
Where C refers to the center of mass
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6/33
For Fixed Axis Rotation about
point O:
 F  ma
 M  Iα
 M  I
M  I α
M  I 
Given:
G
Find:
G
O
O
O
O
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FAS, M = 20 kg,
released from rest
Reaction forces at
pin O
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Example Problem: Textbook 6/33
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Example Problem: Textbook 6/77
94
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24
6/77
Kinetic Energy:
Translation:
T 
1
mv 2
2
T 
1
I O 2
2
T 
1
1
mv 2  I  2
2
2
T 
1
I IC  2
2
Fixed Axis Rotation:
Given:
General Plane Motion:
FAS, released from rest,
=40 s=0.3,
=40,
=0 3 k=0.2
=0 2
aG, friction force
Find:
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Plane Kinetics of Rigid Bodies Work/Energy Methods
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99
Potential Energy: Same as for Particles
Gravity:
h is measured
• All of the work/energy concepts of particle kinetics
apply to kinetics of rigid bodies.
from the datum.
+ if above datum
- if below datum
• However, we must account for additional rotational
effects .
• Recall:
R
ll
T1  V g 1  Ve1  U 1 2  T2  V g 2  Ve 2
Spring:
Or:
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T1  U 1  W1 2  T2  U 2
V g  mgh
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Ve 
1 2
kx
2
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25
Example Problem: Textbook 6/122
Plane Kinetics of Rigid Bodies Impulse/Momentum Method
• All of the Impulse/Momentum concepts of
particle kinetics apply to kinetics of rigid
bodies.
• However, once again, we must account for
additional rotational effects .
• Linear Impulse/Momentum
t2
G 1 r:   Fdt  G 2
G  mv
t1
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6/122
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Angular Momentum:
Angular Impulse:
HG  I 
t2
M
G
dt
t1
Then:
Given:
FAS, W=12 lb
kspring = 3 lb/in
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H G1    M G dt  H G2
t1
kO= 10 in,
1=90, 2=0,
1 = 0
2 = 4 rad/s
M
Find:
t2
For Fixed
Axis Rotation
About O:
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t2
H O  I O  H O1    M O dt  H O2
t1
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26
Example Problem: Textbook 6/173
Vibration and Time Response
• Mechanical and structural systems are
often
oft
n su
subjected
j ct to vibratory
ratory mot
motion.
on.
–
–
–
–
Automobiles on a rough road
Power lines and bridges on a windy day
Aircraft wings experiencing flutter
Buildings during an earthquake
• Here we have a brief introduction to
undamped, free vibration of particles.
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6/173
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Undamped Free Vibration:
Consider what happens when the
spring mounted cart is disturbed
from its equilibrium position a
distance x.
107
From the FBD and
Newton’s Second Law:
F
x
 ma x
O
Or:
 kx  mx
mx  kx  0
Given:
Find:
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FAS
 at t=4s
for each
case
Fig 8/1 Meriam and Kraige
106
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27
mx  kx  0
Let’s define the following:
n  k m
Fig 8/2 Meriam and Kraige
Then:
Displacement

x x  0
x  x0 cos  n t 
2
n
Fig 8/1 Meriam and Kraige
This equation describes simple harmonic
motion. The acceleration is proportional
to the displacement, but of opposite
sign.
Let
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109
A  x0
x0
n
and
Natural Frequency
sin  n t
B
x 0
n
n  k m
Period
 n  2 
n
Then the amplitude is
C
A2  B 2
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Example Problem: Textbook 8/4 - p 601

x   n2 x  0
This is a linear, homogeneous,
second order, differential
equation The solution is as
equation.
follows:
Initial velocity
Fig 8/1 Meriam and Kraige
x  x0 cos  n t 
x0
n
sin  n t
Position
Initial
displacement
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Natural
frequency
Time
110
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28
8/4
8/17
Given: FAS, x0 = -2 in
v0 = 7 in/sec
Given: Weight = 120 lb
Deflection = 0.9 in
Find: n
Find: Amplitude
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Torsional Vibration
Example Problem: Textbook 8/17
   0 cos  n t 
0
sin  n t
n
 n  kt I
kt  GJ
L
I = Mass moment of inertia
G = Shear modulus
J = Polar moment of inertia
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29