1
Dynamics FE ReviewMechanics
Response of mass (body) to mechanical disturbance
StaticsAnalysis of body
at rest
DynamicsAnalysis of body
in motion
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KinematicsGeometry of motion—no concern for forces
that caused motion
KineticsRelation between force, mass, and
motion
Figures and problems taken from the textbook Dynamics, 5th
edition, Meriam and Kraige, Wiley.
Video
Kinematics of Particles• Fundamental equations of motion
ddt
rv d
dt
va
For a particle whose position is defined by the vector r:
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Where v is the instantaneous velocity, a is the instantaneous acceleration, and t is time.
For rectilinear motion - motion in a straight line - where the position is defined by s:
dsvdt
dvadt
dvvdsa
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Note: Instead of s, the position could be defined by x, y, etc.
Example Problem:
Given: Position of a car is described bym.3 23s t t
Find: v(t) and a(t)
Solution: :dsvdt
29 2 m/sv t t
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:dvadt
218 2 m/sa t
2
Example Problem:
Given: Acceleration of a car is given by:2
( ) 3 5 1ta t t m/s2( )t m/
Find: Velocity (v) when t=3 s.
At t = 0, v0 = 4 m/s
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First, look for a fundamental equation that contains both a, t and v.
dsvdt
dvadt
dvvdsa
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( ) tdv a dtThen:
( )tdvadt
So, we start withNote we wrote theacceleration as a(t) toemphasize the fact thata is a function of t.
0
( )0
0 ( )
t
t
v t
v
t
dv a dt
v v a dt
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0
This gives us v as afunction of time or v(t) ( ) 0 ( )
0t t
t
v v a dt
( ) 0 ( )0
t t
t
v v a dt
So for a(t) = 3t2+5t+1, v0 = 4 m/s, and t = 3 s
2
033 2
4 (3 5 1)
3 54
t
v t t dt
t tv t
So for a(t) 3t 5t 1, v0 4 m/s, and t 3 s
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03 2
56.5 m/sv
3
For uniformly accelerated rectilinear (UARM) motion (a=constant) the following equations apply:
0 0( )v v a t t
20
0 0 0( )( )
2a t ts s v t t
2 2 2
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2 20 02v v a s s
THESE ONLY APPLY IF THE ACCELERATION IS CONSTANT!!!
• Curvilinear Motion Using Rectangular Coordinates (x-y)– Useful when the position (r) is given in
rectangular coordinatesFig 2/7 Meriam and Kraige
d x ydt
rv i j
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d x ydt
va i j
– If the acceleration is constant, we can apply the UARM equations in the x and ydirections.
For the x direction For the y direction
02
0 0
22
2
x x x
xx
v v a t
a tx x v t
For the x direction
02
0 0
22
y y y
yy
v v a t
a ty y v t
For the y direction
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2200
2x x xv v a x x 2200
2y y yv v a y y
Note: This is for t0 = 0. If t0 = 0 then replace t with t-t0
– Projectile Motion using Rectangular Coordinates
ax = 0ay = -g
Fig 2/8 Meriam and Kraige
0x xv v
For the x direction For the y direction
0y yv v gt
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0
0 0xx x v t
02
0 0
2200
2
2
y
y y
gty y v t
v v g y y
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Example:
Given: Projectile fired off a cliff as shown
y
x
y
150 m
180 m/s
30ºo ymax
x at impact
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Find: x at impact and ymax
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• Curvilinear Motion Using Normal-Tangential (Path) Coordinates (n-t)– Useful when the path is given, especially
the curvature of the path
Fig 2/9 Meriam and Kraige
en
et
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g g
v tv e2v v
n ta e e2
n tva a v
– Special Case: Circular motion using n-t coordinates
= constant = rAngular position i b given by
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n
v rva r vr
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ta v r
Fig 2/12 Meriam and Kraige
note: and
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Example: Car on the circular part of track
Given: FAS, r = 200 m, v = 50 m/s, at = 2 m/s2
et
r
t
en
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Find: a
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r r r θv e e
For Curvilinear Motion Using Polar Coordinates:
2 2r r r r r θa e e
For circular motion:
00
rr
ere
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Fig P2/144 Meriam and Kraigenote:
and
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
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this force.
• FBD: You must be able to draw good free body diagrams!
mF a (assumes m is constant)
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• Rectangular Coordinates (Cartesian)
yF F
For particle P:mF a
P
Scalar components:
x xF ma
F1
F2 F3
x yma ma F i j
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x x x
y yF ma
• Normal/Tangential Coordinates (Path)
For particle C:mF a
F2F3
Scalar components:2
nvF ma m
n tma ma n tF e e
Fig 2-9 Meriam and Kraige
F1
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n n
t tF ma mv
• Polar Coordinates (Radial/Transverse)
For particle A:mF a
Scalar components:
2r r
F ma m r r
r rma ma F e e
F1
F2
F3
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2F ma m r r Fig 2-13 Meriam and Kraige
– UnitsSI US
Force N lbA l ti / 2 ft/ 2Acceleration m/s2 ft/sec2
Mass kg slug (32.2 lbm)g 9.81 m/s2 32.2 ft/sec2
a = 1 m/s2 a = 1 ft/sec2
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F F = 1 N
m=1 kg
F = 1 lb
m=1 slug
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Example Problem: Textbook 3/1
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3/1
Given: FASv0 = 7 m/s at x0 = 0k = 0.4, m = 50 kg
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Find: t and x when v = 0
Example Problem: Textbook 3/12
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3/12
Given: FAS, W = 100 lba = 5 ft/sec2 up incline
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5 ft/s c up nc n k = 0.25
Find: P
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Example Problem:Textbook 3/50
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3/50
Given: FAS, m = 2 kgvB = 3.5 m/s = 2 4 m
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= 2.4 m
Find: NB and vA such thatNA = 0
Example Problem: Textbook 3/54
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3/54
Given: FAS, W = 0.2 lb = 30 = 3 rad/sec ccw
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= 3 rad/sec ccwr = -4 ft/sec
Find: N
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Example Problem: Textbook 3/56
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3/56
Given: FAS, W = 3000 lb
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,r = 100 ft, v = 35 mi/hr
Find: aN, FN
Work/Energy
• We have been using the direct application of • We have been using the direct application of Newton’s Second Law to solve kinetics problems.
mF a
Forces Acceleration Motion
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Forces Acceleration Motion
This method of solution can be very difficult sometimes!
• Work/Energy methods:
– These methods will make it MUCH EASIERto solve some kinetics problems! to solve some kinetics problems!
– Definition of work: U• Component of force acting in the direction of
motion times the displacement.
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• Units: SI N•m = JUS ft•lb
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Consider a particle moving along the path from A to A
2
1
r
r
U d F rSCALAR!
If we let |dr| = ds
2
1
s
ts
U F ds
1
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Fig 3-2 Meriam and Kraige
Sign convention:Positive if active force (Ft) is in the direction of motion and negative if it is in the opposite direction
– Potential Energy: (V)• Energy available due to position
Gravity: gV mghh is measuredfrom the datum.
+ if above datum- if below datum
datum
Fig 3-6 Meriam and Kraige
Spring: 212eV kx
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Fig 3 6 Meriam and Kraige
- Kinetic Energy: (T) 212
T mv
– Conservative force: Work done by a conservative force is independent of path.• The work only depends on the starting and
ending positions!• When a particle moves under the influence of a p f f
conservative force:
1 2 1 2U V V 1 1 2 2T V T V and
1 2U •Work done by nonconservative forces:
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1 1 1 1 2 2 2 2g e g eT V V U T V V Then:
This is the fundamental equation for applying the work/energy method.
1 1 1 1 2 2 2 2g e g eT V V U T V V
1 1 1 2 2 2T U W T U
The FE reference handbook gives the equation in this form:
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Whenever you have conservative forces doing work -- gravity and springs -- consider using the work/energy method.
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Example Problem: Textbook 3/104
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3/104
Given: FAS, vA = 5 m/s,hA = 0,hB = 0.8 m
Find: vB
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Example Problem: Textbook Sample 3/17
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Sample 3/17
Given: FAS, mg = 6 lb, k = 2 lb/in,unstretched length = 24 in,vA = 0
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Find: vB
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Example Problem: Textbook 3/144
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3/144
Given: FAS, m = 4 kg, vA = 0unstretched length = 24 in
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Find: vB, x at max deformation
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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Define: 2
Linear Impulset
dt F VECTOR!
•Linear Impulse - Linear Momentum
1
Linear Impulset
dt F
Linear Momentumm v G
Then: 2
2 1
t
dt F G G G
VECTOR!
VECTOR!
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1
2 1t
dt F G G G
Linear Impulse = Change of Linear Momentum
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2
1
1 2
t
t
dt G F GRearranging:
Scalar Components: 2t
Scalar Components:
1 2
1
2
1 2
1
x x xt
t
y y yt
G F dt G
G F dt G
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If 1 2If 0 then F G GConservation of Linear Momentum:
Example Problem: Textbook 3/179
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3/179
Given: FAS, For projectile: M=75 g, v1=600 m/sFor block: M=50 kgv1 = 0
Find: E during impact
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Example Problem: Textbook 3/188
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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
3/188
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• Angular Impulse - Angular Momentum– Angular Momentum: H0
• The moment of the linear momentum about a point about a point
O
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O
y
For plane motion in thex-y plane:
Fig 3-11 Meriam and Kraigex
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sin o m
mvr
H r vk
DefineAngularImpulse:
2
1
Angular Impulset
ot
dt M
It can be shown:2
2 1
1
t
o O O Ot
dt M H H H
Rearranging:2
1 2
t
O o Odt H M H
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1 2
1t
1 2If 0 then O O O M H HConservation of
Angular Momentum:
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Example Problem: Textbook 3/227
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3/277
Given: FAS, N1 = 0 N 150
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N2 = 150 rpmT = 20 N
Find: t
Kinetics of Particles -Impact
• Impact: Collision between two bodiesImpact: Collision between two bodies.
Direct Central Impact: Centers of mass located on Line of Impact
LOI
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L ne of Impact (LOI). Velocities in direction of LOI.
LOI
If we have no external impulsive forces, TOTALlinear momentum of the system is conserved.
Before AfterG G
' '1 1 2 2 1 1 2 2m v m v m v m v
Along the LOI:
Coefficient of Restitution (e)
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' '2 1
1 2
Relative velocity after
Relative velocity before
v vev v
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Example Problem: Textbook 3/247
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Gi FAS 0 6
3/247
Given: FAS, e = 0.6
Find: v1’ and v2’
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Plane Kinematics of Rigid Bodies - Plane Motion
• Rigid Body: System of particles for 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
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in parallel planes.– Plane of Motion: Plane that contains the
center of mass.
• Types of Plane Motion:
Translation: All points on the rigid body have the same velocity and
l acceleration. Kinematic concepts from Chapter 2 apply
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– Rotation Concepts:
ddtd
Angular velocity
ddt
d d
Fig 5-2 Meriam and Kraige
Angular acceleration
For = constant:( )t t
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0
2
20 0
( )
21( ) ( )2
O
O O
O
t t
t t t t
– Fixed Axis Rotation:
22
v rva r v
For any point on the rigid body,
n
t
a r vra r
v r ω rIn vector form,
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Fig 5-3 Meriam and Kraige 2
n
t
a ω ω r ra α r
Example Problem: Textbook 5/2
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5/2
Given: FAS, =10 rad/s
Find: vA and aA
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F A an A
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Plane Kinematics of Rigid Bodies -Relative Motion Method
G l Pl M ti b id d • General Plane Motion can be considered as Translation + Rotation
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Relative Motion:
XY Fixedxy Translating with B
or
or
A B AB
A B A A B AB B
A B A A B AB B
r r r
r r r v v v
r r r a a a
Fig. 2-17 Meriam and Kraige
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g g
Relative Motion -Translating Axes (Velocity)
Absolute velocity of A
Absolute velocity of origin of the translating frame at B wrt
A B A B v v v
wrt fixed frame (X,Y)
frame at B wrt fixed frame (X,Y)
R l l f x
y
Y
(TranslatingFrame)
A/B
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If A and B are on the same rigid body, vA/B = rA/B
Relative velocity of A wrt a translating frame (x,y)attached to B. X
(Fixed Frame)Fig 5/6 Meriam and Kraige
Relative Motion -Translating Axes (Acceleration)
y(Translating
Absolute acceleration of A wrt fixed f
Absolute acceleration of origin of the t l ti
x
Y
(Fixed Frame)
( ranslat ngFrame)
A/B
A B A B a a a
frame (X,Y) translating frame at B wrt fixed frame (X,Y)
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X(Fixed Frame)
Relative acceleration of A wrt a translating frame (x,y) attached to B. Due to rotation about B
Fig 5/9 Meriam and Kraige
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Relative Motion -Translating Axes.To summarize:
A B A B
A B A B
or
v v v
v v ω r
x
y
Y
(TranslatingFrame)
A/B A B rel
or v v ω r
A B A B
or
a a a
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X(Fixed Frame)
A B A B A B
A B rel rel
or
a a ω ω r α r
a a ω ω r α r
Example Problem: Textbook 5/120
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5/120
Given: FAS, = 2 rad/s, = 0, a0=3 m/s2
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Find: aA when = 0º, 90º,and 180º
Example Problem: Textbook 5/141
100
8080
60
X
Y 180
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20
5/141
80
100
60Y80
180
Given: FAS, = 10 rad/s
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OA = 10 rad/sCCW
Find: AB
5/141
80
100
60Y80
180
Given: FAS,OA = 10 rad/s
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OACCW
Find: AB
• Relative Motion - Rotating Axes– This method works best when sliding occurs
relative to two rigid bodies.– Consider the following:
P
A
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P
/A B P B A P v v v v
Relative Motion -Rotating Axes (Velocity)
A B rel rel v v ω r v
Or:
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21
Relative Motion -Rotating Axes (Acceleration)
/ A B P B A P a a a a
Or:
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2A B rel rel rel rel a a α r ω ω r ω v a
Or:
Example Problem: Textbook 5/174
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5/174
Given: FAS,
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OA = 10 rad/s CW = 30
Find: BC , arel
Given: FAS,
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OA = 10 rad/s CW = 30
Find: BC , arel
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Given: FAS,
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OA = 10 rad/s CW = 30
Find: BC , arel
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)
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Works well if we know the directions of the velocity vectors of two points on the rigid body
1. Identify directions of velocity vectors of two points.
IC Method Steps:
2. At these two points draw lines
Fig 5/7 Meriam and Kraige
perpendicular to the velocity vectors.
3. These lines intersect at the IC point C.
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A A
B B
v rv r
4. If we know the magnitude of vA or vb you can solve for .
Example Problem: Textbook 5/97
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23
5/97
Given: FAS
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Given: FAS,OB = 0.8 rad/seccw
Find: vA and vC
Plane Kinetics of Rigid Bodies -Newton’s Second Law
• All of the concepts of particle kinetics apply to kinetics of rigid bodies.
• However, we must account for the rotationaleffects of the rigid body.
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• We will use the overbar to indicate a quantity reverenced to the center of mass, G.
:
m
I
F a
M α
For general Plane Motion: Video
:G
G
I
M I
M α
Or:
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C
C C
m
I
F a
M α
Where C refers to the center of mass, shown as G in the figures
If we take moments about an arbitrary point, P, then the moment equation becomes:
P
P
I m
M I mad
M α ρ a
equation becomes:
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P C PC CI m M α ρ a
Or:
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For Fixed Axis Rotation about point O:
m F a
G
G
I
M I
M α
O OI
M α
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O OM I
Example Problem: Textbook 6/33
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6/33
Given: FAS, M = 20 kg,released from rest
Find: Reaction forces atpin O
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Example Problem: Textbook 6/77
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25
6/77
Given: FAS, released from rest, =40 =0 3 =0 2
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=40, s=0.3, k=0.2Find: aG, friction force
Plane Kinetics of Rigid Bodies -Work/Energy Methods
• All of the work/energy concepts of particle kinetics apply to kinetics of rigid bodies.
• However, we must account for additional rotationaleffects .
R ll T V V U T V V
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• Recall: 1 1 1 1 2 2 2 2g e g eT V V U T V V
1 1 1 2 2 2T U W T U
Or:
Kinetic Energy:
Translation: 212
T mv
Fixed Axis Rotation:
General Plane Motion:
212 OT I
2 21 12 2
T mv I
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2 2
212 ICT I
Potential Energy: Same as for Particles
Gravity: gV mghg
h is measuredfrom the datum.
+ if above datum- if below datum
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Spring: 212eV kx
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Example Problem: Textbook 6/122
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6/122
Given: FAS, W=12 lbkspring = 3 lb/in
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spring
kO= 10 in, 1=90, 2=0, 1 = 02 = 4 rad/s
Find: M
Plane Kinetics of Rigid Bodies -Impulse/Momentum Method
• All of the Impulse/Momentum concepts of 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
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• Linear Impulse/Momentum2
1
1 2
t
t
dt G F Gr: mG v
Angular Momentum: GH I
Angular Impulse:2t
GM dt
2
1 2
1
t
G G Gt
H M dt H
1
Gt
Then:
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For FixedAxis RotationAbout O: O OH I
2
1 2
1
t
O O Ot
H M dt H
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Example Problem: Textbook 6/173
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6/173
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Given: FASFind: at t=4s
for each case
Vibration and Time Response• Mechanical and structural systems are
often subjected to vibratory motion.oft n su j ct to ratory mot on.– Automobiles on a rough road– Power lines and bridges on a windy day– Aircraft wings experiencing flutter– Buildings during an earthquake
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• Here we have a brief introduction to undamped, free vibration of particles.
Undamped Free Vibration:Consider what happens when the spring mounted cart is disturbed from its equilibrium position a distance x.
From the FBD and Newton’s Second Law:
x xF maO
kx mx
Or:
0mx kx
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Fig 8/1 Meriam and Kraige
28
0mx kx
k
Let’s define the following:
Fig 8/1 Meriam and Kraige
n k m
Then:2 0nx x
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This equation describes simple harmonic motion. The acceleration is proportional to the displacement, but of opposite sign.
2 0nx x
This is a linear, homogeneous, second order, differential equation The solution is as
Fig 8/1 Meriam and Kraige
equation. The solution is as follows:
00 cos sinn n
n
xx x t t
Initial velocity
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Position
Initialdisplacement Natural
frequency Time
00 cos sinn n
n
xx x t t
Fig 8/2 Meriam and Kraige
n k m 2n
n
Displacement Natural Frequency Period
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n
0A x 0
n
xB
Let and
Then the amplitude is
2 2C A B
Example Problem: Textbook 8/4 - p 601
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29
8/4
Given: FAS, x0 = -2 in v0 = 7 in/sec
Find: Amplitude
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Find: Amplitude
Example Problem: Textbook 8/17
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8/17
Given: Weight = 120 lb Deflection = 0.9 in
Find:
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Find: n
00 cos sinn nt t
Torsional Vibration
n
n tk I
GJk
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tk L
I = Mass moment of inertiaG = Shear modulusJ = Polar moment of inertia