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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

4

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


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

22

n

v rva r vr

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ta v r


note: and

5

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)

6

• 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


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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


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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

9


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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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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


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.

11


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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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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

13

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:


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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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14

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:

15


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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

16


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Gi FAS 0 6

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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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17

– Rotation Concepts:

ddtd

Angular velocity

ddt

d d


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


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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


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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


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Given: FAS, = 2 rad/s, = 0, a0=3 m/s2

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Find: aA when = 0º, 90º,and 180º


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

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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:


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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

22

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


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 .


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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:

24

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


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6/33

Given: FAS, M = 20 kg,released from rest

Find: Reaction forces atpin O

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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

26


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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

27


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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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28

0mx kx

k

Let’s define the following:


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


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


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


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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

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