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King Fahd University of Petroleum &Minerals

Mechanical Engineering

Dynamics ME 201

B

Dr! Meyassar "! #l$%addad

ecture ' 22

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Principle of Linear Impulse and

Momentum

=+2

1

21 vFv

t

t

mdtm

Initial momentum + Sum of all Impulse = Final momentum

{ } 21 mtNtNtmgtFm cc =++++

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Conservation of Linear Momentum for a system

of Particles

=+2

1

21 )(vF)(v

t

t

iiiiimdtm

0

=

21 )v()(v iiii mm

Conservation of linear momentum equation

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Impulse & Averae Force

=+2

1

21 )(vF)(v

t

t

iiiiimdtm

tFRdt avg== Impulse

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Impact

! Impact occurs "#en t"o $odies collide "it#

eac# ot#er durin a very s#ort period of time%

!ypes of impact'Central impact

$lique impact

Line of impact

Plane of impact

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Coefficient of restitution e

=Pdt

Rdte

Coefficient of restitution eis defined as the ratio of the restitution impulse

to the deformation impulse.

Coefficient of restitution eis defined as the ratio of relative velocity after impactto the relative velocity before impact

Coefficient of restitution eis according to the impact velocity, material,size and shape of the colliding body,

Coefficient of restitution e range bet!een 0"#

Coefficient of restitution eis defined along the line of impact only$lastic impact e = #%re"bounce !ith same velocity&

'lastic impact e = 0 %couple or stic( together and move !ith common velocity&

11

22

)()(

)()(

BA

ABe

=

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Procedure for Analysis

! Identify t#e intial velocity * *you may use+

! 1, -1 . 2, -2

! Apply t#e conservation of momentum

alon t#e line of impact/ you "ill et one

equation "it# t"o un0no"n velocity

! se t#e coefficient of restitution to o$tain

a second equation

! olve $ot# equation for final velocities

after t#e impact

= 21 mm

11

22

)()(

)()(

BA

ABe

=

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$lique Impact

Central Impact one )imension*bliue Impact !o )imension

Four un(no!ns 2222 //)(/)( BA

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Procedure for Analysis

1111)(/)(/)(/)(

ByBxAyAx and

-$stablish "ais as line of impact-$stablish y"ais as plane of impact-/esolve the velocity components

along , and y as

-pply the conservation of momentum along the line of impact

-1se the coefficient of restitution to obtain a second euation

--Solve both euation for final velocities along "ais after the impact-he momentum is conserved along the plane of impact2 so

= 21 xx mm

11

22

)()()()(

xBxA

xAxBe

=

21

21

)()(

)()(

ByBy

AyAy

=

=

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Procedure for Analysis

1111 )(/)(/)(/)( ByBxAyAx and

=+

21 xx mm

11

22

)()(

)()(

xBxA

xAxBe

=

+

21

21

)()(

)()(

ByBy

AyAy

=

=

2

21

2

21

2

2

2

22

2

2

2

22

tan

tan

)()()(

)()()(

Bx

By

Ax

Ay

ByBxB

AyAxA

v

v

=

=+=

+=

long the line of the impact only

long the 'lane of the impact only

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ANGULAR MOMENTUM

! For a point o$3ect t#e anular momentum is

r

m

v

nits 4 0.m2/s or sl.ft2/s

It is a vector%

5ere t#e vector is pointin to"ard you%

sin ri#t4#and rule

))(()( mvrHo =

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Angular Impulse and Momentum Principles

=+ 26616 )()( 2

1

HMHt

tdt

== 2

1

2

16impulseAnular

t

t

t

tdtdt rM

=++ 2

1 21

t

tvmrdtFrdtMvmr

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2616

)()( HH =

Conservation of Anular Momentum

2211

mvrvmr =

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$ample #3"4

5=30 Ib

'=%60t& Ib

v6=7t=6 sec.

v#=4 ft8s

(=0.4

=+2

1

21 )()(

t

t

xxx mdtFm

2

2

2

6

77%1768%69688%9

2%:2

76)2)(:6sin76()2(:%626):(

2%:2

76

=++

=++ c

o

c

N

Nt

=6yF

6:6cos76 = IbN oc

sft

IbNc

;2%99

:%9:

2=

=

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$ample #3"4

From rest

v9=7

t=: sec.

=+2

1

21 )()(

t

t

AyA mdtFm

9loc(

2))(:()8)(

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'roblem

m=#6 ;g

Fy=#30 (666()>7%6)(12666()7%1)(17666( =

= sm ;7%62

21 )()()()( !avg! mtFm =++

)7%6)(17666()

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2211 )()()()()( !!cc!!cc mmmm +=++

)1766(2%:2

7%6

21

===

e

kgmkgm BA

)1(16>%6(28%2

)()()()(

inputoflinet#ealonconservedissystemt#eofMomentum

22

22

2211

=+

+=+

+=+

BxAx

BxAx

BxBAxABxBAxA

vv

vv

vmvmvmvm

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!Momentum of particle A/ is conserved alon t#e y a@is/ since noimpulse acts on particle A/

=

=

==

=+=

=+=

1%:622%1

>6>%6tan

7628%1

7%1

tan

;91%122%1>6>%6)(

;=8%17%128%1)(

1

1

22

2

22

2

smv

smv

B

A

==

==

smvvmvm

smvvmvm

ByByBByB

AyAyAAyA

;>6>%6)()()(

;7%1)()()(

221

221

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);(26

21829

79%69%6:2

1

)7(9%6)9)(16)(9%6(:6

2

2

2

9

6

9

6

2

2

9

6

smv

v

vtPt

vsNmdtt

=

=+

=+

=++

%ori&ontalsmoot%

v

t

kgm

mNtM

B

sec9

7

)(:

==

==

?@ample 1741:

21 )()(Momentum&ImpulseAnularofPrinciple

2

1&

t

t && HdtMH =+

=++2

121

t

t BA vmrdtPrdtMvmr

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Pro$lem

m . 966

v1. 2 m;s

M . 6%8 E%m

v2=7

t . : s

=+ 21 )()( ooo HdtMH

F)9%6(:%6G2):(8%6)F2)(9%6(:%6G2 2v=+

smv ;76%=2=

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