LEC 22 (Review 15 051)
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Transcript of LEC 22 (Review 15 051)
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8/9/2019 LEC 22 (Review 15 051)
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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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