LEC 18 ( Impulse & Momentum 042)

8/9/2019 LEC 18 ( Impulse & Momentum 042)

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King Fahd University ofPetroleum & Minerals

Mechanical EngineeringDynamics ME 201

B

Dr! Meyassar "! #l$%addadecture ' 1(


2/16


3/16

== mmaF

=2

1

2

1

vF

t

t

v

vdmdt

Principle of linear impulse and momentum e"uation

=2

1

12 vvF

t

t

mmdt

change in momentumImpulse

#e$ton%s &econd 'a$

dt

dv


4/16

'inear momentum

' ( mv Measure of impact effect

Compare

bullets with small mass and high velocity

Ship with huge mass and slow velocity

L is a vector )) points in same direction as v

*nit: +g.m,s or slug.ft,s

-omponents rectangular coordinates/: '0(mv0 '!(mv! '1(mv1


5/16

'inear Impulse

Measure the effect of force during the time

force acts.

I is a vector "uantit! in the direction of force

*nit #.s or Ib.s

2hen 3 is constant

=2

1

F(t)It

t

dt

)(FI 12 ttc =


6/16

Principle of 'inear Impulse and

Momentum

=+2

1

21 vFv

t

t

mdtm

Initial momentum 4 &um of all Impulse ( 3inal momentum

{ } 21 mtNtNtmgtFm cc =++++


7/16

Principle of 'inear Impulse and Momentum

for a s!stem of Particles

=+2

1

21 )(vF)(v

t

t

iiiiimdtm

=+2

1

2G1G )v(F)v(t

t

mdtm

== iiGi mmmm vv


8/16

&calar 5"uations

=+2

1

21 )()(

t

t

xxx mdtFm

=+2

1

21 )()(

t

t

yyy mdtFm

=+2

1

21 )()(

t

t

zzz mdtFm


9/16

3ree 6od! 7iagram

&elect the particle

5stablish 0 ! frame

5stablish the direction of theparticle initial and finalvelocities

7re$ the impulseand momentdiagrams. Include all theforces acting on the particle%s367 $ill create an impulseeven though some of these

forces $ill do no $or+. 8esolving the vectors along

the 0 ! a0is

9ppl! =+

2

1

2xx1x )(F)(

t

t

mdtm

=+2

1

2yy1y )(F)(

t

tmdtm


10/16

50ample ;)

m((?

#(?

=++ 2

1

21 )()()(

t

t

xxx mdtFm

sm

kgsN o

/1.14

)100()10(45cos2000

2

2

=

=+

( ) =++

2

1

21 )()(

t

tyyy

mdtFm

NN

sNsNsN

c

o

c

840

045sin)10(200)10(981)10(0

=

=++

=0yF


11/16

50ample ;)@

2(;< Ib

P(>(?T(> sec.

=(@ ft,s

+(


12/16

50ample ;)@

3rom rest

v6(?

t(A sec.

=++2

1

21 )()()(

t

t

AyA mdtFm

6loc+ 9

2))(3()6)(81.9(3)6(20 AkgssT =+

6loc+ 6

=++2

1

21 )()()(

t

t

ByB mdtFm

2))(5()6()6)(81.9(50 BkgsTs =+

BA

BA lSS

=

=+

2

2

NT

sm

B

B

2.19

/8.35)( 2

=

=


13/16

Problem ;)B

m(> Mg

3!(;< +#

=(?

h(?t(A s

&tart from rest

( ) =++2

1

21 )()(

t

t

yyy mdtFm

sm /1.16

)10(12)6)(81.9)(10(12)6)(10(1500

2

2

333

=

=+

( )

2

0

/69.2

)6(01.16

sma

a

tavv c

=

+=+=+

( )

ms

s

tatvssc

4.48

)6)(69.2(2

100

2

1

2

2

00

=

++=

++=+


14/16

Problem ;)A

m ( >C Mg

9t rest

=( ?t ( B s

3 ( B )


15/16

Problem ;)D

m ( E; +g

&tart from rest

v ( >


16/16

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LEC 18 ( Impulse & Momentum 042)