kinetics of a particle: impulse and...
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Chungnam National University
Kinetics of a Particle: Impulse and Momentum
Linear momentum
Linear impulse
mºL v
dtº òI F
( )d dm mdt dt
= = =å LF a v
2
12 1 2 1
t
tdt m m= - = -åò F L L v v
Newton’s 2nd law: The resultant of all forces acting on a particle is equal to its time rate of change of linear momentum.
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Principle of Linear Impulse and momentum
2
11 2
t
tm dt m+ =åòv F v
2
1
2
1
2
1
1 2
1 2
1 2
( ) ( )
( ) ( )
( ) ( )
t
x x xt
t
y y yt
t
z z zt
m v F dt m v
m v F dt m v
m v F dt m v
+ =
+ =
+ =
åòåòåò
2
12 1 2 1
t
tdt m m= - = -åò F L L v v
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Linear Impulse and momentum: Examples
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Principle of Linear Impulse and momentum for a system of particles
ii i i
dmdt
+ =å å å vF f
2
11 2( ) ( )
t
G i Gtm dt m+ =åòv F v
( )ii i
d mdt
+ =vF f for particle i
For system of particles
2
121
( ) ( )t
i i i i itm dt m+ =å å åòv F v
G i i
G i i
m m
m m
=
=åå
r r
v vim m=åwhere
For rigid body
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Conservation of Linear Momentum for a system of particles
21( ) ( )i i i im m=å åv v
1 2( ) ( )G G=v v
If the resultant force on a particle is zero during an interval of time, 2
121
( ) ( )t
i i i i itm dt m+ =å å åòv F v
G i im m=åv vFor rigid body, since
When is the resultant force on a particle zero during an interval of time? (1) Particles collide or interact.
(2) External impulse is negligible, when the time period is very short2
11 2
t
A A A Atm dt m+ - =åòv F v
2
11 2
t
B B B Btm dt m+ =åòv F v
1 1 2 2A A B B A A B Bm m m m+ = +v v v v
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Conservation of Linear Momentum: example
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Conservation of Linear Momentum: example
1 1 ( )proj proj Block Block proj Block proj Blockm m m m m m+ = + = +v v v v v
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Impact
Definition of impact: collision between two bodies is characterized by the generation of relatively large contact forces that act over a very shot interval of time.
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Central Impact
1 1 2 2( ) ( ) ( ) ( )A A B B A A B Bm v m v m v m v+ = +
Law of conservation of linear momentum is valid. Why?Internal impulse of deformation and restitution cancel, during collision
How many unknowns in the above equation?
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Central Impact
1( )A A Am dt m- =òv P v
Deformation PeriodFor particle A
For particle B
1( )B B Bm dt m+ =òv P v
1( )A A Adt m m= -òP v v
1( )B B Bdt m m= -òP v v
-
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Central Impact
2( )A A Am dt m- =òv R v
Restitution Period
For particle A
For particle B
2( )A A Adt m m= -òR v v
2( )B B Bm dt m+ =òv R v
2( )B B Bdt m m= -òR v v
-
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Coefficient of Restitution
2
1
( )( )
A
A
Rdt v vev vPdt-
= =-
òò
2
1
( )( )
B
B
Rdt v vev vPdt
-= =
-òò
2 2
1 1
( ) ( )( ) ( )
B A
A B
v vev v
-=
-
Coefficient of restitution: the ratio of the restitution impulse to the deformation impulse.
For particle A
For particle B
(1)
(2)
Eliminate v using eqs. 1 & 2
relative velocity of separationerelative velocity of approach
- =
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Central Impact
1 1 2 2( ) ( ) ( ) ( )A A B B A A B Bm v m v m v m v+ = +
e=1: restitution impulse = deformation impulse
No energy loss – perfectly elastic
e=0: plastic impact 100% energy loss 2 2( ) ( )A Bv v v= =
Summary of Central impact problem
2 2
1 1
( ) ( )( ) ( )
B A
A B
v vev v
-=
-
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Impact example
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Impact example
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Oblique ImpactX direction
1 1 1( ) ( ) cosAx Av v q=1 1 1( ) ( ) cosBx Bv v f=
Y direction1 1 1( ) ( ) sinAy Av v q=
1 1 1( ) ( ) sinBy Bv v f=
2 2 2( ) ( ) cosAx Av v q=
2 2 2( ) ( ) cosBx Bv v f=
2 2 2( ) ( ) sinAy Av v q=
2 2 2( ) ( ) sinBy Bv v f=unknowns
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Oblique ImpactX direction
1 1 2 2( ) ( ) ( ) ( )A Ax B Bx A Ax B Bxm v m v m v m v+ = +
2 2
1 1
( ) ( )( ) ( )
Bx Ax
Ax Bx
v vev v
-=
-
Y direction
1 2( ) ( )A Ay A Aym v m v=
1 2( ) ( )B By B Bym v m v=
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Angular Momentum
( ) ( )( )oH d mv=
The moment of the linear momentum L about O is defined as the angular momentum Ho of particle P about O.
o
o x y z
x y z
m
i j kr r rmv mv mv
= ´
=
H r v
H
Unit: kg(m/s)m=kg(m/s2)ms=Nms
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Relation Between Moment of a Force and Angular Momentum
( )
o
o
m
md m m mdt
=
= ´ = ´
= ´ = ´ + ´
åå åF v
M r F r v
H r v r v r v
&
&
& & &
oo=åM H& =åF L&
The resultant moment about the fixed point O of all forces acting on a particle is equal to its time rate of change of angular momentum of the particle about O.
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System of Particles
( ) ( ) ( )i i i i o´ + ´ =ir F r f H&
( ) ( ) ( )i i i i i o´ + ´ =å å år F r f H&
For the particle i
oo=åM H&
For system of particles
The sum of the moments about point O of all the external forces acting on a system of particles is equal to the time rate of change of the total angular momentum of the system about point O.
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Angular Impulse and Momentum Principles
2
1
2
1
2 1
1 2
( ) ( )
( ) ( )
t
o o ot
t
o o ot
dt
dt
= -
+ =
åòåò
M H H
H M H
Principle of angular impulse and momentum
ooo
ddt
= =å HM H& oodt d=åM H
2 2
1 1
( )t t
ot tangular impulse dt dtº = ´ò òM r F
2
11 2( ) ( )
t
o o otdt+ =å åòH M H
For a particle
For system of particles
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2
1
2
1
1 2
1 2( ) ( )
t
t
t
o o ot
m dt m
dt
+ =
+ =
åòåò
v F v
H M H
2
1
2
1
2
1
1 2
1 2
1 2
( ) ( )
( ) ( )
( ) ( )
t
x x xt
t
y y yt
t
o o ot
m v F dt m v
m v F dt m v
H M dt H
+ =
+ =
+ =
åòåòåò
Vector formulation
Scalar formulation (2D case)
Impulse and Momentum Principles
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Conservation of Angular Momentum
1 2( ) ( )o o=H H
When the angular impulses acting on a particle are all zero during the time t1 to t2, then
1 2( ) ( )o o=å åH HFrom t1 to t2, the particles angular momentum remains constant.
Conservation of angular momentum of a system of particles
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Conservation of Angular Momentum: example
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Conservation of Angular Momentum: example