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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