1

2

m1L

1v

1mv1F3F

2F

Linear Momentum of a Particle:

Problems involve forces, masses, velocities and time can better be solved using principles of impulse and momentum. We have learned that for a particle

( ),ext Rd

m mdt

F F a v

ik

j

1 1m v L

1 2( )I

1mv

2mv

2

1

t

Rt

dt F

RF

2

m

L

2v

Momentum finally, just after the applivation of external Forces

final momentum

1 2Momentum initially :Impulse, [N.s]

before the applivation of external Forces

The initial momentum +acting impulses

I

2 2m v L

1 2( )I

( ) ,Rdt d mv mdv F2 2

1 1

( )t v

Rt v

dt d mv F

1 2

1 22

1

1 2

L Lt

Rt

mv dt mv

I

F

1 2 2 1L I L

Linear Momentum of a Particle:

1 0v 2v

impactF

frF

Find the velocity of the particle just after impact knowing its velocity before impact is zero.

1mv

Find the time required for the particle to reach zero velocity up the hill

2 2

1 1

2 sin( ) (0)t t

rt t

mv mg F m

Now, the impulse of the weight and friction has a significant value it acts for a long time.

Example:

sin( )W W

2

1

t

impactt

F dt 2mv2

1

0

sin( )t

t

mg

2

1

)t

rt

F dt 2

0

mv

Example 6:

The hydraulic braking system for the truck and trailer is applied for the two units. If the brakes are applied uniformly for 5 sec. to bring the rig to a stop from a speed of 20 mile/hr down the 10% grade, determine the force in the coupling between the trailer and the truck. The truck weight is 20,000 lb and the trailer weight is 15,000 lb.

.

20,000 lb

15,000 lb

110

trailerN

( )truckW Sin ,r truckF

truckN

y

xP

15,000 lb

P

P,r trailerF

( )trailerW Cos

( )trailerW Sin

trailerN

y

x

05279

20 20 29.32 /3600

mile ftv ft s

hr s 0.fv

..

5

00

5

0

( 1) ( ) ( 2) ,

20,000 1 10(29.3) (20,000 20,000 ) 0,

32.2 101 101

x x x

for truck

m v F dt m v

P dt

5

00

5

0

( 1) ( ) ( 2) ,

15,000 1 10(29.3) (15,000 15,000 ) 0,

32.2 101 101

x x x

for trailer

m v F dt m v

P dt

4214.9621 14581 ,P

5619.952 19801.98 ,P

5619.95 19801.981 -1520062.6744.21 lb

4214.96 14581 34382.98P

1 5619.951 9834.91210.286

1 4214.9621 34382.98

Example 6:

The hydraulic braking system for the truck and trailer is set to produce equal braking forces for the two units. If the brakes are applied uniformly for 5 sec. to bring the rig to a stop from a speed of 30 km/hr down the 10% grade, determine the force in the coupling between the trailer and the truck. The mass of the truck is 10 Mg and the mass of the trailer is 7.5 Mg.

.

..

Shaded Area

=Impulse

Conservation of linear momentum:

For a particle or system of group of particles, the linear momentum of the system is conserved as long as the no external forces act on the system

Example7:

A ball is moving with a velocity , its momentum is

If the ball is suddenly exploded into two segments and

1v 1mv

2m

3m

1

2

01

1t

t

m dt

L

v F

2

2 2 3 3 Gm m m L

v v v

2 2 3 32 2 3 3 1 2

1 2( ) ;

( )G Gm m

m m m mm m

r r

r r r r

In general, the center of gravity for a group of particles is

; i ii i G i G

i

mm m

m

r

r r r

The velocity of the center of gravity for a group of particles is

( ) ; ( )i Gi i i i i G

d dm m m m

dt dt

L L

r rv v i i

Gi

m

m

v

v

1mvG

3 3m v3rGr

2r1 Gm mv v

Gr

1mv

3 3m v2 2m v

2 2m v

1 1m v

2 2m v

i im v

3 3m v

1rG2r

3r

ir

Where G is the center of gravity for both exploded segments, it is defined according to the following relation.

0.05

1 20

0

, ,xL F d t L ,280 80 10

( 16cos(30) 0 ( )32.2 32.2 xv

,2 12.32 /xv ft s

0.05

10

, 0,yL F dt 80

16sin(30) 0.05[ ( ) ] 0,32.2 boy skateW W N

487N lb

80 .lb

10 .lb30

16 / .ft s

Example 10: The 80-lb boy has taken a running jump from the upper surface and lands on his 10-lb skateboard with a velocity of 16 ft/s in the plane of the figure as shown. If his impact with the skateboard has a time duration of o.o5 s, determine the final speed v along the horizontal surface and the total normal force N excreted by the surface on the skateboard wheels during the impact.

( )A B Gm m v

1 2

1 2

1 1 2 2

1 2

( ) ( ) 0 ( ) ( )

( )( ) 0 ( )( )

A A B B A A B B

A B G A B G

m m m m

m m m m

L L

L L

v v v v

v v

Example 8:

( )A B Gm m v

1( )A Am v1( )B Bm v

1( )A Am v

Before impact

1( )B Bm v1( )Bv

1( )Av

1 1 1( )A A Bm mv 2( )A Am v 2( )B Bm v

2( )A Am v

After impact

2( )Bv

2( )Av

2( )B Bm v

2( )A Am v

Does not change

1 2

1 2

( ) ( )

( ) 0 ( ) ;

x x

x xm m L L

v v

1 2

2

1 2

1( ) ( )

( ) ( )

y y

t

y y

t

m Ndt m L L

v v

1( )m v

During impact

2( )m v

Before impact After impact

N

The linear momentum may be conserved in one direction while it is not in the other direction

Angular Impulse and Angular Momentum

Linear momentum

mL v

Angular momentum

2( ), . /o m kg m s H r v

Perpendicular to plane A which passes through r an L

| | | ( ) | sin( )o m rmv H r v

( ) x y z x y zo

x y z x y z

r r r r r rm m

mv mv mv v v v

i j k i j k

H r v

x

y

z

xmv

ymv

zmv

r

yrxr

zr

[ ] [ ] [ ]

yx z

o y z z y x z z x x y y x

HH H

m r v r v m r v r v m r v r v H i j k

Rate of Change of Angular Momentum

o m

m m

M r v

F a r F r a

0

( ) ( ) ( )o m m m v

H r v r v r v

o o M H ( ) ( ( ) ( ( ) ( '') ; ') ; )o x o y o zM H M H M Ho x o y o z

Angular Impulse and Angular Momentum( )12

( )1 1

;t o

o ot o

o dto od

ddt

H

H

M MH

H H

2

11 2

Initial angular Final angular momentum momentumAngular Impulse

( ) ( )t

ot

dto o MH H

2

11 2( ) ( )

t

oxt

M dtox oxH H 2

11 2( ) ( )

t

oyt

M dtoy oyH H

Conservation of angular Momentum2

1

0t

ot

dt M 1 2 1 2

Initial angular Final angular momentum momentum

( ) ( ) ( ) ( )o o o o H H H H

1 1( )m r × v

Example: Plane Motion,

2

1

t

t

dt r ×F 2 2( )m r × v

1

1 1 1sin

d

mv r 2

1

sin( )t

t

r dtF 2

2 2 2sin

d

mv r 1

2

2

11 2( ) ( )

t

oyt

M dtoy oyH H

Example: Plane Motion,

Problem 3/230:The mechanism is initially at rest, when balls of masses m and 2m strikes and sticks with the end cups, Determine the angular velocity of the mechanism.

, , ,o before o afterH H

5

3

v

l

' '(2 ) ( )(3 ) 2 (3 ) 5 ( )l mv l m v mlv ml v ml l

, ,2 , ,2( ) ( )o m o m before o m o m after H H H H

, ,A,z B z A,z B z

Conservation of momentum around z-axis:H H H H

A A exm dt r v M cos( ) cos( ) (1)B B om r mv r mv r v

A BConservation of energy: E E

A A B BT V T V

Problem 3/247:A particle is released on the smooth inside wall of a cylindrical tank at A with a velocity vo which makes an angle β with the horizontal tangent. When the particle reaches point B a distance h below A, determine the expression for the angle θ made the its velocity with the horizontal tangent at B.

(0) ( ) (2)2 2o B

1 1mv mg mv mg h

2 2

Eleminate v from Eq.1 and Eq.2 to get

2 2

cos( ) cos( )( ) ( )

2 2

o o

o o

v vcos cos

v gh v gh

nr

mvot

z

Mvo,z

mvo,tN

mg

z

Ar

Br

Datum

Problem 3/245:

A small 0.1 kg particle is given a velocity of 2 m/s on the horizontal x-y plane and is guided by the fixed curved rail. Friction is negligible. As the particle crosses the y-axis At A, its velocity is in the x-direction, and as it crosses the x-axis at B, its velocity makes a 60º angle with the x-axis. The radius of curvature of the path at B is 500 mm. Determine the time rate of change of angular momentum Ho of the particle about o at both A and B.

At A: 0, 0

z z

z z

o o o o

o o

M H

M H

M H

At B: , ,z zo o o B o BM N b H N b M H

2 22 22

, 0.1 0.8 , 0.8(0.15) 0.12 / 0.12 . /0.5 z

BB o

vBut N m N H N m kg m s

Impacts:Impact: A collision between two bodies occurs in a relatively short time interval during which the two bodies exerts on each other relatively a large forces

Line of impact

Normal to

line of impact Line of impact: A line normal to both surfaces in contact.

Line of impact

Central impact

Line of impact

Eccentric impact

Central and eccentric impact:

Direct and Oblique impact:

Line of impact

Direct impactLine of impact

Oblique impact

Before

After

During impact

Line of impact

A Am v B Bm v

Line of impact

A Am v B Bm 'v1t *t 2t

Time of

deformation

Time of

restitution

Impact time

P

A B A BSyst. m +m m ' m ' (1)A B A B v v v v

u

(2 )

(2 )

a

b

(3a)

(3b)

A

Body A:

m Av

*

1

Am ,

t

t

dt u P

2

*

Am

t

t

u dt R

B

Body B:

m b v

Am 'A v

*

1

Bm

t

t

dt uP

Bm u 2

*

t

t

dtR Bm 'B v

Direct Central Impact and Coefficient of Restitution:

2

*

*

1

2

*

*

1

' '

Coeff.of Rest.

e =

e =Coeff.of Rest.

'e =

t

t At

At

A B

B A

t

t Bt

Bt

dtu v

v udt

v v

v v

dtv u

u vdt

R

P

R

P

P R R

not re

quire

d

Eliminate from eq.2* and. 3* to get u

' 'e= B A

A B

v v

v v

(4)

e=0: Perfect Plastic collision bodies stick together, and energy loss is maxinum)

: elastic Plastic collision.0<e 1

e=1: Perfect elastic collision (the magnitude of relative velocities before and after impact is the same, the energy loss during impact is zero)

You can use equation 1 and 4 to solve for two unknowns only.

Oblique Impacts: not required

During impact

A(v )t

A(v )n

B(v )t

B(v )n

B B A A B B A Am (v ) m (v ) m (v' ) m (v' ) (1)n n n n

n t

nn

t

B(v )n

A(v )n n

Direct impact

frictionless contact

A A A Am (v ) m (v' ) (3)t t

B A A B[(v ) (v ) ] (v' ) (v' ) (2)n n n ne

B B B Bm (v ) m (v' ) (4)t t

Impact energy loss: is the difference between the total energy before impact and that after impact

Oblique Impacts are Not required

Motion Constraint: The particle is not allowed to move freely.

Bv

( )B nv

( )B tv

cv

xy

Before

n

t

R t

sinR t

sinR t During

1)Motion of the carriage is constraint in the x-direction

2) Momentum is conserved in the x-direction, (no external force acts in this direction)

( ) ' ( ' )c c b b x c c b b xm v m v m v m v

3) Momentum of the ball alone is conserved in the tangential direction

( ) ( ' )b b t b b tm v m v

4) The relation between the relative velocities of the bodies before and after the impact remains valid along the line of impact

( ' ) ( ' )e=

( ) ( )b n c n

c n b n

v v

v v

'Bv ( ' )B nv

( ' ) ( )B t B tv v

After

'cv

not re

quire

d

not re

quire

d

2 2 2 ' 2 '0.05(300)(0.4cos(20 ) 3.2(0.2) (6) 3.2(0.4) 6 (0.05 3.2)(0.4) (3.2)(0.2) ' 2.77 / ,rad s CCW

Find The Energy Loss :before afterE E E

, just after Impact, just after Impact

2 2

,

1 1(0.05 3.2) (0.4)(2.77) (3.2)[(0.2)(2.77)] [3.2(0.2)9.81 (3.2 0.05)(0.4)9.81]

2 20 3.2(0.2)9.81cos( ) (3.2 0.05)(0.4)9.81cos( )

TV

V

At max. height

52.1

Conservation of energy after impact: ' 'T V T max.height, 0T

V

Problem 3/248:

A pendulum consists of two 3.2 kg concentrated masses positioned as shown on a light but rigid bar. The pendulum is swinging through the vertical position with a clockwise angular velocity ω=6 rad./s when a 50-g bullet traveling with velocity v=300 m/s in the direction shown strikes the lower mass and becomes embedded in it. Calculate the angular velocity ω’ which the pendulum has immediately after impact and find the maximum angular deflection θ of the pendulum.

Datum

y

The system angular momentum is conserved during impact:1 2o oH H

Just beforeDuring

Just After

Problem 278:A 2.0 kg sphere is moving horizontally with a velocity of 10 m/s and strikes the inclined face of a 10 kg carriage which is initially at rest and backed by unstretched spring of 1600 N/m. Knowing that the coefficient of restitution between the sphere and the wedge is e = 0.6, determine (a) the angle θ , (b) the rebound velocity v’ of the sphere, (c) the maximum travel δ of the carriage after the impact (c) the energy lost due to the impact.

'v sin(60 ) cos( 30 )Coefficient of Restitution: 0.6= (3)

10cos(30 ) 0o v

': 0; (2)(10) 0 2 cos 10 (1)x oFor All ystem L v v

': 0; (2)(10sin(30 ) 2 sin( 30 ) (2)tFor sphere alone L v

' 'oEquation 3is: 5.196=0.866v 0.866 cos( ) 0.5 sin( )v v

'

solve Eq. 4 and 5 to get

6.04 / , 85.9v m s

' ' ' '

' '

Substitute in Eq. 1 to eleminate v : 5.196=0.866(2+0.2v cos ) 0.866 cos( ) 0.5 sin( )

: 1.039v cos 0.5 sin( )=3.464 (4)

v v

or v

'

Eq. 2 becomes

0.866v sin 0.5cos 5 (5)

o

2 2spring

From Eq. 1 v 2.087 /

1 1For carriage: T+ V 0; (10)(2.087) 1600 0

2 2

m s

0.1650 , 165.0m or mm

Momentum Diagram

not re

quire

d

not re

quire

d