dynamics lecture4.1

8/22/2019 Dynamics Lecture4.1

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Chapter 4 : Kinetics Of A Particle

Work and Energy


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Chapter Objectives

To develop the principle of linear impulse and

momentum for a particle.

To study the conservation of linear momentum for

particles.

To analyze the mechanics of impact.

To introduce the concept of angular impulse and

momentum. To solve problems involving steady fluid streams

and propulsion with variable mass.


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Chapter Outline

Principle of Linear Impulse and Momentum

Principle of Linear Impulse and Momentum for a

System of Particles

Conservation of Linear Impulse for a System ofParticles

Impact

Angular Momentum


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Relation between Momentum of a Force andAngular Momentum

Angular Impulse and Momentum Principles Steady Fluid Streams

Propulsion with a Variable Mass

Chapter Outline


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The equation of motion for a particle of mass mcan be written as

F = ma = mdv/dt

where a and v are both measured from an inertial

frame of reference.

Rearranging the terms and integrating betweenthe limits v = v1 at t = t1 and v = v2 at t = t2

122

1

2

1

2

1

vvFvF mmdtdmdtt

t

v

v

t

t or


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This equation referred to as theprinciple of linearimpulse and momentum.

It provides a direct means of obtaining theparticles final velocity after a specified time periodwhen the initial velocity is known and the forces

acting on the particle are either constant or can be

expressed as a function of time.



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Linear Momentum

Each of the two vectors of the form G = mv is

referred to as the particles linear momentum. The linear-momentum vector has the samedirection as v, and its magnitude mvhas unit of

mass-velocity, kg.m/s



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Linear Impu lse

The integral is referred to as the linear

impulse, which is a vector quantity whichmeasures the effect of a force during the time the

force acts.

The impulse acts in the same direction as theforce, and its magnitude has unit of force-time, N.s

dtFI



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Princ ip le of Linear Impu lse and Momentum

The equation is rewritten in the form

21

2

1

vFvmdtm

t

t

which state that the initial momentum of

the particle at t1 plus the sum of all the

impulses applied to the particle from t1to t2 is equivalent to the final momentum

of the particle at t2



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If the magnitude or direction of a force varies withtime, the impulse is represented on the impulse

diagram as

If the force is constant, the impulse applied to the

particle is Fc(t1t2), and it acts in the samedirection as Fc

2

1

tt

dtF



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Scalar Equation s

Resolving each of the vectors in the equation ofprinciple of linear impulse and momentum into itsx,

y, zcomponents,

21

21

21

)()(

)()(

)()(

2

1

2

1

2

1

z

t

tzz

ytt

yy

x

t

txx

vmdtFvm

vmdtFvm

vmdtFvm



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PROCEDURE FOR ANALYSIS

Free-Bod y Diagram

Establish thex, y, zinertial frame of reference anddraw the particles free-body diagram in order to

account for all the forces that produce impulses onthe particle.

The direction and sense of the particles initial andfinal velocities should be established.

If a vector is unknown, assume that the sense ofits components is in the direction of the positive

inertial coordinate(s).


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As an alternative procedure, draw the impulseand momentum diagrams for the particle.

Princip le of Impu lse and Momentum

In accordance with the established coordinatesystem apply the principle of linear impulse and

momentum,

212

1mvdtFmv

t

t



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If the motion occurs in thex-yplane, the twoscalar components equations can be formulated by

either resolving the vector components ofF from

the free-body diagram, or by using the data on the

impulse and momentum diagrams.

Realize that every force acting on the particlesFBD will create an impulse, even though some of

these forces will do no work.



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Forces that are functions of time must beintegrated to obtain the impulse. The impulse is

equal to the area under the force-time curve.

If the problem involves the dependent motion ofseveral particles, use method to relate theirvelocities. Make sure the positive coordinate

directions used for writing these kinematics

equations are the same as those used for writingthe equations of impulse and momentum.



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Example

The 100-kg stone is originally at rest on the smoothhorizontally surface. If a towing force of 200 N,

acting at an angle of 45, is applied to the stone for10 s, determine the final velocity and the normal

force which the surface exerts on the stone duringthe time interval.


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Free-Body Diagram . Since all forces acting are

constant, the impulses are simply the product of

the force magnitude and 10 s [I = Fc(t2t1)].Princ ip le of Impu lse and Momentum . Resolving

the vectors along thex, y, zaxes,

smvv

vmdtFvm xt

txx

/1.14)100(45cos)10(2000

)()(

2

2

212

1


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NN

N

vmdtFvm

C

C

y

t

tyy

840

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

)()( 212

1

Since no motion occurs in the ydirection, directapplication of the equilibrium equation Fy= 0 givesthe same result forNC


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Example

The 250-N crate is acted upon by a force having a

variable magnitude P= (100t) N. Determine the

crates velocity 2 s afterP has been applied. Theinitial velocity is v1 = 1 m/s down the plane, and the

coefficient of kinetic friction between the crate andthe plane isk= 0.3.


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Free-Body Diagram . The impulse created can be

determined by integrating P= 100tover the 2-s

time interval. The weight, normal force andfrictional force are all constant, so the impulse

created by each of these forces is simply the

magnitude of the force times 2 s.


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Princ ip le of Impu lse and Momentum .

2

2

2

0

21

5.252506.02005.25

81.9

25030sin)2(250)2(3.0)100()1(

81.9

250

)()(2

1

vN

vNdtt

vmdtFvm

C

C

x

t

t xx

+

The equation of equilibrium can be applied in the y

direction030cos250 CN+

NC

= 216.5 N v2

= 13.6m/s


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Example

Block A and B have a mass of 3 kg and 5 kg

respectively. If the system is released from rest,

determine the velocity of block B in 6 s.


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Free-Body Diagram. Since the weight

of each block is constant, the cordtensions will also be constant.

Furthermore, since the mass of pulley D

is neglected, the cord tension TA

= 2TB.

Note that the blocks are both assumed to

be traveling downward in the positive

directions, sA and sB


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Princip le of Impu lse and Momentum .

BlockA:

2

21

))(3()6)(81.9(3)6(20

)()(2

1

AB

A

t

tyA

vT

vmdtFvm

Block B:

2

21

))(5()6()6)(81.9(50

)()(2

1

BB

B

t

tyB

vT

vmdtFvm

(1)

(2)


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Kinemat ics.

lss BA

2

Taking time derivative yields

BA vv 2

As indicated by the negative sign, when B moves

downward A moves upward. Substituting this result

into Eq. 1 and solving Eqs. 1 and 2 yields(vB)2 = 35.8 m/s TB = 19.2 N


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Free-Body Diagram

Establish thex, y, zinertial frame of reference anddraw the FBD for each particle of the system in

order to identify the internal and external forces.

The conservation of linear momentum applied tothe system in a given direction when no external

forces or if non-impulsive forces act on the systemin that direction



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Establish the direction and sense of the particlesinitial and final velocities. If the sense is unknown,

assume it is along a positive inertial coordinate

axis.

As an alternative procedure, draw the impulseand momentum diagrams for each particle of the

system.



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Momentum Equat ions

Apply the principle of linear impulse andmomentum or the conservation of linear

momentum in the appropriate directions

If it is necessary to determine the internal impulseF.dtacting on only one particle of a system, then

the particle must be isolated(free-body diagram),and the principle of linear impulse and momentum

must be applied to the particle.



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After the impulse is calculated, and provided thetime tfor which the impulse acts is known, thenthe average impulsive force Favg can be determined

from Favg= F dt/t.



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Example

The 15-Mg boxcar A is coasting at 1.5 m/s on thehorizontal track when it encounters a 12-mg tank B

coasting at 0.75 m/s toward it. If the cars meet and

couple together, determine (a) the speed of both

cars just after the coupling, and (b) the averageforce between them if the coupling takes place in

0.8 s.


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Part (a) Free-Bod y Diagram. We will consider

both cars as a single system. By inspection,

momentum is conserved in thexdirection since thecoupling force F is internalto the system and will

therefore cancel out. It is assumed both cars, when

coupled, move at v2 in the positive xdirection.


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Conservat ion o f Linear Momentum .

smv

v

vmmvmvm BABBAA

/5.0

)27000()75.0)(12000()5.1)(15000(

)()()(

2

2

211( )

Part (b ) The average (impulsive) coupling forceFavg, can be determined by applying the principle of

linear momentum to eitherone of the cars.

Free-Body Diagram.


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Conservation of Momentum .

kNF

F

vmdtFvm

avg

avg

AAA

8.18

)5.0)(15000()8.0()5.1)(15000(

)( 21

( )


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Example

The 600-kg cannon fires an 4-kg projectile with a

muzzle velocity of 450 m/s relative to the ground. If

firing takes place in 0.03 s, determine (a) the recoil

velocity of the cannon just after firing, and (b) the

average impulsive force acting on the projectile.


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Part (a) Free-Body Diagram . We will consider

the projectile and cannon as a single system, sincethe impulsive forces, F, between the cannon and

projectile are internal to the system and will

therefore cancel from the analysis.

During the time t= 0.03 s, the two recoil springswhich are attached to the support each exert a

non-impulsive forceFs on the cannon. This is

because t is very short, so that during this timethe cannon only moves through a very small

distances.


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Consequently, Fs = ks 0, where kis the springs

stiffness. It may be consluded that momentum forthe system is conserved in the horizontal direction.

We assume that the cannon moves to the left,

while the projectile moves to the right after firing.


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Conservation of Momentum .

smv

v

vmvmvmvm

c

c

ppccppcc

/3)(

)450)(4())(600(00

)()()()(

2

2

2211( )


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Part (b ) The average impulsive force exerted by

the cannon on the projectile can be determined by

applying the principle of linear impulse and

momentum to the projectile.

Princ ip le of Impulse and Momentum .

kNF

F

vmdtFvm

avg

avg

pp

0.60)10(60

)450)(4()03.0(0

)()(

3

21

( )


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Example

The 350-Mg tugboat Tis used to pull the 50-Mg

barge B with a rope R. If the barge is initially at rest

and the tugboat is coasting freely with a velocity of

(vT)1 = 3 m/s while the rope is slack, determine the

velocity of the tugboat directly after the ropebecome taut.


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Free-Bod y Diagram. We will consider the entire

system. The impulsive force created between the

tugboat and the barge is internalto the system,

therefore momentum of the system is conserved

during the instant of towing.


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Alternative procedure of drawing the systemsimpulse and momentum diagrams is as shown

below


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Conservat ion of Momentum . Noting that (vB)2

= (vT)2

smv

vv

vmvmvmvm

T

TT

BBTTBBTT

/62.2)(

))(10(50))(10(3500)3)(10)(350(

)()()()(

2

23

233

2211

This value represents the tugboats velocityjustafterthe towing impulse.


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Example

An 800-kg rigid pile P is driven

into the ground using a 300-kk

hammer H. the hammer falls

from rest at a height y0 = 0.5 m

and strikes the top of the pile.Determine the impulse which

the hammer imparts on the pile

if the pile is surrounded entirely

by loose sand so that afterstriking, the hammer does not

rebound off the pile.


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Conservat ion o f Energy. The velocity at which

the hammer strikes the pile can be determined

using the conservation of energy equation appliedto the hammer.

smv

v

yWvmyWvm

VTVT

H

H

HHHHHH

/13.3)(

0))(300(2

1)5.0)(81.9(3000

)(2

1)(

2

1

1

21

1210

20

1100


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Free-Body Diagram. During the short

time occurringjust before tojust afterthe collision, the weight of the hammer

and pile and the resistance force Fs of

the sand are all non-impulsive. The

impulsive force R is internal to thesystem and therefore cancels.

Consequently, momentum is conserved

in the vertical direction during this short

time.


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Conservat ion of Momentum . Since the

hammer does not rebound off the pile just after thecollision, then (vH)2 = (vP)2 = v2

smv

vvvmvmvmvm pHppHH

/854.0

8003000)13.3)(300()()()(

2

22

2211


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Princ ip le of Impulse and Momentum . The

impulse which the pile imparts to the hammer cannow be determined since v2 is known.

sNdtR

dtR

vmdtFvmH

t

t yHH

683

)854.0)(300()13.3)(300(

)()(21

2

1


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Example

A boy having a mass of 40-kg stands on the back

of a 15-kg toboggan which is originally at rest. If hewalks to the front B and stops, determine the

distance the toboggan moves.


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Free-Body Diagram. Theunknown frictional force of the

boys shoes on the bottom of thetoboggan can be excludedfrom the

analysis if the toboggan and theboy on it are considered as a single

system. In this way the frictional

force F becomes internaland the

conservation of momentum applies


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Conservat ion of Momentum . Since both the

initial and final momenta of the system are zero

(because initial and final velocities are zero), the

systems momentum must also be zero when theboy is at some intermediate point betweenA and

B, thus

0

ttbb vmvm( )


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The 2 unknowns vb and vtrepresent the velocities

of the boy moving to the left and the tobogganmoving to the right. Both are measured from a fixed

inertial reference on the ground.

At any instant the position of point A on thetoboggan and the position of the boy must be

determined by integration. Since v= ds/dt, then

mbdsb + mtdst= 0


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Assuming the initial position of point A to be at the

origin, then at the final position we havembsb +mtst= 0. Since sb+ st= 2 m,

mmm

ms

smsm

tb

bt

tttb

45.11540

)40(22

0)2(


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Impact

Impactoccurs when two bodies collide with eachother during a very shortperiod of time, causing

relatively large (impulsive) forces to be exerted

between the bodies.

There are two types of impact.

Central impactoccurs when the direction ofmotion of the mass centers of the two colliding

particles is along the line of impact.


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Oblique impactoccurs when one or both of theparticles is at an angle with the line of impact.

The line of impact passes through the masscenters of the particles.

Impact


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Central Impact. Consider the central impact oftwo smooth particles A and B,

The particles have the initial momenta as shown.Provided (v

A

)1

> (vB

)1

, collision will eventually

occur.

Impact


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During the collision the particles must be thoughtof as deformable or non-rigid. The particles willundergo aperiod of deformable such that they

exert an equal but opposite deformation impulse

Pdton each other.

Impact


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Only at the instant ofmaximum deformation will

both particles move with a common velocity v,since their relative motion is zero.

Impact

I t


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Afterward aperiod of restitution occurs, in whichcase the particles will either return to their originalshape or remain permanently deformed. The equal

but opposite restitution impulseRdtpushes theparticle apart from one another. In reality, the

physical properties of any two bodies are such that

the deformation impulse is always greaterthat that

of restitution, i.e. Pdt> Rdt.

Impact


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Just after the separation the particles will have thefinal momenta, where (vB)2 > (vA)2

Impact


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The ratio of the restitution impulse to thedeformation impulse is called the coefficient ofrestitution.

The coefficient of restitution can be expressed interms of the particles initial and final velocities,

11

22)()()()(

BA

ABvvvve

Impact


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Coefficient of Restitution. With reference to the case where the centralimpact of two smooth particles A and B, it is seen

that the eqn for the coefficient of restitution states

that e is equal to the ratio of the relative velocity of

the particles separationjust after impact(vB)2(vA)2, to the relative velocity of the particlesjust

before impact, (vA)1 (vB)1e has a value between zero and one.

Impact

I t


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Elastic Impact.

If the collision between the two particles isperfectly elastic(e = 1), the deformation impulse

(Pdt) is equal and opposite to the restitution

impulse (Rdt).

Plastic (inelastic) Impact.

There is no restitution impulse given to the

particles (Rdt = 0), so that after collision bothparticles couple or stick togetherand move with a

common velocity

Impact



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(CENTRAL IMPACT)

The final velocities of the two smooth particles are

to be determinedjust afterthey are subjected to

direct central impact. Provided the coefficient of

restitution, the mass of each particle and eachparticles initial velocityjust before impact areknown, the solution to the problem can be obtained

using the following two equations:

The conservation of momentum applies to thesystem of particles, mv1 = mv2



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The coefficient of restitution relates the relativevelocities of the particles along the line of impact,

just before and just after collision.

When applying these two equations, the sense of

an unknown velocity can be assumed. If the

solution yields a negative magnitude, the velocity is

in the opposite sense.


(CENTRAL IMPACT)


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Impact

Oblique Impact.

When oblique impact occurs between two smooth

particles, the particles move away from each otherwith velocities having unknown directions and

unknown magnitudes.

Provided the initial velocities are known, fourunknown are present in the problem.

I


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These unknown are represented either as (vA)2,(vB)2, 2 and 2, or as thexand ycomponents of

the final velocities.

Impact



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(OBLIQUE IMPACT)

If the yaxis is established within the plane ofcontact and thexaxis along the line of impact, the

impulsive forces of deformation and restitution act

only in the x direction.



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Resolving the velocity or momentum vectors into

components along the x and y axes, it is possible

to write four independent scalar equations in order

to determine (vAx)2, (vAy)2, (vBx)2 and (vBy)2

Momentum of the system is conserved along theline of impact, x axis so that m(vx)1 = m(vx)2.

The coefficient of restitution, e, relates therelative-velocity components of the particles along

the line of impact(xaxis).


(OBLIQUE IMPACT)



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Momentum of particleA is conserved along the yaxis, perpendicular to the line of impact, since no

impulse acts on particleA in this direction.

Momentum of particle B is conserved along the yaxis, perpendicular to the line of impact, since no

impulse acts on particle B in this direction.


(OBLIQUE IMPACT)

Example


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Example

The bagA, having a mass of 6 kg is released from

rest at the position = 0. After falling to = 90, isstrikes an 18 kg box B. If the coefficient of

restitution between the bag and the box is e = 0.5,

determine the velocities of the bag and box just

after impact and the loss of energy during collision.


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Conservat ion o f Energy. With the datum at =

0, we have

smv

v

VTVT

A

A

/43.4)(

)1)(81.9(6))(6(2

100

1

21

1100


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Conservation of Momentum . After impact, we

will assumeA and B travel to the left.

22

22

2211

)(343.4)(

)(6))(18()43.4)(6(0)()()()(

BA

AB

AABBAABB

vv

vvvmvmvmvm

( )


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Conservat ion of Rest i tut ion. Realizing that

for separation to occur after collision (vB)2 > (vA)2,

smv

smsmv

vv

vv

vve

BA

BA

BA

AB

/66.1)(

/554.0/554.0)(

215.2)()(

)()(

)()(

22

22

11

22( )

Solving the two equations

simultaneously,


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Loss o f Energy. Applying the principle of wrk

and energy to the bag and box just before and after

collision, we have

J

U

TTU

15.33

)33.4)(6(2

1)544.0)(6(

2

1)66.1)(18(

2

1 22221

1221

Example


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Example

The ball B has a mass of 1.5 kg and is suspended

from the ceiling by a 1 m long elastic cord. If the

cord is stretched downward 0.25 m and the ball is

released from rest, determine how far the cord

stretched after the ball rebounded from the ceiling.k = 800 N/m, e = 0.8


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Conservat ion o f Energy. With the datum as

shown, realizing that initially y= y0 = (1 + 0.25) m =1.25 m, we have

smv

v

vmksyWvm

VTVT

B

B

BBB

/97.2)(

))(5.1(2

1

)25.0)(800(2

1

)25.1)(81.9(5.10

0)(2

1

2

1)(

2

1

1

2

1

2

21

20

20

1100


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Conservation of Rest i tut ion.

smsmv

v

vv

vve

B

BBA

AB

/37.2/37.2)(

97.20

0)(8.0

)()(

)()(

2

211

22( )


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Conservat ion o f Energy. The maximum stretch

s3 in the cord may be determined by applying the

conservation of energy equation to the ball justafter the collision. Assuming that y= y3 = (1 + s3) m

094.1872.14400

)800(2

1

)1)(5.1(81.90)37.2)(5.1(2

1

2

1)(

2

10)(

2

1

323

2

33

2

233

23

22

3322

ss

ss

ksyWvmvm

VTVT

BBB


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Solving the quadratic equation for the positive root

yields,

ms 237.03

Example


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Example

Two smooth disksA and B, having mass of 1 kg

and 2 kg respectively, collide with the velocities

shown. If the coefficient of restitution for the disks

is e = 0.75, determine thexand ycomponents of

the final velocity of each disk just after collision.


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Solut ion

Resolving each of the initial velocities into x and y

components, we have

smv

smv

smv

smv

By

Bx

Ay

Ax

/707.045sin1)(

/707.045cos1)(

/50.130sin3)(

/60.230cos3)(

1

1

1

1


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The four unknown velocity components after

collision are assumed to act in the positive

directions. Since the impact occurs only in the x

direction (line of impact), the conservation ofmomentum forboth disks can be applied in this

direction.


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Conservation of x Momentum.

18.1)(2)(

)(2)(1)707.0(2)60.2(1

)()()()(

22

22

2211

BxAx

BxAx

BxBAxABxBAxA

vv

vv

vmvmvmvm( )

C ti f ( ) R t i t t i B th di k


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Conservat ion of (x) Rest i tu t ion . Both disks are

assumedto have components of velocity in the +x

direction after collision,

48.2)()(

)07.0(60.2

)()(75.0

)()(

)()(

22

22

11

22

AxBx

AxBx

BxAx

AxBx

vv

vv

vv

vve( )


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Solving the two simultaneous equations,

smv

smsmv

Bx

Ax

/22.1)(

/26.1/26.1)(

2

2

C ti f M t Th


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Conservation of y Momentum. The

momentum ofeach diskis conservedin the y

direction (plane of contact), since the disks aresmooth and therefore no external impulse acts in

this direction.

smsmv

vmvm

smv

vmvm

By

ByBByB

Ay

AyAAyA

/707.0/707.0)(

)()(

/5.1)(

)()(

2

21

2

21( )

( )

Angular Momentum


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g

The angular momentum, HO, of a particle about

point Ois defined as the moment of the particleslinear momentum about O.

Scalar Formulation.

If a particle is moving along acurve lying in thex-yplane, the

angular momentum at any instant

can be determine about point O byusing a scalar formulation.

Angular Momentum


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The magnitude ofHO is

))(()( mvdH zO

dis the moment arm or perpendicular distancefrom O to the line of action ofmv.

Common for (HO)z is kg.m2/s

The direction ofHO is defined by the right-handrule.

Angular Momentum

Angular Momentum


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Vector Formulation. If the particle is moving along aspace curve, the vector cross

product can be used to determine

the angular momentum about O.In this case,

vrH mO

rdenotes a position vector drawn from point O to

the particle P. HO isperpendicularto the shaded

plane containing rand mv.

Angular Momentum

A l M t


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To evaluate the cross product, rand mv shouldbe expressed in terms of their Cartesiancomponents, so that the angular momentum is

determined by evaluating the determinant:

zyx

zyxOmvmvmv

rrr

kji

H

Angular Momentum

Relation Between Moment of a Force


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and Angular Momentum

The moments about Point O of all forces actingon the particle may by related to the particlesangular momentum by using the equation of

motion. If the mass of the particle is constant, wemay write

vF m



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vrFrM mO

The moments of the forces about point O can beobtained by performing a cross-product

multiplication of each side of this equation by the

position vectorr, which is measured in thex, y, zinertial frame of reference,

The derivation ofrx mv can be written as

vrvrvrH mmmdt

dO )(




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0)( rrvr mm Since the equation becomes

OO HM

This equation states that the resultant momentumabout point O of all the forces acting on the particle

is equal to the time rate of change of the particlesangular momentum about point O.





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The result is also similar to

GF

Here G = mv, so the resultant force acting on theparticle is equal to the time rate of change of the

particles linear momentum.



Example


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Example

The box has a mass m and is traveling down the

smooth circular ramp such that when it is at theangle it is a speed v. Determine its angular

momentum about point O at this instant and the

rate of increase in its speed, i.e., at.

Example


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Solution

Since v is tangent to the path, the angular

momentum isrmvHO

From the free-body diagram of the block,

it is seen that only the weight W = mg

contributes a moment about O

)()sin(; rmvdt

drmgHM OO

Example


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Since rand m are constant,

sin

sin

gdt

dv

dt

dvrmmgr

Angular Impulse and Momentum


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g p

Principles

Principle of Angular Impulse and Momentum

We have MOdt= dHO and integrated, assumingat time t = t

1

, HO

= (HO

)1

and time t = t2

, HO

= (HO

)2

21

12

)()(

)()(

2

1

2

1

O

t

t OO

OO

t

tO

dt

dt

HMH

HHM

or

This equation is referred to as theprinciple ofangular impulse and momentum.



99/130

The initial and final angular momenta (HO)1 and(HO)2 are defined as the moment of the linear

momentum of the particle (HO = rx mv) at the

instant t1 and t2 respectively.

The second term on the left side, MOdt, iscalled the angular impulse. It is determined by

integrating, w.r.t time, the moments of all forcesacting on the particle over the time period t1 to t2.


Principles



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Since the moment of a force about point O is MO= rx F, the angular impulse may be expressed in

vector form as

2

1

2

1

)(t

t

t

tO dtdt FrMAngular impulse

The principle of angular impulse and momentumfor a system of particles may be written as

21 )()(2

1O

t

tOO dt HMH


Principles



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Vector Formulation

Using impulse and momentum principles, it ispossible to write which define the particles motion,

212

1

vFv mdtmt

t

21 )()(2

1O

t

tOO dt HMH


Principles



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Scalar Formulation

The above equations may be expressed in x, y, zcomponent form. If the particle is confined to move

in the x-y plane, three independent equations maybe written to express the motion,

21 )()(2

1

x

t

t

xx vmdtFvm

21 )()(2

1O

t

tOO HdtMH

21 )()(2

1y

t

tyy vmdtFvm

g p

Principles



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Conservation of Angular Momentum

When the angular impulse acting on a particle areall zero during the time t1 to t2, it may be written as

21 )()( OO HH

This equation is known as the conservation of

angular momentum. It states that from t1 to t2 theparticles angular momentum remain constant.

If no external impulse is applied to the particle,both linear and angular momentum is conserved.

g p

Principles



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In some cases, the particles angular momentumwill be conserved and linear momentum may not.

This occurs when the particle is subjected onlytoa central force.


Principles



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The impulsive central force F is always directedtoward point O as the particle moves along the

path.

The angular impulse (moment) created by Fabout zaxis passing through point O is always

zero, and therefore angular momentum of the

particle is conserved about this axis. The conservation of angular momentum for asystem of particles,

21 )()( OO HH


Principles



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Free-Body DiagramDraw the particles FBD in order to determine anyaxis about which angular momentum is conserved.

For this to occur, the moments of the forces (or

impulse) must be parallel or pass through the axis

so as to create zero moment throughout the time

period t1 to t2.

The direction and sense of the particles initial andfinal velocities should be established



107/130

Momentum Equations. Apply the principle of angular impulse andmomentum,

21 )()(2

1O

t

tOO dt HMH

Or if appropriate, apply the conservation ofangular momentum,

21 )()( OO HH

Example


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The 5 kg block rests on the smooth horizontal

plate. It is attached at A to a slender rod ofnegligible mass. The rod is attached to a ball-and-

socket join at B. If the moment M = (3t) N.m where

t is in seconds, is applied to the rod and a

horizontal force P = 10 N is applied to the block,

determine the speed of the block in 4 s starting

from rest.


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Free-Bod y Diagram. If we consider the system of

both rod and block, then the resultant forcereaction FB at the ball-and-socket can be

eliminated from the analysis by applying principle

of angular impulse and momentum about the z

axis. If this is done, the angular impulses createdby the weight and normal reaction NA are also

eliminated, since they act parallel to zaxis and

therefore create zero moment about this axis.

Princip le of Angular Impu lse and Momentum


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Princip le of Angular Impu lse and Momentum .

smv

vdtt

HtPrdtMH

HdtMH

A

A

zBA

t

tz

ztt

zz

/20)(

)4.0()(5)4)(10)(4.0(30

)()()(

)()(

2

2

4

0

21

21

2

1

2

1

Example


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Example

The 0.4 kg ball B is attached to a cord which passes

through a hole atA in a smooth table. When the ballis r1 = 0.5 m from the hole, it is rotating around in a

circle such that its speed is v1 = 1.2 m/s. By applying

a force F the cord is pulled downward through the

hole with a constant speed vc= 2 m/s. Determine (a)

the speed of the ball at the instant it is r2 = 0.2 m

from the hole, and (b) the amount of work done by F

in shortening the radial distance from r1 to r2.


112/130


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Part (a) Free-Body Diagram . As the ball

moves from r1 to r2, the cord force F on the ball

always passes through the zaxis, and the weight

and NB are parallel to it. Hence the moments, or

angular impulses created by these forces, are allzero about this axis. Therefore, the conservation of

angular momentum applies about the zaxis.

C ti f A l M t Th b ll


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Conservat ion of Angu lar Momentum . The ballsvelocity v2 is resolved into two components. The

radial component, 2 m/s, is known; however, itproduces zero angular momentum about the zaxis.

Thus

smv

v

vmrvmr BB

/3

)4.0)(2.0()2.1)(4.0)(5.0(

2

2

2211

21

HH

The speed of the ball is thus


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The speed of the ball is thus

sm

v

/606.3

)2()0.3( 222

Part (b). The only force that does work on the

ball is F. The initial and final kinetic energies of the

ball can be determine so that from the principle of

work and energy,


116/130

JU

U

TUT

F

F

313.2

)606.3)(4.0(

2

1)2.1)(4.0(

2

1 22

2211

Example


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The 2 kg disk rests on a smooth horizontal surface

and is attached to an elastic cord that has a stiffnesskc= 20 N/m and is initially unstretched. If the disk is

given a velocity (vD)1 = 1.5 m/s, perpendicular to the

cord, determine the rate at which the cord is being

stretched and the speed of the disk at the instant the

cord is stretched 0.2 m.

Free-Body Diagram . After the disk has been


118/130

launched, it slides along the path. By inspection,

angular momentum about point O is conserved,since none of the forces produce an angular

impulse about this axis. Also, when the distance is

0.7 m, only the component (vD)2 produces angular

momentum of the disk about O.

C ti f A l M t Th


119/130

Conservat ion of Angu lar Momentum . The

component (vD)2 can be obtained by applying the

conservation of angular momentum about O

smv

vvmrvmr

D

D

DDDD

OO

/07.1)(

))(2)(7.0()5.1)(2)(5.0()()(

)()(

2

2

2211

21

HH


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Conservat ion o f Energy. Applying the

conservation of energy equation at the point wherethe disk was launched and at the point where the

cord is stretched 0.2 m.

smv

v

VTVT

D

D

/36.1)(

)2.0)(20(2

1))(2(

2

10)5.1)(2(

2

1

2

222

2

2211


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Having determine (vD)2 and its component (vD)2,

the rate of stretch of the cord (vD)2 is determinedfrom the Pythagorean theorem.

sm

vvvDDD

/838.0

)07.1()36.1(

)()()(

22

2

2

2

22

CHAPTER REVIEW


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Impulse

An impulse that acts on the particle is defined by

dtFI

Graphically this represents the area under the F-tdiagram. If the force is constant, then the

impulse becomes

)( 12 ttc FI

CHAPTER REVIEW


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Princ ip le of Impulse and Momentum

When the equation of motion, F =ma, and thekinematic equation, a = dv/dt, are combined, we

obtain the principle of impulse and momentum.

212

1

vFv mdtm tt

The initial momentum of the particle, mv1, plus all

of the impulses that are applied to the particleduring the time t1 to t2, Fdt, equal the finalmomentum mv2 of the particle.

CHAPTER REVIEW


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This is a vector equation that can be resolvedinto components and is used to solve problems

that involve force, velocity and time.

For application, the free-body diagram should bedrawn in order to account for all the impulses that

act on the particle.

CHAPTER REVIEW


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Conservat ion of L inear Momentum

If the principle of impulse and momentum isapplied to the system of particles, then the

collisions between the particles produce internal

impulse that are equal, opposite and collinear, andtherefore cancel out from the equation.

If an external impulse is small, that is, the force is

small and the time is short, then the impulse can beclassified as non-impulsive and can be neglected.

CHAPTER REVIEW


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Consequently, momentum for the system ofparticles is conserved, and so

21 ii mm vv

This equation is useful for finding the finalvelocity of a particle when internal impulses are

exerted between two particles.

If the internal impulse is to be determined, thenone of the particles is isolated and the principle of

impulse and momentum is applied to this system.

CHAPTER REVIEW


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Impact

When two particles collide (A and B), the internalimpulse between them is equal, opposite, and

collinear.

Consequently, the conversation of momentum forthis system applies along the line of impact.

If the final velocities are unknown, a secondequation is needed for solution, and we use the

coefficient of restitution, e.

2211 BBAABBAA

vmvmvmvm

CHAPTER REVIEW


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11

22

)()()()(

BA

AB

vvvve

If the collision is elastic, no energy is lost and e =1. For a plastic collision e = 0

If the impact is oblique, then conservation ofmomentum for the system and the coefficient of

restitution equation apply along the line of impact.

Conservation of momentum for each particleapplies perpendicular to this line.

CHAPTER REVIEW


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Princ ip le of Angu lar Impu lse and Momentum

The momentum of the linear momentum about anaxis (z) is called the angular momentum. Its

magnitude is

))(()( mvdH zO

In three dimensions, the cross product is used

vrH mO

CHAPTER REVIEW


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The principle of angular impulse and momentumis derived from taking moments of the equation ofmotion about inertial axis, using a = dv/dt. The

result is

21 )()(

2

1 O

t

t OO dt HMH This equation is used to eliminate unknownimpulses by summing the moments about an axis

through which the lines of action of these impulses

produce no moment

dynamics lecture4.1

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