chapter 4 linear impulse and momentum

8/10/2019 Chapter 4 Linear Impulse and Momentum

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This lecture note is taken and modified from Engineering Mechanics -Dynamics, R.C.Hibbeler, Prentice Hall whichcopyright belongs to Pearson Education South Asia Pte Ltd.

Trimester 2 2014/2015

Chapter 4Kinematics of a Particle:

Impulse and Momentum


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Chapter Objectives

Principle of linear impulse and momentum for a particle

Conservation of linear momentum for particles

Analyze the mechanics of impact

Concept of angular impulse and momentum

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Chapter Outline

1. Principle of Linear Impulse and Momentum

2. Principle of Linear Impulse and Momentum for a System of

Particles

3. Conservation of Linear Impulse for a System of Particles4. Impact

5. Angular Momentum

6. Relation between Momentum of a Force and Angular

Momentum7. Principle of Angular Impulse and Momentum

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15.1 Principle of Linear Impulse and

Momentum

Equation of motion for a particle of mass m isF= ma= mdv/dt

Rearranging the terms and integrating between the

limits v= v1at t = t1and v= v2at t = t2

Referred to as the

principle of linear impulseand momentum

12

2

1

2

1

2

1

vvFvF mmdtdmdtt

t

v

v

t

t

4


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Momentum

L inear Momentum Vectors of the form L = mvis called

l inear momentum

Magnitude mv has unit of mass-velocity, kg.m/s

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Momentum

L inear Impulse is called linear impulse, and measure the

effect of a force during the time the force acts

The impulse acts in the same direction as the force,

magnitude has unit of force-time, N.s

dtFI

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Momentum

Principle of L inear Impulse and Momentum The equation is rewritten in the form

Principle of l inear impulse and momentumin itsx, y,

zcomponents is

212

1

vFv mdtmt

t

21

21

21

)()(

)()(

)()(

2

1

2

1

2

1

z

t

t

zz

y

t

t yy

x

t

t xx

vmdtFvm

vmdtFvm

vmdtFvm

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Momentum

Procedures for AnalysisFree-Body Diagram

Establish thex, y, zinertial frame of reference and

draw FBD

Establish direction and sense of initial and final

velocities

Assume the sense of vector components in the

direction of the positive inertial coordinates

Draw the impulse and momentum diagrams for the

particle

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Momentum

Procedures for AnalysisPrinciple of Impulse and Momentum

Apply the principle of linear impulse andmomentum,

If the motion occurs in thex-yplane, itcan resolve the vector components of F

Every force acting on the particles FBDcan create an impulse

The impulse is equal to the area under theforce-time curve

21 21

mvdtFmvt

t

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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 for 10 s, determinethe final velocity and the normal force which the surface

exerts on the stone during the time interval.

Example 15.1

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SolutionFree-Body Diagram

Since all forces acting are constant, the impulses are

I= Fc(t2t1)

Principle of Impulse and Momentum

Resolving the vectors along thex, y, zaxes,

Example 15.1

m/s1.14)100(45cos)10(2000

)()(

22

21

2

1

vv

vmdtFvm xt

t xx

N840045sin)10(200)10(981)10(0

)()( 212

1

CC

y

t

t yy

NN

vmdtFvm

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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.

Example 15.3

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Since weight of each block is constant,

the cord tensions will also be constant.

Since mass of pulleyDis neglected,

the cord tension is TA= 2TB.

Example 15.3

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SolutionPrinciple of Impulse and Momentum

BlockA:

BlockB:

Example 15.3

1))(3()6)(81.9(3)6(20

)()(

2

21

2

1

AB

A

t

t yA

vT

vmdtFvm

2))(5()6()6)(81.9(50

)()(

2

21

2

1

BB

B

t

t yB

vT

vmdtFvm

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SolutionKinematics

We have

Taking time derivative yields

When B moves downward A moves upward.

Sub this result into Eq. 1 and solving Eqs. 1 and 2 yields

(vB)2= 35.8 m/s and TB= 19.2 N

Example 15.3

lss BA 2

BA vv 2

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15.2 Principle of Linear Impulse and Momentum for

a System of Particles

Principle of linear impulse and momentum for asystem of moving particles is

Internal forces fiacting between particles do not

appear with this summation,

dt

dm i

ii

vF

21

2

1 ii

t

t iii mdtm vFv

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15.3 Conservation of Linear Momentum

for a System of Particles

When the sum of the external impulsesacting on asystem of particles iszero, the equation is

This is called conservation of linear momentum

The total momentum for a system of particles remains

constant during the time period t1to t2

21 iiii mm vv

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Internal impulsesfor the system will always cancelout as they occur in equal but opposite collinear pairs

The forces causing negligible impulses are called non-

impulsive forces

Forces that are very large and act for a very short

period of time are called impulsive forces

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Procedure for AnalysisFree-Body Diagram

Establish thex, y, zinertial frame of reference and

draw the FBD

Apply conservation of linear momentum in a given

direction

Establish the direction and sense of the particles

initial and final velocities.

Draw the impulse and momentum diagrams for each

particle of the system

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Procedure for AnalysisMomentum Equations

Apply the principle of linear impulse and momentum

or the conservation of linear momentum

Determine the internal impulse F.dt acting on onlyone particle of a system

Average impulsive forceFavgcan be determined from

Favg= F dt/t.

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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 average force between them if the coupling takes place

in 0.8 s.

Example 15.4

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SolutionPart (a)

Free-Body Diagram.

Consider bothcars as a single system.

Momentum is conserved in thexdirection since the

coupling force Fis internalto the system.

Example 15.4

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SolutionConservation of L inear Momentum

Part (b)

Average (impulsive) coupling forceFavg, can be

determined by applying the principle of linear momentum

to eitherone of the cars

Example 15.4

smv

v

vmmvmvm BABBAA

/5.0

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

)()()(

2

2

211( )

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SolutionPart (b)

Conservation of M omentum

Example 15.4

kNF

F

vmdtFvm

avg

avg

AAA

8.18

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

)( 21

( )

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The bumper carsA andB each have a mass of 150 kg andare coasting with the velocities shown before they freely

collide head on. If no energy is lost during the collision,

determine their velocities after collision.

Example 15.6

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The cars will be considered as a single system.

The free-body diagram is shown.

Conservation of Momentum

Example 15.6

11

150)150(2150)3)(150(

)()()(

22

22

2211

BA

BA

BBAABBAA

vv

vv

vmvmvmvm

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SolutionConservation of Energy

Sub Eq. (1) into (2), we get

Since refer before the collision, hence

and

Example 15.6

213

002

2

2

2

2

22

12

22

12

12

12

12

1

2211

BA

BBAABBAA

vvvmvmvmvm

VTVT

m/s2orm/s2andm/s3 22 BB vv

27

m/s22 Bv

m/s32

B

v m/s22 Av


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An 800-kg rigid pile P is driven into the ground using a300-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 after striking,the hammer does not rebound off the pile.

Example 15.7

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Velocity can be determined using the conservation of

energy equation applied to the hammer.

Example 15.7

m/s13.3)(

0))(300(21)5.0)(81.9(3000

)(2

1)(

2

1

1

2

1

1

2

10

2

0

1100

H

H

HHHHHH

v

v

yWvmyWvm

VTVT

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Weight of the hammer and pile and the resistance force Fs

of the sand are all non-impulsive,

Momentum is conserved in the vertical direction during

this short time.

Example 15.7

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SolutionConservation of Momentum

Since the hammer does not rebound off the pile just after

the collision, then (vH)2= (vP)2= v2

Example 15.7

m/s8542.0

8003000)13.3)(300(

)()()(

2

22

2211

v

vv

vmvmvmvm pHppHH

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SolutionPrinciple of Impulse and Momentum

The impulse which the pile imparts to the hammer can

now be determined since v2is known.

Example 15.7

sN683

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

)()( 212

1

dtR

dtR

vmdtFvm Ht

t yHH

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15.4 Impact

Impact occurs when two bodies collide and causeimpulsive forces to be exerted between them

1. Central impactoccurs when the direction of motion of

the mass centers of the two colliding particles is along

the line of impact

2. Oblique impactoccurs when one or both of the

particles is at an angle with the line of impact

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15.4 Impact

Central Impact When (vA)1> (vB)1, collision will occur

During collision, particles undergo aperiod of

deformable

Only at maximum deformationwill both particles have

common velocity as their relative motion is zero

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15.4 Impact

Central Impact Afterward aperiod of restitution occurs, particles will

return to their original shape or remain permanently

deformed

Equal but opposite restitution impulseRdtpushesthe particle apart from one another

After separation the particles will have the final

momentum, where (vB)2> (vA)2

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15.4 Impact

Central Impact Ratio of the restitution impulse to the deformation

impulse is called the coefficient of restitution

Can be expressed in terms of the particles initial andfinal velocities,

11

22

)()(

)()(

BA

AB

vv

vve

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15.4 Impact

Coefficient of Restitution ehas a value between zero and one

Elastic Impact (e = 1) Deformation impulse (Pdt) is equal and opposite to

the restitution impulse (Rdt)

Plastic (inelastic) Impact (e = 0)

After collision both particles couple or stick together

and move with a common velocity

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15.4 Impact

Procedure for Analysis (Central Impact) The conservation of momentum applies to the system

of particles, mv1= mv2

Coefficient of restitution relates the relative velocities

of the particles along the line of impact, just before

and just after collision

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15.4 Impact

Oblique Impact When oblique impact occurs, particles move away

from each other with velocities having unknown

directions and unknown magnitudes

Provided the initial velocities are known, four

unknown are present in the problem

Unknown are (vA)2, (vB)2, 2and 2, orxandy

components of the final velocities

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15.4 Impact

Oblique Impact Wheny axis is established within the plane of contact

and thexaxis along the line of impact, the impulsive

forces of deformation and restitution act only in the x

direction

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15.4 Impact

Procedure for Analysis Momentum of the system is conserved along the line

of impact, x axis so that m(vx)1= m(vx)2 The coefficient of restitution, e, relates the relative-

velocity components of the particles along the line ofimpact(xaxis).

Momentum of particleAis conserved along theyaxis,perpendicular to the line of impact, since no impulse

acts on particleAin this direction. Momentum of particleBis conserved along theyaxis,

perpendicular to the line of impact, since no impulseacts on particleBin this direction.

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The bagA, having a mass of 6 kg is released from rest atthe position = 0. After falling to = 90, is strikes an18 kg boxB. 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 energyduring collision.

Example 15.9

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At = 0, we have

Conservation of Momentum

After impact,AandBassume travel to the left

Example 15.9

smv

v

VTVT

A

A

/43.4)(

0))(6(21181.960

1

2

1

1100

2222

2211

)(343.4)()(6))(18()43.4)(6(0

)()()()(

BAAB

AABBAABB

vvvv

vmvmvmvm

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SolutionConservation of Resti tution

Realizing that for separation to occur after collision (vB)2

> (vA)2,

Solving the two equations,

Example 15.9

215.2)()()()(

)()( 22

11

22

BA

BA

AB vvvv

vve

m/s66.1)(

m/s554.0m/s554.0)(

2

2

B

A

v

v

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SolutionLoss of Energy

Applying the principle of work and energy to the bag and

box just before and after collision,

Example 15.9

J

U

TTU

15.33

)33.4)(6(

2

1)544.0)(6(

2

1)66.1)(18(

2

1 22221

1221

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Two smooth disksAandB, having mass of 1 kg and 2 kgrespectively, collide with the velocities shown. If the

coefficient of restitution for the disks is e= 0.75,

determine thexandycomponents of the final velocity of

each disk just after collision.

Example 15.11

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SolutionResolving each of the initial velocities into x and y

components, we have

Example 15.9

smv

smv

smvsmv

By

Bx

Ay

Ax

/707.045sin1)(

/707.045cos1)(

/50.130sin3)(/60.230cos3)(

1

1

1

1

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SolutionConservation of x Momentum

We have

Conservation of (x) Resti tution

We have

Example 15.9

18.1)(2)()(2)(1)707.0(2)60.2(1

)()()()(

2222

2211

BxAxBxAx

BxBAxABxBAxA

vvvv

vmvmvmvm

48.2)()()07.0(60.2

)()(75.0

)()(

)()(

2222

11

22

AxBxAxBx

BxAx

AxBx

vvvv

vv

vve

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SolutionSolving the two simultaneous equations,

Conservation of y Momentum

The momentum of each diskis conservedin the y

direction (plane of contact),

Example 15.9

smv

smsmv

Bx

Ax

/22.1)(

/26.1/26.1)(

2

2

smsmvvmvm

smvvmvm

ByByBByB

AyAyAAyA

/707.0/707.0)()()(

/5.1)()()(

221

221

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15.5 Angular Momentum

Angular momentum, HO, of a particle about point Oisdefined as the moment of the particles linearmomentum about O

Scalar Formulation

If a particle is moving along a curve, the angular

momentum can be determine by using a scalar

formulation

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Scalar Formulation The magnitudeof HOis

dis the moment arm or perpendicular distance from O

to the line of action of mv.

Common for (HO)zis kg.m2/s

The directionof HOis defined by the right-hand rule.

))(()( mvdH zO

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Vector Formulation If the particle is moving along a space curve, the

vector cross product is used to determine the angular

momentumabout O

We have

Angular momentum is determined by evaluating the

determinant

vrH mO

zyx

zyxO

mvmvmv

rrr

kji

H

52

15 6 Relation Between Moment of a Force and


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15.6 Relation Between Moment of a Force and

Angular Momentum

The moments about Point Oof all forces acting isrelated by

The moments of the forces

about point Ocan be obtained by

Derivation of rx mvcan be written as

Since therefore

vF m

vrFrM mO

vrvrvrH mmmdt

dO )(

0)( rrvr mm

OO HM 53



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Angular Momentum

The equation states that resultant momentum aboutpoint O of all the forces acting on the particle is equal

to the time rate of change of the particles angular

momentum about point O

The result is also similar to

System of Particles

Forces acting on arbitrary ith particle of the system

consist of a resultant external forceFi, and a resultantinternal forcefi

LF

Oiiiii )()()( HfrFr 54



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Angular Momentum

System of Particles Similar equations can be written for each of the other

particles of the system

The second term is sero since the internal forces occur

in equal but opposite collinear pairs, and hence the

moment of each pair about point Ois zero

Oiiiii )()()( HfrFr

OO HM 55


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The box has a mass m and is traveling down the smoothcircular ramp such that when it is at the angle 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 15.12

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SolutionSince vis tangent to the path, the angular momentum is

From the FBD, the weight W = mgcontributes a momentabout O

Since rand mare constant,

Example 15.12

rmvHO

)()sin(; rmvdt

drmgHM OO

sinsin gdt

dv

dt

dvrmmgr

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Angular Momentum

Principle of Angular Impulse and Momentum We have MOdt= dHOand integrated, assuming at

time t = t1, HO= (HO)1and time t = t2, HO= (HO)2

This equation is referred to as theprinciple of angular

impulse and momentum

21

12

)()(

)()(

2

1

2

1

O

t

t OO

OO

t

t

O

dt

dt

HMH

HHM

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Angular Momentum

Principle of Angular Impulse and Momentum Since the moment of a force about point Ois MO= r

x F, the angular impulse may be expressed in vector

form as

The principle of angular impulse and momentum for a

system of particles may be written as

2

1

2

1

)(tt

t

t O dtdt FrMAngular impulse

21 )()(2

1O

t

tOO dt HMH

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Angular Momentum

Vector Formulation Using impulse and momentum principles, it is

possible to write which define the particles motion,

212

1

vFv mdtmt

t

21 )()(2

1

O

t

t OO dt HMH

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Angular Momentum

Scalar Formulation If the particle is confined to move in the x-y plane,

three independent equations may be written to express

the motion,

21 )()(2

1x

t

t xx vmdtFvm

21 )()(2

1O

t

t OO HdtMH

21 )()(2

1

y

t

t yy vmdtFvm

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Angular Momentum

Conservation of Angular Momentum When the angular impulse acting on a particle are all

zero during the time t1to t2, it may be written as

It is known as the conservation of angular momentum.

If no external impulse is applied to the particle, both

linear and angular momentum is conserved.

21 )()( OO HH

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5.7 e at o etwee o e t o a o ce a d

Angular Momentum

Conservation of Angular Momentum In some cases, the particles angular momentum will

be conserved and linear momentum may not.

This occurs when the particle is subjected onlyto a

central force.

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Angular Momentum

Procedure for AnalysisFree-Body Diagram

Draw the particles FBD where angular momentum isconserved.

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

Momentum Equations

Apply the principle of angular impulseand momentum,

2121 )()(or)()(2

1OOO

t

t OO dt HHHMH

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The 0.4 kg ballBis attached to a cord which passesthrough a hole atAin a smooth table. When the ball is 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 Fthe

cord is pulled downward through the hole with a constantspeed vc= 2 m/s.

Example 15.14

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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

Fin shortening the radial distance from r1to r2.

Example 15.14

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SolutionPart (a) Free-Body Diagram

As the ball the cord force Fon the ball passes through the

zaxis.

Weight and NBare parallel and the conservation ofangular momentum applies about thezaxis.

Conservation of Angular Momentum

Example 15.14

m/s3)4.0)(2.0()2.1)(4.0)(5.0( 22

2211

21

vv

vmrvmr BB

HH

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SolutionConservation of Angular Momentum

The speed of the ball is thus

Part (b)

The only force that does work on the ball is F.

Example 15.14

m/s606.3)2()0.3( 222 v

JUU

TUT

FF 313.2)606.3)(4.0(2

1)2.1)(4.0(

2

1 22

2211

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END

chapter 4 linear impulse and momentum

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