CHAPTER 13site.iugaza.edu.ps/mhaiba/files/2012/01/CH-13-Kinetics-of...CHAPTER 13 Kinetics of...

CHAPTER 13 Kinetics of Particles:

Energy and Momentum

Methods

Friday, February 28, 2014

Dr. Mohammad Suliman Abuhaiba, PE 1

Chapter OutLine


Dr. Mohammad Suliman Abuhaiba, PE

2

1. Introduction

2. Work of a Force

3. Principle of Work & Energy

4. Applications of the Principle

of Work & Energy

5. Power and Efficiency

6. Potential Energy

7. Conservative Forces

8. Conservation of Energy

10. Motion Under a Conservative

Central Force

11. Principle of Impulse and

Momentum

12. Impulsive Motion

13. Impact

14. Direct Central Impact

15. Oblique Central Impact

16. Problems Involving Energy and

Momentum

Introduction



3

• Chapter 12 deals with the motion of particles

through the fundamental equation of motion,

amF

Introduction



4

Chapter 13: two additional methods of

analysis:

Method of work & energy: directly relates force,

mass, velocity and displacement.

Method of impulse &momentum: directly relates

force, mass, velocity, and time.

Work of a Force



5

= particle displacement rd

Work of the force is

dzFdyFdxF

dsFrdFdU

zyx

cos

Work is a scalar quantity: has

magnitude & sign but not direction.

Units of work = Length × Force

1 J = (1N)(1m) 1 ft.lb = 1.356 J

Work of a Force



6

Work of a force during a

finite displacement,

2

1

2

1

2

1

2

1

cos21

A

A

zyx

s

s

t

s

s

A

A

dzFdyFdxFdsF

dsFrdFU

Work = area under Ft vs. s

curve

Work of a Force



7

• Work of a constant force in rectilinear motion,

xFU cos21

Work of a Force - gravity



8

Work of the force of gravity,

yWyyWdyWU

dyWdzFdyFdxFdU

y

y

zyx

1221

2

1

Work of weight is positive when y < 0, i.e.,

when weight moves down

Work of a Force - spring



9

Magnitude of force exerted by a spring is

proportional to deflection,

lb/in.or N/mconstant spring

k

kxF

Work of force exerted by spring,

222

1212

121

2

1

kxkxdxkxU

dxkxdxFdU

x

x

Work of a Force - spring



10

Work of force exerted by spring is positive when

x2 < x1, i.e., when spring is returning to its

undeformed position.

Work of force exerted by spring = negative of area

under F vs. x curve,

xFFU 2121

21

Work of a Force - gravitational



11

Work of a gravitational force (particle M occupies

fixed position O while particle m follows path

shown),

)11

(12

221

2

2

1

rrGMm

drr

MmGU

drr

MmGFdrdU

r

r

Forces which do not do work (ds = 0 or cos 0:

Weight of a body when its center of gravity

moves horizontally

Reaction at a roller moving along its track

Reaction at frictionless surface when body in

contact moves along surface

Reaction at frictionless pin supporting rotating

body

Work of a Force



12

Particle Kinetic Energy: Principle

of Work & Energy



13

dvmvdsF

ds

dvmv

dt

ds

ds

dvm

dt

dvmmaF

t

tt

• Consider a particle of mass m acted upon by force F

• Integrating from A1 to A2

2

21

1221

2

1212

221

2

1

2

1

mvTTTU

mvmvdvvmdsF

v

v

s

s

t

Particle Kinetic Energy: Principle

of Work & Energy



14

Work of force = change in kinetic energy

of the particle F

Units of work and kinetic energy are the same:

JmNms

mkg

s

mkg

2

22

21

mvT

Applications of the Principle of

Work and Energy



15

Determine velocity of bob at A2

Force acts normal to path and does no work P

glv

vg

WWl

TUT

2

2

10

2

22

2211

Velocity found without determining expression for

acceleration and integrating

All quantities are scalars and can be added directly Forces which do no work are eliminated from problem


Work and Energy



16

Principle of work & energy cannot be applied to

directly determine acceleration of bob

Calculating tension in the cord requires application

of Newton’s 2nd law


Work and Energy



17

As the bob passes through A2

Wl

gl

g

WWP

l

v

g

WWP

amF nn

32

22

glv 22

Power and Efficiency



18

Power = rate at which work is done

vFdt

rdF

dt

dUPower

Units of power:

W746s

lbft550 hp 1or

s

mN 1

s

J1 (watt) W 1

inputpower

outputpower

input work

koutput worefficiency

Sample Problem 13.1



19

An automobile weighing 4000 lb is driven down a 5o

incline at a speed of 60 mph when the brakes are

applied causing a constant total breaking force of

1500 lb.

Determine the distance traveled by the automobile

as it comes to a stop.

Solution:

Evaluate the change in kinetic energy

lbft481000882.324000

sft88s 3600

h

mi

ft 5280

h

mi60

2

212

121

1

1

mvT

v

00 22 Tv

Sample Problem 13.1



20

0lb1151lbft481000

2211

x

TUT

ft 418x

Determine the distance required for work to equal

the kinetic energy change.

xxxU lb11515sinlb4000lb150021

Sample Problem 13.1



21

Sample Problem 13.2



22

Two blocks are joined by an

inextensible cable as shown.

If the system is released

from rest, determine the

velocity of block A after it

has moved 2 m. Assume

that the coefficient of

friction between block A

and the plane is mk = 0.25

and that the pulley is

weightless and frictionless.

Sample Problem 13.2



23

Solution:

• Apply the principle of work &

energy separately to blocks A & B

2

21

2

21

2211

2

kg200m2N490m2

m2m20

:

N490N196225.0

N1962sm81.9kg200

vF

vmFF

TUT

WNF

W

C

AAC

AkAkA

A

mm



24

221

221

2211

2

kg300m2N2940m2

m2m20

:

N2940sm81.9kg300

vF

vmWF

TUT

W

c

BBc

B

Sample Problem 13.2

Sample Problem 13.2



25

When the two relations are combined, the work of

the cable forces cancel. Solve for the velocity.

221 kg200m2N490m2 vFC

221 kg300m2N2940m2 vFc

221

221

kg500J 4900

kg300kg200m2N490m2N2940

v

v

sm 43.4v

Sample Problem 13.3



26

A spring is used to stop a 60 kg package which is sliding on a

horizontal surface. The spring has a constant k = 20 kN/m and

is held by cables so that it is initially compressed 120 mm. The

package has a velocity of 2.5 m/s in the position shown and the

maximum deflection of the spring is 40 mm. Determine

a. Coefficient of kinetic friction between the package &

surface

b. Velocity of the package as it passes again through the

position shown.

Sample Problem

13.3



27

Solution:

Apply principle of work and energy

between initial position and the point

at which spring is fully compressed.

0

J5.187sm5.2kg60

2

2

212

121

1

T

mvT

k

k

kfxWU

m

m

m

J377

m640.0sm81.9kg60 2

21

Sample Problem

13.3



28

J0.112

m040.0N3200N2400

N3200

m160.0mkN20

N2400

m120.0mkN20

21

maxmin21

21

0max

0min

xPPU

xxkP

kxP

e

J112J377

212121

k

efUUU

m

0J112J 377-J5.187

:2211

k

TUT

m

20.0km

Sample Problem 13.3



29

Apply the principle of work and energy for the

rebound of the package.

232

1232

132 kg600 vmvTT

J36.5J112J377

323232

k

efUUU

m

2

321

3322

kg60J5.360

:

v

TUT

sm103.13 v

Sample Problem 13.4



30

A 2000 lb car starts from rest at point 1 and moves

without friction down the track shown. Determine:

a. the force exerted by the track on the car at point 2

b. min safe value of radius of curvature at point 3

Sample Problem 13.4



31

Solution:

• Apply principle of work & energy to determine

velocity at point 2.

sft8.50

ft2.32ft402ft402

2

1ft400:

ft40

2

10

2

22

2

2

22211

21

2

2

2

221

21

v

sgv

vg

WWTUT

WU

vg

WmvTT

Apply Newton’s 2nd law to find normal force by the

track at point 2

:nn amF

lbWN

g

g

W

v

g

WamNW n

000,105

ft20

ft402

2

2

2

Sample Problem 13.4



32

Sample Problem 13.4



33

Apply principle of work & energy to determine

velocity at point 3

sft1.40sft2.32ft252ft252

2

1ft250

323

233311

vgv

vg

WWTUT

Sample Problem 13.4



34

Apply Newton’s 2nd law to find min radius of

curvature at point 3 such that a positive normal force

is exerted by the track.

:nn amF

33

23 ft252

g

g

Wv

g

W

amW n

ft503

Sample Problem 13.5



35

The dumbwaiter D and its load have a

combined weight of 600 lb, while the

counterweight C weighs 800 lb.

Determine the power delivered by the

electric motor M when the dumbwaiter

a. is moving up at a constant speed of 8

ft/s

b. has an instantaneous velocity of 8 ft/s

and an acceleration of 2.5 ft/s2, both

directed upwards.

Sample Problem 13.5



36

In first case, bodies are in uniform motion.

Determine force exerted by motor cable from

conditions for static equilibrium.

Free-body C:

:0 yF

lb 4000lb8002 TT

Sample Problem 13.5



37

slbft1600

sft8lb 200

DFvPower

hp 91.2slbft550

hp 1slbft1600

Power

Free-body D:

:0 yF

lb 200lb 400lb 600lb 600

0lb 600

TF

TF

Sample Problem 13.5



38

2nd case: both bodies are accelerating.

Apply Newton’s 2nd law to each body to

determine the required motor cable force.

2212 sft25.1sft5.2 DCD aaa

Free-body C: :CCy amF

lb5.38425.12.32

8002800 TT

Sample Problem 13.5



39

Free-body D: :DDy amF

lb 1.2626.466005.384

5.22.32

600600

FF

TF

slbft2097

sft8lb 1.262

DFvPower

hp 81.3

slbft550

hp 1slbft2097

Power

Home Work Assignment 13-1

1, 8, 15, 22, 28, 43, 50

Due Saturday 8/3/2014



40

Potential Energy Friday, February 28, 2014


41

2121 yWyWU

Work of force of gravity , W

Work is independent of path

followed; depends only on

initial and final values of Wy.

WyVg

2121 gg VVU



42

Choice of datum from which elevation y is measured

is arbitrary.



43

For a space vehicle, variation of the force of

gravity with distance from the center of earth

should be considered.

Work of a gravitational force,

1221

r

GMm

r

GMmU

Potential energy Vg when the

variation in the force of gravity

can not be neglected,

r

WR

r

GMmVg

2



44

Work of force exerted by a spring

depends only on initial & final

deflections of the spring,

222

1212

121 kxkxU

2121

221

ee

e

VVU

kxV

Conservative

Forces



45

If the work of force is

independent of the path

followed by its point of

application.

22211121 ,,,, zyxVzyxVU

Such forces are described as

conservative forces.

For any conservative force

applied on a closed path,

0 rdF

Conservative

Forces



46

Elementary work corresponding

to displacement between two

neighboring points,

zyxdV

dzzdyydxxVzyxVdU

,,

,,,,

Vz

V

y

V

x

VF

dzz

Vdy

y

Vdx

x

VdzFdyFdxF zyx

grad

Conservation of

Energy



47

Work of a conservative force,

2121 VVU

Concept of work & energy,

1221 TTU

constant

2211

VTE

VTVT

WVT

WVT

11

11 0

WVT

VWgg

WmvT

22

2222

12 02

2

1

Conservation of

Energy



48

When a particle moves under the action of

conservative forces, total mechanical energy is

constant.

Friction forces are not conservative. Total

mechanical energy of a system involving friction

decreases.

Mechanical energy is dissipated by friction into

thermal energy. Total energy is constant.

Motion Under a Conservative Central Force



49

Both the principle of conservation of angular

momentum & the principle of conservation of

energy may be applied.

sinsin 000 rmvmvr

r

GMmmv

r

GMmmv

VTVT

221

0

202

1

00

Sample Problem 13.6



50

A 20 lb collar slides without

friction along a vertical rod as

shown. The spring attached to

the collar has an undeflected

length of 4 in. and a constant

of 3 lb/in.

If the collar is released from

rest at position 1, determine its

velocity after it has moved 6

in. to position 2.

Sample Problem 13.6



51

Solution:

Apply principle of conservation of energy between

positions 1 & 2.

Position 1:

0 lb,ft20lbin.24

lbin.24in. 4in. 8in.lb3

11

2

212

121

TVVV

kxV

ge

e

Sample Problem 13.6



52

Position 2:

22

22

222

12

2

2

212

221

311.02.32

20

2

1

lbft 5.5lbin. 6612054

lbin. 120in. 6lb 20

lbin.54in. 4in. 01in.lb3

vvmvT

VVV

WyV

kxV

ge

g

e

Conservation of Energy:

lbft 5.50.311lbft 20 22

2211

v

VTVT

sft91.42v

Sample Problem 13.7



53

The 0.5 lb pellet is pushed against the spring and

released from rest at A. Neglecting friction,

determine the smallest deflection of the spring for

which the pellet will travel around the loop and

remain in contact with the loop at all times.

Sample Problem 13.7



54

Solution:

Setting the force exerted by the loop to zero, solve

for min velocity at D.

:nn maF

222

2

sft4.64sft32.2ft 2

rgv

rvmmgmaW

D

Dn

Apply principle of conservation of energy

between points A & D.

0 ,18ftlb36

0

1

22

21

2

21

1

Txx

kxVVV ge

Sample Problem 13.7



55

lbft5.0

sft4.64sft2.32

lb5.0

2

1

lbft2

ft4lb5.00

22

2

2

21

2

2

D

ge

mvT

WyVVV

25.0180 2

2211

x

VTVT

in. 47.4ft 3727.0 x


55, 61, 67, 76, 95

Due Wednesday

12/3/2014



56

221121 , of impulse 2

1

vmvmFdtF

t

t

ImpImp

12

2

1

, vmvmdtFvmddtF

t

t

Principle of Impulse & Momentum



57

From Newton’s 2nd law,

vmvmdt

dF

linear momentum

Principle of Impulse & Momentum



58

Final momentum of particle = vector sum of its

initial momentum and the impulse of force during

the time interval.

Units for the impulse of a force are

smkgssmkgsN 2

Impulsive Motion



59

Impulsive force: Force acting on a particle during

a very short time interval that is large enough to

cause a significant change in momentum

When impulsive forces act on a particle,

21 vmtFvm

Impulsive Motion



60

21 vmtFvm

When a baseball is struck by a bat, contact occurs

over a short time interval but force is large

enough to change sense of ball motion.

Non-impulsive forces: forces for which the

impulse is small and therefore, may be neglected.

Sample Problem 13.10



61

An automobile weighing 4000 lb is driven down a 5o

incline at a speed of 60 mph when the brakes are

applied, causing a constant total braking force of

1500 lb.

Determine the time required for the automobile to

come to a stop.




62

Solution:

Apply the principle of impulse & momentum

2211 vmvm

Imp

015005sin4000sft882.32

4000

05sin1

tt

FttWmv

Taking components parallel to the incline,

s49.9t




63

A 4 oz baseball is pitched

with a velocity of 80 ft/s.

After the ball is hit by the

bat, it has a velocity of

120 ft/s in the direction

shown. If the bat and

ball are in contact for

0.015 s, determine the

average impulsive force

exerted on the ball during

the impact.

Sample

Problem 13.11



64

Solution:

Apply the principle of impulse

& momentum in terms of

horizontal & vertical component

equations.

2211 vmvm

Imp

x

y

x component:

lb89

40cos1202.32

16415.080

2.32

164

40cos21

x

x

x

F

F

mvtFmv

Sample

Problem 13.11



65

x

y

y component:

lb9.39

40cos1202.32

16415.0

40sin0 2

y

y

y

F

F

mvtF

lb5.97,lb9.39lb89 FjiF




66

A 10 kg package drops from a

chute into a 24 kg cart with a

velocity of 3 m/s. Knowing

that the cart is initially at rest

and can roll freely, determine

a. the final velocity of the cart

b. the impulse exerted by the

cart on the package

c. the fraction of the initial

energy lost in the impact




67

Solution:

Apply the principle of impulse & momentum to

package-cart system to determine the final velocity.

2211 vmmvm cpp

Imp

x

y

x components: 2

21

kg 25kg 1030cosm/s 3kg 10

030cos

v

vmmvm cpp

m/s 742.02 v




68

Apply same principle to package alone to determine

impulse exerted on it from change in its momentum.

x

y

2211 vmvm pp

Imp

x components:

2

21

kg 1030cosm/s 3kg 10

30cos

vtF

vmtFvm

x

pxp

sN56.18 tFx




69

x

y

y components:

030sinm/s 3kg 10

030sin1

tF

tFvm

y

yp

sN15 tFy

sN 9.23

sN 51sN 56.1821

tF

jitF

Imp




70

To determine the fraction of energy lost,

J 63.9sm742.0kg 25kg 10

J 45sm3kg 10

2

212

221

2

2

212

121

1

vmmT

vmT

cp

p

786.0J 45

J 9.63J 45

1

21

T

TT


119, 129, 136, 149, 154

Due Monday 18/3/2014



71

Impact Friday, February 28, 2014


72

• Impact: Collision between two

bodies which occurs during a

small time interval and during

which the bodies exert large

forces on each other.

• Line of Impact: Common

normal to the surfaces in

contact during impact.

Direct Central Impact

Oblique Central Impact



73

• Central Impact: Impact for

which the mass centers of the

two bodies lie on the line of

impact; otherwise, it is an

eccentric impact..

• Direct Impact: Impact for

which the velocities of the two

bodies are directed along the

line of impact.





74

• Oblique Impact: Impact for

which one or both of the bodies

move along a line other than the

line of impact.



Direct Central

Impact



75

Bodies moving in same

straight line, vA > vB .

Upon impact the bodies

undergo a period of

deformation, at end of which,

they are in contact and moving

at a common velocity.

A period of restitution follows

during which the bodies either

regain their original shape or

remain permanently deformed.

Direct Central

Impact



76

Total momentum of the two

body system is preserved,

BBAABBAA vmvmvmvm

A second relation between

final velocities is required.




77

• Period of deformation: umPdtvm AAA

• Period of restitution: AAA vmRdtum

10

e

uv

vu

Pdt

Rdtnrestitutio of tcoefficien e

A

A




78

A similar analysis of particle B yields B

B

vu

uv e

Combining the relations leads to the desired

second relation between the final velocities.

BAAB vvevv

Perfectly plastic impact, e = 0: vvv AB

vmmvmvm BABBAA

Perfectly elastic impact, e = 1:

Total energy and total momentum conserved.

BAAB vvvv




79

No tangential impulse component; tangential

component of momentum for each particle is

conserved. tBtBtAtA vvvv

Normal component of total momentum of the two

particles is conserved.

nBBnAAnBBnAA vmvmvmvm




80

Final velocities are unknown in magnitude &

direction.

Four equations are required.

Normal components of relative velocities before

and after impact are related by the coefficient of

restitution.

nBnAnAnB vvevv




81

Block constrained to move along horizontal surface.

Impulses from internal forces along the n

axis and from external force exerted by

horizontal surface and directed along the vertical to

the surface.

FF

and

extF




82

Tangential momentum of ball is conserved. tBtB vv

Total horizontal momentum of block & ball is

conserved.

xBBAAxBBAA vmvmvmvm

Normal component of relative velocities of block

and ball are related by coefficient of restitution.

nBnAnAnB vvevv

Problems Involving Energy and

Momentum



83

Three methods for analysis of kinetics problems:

1. Direct application of Newton’s 2nd law

2. Method of work & energy

3. Method of impulse & momentum




84

A ball is thrown against a

frictionless, vertical wall.

Immediately before the ball

strikes the wall, its velocity

has a magnitude v and forms

angle of 30o with the

horizontal. Knowing that

e = 0.90, determine the

magnitude and direction of

the velocity of the ball as it

rebounds from the wall.




85

Component of ball momentum

tangential to wall is conserved.

vvv tt 500.0

Solution:

Resolve ball velocity into components

parallel & perpendicular to wall

vvv

vvv

t

n

500.030sin

866.030cos

n

t




86

Apply coefficient of restitution

relation with zero wall velocity.

vvv

vev

n

nn

779.0866.09.0

00

n

t

7.32500.0

779.0tan926.0

500.0779.0

1vv

vvv tn




87

The magnitude & direction of the velocities of two

identical frictionless balls before they strike each other

are as shown. Assuming e = 0.9, determine the

magnitude & direction of the velocity of each ball after

the impact.

Solution:

Resolve ball velocities into components normal and

tangential to the contact plane.

sft0.2630cos AnA vv

sft0.1530sin AtA vv

sft0.2060cos BnB vv

sft6.3460sin BtB vv




88

• Tangential component of momentum for each ball is

conserved. sft0.15tAtA vv

sft6.34tBtB vv

Total normal component of

momentum of the two ball system

is conserved.

0.6

0.200.26

nBnA

nBnA

nBBnAAnBBnAA

vv

vmvmmm

vmvmvmvm




89




90

Normal relative velocities of

the balls are related by the

coefficient of restitution.

4.41

0.200.2690.0

nBnAnBnA vvevv

Solve the last two equations simultaneously for

the normal velocities of the balls after the impact.

sft7.17nAv sft7.23

nBv




91

6.557.23

6.34tansft9.41

6.347.23

3.407.17

0.15tansft2.23

0.157.17

1

1

B

ntB

A

ntA

v

v

v

v

t

n




92

Ball B is hanging from an

inextensible cord. An identical ball

A is released from rest when it is

just touching the cord and acquires

a velocity v0 before striking ball B.

Assuming perfectly elastic impact

(e = 1) and no friction, determine

the velocity of each ball

immediately after impact.

Sample Problem

13.16



93

Solution:

Determine orientation of

impact line of action.

30

5.02

sin

r

r

Momentum component of

ball A tangential to the

contact plane is

conserved.

0

0

5.0

030sin

vv

vmmv

vmtFvm

tA

tA

AA




94

0

0

433.05.0

30sin30cos5.00

30sin30cos0

vvv

vvv

vmvmvm

vmvmtTvm

BnA

BnA

BnAtA

BAA

Momentum x-component of the two ball system is

conserved.




95

Relative velocities along the

line of action before and after

the impact are related by the

coefficient of restitution.

0

0

866.05.0

030cos30sin

vvv

vvv

vvevv

nAB

nAB

nBnAnAnB

Solve last two expressions

for the velocity of ball A

along the line of action and

the velocity of ball B which

is horizontal.

00 693.0520.0 vvvv BnA

0

10

00

693.0

1.16301.46

1.465.0

52.0tan721.0

520.05.0

vv

vv

vvv

B

A

ntA




96




97

A 30 kg block is dropped

from a height of 2 m onto

the 10 kg pan of a spring

scale. Assuming the impact

to be perfectly plastic,

determine the maximum

deflection of the pan. The

spring constant is k = 20

kN/m.




98

Solution:

Apply principle of conservation of energy to

determine velocity of block at instant of impact.

sm26.6

030 J 5880

030

J 588281.9300

2

2

221

2211

2

2

2212

221

2

11

A

A

AAA

A

v

v

VTVT

VvvmT

yWVT




99

Determine velocity after impact: total momentum of

the block and pan is conserved.

sm70.41030026.630 33

322

vv

vmmvmvm BABBAA

Sample Problem

13.17



100

Initial spring deflection

due to pan weight:

m1091.4

1020

81.910 3

33

k

Wx B

Apply principle of conservation

of energy to determine

maximum deflection of spring.

2

43

213

4

24

3

21

34

242

14

4

233

212

321

3

2

212

321

3

10201091.4392

1020392

0

J 241.01091.410200

J 4427.41030

xx

xxx

kxhWWVVV

T

kx

VVV

vmmT

BAeg

eg

BA




101

m 230.0

10201091.43920241.0442

4

24

3

213

4

4433

x

xx

VTVT

m 1091.4m 230.0 334

xxh m 225.0h


155, 162, 168, 175, 189

Due Saturday 22/3/2014



102

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