king fahd university of petroleum & minerals

King Fahd University of Petroleum & Minerals

Mechanical EngineeringDynamics ME 201

BYDr. Meyassar N. Al-Haddad

Lecture # 36

Chapter 18Planar Kinetics of Rigid Bodies:

Work and Energy

Kinetic Energy

ji

jikji

vvv

])[(])[(

)()()(

)1(2

12

1

/

2

2

ωxvyv

yxωvv

vmdT

vmdT

yPxP

yPxP

PiPi

m i

ii

0 )(

2

1)()(

2

12

1)()(

2

1

)1()2(

)2()(2)(2

)(2)()(2)(

])[(])[(

22

222

222

222222

222

xycentermassP

ωImxωvmyωvvm

dmrωdmxωvdmyωvvdmT

rωxvyvv

xωxvvyωyvv

xvyvv

G

pyPxPP

mmyPmxPPm

yPxPP

yPyPxPxP

yPxPiii

vv

22

2

1

2

1ωIvmT GG

2

2

1GvmT 22

2

1

2

1ωIvmT GG

energyKineticenergyKinetic

RotationalnalTranslatio2

1

2

1 22 ωIvmT GG ωrv GG

222

2

1

2

1ωIωrmT GG

22 )(2

1ωrmI GG

2

2

1ωIO

Translation Rotation General

Rotation

2

2

1OOIT

22 )2

1(

2

1OmrT

General Plane Motion (Translation and Rotation)

22

2

1

2

1ωIvmT BB

Example

K.E. of the Body of Bike and man ( translational )

2)(2

1 manBManB mmT

K.E of the wheel ( Translational + Rotational )

22

2

1

2

1 Owheel ImT

Total K.E. = 2 Twheel +T B+Man

Total K.E? .

Example 18-1 mB=6 kgmD=10 kgmC=12 kgNo slippingTotal K.E = ?

Block (Translation)

J92.1)/8.0)(6(2

1

2

1 22 smkgmT BBB

Disk (Rotation)222 )

2

1(

2

1

2

1DDDAAD rmIT

J60.1)/8]()1.0)(10(2

1[

2

1 22 sradmkgTD

Cylinder ( Translation &Rotation)22222 )

2

1(

2

1

2

1

2

1CCCGCGGC rmmImT

J44.1)/4]()1.0)(12(2

1[

2

1)/4.0)(10(

2

1 222 sradmkgsmkgTC

CDBTotal TTTT J96.444.16.192.1 TotalT

sradrD

BD /8

1.0

8.0

sradrC

BC /4

2.0

8.0

2

ProblemwAB=wCD=10 IbwBC=20 IbTotal K.E = ?

ICCICB rr //

ft/s5)1)(/5( ftsradCGB

sradft

sft

rCD

CCD /5

1

/5

ft.Ib294.1)5]()1)(2.32

10(

3

1[

2

1

2

1 222 ABAAB IT

ft.Ib764.70)5)](2.32

20[

2

1

2

1

2

1 222 BCGGBC ImT

ft.Ib294.1)5]()1)(2.32

10(

3

1[

2

1

2

1 222 CDDCD IT

ft.Ib4.10294.1764.7294.1 TT

The work of a force

forcevariableaofWork

dsθFUsF cos

forceconstantaofWork

sθFU ccF )cos(

::

::ntdisplacemevertical:

weightaofWork

W

W

U

Uy

yWUW

• Work of a Spring Force

)2

1

2

1( 2

12

2 ksksU s

12 sswhere

Force Perpendicular to the displacement-Weight-Normal reaction

Force That Do No Work

Rolling resistance force Fr

Fr acts at point which has zero velocityThe point is not displaced

Pin Reaction

Ot

On

Oy

Ox

Pin Reaction Do No workAct on zero velocity and create no displacement

The Work of a couple

-Do work only when the body undergoes a rotation

-Translation : positive work cancels the negative work

-Rotation

θdM

θdrFθdr

Fθdr

FdUM

)(

22

)( 122

1θθMdθMU

θ

θM

dθr

Sd θ 2

m =10 kg M = 50 NmP = 80N, unstretched 0.5 m

total work? 90~0θ

J2.147)5.1)(81.9(10

Example 18-2

)(2

1)

2

1

2

1( 2

12

22

12

2 sskksksU s ms

ms

25.25.0)275.0(

25.05.075.0

2

1

SPUP .

J5280 PSMWAA UUUUUUUyx

J75)25.025.2(302

1 22 SU

mπrπrπ

ArchS 712.42

3

24

2)(

JmNUP 377712.480

J5.78)02

(50

)( 12 θθMUM

)( ygmUW