CE

E27

1:A

pplie

dM

echa

nics

II,D

ynam

ics

–Le

ctur

e25

:Ch.

17,S

ec.4

-5–

Pro

f.A

lber

tS.K

im

Civ

ilan

dE

nviro

nmen

talE

ngin

eerin

g,U

nive

rsity

ofH

awai

iatM

anoa

Dat

e:__

____

____

____

____

1/3

6

EQ

UAT

ION

SO

FM

OTI

ON

:RO

TATI

ON

AB

OU

TA

FIX

ED

AX

IS

Toda

y’s

obje

ctiv

es:S

tude

nts

will

beab

leto

1A

naly

zeth

epl

anar

kine

tics

ofa

rigid

body

unde

rgoi

ngro

tatio

nalm

otio

n.

In-c

lass

activ

ities

:•

Rea

ding

Qui

z•

App

licat

ions

•R

otat

ion

Abo

utan

Axi

s•

Equ

atio

nsof

Mot

ion

•C

once

ptQ

uiz

•G

roup

Pro

blem

Sol

ving

•A

ttent

ion

Qui

z

2/3

6

RE

AD

ING

QU

IZ

1In

rota

tiona

lmot

ion,

the

norm

alco

mpo

nent

ofac

cele

ratio

nat

the

body

’sce

nter

ofgr

avity

(G)i

sal

way

s.

(a)

zero

(b)

tang

entt

oth

epa

thof

mot

ion

ofG

(c)

dire

cted

from

Gto

war

dth

ece

nter

ofro

tatio

n(d

)di

rect

edfro

mth

ece

nter

ofro

tatio

nto

war

dG

AN

S:(

c)2

Ifa

rigid

body

rota

tes

abou

tpoi

ntO

,the

sum

ofth

em

omen

tsof

the

exte

rnal

forc

esac

ting

onth

ebo

dyab

out

poin

tOeq

uals

?(a

)IGα

(b)IOα

(c)maG

(d)maO

AN

S:(

b)

3/3

6

AP

PLI

CAT

ION

S

Pin

at th

e ce

nter

of

rota

tion.

•Th

ecr

ank

onth

eoi

l-pum

prig

unde

rgoe

sro

tatio

nab

outa

fixed

axis

,cau

sed

byth

edr

ivin

gto

rque

,M

,fro

ma

mot

or.

•A

sth

ecr

ank

turn

s,a

dyna

mic

reac

tion

ispr

oduc

edat

the

pin.

This

reac

tion

isa

func

tion

ofan

gula

rve

loci

ty,a

ngul

arac

cele

ratio

n,an

dth

eor

ient

atio

nof

the

cran

k.•

Ifth

em

otor

exer

tsa

cons

tant

torq

ueM

onth

ecr

ank,

does

the

cran

ktu

rnat

aco

nsta

ntan

gula

rvel

ocity

?Is

this

desi

rabl

efo

rsuc

ha

mac

hine

?

4/3

6

AP

PLI

CAT

ION

S(c

ontin

ued)

•Th

epe

ndul

umof

the

Cha

rpy

impa

ctm

achi

neis

rele

ased

from

rest

whe

nθ=

0◦.I

tsan

gula

rve

loci

ty(ω

)beg

ins

toin

crea

se.

•C

anw

ede

term

ine

the

angu

lar

velo

city

whe

nit

isin

verti

cal

posi

tion?

•O

nw

hich

prop

erty

(P)o

fthe

pend

ulum

does

the

angu

lar

acce

lera

tion

(α)d

epen

d?•

Wha

tis

the

rela

tions

hip

betw

een

Pan

dα

?

5/3

6

EQ

UAT

ION

SO

FM

OTI

ON

:RO

TATI

ON

AB

OU

TA

FIX

ED

AX

IS(S

ectio

n17

.4)

•W

hen

arig

idbo

dyro

tate

sab

outa

fixed

axis

perp

endi

cula

rto

the

plan

eof

the

body

atpo

intO

,the

body

’sce

nter

ofgr

avity

Gm

oves

ina

circ

ular

path

ofra

dius

rG

.Thu

s,th

eac

cele

ratio

nof

poin

tGca

nbe

repr

esen

ted

bya

tang

entia

lcom

pone

nt(a

G) t=

rGα

and

ano

rmal

com

pone

nt(a

G) n

=rGω2.

•S

ince

the

body

expe

rienc

esan

angu

lara

ccel

erat

ion,

itsin

ertia

crea

tes

am

omen

tofm

agni

tude

,IGα

,equ

alto

the

mom

ento

fthe

exte

rnal

forc

esab

outp

oint

G.T

hus,

the

scal

areq

uatio

nsof

mot

ion

can

best

ated

as:

ΣFn=

m(a

G) n

=mrGω2

ΣFt=

m(a

G) t

=mrGα

MG

=IGα

6/3

6

EQ

UAT

ION

SO

FM

OTI

ON

(con

tinue

d)

•N

ote

that

theΣM

Gm

omen

tequ

atio

nm

aybe

repl

aced

bya

mom

ents

umm

atio

nab

outa

nyar

bitra

rypo

int.

Sum

min

gth

em

omen

tabo

utth

ece

nter

ofro

tatio

nO

yiel

dsΣM

O=

IGα+rGm(a

G) t=

[IG+m(r

G)2]α

•Fr

omth

epa

ralle

laxi

sth

eore

m,I

O=

IG+

m(r

G)2

,th

eref

ore

the

term

inpa

rent

hese

sre

pres

ents

IO

.

•C

onse

quen

tly,w

eca

nw

rite

the

thre

eeq

uatio

nsof

mot

ion

fort

hebo

dyas

:

ΣFn=

m(a

G) n

=mrGω2

ΣFt=

m(a

G) t

=mrGα

MO

=IOα

7/3

6

PR

OC

ED

UR

EFO

RA

NA

LYS

IS

•P

robl

ems

invo

lvin

gth

eki

netic

sof

arig

idbo

dyro

tatin

gab

outa

fixed

axis

can

beso

lved

usin

gth

efo

llow

ing

proc

ess.

1E

stab

lish

anin

ertia

lcoo

rdin

ate

syst

eman

dsp

ecify

the

sign

and

dire

ctio

nof

(aG) n

and(a

G) t

.2

Dra

wa

free

body

diag

ram

acco

untin

gfo

rall

exte

rnal

forc

esan

dco

uple

s.S

how

the

resu

lting

iner

tiafo

rces

and

coup

le(ty

pica

llyon

ase

para

teki

netic

diag

ram

).3

Com

pute

the

mas

smom

ento

fine

rtiaIG

orIO

.4

Writ

eth

eth

ree

equa

tions

ofm

otio

nan

did

entif

yth

eun

know

ns.S

olve

fort

heun

know

ns.

5U

seki

nem

atic

sif

ther

ear

em

ore

than

thre

eun

know

ns(s

ince

the

equa

tions

ofm

otio

nal

low

foro

nly

thre

eun

know

ns).

8/3

6

DE

PE

ND

EN

TM

OTI

ON

EX

AM

PLE

•G

iven

:The

unifo

rmsl

ende

rrod

has

am

ass

of15

kg.

•Fi

nd:

The

reac

tions

atth

epi

nO

and

the

angu

lara

ccel

erat

ion

ofth

ero

dju

staf

tert

heco

rdis

cut.

•P

lan:

Sin

ceth

em

ass

cent

er,G

,mov

esin

aci

rcle

ofra

dius

0.15

m,i

t’sac

cele

ratio

nha

sa

norm

alco

mpo

nent

tow

ardO

and

ata

ngen

tialc

ompo

nent

actin

gdo

wnw

ard

and

perp

endi

cula

rtorG

.App

lyth

epr

oble

mso

lvin

gpr

oced

ure.

9/3

6

EX

AM

PLE

(Sol

utio

n)

•FB

Dan

dK

inet

icD

iagr

am

= r G

•E

quat

ions

ofm

otio

n:

(+ →)Σ

Fn=

man=

mrGω2

⇒O

x=

0N

(+↓)ΣFt=

mat=

mrGα

⇒−O

y+15

(9.81)

=15

(0.15)α

(�+)Σ

MO

=IGα+mrGα(r

G)

⇒(0.15)15

(9.81)

=IGα+m(r

G)2

•U

sing

IG=

(ml2)/12

andrG=

(0.15)

,we

can

writ

e:IGα+m(r

G)2α=

[(15

×0.92)/12

+15

(0.15)

2]α

=1.35α

10/3

6

EX

AM

PLE

(con

tinue

d)

•A

fters

ubst

itutin

g:

22.07=

1.35α⇒

α=

16.4rad/s2

•Fr

omth

ese

cond

equa

tion

inth

ela

stpa

ge:

−O

y+15

(9.81)

=15

(0.15)α

Oy

=15

(9.81)

−15

(0.15)16

.4=

110N

11/3

6

CO

NC

EP

TQ

UIZ

1If

arig

idba

rofl

engt

hl

isre

leas

edfro

mre

stin

the

horiz

onta

lpos

ition

(θ=

0),t

hem

agni

tude

ofits

angu

lar

acce

lera

tion

isat

max

imum

whe

n (a)θ=

0

(b)θ=

90o

(c)θ=

180o

(d)θ=

0oan

dθ=

180o

AN

S:(

d)2

Inth

eab

ove

prob

lem

,whe

nθ=

90o,t

heho

rizon

tal

com

pone

ntof

the

reac

tion

atpi

nA

is.

(a)

zero

(b)mg

(c)

1 2mω2

(d)

Non

eof

the

abov

eA

NS

:(a)

12/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

•G

iven

:Wsp

here=

30lb

,W

rod=

10lb

•Fi

nd:T

here

actio

nat

the

pinO

just

afte

rthe

cord

AB

iscu

t.•

Pla

n:D

raw

the

free

body

diag

ram

and

kine

ticdi

agra

mof

the

rod

and

sphe

reas

one

unit.

Then

appl

yth

eeq

uatio

nsof

mot

ion.

13/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

(Sol

utio

n)

•FB

Dan

dki

netic

diag

ram

;30lb

=

OyO

x

10lb

mrod(1.0α)

IG,rodα

mrod(1.0)(0)

2IG,sphereαm

sphere(3)(0)

2

msp

here(3α)

•E

quat

ions

ofm

otio

nfo

rΣFn=

m(a

G) n

:

Ox

=(30/

32.2)(3)(0)2

+(10/

32.2)(1.0)(0)2

⇒O

x=

0lb

14/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

(con

tinue

d)

30lb

=

OyO

x

10lb

mrod(1.0α)

IG,rodα

mrod(1.0)(0)

2IG,sphereαm

sphere(3)(0)

2

msp

here(3α)

•ΣFt=

m(a

G) t

:⇒O

y=

40−3.10

6αfro

m

−O

y+30

+10

=(30/

32.2)(3α

)+(10/

32.2)(1.0α

)

•M

O=

Ioα

:⇒10

0=

9.17

2αfro

m

30(3.0)+10

(1.0)

=[0.4(30/

32.2)(1)

2+(30/

32.2)(3)

2] sphereα

+[(1/

12)(10/32.2)(2)

2+(10/

32.2)(1)

2] rodα

•Th

eref

ore,

α=

10.9rad/s2

andO

y=

6.14

lb15

/36

CO

NC

EP

TQ

UIZ

(p.4

30)

1A

drum

ofm

assm

isse

tint

om

otio

nin

two

way

s:(a

)by

aco

nsta

nt40

Nfo

rce

(left)

,and

,(b)

bya

bloc

kof

wei

ght4

0N

(rig

ht).

Ifαa

andαb

repr

esen

tthe

angu

lara

ccel

erat

ion

ofth

edr

umin

each

case

,sel

ectt

hetru

est

atem

ent.

(a)

(a)αa>

αb

(b)αa<

αb

(c)αa=

αb

(d)

Non

eof

the

abov

e.

2In

case

(b)a

bove

(rig

htfig

ure)

,wha

tis

the

tens

ionT

inth

eca

ble?

(b)

(a)T

=40

N(b

)T

<40

N(c

)T

>40

N(d

)N

one

ofth

eab

ove.

16/3

6

EQ

UAT

ION

SO

FM

OTI

ON

:GE

NE

RA

LP

LAN

EM

OTI

ON

Toda

y’s

obje

ctiv

es:S

tude

nts

will

beab

leto

1A

naly

zeth

epl

anar

kine

tics

ofa

rigid

body

unde

rgoi

ngge

nera

lpla

nem

otio

n.

In-c

lass

activ

ities

:•

Rea

ding

Qui

z•

App

licat

ions

•E

quat

ions

ofM

otio

n•

Fric

tiona

lRol

ling

Pro

blem

s•

Con

cept

Qui

z•

Gro

upP

robl

emS

olvi

ng•

Atte

ntio

nQ

uiz

17/3

6

RE

AD

ING

QU

IZ

1If

adi

skro

llson

aro

ugh

surfa

cew

ithou

tslip

ping

,the

acce

lera

tion

atth

ece

nter

ofgr

avity

(G)w

illan

dth

efri

ctio

nfo

rce

will

be.

(a)

notb

eeq

ualt

oαr;l

ess

than

µsN

(b)

beeq

ualt

oαr;e

qual

toµkN

(c)

beeq

ualt

oαr;l

ess

than

µsN

(d)

Non

eof

the

abov

eA

NS

:(c)

2If

arig

idbo

dyex

perie

nces

gene

ralp

lane

mot

ion,

the

sum

ofth

em

omen

tsof

exte

rnal

forc

esac

ting

onth

ebo

dyab

out

any

poin

tPis

equa

lto

.(a

)IPα

(b)IPα+

maP

(c)maG

(d)IGα+

rGP×maP

AN

S:(

a)

18/3

6

AP

PLI

CAT

ION

S

•A

sth

eso

ilco

mpa

ctor

acce

lera

tes

forw

ard,

the

front

rolle

rexp

erie

nces

gene

ralp

lane

mot

ion

(bot

htra

nsla

tion

and

rota

tion)

.•

Wha

tare

the

load

sex

perie

nced

byth

ero

llers

haft

orbe

arin

gs?

•Th

efo

rces

show

non

the

rolle

r’sFB

Dca

use

the

acce

lera

tions

show

non

the

kine

ticdi

agra

m.I

sth

epo

intA

theIC

?

19/3

6

AP

PLI

CAT

ION

S(c

ontin

ued)

•Th

ela

wn

rolle

ris

push

edfo

rwar

dw

itha

forc

eof

200N

whe

nth

eha

ndle

isat

45◦ .

•H

owca

nw

ede

term

ine

itstra

nsla

tion

acce

lera

tion

and

angu

lara

ccel

erat

ion?

•D

oes

the

acce

lera

tion

depe

ndon

the

coef

ficie

nts

ofst

atic

and

kine

ticfri

ctio

n?20

/36

AP

PLI

CAT

ION

S(c

ontin

ued)

•D

urin

gan

impa

ct,t

hece

nter

ofgr

avity

ofth

iscr

ash

dum

my

will

dece

lera

tew

ithth

eve

hicl

e,bu

tal

soex

perie

nce

anot

her

acce

lera

tion

due

toits

rota

tion

abou

tpoi

ntA

.•

Why

?

•H

owca

nen

gine

ers

use

this

info

rmat

ion

tode

term

ine

the

forc

esex

erte

dby

the

seat

belt

ona

pass

enge

rdur

ing

acr

ash?

21/3

6

GE

NE

RA

LP

LAN

EM

OTI

ON

(Sec

.17.

5)

•W

hen

arig

idbo

dyis

subj

ecte

dto

exte

rnal

forc

esan

dco

uple

-m

omen

ts,i

tcan

unde

rgo

both

trans

latio

nalm

otio

nan

dro

tatio

nal

mot

ion.

This

com

bina

tion

isca

lled

gene

ralp

lane

mot

ion.

•U

sing

anx−y

iner

tialc

oord

inat

esy

stem

,the

equa

tions

ofm

otio

nsab

outt

hece

nter

ofm

ass,

G,m

aybe

writ

ten

as: ΣFx

=m(a

G) x

(1)

ΣFy

=m(a

G) y

(2)

MG

=IGα

(3) 22

/36

GE

NE

RA

LP

LAN

EM

OTI

ON

(con

tinue

d)

•S

omet

imes

,itm

aybe

conv

enie

ntto

writ

eth

em

omen

tequ

atio

nab

outa

poin

tPot

hert

hanG

.The

nth

eeq

uatio

nsof

mot

ion

are

writ

ten

asfo

llow

s:

ΣFx=

m(a

G) x

(4)

ΣFy=

m(a

G) y

(5)

MP=

Σ(M

k) P

(6)

•In

this

case

,Σ(M

k) P

repr

esen

tsth

esu

mof

the

mom

ents

ofIGα

and

r G/P×

maG

abou

tpoi

ntP

.

23/3

6

FRIC

TIO

NA

LR

OLL

ING

PR

OB

LEM

S

•W

hen

anal

yzin

gth

ero

lling

mot

ion

ofw

heel

s,cy

linde

rs,o

rdi

sks,

itm

ayno

tbe

know

nif

the

body

rolls

with

outs

lippi

ngor

ifit

slid

esas

itro

lls. •

Fore

xam

ple,

cons

ider

adi

skw

ithm

assm

and

radi

usr,s

ubje

cted

toa

know

nfo

rceP

.•

The

equa

tions

ofm

otio

nw

illbe

:ΣFx=

m(a

G) x

⇒P

−F

=maG

ΣFy=

m(a

G) y

⇒N

−mg=

0ΣM

G=

IGα⇒

Fr=

IGα

•Th

ere

are

4un

know

ns(F

,N,a,a

ndaG

)in

thes

eth

ree

equa

tions

.

24/3

6

FRIC

TIO

NA

LR

OLL

ING

PR

OB

LEM

S(c

ontin

ued)

•H

ence

,we

have

tom

ake

anas

sum

ptio

nto

prov

ide

anot

her

equa

tion.

Then

,we

can

solv

efo

rth

eun

know

ns.

•Th

e4t

heq

uatio

nca

nbe

obta

ined

from

the

slip

orno

n-sl

ipco

nditi

onof

the

disk

.

•C

ase

1:A

ssum

eno

slip

ping

and

useaG=

αr

asth

e4t

h

equa

tion

and

DO

NO

Tus

eFf=

µsN

.Afte

rsol

ving

,you

will

need

tove

rify

that

the

assu

mpt

ion

was

corr

ectb

ych

ecki

ngifFf≤

µsN

.•

Cas

e2:

Ass

ume

slip

ping

and

useFf=

µkN

asth

e4t

h

equa

tion.

Inth

isca

se,a

G�=

αr.

25/3

6

PR

OC

ED

UR

EFO

RA

NA

LYS

IS

Pro

blem

sin

volv

ing

the

kine

tics

ofa

rigid

body

unde

rgoi

ngge

nera

lpla

nem

otio

nca

nbe

solv

edus

ing

the

follo

win

gpr

oced

ure.

1.E

stab

lish

thex−y

iner

tialc

oord

inat

esy

stem

.Dra

wbo

thth

efre

ebo

dydi

agra

man

dki

netic

diag

ram

fort

hebo

dy.

2.S

peci

fyth

edi

rect

ion

and

sens

eof

the

acce

lera

tion

ofth

em

ass

cent

er,a

G,a

ndth

ean

gula

racc

eler

atio

n,α

,oft

hebo

dy.I

fnec

essa

ry,c

ompu

teth

ebo

dy’s

mas

sm

omen

tof

iner

tiaIG

.3.

Ifth

em

omen

tequ

atio

nM

P=

Σ(M

k) P

isus

ed,u

seth

eki

netic

diag

ram

tohe

lpvi

sual

ize

the

mom

ents

deve

lope

dby

the

com

pone

ntsm(a

G) x

,m(a

G) y

,and

IGα

.4.

App

lyth

eth

ree

equa

tions

ofm

otio

n.

26/3

6

PR

OC

ED

UR

EFO

RA

NA

LYS

IS(c

ontin

ued)

5.Id

entif

yth

eun

know

ns.I

fnec

essa

ry(i.

e.,t

here

are

four

unkn

owns

),m

ake

your

slip

orno

-slip

assu

mpt

ion

(typi

cally

nosl

ippi

ng,o

rthe

use

ofaG=

αr,i

sas

sum

edfir

st).

6.U

seki

nem

atic

equa

tions

asne

cess

ary

toco

mpl

ete

the

solu

tion.

7.If

asl

ip-n

osl

ipas

sum

ptio

nw

asm

ade,

chec

kits

valid

ity!!!

Key

poin

tsto

cons

ider

1B

eco

nsis

tent

inus

ing

the

assu

med

dire

ctio

ns.T

hedi

rect

ion

ofaG

mus

tbe

cons

iste

ntw

ithα

.2

IfFf=

µkN

isus

ed,F

fm

usto

ppos

eth

em

otio

n.A

sa

test

,as

sum

eno

frict

ion

and

obse

rve

the

resu

lting

mot

ion.

This

may

help

visu

aliz

eth

eco

rrec

tdire

ctio

nof

Ff.

27/3

6

DE

PE

ND

EN

TM

OTI

ON

EX

AM

PLE

•G

iven

:Asp

oolh

asa

mas

sof

200

kgan

da

radi

usof

gyra

tion

(kG

)of0

.3m

.The

coef

ficie

ntof

kine

ticfri

ctio

nbe

twee

nth

esp

oola

ndth

egr

ound

isµk=

0.1.

•Fi

nd:

The

angu

lara

ccel

erat

ion

(α)o

fthe

spoo

land

the

tens

ion

inth

eca

ble.

•P

lan:

Focu

son

the

spoo

l.Fo

llow

the

solu

tion

proc

edur

e(d

raw

aFB

D,e

tc.)

and

iden

tify

the

unkn

owns

.

28/3

6

EX

AM

PLE

(Sol

utio

n)

•Th

efre

ebo

dydi

agra

man

dki

netic

diag

ram

fort

hebo

dyar

e:

I G a

ma G

=

1962

N

•E

quat

ions

ofm

otio

n:

ΣFy=

m(a

G) y

:N

B−

(200

kg)(9.8m/s2)=

0

⇒N

B=

1962

N(7

)

29/3

6

EX

AM

PLE

(con

tinue

d)

•N

ote

that

aG=

(0.4)α

.Why

?•ΣFx=

m(a

G) x

:

T−0.1N

B=

200a

G=

200(0.4)α

T−19

6.2

=80α

•ΣM

G=

IGa:

450−

T(0.4)−0.1N

B(0.6)

=20

(0.3)2α

450−T(0.4)−

196.2(0.6)

=1.8α

•S

olvi

ngth

ese

two

equa

tions

,we

get

α=

7.50

rad/s2

andT

=79

7N30

/36

CO

NC

EP

TQ

UIZ

1A

n80

kgsp

ool(kG=

0.3meter

)is

ona

roug

hsu

rface

and

aca

ble

exer

tsa30

Nlo

adto

the

right

.The

frict

ion

forc

eat

Aac

tsto

the

and

theaG

shou

ldbe

dire

cted

toth

e.

(a)

right

,lef

t

(b)

left,

right

(c)

right

,rig

ht

(d)

left,

left

AN

S:(

c)

0.2

m A

30N

α

0.75

m

G

2Fo

rthe

situ

atio

nab

ove,

the

mom

ente

quat

ion

abou

tGis

?(a

)0.75(F

fA)−0.2(30)=

−(80)(0.3

2)α

(b)−0.2(30)=

−(80)(0.3

2)α

(c)0.75(F

fA)−0.2(30)=

−(80)(0.3

2)α

+80

aG

(d)

Non

eof

the

abov

eA

NS

:(a)

31/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

•G

iven

:A80

kgla

wn

rolle

rhas

ara

dius

ofgy

ratio

nof

kG=

0.17

5meter

.Iti

spu

shed

forw

ard

with

afo

rce

of20

0N.

•Fi

nd:T

hean

gula

racc

eler

atio

nif

µs=

0.12

andµk=

0.1.

•P

lan:

Follo

wth

epr

oble

mso

lvin

gpr

oced

ure.

•S

olut

ion:

The

mom

ento

fine

rtia

ofth

ero

llera

bout

Gis

IG=

m(k

G)2

=(80)(0.175

)2=

2.45

kg·m

2

32/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

(con

tinue

d)

•FB

D:

•E

quat

ions

ofm

otio

n:

ΣFx

=m(a

G) x

FA−20

0cos45

=80aG

(8)

ΣFy

=m(a

G) y

NA−78

4.8−20

0sin45

=0

(9)

ΣM

G=

IGα

−0.2F

A=

2.45α

(10)

•W

eha

ve4

unkn

owns

:NA,FA,aG

andα

.•

Ano

ther

equa

tion

isne

eded

toal

low

solv

ing

fort

heun

know

ns.

33/3

6

GR

OU

PP

RO

BLE

MS

OLV

ING

(con

tinue

d)

•Th

eth

ree

equa

tions

we

have

now

are:

FA−20

0cos45

◦=

80aG

NA−78

4.8−20

0sin45

◦=

0

−0.2F

A=

2.45α

•Fi

rst,

assu

me

the

whe

elis

not s

lippi

ng.

Thus

,we

can

writ

eaG=

rα=

0.2α

•N

owso

lvin

gth

efo

ureq

uatio

nsyi

elds

:NA=

926.2N

,FA=

61.4N

,α=

−5.01

rad/s2

,and

aG=

−1.0m/s2

•Th

eno

-slip

assu

mpt

ion

mus

tbe

chec

ked.

IsFA(=

61.4N)≤

µsN

A(=

111.1N

)?

•Ye

s,th

eref

ore,

the

whe

elro

llsw

ithou

tslip

.

34/3

6

CO

NC

EP

TQ

UIZ

1A

slen

der1

00kg

beam

issu

spen

ded

bya

cabl

e.Th

em

omen

tequ

atio

nab

outp

oint

Ais

?

(a)3(10)=

1/12(100)(42)α

(b)3(10)=

1/3(100)(4

2)α

(c)3(10)=

1/12(100)(42)α

+(100

aGx)(2)

(d)

Non

eof

the

abov

e.

AN

S:(

c)

4m3m

10N

A

2S

elec

tthe

equa

tion

that

best

repr

esen

tsth

e‘n

o-sl

ip’

assu

mpt

ion.

(a)Ff=

µsN

(b)Ff=

µkN

(c)aG=

rα

(d)

Non

eof

the

abov

e.A

NS

:(c)

35/3

6

note

36/3

6