M F

E Uncr PPla(ith(imco

P

w

θ

th

ME 352 - Ma

Fall Semeste

EXAM 1. O

Use the blanecessary, yorib sheet to

Problem 1 (2Part I. (i) Deabel the loweii) Define vehe direction oiii) Write th

mechanism; aonstraints th

Part II. For a

where 2 5R =

9θ for the Ne

he correction

achine Desig

er 2010

PEN BOOK

nk paper pou can use tthe end of p

25 Points). etermine theer pairs and tectors that arof each vect

he vector loand (b) the k

hen write the

a particular p

cm, 2 45θ =

ewton-Raph

ns 9*RΔ and

gn I

K AND CLO

provided forthe given fiproblem 1. A

e mobility othe higher pre suitable fotor on Figureoop equationknown quanconstraint e

planar mecha

5 ,° 1 8 cmR =

son techniqu

9*θΔ (to fou

OSED NOT

r your solugures to shAny work t

of the mechaairs on the fi

for a complete 1. n(s) for thisntities, the uequation(s).

Figure 1. A

anism, the v

IR2√

m, and 1θ =

ue are 9*R =

ur decimal p

1

Name

Lab.

TES.

utions. Writow vectors hat cannot

anism shownfigure. te kinematic

s mechanismnknown var

Planar Mec

ector loop eq

??R R9 1

√√− − =

0 .° The init

6 cm= and θ

laces) that a

e of Student

Div. Numb

We

te on one sand instantbe followed

n in Figure

c analysis of

m and identriables, and

chanism.

quation can

0=

tial estimates

9* 135 .θ = ° C

are to be used

t_________

er________

ednesday, Se

side of the t centers. A

d will assum

1. Clearly n

f the mechan

ify: (a) suitany constra

be written a

s of the posi

Calculate the

d in the next

___________

___________

eptember 29

paper onlyAttach your me to be wro

number each

nism. Label

table input(sints. If you

as

ition variable

e numerical

t iteration.

________

________

9th, 2010

y. Where one page

ong.

h link and

and show

s) for the identified

es 9R and

values of

M F Pin(iof(iva(ith(iG

ME 352 - Ma

Fall Semeste

Problem 2 (2nput link 2 isi) On Figuref the mechanii) Write theariables, andiii) Calculatehe vector froiv) Determin

Give the mag

achine Desig

er 2010

25 Points). Fs 2 12 kω = −e 2, draw andnism. e vector lood any constrae the numeri

om the grounne the angulagnitude and t

gn I

For the mechk rad / s . Thd clearly lab

op equationaints. If you ical values o

nd pivot 4O ar velocity othe direction

hanism in thhe length of tbel all the ve

ns. Clearly iidentified co

of the first-oto point A o

of link 4 andof each vec

Figure 2. A

2

Name

Lab.

he position sthe ground leectors that ar

indicate the onstraints thorder kinemaon link 3). d the velocitytor.

A planar mech

e of Student

Div. Numb

shown in Figength is 2O Ore required f

input, the hen write theatic coeffici y of point A

hanism.

t_________

er________

gure 2, the a4O 10 cm.=

for a comple

known quae constraint eients 4θ ′ and

A relative to

___________

___________

angular veloc ete kinemati

antities, the equations. d 43R′ (wher

the ground p

________

________

city of the

c analysis

unknown

re 43R is

pivot 4O .

M F Pco(i(i

(i(i

ME 352 - Ma

Fall Semeste

Problem 3 (2ounterclockwi) List the prii) Show the

Using theiii) The first-iv) The magn

achine Desig

er 2010

25 Points). Fwise with animary instanlocations of

e locations of-order kinemnitudes and d

gn I

For the mechn angular velnt centers andf all instant cf the instant

matic coefficidirections of

Figure 3.

hanism in thlocity 2 9ω =d the secondcenters on Ficenters, deteients of linkf the angular

The Planar M

3

Name

Lab.

he position s9 rad / s. Thedary instant cigure 3. Pleaermine: s 3, 4, and lir velocities o

Mechanism.

e of Student

Div. Numb

shown in Fige mechanismcenters for thase use the K

ink 7. of links 3 and

Drawn full

t_________

er________

gure 3, the inm is drawn fuhe mechanis

Kennedy circ

d 4.

scale.

___________

___________

nput link 2 iull scale in thsm. cle shown be

________

________

is rotating he figure.

elow.

M F Pcocian(iof(iva(i(i(v

ME 352 - Ma

Fall Semeste

Problem 4 (2ounterclockwircular grounnd 3 are 2O Ai) On Figuref the mechanii) Write theariables, andiii) Determiniv) The anguv) The angul

achine Desig

er 2010

25 Points). Fwise with annd surface aA AB 8 c= =

e 4, draw andnism. e vector lood any constrane the first-oular velocity lar velocity o

gn I

For the mechn angular velat point C. Tcm. d clearly lab

op equationaints. If you rder kinemaof link 3. Gi

of link 4. Giv

F

hanism in thlocity 2ω =he radius of

bel all the ve

ns. Clearly iidentified co

atic coefficieive the magnve the magn

Figure 4. The4

Name

Lab.

he position s15 rad / s. Thf the ground

ectors that ar

indicate the onstraints th

ents of links nitude and th

nitude and th

e Planar Me

e of Student

Div. Numb

shown in Fighe wheel, lin

d link is 1ρ =

re required f

input, the hen write the3 and 4. he direction

he direction o

echanism.

t_________

er________

gure 4, the innk 4, is in ro10 cm= and

for a comple

known quae constraint e

of this vectoof this vector

___________

___________

nput link 2 iolling contacthe lengths

ete kinemati

antities, the equations.

or. r.

________

________

is rotating ct with the of links 2

c analysis

unknown

S(i

F

Su

T

Nm

OLUTION i) The labels

The Kutzb

or the wheel

ubstituting E

Therefore, the

Note that if thmobility of th

TO PROBL on the link

Fi

bach mobilit

l rolling with

Equation (2)

e mobility of

he wheel is he mechanism

LEM 1. numbers and

gure 1. Num

ty criterion c

hout slipping

n 7=

into Equatio

m

f the mechan

rolling and m is

m =

d the joint ty

mber of links

can be writte

m 3(n=

g on the grou

17, j 8=

on (1) gives

m 3(7 1)= − −

nism is

m

slipping on

3(7 1) 2(7− −

5

ypes in the m

and joint ty

en as

1n 1) 2( j )− − −

und link then

, and

2(8) 1(1)− − =

m 1=

the ground

7) 1(2) 18− =

mechanism a

ypes in the m

21( j )−

n

2j 1=

18 16 1= − −

link then 1j

8 14 2 2− − =

are shown on

mechanism.

7= and 2j

2

n Figure 1.

2.= In this

(1)

(2)

(3a)

(3b)

s case, the

(4)

(i

(i

an

(a(banT

ii) The vecto

iii) Three vec

nd

a) The input b) The knownd θ7.

The unknownThere are

ors for a kine

ctor loop equ

for the mechwn quantities

n variables arfour constra

9 14 9θ = θ −

ematic analy

Figure

uations are n

hanism is ths are the sev

re the three laint equation

0 ,° 2θ

ysis of the me

2. The vecto

necessary for

I2R R√ √

+

33 2

CR R√ √Ι

−

I2R R√

−

e angular poven lengths R

lengths R14, ns, namely

26 2,= θ

6

echanism are

ors for a kine

r the kinema

3 14R R√? ?√ √

− −

2 26

? ?CR R

Ι √+ +

5533C C

R R√ √

+ −

osition of linR2, R3, R33,

R26, and R7,

33 3,θ = θ

e shown in F

ematic analy

atic analysis

09C

R√

=

6 5

? ?0R R

√− =

?07R

√− =

nk 2, that is, θR6, R5, R55

, and the thr

and

Figure 2.

ysis.

of this mech

θ2. 5, and R9, an

ee angles θ3

55θ = θ

hanism, nam

nd the two a

, θ5, and θ6.

5θ + γ

mely

(5a)

(5b)

(5c)

angles θ14,

(6)

7

Since there is rolling contact between link 4 (the pinion) and the ground link 1 (the rack) then the rolling constraint equation can be written as

14 4 4 14R ( )± Δ = ρ Δθ −Δθ (7a)

The positive sign must be used for the mechanism in the given position, that is, the vector 14R is getting longer for a counterclockwise rotation of link 4, or the vector is getting shorter for a clockwise rotation of link 4. Therefore, Equation (7a) can be written as

14 4 4 14R ( )+ Δ = ρ Δθ −Δθ (7b) (iv) From the given VLE, the X and Y components of the error vector ε can be written as

* *2 2 32 32 1 1 XR cos R cos R cosθ − θ − θ = ε (1a)

and * *

2 2 32 32 1 1 YR sin R sin R sinθ − θ − θ = ε (1b) Taking the partial derivates of Equation (1a) with respect to the estimates of the of the position variables gives

*X32*

32cos

R∂ε

= − θ∂

and * *X32 32*

32R sin∂ε

= + θ∂θ

(2a)

Also, taking the partial derivates of Equation (8b) with respect to the estimates of the position variables gives

*Y32*

32sin

R∂ε

= − θ∂

and * *Y32 32*

32R cos∂ε

= − θ∂θ

(2b)

From Cramers rule, the coefficient matrix can be written as

X X* *32 32

Y Y* *32 32

R

R

∂ε ∂ε⎡ ⎤⎢ ⎥∂ ∂θ⎢ ⎥⎢ ⎥∂ε ∂ε⎢ ⎥∂ ∂θ⎢ ⎥⎣ ⎦

(3a)

Substituting Equations (2) into Equation (3a), the coefficient matrix is

* * *32 32 32* * *32 32 32

cos R sin

sin R cos

⎡ ⎤− θ + θ⎢ ⎥⎢ ⎥− θ − θ⎣ ⎦

(3b)

The matrix equation for the corrections to the position variables can be written as

* * * *32 32 32 32 X* * * * Y32 32 32 32

cos R sin R

sin R cos

⎡ ⎤ ⎡ ⎤− θ + θ Δ − ε⎡ ⎤⎢ ⎥ ⎢ ⎥ = ⎢ ⎥− ε⎢ ⎥ ⎢ ⎥ ⎣ ⎦− θ − θ Δθ⎣ ⎦ ⎣ ⎦

(4)

8

Using Cramer’s rule, the corrections can be written from Equation (11) as

* * * ** X 32 32 Y 32 3232

R cos R sinRDET

ε θ + ε θΔ = (5a)

and * *

* 32 Y 32 X32

cos sinDET

θ ε − θ εΔθ = (5b)

The determinant of the coefficient matrix, see Equation (3b), is

* * * * * * *32 32 32 32 32 32 32DET ( cos )( R cos ) ( sin )( R sin ) R= − θ − θ − − θ + θ = (6)

The initial estimates of the position variables are *9 6 cmR = and *

9 135 .θ = ° Substituting this information and the given data into Equations (1), the X and Y components of the error vector are

X 5cos45 6cos135 8cos0 0.2218 cmε = °− °− ° = − (7a) and

Y 5sin 45 6sin135 8sin 0 0.7071cmε = °− °− ° = − (7b) Also, substituting the initial estimates of the position variables into Equation (6), the determinant is

*32DET R 6 cm= = (8)

Substituting Equations (7) and (8) into Equations (5), the corrections can be written as

*32

( 0.2218 cm)(6 cm)cos135 ( 0.7071 cm)(6 cm)sin135R6 cm

− °+ − °Δ = (9a)

and *32

cos135 ( 0.7071cm) sin135 ( 0.2218 cm)6 cm

° − − ° −Δθ = (9b)

Therefore, the corrections that are required for the next iteration are

2 2*32

0.9410 cm 3.0000 cmR 0.3432 cm6 cm−

Δ = = − (10a)

and *32

0.5000 cm 0.1568 cm 0.1095 rad6 cm+

Δθ = = + (10b)

S(i

an

TR

F

(i

an

OLUTION i) The vector

There are

nd the secon

The input is tR43, R45, and

rom the geo

ii) The X and

nd

TO PROBLrs that are su

Figu

two indepen

nd vector loo

he angular pR5 and the a

metry of the

d Y compon

LEM 2 uitable for a v

ure 1. The v

ndent vector

op equation i

position of thangular posit

e mechanism

43R =

nents of Equa

velocity ana

ectors for a v

r loops for th

IR2√

is I

R2√

he input linktion of link 4

43θ =

m, the length

45 2 cR R= = =

ation (1a) are

2 2cosR Rθ −

2 2sinR Rθ −

9

alysis of the m

velocity ana

his mechanis

??R R43 1

√√− −

45

? ?R R5C √

+ −

k 2, that is, θ24, namely, θ4

45 4θ θ= =

of the input

5 10cos30 3

=°

e

43 4cosR Rθ −

43 4 1sinR Rθ −

mechanism

alysis of the m

sm. The first

01√

=

0=

2. The four u4. The constr

link is

cm 5.77 cm=

1 1cos 0R θ =

1 1sin 0θ =

are as shown

mechanism.

t vector loop

unknown varraint equatio

m

n in Figure 1

p equation is

riables are thon is

1.

(1a)

(1b)

he lengths

(2)

(3)

(4a)

(4b)

10

Differentiating Equations (4) with respect to the input position gives

2 2 43 4 43 4 4sin cos sin 0R R Rθ θ θ θ′ ′− − + = (5a) and

2 2 43 4 4 4 4cos sin sin 0R R Rθ θ θ θ′ ′− − = (5b) Writing Equations (5) in matrix form gives

4 43 4 43 2 2

4 43 4 4 2 2

cos sin sinsin cos cos

R R RR R

θ θ θθ θ θ θ

′− + +⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′− − −⎣ ⎦ ⎣ ⎦⎣ ⎦

(6)

Solving Equation (6) using Cramer’s rule, the first-order kinematic coefficients can be written as

2 43 4 243

sin( )R RR

DETθ θ−′ = (7a)

and 2 4 2

4cos( )R

DETθ θ

θ−′ = (7b)

where the determinant is 43DET R= (8)

Substituting Equation (8) into Equations (7), the first-order kinematic coefficients can be written as

2 43 4 243 2 4 2

43

sin( )sin( )

R RR R

Rθ θ

θ θ−′ = = − (9a)

and 2 4 2

4 4 243

cos( ) cos( )RRθ θ

θ θ θ−′ = = − (9b)

Substituting the given data into Equations (9), the first-order kinematic coefficients for the mechanism are

43 cm10 sin(150 30 ) 5 cm/rad3

R′ = ° − ° = + (10a)

The positive sign indicates that the vector 43R is getting shorter for a clockwise rotation of the input link and

4 cos(150 30 ) 0.5 rad/radθ ′ = ° − ° = − (11) The negative sign indicates that link 4 is rotating counterclockwise (for the input rotating clockwise). (iv) 5 Points. The angular velocity of link 4 is

3 4 2 0.5 ( 12 rad/s) 6 rad/sω θ ω′= = − − = + (12) The angular velocity of link 4 is counterclockwise.

The velocity of point A relative to the ground pivot 4O is

34 43 2 5 cm / rad ( 12 rad/s) 60 cm/sV R ω′= = + − = − (13)

This velocity is acting towards the ground pivot 4O .

S

(ian(i1li2in3li4li5li6li7li8li9li10li

OLUTION The numb

i) 5 points. Tnd 10 secondii) 5 Points. . Connect innes gives the. Connect inntersection o. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the0. Connect ines gives the

TO PROBLber of instant

There are 11dary instant The procedu

nstant centere instant cennstant cente

of these two lnstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant ceninstant centee instant cen

LEM 3 t centers for

1 primary incenters, nam

ure to locate rs I12 and I2nter I15. rs I15 and Ilines gives thrs I12 and I1nter I26. rs I12 and I1nter I27. rs I16 and I1nter I67. rs I14 and I1nter I47. rs I14 and I1nter I46. rs I13 and I1nter I36. rs I13 and I1nter I37. ers I13 and Inter I35.

this mechan

nN =

nstant centersmely, I15, I56,

the seconda5. Connect i

16. Draw a he instant ce6. Connect i

7. Connect i

7. Connect i

7. Connect i

6. Connect i

6. Connect i

7. Connect i

15. Connect

Figure 1. T11

nism is

( 1) 7(72 2

n − −=

s, namely, I I26, I27, I67,

ary instant ceinstant cente

line perpenenter I56. instant cente

instant cente

instant cente

instant cente

instant cente

instant cente

instant cente

instant cent

he Kennedy

1) 212− =

12, I13, I14, I1I47, I46, I36, Ienters (see thers I17 and I

ndicular to th

ers I25 and I

ers I25 and I

ers I56 and I

ers I45 and I

ers I47 and I

ers I34 and I

ers I34 and I

ters I37 and I

y Circle.

16, I17, I23, I2I37, and I35. he Kennedy I57. The inte

he slot thro

I56. The inte

I57. The inte

I57. The inte

I57. The inte

I67. The inte

I46. The inte

I47. The inte

I57. The inte

24, I25, I34, I4 circle) is:

ersection of

ugh the con

ersection of

ersection of

ersection of

ersection of

ersection of

ersection of

ersection of

ersection of

(1)

45, and I57;

these two

ntact. The

these two

these two

these two

these two

these two

these two

these two

these two

T

(i

wm

T

TarT

w

The construct

iii) 5 Points.

where the pomeasurement

Therefore, the

The negative rm). Therefo

The rolling co

where the neg

tion of the se

. The rolling

ositive sign ts and the kn

e first-order

sign indicatore, gear 4 isontact equati

gative sign m

econdary ins

Figure 2. T

g contact equ

must be unown kinema

kinematic c

tes that gear s rotating in tion between

must be used

stant centers

The location

uation betwe

1

4

ρρ

± =

used here beatic coefficie

4 cm1cm

+

oefficient of

4θ′ = −

4 is rotatingthe clockwis gear 3 and g

4

3

ρρ

± =

d here becaus

12

is shown in

of the prima

en gear 4 an

4 2

1 2

θ θθ θ′ ′−

=′ ′−

ecause of thents into Equ

4 1mm 0 1

θ ′ −=

−

f gear 4 is

3 rad/rad−

g in the oppse direction. gear 4 can b

3 2

4 2

θ θθ θ′ ′−

=′ ′−

se of externa

n Figure 2.

ary instant c

nd the fixed g

he internal uation (2a) g

osite directi e written as

al rolling con

centers.

gear 1 can be

rolling contgives

ion to the inp

ntact.

e written as

tact. Substit

put link 2 (t

(2a)

tuting the

(2b)

(2c)

that is, the

(3a)

13

Substituting the measurements and the known kinematic coefficients into Equation (3a) gives

3

4

11cm2 cm 1

θθ′ −

− =′ −

(3b)

Substituting Equation (2c) into Equation (3b), the first-order kinematic coefficient of gear 3 is

3 3 rad/radθ′ = + (3c) The positive sign indicates that gear 3 is rotating in the same direction to the input link 2 (that is, the arm). Therefore, gear 3 is rotating in the counterclockwise direction. Alternative approach. The first-order kinematic coefficient for link j can be written as

12 2

1 2

jj

j j

I II I

θ ′ = (4a)

Therefore, the first-order kinematic coefficient for gear 4 can be written as

12 244

14 24

I II I

θ ′ = (4b)

The distances required to obtain the first-order kinematic coefficient of link 4 are measured as

12 24 3 cmI I = and 14 24 1 cmI I = (5a) Substituting Equation (5a) into Equation (4b), the first-order kinematic coefficient of link 4 is

43 cm 3 rad/rad1 cm

θ ′ = = (5b)

Note that the relative instant center 24I is between the two absolute instant centers 12 14and .I I Therefore, the first-order kinematic coefficient for gear 4 must be negative, that is

4 3 rad/radθ ′ = − (5c) Similarly, the first-order kinematic coefficient for gear 3 can be written as

12 233

13 23

I II I

θ ′ = (6)

Note in this case that the instant centers 12 23 13, andI I I are coincident, see Figure 2. Therefore, the first-order kinematic coefficient of link 3 cannot be obtained from this equation.

Consider the point of contact between gears 3 and 4. Denote this point as A (note that point A on gear 3 is coincident with point A on gear 4 (that is, point A is the instant center 34I ). Therefore, the velocity of point A can be written as

A A3 4V V= (7a)

which can be written as 3 13 34 4 14 34( ) ( )I I I Iω ω= (7b)

14

Dividing both sides of Equation (7b) by the input angular velocity 2ω gives

3 13 34 4 14 34( ) ( )I I I Iθ θ′ ′= (7c) Then rearranging this equation, the first-order kinematic coefficient of link 3 can be written as

14 343 4

13 34

( )I II I

θ θ′ ′= (7d)

The lengths necessary to solve for the first-order kinematic coefficient of link 3 are measured to be

13 34 2 cmI I = and 14 34 2 cmI I = (8a) Substituting Equations (5c) and (8a) into Equation (7d) gives

32 3( )( ) 3 rad/rad2 1

θ ′ = ± − = ± (8b)

Note that the relative instant center 34I is between the absolute instant centers 13 14, and .I I Therefore, the first-order kinematic coefficient for gear 3 has the opposite sign as the first-order kinematic coefficient of gear 4. Therefore, the first-order kinematic coefficient of link 3 must be positive, that is

3 3 rad/radθ ′ = + (9)

The first-order kinematic coefficient for a slider, say link k, can be written as

12 2j jR I I′ = (10a) Therefore, the first-order kinematic coefficient for link 7 can be written as

7 12 27R I I′ = (10b) The distance required for the first-order kinematic coefficient of link 7 is measured as

12 27 4.75 cmI I = (11) Substituting Equation (11) into Equation (10b), the first-order kinematic coefficient for link 7 is

7 4.75 cmR′ = − (12)

The negative sign must be used here because the vector 7R is getting shorter for a positive change in the position of the input link 2, see Figure 4. In other words, link 7 is moving upward for a positive change in the rotation of the input link 2. (iv) 5 points. The angular velocity of gear 3 can be written as

3 3 2ω θ ω′= (13a) Substituting Equation (9) and the input angular velocity into Equation (13) gives

3 ( 3)( 9) 27 rad/sω = + + = + (13b) The positive sign indicates that gear 3 is rotating counterclockwise.

Su

T

The angul

ubstituting E

The negative

lar velocity o

Equation (6)

sign indicat

F

of gear 4 can

and the inpu

es that gear

Figure 4. A s

n be written

ut angular ve

4 (ω =

4 is rotating

suitable choi

15

as

4 4 2ω θ ω′=

elocity into E

( 3)( 9)− + =

g clockwise.

ice of vector

Equation (15

27 rad/s−

rs for the me

5) gives

echanism.

(14a)

(14b)

S

tr

U

T

Su

(i

T

T

an

D

an

T

OLUTION Note that

riangle, that

Using the sum

Therefore, the

ubtracting th

i) 5 points. A

The vector lo

The X and Y

nd

Differentiting

nd

Then writing

TO PROBLt the lengthsis

m of the inter

e triangle O2

he length of

A suitable ch

op equation

components

g Equation (2

Equations (3

LEM 4. s of link 2

rior angles o

2AB is an eq

O2B from th

hoice of vect

Figu

can be writt

s of Equation

2) with respe

3) in matrix

−⎡⎢+⎣

and link 3

2AO

of a triangle,

2 18O AB =

quilateral tria

2O A =

he radius of t

4ρ ρ=

tors for the m

ure 1. The ve

ten as

C2R R

√ √+

n (1) can be

2 2cosR Rθ +

2 2sinR Rθ +

ect to the inp

2 2sinR θ− −

2 2cosR θ+ +

form gives

3 3

3 3

sincos

RR R

θθ

− ++ −

16

are equal, t

2B O BA= =

280 AO B° −

angle and

2AB O B= = =

the ground l

1 2 2 cO Bρ − =

mechanism a

ctors for the

9

? ?03R R

√ √− =

written as

3 3 9cosR Rθ −

3 3 9sin sR Rθ −

put position

3 3 3sinR θ θ ′ +

3 3 3cosR θ θ ′+ −

9 9 3

9 9 9

sincos

RR

θ θθ θ

′⎤ ⎡⎥ ⎢ ′⎦ ⎣

therefore, th

60= °

2 6O BA− =

8 cm

link then the

cm

are shown in

e mechanism

9cos 0θ =

9sin 0θ =

θ2 gives

9 9 9sinR θ θ ′ =

9 9 9cosR θ θ ′− =

3 2

9 2

sincos

RR

θ ′ +⎤ ⎡=⎥ ⎢′ −⎣⎦

he triangle O

60°

radius of lin

n Figure 1.

m.

0=

0=

2

2

nsθθ

⎤⎥⎦

O2AB is an

nk 4 is

isosceles

(1a)

(1b)

(1c)

(1d)

(1)

(2a)

(2b)

(3a)

(3b)

(4)

17

Solving Equation (4) using Cramer’s rule, the first-order kinematic coefficients can be written as

2 9 2 93

sin( )R RDET

θ θθ

− −′ = (5a)

and 2 3 2 3

9sin( )R RDET

θ θθ

− −′ = (5b) where the determinant of the coefficient matrix can be written as

3 9 3 9sin( )DET R R θ θ= − (6) Substituting Equation (6) into Equations (5), the first-order kinematic coefficients can be written as

2 2 93

3 3 9

sin( )sin( )

RR

θ θθ

θ θ− −′ =

− (7a)

and 2 2 3

99 3 9

sin( )sin( )

RR

θ θθ

θ θ− −′ =

− (7b)

Substituting the position data into Equations (7), the first-order kinematic coefficients are

38 cm sin( 60 0 ) 1 rad/rad8 cm sin(60 0 )

θ − − ° − °′ = = +° − °

(8a)

and

98 cm sin( 60 60 ) 1 rad/rad8 cm sin(60 0 )

θ − − ° − °′ = = +° − °

(8b)

The positive sign for the first-order kinematic coefficients in Equations (8) indicate that links 3 and 9 are rotating in the same angular velocity as the input link 2. The answers in Equations (8) also imply that links 3 and 9 have the same angular velocities as the input link 2. Note. For the given input position and link dimensions the answers in Equations (8) should be intuitively obvious. Since the lengths of O2A, AB, and O2B are equal then O2AB is a rigid equilateral triangle rotating about the ground pivot O2. Since the triangle experiences rigid body motion then the angular velocity of this triangle must be same as the angular velocity of link 2. Therefore, the first-order kinematic coefficients of links 3 and 9 must be 1. (ii) 5 points. To determine the angular velocity of link 4, consider the rolling contact equation between links 1 and 4. Recall that the general form of the rolling contact equation between a gear and a pinion can be written as

gear pinion arm

pinion gear arm

ρ θ θρ θ θ

Δ − Δ± =

Δ − Δ (9)

where the positive sign is used for internal contact and the negative sign is used for external contact.

For the given mechanism the gear is link 1, the pinion is link 4, and the arm is link 2. Substituting this notation into Equation (9) gives

1 4 2

4 1 2

ρ θ θρ θ θ

Δ − Δ+ =

Δ − Δ (10)

18

Differentiating Equation (10) with respect to the input position θ2 gives

1 4

4

10 1

ρ θρ

′ −+ =

− (11)

Solving Equation (11) for the first-order kinematic coefficient of link 4 gives

14

41 ρ

θρ

′ = − (12)

Then substituting the radius of the ground and the radius of link 4 into Equation (12) gives

410 cm1 4 rad/rad2 cm

θ ′ = − = − (13)

(iii) 5 points. The angular velocity of link 3 can be written as

3 3 2ω θ ω′= (14a) Substituting the known values into Equation (14a) gives

3 ( 1 rad/rad)( 15 rad/s) 15 rad/sω = + + = + (14b) The positive sign indicates that the angular velocity of link 3 is counterclockwise. (iv) 5 points. The angular velocity of link 4 can be written as

4 4 2ω θ ω′= (15a) Substituting the known values into Equation (15a) gives

4 ( 4 rad/rad)( 15 rad/s) 60 rad/sω = − + = − (15b) The negative sign indicates that the angular velocity of link 4 is clockwise.